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https://stats.stackexchange.com/questions/74561/how-to-create-forecast-data-prediction-interval-bands | # How to create forecast data prediction interval bands
I have seasonal data from which I create forecasts. The steps I perform are: deseasonalizing the data, finding the linear regression for the deseasonalized points, predicting a few points from the linear regression and adding seasonality to the predicted values to get forecast data. My input is quite sinusoidal so all works well.
The problem is that the more in the future you predict, the more prediction errors increase. I'd like to show that on a chart, but I am not sure how to calculate these errors. I was thinking something like prediction interval bands for forecast data (whatever they are called). These bands would increase the further you predict in the future.
Here are some images that show what I'm trying to do: sample bands image1 sample bands image2
My question is what is the name for these bands? (then I can do a google search for it) I'd also appreciate the formulas needed for the band calculations. I'm guessing there is a standard deviation in there somewhere.
I've looked at confidence interval, but that seems to be for the data already present, not for the forecast data.
• Are you sure that your time series is deterministic in time? I.e., $Y_t=at+\epsilon_t$ after deseasonalizing, or have you considered a random walk with fixed drift: $Y_{t+1}=Y_t+\delta+\epsilon_t$? Where $\delta$ is a fixed "drift".The differnece is that the second form has much more variance and will not fall on a neat line, but it will increase over time. Just something to consider, as that info will help us recommend prediction bands. – user31668 Nov 4 '13 at 19:25
• @Eupraxis1981 The data is from google analytics so I am guessing it is deterministic in time. BTW my stats background is minimal. – Adrian Nov 4 '13 at 21:27
• Thanks for clarifying. I was asking about the specific data you are modeling. Is there a reason to believe that the linear trend model is appropriate. I.e., would you expect "regression towards the mean" at time $t>t_0$ if you saw a large/small value at $t_0$? Or, would you expect the trend to continue from the last observed point? – user31668 Nov 4 '13 at 21:42
• @Eupraxis1981 yes the linear trend is fine. – Adrian Nov 5 '13 at 16:30
## 1 Answer
Based on OP's comment that deseasonalized time series is a linear trend (t is a true predictor), then you will either want the Prediction Interval for linear regression (if you are trying to predict 1 time period ahead), or tolerance intervals if you are trying to capture a specific proportion of future measurements.
If the residuals from your linear fit to the deseasonalized data are approximately normal, then there are nice formulas for as you will see in the above links: Also, eee this other CrossValidated Post.
You would then re-seasonalize these intervals/bands to get your actual forcasts.
• This is a great answer. The links are very useful. – Adrian Nov 5 '13 at 19:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8201520442962646, "perplexity": 862.5337522597531}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141750841.83/warc/CC-MAIN-20201205211729-20201206001729-00393.warc.gz"} |
https://www.studyadda.com/notes/6th-class/mathematics/arithmetic/arithmetic/6161 | # 6th Class Mathematics Arithmetic ARITHMETIC
ARITHMETIC
Category : 6th Class
Learning Objective
• To understand the term fraction and its types (proper, improper mixed, equivalent, like and unlike fractions)
• To learn how to odd. Subtract multiply and divide proper, improper and mixed factions.
• To understand the term decimal and representation of decimals on number line.
• To learn how to compare decimal.
• To learn how to add, subtract multiply and divide decimals.
• To understand the terms ratio and proportion.
• To learn how to find the value of one unit by using unitary method.
FRACTION
Fraction is a method for representing the parts of a whole number. In the fraction, $\frac{a}{b},$ a is the numerator and b is the denominator.
Example: $\frac{2}{3},\,\frac{7}{8},\,\frac{3}{7},\,\frac{4}{9}]\ etc. TYPES OF FRACTION PROPER FRACTIONS In a proper fraction, the numerator is always smaller than the denominator. For example: \[\frac{1}{4},\,\frac{3}{5}$ etc.
IMPROPER FRACTIONS
In an improper fraction the numerator is greater than the denominator.
For example: $\frac{5}{3},\,\frac{7}{4}$ is
Fractions in the form of $1\frac{1}{4}$ or $2\frac{1}{2}$ are know as mixed fractions.
Let us represent mixed fraction by using figures.
EQUIVALENT FRACTIONS
Equivalent fractions represent same part of the whole.
For example $\frac{1}{2}\,=\frac{2}{4}\,=\frac{3}{6}\,=\frac{4}{8}\,=\frac{5}{10}$
We can find more equivalent fractions by multiplying or dividing the numerator and the denominator by the same number.
SIMPLEST FORM OF A FRACTION
A fraction is said to be in the simplest form (or lowest form) if its numerator and denominator have no common factor except 1.
The easiest way to find the simplest form of a fraction is to divide the numerator and denominator by their HCF.
For example: To reduce $\frac{125}{225},$ find their HCF.
HCF of 125 and 225 =25
$\therefore \,\,\,\,\,\,\frac{125}{225}\div \frac{25}{25}=\frac{5}{9}$ simplest form
To reduce $\frac{36}{72}$
H.C.F of 36 and 72 = 36
$\therefore \,\,\,\,\,\frac{36}{72}\div \frac{36}{36}=\frac{1}{2}$ simplest form.
LIKE FRACTIONS
Fractions with the same denominator are called like fractions.
$\frac{4}{13},\,\frac{3}{13},\,\frac{12}{13},\,\frac{9}{13}$ are examples of like fractions.
UNLIKE FRACTIONS
Fractions like $\frac{1}{5},\,\frac{2}{3},\,\frac{3}{4}$ have different denominators are called unlike fractions.
Fraction on the number line
Let us draw a number line and mark $\frac{3}{4}$ on it. $\frac{3}{4}$ is greater than 0 and less that 1. As $\frac{3}{4}$ means 3 part out of 4, we will divide the gap between 0 and 1 into four equal parts, and mark $\frac{1}{4},\,\frac{2}{4},\,\frac{3}{4},\,\frac{4}{4}(=1)$ as shown below.
COMPARING FRACTIONS
COMPARING LIKE FRACTIONS
To compare like fractions like $\frac{2}{5},\,\frac{7}{5},\,\frac{1}{5},\,\frac{3}{5},\,\frac{9}{5}$ we compare the numerators only $1<2<3<7<9$
$\frac{1}{5},\,\frac{2}{5},\,\frac{3}{5},\,\frac{7}{5},\,\frac{9}{5}$
To compare unlike fractions like $\frac{2}{3},\,\frac{3}{4},\,\frac{5}{7}$. We must first convert then to like fraction as follows:
(i) Find the L.C.M of denominators 3, 4 and 7 i.e., L.C.M of 3, 4 and 7 is 84.
(ii) Make each denominator 84.
$\frac{2\times 28}{3\times 28}=\frac{56}{84}$
$\frac{3\times 21}{4\times 21}\,=\frac{63}{84}$
$\frac{5\times 12}{7\times 12}\,=\frac{60}{84}$
(iii) Now, we compare the numerators of these like fractions
$\frac{56}{84},\,\frac{63}{84},\,\frac{60}{84}$
$\Rightarrow \,\,\frac{56}{84},\,\frac{60}{84},\,\frac{63}{84}$
i.e., $\frac{2}{3},\,\frac{5}{7},\,\frac{3}{4}$
ADDITION AND SUBTRACTION OF LIKE FRACTIONS
Addition and subtraction of like fraction is very simple as they have same denominator.
For example:
$\frac{2}{10}+\frac{9}{10}=\frac{11}{10}$ (add the numerators)
$\frac{9}{11}-\frac{5}{11}=\frac{4}{11}$ (subtract the numerators)
ADDITION AND SUBTRACTION OF UNLIKE FRACTIONS
To add or subtract unlike fractions, we should first find their equivalent fractions with the same denominator
For example:
To add $\frac{3}{2}$ and $\frac{4}{3},$ we find the LCM of the denominators 2 and 3 which is 6
Thus $\frac{3}{2}\times \frac{3}{3}=\frac{9}{6}$ (add the numerators)
Similarly, to subtract $\frac{1}{3}$ from $\frac{8}{7}$ i.e., $\frac{8}{7}-\frac{1}{3}$
LCM of 7 and 3 =21
$\frac{8}{7}\times \frac{3}{3}\,=\frac{24}{21}$ and $\frac{1}{3}\,\times \frac{7}{7}=\frac{7}{21}$
Thus $\frac{24}{21}\,-\frac{7}{21}\,=\frac{17}{21}$ (subtract the numerators)
ADDITION AND SUBTRACTION OF MIXED FRACTIONS
Method I: Convert mixed fraction into improper fraction and add as in the case of unlike fraction.
For example: Add: $3\frac{3}{5}$ and $2\frac{5}{6}$
$3\frac{3}{5}+2\frac{5}{6}=\frac{18}{5}+\frac{17}{6}=\frac{18\times 6}{5\times 6}\,+\frac{17\times 5}{6\times 5}$
$=\frac{108}{30}+\frac{85}{30}$ (L.C.M of 5, 6 is 30)
$\frac{193}{30}=6\frac{13}{30}$
Subtract: $3\frac{1}{4}-1\frac{1}{6}$
$3\frac{1}{4}\,-1\frac{1}{6}\,=\frac{13}{4}-\frac{7}{6}$
$=\frac{13\times 3}{4\times 3}-\frac{7\times 2}{6\times 2}$
$=\frac{39}{12}-\frac{14}{12}$
$=\frac{25}{12}=2\,\frac{1}{12}$
Method II
The other method is to add the whole parts and proper fractions separately.
For example:
Add: $3\frac{3}{5}+2\frac{5}{6}$
$3\frac{3}{5}+2\frac{5}{6}\,=3+2+\frac{3}{5}+\frac{5}{6}\,=5+\frac{3}{5}+\frac{5}{6}$
Now,
$\frac{3}{5}+\frac{5}{6}\,=\frac{3\times 6}{5\times 6}\,+\frac{5\times 5}{\,6\times 5}\,=\frac{18}{30}+\frac{25}{30}=\frac{43}{30}\,=1\frac{13}{30}$
$\therefore \,\,3+2+\frac{3}{5}+\frac{5}{6}\,=5+1\frac{13}{30}\,=5+1+\frac{13}{30}\,=6\frac{13}{30}$
Subtract: $3\frac{1}{4}\,-1\frac{1}{6}$
$3\frac{1}{4}\,-1\frac{1}{6}=3-1+\frac{1}{4}-\frac{1}{6}$
Consider, $\frac{1}{4}-\frac{1}{6}=\frac{1\times 3}{4\times 3}-\frac{1\times 2}{6\times 2}$
$=\frac{3}{12}-\frac{2}{12}=\frac{1}{12}$
$\therefore \,\,3-1\,+\frac{1}{4}\,-\frac{1}{6}\,=2+\frac{1}{12}=2\frac{1}{12}$
DECIMALS
Decimal numbers or simply decimals are the fractions with denominators 10, 100, 1000 etc.
For example: $\frac{7}{10},\,\frac{15}{100},\,\frac{21}{1000}$ etc.
The number before the decimal point is called the whole part or integral part, whereas the number after the decimal point is called the decimal part.
For example: In 12.73
Whole part is 12 and decimal part is 73
We read 12.73 as twelve point seven three.
LIKE DECIMALS
Like decimals have an equal number of digits to the right of the decimal point.
For example:
$13.$ and $4.$ are like decimals.
UNLIKE DECIMALS
The decimals having the different number of decimal places are called unlike decimals.
For example:
$2.$ $1.$ $2.$
REPRESENTATION OF DECIMALS ON THE NUMBER LINE
REPRESENTATION OF 4.2 ON THE NUMBER LINE
Clearly 4.2 lies between 4 and 5 in Fig. (i).
Take a magnified look of the line segment between 4 and 5 and divide it into 10 equal parts and mark each point of division between 4 and 5 as shown in the Fig. (ii).
We can see in Fig. (ii), 4.2 is represented by the second mark of division after 4 in between 4 and 5.
FRACTIONS AS DECIMALS
To express given fraction into decimals, we follow the following steps:
Case I: Fractions whose denominators are of 10, 100, 1000 etc.
Step (i) Count the number of zeroes in denominator.
Step (ii) Place the decimal point in numerator so that the number of digits on right of decimal point becomes equal to the number of zeroes in denominator.
Step (iii) In case the number of digits in numerator is less than the number of zeroes in denominator, we place zero just right to decimal.
For example: In $\frac{14}{1000},$ number of zeros in denominator is 3
So, $\frac{14}{1000}=0.014$
Case II: Fractions whose denominators are not 10, 100, 1000 etc.
Divide the numerator by the denominator and write the quotient in decimal form.
DECIMALS AS FRACTION
(i) To express the given decimals into fraction, we follow the following steps;
Step (i) Write the decimal without the decimal point as the numerator of the fraction.
Step (ii) Write the denominator of the fraction by inserting as many zeros on the right of 1 as the number of decimal places in the given decimals.
Step (iii) Simplify the fraction and write the fraction in the lowest form.
For example:
Express 0.038 as fraction $0.038\,=\frac{38}{1000}\,=\frac{19}{500}$
COMPARING DECIMALS
TO COMPARE LIKE DECIMALS
To compare like decimals follow these steps.
1. Compare the whole part of the decimal number. The number with greater whole part will be greater they are same, then go to the second step.
2. Compare the tenths place digits. The number with greater tenths digit will be greater. If they are also same, then go to the third step.
3. Compare the digit in the hundredths place. The number with greater hundredths digit will be greater. If they are also equal, then compare the thousandths place and so on.
For example:
To compare 0.275, 2.34, 4.67, 2.24,4.67 > 2.34 > 2.24 > 0.275.
TO COMPARE UNLIKE DECIMALS
Like decimals and follow the steps to compare like decimals.
For example: To compare 4.2 and 4.27 first convert them to like decimals 4.20 and 4.27, 4.27 > 4.20 (0. 7 > 0)
USES OF DECIMALS
(I) IN MONEY VALUE
We know that 100 paise = Rs 1
Therefore, 1 paisa $=\frac{1}{100}=\text{Rs}\,0.01$
So 45paisa $=\frac{45}{100}=\text{Rs}\,0.45$
(II) IN MEASURES OF LENGTH
1km = 1000m $1m\,=\frac{1}{1000}km$
1m = 100 cm $1cm=\frac{1}{100}m$
1dm =10 cm $1cm=\frac{1}{10}\,dm$
1 cm = 10 mm $1mm=\frac{1}{10}\,cm$
For example:
$7m=\frac{7}{1000}\,=0.007km$
(III) IN MEASURES OF WEIGHT
Same as discussed in length (above) just use 'g' instead of 'm'.
1kg = 1000gm
1gm = 1000mg
i.e., $1\,g\,=\frac{1}{1000}\,kg$
$1mg\,=\frac{1}{1000}\,gm$
For example: $7gm\,=\frac{7}{1000}\,kg\,=0.007\,gm$
OPERATIONS OF DECIMALS
Step (i) Convert the given decimals to like decimals.
Step (ii) Write the decimals in columns with the decimal points directly below each other so that tenths come under tenths hundredths come under hundredths and so on.
Step (iv) Place the decimal point in the answer directly below the other decimal points.
For Example: Add 12.73, 4.7, 1.074
12.730
+ 4.700
+ 1.074
_______
18.504
_______
9.23
+ 4.75
+ 8.10
______
22.08
______
SUBTRACTION OF DECIMALS
We may follow the following steps to subtract a decimal number from another decimal number.
Step (i) Convert the given decimals to like decimals.
Step (ii) Write the decimals in columns with decimal points directly below each other.
Step (iii) Subtract as we subtract whole numbers starting from right.
Step (iv) Place the decimal point in the difference directly below the other decimal points.
For example: Subtract 10.205 from 20.05
20.050
- 10.250
_______
9.800
_______
RATIO
Ratio: If a and $b(b\ne 0)$ are two quantities of the same kind, then the fraction a- is called the ratio of a to b.
• For a ratio the two quantities must be in the same unit.
• Two ratios are equivalent, if the fractions corresponding to them are equivalent.
• Ratio is expressed in its simplest form cannot be further simplified.
COMPARISON OF RATIOS
To compare two given ratios, we follow the following steps:
(i) Express the given ratio in the form of a fraction.
(ii) Convert the fraction in its simplest form.
(iii) Find the L.C.M of the denominators of the simplest form of the fraction obtained in above step.
(iv) Divide the L.C.M obtained in step (iii) by the donominator of first fraction to get a number a (say).
Now, multiply the numerator and denominator of the fraction by x. Apply the same procedure for other fraction.
(v) Compare the numerators of the fractions obtained in step (v).
(vi) Having the same denominators.
The fraction having larger numerator will be larger than the other.
For example:
Compare the ratios 7 : 12 and 5:8
we have, $7:12\,=\frac{7}{12}$ and $5:8=\frac{5}{8}$
(given ratios are in their simplest form)
Now, L.C.M. of 12 and 8 is 24.
Making the denominators equal to 24 of each fraction, we have
$\frac{7\times 2}{12\times 2}\,=\frac{14}{24}\,$ and $\frac{5\times 3}{\,8\times 3}\,=\frac{15}{24}$
Clearly,
15 > 14
$\therefore \,\,\,\frac{15}{24}>\frac{14}{24}$
$\Rightarrow \,\,\,\frac{5}{8}\,>\frac{7}{12}$
EQUIVALENT RATIOS
A ratio obtained by multiplying or dividing the numerator and denominator by the same number is called an equivalent ratio.
For example:
Consider the ratio 5 : 7
We have,
$\frac{5}{7}=\frac{5\times 2}{7\times 2}\,=\frac{10}{14}$
$\frac{5}{7}\,=\frac{5\times 3}{7\times 3}\,=\frac{15}{21}$ and so on.
also, $\frac{10}{14}=\frac{10\div 2}{14\div 2}=\frac{5}{7}$
$\frac{10}{14},\,\frac{15}{21},\,\frac{20}{28}$ etc. are equivalent to the ratio $\frac{5}{7}$
PROPORTION
PROPORTION
Proportion is defined as an equality of two ratios,
Four (non-zero) quantities of the same kind a, b, c and d are said to be in proportion if the ratio of a to b is equal to the ratio of c to d
i.e., if $\frac{a}{b}=\frac{c}{d}$
we can write as a : b : : c : d
a, b, c, d are in proportion if ad = be
The (non-zero) quantities of the same kind a, b, c, d, e, f,... are said to be in continued proportion.
$\frac{a}{b}=\frac{b}{c}\,=\frac{c}{d}=\frac{d}{e}=$
If a, b, c are in continued proportion, then b is called mean proportional of a and c.
If a, b, c are in continued proportion then c is called the third proportional.
For example: Check if 3,4, 6 and 12 are in proportion we have, a = 3, b = 4, c = 6 and d = 12
$\frac{a}{b}=\frac{3}{4}$ and $\frac{c}{d}=\frac{6}{12}\,=\frac{1}{2}$
Clearly, $\frac{a}{b}\ne \frac{c}{d}$
$\therefore \,\,\,\,\,\,3:4\ne 6:12$
Hence, 3, 4, 6 and 12 are not in proportion.
UNITARY METHOD
The method in which first we find the value of one unit and then the value of required number of units by multiplying the value of one unit with the number of required units.
For example:
A car travels 240 km in 4 hours.
How for does it travel in 7 hours?
Solution
We have,
Distance travelled in 4 hours = 240 km
\Distance travelled in 1 hour $=\left( \frac{240}{4} \right)$ km = 60 km
Hence, the distance travelled in 7 hours = (60 x 7) = 420 km
For example:
The cost of four dozens of mangoes is Rs 150. What will be the cost of such 12 dozens of mangoes?
Solution: We know
cost of 1 dozen mangoes $=\frac{150}{4}$
cost of 12 dozen of mangoes $=\frac{150}{\bcancel{4}}\times {{\bcancel{12}}^{3}}$
= Rs450
#### Other Topics
##### Notes - Arithmetic
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https://stats.stackexchange.com/questions/259479/if-i-toss-a-biased-coin-4-times-and-i-bet-on-one-side-twice-then-the-other-side | # If I toss a biased coin 4 times, and I bet on one side twice then the other side twice, does it matter which side I start on?
Steven and I have a biased coin. The coin has a 90% chance to show heads, and a 10% chance to show tails. We flip the coin in the same way 4 times. Steven picks heads for the first two rounds and I pick tails for the first two rounds. Then I choose heads for the last two rounds and Steven chooses tails for the last two rounds.
In order to win, one of us must succeed 3 times. Is one of us more likely to win?
My calculation shows that we each have a 0.1557 chance of winning. My friend is arguing that I am wrong based on absorbing markov chains.
Here's what I've done:
https://www.scribd.com/document/338158368/Swap-Side-Balance
Am I crazy or is this really obvious that it doesn't matter which side you start on?
• Which player does your friend claim will win more often and why? – whuber Oct 19 '17 at 14:44
Before looking at the precise numbers, this looks essentially symmetric if you toss four times and so each player has the same probability of winning. Stopping after somebody has three should not change this
Now looking at the probabilities, Steven wins if:
• HHH probability $0.9^3=0.081$
• HHTH probability $0.9^3\times 0.1 = 0.0729$
• HTHH probability $0.9 \times 0.1^3=0.0009$
• THHH probability $0.9 \times 0.1^3=0.0009$
which is $0.1557$ as you say
You win if
• TTT probability $0.9 \times 0.1^2=0.009$
• TTHT probability $0.9 \times 0.1^3=0.0009$
• THTT probability $0.9^3\times 0.1 = 0.0729$
• HTTT probability $0.9^3\times 0.1 = 0.0729$
which is also $0.1557$
You can calculate the probability of a two-two tie as $0.9^4+4\times 0.9^2\times 0.1^2+0.1^4=0.6886$, making the total probability $1$
Check out the 'Monty Hall' problem ( https://betterexplained.com/articles/understanding-the-monty-hall-problem/ )
If you assume you know, in advance, what will be the odds then you will get problematic logic. In this case, you assume both of you know this coin is biased, and what the odds are. Thus, the 'betting strategy' makes no difference: either of you can choose the first two (or last two) tosses, and things will be equal.
Things come out very differently if either (1) only one of you knows the facts, or (2) neither of you knows the facts.
People have written long papers on the Monty Hall problem. In the end - all math aside - it comes down to the fact he knows something you do not. A great deal of the field of Information Theory is about dealing with the balance of the known and unknown. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8802645802497864, "perplexity": 720.2200215900679}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347432521.57/warc/CC-MAIN-20200603081823-20200603111823-00558.warc.gz"} |
https://registration.mcs.cmu.edu/event/1/contributions/6/ | Jun 2 – 7, 2019
Carnegie Mellon University
America/New_York timezone
## Properties of hadrons in medium in the context of Fermi Liquid Model and diquarks.
Jun 4, 2019, 4:30 PM
30m
Rangos 2
### Speaker
The Landau Fermi liquid theory is an phenomenological approach to strongly interacting normal Fermi system at small excitation energies. It is a model which suggest a point to point correspondence between low energy excitation of non interacting Fermi gas. The model has been widely used to study the properties of liquid He-3, electron in metal and nuclear matter. It gives an effective description of low energy elementary excitations like the quasi particles in crystal lattice. The model is found to be successful in describing some aspects of QCD, quark and hadronic matter also. In the current work a Fermi liquid model for hadrons has been suggested for the hadrons in medium. The hadrons are supposed to behave like quasi particle as Fermi excitation while in the medium and the effective mass of the hadrons have been estimated using Fermi liquid model. Considering a momentum dependent potential like $V(r,p^{2}) =V^{'}e^{-\gamma(\frac{p^{2}}{m}})\upsilon(r)$ inside the medium to describe the interaction, the effective masses of the hadrons are estimated.Compressibility, specific heats, density of states in medium have been studied. We have extracted the values of available well depths which give the idea about the binding energy of the particles in medium. We have also studied the masses of exotic baryons in the framework of diquark formalism. A quasi particle model of diquark has been suggested in an analogy with composite fermion and subsequently used to compute the masses of baryons like $\lambda_{b}^{0}$, $\Sigma_{b}^{0}$, $\Xi_{b}^{0}$, $\Xi_{cc}^{+}$, $\Omega_{cc}^{+}$, $\Omega_{cb}^{0}$,$\Omega_{ccc}$, $\Omega_{bbb}$. Using a density and momenum dependent potential of the form $V_p = \frac{\rho/\rho_{0}}{1+ {(\frac{p}{\lambda})}^2}$ at p= $p_{f}$, the mass of the diquark have been estimated. The results are found to be very interesting and compared with the other theoretical and experimental studies available in literature. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9507775902748108, "perplexity": 484.65560940981055}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711108.34/warc/CC-MAIN-20221206124909-20221206154909-00570.warc.gz"} |
http://mathhelpforum.com/advanced-applied-math/225822-archimedean-property.html | 1. ## Archimedean property
So there is this statement, claiming that N (Natural numbers) is not bounded above in R, which is really easy to prove, however after proving that they drew a conclusion as I saw in my lecture notes,
Conclusion; For every epsilon (I will write epsilon as e because I don't know how else to write it) greater than zero there exist n in N with 0<1/n<e
How did he drew this conclusion from the fact that N is not bounded above in R? From what I understand what this says , for every epsilon in R (that is greater than 0) you can find a natural number n that is bigger than him. Is this logic correct?
2. ## Re: Archimedean property
Originally Posted by davidciprut
Conclusion; For every epsilon (I will write epsilon as e because I don't know how else to write it) greater than zero there exist n in N with 0<1/n<e How did he drew this conclusion from the fact that N is not bounded above in R?
Using LaTeX you can enter [TEX]\forall\varepsilon>0 [/TEX] gives $\displaystyle \forall\varepsilon>0$
If $\displaystyle \mathbb{N}$ is not bounded above then $\displaystyle \frac{1}{\varepsilon}$ is not an upper bound.
So $\displaystyle \exists n\in\mathbb{N}$ such that $\displaystyle n>\frac{1}{\varepsilon}$. Can you see how it works? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9891591668128967, "perplexity": 305.2671156258104}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257646375.29/warc/CC-MAIN-20180319042634-20180319062634-00775.warc.gz"} |
https://www.tiborstiluslapja.hu/profile/Delphi-2014-R3-Keygen-15-Latest-2022/profile | Csatlakozás dátuma: 2022. máj. 13.
###### Névjegy
Delphi 2014 R3 Keygen 15 [Latest] 2022
new york city, delphi 2014 r3 keygen 31 suisun city,delphi 2014 r3 keygen with word pad support,delphi 2014 r3 keygen may,delphi 2014 r3 keygen bbpress city,delphi 2014 r3 keygen asus city,delphi 2014 r3 keygen for mac.August 25th Saw most of the recent SFG post on P300, so I’ll comment briefly. The points I take issue with are: First, the P300 latency is on the order of 200ms, which is a very long time. Rather than interpret a peak latency at 200ms as the result of a predictable response time (400ms in the classic experiment by Sutton et al. (1988), for example), we need to keep in mind that it is a very rare phenomenon, not to be expected at all times. People respond to stimuli in the environment with certain relative probabilities, but the response itself is only a potential of action, it is not a certainty. If we assume, in the absence of data to the contrary, that this means that the P300 is a “slow” process, it means nothing to the overall problem of driving attention. In fact, it means the opposite. All that slow latency indicates is a large need for attention. Second, the latency of the P300 is the result of a complex process, and it can vary a great deal between individuals and over the course of time, depending on how much prefrontal cortex is functioning. Although in many conditions we can predict where the P300 will be by its latency, and so we can say that it is a “pre-cognitive” attention signal, the question of whether it is cognitive, or a mere potential of action, is something that we don’t know and don’t have a way of knowing at this time. We can say that the higher the P300 latency, the lower the odds of any action taking place. We can say that the more likely it is that a stimulus will lead to an action, the lower the latency. But we cannot say that all subjects with high P300 latency will ignore the stimulus, nor that all subjects with low P300 latency will attend to the stimulus. As we just saw in the SFG post, and in the comments of Seth and Marieke, the P300 can come and go, peak and trough, without any overall change in driving attention.
44926395d7 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8848583102226257, "perplexity": 1208.799894625056}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710953.78/warc/CC-MAIN-20221204004054-20221204034054-00123.warc.gz"} |
https://getrevising.co.uk/revision-tests/physics_ac_circuits | Physics- Ac circuits
HideShow resource information
• Created by: FireDwarf
• Created on: 23-11-13 12:09
What is a ac current?
An ac current is when a current repeatly reverses its direction. In one cycle, the carrier moves one way in the circuit and then reverse direction then re-reverse direction.
1 of 11
What is the freqency?
The number of complete cycles it passes through each second (hz)
2 of 11
Peak value?
Maximum current/pd which is the same in either direction.
3 of 11
What does time period mean?
Time for one complete cycle.
4 of 11
Why do we use ac?
Because a larger V is more efficent then larger current because less power is dissipated.
5 of 11
What is the freqency and mean voltage of the UK?
50hz and 230V
6 of 11
What does RMS mean?
It is the root mean square value and is the value of DC current that would give the same heating effect as the AC in the same resistor.
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What are the formulas for RMS?
Prms= Po (Peak Power)/ square root of 2 (same for Irms) Power = Irms(^2)R, Vrms = Irms x R
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heating effect of an AC current?
an electric heater at a very low frequency would be going cold then hot then cold then hot due to the alternating current (switching directions).
9 of 11
How does an ossiliscope work?
Electron gun emits an electron at a flurescent screen. Posistion of the spot of light is effected by the P.D across either of the deflecting plates. With no pd, the spot remains in the same space. Displacement is proportional to the applied p.d
10 of 11
Therefore how can it be used as a volt meter?
As the displacement of the beam is proprotional to the p.d applied.
11 of 11
Other cards in this set
Card 2
Front
What is the freqency?
Back
The number of complete cycles it passes through each second (hz)
Peak value?
Card 4
Front
What does time period mean?
Card 5
Front
Why do we use ac? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8782323598861694, "perplexity": 3789.618537746322}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698542686.84/warc/CC-MAIN-20161202170902-00057-ip-10-31-129-80.ec2.internal.warc.gz"} |
http://peeterjoot.com/tag/electromagnetic-field-strength/ | ## Potential solutions to the static Maxwell’s equation using geometric algebra
When neither the electromagnetic field strength $$F = \BE + I \eta \BH$$, nor current $$J = \eta (c \rho – \BJ) + I(c\rho_m – \BM)$$ is a function of time, then the geometric algebra form of Maxwell’s equations is the first order multivector (gradient) equation
\label{eqn:staticPotentials:20}
While direct solutions to this equations are possible with the multivector Green’s function for the gradient
\label{eqn:staticPotentials:40}
G(\Bx, \Bx’) = \inv{4\pi} \frac{\Bx – \Bx’}{\Norm{\Bx – \Bx’}^3 },
the aim in this post is to explore second order (potential) solutions in a geometric algebra context. Can we assume that it is possible to find a multivector potential $$A$$ for which
\label{eqn:staticPotentials:60}
is a solution to the Maxwell statics equation? If such a solution exists, then Maxwell’s equation is simply
\label{eqn:staticPotentials:80}
which can be easily solved using the scalar Green’s function for the Laplacian
\label{eqn:staticPotentials:240}
G(\Bx, \Bx’) = -\inv{\Norm{\Bx – \Bx’} },
a beastie that may be easier to convolve than the vector valued Green’s function for the gradient.
It is immediately clear that some restrictions must be imposed on the multivector potential $$A$$. In particular, since the field $$F$$ has only vector and bivector grades, this gradient must have no scalar, nor pseudoscalar grades. That is
\label{eqn:staticPotentials:100}
This constraint on the potential can be avoided if a grade selection operation is built directly into the assumed potential solution, requiring that the field is given by
\label{eqn:staticPotentials:120}
However, after imposing such a constraint, Maxwell’s equation has a much less friendly form
\label{eqn:staticPotentials:140}
Luckily, it is possible to introduce a transformation of potentials, called a gauge transformation, that eliminates the ugly grade selection term, and allows the potential equation to be expressed as a plain old Laplacian. We do so by assuming first that it is possible to find a solution of the Laplacian equation that has the desired grade restrictions. That is
\label{eqn:staticPotentials:160}
\begin{aligned}
\end{aligned}
for which $$F = \spacegrad A’$$ is a grade 1,2 solution to $$\spacegrad F = J$$. Suppose that $$A$$ is any formal solution, free of any grade restrictions, to $$\spacegrad^2 A = J$$, and $$F = \gpgrade{\spacegrad A}{1,2}$$. Can we find a function $$\tilde{A}$$ for which $$A = A’ + \tilde{A}$$?
Maxwell’s equation in terms of $$A$$ is
\label{eqn:staticPotentials:180}
\begin{aligned}
J
\end{aligned}
or
\label{eqn:staticPotentials:200}
This non-homogeneous Laplacian equation that can be solved as is for $$\tilde{A}$$ using the Green’s function for the Laplacian. Alternatively, we may also solve the equivalent first order system using the Green’s function for the gradient.
\label{eqn:staticPotentials:220}
Clearly $$\tilde{A}$$ is not unique, as we can add any function $$\psi$$ satisfying the homogeneous Laplacian equation $$\spacegrad^2 \psi = 0$$.
In summary, if $$A$$ is any multivector solution to $$\spacegrad^2 A = J$$, that is
\label{eqn:staticPotentials:260}
A(\Bx)
= \int dV’ G(\Bx, \Bx’) J(\Bx’)
= -\int dV’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} },
then $$F = \spacegrad A’$$ is a solution to Maxwell’s equation, where $$A’ = A – \tilde{A}$$, and $$\tilde{A}$$ is a solution to the non-homogeneous Laplacian equation or the non-homogeneous gradient equation above.
### Integral form of the gauge transformation.
Additional insight is possible by considering the gauge transformation in integral form. Suppose that
\label{eqn:staticPotentials:280}
A(\Bx) = -\int_V dV’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} } – \tilde{A}(\Bx),
is a solution of $$\spacegrad^2 A = J$$, where $$\tilde{A}$$ is a multivector solution to the homogeneous Laplacian equation $$\spacegrad^2 \tilde{A} = 0$$. Let’s look at the constraints on $$\tilde{A}$$ that must be imposed for $$F = \spacegrad A$$ to be a valid (i.e. grade 1,2) solution of Maxwell’s equation.
\label{eqn:staticPotentials:300}
\begin{aligned}
F
&=
-\int_V dV’ \lr{ \spacegrad \inv{\Norm{\Bx – \Bx’} } } J(\Bx’)
&=
\int_V dV’ \lr{ \spacegrad’ \inv{\Norm{\Bx – \Bx’} } } J(\Bx’)
&=
\int_V dV’ \spacegrad’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} } – \int_V dV’ \frac{\spacegrad’ J(\Bx’)}{\Norm{\Bx – \Bx’} }
&=
\int_{\partial V} dA’ \ncap’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} } – \int_V \frac{\spacegrad’ J(\Bx’)}{\Norm{\Bx – \Bx’} }
\end{aligned}
Where $$\ncap’ = (\Bx’ – \Bx)/\Norm{\Bx’ – \Bx}$$, and the fundamental theorem of geometric calculus has been used to transform the gradient volume integral into an integral over the bounding surface. Operating on Maxwell’s equation with the gradient gives $$\spacegrad^2 F = \spacegrad J$$, which has only grades 1,2 on the left hand side, meaning that $$J$$ is constrained in a way that requires $$\spacegrad J$$ to have only grades 1,2. This means that $$F$$ has grades 1,2 if
\label{eqn:staticPotentials:320}
= \int_{\partial V} dA’ \frac{ \gpgrade{\ncap’ J(\Bx’)}{0,3} }{\Norm{\Bx – \Bx’} }.
The product $$\ncap J$$ expands to
\label{eqn:staticPotentials:340}
\begin{aligned}
\ncap J
&=
&=
\ncap \cdot (-\eta \BJ) + \gpgradethree{\ncap (-I \BM)} \\
&=- \eta \ncap \cdot \BJ -I \ncap \cdot \BM,
\end{aligned}
so
\label{eqn:staticPotentials:360}
=
-\int_{\partial V} dA’ \frac{ \eta \ncap’ \cdot \BJ(\Bx’) + I \ncap’ \cdot \BM(\Bx’)}{\Norm{\Bx – \Bx’} }.
Observe that if there is no flux of current density $$\BJ$$ and (fictitious) magnetic current density $$\BM$$ through the surface, then $$F = \spacegrad A$$ is a solution to Maxwell’s equation without any gauge transformation. Alternatively $$F = \spacegrad A$$ is also a solution if $$\lim_{\Bx’ \rightarrow \infty} \BJ(\Bx’)/\Norm{\Bx – \Bx’} = \lim_{\Bx’ \rightarrow \infty} \BM(\Bx’)/\Norm{\Bx – \Bx’} = 0$$ and the bounding volume is taken to infinity.
# References
## Generalizing Ampere’s law using geometric algebra.
The question I’d like to explore in this post is how Ampere’s law, the relationship between the line integral of the magnetic field to current (i.e. the enclosed current)
\label{eqn:flux:20}
\oint_{\partial A} d\Bx \cdot \BH = -\int_A \ncap \cdot \BJ,
generalizes to geometric algebra where Maxwell’s equations for a statics configuration (all time derivatives zero) is
\label{eqn:flux:40}
where the multivector fields and currents are
\label{eqn:flux:60}
\begin{aligned}
F &= \BE + I \eta \BH \\
J &= \eta \lr{ c \rho – \BJ } + I \lr{ c \rho_\txtm – \BM }.
\end{aligned}
Here (fictitious) the magnetic charge and current densities that can be useful in antenna theory have been included in the multivector current for generality.
My presumption is that it should be possible to utilize the fundamental theorem of geometric calculus for expressing the integral over an oriented surface to its boundary, but applied directly to Maxwell’s equation. That integral theorem has the form
\label{eqn:flux:80}
\int_A d^2 \Bx \boldpartial F = \oint_{\partial A} d\Bx F,
where $$d^2 \Bx = d\Ba \wedge d\Bb$$ is a two parameter bivector valued surface, and $$\boldpartial$$ is vector derivative, the projection of the gradient onto the tangent space. I won’t try to explain all of geometric calculus here, and refer the interested reader to [1], which is an excellent reference on geometric calculus and integration theory.
The gotcha is that we actually want a surface integral with $$\spacegrad F$$. We can split the gradient into the vector derivative a normal component
\label{eqn:flux:160}
so
\label{eqn:flux:100}
=
\int_A d^2 \Bx \boldpartial F
+
\int_A d^2 \Bx \ncap \lr{ \ncap \cdot \spacegrad } F,
so
\label{eqn:flux:120}
\begin{aligned}
\oint_{\partial A} d\Bx F
&=
\int_A d^2 \Bx \lr{ J – \ncap \lr{ \ncap \cdot \spacegrad } F } \\
&=
\int_A dA \lr{ I \ncap J – \lr{ \ncap \cdot \spacegrad } I F }
\end{aligned}
This is not nearly as nice as the magnetic flux relationship which was nicely split with the current and fields nicely separated. The $$d\Bx F$$ product has all possible grades, as does the $$d^2 \Bx J$$ product (in general). Observe however, that the normal term on the right has only grades 1,2, so we can split our line integral relations into pairs with and without grade 1,2 components
\label{eqn:flux:140}
\begin{aligned}
&=
\int_A dA \gpgrade{ I \ncap J }{0,3} \\
&=
\int_A dA \lr{ \gpgrade{ I \ncap J }{1,2} – \lr{ \ncap \cdot \spacegrad } I F }.
\end{aligned}
Let’s expand these explicitly in terms of the component fields and densities to check against the conventional relationships, and see if things look right. The line integrand expands to
\label{eqn:flux:180}
\begin{aligned}
d\Bx F
&=
d\Bx \lr{ \BE + I \eta \BH }
=
d\Bx \cdot \BE + I \eta d\Bx \cdot \BH
+
d\Bx \wedge \BE + I \eta d\Bx \wedge \BH \\
&=
d\Bx \cdot \BE
– \eta (d\Bx \cross \BH)
+ I (d\Bx \cross \BE )
+ I \eta (d\Bx \cdot \BH),
\end{aligned}
the current integrand expands to
\label{eqn:flux:200}
\begin{aligned}
I \ncap J
&=
I \ncap
\lr{
\frac{\rho}{\epsilon} – \eta \BJ + I \lr{ c \rho_\txtm – \BM }
} \\
&=
\ncap I \frac{\rho}{\epsilon} – \eta \ncap I \BJ – \ncap c \rho_\txtm + \ncap \BM \\
&=
\ncap \cdot \BM
+ \eta (\ncap \cross \BJ)
– \ncap c \rho_\txtm
+ I (\ncap \cross \BM)
+ \ncap I \frac{\rho}{\epsilon}
– \eta I (\ncap \cdot \BJ).
\end{aligned}
We are left with
\label{eqn:flux:220}
\begin{aligned}
\oint_{\partial A}
\lr{
d\Bx \cdot \BE + I \eta (d\Bx \cdot \BH)
}
&=
\int_A dA
\lr{
\ncap \cdot \BM – \eta I (\ncap \cdot \BJ)
} \\
\oint_{\partial A}
\lr{
– \eta (d\Bx \cross \BH)
+ I (d\Bx \cross \BE )
}
&=
\int_A dA
\lr{
\eta (\ncap \cross \BJ)
– \ncap c \rho_\txtm
+ I (\ncap \cross \BM)
+ \ncap I \frac{\rho}{\epsilon}
-\PD{n}{} \lr{ I \BE – \eta \BH }
}.
\end{aligned}
This is a crazy mess of dots, crosses, fields and sources. We can split it into one equation for each grade, which will probably look a little more regular. That is
\label{eqn:flux:240}
\begin{aligned}
\oint_{\partial A} d\Bx \cdot \BE &= \int_A dA \ncap \cdot \BM \\
\oint_{\partial A} d\Bx \cross \BH
&=
\int_A dA
\lr{
– \ncap \cross \BJ
+ \frac{ \ncap \rho_\txtm }{\mu}
– \PD{n}{\BH}
} \\
\oint_{\partial A} d\Bx \cross \BE &=
\int_A dA
\lr{
\ncap \cross \BM
+ \frac{\ncap \rho}{\epsilon}
– \PD{n}{\BE}
} \\
\oint_{\partial A} d\Bx \cdot \BH &= -\int_A dA \ncap \cdot \BJ \\
\end{aligned}
The first and last equations could have been obtained much more easily from Maxwell’s equations in their conventional form more easily. The two cross product equations with the normal derivatives are not familiar to me, even without the fictitious magnetic sources. It is somewhat remarkable that so much can be packed into one multivector equation:
\label{eqn:flux:260}
\oint_{\partial A} d\Bx F
=
I \int_A dA \lr{ \ncap J – \PD{n}{F} }.
# References
[1] A. Macdonald. Vector and Geometric Calculus. CreateSpace Independent Publishing Platform, 2012. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9967906475067139, "perplexity": 3152.3234565702055}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496671363.79/warc/CC-MAIN-20191122143547-20191122172547-00073.warc.gz"} |
https://ec.gateoverflow.in/1522/gate2020-ec-1 | If $v_{1},v_{2}, \dots ,v_{6}$ are six vectors in $\mathbb{R}^{4}$ , which one of the following statements is $\text{FALSE}$?
1. It is not necessary that these vectors span $\mathbb{R}^{4}$.
3. Any four of these vectors form a basis for $\mathbb{R}^{4}$.
4. If $\left \{ v_{1},v_{3},v_{5},v_{6} \right\}$ spans $\mathbb{R}^{4}$ , then it forms a basis for $\mathbb{R}^{4}$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9738558530807495, "perplexity": 391.35277514013103}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107874026.22/warc/CC-MAIN-20201020162922-20201020192922-00662.warc.gz"} |
http://www.maths.ox.ac.uk/events/past/627/1990--now?field_seminar_date_value_op=%3C&field_seminar_date_value%5Bvalue%5D=&field_seminar_date_value%5Bmin%5D=&field_seminar_date_value%5Bmax%5D=&page=9 | # Past Applied Analysis and Mechanics Seminar
2 February 2004
17:00
Daniel Faraco
Abstract
Recently Friesecke, James and Muller established the following quantitative version of the rigidity of SO(n) the group of special orthogonal matrices. Let U be a bounded Lipschitz domain. Then there exists a constant C(U) such that for any mapping v in the L2-Sobelev space the L^2-distance of the gradient controlls the distance of v a a single roation. This interesting inequality is fundamental in several problems concerning dimension reduction in nonlinear elasticity. In this talk, we will present a joint work with Muller and Zhong where we investigate an analagous quantitative estimate where we replace SO(n) by an arbitrary smooth, compact and SO(n) invariant subset of the conformal matrices E. The main novelty is that exact solutions to the differential inclusion Df(x) in E a.e.x in U are not necessarily affine mappings.
• Applied Analysis and Mechanics Seminar
26 January 2004
17:00
Jonathan Bevan
Abstract
Using a technique explored in unpublished work of Ball and Mizel I shall show that already in 2 and 3 dimensions there are vectorfields which are singular minimizers of integral functionals whose integrand is strictly polyconvex and depends on the gradient of the map only. The analysis behind these results gives rise to an interesting question about the relationship between the regularity of a polyconvex function and that of its possible convex representatives. I shall indicate why this question is interesting in the context of the regularity results above and I shall answer it in certain cases.
• Applied Analysis and Mechanics Seminar
7 January 2004
16:00
N Kikuchi
Abstract
• Applied Analysis and Mechanics Seminar
1 December 2003
17:00
Jan Maly
Abstract
• Applied Analysis and Mechanics Seminar
17 November 2003
17:00
Dr Andrew Lorent
Abstract
Take any region omega and let function u defined inside omega be the distance from the boundary, u solves the iconal equation \lt|Du\rt|=1 with boundary condition zero. Functional u is also conjectured (in some cases proved) to be the "limiting minimiser" of various functionals that arise models of blistering and micro magnetics. The precise formulation of these problems involves the notion of gamma convergence. The Aviles Giga functional is a natural "second order" generalisation of the Cahn Hilliard model which was one of the early success of the theory of gamma convergence. These problems turn out to be surprisingly rich with connections to a number of areas of pdes. We will survey some of the more elementary results, describe in detail of one main problems in field and state some partial results.
• Applied Analysis and Mechanics Seminar
10 November 2003
17:00
Viet Ha Hoang
Abstract
• Applied Analysis and Mechanics Seminar
3 November 2003
17:00
Carlos Mora-Corral
Abstract
• Applied Analysis and Mechanics Seminar
13 October 2003
17:00
David Schaeffer
Abstract
• Applied Analysis and Mechanics Seminar | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9337983131408691, "perplexity": 1346.9350865969798}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267156192.24/warc/CC-MAIN-20180919102700-20180919122700-00297.warc.gz"} |
https://www.physicsforums.com/threads/potential-energy-of-an-arrangement-of-point-charges.796375/ | # Potential energy of an arrangement of point charges
1. Feb 6, 2015
### acdurbin953
1. The problem statement, all variables and given/known data
What is the potential energy of the system composed of the three charges q1, q3, and q4, when q1 is at point R? Define the potential energy to be zero at infinity.
Charges are arranged in a triangle, with q3 and q4 located up and down 2.2 cm from the origin, and q1 on the x axis at a distance R = 3.5 cm from the origin.
q1 = 1.9uC
q3 = q4 = -1.8 uC
distance from q1 to each of the other 2 charges is 4.1 cm
2. Relevant equations
U-kqq/r
3. The attempt at a solution
U1 = [(9E9)(-1.8E-6)(1.9E6)]/0.041 = -0.751
Since there is symmetry I doubled U1 to find U = -1.501, however that answer is wrong. Where am I messing up?
2. Feb 6, 2015
### acdurbin953
I just figured out that you have to also considered the two point charges potential energy on each other. Case closed! Sorry for the (unnecessary) post. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.880260169506073, "perplexity": 770.8304214851963}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257648431.63/warc/CC-MAIN-20180323180932-20180323200932-00227.warc.gz"} |
http://mathhelpforum.com/calculus/28099-equation-motion-particle-where-s-meters-t-seconds-gi-print.html | # The equation of motion of a particle, where s is in meters and t is in seconds, is gi
• Feb 12th 2008, 01:03 PM
plstevens
The equation of motion of a particle, where s is in meters and t is in seconds, is gi
s=7t^3-3t
(a)Find the velocity and acceleration as functions of t.
v(t)=
a(t)=
(b) Find the acceleration after 7 seconds.
_______m/s^2
(c) Find the acceleration when the velocity is 0. (Round the answer to one decimal place) _______m/s^2
• Feb 12th 2008, 02:11 PM
CaptainBlack
Quote:
Originally Posted by plstevens
s=7t^3-3t
(a)Find the velocity and acceleration as functions of t.
v(t)=
a(t)=
(b) Find the acceleration after 7 seconds.
_______m/s^2
(c) Find the acceleration when the velocity is 0. (Round the answer to one decimal place) _______m/s^2
$v(t)=\frac{ds}{dt}(t)$
and:
$a(t)=\frac{dv}{dt}(t)=\frac{d^2 s}{dt^2}(t)$
RonL
• Feb 12th 2008, 06:01 PM
plstevens
so what do i do for b and c
• Feb 12th 2008, 06:41 PM
topsquark
Quote:
Originally Posted by plstevens
so what do i do for b and c
b) You have your function for a(t) so what is a(7)?
c) When is v(t) = 0? Find that value of t and then find the acceleration at that time.
-Dan
• Feb 12th 2008, 06:50 PM
plstevens
i don't mean to be a dumby but i still don't get it
• Feb 12th 2008, 07:45 PM
topsquark
Quote:
Originally Posted by plstevens
i don't mean to be a dumby but i still don't get it
If I gave you a function, for example, a(t) = 8t + 3 what would you tell me the value of a(7) is? Part b is exactly this, but of course with the function you derived in part a).
-Dan
• Feb 12th 2008, 07:47 PM
plstevens
I'm so sorry but thats not making sense
• Feb 12th 2008, 07:54 PM
topsquark
Quote:
Originally Posted by plstevens
I'm so sorry but thats not making sense
In part a) your position function is given as $s(t) = 7t^3 - 3t$. From this you get that the acceleration function is
$a(t) = 42t$
What is the value of a(7), the value of the function a(t) when t = 7?
If you can't answer that, then I suggest you have a long talk with your instructor.
-Dan | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9406499266624451, "perplexity": 895.2331551248528}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891814833.62/warc/CC-MAIN-20180223194145-20180223214145-00107.warc.gz"} |
http://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/Lévy+hierarchy | foundations
# The Lévy hierarchy
## Idea
In logic, model theory, and set theory, the Lévy hierarchy is a stratification of formulas, definable sets, and (definable) classes according to the complexity of the unbounded quantifiers.
## Definition
###### Definition
We define classes of formulas $\Sigma_n$, $\Pi_n$, and $\Delta_n$ by induction on $n$.
• A formula is $\Sigma_0$ iff it is $\Pi_0$ iff it is $\Delta_0$, by definition if it is equivalent to a formula all of whose quantifiers are bounded, i.e. of the form $\forall x\in A$ or $\exists x\in A$.
• A formula is $\Sigma_{n+1}$ if it is equivalent to one of the form $\exists \vec{x}. \phi$, where $\vec{x}$ is a list of variables and $\phi$ is $\Pi_n$.
• A formula is $\Pi_{n+1}$ if it is equivalent to one of the form $\forall \vec{x}. \phi$, where $\vec{x}$ is a list of variables and $\phi$ is $\Sigma_n$.
• A formula is $\Delta_n$ if it is both $\Sigma_n$ and $\Pi_n$.
A class is given one of these labels if it can be defined by a formula which has that label.
###### Remark
The notation “$\Sigma$” and “$\Pi$” can be explained by the fact that the existential quantifier is related to a dependent sum, while the universal quantifier is related to the dependent product.
###### Remark
These definitions are most useful in classical mathematics, in which every formula is equivalent to one all of whose unbounded quantifiers are in the front, that is, a formula in prenex normal form?, so that every formula belongs to some $\Sigma_n$ or $\Pi_n$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 26, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9994100332260132, "perplexity": 167.81380824022992}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320301720.45/warc/CC-MAIN-20220120035934-20220120065934-00342.warc.gz"} |
https://www.myopenmath.com/course/public.php?cid=8812&folder=0-10-4 | ## Section 4.3: Linear Equations in Standard Form
Textbook Section 4.3: Linear Equations in Standard Form
Textbook Section 4.3: Linear Equations in Standard Form
Learning Objectives
• Write equivalent equations in standard form.
• Find the slope and y−intercept from an equation in standard form.
• Write equations in standard form from a graph.
• Solve real-world problems using linear models in standard form.
Section 4.3 Videos
This is the publicly accessible content from a course on MyOpenMath. There may be additional content available by logging in | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9598469138145447, "perplexity": 3553.9075807934796}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891816370.72/warc/CC-MAIN-20180225110552-20180225130552-00187.warc.gz"} |
https://research.tue.nl/en/publications/a-sequential-least-squares-algorithm-for-armax-dynamic-network-id | # A sequential least squares algorithm for ARMAX dynamic network identification
Harm H.M. Weerts, Miguel Galrinho, Giulio Bottegal, Håkan Hjalmarsson, Paul M.J.Van den Hof
Research output: Contribution to journalConference articleAcademicpeer-review
### Abstract
Identification of dynamic networks in prediction error setting often requires the solution of a non-convex optimization problem, which can be difficult to solve especially for large-scale systems. Focusing on ARMAX models of dynamic networks, we instead employ a method based on a sequence of least-squares steps. For single-input single-output models, we show that the method is equivalent to the recently developed Weighted Null Space Fitting, and, drawing from the analysis of that method, we conjecture that the proposed method is both consistent as well as asymptotically efficient under suitable assumptions. Simulations indicate that the sequential least squares estimates can be of high quality even for short data sets.
Language English 844-849 6 IFAC-PapersOnLine 51 15 10.1016/j.ifacol.2018.09.119 Published - 1 Jan 2018
### Fingerprint
Large scale systems
### Keywords
• dynamic networks
• identification algorithm
• least squares
• System identification
### Cite this
@article{2fe0037221b54321858b0b9aa01c1dd6,
title = "A sequential least squares algorithm for ARMAX dynamic network identification",
abstract = "Identification of dynamic networks in prediction error setting often requires the solution of a non-convex optimization problem, which can be difficult to solve especially for large-scale systems. Focusing on ARMAX models of dynamic networks, we instead employ a method based on a sequence of least-squares steps. For single-input single-output models, we show that the method is equivalent to the recently developed Weighted Null Space Fitting, and, drawing from the analysis of that method, we conjecture that the proposed method is both consistent as well as asymptotically efficient under suitable assumptions. Simulations indicate that the sequential least squares estimates can be of high quality even for short data sets.",
keywords = "dynamic networks, identification algorithm, least squares, System identification",
author = "Weerts, {Harm H.M.} and Miguel Galrinho and Giulio Bottegal and H{\aa}kan Hjalmarsson and {den Hof}, {Paul M.J.Van}",
year = "2018",
month = "1",
day = "1",
doi = "10.1016/j.ifacol.2018.09.119",
language = "English",
volume = "51",
pages = "844--849",
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}
A sequential least squares algorithm for ARMAX dynamic network identification. / Weerts, Harm H.M.; Galrinho, Miguel; Bottegal, Giulio; Hjalmarsson, Håkan; den Hof, Paul M.J.Van.
In: IFAC-PapersOnLine, Vol. 51, No. 15, 01.01.2018, p. 844-849.
Research output: Contribution to journalConference articleAcademicpeer-review
TY - JOUR
T1 - A sequential least squares algorithm for ARMAX dynamic network identification
AU - Weerts,Harm H.M.
AU - Galrinho,Miguel
AU - Bottegal,Giulio
AU - den Hof,Paul M.J.Van
PY - 2018/1/1
Y1 - 2018/1/1
N2 - Identification of dynamic networks in prediction error setting often requires the solution of a non-convex optimization problem, which can be difficult to solve especially for large-scale systems. Focusing on ARMAX models of dynamic networks, we instead employ a method based on a sequence of least-squares steps. For single-input single-output models, we show that the method is equivalent to the recently developed Weighted Null Space Fitting, and, drawing from the analysis of that method, we conjecture that the proposed method is both consistent as well as asymptotically efficient under suitable assumptions. Simulations indicate that the sequential least squares estimates can be of high quality even for short data sets.
AB - Identification of dynamic networks in prediction error setting often requires the solution of a non-convex optimization problem, which can be difficult to solve especially for large-scale systems. Focusing on ARMAX models of dynamic networks, we instead employ a method based on a sequence of least-squares steps. For single-input single-output models, we show that the method is equivalent to the recently developed Weighted Null Space Fitting, and, drawing from the analysis of that method, we conjecture that the proposed method is both consistent as well as asymptotically efficient under suitable assumptions. Simulations indicate that the sequential least squares estimates can be of high quality even for short data sets.
KW - dynamic networks
KW - identification algorithm
KW - least squares
KW - System identification
UR - http://www.scopus.com/inward/record.url?scp=85054462289&partnerID=8YFLogxK
U2 - 10.1016/j.ifacol.2018.09.119
DO - 10.1016/j.ifacol.2018.09.119
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SP - 844
EP - 849
JO - IFAC-PapersOnLine
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http://physics.wikia.com/wiki/Hooke%27s_law | ## FANDOM
150 Pages
Hooke's law is a law which states that the force, F, required to bend a spring (or some other elastic object) is directly proportional to the distance X by some constant k, known as the stiffness constant.
$F=kX$
This is sometimes written as
$F=-kX$
In this case, F is equal to the force with which the spring pushes back.
By integrating with respect to x, we can find the work needed to compress or stretch a spring a given distance and the potential energy stored in said spring.
$W = E_p = \frac{1}{2} k x^2$
Hooke's law is only an approximation, as all materials will deform past a certain point (called the elastic limit). In fact, many objects deviate from Hooke's law well before their elastic limits. However, for most cases, Hooke's law is fairly accurate.
## Harmonic motionEdit
If a weight is attached to a spring and the spring is stretched or compressed released, the motion can be described as
$x=A \cos ( \sqrt{ \tfrac{k}{m}} t )$
where A is the amplitude, or how far the spring is stretched, k is the stiffness constant, and m is the mass of the weight. By taking the first and second derivatives, the speed and acceleration can be found.
The total energy of the system is equal to
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http://mathhelpforum.com/algebra/220898-addition-fractions-2.html | # Thread: Addition with fractions
1. ## Re: Addition with fractions
[QUOTE=Plato;794073]Why do you complicate this so?
What a ridiculous thing to say... I made it complicated because I didn't know what I was doing, also your post doesn't help at all, you have not told me what you did the arrive at that answer. HallsOfIvy, topsquark and emakarov - Thank you for explaining.
2. ## Re: Addition with fractions
uperkurk...
$\displaystyle \frac{2}{5}$ is a fraction.
$\displaystyle \frac{3}{x(x+2)}$ is a fraction.
$\displaystyle \frac{2x^2 + 5x + 3}{x^2 - 10x + 25}$ is a fraction.
Don't let the complicated terms scare you. Essentially, they're just fractions. That means they have to follow all the rules of fractions. So whatever you'd normally do with $\displaystyle \frac{2}{3} + \frac{4}{5}$, you apply the same set of rules with $\displaystyle \frac{3x}{x-2} + \frac{4}{x+3}$.
3. ## Re: Addition with fractions
Originally Posted by uperkurk
Originally Posted by Plato
Why do you complicate this so?...
What a ridiculous thing to say... I made it complicated because I didn't know what I was doing, also your post doesn't help at all, you have not told me what you did the arrive at that answer. HallsOfIvy, topsquark and emakarov - Thank you for explaining.
I'm certain Plato meant no offense...he was simply demonstrating what is sometimes referred to as "cross multiplying."
If you are given an equation of the form:
$\displaystyle \frac{a}{b}=\frac{c}{d}$
Then, by this process, you may take as one side of the equation the product of the numerator on one side and the denominator on the other, and then then other side of the equation is the product of the other numerator-denominator pair. So, the equation above can then be written as:
$\displaystyle ad=bc$
4. ## Re: Addition with fractions
Fractional quantities are a by-product of wanting number systems "complete under division (except for 0)". As a result, multiplying with fractions is EASY, but adding them is HARD. The general rule is, of course:
a/b + c/d = (ad +bc)/(bd)
Sometimes, that is as good as it gets...it's not always possible to "factor out common factors" (although textbook problems often feature this, misleading students into an overly optimistic view of how things will turn out).
The equation:
a/b = c/d is often taken by definition to MEAN:
ad = bc...it's how we can tell that 1/2 is the same fraction as 2/4 (...because 1*4 = 2*2, see?). In other words "fractional expressions" aren't UNIQUE, we can always multiply top and bottom by the same thing (because a/a = 1, if a isn't 0), to get a "different-looking fraction". "Cancelling" is this process "in reverse" we are DIVIDING by a/a (and when you DIVIDE by b/c, you multiply by c/b (the reciprocal), and the reciprocal of a/a is, strangely enough, a/a again).
I've always felt that too much emphasis is placed upon putting things in "simplest form". It doesn't make an answer any more or less CORRECT, it just takes up less space on a piece of paper (and it might simplify calculations if you have to USE that answer later on). Fractions are often messy, and ugly-looking beasts. It may not seem so to you, dear uperkurk, but the problems you are tasked with solving have been tailored to shield you from the full blunt force of what fractional expressions can be.
5. ## Re: Addition with fractions
What you have to remember, uperkurk, is that the denominator has to be the same in all terms of the equation. It's no different from when you learned to add fractions in Arithmetic.
Example: $\displaystyle \displaystyle\frac{1}{4} + 1 +\displaystyle\frac{2}{3}$
$\displaystyle \displaystyle\frac{1}{4} = \displaystyle\frac{1(3)}{4(3)} = \displaystyle\frac{3}{12}$
$\displaystyle \displaystyle\frac{2}{3} = \displaystyle\frac{2(4)}{3(4)} = \displaystyle\frac{8}{12}$
$\displaystyle 1= \displaystyle\frac{1(3)(4)}{1(3)(4)} = \displaystyle\frac{12}{12}$
$\displaystyle \displaystyle\frac{1}{4} + 1 + \displaystyle\frac{2}{3} = \displaystyle\frac{3}{12} +\displaystyle\frac{12}{12} + \displaystyle\frac{8}{12} = \displaystyle\frac{23}{12}$
You are doing the same thing when adding in algebraic equations. I believe this was explained in earlier posts, but I thought that one more example would help.
Page 2 of 2 First 12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.913118302822113, "perplexity": 962.0076869049686}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267860089.13/warc/CC-MAIN-20180618070542-20180618090542-00522.warc.gz"} |
https://www.physicsforums.com/threads/angle-of-airfoil-to-fuselage.771434/ | # Angle of airfoil to fuselage?
1. Sep 17, 2014
### WK95
For a given airfoil, the amount of lift and drag and thus the L/D ratio vary depending on it's angle towards the airflow.
In general, for small aircraft with a straight wing shape (I say so because that seems to be the simplest case since the airfoil shape will be constant throughout the length of the wing), at what angle of attack is the wing attached to the rest of the plane or fuselage.
Forgive me if I misused any terms or lack certain details. I'm not an aerospace engineer (yet).
2. Sep 20, 2014
### FactChecker
Similar Discussions: Angle of airfoil to fuselage? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9044695496559143, "perplexity": 1497.854315822137}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187822851.65/warc/CC-MAIN-20171018085500-20171018105500-00204.warc.gz"} |
https://www.mapleprimes.com/users/johan162/questions | ## 4 Badges
13 years, 191 days
## Difference between latex() command and "...
Maple 2021
It seem like there is an (unwanted) difference between the latex() command and the newly introduced "Copy as LaTeX" in handling of \left and \right delimiters. A trivial example
>latex(sin(b/a));
\sin \! \left(\frac{b}{a}\right)
but if I instead use the new "Copy as LaTeX" command I get the result
\sin(\frac{b}{a})
These should of course be the same and I consider the first one to be the better conversion. It seems a very strange problem since it surely must be the same library routine that does the conversion in both cases? From a small set of examples I tried on it seem that the "Copy as .." command never uses \left and \right which in several cases is absolutely necessary to get acceptable result.
## Color of minortick gridlines?...
Maple
Version: Maple 2020.1
When i set the color for the gridlines it only seems to be applied for the major-tick gridlines as the following trivial example shows:
plot(sin(t), t = 0 .. Pi, axes = frame, background = "#303030", color = "Orange", axis = [gridlines = [color = "#707070", linestyle = dot]])
I assume it must also be possible to also specify the color for the minor tick-marks gridlines?
The obvious (?) variant "axis=[gridlines = [color = ["#707070", "#707070"] , ... " just seems to crash maple (nothing happens when the plot() expression is evaluated).
I'm unable to find anything in the documentation regarding this and it only seems to imply that the color should be applied to both major & minor gridlines which is not the case.
?
## Repeatable Kernel crash using 2D-math bu...
Maple 2020
(Context: As part of a EU consumer watchdog report I've been asked to re-validate a number of publically stated APR rates for various consumer loan. )
(Apologize if this should have been posted to some bug-tracker but I was unable to find such a forum.)
System: MacOS 10.14.6, Maple 2020.1
Summary: Some simple exponential summations entered in 2D Input seems to crash the kernel and it is dependent on the numerical value of the exponent. The same expressions entered in 1D plaintext Maple Notation works fine. The numerically/expression evaluations are also significantly slower in 2D Input in a Document (x10) as compared to 1D Maple Notation in a Worksheet.
The attached worksheet is a "killer" worksheet and will on OSX 10.14.6 + Maple 2020.1 kill the kernel connection (crash the kernel).
kernel-crash.mw
Example:
A trivial example, entering 2D Math (assume PV & C are positive real numbers), say
will cause a lost connection to the kernel after 40-50s entering the epression and the UI being busy (unclear what it is doing since no real calculation is performed) which I assume is a sign that the kernel crashed. It seems to dependent on the exponent in the divisor. So for example the following variation will not crash the kernel
Doing the exact same calculation with a worksheet in old plain maple notation both varianta are both significantly faster in the numerical operation (solving for 'r') and assigning 'eq2' and never crashes (regardless of numerical value of exponent).
The workaround is of course obvious but it would be nice if this bug could be adressed.
It seems that whenever I try to give the 2D-UI a chance (since it is actally easier to visually view complex expressions) something always comes back and bites me...
Update: The crash is only repeatable with the "sum()" command and not the "add()" command.
Page 1 of 1
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8£º30-8£º40 ¿ªÄ»Ê½ 8£º40-9£º40 ±¨¸æÌâÄ¿£º Harmonic mappings for bounded distortion shape deformation and interpolation ±¨¸æÈË£º ³ÂÈÊ½Ü µÂ¹úÂí¿Ë˹ÆÕÀʿ˼ÆËã»úÑо¿Ëù ±¨¸æÕªÒª£º Harmonic mappings are extensively used in geometry processing applications to produce visually appealing deformations. We establish the sufficient and necessary conditions for a harmonic planar mapping to have bounded distortion. Our key observation is that these conditions relate solely to the boundary behavior of the mapping. This leads to an efficient and accurate algorithm that supports handle-based interactive shape-and-image deformation and is demonstrated to outperform other state-of-the-art methods. The particular structure of harmonic mappings further allows efficient shape interpolation. Given the closed-form expressions for the interpolants, our interpolation algorithm runs embarrassingly in parallel and is orders of magnitude faster than state-of-the-art methods due to its simplicity, yet it produces mappings that are superior to those existing techniques due to guaranteed bounds on geometric distortions. 9£º50-10£º50 ±¨¸æÌâÄ¿£º Weak approximation for 0-cycles on products of varieties ±¨¸æÈË£º ÁºÓÀì÷ °ÍÀèÆß´ó ±¨¸æÕªÒª£º We consider the sequence for smooth projective varieties over number fields. Its exactness is conjectured by Colliot-Th¡äen`ene{Sansuc and Kato{Saito, it means roughly that the Brauer{Manin obstruction is the only obstruction to weak approximation for 0-cycles. We work on the compatibility of the exactness for products of varieties. Assume that (E) is exact for a rationally connected variety X (after all finite extensions of the base field). One may ask the question : For which family of varieties Y the sequence (E) is exact for X ¡Á Y ? When Y is a smooth projective curve with the finiteness of the Tate{Shafarevich group of its jacobian assumed, Harpaz and Wittenberg give a positive answer to the question (for much more general fibrations rather than only for products). We will talk about the case where Y is a smooth compactification of a homogeneous space. 11£º00-12£º00 ±¨¸æÌâÄ¿£º The monodromy theorem for compact Kähler manifolds and smooth quasi-projective varieties ±¨¸æÈË£º ÁõÓÀÇ¿ KU Leuven ±¨¸æÕªÒª£º Given any connected topological space X, assume that there exists an epimorphism from the fundamental group of X to the free ableian group Z. The deck transformation group Z acts on the associated infinite cyclic cover of X, hence on the homology group of the covering space with complex coefficients. This action induces a linear automorphism on the torsion part of the homology group as a module over the complex Laurent polynomial ring, which is a finite dimensional complex vector space. We study the sizes of the Jordan blocks of this linear automorphism. When X is a compact K\"ahler manifold, we show that all the Jordan blocks are of size one. When X is a smooth complex quasi-projective variety, we give an upper bound on the sizes of the Jordan blocks, which is an analogue of the Monodromy Theorem for the local Milnor fibration. This is a joint work with Nero Budur and Botong Wang. 12£º00-14£º00 Lunch & break 14:00-15:00 ±¨¸æÌâÄ¿£º Boundary $C^{1,\alpha}$ regularity of Potential functions in Optimal transportation ±¨¸æÈË£º ³ÂÊÀ±þ °ÄÖÞ¹úÁ¢´óѧ ±¨¸æÕªÒª£º We provide a different proof for the global $C^{1,\alpha}$ regularity of potential functions in optimal transport problem, which was originally proved by Caffarelli. Moreover, our method applies to a more general class of domains. This is based on a joint work with Elina Andriyanova. 15£º10-16£º10 ±¨¸æÌâÄ¿£º Scaling Limits of Critical Inhomogeneous Random Graphs ±¨¸æÈË£º Minmin Wang Institut Henri-Poincar¨¦ and University of Bath ±¨¸æÕªÒª£º Branching processes are known to be useful tools in the study of random graphs, in particular in understanding the appearance of a phase transition in the sizes of the largest clusters of the graphs. Recently, growing interests are paid to inhomogeneous random graphs. In this talk, we look at one particular model of such graphs, called the Poisson random graph, where edges are formed with probabilities proportional to some prescribes weights on the vertices. One challenge in the study of inhomogeneous random graphs is to describe the geometry of the graphs around the critical point. In the case of Poisson random graph, we obtain a simple representation of the graph using Galton-Watson trees (genealogy trees of branching processes). Relying on this representation and previous works of Duquesne & Le Gall on the convergence of Galton-Watson trees, we prove that in the critical window, the scaling limits of the largest components of the Poisson random graphs are a collection of almost-tree-like compact metric spaces, which can be constructed explicitly from the so-called Levy processes without replacement. Based on a joint work with Nicolas Broutin and Thomas Duquesne. 16£º20-17£º20 ±¨¸æÌâÄ¿£º Computing with functions ±¨¸æÈË£º Kuan Xu, University of Kent ±¨¸æÕªÒª£º In very recent years, the idea of computing with functions has been sparkled by the endeavor of the Chebfun project. Polynomial approximations of functions enables functional operations and operators of all kinds to be numerically approximated using fast, accurate, and robust algorithms based on these approximations, giving people the feel of symbolic computing but with the lightning speed of numerical computation. The idea of computing with functions has also extended to the solution of differential and integral equations. In this talk, the core idea of computing with functions and the underlying mathematics of Chebfun project will be discussed with the aid of hands-on-keyboard demos. 17£º30- Dinner
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¡¡ | | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8860976099967957, "perplexity": 624.4882545921505}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917121000.17/warc/CC-MAIN-20170423031201-00424-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://bird.bcamath.org/handle/20.500.11824/20/browse?type=subject&value=Variational+formulation | Now showing items 1-1 of 1
• #### A multilayer method for the hydrostatic Navier-Stokes equations: A particular weak solution
(Journal of Scientific Computing, 2014-12-31)
In this work we present a multilayer approach to the solution of non-stationary 3D Navier-Stokes equations. We use piecewise smooth weak solutions. We approximate the velocity by a piecewise constant (in z) horizontal ... | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9693530201911926, "perplexity": 1163.5224042876266}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578529813.23/warc/CC-MAIN-20190420120902-20190420142902-00405.warc.gz"} |
https://askdev.io/questions/555/the-absolutely-no-part-of-tensor-items-of-rated-mods | # The absolutely no part of tensor items of rated mods
I assure there is a very easy reference on this, but also for some factor I can not locate it. If you can aim me to a reference or simply write a brief evidence for me, I would certainly be really satisfied.
Offered a rated ring $R_{\bullet}$ and also a localization $R_{\bullet}^{*}$. We additionally have actually a rated $R_{\bullet}$-mod, $M_{\bullet}$.
So what I need to know; is $\left(R_{\bullet}^{*}\otimes M_{\bullet}\right)_0=\left(R_{\bullet}^{*}\right)_0\otimes \left(M_{\bullet}\right)_0$?
0
2019-05-04 16:21:47
Source Share
Answers: 1
For a counterexample, take $R=k[t]$ with its common grading and also $M=R(1)$, the free component of ranking one created in level $1$.
0
2019-05-08 01:49:44
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http://riotorto.users.sourceforge.net/Maxima/descriptive/pca/index.html | # Principal Components Analysis
## The data
File wind.data contains daily average wind speeds at 5 meteorological stations in the Republic of Ireland (This is part of a data set taken at 12 meteorological stations. The original file was downloaded from the StatLib Data Repository and its analysis is discused in
Haslett, J., Raftery, A. E. (1989) Space-time Modelling with Long-memory Dependence: Assessing Ireland's Wind Power Resource, with Discussion. Applied Statistics 38, 1-50.
To run these examples, you must load package descriptive and the data:
load("descriptive")$s2 : read_matrix (file_search ("wind.data"))$
Before reducing the dimensionality, we need to know the true dimension of the multivariate variable,
p: length(first(s2));
$5$
Sample size,
n: length(s2);
$100$
Let's calculate the multivariate mean,
mean(s2);
$\left[ 9.948 , 10.16 , 10.87 , 15.72 , 14.84 \right]$
The variances,
var(s2);
$\left[ 17.22 , 14.99 , 15.48 , 32.18 , 24.42 \right]$
Since data are of similar nature and scale, we don't need to standardize them.
The multivariate scatter plot to see if variables are correlated.
set_draw_defaults(
xlabel = "",
ylabel = "")$scatterplot( s2, nclasses = 5, fill_color = blue, fill_density = 0.3, color = red, point_size = 1/2, dimensions = [850,750], xtics = 5)$
There seems to be linear correlation among all variables, so we can calculate the correlation matrix,
cor(s2);
$\pmatrix{1.0&0.8476&0.8804&0.824&0.752\cr 0.8476&1.0&0.8736&0.6903& 0.7825\cr 0.8804&0.8736&1.0&0.7764&0.8323\cr 0.824&0.6903&0.7764&1.0 &0.7294\cr 0.752&0.7825&0.8323&0.7294&1.0\cr }$
## The model
In order to see if we can reduce the dimensionality of this sample, we calculate the principal components,
pc: principal_components(s2);
$\left[ \left[ 87.57 , 8.753 , 5.515 , 1.889 , 1.613 \right] , \left[ 83.13 , 8.31 , 5.235 , 1.793 , 1.531 \right] , \\ \pmatrix{ 0.4149&0.03379&-0.4757&-0.581&-0.5126\cr 0.369&-0.3657&-0.4298& 0.7237&-0.1469\cr 0.3959&-0.2178&-0.2181&-0.2749&0.8201\cr 0.5548& 0.7744&0.1857&0.2319&0.06498\cr 0.4765&-0.4669&0.712&-0.09605&- 0.1969\cr } \right]$
The returned list contains three elements, the first of them is a list with the variances of the principal components in descending order. We see that the first variance is much greater than the second one:
first(pc);
$\left[ 87.57 , 8.753 , 5.515 , 1.889 , 1.613 \right]$
The second element is the list with percentages of total variation explained by each principal component:
second(pc);
$\left[ 83.13 , 8.31 , 5.235 , 1.793 , 1.531 \right]$
We can accumulate percentages to help to decide how many variables we are going to take for further analysis. We see that with the two first principal componentes we explain about 91.44% of total variance:
block([ap: copy(pc[2])],
for k:2 thru length(ap) do ap[k]: ap[k]+ap[k-1],
ap);
$\left[ 83.13 , 91.44 , 96.68 , 98.47 , 100.0 \right]$
We can also plot a Pareto chart showing the percentages graphically:
draw2d(
fill_density = 0.2,
apply(bars, makelist([k, pc[2][k], 1/2], k, p)),
points_joined = true,
point_type = filled_circle,
point_size = 2,
points(makelist([k, pc[2][k]], k, p)),
xlabel = "Variances",
ylabel = "Percentages",
xtics = setify(makelist([concat("PC",k),k], k, p))) $ Finally, the last element returned by function principal_components is the rotation matrix: rot: third(pc); $\pmatrix{0.4149&0.03379&-0.4757&-0.581&-0.5126\cr 0.369&-0.3657&- 0.4298&0.7237&-0.1469\cr 0.3959&-0.2178&-0.2181&-0.2749&0.8201\cr 0.5548&0.7744&0.1857&0.2319&0.06498\cr 0.4765&-0.4669&0.712&-0.09605 &-0.1969\cr }$ These are the coefficients we need to calculate the p principal components $Y_1, Y_2, Y_3, \ldots, Yp$ from the original variables $X_1, X_2, X_3, \ldots, Xp.$ If the observed data corresponding to one metheorological station is the vector $(x_1, x_2, x_3, x_4, x_5),$ the first principal component is calculated this way: matrix([x1, x2, x3, x4, x5]) . col(rot,1); $0.4765\,{\it x_5}+0.5548\,{\it x_4}+0.3959\,{\it x_3}+0.369\, {\it x_2}+0.4149\,{\it x_1}$ ... the second matrix([x1, x2, x3, x4, x5]) . col(rot,2); $-0.4669\,{\it x_5}+0.7744\,{\it x_4}-0.2178\,{\it x_3}-0.3657\, {\it x_2}+0.03379\,{\it x_1}$ ... and so on. In general, the transformed multivariate sample is calculated this way: new: s2 . rot$
And if we want to use the first two components, we remove the last three columns from the transformed sample:
new2: submatrix(new, 3,4,5);
$\pmatrix{30.66&0.9512\cr 27.93&1.78\cr 22.64&-0.567\cr 13.44&-4.203 \cr 23.64&-1.542\cr 18.23&-3.966\cr 24.12&0.1813\cr 27.78&-0.2169 \cr 25.23&-1.561\cr 20.34&-3.698\cr 28.55&4.444\cr 29.73&-1.084\cr 8.702&-0.7874\cr 8.754&2.462\cr 14.22&-0.04463\cr 21.12&1.576\cr 28.8&-0.469\cr 30.19&-10.2\cr 10.48&0.4667\cr 11.37&0.1515\cr 19.18& -4.55\cr 18.99&0.08038\cr 32.9&-4.489\cr 44.54&-2.166\cr 27.14&- 2.229\cr 43.18&-5.436\cr 49.67&-3.064\cr 33.06&0.8554\cr 50.74&- 0.9644\cr 32.28&-0.2534\cr 24.75&-2.612\cr 14.98&-0.08559\cr 18.81& 1.342\cr 16.99&-0.3552\cr 28.55&0.4827\cr 34.3&0.8969\cr 45.56& 0.7853\cr 30.51&-3.537\cr 42.18&0.6286\cr 39.6&0.9222\cr 32.79&- 0.9442\cr 39.74&-1.838\cr 37.8&5.232\cr 42.83&4.742\cr 30.05&-2.859 \cr 24.46&3.398\cr 39.77&-1.495\cr 34.39&-1.734\cr 29.94&2.494\cr 40.82&6.545\cr 25.03&0.6239\cr 17.42&3.25\cr 27.6&-0.2265\cr 42.01&- 0.1073\cr 16.71&-0.5108\cr 30.1&-0.8958\cr 45.95&-3.996\cr 32.98& 0.9275\cr 33.2&0.4342\cr 32.0&3.716\cr 30.16&1.459\cr 35.01&4.616 \cr 19.65&1.3\cr 26.0&-3.86\cr 14.44&2.485\cr 21.26&4.471\cr 27.66& 1.426\cr 31.58&-1.637\cr 23.97&-2.048\cr 36.51&3.325\cr 40.06&-3.581 \cr 33.96&-0.7837\cr 30.13&3.338\cr 33.26&4.878\cr 23.83&6.641\cr 38.61&1.058\cr 36.82&-3.485\cr 20.68&-0.9488\cr 32.66&-6.451\cr 22.35&-4.885\cr 19.41&-4.469\cr 27.15&-1.928\cr 27.42&-1.599\cr 37.65&-0.6065\cr 34.42&-6.523\cr 17.13&-5.456\cr 31.67&-0.5124\cr 36.0&-2.487\cr 31.92&-1.82\cr 23.97&-2.088\cr 20.15&0.6859\cr 19.49& -1.63\cr 18.92&0.6379\cr 23.3&1.388\cr 26.37&1.804\cr 19.06&-0.8494 \cr 12.86&-0.1988\cr 22.46&-1.34\cr 24.96&-5.134\cr 15.1&-2.422\cr }$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8520335555076599, "perplexity": 4531.464460200841}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501170249.75/warc/CC-MAIN-20170219104610-00168-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://plainmath.net/90439/summing-odd-fractions-to-one-from-the-li | # Summing Odd Fractions to One From the list 1/3,1/5,1/7,1/9,1/11..... is it possible to chose a limited number of terms that sum to one? This can be done with even fractions: 1/2,1/4,1/8,1/12,1/24
Summing Odd Fractions to One
From the list $\frac{1}{3},\frac{1}{5},\frac{1}{7},\frac{1}{9},\frac{1}{11}$..... is it possible to chose a limited number of terms that sum to one? This can be done with even fractions: $\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{12},\frac{1}{24}$
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Such a representation of a fraction as the sum of fractions with numerator 1 and different denominators is called Egyption fraction, because that was the way fractions were written in ancient Egypt. It's clear that for 1, we must have an odd number of summands, because otherwise the numerator of the sum would be even and the denominator odd. As it turns out, the minimal number is 9, and there are the following 5 solutions:
$\begin{array}{rl}1& =\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{15}+\frac{1}{35}+\frac{1}{45}+\frac{1}{231}\\ 1& =\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{15}+\frac{1}{21}+\frac{1}{231}+\frac{1}{315}\\ 1& =\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{15}+\frac{1}{33}+\frac{1}{45}+\frac{1}{385}\\ 1& =\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{15}+\frac{1}{21}+\frac{1}{165}+\frac{1}{693}\\ 1& =\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{15}+\frac{1}{21}+\frac{1}{135}+\frac{1}{10395}\end{array}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 26, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9604037404060364, "perplexity": 433.2758053387965}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337421.33/warc/CC-MAIN-20221003133425-20221003163425-00308.warc.gz"} |
https://wiki.cimec.unitn.it/tiki-index.php?page=Frequency-domain+analysis | # Frequency-domain analysis
Following Fourier ideas, EEG signals can be represented in the time domain (when recording the brain activity from scalp EEG surface electrodes) or alternatively in the frequency domain (when decomposing time-domain signals into weighted sums of sine and cosine functions). In the frequency-domain analysis (but also in the time-frequency domain analysis), EEG signals are represented in terms of oscillatory components, each of which is believed to isolate the activity of localized neuronal populations. To describe any oscillatory component, we estimate its amplitude and phase. The amplitude refers to the magnitude of an oscillation and can be measured in units of signal amplitude (µV), in power (µV2) or in decibel units (20*log10(µV) or 10*log10(µV2)). The phase refers to the position of a point in time on a waveform cycle and is commonly measured in degrees or radians.
The Fourier Transform is usually employed to represent a signal in the frequency-domain. Especially after the introduction of the Fast Fourier Transform (a very fast and efficient algorithm to calculate the Fourier Transform), the Fourier Transform has become one of the most used tools in different scientific disciplines.
Let us consider a digital signal x[n], n = 0,…,N-1, which has been obtained from a continuous signal x(t) by sampling at equal time intervals Δt (i.e., with a sampling frequency fs = 1/Δt). The discrete Fourier Transform is given by the following formula:
The signal x[n] can be reconstructed with the inverse discrete Fourier Transform:
The Fourier coefficients X[k] are complex numbers that can be represented both in Cartesian and polar forms.
where XR and XI denote the real and imaginary parts in the Cartesian representation, whereas |X[k]| and φ denote the amplitude and phase in the polar representation.
If we restrict our attention only to signal power, from the complex Fourier coefficients we can compute the periodogram as:
where X[k] denotes the complex conjugation of X[k].
The periodogram is a raw estimation of the power spectrum of the signal, provided that the signal is stationary (i.e., the main characteristics of the signal, such as its mean and variance do not change over time).
To illustrate why the periodogram is a raw estimate of the power spectrum, let us consider the following example.
##### Figure 4.1: Spectral leakage
The sinusoid on the left panel of Figure 4.1 has an exact number of cycles in the 2 s period of the signal. Its corresponding periodogram (on the left bottom side) shows a single peak at 3 Hz. On the other hand, the sinusoid on the right has a non-integer number of cycles and its periodogram shows an activity that is spread between 2 and 5 Hz. This phenomenon, consisting of smearing the power spectrum estimation, is called leakage (for further details, see Oppenheim and Schaffer, 1999). A simple way to avoid leakage would be to take an integer number of cycles. However, in general it is not possible to define a single periodicity because real signals contain activity at multiple frequencies. A useful approach for reducing the leakage effects is to use an appropriate window function that tapers the borders of the signal. This approach is referred to as a ‘windowing’. Several window functions have been proposed and their advantages and disadvantages depend on the specific application. Among the most popular window functions are the Hanning, Hamming, Barlett and Blackman.
Another reason why the periodogram is a raw estimate of the power spectrum is the following. Because the variance of the periodogram does not go to zero, even in the limit of an infinite sample size, the periodogram is not a statistically consistent estimate of the power spectrum. For large sample sizes, moreover, the periodogram tends to vary rapidly with frequency, and the resulting power spectrum tends to look like a random pattern. To obtain a smoother estimate of the power spectrum, Barlett (1953) proposed to divide the signal into segments, to calculate the periodogram for each segment, and then to average the periodograms. Later on, Welch (1967) showed that better estimates can be achieved by using half-overlapping segments.
The number of segments to be used for averaging the periodograms depends on the specific application. Ideally, it should be more than 30. As regards the length of the segments, it is noteworthy to point out that the larger the segments, the better the frequency resolution (the frequency resolution Δf is defined by the formula Δf = 1 / (N*Δt), where N is the number of data points and Δt is the time resolution; note that N*Δt is the duration of the signal). However, it is advisable to use segments that are not too long, so that the signal can be considered stationary to a first approximation.
Now we will upload the dataset that we employed to illustrate the preprocessing steps, and we will estimate the power spectrum using the Welch method (click here to download all datasets).
```%% Initializations
dataPath = 'C:\Users\CIMeC\Desktop\wikiDatasets\'; % It depends on where
% the wikiDatasets
% folder is stored
fileName = 'S01RestPreprocessed.set';
dataSet = fullfile(dataPath,fileName);```
###### ↑ Go up
```%% Defining trials (or epochs)
% Reading the data as one long continuous segment
cfg = [];
cfg.dataset = dataSet; % file path
data = ft_preprocessing(cfg);
% Segmenting the data into trials
cfg = [];
cfg.length = 2; % seconds
cfg.overlap = 0.5; % value between 0 and 1 (percent)
dataSeg = ft_redefinetrial(cfg,data);
% Reading the events from the data
% Removing trials containing discontinuities
trialsToReject = zeros(length(dataSeg.sampleinfo),1);
for eventId = 1:length(event)
if isempty(event(eventId).type)
tmpSample = event(eventId).sample;
for sampleInfoId = 1:length(dataSeg.sampleinfo)
if (tmpSample - dataSeg.sampleinfo(sampleInfoId,1)) * ...
(tmpSample - dataSeg.sampleinfo(sampleInfoId,2)) < 0
trialsToReject(sampleInfoId) = sampleInfoId;
end
end
end
end
trialsToReject = trialsToReject(trialsToReject ~= 0);
trialsToKeep = 1:length(dataSeg.sampleinfo); % initialize vector with all
% trial indices
trialsToKeep(trialsToReject) = []; % create trial indices to keep
cfg = [];
cfg.trials = trialsToKeep;
dataSeg = ft_selectdata(cfg,dataSeg);```
###### ↑ Go up
```%% Frequency-domain analysis
% Computing power spectrum for each trial
cfg = [];
cfg.foilim = [0 45];
cfg.method = 'mtmfft';
cfg.taper = 'hanning';
cfg.keeptrials = 'yes';
cfg.channel = {'all' '-AFp9' '-AFp10'}; % all but ocular channels
freq = ft_freqanalysis(cfg,dataSeg);
% Computing the average over trials
cfg = [];
%cfg.keeptrials = 'yes';
freqAvg = ft_freqdescriptives(cfg,freq);```
###### ↑ Go up
```%% Plot PSD
freqAvg.powSpctrmDB = 10*log10(freqAvg.powspctrm); % convert to
% decibel (dB)
figure; plot(freqAvg.freq,freqAvg.powSpctrmDB')
set(gca,'ylim',[-30 15])
xlabel('Frequency (Hz)')
ylabel('Power spectral density (dB/Hz)')
% Defining channel layout
templateLayout = 'EEG1005.lay'; % one of the template layouts
% included in FieldTrip
cfg = [];
cfg.layout = which(templateLayout);
layout = ft_prepare_layout(cfg);
% Increasing layout width and height
if strcmp(templateLayout,'EEG1005.lay')
layout.width = 0.07 * ones(length(layout.width),1);
layout.height = 0.04 * ones(length(layout.height),1);
end
% Topographic plot
cfg = [];
cfg.xlim = [8 12]; % alpha band
cfg.zlim = [-13 -1];
cfg.parameter = 'powSpctrmDB';
cfg.marker = 'off';
cfg.highlight = 'off';
cfg.comment = 'auto';
cfg.layout = layout;
cfg.colorbar = 'yes'; % or 'southoutside'
cfg.colormap = jet;
figure; ft_topoplotER(cfg,freqAvg);
c = colorbar;
c.LineWidth = 1;
c.FontSize = 14;
title(c,'dB')```
###### ↑ Go up
Analogously to the ft_timelockgrandaverage function, the ft_freqgrandaverage allows computing the grand average spectra, including all individual subjects’ spectrum as an input argument.
Differences in spectral power between conditions can be statistically tested using the function ft_freqstatistic. The configuration structure remains largely the same, as compared to the statistical evaluation of time-domain data described above. Again, we can use the function ft_topoplotER to show the outcome of the statistical comparisons (see the previous section).
#### References
• Oppenheim, A., & Schaffer, R. (1999). Discrete-time signal processing. Prentice Hall, London.
• Barlett, M.S. (1953). An introduction to stochastic processes with special reference to methods and applications. Cambridge University Press, MA.
• Welch, P.D. (1967). The use of Fast Fourier Transform for the estimation of power spectra: a method based on time-averaging over short modified periodograms. IEEE Trans Audio-Electroacust, 15, 70-73.
###### ↑ Go up
Created by daniele.patoner. Last Modification: Monday 11 of March, 2019 13:12:34 CET by tommaso.mastropasqua.
### Tiki Assistant
Thank you for installing Tiki! | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9524655342102051, "perplexity": 663.9305957009115}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578681624.79/warc/CC-MAIN-20190425034241-20190425060241-00552.warc.gz"} |
http://www.thestudentroom.co.uk/showthread.php?t=2009419&p=37710120 | You are Here: Home
# CCEA Weber Definition Tweet
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1. CCEA Weber Definition
Does anybody know how the Weber is defined for CCEA A2 Physics? The spec says we need to know it but I can't find it anywhere in my notes.
2. Re: CCEA Weber Definition
It should be defined the same way as anywhere else, I would hope
"The weber is the magnetic flux which induces an emf of one volt in a coil of one turn when the flux is reduced uniformly to zero in one second."
(Or some similar equivalent wording expressing the same idea. It's based on Faraday's Law.)
3. Re: CCEA Weber Definition
(Original post by Stonebridge)
It should be defined the same way as anywhere else, I would hope
"The weber is the magnetic flux which induces an emf of one volt in a coil of one turn when the flux is reduced uniformly to zero in one second."
(Or some similar equivalent wording expressing the same idea. It's based on Faraday's Law.)
Thanks
4. Re: CCEA Weber Definition
For OCR anyway - the weber is based on the definition of magnetic flux.
1 Weber is the magnetic flux when there is a magnetic flux density of 1 Tesla perpendicular to a coil of area 1 m^2.
ie 1Wb = 1T x 1m^2.
(of course this comes to the same thing given the definitions of the Tesla and the Volt)
Last edited by teachercol; 23-05-2012 at 09:42. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9103463292121887, "perplexity": 2844.9194724183603}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368705097259/warc/CC-MAIN-20130516115137-00006-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://de.maplesoft.com/support/help/maple/view.aspx?path=GraphTheory%2FIsForest | GraphTheory - Maple Programming Help
Home : Support : Online Help : Mathematics : Discrete Mathematics : Graph Theory : GraphTheory Package : GraphTheory/IsForest
GraphTheory
IsForest
Calling Sequence IsForest(G)
Parameters
G - graph
Description
• The IsForest command returns true if the input graph is a forest or false otherwise. A forest is a graph whose connected components are all trees.
Examples
> $\mathrm{with}\left(\mathrm{GraphTheory}\right):$
> $F≔\mathrm{Graph}\left(\left[1,2,3,4,5,6\right],\left\{\left\{1,2\right\},\left\{2,3\right\},\left\{5,6\right\}\right\}\right)$
${F}{≔}{\mathrm{Graph 1: an undirected unweighted graph with 6 vertices and 3 edge\left(s\right)}}$ (1)
> $\mathrm{IsForest}\left(F\right)$
${\mathrm{true}}$ (2)
> $C≔\mathrm{ConnectedComponents}\left(F\right)$
${C}{≔}\left[\left[{1}{,}{2}{,}{3}\right]{,}\left[{4}\right]{,}\left[{5}{,}{6}\right]\right]$ (3)
> $\mathrm{seq}\left(\mathrm{IsTree}\left(\mathrm{InducedSubgraph}\left(F,i\right)\right),i=C\right)$
${\mathrm{true}}{,}{\mathrm{true}}{,}{\mathrm{true}}$ (4)
> $\mathrm{NumberOfVertices}\left(F\right)-\mathrm{NumberOfEdges}\left(F\right)-\mathrm{nops}\left(\mathrm{ConnectedComponents}\left(F\right)\right)$
${0}$ (5) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 11, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9019916653633118, "perplexity": 2746.6942880718366}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600401601278.97/warc/CC-MAIN-20200928135709-20200928165709-00411.warc.gz"} |
http://mathhelpforum.com/statistics/2015-question.html | Math Help - Question
1. Question
A car dealer has 8 Red, 11 Gray, and 9 Blue cars in stock Ten cars are randomly chosen to be displayed in from of the dealership. Find the probability that 4 are red and 6 are blue; 3 are red, 3 are blue and 4 are gray; 5 are gray and non are blue; all 10 are gray.
2. Originally Posted by batman123
A car dealer has 8 Red, 11 Gray, and 9 Blue cars in stock Ten cars are randomly chosen to be displayed in from of the dealership. Find the probability that 4 are red and 6 are blue; 3 are red, 3 are blue and 4 are gray; 5 are gray and non are blue; all 10 are gray.
Remember that probability is favorable outcomes divided by possible outcomes.
1)4 Red and 6 Blue: There are 8 red he is chosing 4 thus there are $_8C_4=70$ different ways. There are 9 blue he is chosing 6 thus there are $_9C_6=_9C_3=84$ different ways. Thus there are a complete total of $70\times 84=5880$ ways of chosing these types of cars. This is the favorable outcomes, now we find the possible outcomes. There are $28$ cars altogether and he is taking 10, thus there are $_{28}C_{10}=13123110$ different ways. Thus the probability is the ratio thus,
$p=\frac{5880}{13123110}$.
Everything else is solved similarly. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9570194482803345, "perplexity": 483.3525586807966}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500832538.99/warc/CC-MAIN-20140820021352-00063-ip-10-180-136-8.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/136108/automorphisms-and-normal-bundle | # automorphisms and normal bundle
Let $X$ be a smooth projective algebraic variety and $Z$ a smooth closed subvariety of $X$. Let $f: X \to X$ be an automorphism of $X$ such that $f(Z)\subset Z$.
How does $f$ act on the normal bundle $N_{Z/X}$?
-
$N_{Z/X} = TX|_Z/TZ$. By your assumption, $Tf:TX\to TX$ maps the restriction $TX|_Z$ to itself, and also $TZ\to TZ$, with the same foot-point mapping $Z\to Z$. Thus it induces $TX|_Z/TZ\to TX|_Z/TZ$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9947547912597656, "perplexity": 77.09550432271219}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931008289.40/warc/CC-MAIN-20141125155648-00207-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://socratic.org/questions/how-do-you-find-the-roots-real-and-imaginary-of-y-2x-2-15x-4x-3-2-using-the-quad | Algebra
Topics
# How do you find the roots, real and imaginary, of y= 2x^2-15x-(4x+3)^2 using the quadratic formula?
Jun 30, 2016
x~~-1.69" and "x~~-0.38" to 2 decimal places")
#### Explanation:
You need to add like terms such that you and up with equation form:$\text{ } y = a {x}^{2} + b x + c$
To do this expand the brackets and simplify.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Expanding the brackets}}$
Note that ${\left[- \left(4 x + 3\right)\right]}^{2} \text{ is not the same as } - \left[{\left(4 x + 3\right)}^{2}\right]$
The condition we have is$\text{ } - \left[{\left(4 x + 3\right)}^{2}\right]$
$\textcolor{b l u e}{\left(4 x + 3\right)} \textcolor{b r o w n}{\left(4 x + 3\right)}$
$\textcolor{b r o w n}{\textcolor{b l u e}{4 x} \left(4 x + 3\right) \textcolor{b l u e}{+ 3} \left(4 x + 3\right)}$
$16 {x}^{2} + 12 x \text{ "+12x+9" "=" } \textcolor{g r e e n}{16 {x}^{2} + 24 x + 9}$
,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Substituting the expanded brackets and solving}}$
$\implies y = 2 {x}^{2} - 15 x - \left(\textcolor{g r e e n}{16 {x}^{2} + 24 x + 9}\right)$
Multiply everything inside the brackets by -1 and group like terms
$\implies y = 2 {x}^{2} - 16 {x}^{2} - 15 x - 24 x - 9$
$\implies y = - 14 {x}^{2} - 29 x - 9$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
From standard form we have:
$a = - 14$
$b = - 29$
$c = - 9$
Thus
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
becomes" "x=(+29+-sqrt((-29)^2-4(-14)(-9)))/(2(-14)
$x = \frac{29 \pm \sqrt{841 - 504}}{- 28}$
$x = \frac{29 \pm \sqrt{337}}{- 28} \text{ "larr 337" is a prime number}$
$\textcolor{b l u e}{x \approx - 1.69 \text{ and "x~~-0.38" to 2 decimal places}}$
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https://terrytao.wordpress.com/category/mathematics/mathds/page/2/ | You are currently browsing the category archive for the ‘math.DS’ category.
I’ve just uploaded to the arXiv my paper “Failure of the ${L^1}$ pointwise and maximal ergodic theorems for the free group“, submitted to Forum of Mathematics, Sigma. This paper concerns a variant of the pointwise ergodic theorem of Birkhoff, which asserts that if one has a measure-preserving shift map ${T: X \rightarrow X}$ on a probability space ${X = (X,\mu)}$, then for any ${f \in L^1(X)}$, the averages ${\frac{1}{N} \sum_{n=1}^N f \circ T^{-n}}$ converge pointwise almost everywhere. (In the important case when the shift map ${T}$ is ergodic, the pointwise limit is simply the mean ${\int_X f\ d\mu}$ of the original function ${f}$.)
The pointwise ergodic theorem can be extended to measure-preserving actions of other amenable groups, if one uses a suitably “tempered” Folner sequence of averages; see this paper of Lindenstrauss for more details. (I also wrote up some notes on that paper here, back in 2006 before I had started this blog.) But the arguments used to handle the amenable case break down completely for non-amenable groups, and in particular for the free non-abelian group ${F_2}$ on two generators.
Nevo and Stein studied this problem and obtained a number of pointwise ergodic theorems for ${F_2}$-actions ${(T_g)_{g \in F_2}}$ on probability spaces ${(X,\mu)}$. For instance, for the spherical averaging operators
$\displaystyle {\mathcal A}_n f := \frac{1}{4 \times 3^{n-1}} \sum_{g \in F_2: |g| = n} f \circ T_g^{-1}$
(where ${|g|}$ denotes the length of the reduced word that forms ${g}$), they showed that ${{\mathcal A}_{2n} f}$ converged pointwise almost everywhere provided that ${f}$ was in ${L^p(X)}$ for some ${p>1}$. (The need to restrict to spheres of even radius can be seen by considering the action of ${F_2}$ on the two-element set ${\{0,1\}}$ in which both generators of ${F_2}$ act by interchanging the elements, in which case ${{\mathcal A}_n}$ is determined by the parity of ${n}$.) This result was reproven with a different and simpler proof by Bufetov, who also managed to relax the condition ${f \in L^p(X)}$ to the weaker condition ${f \in L \log L(X)}$.
The question remained open as to whether the pointwise ergodic theorem for ${F_2}$-actions held if one only assumed that ${f}$ was in ${L^1(X)}$. Nevo and Stein were able to establish this for the Cesáro averages ${\frac{1}{N} \sum_{n=1}^N {\mathcal A}_n}$, but not for ${{\mathcal A}_n}$ itself. About six years ago, Assaf Naor and I tried our hand at this problem, and was able to show an associated maximal inequality on ${\ell^1(F_2)}$, but due to the non-amenability of ${F_2}$, this inequality did not transfer to ${L^1(X)}$ and did not have any direct impact on this question, despite a fair amount of effort on our part to attack it.
Inspired by some recent conversations with Lewis Bowen, I returned to this problem. This time around, I tried to construct a counterexample to the ${L^1}$ pointwise ergodic theorem – something Assaf and I had not seriously attempted to do (perhaps due to being a bit too enamoured of our ${\ell^1(F_2)}$ maximal inequality). I knew of an existing counterexample of Ornstein regarding a failure of an ${L^1}$ ergodic theorem for iterates ${P^n}$ of a self-adjoint Markov operator – in fact, I had written some notes on this example back in 2007. Upon revisiting my notes, I soon discovered that the Ornstein construction was adaptable to the ${F_2}$ setting, thus settling the problem in the negative:
Theorem 1 (Failure of ${L^1}$ pointwise ergodic theorem) There exists a measure-preserving ${F_2}$-action on a probability space ${X}$ and a non-negative function ${f \in L^1(X)}$ such that ${\sup_n {\mathcal A}_{2n} f(x) = +\infty}$ for almost every ${x}$.
To describe the proof of this theorem, let me first briefly sketch the main ideas of Ornstein’s construction, which gave an example of a self-adjoint Markov operator ${P}$ on a probability space ${X}$ and a non-negative ${f \in L^1(X)}$ such that ${\sup_n P^n f(x) = +\infty}$ for almost every ${x}$. By some standard manipulations, it suffices to show that for any given ${\alpha > 0}$ and ${\varepsilon>0}$, there exists a self-adjoint Markov operator ${P}$ on a probability space ${X}$ and a non-negative ${f \in L^1(X)}$ with ${\|f\|_{L^1(X)} \leq \alpha}$, such that ${\sup_n P^n f \geq 1-\varepsilon}$ on a set of measure at least ${1-\varepsilon}$. Actually, it will be convenient to replace the Markov chain ${(P^n f)_{n \geq 0}}$ with an ancient Markov chain ${(f_n)_{n \in {\bf Z}}}$ – that is to say, a sequence of non-negative functions ${f_n}$ for both positive and negative ${f}$, such that ${f_{n+1} = P f_n}$ for all ${n \in {\bf Z}}$. The purpose of requiring the Markov chain to be ancient (that is, to extend infinitely far back in time) is to allow for the Markov chain to be shifted arbitrarily in time, which is key to Ornstein’s construction. (Technically, Ornstein’s original argument only uses functions that go back to a large negative time, rather than being infinitely ancient, but I will gloss over this point for sake of discussion, as it turns out that the ${F_2}$ version of the argument can be run using infinitely ancient chains.)
For any ${\alpha>0}$, let ${P(\alpha)}$ denote the claim that for any ${\varepsilon>0}$, there exists an ancient Markov chain ${(f_n)_{n \in {\bf Z}}}$ with ${\|f_n\|_{L^1(X)} = \alpha}$ such that ${\sup_{n \in {\bf Z}} f_n \geq 1-\varepsilon}$ on a set of measure at least ${1-\varepsilon}$. Clearly ${P(1)}$ holds since we can just take ${f_n=1}$ for all ${n}$. Our objective is to show that ${P(\alpha)}$ holds for arbitrarily small ${\alpha}$. The heart of Ornstein’s argument is then the implication
$\displaystyle P(\alpha) \implies P( \alpha (1 - \frac{\alpha}{4}) ) \ \ \ \ \ (1)$
for any ${0 < \alpha \leq 1}$, which upon iteration quickly gives the desired claim.
Let’s see informally how (1) works. By hypothesis, and ignoring epsilons, we can find an ancient Markov chain ${(f_n)_{n \in {\bf Z}}}$ on some probability space ${X}$ of total mass ${\|f_n\|_{L^1(X)} = \alpha}$, such that ${\sup_n f_n}$ attains the value of ${1}$ or greater almost everywhere. Assuming that the Markov process is irreducible, the ${f_n}$ will eventually converge as ${n \rightarrow \infty}$ to the constant value of ${\|f_n\|_{L^1(X)}}$, in particular its final state will essentially stay above ${\alpha}$ (up to small errors).
Now suppose we duplicate the Markov process by replacing ${X}$ with a double copy ${X \times \{1,2\}}$ (giving ${\{1,2\}}$ the uniform probability measure), and using the disjoint sum of the Markov operators on ${X \times \{1\}}$ and ${X \times \{2\}}$ as the propagator, so that there is no interaction between the two components of this new system. Then the functions ${f'_n(x,i) := f_n(x) 1_{i=1}}$ form an ancient Markov chain of mass at most ${\alpha/2}$ that lives solely in the first half ${X \times \{1\}}$ of this copy, and ${\sup_n f'_n}$ attains the value of ${1}$ or greater on almost all of the first half ${X \times \{1\}}$, but is zero on the second half. The final state of ${f'_n}$ will be to stay above ${\alpha}$ in the first half ${X \times \{1\}}$, but be zero on the second half.
Now we modify the above example by allowing an infinitesimal amount of interaction between the two halves ${X \times \{1\}}$, ${X \times \{2\}}$ of the system (I mentally think of ${X \times \{1\}}$ and ${X \times \{2\}}$ as two identical boxes that a particle can bounce around in, and now we wish to connect the boxes by a tiny tube). The precise way in which this interaction is inserted is not terribly important so long as the new Markov process is irreducible. Once one does so, then the ancient Markov chain ${(f'_n)_{n \in {\bf Z}}}$ in the previous example gets replaced by a slightly different ancient Markov chain ${(f''_n)_{n \in {\bf Z}}}$ which is more or less identical with ${f'_n}$ for negative times ${n}$, or for bounded positive times ${n}$, but for very large values of ${n}$ the final state is now constant across the entire state space ${X \times \{1,2\}}$, and will stay above ${\alpha/2}$ on this space.
Finally, we consider an ancient Markov chain ${F_n}$ which is basically of the form
$\displaystyle F_n(x,i) \approx f''_n(x,i) + (1 - \frac{\alpha}{2}) f_{n-M}(x) 1_{i=2}$
for some large parameter ${M}$ and for all ${n \leq M}$ (the approximation becomes increasingly inaccurate for ${n}$ much larger than ${M}$, but never mind this for now). This is basically two copies of the original Markov process in separate, barely interacting state spaces ${X \times \{1\}, X \times \{2\}}$, but with the second copy delayed by a large time delay ${M}$, and also attenuated in amplitude by a factor of ${1-\frac{\alpha}{2}}$. The total mass of this process is now ${\frac{\alpha}{2} + \frac{\alpha}{2} (1 -\frac{\alpha}{2}) = \alpha (1 - \alpha/4)}$. Because of the ${f''_n}$ component of ${F_n}$, we see that ${\sup_n F_n}$ basically attains the value of ${1}$ or greater on the first half ${X \times \{1\}}$. On the second half ${X \times \{2\}}$, we work with times ${n}$ close to ${M}$. If ${M}$ is large enough, ${f''_n}$ would have averaged out to about ${\alpha/2}$ at such times, but the ${(1 - \frac{\alpha}{2}) f_{n-M}(x)}$ component can get as large as ${1-\alpha/2}$ here. Summing (and continuing to ignore various epsilon losses), we see that ${\sup_n F_n}$ can get as large as ${1}$ on almost all of the second half of ${X \times \{2\}}$. This concludes the rough sketch of how one establishes the implication (1).
It was observed by Bufetov that the spherical averages ${{\mathcal A}_n}$ for a free group action can be lifted up to become powers ${P^n}$ of a Markov operator, basically by randomly assigning a “velocity vector” ${s \in \{a,b,a^{-1},b^{-1}\}}$ to one’s base point ${x}$ and then applying the Markov process that moves ${x}$ along that velocity vector (and then randomly changing the velocity vector at each time step to the “reduced word” condition that the velocity never flips from ${s}$ to ${s^{-1}}$). Thus the spherical average problem has a Markov operator interpretation, which opens the door to adapting the Ornstein construction to the setting of ${F_2}$ systems. This turns out to be doable after a certain amount of technical artifice; the main thing is to work with ${F_2}$-measure preserving systems that admit ancient Markov chains that are initially supported in a very small region in the “interior” of the state space, so that one can couple such systems to each other “at the boundary” in the fashion needed to establish the analogue of (1) without disrupting the ancient dynamics of such chains. The initial such system (used to establish the base case ${P(1)}$) comes from basically considering the action of ${F_2}$ on a (suitably renormalised) “infinitely large ball” in the Cayley graph, after suitably gluing together the boundary of this ball to complete the action. The ancient Markov chain associated to this system starts at the centre of this infinitely large ball at infinite negative time ${n=-\infty}$, and only reaches the boundary of this ball at the time ${n=0}$.
An extremely large portion of mathematics is concerned with locating solutions to equations such as
$\displaystyle f(x) = 0$
or
$\displaystyle \Phi(x) = x \ \ \ \ \ (1)$
for ${x}$ in some suitable domain space (either finite-dimensional or infinite-dimensional), and various maps ${f}$ or ${\Phi}$. To solve the fixed point iteration equation (1), the simplest general method available is the fixed point iteration method: one starts with an initial approximate solution ${x_0}$ to (1), so that ${\Phi(x_0) \approx x_0}$, and then recursively constructs the sequence ${x_1, x_2, x_3, \dots}$ by ${x_n := \Phi(x_{n-1})}$. If ${\Phi}$ behaves enough like a “contraction”, and the domain is complete, then one can expect the ${x_n}$ to converge to a limit ${x}$, which should then be a solution to (1). For instance, if ${\Phi: X \rightarrow X}$ is a map from a metric space ${X = (X,d)}$ to itself, which is a contraction in the sense that
$\displaystyle d( \Phi(x), \Phi(y) ) \leq (1-\eta) d(x,y)$
for all ${x,y \in X}$ and some ${\eta>0}$, then with ${x_n}$ as above we have
$\displaystyle d( x_{n+1}, x_n ) \leq (1-\eta) d(x_n, x_{n-1} )$
for any ${n}$, and so the distances ${d(x_n, x_{n-1} )}$ between successive elements of the sequence decay at at least a geometric rate. This leads to the contraction mapping theorem, which has many important consequences, such as the inverse function theorem and the Picard existence theorem.
A slightly more complicated instance of this strategy arises when trying to linearise a complex map ${f: U \rightarrow {\bf C}}$ defined in a neighbourhood ${U}$ of a fixed point. For simplicity we normalise the fixed point to be the origin, thus ${0 \in U}$ and ${f(0)=0}$. When studying the complex dynamics ${f^2 = f \circ f}$, ${f^3 = f \circ f \circ f}$, ${\dots}$ of such a map, it can be useful to try to conjugate ${f}$ to another function ${g = \psi^{-1} \circ f \circ \psi}$, where ${\psi}$ is a holomorphic function defined and invertible near ${0}$ with ${\psi(0)=0}$, since the dynamics of ${g}$ will be conjguate to that of ${f}$. Note that if ${f(0)=0}$ and ${f'(0)=\lambda}$, then from the chain rule any conjugate ${g}$ of ${f}$ will also have ${g(0)=0}$ and ${g'(0)=\lambda}$. Thus, the “simplest” function one can hope to conjugate ${f}$ to is the linear function ${z \mapsto \lambda z}$. Let us say that ${f}$ is linearisable (around ${0}$) if it is conjugate to ${z \mapsto \lambda z}$ in some neighbourhood of ${0}$. Equivalently, ${f}$ is linearisable if there is a solution to the Schröder equation
$\displaystyle f( \psi(z) ) = \psi(\lambda z) \ \ \ \ \ (2)$
for some ${\psi: U' \rightarrow {\bf C}}$ defined and invertible in a neighbourhood ${U'}$ of ${0}$ with ${\psi(0)=0}$, and all ${z}$ sufficiently close to ${0}$. (The Schröder equation is normalised somewhat differently in the literature, but this form is equivalent to the usual form, at least when ${\lambda}$ is non-zero.) Note that if ${\psi}$ solves the above equation, then so does ${z \mapsto \psi(cz)}$ for any non-zero ${c}$, so we may normalise ${\psi'(0)=1}$ in addition to ${\psi(0)=0}$, which also ensures local invertibility from the inverse function theorem. (Note from winding number considerations that ${\psi}$ cannot be invertible near zero if ${\psi'(0)}$ vanishes.)
We have the following basic result of Koenigs:
Theorem 1 (Koenig’s linearisation theorem) Let ${f: U \rightarrow {\bf C}}$ be a holomorphic function defined near ${0}$ with ${f(0)=0}$ and ${f'(0)=\lambda}$. If ${0 < |\lambda| < 1}$ (attracting case) or ${1 < |\lambda| < \infty}$ (repelling case), then ${f}$ is linearisable near zero.
Proof: Observe that if ${f, \psi, \lambda}$ solve (2), then ${f^{-1}, \psi^{-1}, \lambda^{-1}}$ solve (2) also (in a sufficiently small neighbourhood of zero). Thus we may reduce to the attractive case ${0 < |\lambda| < 1}$.
Let ${r>0}$ be a sufficiently small radius, and let ${X}$ denote the space of holomorphic functions ${\psi: B(0,r) \rightarrow {\bf C}}$ on the complex disk ${B(0,r) := \{z \in {\bf C}: |z| < r \}}$ with ${\psi(0)=0}$ and ${\psi'(0)=1}$. We can view the Schröder equation (2) as a fixed point equation
$\displaystyle \psi = \Phi(\psi)$
where ${\Phi: X' \rightarrow X}$ is the partially defined function on ${X}$ that maps a function ${\psi: B(0,r) \rightarrow {\bf C}}$ to the function ${\Phi(\psi): B(0,r) \rightarrow {\bf C}}$ defined by
$\displaystyle \Phi(\psi)(z) := f^{-1}( \psi( \lambda z ) ),$
assuming that ${f^{-1}}$ is well-defined on the range of ${\psi(B(0,\lambda r))}$ (this is why ${\Phi}$ is only partially defined).
We can solve this equation by the fixed point iteration method, if ${r}$ is small enough. Namely, we start with ${\psi_0: B(0,r) \rightarrow {\bf C}}$ being the identity map, and set ${\psi_1 := \Phi(\psi_0), \psi_2 := \Phi(\psi_1)}$, etc. We equip ${X}$ with the uniform metric ${d( \psi, \tilde \psi ) := \sup_{z \in B(0,r)} |\psi(z) - \tilde \psi(z)|}$. Observe that if ${d( \psi, \psi_0 ), d(\tilde \psi, \psi_0) \leq r}$, and ${r}$ is small enough, then ${\psi, \tilde \psi}$ takes values in ${B(0,2r)}$, and ${\Phi(\psi), \Phi(\tilde \psi)}$ are well-defined and lie in ${X}$. Also, since ${f^{-1}}$ is smooth and has derivative ${\lambda^{-1}}$ at ${0}$, we have
$\displaystyle |f^{-1}(z) - f^{-1}(w)| \leq (1+\varepsilon) |\lambda|^{-1} |z-w|$
if ${z, w \in B(0,r)}$, ${\varepsilon>0}$ and ${r}$ is sufficiently small depending on ${\varepsilon}$. This is not yet enough to establish the required contraction (thanks to Mario Bonk for pointing this out); but observe that the function ${\frac{\psi(z)-\tilde \psi(z)}{z^2}}$ is holomorphic on ${B(0,r)}$ and bounded by ${d(\psi,\tilde \psi)/r^2}$ on the boundary of this ball (or slightly within this boundary), so by the maximum principle we see that
$\displaystyle |\frac{\psi(z)-\tilde \psi(z)}{z^2}| \leq \frac{1}{r^2} d(\psi,\tilde \psi)$
on all of ${B(0,r)}$, and in particular
$\displaystyle |\psi(z)-\tilde \psi(z)| \leq |\lambda|^2 d(\psi,\tilde \psi)$
on ${B(0,\lambda r)}$. Putting all this together, we see that
$\displaystyle d( \Phi(\psi), \Phi(\tilde \psi)) \leq (1+\varepsilon) |\lambda| d(\psi, \tilde \psi);$
since ${|\lambda|<1}$, we thus obtain a contraction on the ball ${\{ \psi \in X: d(\psi,\psi_0) \leq r \}}$ if ${\varepsilon}$ is small enough (and ${r}$ sufficiently small depending on ${\varepsilon}$). From this (and the completeness of ${X}$, which follows from Morera’s theorem) we see that the iteration ${\psi_n}$ converges (exponentially fast) to a limit ${\psi \in X}$ which is a fixed point of ${\Phi}$, and thus solves Schröder’s equation, as required. $\Box$
Koenig’s linearisation theorem leaves open the indifferent case when ${|\lambda|=1}$. In the rationally indifferent case when ${\lambda^n=1}$ for some natural number ${n}$, there is an obvious obstruction to linearisability, namely that ${f^n = 1}$ (in particular, linearisation is not possible in this case when ${f}$ is a non-trivial rational function). An obstruction is also present in some irrationally indifferent cases (where ${|\lambda|=1}$ but ${\lambda^n \neq 1}$ for any natural number ${n}$), if ${\lambda}$ is sufficiently close to various roots of unity; the first result of this form is due to Cremer, and the optimal result of this type for quadratic maps was established by Yoccoz. In the other direction, we have the following result of Siegel:
Theorem 2 (Siegel’s linearisation theorem) Let ${f: U \rightarrow {\bf C}}$ be a holomorphic function defined near ${0}$ with ${f(0)=0}$ and ${f'(0)=\lambda}$. If ${|\lambda|=1}$ and one has the Diophantine condition ${\frac{1}{|\lambda^n-1|} \leq C n^C}$ for all natural numbers ${n}$ and some constant ${C>0}$, then ${f}$ is linearisable at ${0}$.
The Diophantine condition can be relaxed to a more general condition involving the rational exponents of the phase ${\theta}$ of ${\lambda = e^{2\pi i \theta}}$; this was worked out by Brjuno, with the condition matching the one later obtained by Yoccoz. Amusingly, while the set of Diophantine numbers (and hence the set of linearisable ${\lambda}$) has full measure on the unit circle, the set of non-linearisable ${\lambda}$ is generic (the complement of countably many nowhere dense sets) due to the above-mentioned work of Cremer, leading to a striking disparity between the measure-theoretic and category notions of “largeness”.
Siegel’s theorem does not seem to be provable using a fixed point iteration method. However, it can be established by modifying another basic method to solve equations, namely Newton’s method. Let us first review how this method works to solve the equation ${f(x)=0}$ for some smooth function ${f: I \rightarrow {\bf R}}$ defined on an interval ${I}$. We suppose we have some initial approximant ${x_0 \in I}$ to this equation, with ${f(x_0)}$ small but not necessarily zero. To make the analysis more quantitative, let us suppose that the interval ${[x_0-r_0,x_0+r_0]}$ lies in ${I}$ for some ${r_0>0}$, and we have the estimates
$\displaystyle |f(x_0)| \leq \delta_0 r_0$
$\displaystyle |f'(x)| \geq \eta_0$
$\displaystyle |f''(x)| \leq \frac{1}{\eta_0 r_0}$
for some ${\delta_0 > 0}$ and ${0 < \eta_0 < 1/2}$ and all ${x \in [x_0-r_0,x_0+r_0]}$ (the factors of ${r_0}$ are present to make ${\delta_0,\eta_0}$ “dimensionless”).
Lemma 3 Under the above hypotheses, we can find ${x_1}$ with ${|x_1 - x_0| \leq \eta_0 r_0}$ such that
$\displaystyle |f(x_1)| \ll \delta_0^2 \eta_0^{-O(1)} r_0.$
In particular, setting ${r_1 := (1-\eta_0) r_0}$, ${\eta_1 := \eta_0/2}$, and ${\delta_1 = O(\delta_0^2 \eta_0^{-O(1)})}$, we have ${[x_1-r_1,x_1+r_1] \subset [x_0-r_0,x_0+r_0] \subset I}$, and
$\displaystyle |f(x_1)| \leq \delta_1 r_1$
$\displaystyle |f'(x)| \geq \eta_1$
$\displaystyle |f''(x)| \leq \frac{1}{\eta_1 r_1}$
for all ${x \in [x_1-r_1,x_1+r_1]}$.
The crucial point here is that the new error ${\delta_1}$ is roughly the square of the previous error ${\delta_0}$. This leads to extremely fast (double-exponential) improvement in the error upon iteration, which is more than enough to absorb the exponential losses coming from the ${\eta_0^{-O(1)}}$ factor.
Proof: If ${\delta_0 > c \eta_0^{C}}$ for some absolute constants ${C,c>0}$ then we may simply take ${x_0=x_1}$, so we may assume that ${\delta_0 \leq c \eta_0^{C}}$ for some small ${c>0}$ and large ${C>0}$. Using the Newton approximation ${f(x_0+h) \approx f(x_0) + h f'(x_0)}$ we are led to the choice
$\displaystyle x_1 := x_0 - \frac{f(x_0)}{f'(x_0)}$
for ${x_1}$. From the hypotheses on ${f}$ and the smallness hypothesis on ${\delta}$ we certainly have ${|x_1-x_0| \leq \eta_0 r_0}$. From Taylor’s theorem with remainder we have
$\displaystyle f(x_1) = f(x_0) - \frac{f(x_0)}{f'(x_0)} f'(x_0) + O( \frac{1}{\eta_0 r_0} |\frac{f(x_0)}{f'(x_0)}|^2 )$
$\displaystyle = O( \frac{1}{\eta_0 r_0} (\frac{\delta_0 r_0}{\eta_0})^2 )$
and the claim follows. $\Box$
We can iterate this procedure; starting with ${x_0,\eta_0,r_0,\delta_0}$ as above, we obtain a sequence of nested intervals ${[x_n-r_n,x_n+r_n]}$ with ${f(x_n)| \leq \delta_n}$, and with ${\eta_n,r_n,\delta_n,x_n}$ evolving by the recursive equations and estimates
$\displaystyle \eta_n = \eta_{n-1} / 2$
$\displaystyle r_n = (1 - \eta_{n-1}) r_{n-1}$
$\displaystyle \delta_n = O( \delta_{n-1}^2 \eta_{n-1}^{-O(1)} )$
$\displaystyle |x_n - x_{n-1}| \leq \eta_{n-1} r_{n-1}.$
If ${\delta_0}$ is sufficiently small depending on ${\eta_0}$, we see that ${\delta_n}$ converges rapidly to zero (indeed, we can inductively obtain a bound of the form ${\delta_n \leq \eta_0^{C (2^n + n)}}$ for some large absolute constant ${C}$ if ${\delta_0}$ is small enough), and ${x_n}$ converges to a limit ${x \in I}$ which then solves the equation ${f(x)=0}$ by the continuity of ${f}$.
As I recently learned from Zhiqiang Li, a similar scheme works to prove Siegel’s theorem, as can be found for instance in this text of Carleson and Gamelin. The key is the following analogue of Lemma 3.
Lemma 4 Let ${\lambda}$ be a complex number with ${|\lambda|=1}$ and ${\frac{1}{|\lambda^n-1|} \ll n^{O(1)}}$ for all natural numbers ${n}$. Let ${r_0>0}$, and let ${f_0: B(0,r_0) \rightarrow {\bf C}}$ be a holomorphic function with ${f_0(0)=0}$, ${f'_0(0)=\lambda}$, and
$\displaystyle |f_0(z) - \lambda z| \leq \delta_0 r_0 \ \ \ \ \ (3)$
for all ${z \in B(0,r_0)}$ and some ${\delta_0>0}$. Let ${0 < \eta_0 \leq 1/2}$, and set ${r_1 := (1-\eta_0) r_0}$. Then there exists an injective holomorphic function ${\psi_0: B(0, r_1) \rightarrow B(0, r_0)}$ and a holomorphic function ${f_1: B(0,r_1) \rightarrow {\bf C}}$ such that
$\displaystyle f_0( \psi_1(z) ) = \psi_1(f_1(z)) \ \ \ \ \ (4)$
for all ${z \in B(0,r_1)}$, and such that
$\displaystyle |\psi_1(z) - z| \ll \delta_0 \eta_0^{-O(1)} r_1$
and
$\displaystyle |f_1(z) - \lambda z| \leq \delta_1 r_1$
for all ${z \in B(0,r_1)}$ and some ${\delta_1 = O(\delta_0^2 \eta_0^{-O(1)})}$.
Proof: By scaling we may normalise ${r_0=1}$. If ${\delta_0 > c \eta_0^C}$ for some constants ${c,C>0}$, then we can simply take ${\psi_1}$ to be the identity and ${f_1=f_0}$, so we may assume that ${\delta_0 \leq c \eta_0^C}$ for some small ${c>0}$ and large ${C>0}$.
To motivate the choice of ${\psi_1}$, we write ${f_0(z) = \lambda z + \hat f_0(z)}$ and ${\psi_1(z) = z + \hat \psi(z)}$, with ${\hat f_0}$ and ${\hat \psi_1}$ viewed as small. We would like to have ${f_0(\psi_1(z)) \approx \psi_1(\lambda z)}$, which expands as
$\displaystyle \lambda z + \lambda \hat \psi_1(z) + \hat f_0( z + \hat \psi_1(z) ) \approx \lambda z + \hat \psi_1(\lambda z).$
As ${\hat f_0}$ and ${\hat \psi}$ are both small, we can heuristically approximate ${\hat f_0(z + \hat \psi_1(z) ) \approx \hat f_0(z)}$ up to quadratic errors (compare with the Newton approximation ${f(x_0+h) \approx f(x_0) + h f'(x_0)}$), and arrive at the equation
$\displaystyle \hat \psi_1(\lambda z) - \lambda \hat \psi_1(z) = \hat f_0(z). \ \ \ \ \ (5)$
This equation can be solved by Taylor series; the function ${\hat f_0}$ vanishes to second order at the origin and thus has a Taylor expansion
$\displaystyle \hat f_0(z) = \sum_{n=2}^\infty a_n z^n$
and then ${\hat \psi_1}$ has a Taylor expansion
$\displaystyle \hat \psi_1(z) = \sum_{n=2}^\infty \frac{a_n}{\lambda^n - \lambda} z^n.$
We take this as our definition of ${\hat \psi_1}$, define ${\psi_1(z) := z + \hat \psi_1(z)}$, and then define ${f_1}$ implicitly via (4).
Let us now justify that this choice works. By (3) and the generalised Cauchy integral formula, we have ${|a_n| \leq \delta_0}$ for all ${n}$; by the Diophantine assumption on ${\lambda}$, we thus have ${|\frac{a_n}{\lambda^n - \lambda}| \ll \delta_0 n^{O(1)}}$. In particular, ${\hat \psi_1}$ converges on ${B(0,1)}$, and on the disk ${B(0, (1-\eta_0/4))}$ (say) we have the bounds
$\displaystyle |\hat \psi_1(z)|, |\hat \psi'_1(z)| \ll \delta_0 \sum_{n=2}^\infty n^{O(1)} (1-\eta_0/4)^n \ll \eta_0^{-O(1)} \delta_0. \ \ \ \ \ (6)$
In particular, as ${\delta_0}$ is so small, we see that ${\psi_1}$ maps ${B(0, (1-\eta_0/4))}$ injectively to ${B(0,1)}$ and ${B(0,1-\eta_0)}$ to ${B(0,1-3\eta_0/4)}$, and the inverse ${\psi_1^{-1}}$ maps ${B(0, (1-\eta_0/2))}$ to ${B(0, (1-\eta_0/4))}$. From (3) we see that ${f_0}$ maps ${B(0,1-3\eta_0/4)}$ to ${B(0,1-\eta_0/2)}$, and so if we set ${f_1: B(0,1-\eta_0) \rightarrow B(0,1-\eta_0/4)}$ to be the function ${f_1 := \psi_1^{-1} \circ f_0 \circ \psi_1}$, then ${f_1}$ is a holomorphic function obeying (4). Expanding (4) in terms of ${\hat f_0}$ and ${\hat \psi_1}$ as before, and also writing ${f_1(z) = \lambda z + \hat f_1(z)}$, we have
$\displaystyle \lambda z + \lambda \hat \psi_1(z) + \hat f_0( z + \hat \psi_1(z) ) = \lambda z + \hat f_1(z) + \hat \psi_1(\lambda z + \hat f_1(z))$
for ${z \in B(0, 1-\eta_0)}$, which by (5) simplifies to
$\displaystyle \hat f_1(z) = \hat f_0( z + \hat \psi_1(z) ) - \hat f_0(z) + \hat \psi_1(\lambda z) - \hat \psi_1(\lambda z + \hat f_1(z)).$
From (6), the fundamental theorem of calculus, and the smallness of ${\delta_0}$ we have
$\displaystyle |\hat \psi_1(\lambda z) - \hat \psi_1(\lambda z + \hat f_1(z))| \leq \frac{1}{2} |\hat f_1(z)|$
and thus
$\displaystyle |\hat f_1(z)| \leq 2 |\hat f_0( z + \hat \psi_1(z) ) - \hat f_0(z)|.$
From (3) and the Cauchy integral formula we have ${\hat f'_0(z) = O( \delta_0 \eta_0^{-O(1)})}$ on (say) ${B(0,1-\eta_0/4)}$, and so from (6) and the fundamental theorem of calculus we conclude that
$\displaystyle |\hat f_1(z)| \ll \delta_0^2 \eta_0^{-O(1)}$
on ${B(0,1-\eta_0)}$, and the claim follows. $\Box$
If we set ${\eta_0 := 1/2}$, ${f_0 := f}$, and ${\delta_0>0}$ to be sufficiently small, then (since ${f(z)-\lambda z}$ vanishes to second order at the origin), the hypotheses of this lemma will be obeyed for some sufficiently small ${r_0}$. Iterating the lemma (and halving ${\eta_0}$ repeatedly), we can then find sequences ${\eta_n, \delta_n, r_n > 0}$, injective holomorphic functions ${\psi_n: B(0,r_n) \rightarrow B(0,r_{n-1})}$ and holomorphic functions ${f_n: B(0,r_n) \rightarrow {\bf C}}$ such that one has the recursive identities and estimates
$\displaystyle \eta_n = \eta_{n-1} / 2$
$\displaystyle r_n = (1 - \eta_{n-1}) r_{n-1}$
$\displaystyle \delta_n = O( \delta_{n-1}^2 \eta_{n-1}^{-O(1)} )$
$\displaystyle |\psi_n(z) - z| \ll \delta_{n-1} \eta_{n-1}^{-O(1)} r_n$
$\displaystyle |f_n(z) - \lambda z| \leq \delta_n r_n$
$\displaystyle f_{n-1}( \psi_n(z) ) = \psi_n(f_n(z))$
for all ${n \geq 1}$ and ${z \in B(0,r_n)}$. By construction, ${r_n}$ decreases to a positive radius ${r_\infty}$ that is a constant multiple of ${r_0}$, while (for ${\delta_0}$ small enough) ${\delta_n}$ converges double-exponentially to zero, so in particular ${f_n(z)}$ converges uniformly to ${\lambda z}$ on ${B(0,r_\infty)}$. Also, ${\psi_n}$ is close enough to the identity, the compositions ${\Psi_n := \psi_1 \circ \dots \circ \psi_n}$ are uniformly convergent on ${B(0,r_\infty/2)}$ with ${\Psi_n(0)=0}$ and ${\Psi'_n(0)=1}$. From this we have
$\displaystyle f( \Psi_n(z) ) = \Psi_n(f_n(z))$
on ${B(0,r_\infty/4)}$, and on taking limits using Morera’s theorem we obtain a holomorphic function ${\Psi}$ defined near ${0}$ with ${\Psi(0)=0}$, ${\Psi'(0)=1}$, and
$\displaystyle f( \Psi(z) ) = \Psi(\lambda z),$
obtaining the required linearisation.
Remark 5 The idea of using a Newton-type method to obtain error terms that decay double-exponentially, and can therefore absorb exponential losses in the iteration, also occurs in KAM theory and in Nash-Moser iteration, presumably due to Siegel’s influence on Moser. (I discuss Nash-Moser iteration in this note that I wrote back in 2006.)
The von Neumann ergodic theorem (the Hilbert space version of the mean ergodic theorem) asserts that if ${U: H \rightarrow H}$ is a unitary operator on a Hilbert space ${H}$, and ${v \in H}$ is a vector in that Hilbert space, then one has
$\displaystyle \lim_{N \rightarrow \infty} \frac{1}{N} \sum_{n=1}^N U^n v = \pi_{H^U} v$
in the strong topology, where ${H^U := \{ w \in H: Uw = w \}}$ is the ${U}$-invariant subspace of ${H}$, and ${\pi_{H^U}}$ is the orthogonal projection to ${H^U}$. (See e.g. these previous lecture notes for a proof.) The same proof extends to more general amenable groups: if ${G}$ is a countable amenable group acting on a Hilbert space ${H}$ by unitary transformations ${T^g: H \rightarrow H}$ for ${g \in G}$, and ${v \in H}$ is a vector in that Hilbert space, then one has
$\displaystyle \lim_{N \rightarrow \infty} \mathop{\bf E}_{g \in \Phi_N} T^g v = \pi_{H^G} v \ \ \ \ \ (1)$
for any Folner sequence ${\Phi_N}$ of ${G}$, where ${H^G := \{ w \in H: T^g w = w \hbox{ for all }g \in G \}}$ is the ${G}$-invariant subspace, and ${\mathop{\bf E}_{a \in A} f(a) := \frac{1}{|A|} \sum_{a \in A} f(a)}$ is the average of ${f}$ on ${A}$. Thus one can interpret ${\pi_{H^G} v}$ as a certain average of elements of the orbit ${Gv := \{ T^g v: g \in G \}}$ of ${v}$.
In a previous blog post, I noted a variant of this ergodic theorem (due to Alaoglu and Birkhoff) that holds even when the group ${G}$ is not amenable (or not discrete), using a more abstract notion of averaging:
Theorem 1 (Abstract ergodic theorem) Let ${G}$ be an arbitrary group acting unitarily on a Hilbert space ${H}$, and let ${v}$ be a vector in ${H}$. Then ${\pi_{H^G} v}$ is the element in the closed convex hull of ${Gv := \{ T^g v: g \in G \}}$ of minimal norm, and is also the unique element of ${H^G}$ in this closed convex hull.
I recently stumbled upon a different way to think about this theorem, in the additive case ${G = (G,+)}$ when ${G}$ is abelian, which has a closer resemblance to the classical mean ergodic theorem. Given an arbitrary additive group ${G = (G,+)}$ (not necessarily discrete, or countable), let ${{\mathcal F}}$ denote the collection of finite non-empty multisets in ${G}$ – that is to say, unordered collections ${\{a_1,\dots,a_n\}}$ of elements ${a_1,\dots,a_n}$ of ${G}$, not necessarily distinct, for some positive integer ${n}$. Given two multisets ${A = \{a_1,\dots,a_n\}}$, ${B = \{b_1,\dots,b_m\}}$ in ${{\mathcal F}}$, we can form the sum set ${A + B := \{ a_i + b_j: 1 \leq i \leq n, 1 \leq j \leq m \}}$. Note that the sum set ${A+B}$ can contain multiplicity even when ${A, B}$ do not; for instance, ${\{ 1,2\} + \{1,2\} = \{2,3,3,4\}}$. Given a multiset ${A = \{a_1,\dots,a_n\}}$ in ${{\mathcal F}}$, and a function ${f: G \rightarrow H}$ from ${G}$ to a vector space ${H}$, we define the average ${\mathop{\bf E}_{a \in A} f(a)}$ as
$\displaystyle \mathop{\bf E}_{a \in A} f(a) = \frac{1}{n} \sum_{j=1}^n f(a_j).$
Note that the multiplicity function of the set ${A}$ affects the average; for instance, we have ${\mathop{\bf E}_{a \in \{1,2\}} a = \frac{3}{2}}$, but ${\mathop{\bf E}_{a \in \{1,2,2\}} a = \frac{5}{3}}$.
We can define a directed set on ${{\mathcal F}}$ as follows: given two multisets ${A,B \in {\mathcal F}}$, we write ${A \geq B}$ if we have ${A = B+C}$ for some ${C \in {\mathcal F}}$. Thus for instance we have ${\{ 1, 2, 2, 3\} \geq \{1,2\}}$. It is easy to verify that this operation is transitive and reflexive, and is directed because any two elements ${A,B}$ of ${{\mathcal F}}$ have a common upper bound, namely ${A+B}$. (This is where we need ${G}$ to be abelian.) The notion of convergence along a net, now allows us to define the notion of convergence along ${{\mathcal F}}$; given a family ${x_A}$ of points in a topological space ${X}$ indexed by elements ${A}$ of ${{\mathcal F}}$, and a point ${x}$ in ${X}$, we say that ${x_A}$ converges to ${x}$ along ${{\mathcal F}}$ if, for every open neighbourhood ${U}$ of ${x}$ in ${X}$, one has ${x_A \in U}$ for sufficiently large ${A}$, that is to say there exists ${B \in {\mathcal F}}$ such that ${x_A \in U}$ for all ${A \geq B}$. If the topological space ${V}$ is Hausdorff, then the limit ${x}$ is unique (if it exists), and we then write
$\displaystyle x = \lim_{A \rightarrow G} x_A.$
When ${x_A}$ takes values in the reals, one can also define the limit superior or limit inferior along such nets in the obvious fashion.
We can then give an alternate formulation of the abstract ergodic theorem in the abelian case:
Theorem 2 (Abelian abstract ergodic theorem) Let ${G = (G,+)}$ be an arbitrary additive group acting unitarily on a Hilbert space ${H}$, and let ${v}$ be a vector in ${H}$. Then we have
$\displaystyle \pi_{H^G} v = \lim_{A \rightarrow G} \mathop{\bf E}_{a \in A} T^a v$
in the strong topology of ${H}$.
Proof: Suppose that ${A \geq B}$, so that ${A=B+C}$ for some ${C \in {\mathcal F}}$, then
$\displaystyle \mathop{\bf E}_{a \in A} T^a v = \mathop{\bf E}_{c \in C} T^c ( \mathop{\bf E}_{b \in B} T^b v )$
so by unitarity and the triangle inequality we have
$\displaystyle \| \mathop{\bf E}_{a \in A} T^a v \|_H \leq \| \mathop{\bf E}_{b \in B} T^b v \|_H,$
thus ${\| \mathop{\bf E}_{a \in A} T^a v \|_H^2}$ is monotone non-increasing in ${A}$. Since this quantity is bounded between ${0}$ and ${\|v\|_H}$, we conclude that the limit ${\lim_{A \rightarrow G} \| \mathop{\bf E}_{a \in A} T^a v \|_H^2}$ exists. Thus, for any ${\varepsilon > 0}$, we have for sufficiently large ${A}$ that
$\displaystyle \| \mathop{\bf E}_{b \in B} T^b v \|_H^2 \geq \| \mathop{\bf E}_{a \in A} T^a v \|_H^2 - \varepsilon$
for all ${B \geq A}$. In particular, for any ${g \in G}$, we have
$\displaystyle \| \mathop{\bf E}_{b \in A + \{0,g\}} T^b v \|_H^2 \geq \| \mathop{\bf E}_{a \in A} T^a v \|_H^2 - \varepsilon.$
We can write
$\displaystyle \mathop{\bf E}_{b \in A + \{0,g\}} T^b v = \frac{1}{2} \mathop{\bf E}_{a \in A} T^a v + \frac{1}{2} T^g \mathop{\bf E}_{a \in A} T^a v$
and so from the parallelogram law and unitarity we have
$\displaystyle \| \mathop{\bf E}_{a \in A} T^a v - T^g \mathop{\bf E}_{a \in A} T^a v \|_H^2 \leq 4 \varepsilon$
for all ${g \in G}$, and hence by the triangle inequality (averaging ${g}$ over a finite multiset ${C}$)
$\displaystyle \| \mathop{\bf E}_{a \in A} T^a v - \mathop{\bf E}_{b \in A+C} T^b v \|_H^2 \leq 4 \varepsilon$
for any ${C \in {\mathcal F}}$. This shows that ${\mathop{\bf E}_{a \in A} T^a v}$ is a Cauchy sequence in ${H}$ (in the strong topology), and hence (by the completeness of ${H}$) tends to a limit. Shifting ${A}$ by a group element ${g}$, we have
$\displaystyle \lim_{A \rightarrow G} \mathop{\bf E}_{a \in A} T^a v = \lim_{A \rightarrow G} \mathop{\bf E}_{a \in A + \{g\}} T^a v = T^g \lim_{A \rightarrow G} \mathop{\bf E}_{a \in A} T^a v$
and hence ${\lim_{A \rightarrow G} \mathop{\bf E}_{a \in A} T^a v}$ is invariant under shifts, and thus lies in ${H^G}$. On the other hand, for any ${w \in H^G}$ and ${A \in {\mathcal F}}$, we have
$\displaystyle \langle \mathop{\bf E}_{a \in A} T^a v, w \rangle_H = \mathop{\bf E}_{a \in A} \langle v, T^{-a} w \rangle_H = \langle v, w \rangle_H$
and thus on taking strong limits
$\displaystyle \langle \lim_{A \rightarrow G} \mathop{\bf E}_{a \in A} T^a v, w \rangle_H = \langle v, w \rangle_H$
and so ${v - \lim_{A \rightarrow G} \mathop{\bf E}_{a \in A} T^a v}$ is orthogonal to ${H^G}$. Combining these two facts we see that ${\lim_{A \rightarrow G} \mathop{\bf E}_{a \in A} T^a v}$ is equal to ${\pi_{H^G} v}$ as claimed. $\Box$
To relate this result to the classical ergodic theorem, we observe
Lemma 3 Let ${G}$ be a countable additive group, with a F{\o}lner sequence ${\Phi_n}$, and let ${f_g}$ be a bounded sequence in a normed vector space indexed by ${G}$. If ${\lim_{A \rightarrow G} \mathop{\bf E}_{a \in A} f_a}$ exists, then ${\lim_{n \rightarrow \infty} \mathop{\bf E}_{a \in \Phi_n} f_a}$ exists, and the two limits are equal.
Proof: From the F{\o}lner property, we see that for any ${A}$ and any ${\varepsilon>0}$, the averages ${\mathop{\bf E}_{a \in \Phi_n} f_a}$ and ${\mathop{\bf E}_{a \in A+\Phi_n} f_a}$ differ by at most ${\varepsilon}$ in norm if ${n}$ is sufficiently large depending on ${A}$, ${\varepsilon}$ (and the ${f_a}$). On the other hand, by the existence of the limit ${\lim_{A \rightarrow G} \mathop{\bf E}_{a \in A} f_a}$, the averages ${\mathop{\bf E}_{a \in A} f_a}$ and ${\mathop{\bf E}_{a \in A + \Phi_n} f_a}$ differ by at most ${\varepsilon}$ in norm if ${A}$ is sufficiently large depending on ${\varepsilon}$ (regardless of how large ${n}$ is). The claim follows. $\Box$
It turns out that this approach can also be used as an alternate way to construct the GowersHost-Kra seminorms in ergodic theory, which has the feature that it does not explicitly require any amenability on the group ${G}$ (or separability on the underlying measure space), though, as pointed out to me in comments, even uncountable abelian groups are amenable in the sense of possessing an invariant mean, even if they do not have a F{\o}lner sequence.
Given an arbitrary additive group ${G}$, define a ${G}$-system ${({\mathrm X}, T)}$ to be a probability space ${{\mathrm X} = (X, {\mathcal X}, \mu)}$ (not necessarily separable or standard Borel), together with a collection ${T^g: X \rightarrow X}$ of invertible, measure-preserving maps, such that ${T^0}$ is the identity and ${T^g T^h = T^{g+h}}$ (modulo null sets) for all ${g,h \in G}$. This then gives isomorphisms ${T^g: L^p({\mathrm X}) \rightarrow L^p({\mathrm X})}$ for ${1 \leq p \leq \infty}$ by setting ${T^g f(x) := f(T^{-g} x)}$. From the above abstract ergodic theorem, we see that
$\displaystyle {\mathbf E}( f | {\mathcal X}^G ) = \lim_{A \rightarrow G} \mathop{\bf E}_{a \in A} T^g f$
in the strong topology of ${L^2({\mathrm X})}$ for any ${f \in L^2({\mathrm X})}$, where ${{\mathcal X}^G}$ is the collection of measurable sets ${E}$ that are essentially ${G}$-invariant in the sense that ${T^g E = E}$ modulo null sets for all ${g \in G}$, and ${{\mathbf E}(f|{\mathcal X}^G)}$ is the conditional expectation of ${f}$ with respect to ${{\mathcal X}^G}$.
In a similar spirit, we have
Theorem 4 (Convergence of Gowers-Host-Kra seminorms) Let ${({\mathrm X},T)}$ be a ${G}$-system for some additive group ${G}$. Let ${d}$ be a natural number, and for every ${\omega \in\{0,1\}^d}$, let ${f_\omega \in L^{2^d}({\mathrm X})}$, which for simplicity we take to be real-valued. Then the expression
$\displaystyle \langle (f_\omega)_{\omega \in \{0,1\}^d} \rangle_{U^d({\mathrm X})} := \lim_{A_1,\dots,A_d \rightarrow G}$
$\displaystyle \mathop{\bf E}_{h_1 \in A_1-A_1,\dots,h_d \in A_d-A_d} \int_X \prod_{\omega \in \{0,1\}^d} T^{\omega_1 h_1 + \dots + \omega_d h_d} f_\omega\ d\mu$
converges, where we write ${\omega = (\omega_1,\dots,\omega_d)}$, and we are using the product direct set on ${{\mathcal F}^d}$ to define the convergence ${A_1,\dots,A_d \rightarrow G}$. In particular, for ${f \in L^{2^d}({\mathrm X})}$, the limit
$\displaystyle \| f \|_{U^d({\mathrm X})}^{2^d} = \lim_{A_1,\dots,A_d \rightarrow G}$
$\displaystyle \mathop{\bf E}_{h_1 \in A_1-A_1,\dots,h_d \in A_d-A_d} \int_X \prod_{\omega \in \{0,1\}^d} T^{\omega_1 h_1 + \dots + \omega_d h_d} f\ d\mu$
converges.
We prove this theorem below the fold. It implies a number of other known descriptions of the Gowers-Host-Kra seminorms ${\|f\|_{U^d({\mathrm X})}}$, for instance that
$\displaystyle \| f \|_{U^d({\mathrm X})}^{2^d} = \lim_{A \rightarrow G} \mathop{\bf E}_{h \in A-A} \| f T^h f \|_{U^{d-1}({\mathrm X})}^{2^{d-1}}$
for ${d > 1}$, while from the ergodic theorem we have
$\displaystyle \| f \|_{U^1({\mathrm X})} = \| {\mathbf E}( f | {\mathcal X}^G ) \|_{L^2({\mathrm X})}.$
This definition also manifestly demonstrates the cube symmetries of the Host-Kra measures ${\mu^{[d]}}$ on ${X^{\{0,1\}^d}}$, defined via duality by requiring that
$\displaystyle \langle (f_\omega)_{\omega \in \{0,1\}^d} \rangle_{U^d({\mathrm X})} = \int_{X^{\{0,1\}^d}} \bigotimes_{\omega \in \{0,1\}^d} f_\omega\ d\mu^{[d]}.$
In a subsequent blog post I hope to present a more detailed study of the ${U^2}$ norm and its relationship with eigenfunctions and the Kronecker factor, without assuming any amenability on ${G}$ or any separability or topological structure on ${{\mathrm X}}$.
The 2014 Fields medallists have just been announced as (in alphabetical order of surname) Artur Avila, Manjul Bhargava, Martin Hairer, and Maryam Mirzakhani (see also these nice video profiles for the winners, which is a new initiative of the IMU and the Simons foundation). This time four years ago, I wrote a blog post discussing one result from each of the 2010 medallists; I thought I would try to repeat the exercise here, although the work of the medallists this time around is a little bit further away from my own direct area of expertise than last time, and so my discussion will unfortunately be a bit superficial (and possibly not completely accurate) in places. As before, I am picking these results based on my own idiosyncratic tastes, and they should not be viewed as necessarily being the “best” work of these medallists. (See also the press releases for Avila, Bhargava, Hairer, and Mirzakhani.)
Artur Avila works in dynamical systems and in the study of Schrödinger operators. The work of Avila that I am most familiar with is his solution with Svetlana Jitormiskaya of the ten martini problem of Kac, the solution to which (according to Barry Simon) he offered ten martinis for, hence the name. (The problem had also been previously posed in the work of Azbel and of Hofstadter.) The problem involves perhaps the simplest example of a Schrödinger operator with non-trivial spectral properties, namely the almost Mathieu operator ${H^{\lambda,\alpha}_\omega: \ell^2({\bf Z}) \rightarrow \ell^2({\bf Z})}$ defined for parameters ${\alpha,\omega \in {\bf R}/{\bf Z}}$ and ${\lambda>0}$ by a discrete one-dimensional Schrödinger operator with cosine potential:
$\displaystyle (H^{\lambda,\alpha}_\omega u)_n := u_{n+1} + u_{n-1} + 2\lambda (\cos 2\pi(\theta+n\alpha)) u_n.$
This is a bounded self-adjoint operator and thus has a spectrum ${\sigma( H^{\lambda,\alpha}_\omega )}$ that is a compact subset of the real line; it arises in a number of physical contexts, most notably in the theory of the integer quantum Hall effect, though I will not discuss these applications here. Remarkably, the structure of this spectrum depends crucially on the Diophantine properties of the frequency ${\alpha}$. For instance, if ${\alpha = p/q}$ is a rational number, then the operator is periodic with period ${q}$, and then basic (discrete) Floquet theory tells us that the spectrum is simply the union of ${q}$ (possibly touching) intervals. But for irrational ${\alpha}$ (in which case the spectrum is independent of the phase ${\theta}$), the situation is much more fractal in nature, for instance in the critical case ${\lambda=1}$ the spectrum (as a function of ${\alpha}$) gives rise to the Hofstadter butterfly. The “ten martini problem” asserts that for every irrational ${\alpha}$ and every choice of coupling constant ${\lambda > 0}$, the spectrum is homeomorphic to a Cantor set. Prior to the work of Avila and Jitormiskaya, there were a number of partial results on this problem, notably the result of Puig establishing Cantor spectrum for a full measure set of parameters ${(\lambda,\alpha)}$, as well as results requiring a perturbative hypothesis, such as ${\lambda}$ being very small or very large. The result was also already known for ${\alpha}$ being either very close to rational (i.e. a Liouville number) or very far from rational (a Diophantine number), although the analyses for these two cases failed to meet in the middle, leaving some cases untreated. The argument uses a wide variety of existing techniques, both perturbative and non-perturbative, to attack this problem, as well as an amusing argument by contradiction: they assume (in certain regimes) that the spectrum fails to be a Cantor set, and use this hypothesis to obtain additional Lipschitz control on the spectrum (as a function of the frequency ${\alpha}$), which they can then use (after much effort) to improve existing arguments and conclude that the spectrum was in fact Cantor after all!
Manjul Bhargava produces amazingly beautiful mathematics, though most of it is outside of my own area of expertise. One part of his work that touches on an area of my own interest (namely, random matrix theory) is his ongoing work with many co-authors on modeling (both conjecturally and rigorously) the statistics of various key number-theoretic features of elliptic curves (such as their rank, their Selmer group, or their Tate-Shafarevich groups). For instance, with Kane, Lenstra, Poonen, and Rains, Manjul has proposed a very general random matrix model that predicts all of these statistics (for instance, predicting that the ${p}$-component of the Tate-Shafarevich group is distributed like the cokernel of a certain random ${p}$-adic matrix, very much in the spirit of the Cohen-Lenstra heuristics discussed in this previous post). But what is even more impressive is that Manjul and his coauthors have been able to verify several non-trivial fragments of this model (e.g. showing that certain moments have the predicted asymptotics), giving for the first time non-trivial upper and lower bounds for various statistics, for instance obtaining lower bounds on how often an elliptic curve has rank ${0}$ or rank ${1}$, leading most recently (in combination with existing work of Gross-Zagier and of Kolyvagin, among others) to his amazing result with Skinner and Zhang that at least ${66\%}$ of all elliptic curves over ${{\bf Q}}$ (ordered by height) obey the Birch and Swinnerton-Dyer conjecture. Previously it was not even known that a positive proportion of curves obeyed the conjecture. This is still a fair ways from resolving the conjecture fully (in particular, the situation with the presumably small number of curves of rank ${2}$ and higher is still very poorly understood, and the theory of Gross-Zagier and Kolyvagin that this work relies on, which was initially only available for ${{\bf Q}}$, has only been extended to totally real number fields thus far, by the work of Zhang), but it certainly does provide hope that the conjecture could be within reach in a statistical sense at least.
Martin Hairer works in at the interface between probability and partial differential equations, and in particular in the theory of stochastic differential equations (SDEs). The result of his that is closest to my own interests is his remarkable demonstration with Jonathan Mattingly of unique invariant measure for the two-dimensional stochastically forced Navier-Stokes equation
$\displaystyle \partial_t u + (u \cdot \nabla u) = \nu \Delta u - \nabla p + \xi$
$\displaystyle \nabla \cdot u = 0$
on the two-torus ${({\bf R}/{\bf Z})^2}$, where ${\xi}$ is a Gaussian field that forces a fixed set of frequencies. It is expected that for any reasonable choice of initial data, the solution to this equation should asymptotically be distributed according to Kolmogorov’s power law, as discussed in this previous post. This is still far from established rigorously (although there are some results in this direction for dyadic models, see e.g. this paper of Cheskidov, Shvydkoy, and Friedlander). However, Hairer and Mattingly were able to show that there was a unique probability distribution to almost every initial data would converge to asymptotically; by the ergodic theorem, this is equivalent to demonstrating the existence and uniqueness of an invariant measure for the flow. Existence can be established using standard methods, but uniqueness is much more difficult. One of the standard routes to uniqueness is to establish a “strong Feller property” that enforces some continuity on the transition operators; among other things, this would mean that two ergodic probability measures with intersecting supports would in fact have a non-trivial common component, contradicting the ergodic theorem (which forces different ergodic measures to be mutually singular). Since all ergodic measures for Navier-Stokes can be seen to contain the origin in their support, this would give uniqueness. Unfortunately, the strong Feller property is unlikely to hold in the infinite-dimensional phase space for Navier-Stokes; but Hairer and Mattingly develop a clean abstract substitute for this property, which they call the asymptotic strong Feller property, which is again a regularity property on the transition operator; this in turn is then demonstrated by a careful application of Malliavin calculus.
Maryam Mirzakhani has mostly focused on the geometry and dynamics of Teichmuller-type moduli spaces, such as the moduli space of Riemann surfaces with a fixed genus and a fixed number of cusps (or with a fixed number of boundaries that are geodesics of a prescribed length). These spaces have an incredibly rich structure, ranging from geometric structure (such as the Kahler geometry given by the Weil-Petersson metric), to dynamical structure (through the action of the mapping class group on this and related spaces), to algebraic structure (viewing these spaces as algebraic varieties), and are thus connected to many other objects of interest in geometry and dynamics. For instance, by developing a new recursive formula for the Weil-Petersson volume of this space, Mirzakhani was able to asymptotically count the number of simple prime geodesics of length up to some threshold ${L}$ in a hyperbolic surface (or more precisely, she obtained asymptotics for the number of such geodesics in a given orbit of the mapping class group); the answer turns out to be polynomial in ${L}$, in contrast to the much larger class of non-simple prime geodesics, whose asymptotics are exponential in ${L}$ (the “prime number theorem for geodesics”, developed in a classic series of works by Delsart, Huber, Selberg, and Margulis); she also used this formula to establish a new proof of a conjecture of Witten on intersection numbers that was first proven by Kontsevich. More recently, in two lengthy papers with Eskin and with Eskin-Mohammadi, Mirzakhani established rigidity theorems for the action of ${SL_2({\bf R})}$ on such moduli spaces that are close analogues of Ratner’s celebrated rigidity theorems for unipotently generated groups (discussed in this previous blog post). Ratner’s theorems are already notoriously difficult to prove, and rely very much on the polynomial stability properties of unipotent flows; in this even more complicated setting, the unipotent flows are no longer tractable, and Mirzakhani instead uses a recent “exponential drift” method of Benoist and Quint as a substitute. Ratner’s theorems are incredibly useful for all sorts of problems connected to homogeneous dynamics, and the analogous theorems established by Mirzakhani, Eskin, and Mohammadi have a similarly broad range of applications, for instance in counting periodic billiard trajectories in rational polygons.
As laid out in the foundational work of Kolmogorov, a classical probability space (or probability space for short) is a triplet ${(X, {\mathcal X}, \mu)}$, where ${X}$ is a set, ${{\mathcal X}}$ is a ${\sigma}$-algebra of subsets of ${X}$, and ${\mu: {\mathcal X} \rightarrow [0,1]}$ is a countably additive probability measure on ${{\mathcal X}}$. Given such a space, one can form a number of interesting function spaces, including
• the (real) Hilbert space ${L^2(X, {\mathcal X}, \mu)}$ of square-integrable functions ${f: X \rightarrow {\bf R}}$, modulo ${\mu}$-almost everywhere equivalence, and with the positive definite inner product ${\langle f, g\rangle_{L^2(X, {\mathcal X}, \mu)} := \int_X f g\ d\mu}$; and
• the unital commutative Banach algebra ${L^\infty(X, {\mathcal X}, \mu)}$ of essentially bounded functions ${f: X \rightarrow {\bf R}}$, modulo ${\mu}$-almost everywhere equivalence, with ${\|f\|_{L^\infty(X, {\mathcal X}, \mu)}}$ defined as the essential supremum of ${|f|}$.
There is also a trace ${\tau = \tau_\mu: L^\infty(X, {\mathcal X}, \mu) \rightarrow {\bf C}}$ on ${L^\infty}$ defined by integration: ${\tau(f) := \int_X f\ d\mu}$.
One can form the category ${\mathbf{Prb}}$ of classical probability spaces, by defining a morphism ${\phi: (X, {\mathcal X}, \mu) \rightarrow (Y, {\mathcal Y}, \nu)}$ between probability spaces to be a function ${\phi: X \rightarrow Y}$ which is measurable (thus ${\phi^{-1}(E) \in {\mathcal X}}$ for all ${E \in {\mathcal Y}}$) and measure-preserving (thus ${\mu(\phi^{-1}(E)) = \nu(E)}$ for all ${E \in {\mathcal Y}}$).
Let us now abstract the algebraic features of these spaces as follows; for want of a better name, I will refer to this abstraction as an algebraic probability space, and is very similar to the non-commutative probability spaces studied in this previous post, except that these spaces are now commutative (and real).
Definition 1 An algebraic probability space is a pair ${({\mathcal A}, \tau)}$ where
• ${{\mathcal A}}$ is a unital commutative real algebra;
• ${\tau: {\mathcal A} \rightarrow {\bf R}}$ is a homomorphism such that ${\tau(1)=1}$ and ${\tau( f^2 ) \geq 0}$ for all ${f \in {\mathcal A}}$;
• Every element ${f}$ of ${{\mathcal A}}$ is bounded in the sense that ${\sup_{k \geq 1} \tau( f^{2k} )^{1/2k} < \infty}$. (Technically, this isn’t an algebraic property, but I need it for technical reasons.)
A morphism ${\phi: ({\mathcal A}_1, \tau_1) \rightarrow ({\mathcal A}_2, \tau_2)}$ is a homomorphism ${\phi^*: {\mathcal A}_2 \rightarrow {\mathcal A}_1}$ which is trace-preserving, in the sense that ${\tau_1(\phi^*(f)) = \tau_2(f)}$ for all ${f \in {\mathcal A}_2}$.
For want of a better name, I’ll denote the category of algebraic probability spaces as ${\mathbf{AlgPrb}}$. One can view this category as the opposite category to that of (a subcategory of) the category of tracial commutative real algebras. One could emphasise this opposite nature by denoting the algebraic probability space as ${({\mathcal A}, \tau)^{op}}$ rather than ${({\mathcal A},\tau)}$; another suggestive (but slightly inaccurate) notation, inspired by the language of schemes, would be ${\hbox{Spec}({\mathcal A},\tau)}$ rather than ${({\mathcal A},\tau)}$. However, we will not adopt these conventions here, and refer to algebraic probability spaces just by the pair ${({\mathcal A},\tau)}$.
By the previous discussion, we have a covariant functor ${F: \textbf{Prb} \rightarrow \textbf{AlgPrb}}$ that takes a classical probability space ${(X, {\mathcal X}, \mu)}$ to its algebraic counterpart ${(L^\infty(X, {\mathcal X},\mu), \tau_\mu)}$, with a morphism ${\phi: (X, {\mathcal X}, \mu) \rightarrow (Y, {\mathcal Y}, \nu)}$ of classical probability spaces mapping to a morphism ${F(\phi): (L^\infty(X, {\mathcal X},\mu), \tau_\mu) \rightarrow (L^\infty(Y, {\mathcal Y},\nu), \tau_\nu)}$ of the corresponding algebraic probability spaces by the formula
$\displaystyle F(\phi)^* f := f \circ \phi$
for ${f \in L^\infty(Y, {\mathcal Y}, \nu)}$. One easily verifies that this is a functor.
In this post I would like to describe a functor ${G: \textbf{AlgPrb} \rightarrow \textbf{Prb}}$ which partially inverts ${F}$ (up to natural isomorphism), that is to say a recipe for starting with an algebraic probability space ${({\mathcal A}, \tau)}$ and producing a classical probability space ${(X, {\mathcal X}, \mu)}$. This recipe is not new – it is basically the (commutative) Gelfand-Naimark-Segal construction (discussed in this previous post) combined with the Loomis-Sikorski theorem (discussed in this previous post). However, I wanted to put the construction in a single location for sake of reference. I also wanted to make the point that ${F}$ and ${G}$ are not complete inverses; there is a bit of information in the algebraic probability space (e.g. topological information) which is lost when passing back to the classical probability space. In some future posts, I would like to develop some ergodic theory using the algebraic foundations of probability theory rather than the classical foundations; this turns out to be convenient in the ergodic theory arising from nonstandard analysis (such as that described in this previous post), in which the groups involved are uncountable and the underlying spaces are not standard Borel spaces.
Let us describe how to construct the functor ${G}$, with details postponed to below the fold.
1. Starting with an algebraic probability space ${({\mathcal A}, \tau)}$, form an inner product on ${{\mathcal A}}$ by the formula ${\langle f, g \rangle := \tau(fg)}$, and also form the spectral radius ${\rho(f) :=\lim_{k \rightarrow \infty} \tau(f^{2^k})^{1/2^k}}$.
2. The inner product is clearly positive semi-definite. Quotienting out the null vectors and taking completions, we arrive at a real Hilbert space ${L^2 = L^2({\mathcal A},\tau)}$, to which the trace ${\tau}$ may be extended.
3. Somewhat less obviously, the spectral radius is well-defined and gives a norm on ${{\mathcal A}}$. Taking ${L^2}$ limits of sequences in ${{\mathcal A}}$ of bounded spectral radius gives us a subspace ${L^\infty = L^\infty({\mathcal A},\tau)}$ of ${L^2}$ that has the structure of a real commutative Banach algebra.
4. The idempotents ${1_E}$ of the Banach algebra ${L^\infty}$ may be indexed by elements ${E}$ of an abstract ${\sigma}$-algebra ${{\mathcal B}}$.
5. The Boolean algebra homomorphisms ${\delta_x: {\mathcal B} \rightarrow \{0,1\}}$ (or equivalently, the real algebra homomorphisms ${\iota_x: L^\infty \rightarrow {\bf R}}$) may be indexed by elements ${x}$ of a space ${X}$.
6. Let ${{\mathcal X}}$ denote the ${\sigma}$-algebra on ${X}$ generated by the basic sets ${\overline{E} := \{ x \in X: \delta_x(E) = 1 \}}$ for every ${E \in {\mathcal B}}$.
7. Let ${{\mathcal N}}$ be the ${\sigma}$-ideal of ${{\mathcal X}}$ generated by the sets ${\bigcap_n \overline{E_n}}$, where ${E_n \in {\mathcal B}}$ is a sequence with ${\bigcap_n E_n = \emptyset}$.
8. One verifies that ${{\mathcal B}}$ is isomorphic to ${{\mathcal X}/{\mathcal N}}$. Using this isomorphism, the trace ${\tau}$ on ${L^\infty}$ can be used to construct a countably additive measure ${\mu}$ on ${{\mathcal X}}$. The classical probability space ${(X, {\mathcal X}, \mu)}$ is then ${G( {\mathcal A}, \tau )}$, and the abstract spaces ${L^2, L^\infty}$ may now be identified with their concrete counterparts ${L^2(X, {\mathcal X}, \mu)}$, ${L^\infty(X, {\mathcal X}, \mu)}$.
9. Every algebraic probability space morphism ${\phi: ({\mathcal A}_1,\tau_1) \rightarrow ({\mathcal A}_2,\tau_2)}$ generates a classical probability morphism ${G(\phi): (X_1, {\mathcal X}_1, \mu_1) \rightarrow (X_2, {\mathcal X}_2, \mu_2)}$ via the formula
$\displaystyle \delta_{G(\phi)(x_1)}( E_2 ) = \delta_{x_1}( \phi^*(E_2) )$
using a pullback operation ${\phi^*}$ on the abstract ${\sigma}$-algebras ${{\mathcal B}_1, {\mathcal B}_2}$ that can be defined by density.
Remark 1 The classical probability space ${X}$ constructed by the functor ${G}$ has some additional structure; namely ${X}$ is a ${\sigma}$-Stone space (a Stone space with the property that the closure of any countable union of clopen sets is clopen), ${{\mathcal X}}$ is the Baire ${\sigma}$-algebra (generated by the clopen sets), and the null sets are the meager sets. However, we will not use this additional structure here.
The partial inversion relationship between the functors ${F: \textbf{Prb} \rightarrow \textbf{AlgPrb}}$ and ${G: \textbf{AlgPrb} \rightarrow \textbf{Prb}}$ is given by the following assertion:
1. There is a natural transformation from ${F \circ G: \textbf{AlgPrb} \rightarrow \textbf{AlgPrb}}$ to the identity functor ${I: \textbf{AlgPrb} \rightarrow \textbf{AlgPrb}}$.
More informally: if one starts with an algebraic probability space ${({\mathcal A},\tau)}$ and converts it back into a classical probability space ${(X, {\mathcal X}, \mu)}$, then there is a trace-preserving algebra homomorphism of ${{\mathcal A}}$ to ${L^\infty( X, {\mathcal X}, \mu )}$, which respects morphisms of the algebraic probability space. While this relationship is far weaker than an equivalence of categories (which would require that ${F \circ G}$ and ${G \circ F}$ are both natural isomorphisms), it is still good enough to allow many ergodic theory problems formulated using classical probability spaces to be reformulated instead as an equivalent problem in algebraic probability spaces.
Remark 2 The opposite composition ${G \circ F: \textbf{Prb} \rightarrow \textbf{Prb}}$ is a little odd: it takes an arbitrary probability space ${(X, {\mathcal X}, \mu)}$ and returns a more complicated probability space ${(X', {\mathcal X}', \mu')}$, with ${X'}$ being the space of homomorphisms ${\iota_x: L^\infty(X, {\mathcal X}, \mu) \rightarrow {\bf R}}$. while there is “morally” an embedding of ${X}$ into ${X'}$ using the evaluation map, this map does not exist in general because points in ${X}$ may well have zero measure. However, if one takes a “pointless” approach and focuses just on the measure algebras ${({\mathcal X}, \mu)}$, ${({\mathcal X}', \mu')}$, then these algebras become naturally isomorphic after quotienting out by null sets.
Remark 3 An algebraic probability space captures a bit more structure than a classical probability space, because ${{\mathcal A}}$ may be identified with a proper subset of ${L^\infty}$ that describes the “regular” functions (or random variables) of the space. For instance, starting with the unit circle ${{\bf R}/{\bf Z}}$ (with the usual Haar measure and the usual trace ${\tau(f) = \int_{{\bf R}/{\bf Z}} f}$), any unital subalgebra ${{\mathcal A}}$ of ${L^\infty({\bf R}/{\bf Z})}$ that is dense in ${L^2({\bf R}/{\bf Z})}$ will generate the same classical probability space ${G( {\mathcal A}, \tau )}$ on applying the functor ${G}$, namely one will get the space ${({\bf R}/{\bf Z})'}$ of homomorphisms from ${L^\infty({\bf R}/{\bf Z})}$ to ${{\bf R}}$ (with the measure induced from ${\tau}$). Thus for instance ${{\mathcal A}}$ could be the continuous functions ${C( {\bf R}/{\bf Z} )}$, the Wiener algebra ${A({\bf R}/{\bf Z})}$ or the full space ${L^\infty({\bf R}/{\bf Z})}$, but the classical space ${G( {\mathcal A}, \tau )}$ will be unable to distinguish these spaces from each other. In particular, the functor ${F \circ G}$ loses information (roughly speaking, this functor takes an algebraic probability space and completes it to a von Neumann algebra, but then forgets exactly what algebra was initially used to create this completion). In ergodic theory, this sort of “extra structure” is traditionally encoded in topological terms, by assuming that the underlying probability space ${X}$ has a nice topological structure (e.g. a standard Borel space); however, with the algebraic perspective one has the freedom to have non-topological notions of extra structure, by choosing ${{\mathcal A}}$ to be something other than an algebra ${C(X)}$ of continuous functions on a topological space. I hope to discuss one such example of extra structure (coming from the Gowers-Host-Kra theory of uniformity seminorms) in a later blog post (this generalises the example of the Wiener algebra given previously, which is encoding “Fourier structure”).
A small example of how one could use the functors ${F, G}$ is as follows. Suppose one has a classical probability space ${(X, {\mathcal X}, \mu)}$ with a measure-preserving action of an uncountable group ${\Gamma}$, which is only defined (and an action) up to almost everywhere equivalence; thus for instance for any set ${E}$ and any ${g, h \in \Gamma}$, ${T^{gh} E}$ and ${T^g T^h E}$ might not be exactly equal, but only equal up to a null set. For similar reasons, an element ${E}$ of the invariant factor ${{\mathcal X}^\Gamma}$ might not be exactly invariant with respect to ${\Gamma}$, but instead one only has ${T^g E}$ and ${E}$ equal up to null sets for each ${g \in \Gamma}$. One might like to “clean up” the action of ${\Gamma}$ to make it defined everywhere, and a genuine action everywhere, but this is not immediately achievable if ${\Gamma}$ is uncountable, since the union of all the null sets where something bad occurs may cease to be a null set. However, by applying the functor ${F}$, each shift ${T^g: X \rightarrow X}$ defines a morphism ${T^g: L^\infty(X, {\mathcal X}, \mu) \rightarrow L^\infty(X, {\mathcal X}, \mu)}$ on the associated algebraic probability space (i.e. the Koopman operator), and then applying ${G}$, we obtain a shift ${T^g: X' \rightarrow X'}$ on a new classical probability space ${(X', {\mathcal X}', \mu')}$ which now gives a genuine measure-preserving action of ${\Gamma}$, and which is equivalent to the original action from a measure algebra standpoint. The invariant factor ${{\mathcal X}^\Gamma}$ now consists of those sets in ${{\mathcal X}'}$ which are genuinely ${\Gamma}$-invariant, not just up to null sets. (Basically, the classical probability space ${(X', {\mathcal X}', \mu')}$ contains a Boolean algebra ${\overline{\mathcal B}}$ with the property that every measurable set ${A \in {\mathcal X}'}$ is equivalent up to null sets to precisely one set in ${\overline{\mathcal B}}$, allowing for a canonical “retraction” onto ${\overline{\mathcal B}}$ that eliminates all null set issues.)
More indirectly, the functors ${F, G}$ suggest that one should be able to develop a “pointless” form of ergodic theory, in which the underlying probability spaces are given algebraically rather than classically. I hope to give some more specific examples of this in later posts.
There are a number of ways to construct the real numbers ${{\bf R}}$, for instance
• as the metric completion of ${{\bf Q}}$ (thus, ${{\bf R}}$ is defined as the set of Cauchy sequences of rationals, modulo Cauchy equivalence);
• as the space of Dedekind cuts on the rationals ${{\bf Q}}$;
• as the space of quasimorphisms ${\phi: {\bf Z} \rightarrow {\bf Z}}$ on the integers, quotiented by bounded functions. (I believe this construction first appears in this paper of Street, who credits the idea to Schanuel, though the germ of this construction arguably goes all the way back to Eudoxus.)
There is also a fourth family of constructions that proceeds via nonstandard analysis, as a special case of what is known as the nonstandard hull construction. (Here I will assume some basic familiarity with nonstandard analysis and ultraproducts, as covered for instance in this previous blog post.) Given an unbounded nonstandard natural number ${N \in {}^* {\bf N} \backslash {\bf N}}$, one can define two external additive subgroups of the nonstandard integers ${{}^* {\bf Z}}$:
• The group ${O(N) := \{ n \in {}^* {\bf Z}: |n| \leq CN \hbox{ for some } C \in {\bf N} \}}$ of all nonstandard integers of magnitude less than or comparable to ${N}$; and
• The group ${o(N) := \{ n \in {}^* {\bf Z}: |n| \leq C^{-1} N \hbox{ for all } C \in {\bf N} \}}$ of nonstandard integers of magnitude infinitesimally smaller than ${N}$.
The group ${o(N)}$ is a subgroup of ${O(N)}$, so we may form the quotient group ${O(N)/o(N)}$. This space is isomorphic to the reals ${{\bf R}}$, and can in fact be used to construct the reals:
Proposition 1 For any coset ${n + o(N)}$ of ${O(N)/o(N)}$, there is a unique real number ${\hbox{st} \frac{n}{N}}$ with the property that ${\frac{n}{N} = \hbox{st} \frac{n}{N} + o(1)}$. The map ${n + o(N) \mapsto \hbox{st} \frac{n}{N}}$ is then an isomorphism between the additive groups ${O(N)/o(N)}$ and ${{\bf R}}$.
Proof: Uniqueness is clear. For existence, observe that the set ${\{ x \in {\bf R}: Nx \leq n + o(N) \}}$ is a Dedekind cut, and its supremum can be verified to have the required properties for ${\hbox{st} \frac{n}{N}}$. $\Box$
In a similar vein, we can view the unit interval ${[0,1]}$ in the reals as the quotient
$\displaystyle [0,1] \equiv [N] / o(N) \ \ \ \ \ (1)$
where ${[N]}$ is the nonstandard (i.e. internal) set ${\{ n \in {\bf N}: n \leq N \}}$; of course, ${[N]}$ is not a group, so one should interpret ${[N]/o(N)}$ as the image of ${[N]}$ under the quotient map ${{}^* {\bf Z} \rightarrow {}^* {\bf Z} / o(N)}$ (or ${O(N) \rightarrow O(N)/o(N)}$, if one prefers). Or to put it another way, (1) asserts that ${[0,1]}$ is the image of ${[N]}$ with respect to the map ${\pi: n \mapsto \hbox{st} \frac{n}{N}}$.
In this post I would like to record a nice measure-theoretic version of the equivalence (1), which essentially appears already in standard texts on Loeb measure (see e.g. this text of Cutland). To describe the results, we must first quickly recall the construction of Loeb measure on ${[N]}$. Given an internal subset ${A}$ of ${[N]}$, we may define the elementary measure ${\mu_0(A)}$ of ${A}$ by the formula
$\displaystyle \mu_0(A) := \hbox{st} \frac{|A|}{N}.$
This is a finitely additive probability measure on the Boolean algebra of internal subsets of ${[N]}$. We can then construct the Loeb outer measure ${\mu^*(A)}$ of any subset ${A \subset [N]}$ in complete analogy with Lebesgue outer measure by the formula
$\displaystyle \mu^*(A) := \inf \sum_{n=1}^\infty \mu_0(A_n)$
where ${(A_n)_{n=1}^\infty}$ ranges over all sequences of internal subsets of ${[N]}$ that cover ${A}$. We say that a subset ${A}$ of ${[N]}$ is Loeb measurable if, for any (standard) ${\epsilon>0}$, one can find an internal subset ${B}$ of ${[N]}$ which differs from ${A}$ by a set of Loeb outer measure at most ${\epsilon}$, and in that case we define the Loeb measure ${\mu(A)}$ of ${A}$ to be ${\mu^*(A)}$. It is a routine matter to show (e.g. using the Carathéodory extension theorem) that the space ${{\mathcal L}}$ of Loeb measurable sets is a ${\sigma}$-algebra, and that ${\mu}$ is a countably additive probability measure on this space that extends the elementary measure ${\mu_0}$. Thus ${[N]}$ now has the structure of a probability space ${([N], {\mathcal L}, \mu)}$.
Now, the group ${o(N)}$ acts (Loeb-almost everywhere) on the probability space ${[N]}$ by the addition map, thus ${T^h n := n+h}$ for ${n \in [N]}$ and ${h \in o(N)}$ (excluding a set of Loeb measure zero where ${n+h}$ exits ${[N]}$). This action is clearly seen to be measure-preserving. As such, we can form the invariant factor ${Z^0_{o(N)}([N]) = ([N], {\mathcal L}^{o(N)}, \mu\downharpoonright_{{\mathcal L}^{o(N)}})}$, defined by restricting attention to those Loeb measurable sets ${A \subset [N]}$ with the property that ${T^h A}$ is equal ${\mu}$-almost everywhere to ${A}$ for each ${h \in o(N)}$.
The claim is then that this invariant factor is equivalent (up to almost everywhere equivalence) to the unit interval ${[0,1]}$ with Lebesgue measure ${m}$ (and the trivial action of ${o(N)}$), by the same factor map ${\pi: n \mapsto \hbox{st} \frac{n}{N}}$ used in (1). More precisely:
Theorem 2 Given a set ${A \in {\mathcal L}^{o(N)}}$, there exists a Lebesgue measurable set ${B \subset [0,1]}$, unique up to ${m}$-a.e. equivalence, such that ${A}$ is ${\mu}$-a.e. equivalent to the set ${\pi^{-1}(B) := \{ n \in [N]: \hbox{st} \frac{n}{N} \in B \}}$. Conversely, if ${B \in [0,1]}$ is Lebesgue measurable, then ${\pi^{-1}(B)}$ is in ${{\mathcal L}^{o(N)}}$, and ${\mu( \pi^{-1}(B) ) = m( B )}$.
$\displaystyle [0,1] \equiv Z^0_{o(N)}( [N] )$
of (1).
Proof: We first prove the converse. It is clear that ${\pi^{-1}(B)}$ is ${o(N)}$-invariant, so it suffices to show that ${\pi^{-1}(B)}$ is Loeb measurable with Loeb measure ${m(B)}$. This is easily verified when ${B}$ is an elementary set (a finite union of intervals). By countable subadditivity of outer measure, this implies that Loeb outer measure of ${\pi^{-1}(E)}$ is bounded by the Lebesgue outer measure of ${E}$ for any set ${E \subset [0,1]}$; since every Lebesgue measurable set differs from an elementary set by a set of arbitrarily small Lebesgue outer measure, the claim follows.
Now we establish the forward claim. Uniqueness is clear from the converse claim, so it suffices to show existence. Let ${A \in {\mathcal L}^{o(N)}}$. Let ${\epsilon>0}$ be an arbitrary standard real number, then we can find an internal set ${A_\epsilon \subset [N]}$ which differs from ${A}$ by a set of Loeb measure at most ${\epsilon}$. As ${A}$ is ${o(N)}$-invariant, we conclude that for every ${h \in o(N)}$, ${A_\epsilon}$ and ${T^h A_\epsilon}$ differ by a set of Loeb measure (and hence elementary measure) at most ${2\epsilon}$. By the (contrapositive of the) underspill principle, there must exist a standard ${\delta>0}$ such that ${A_\epsilon}$ and ${T^h A_\epsilon}$ differ by a set of elementary measure at most ${2\epsilon}$ for all ${|h| \leq \delta N}$. If we then define the nonstandard function ${f_\epsilon: [N] \rightarrow {}^* {\bf R}}$ by the formula
$\displaystyle f(n) := \hbox{st} \frac{1}{\delta N} \sum_{m \in [N]: m \leq n \leq m+\delta N} 1_{A_\epsilon}(m),$
then from the (nonstandard) triangle inequality we have
$\displaystyle \frac{1}{N} \sum_{n \in [N]} |f(n) - 1_{A_\epsilon}(n)| \leq 3\epsilon$
(say). On the other hand, ${f}$ has the Lipschitz continuity property
$\displaystyle |f(n)-f(m)| \leq \frac{2|n-m|}{\delta N}$
and so in particular we see that
$\displaystyle \hbox{st} f(n) = \tilde f( \hbox{st} \frac{n}{N} )$
for some Lipschitz continuous function ${\tilde f: [0,1] \rightarrow [0,1]}$. If we then let ${E_\epsilon}$ be the set where ${\tilde f \geq 1 - \sqrt{\epsilon}}$, one can check that ${A_\epsilon}$ differs from ${\pi^{-1}(E_\epsilon)}$ by a set of Loeb outer measure ${O(\sqrt{\epsilon})}$, and hence ${A}$ does so also. Sending ${\epsilon}$ to zero, we see (from the converse claim) that ${1_{E_\epsilon}}$ is a Cauchy sequence in ${L^1}$ and thus converges in ${L^1}$ for some Lebesgue measurable ${E}$. The sets ${A_\epsilon}$ then converge in Loeb outer measure to ${\pi^{-1}(E)}$, giving the claim. $\Box$
Thanks to the Lebesgue differentiation theorem, the conditional expectation ${{\bf E}( f | Z^0_{o(N)}([N]))}$ of a bounded Loeb-measurable function ${f: [N] \rightarrow {\bf R}}$ can be expressed (as a function on ${[0,1]}$, defined ${m}$-a.e.) as
$\displaystyle {\bf E}( f | Z^0_{o(N)}([N]))(x) := \lim_{\epsilon \rightarrow 0} \frac{1}{2\epsilon} \int_{[x-\epsilon N,x+\epsilon N]} f\ d\mu.$
By the abstract ergodic theorem from the previous post, one can also view this conditional expectation as the element in the closed convex hull of the shifts ${T^h f}$, ${h = o(N)}$ of minimal ${L^2}$ norm. In particular, we obtain a form of the von Neumann ergodic theorem in this context: the averages ${\frac{1}{H} \sum_{h=1}^H T^h f}$ for ${H=O(N)}$ converge (as a net, rather than a sequence) in ${L^2}$ to ${{\bf E}( f | Z^0_{o(N)}([N]))}$.
If ${f: [N] \rightarrow [-1,1]}$ is (the standard part of) an internal function, that is to say the ultralimit of a sequence ${f_n: [N_n] \rightarrow [-1,1]}$ of finitary bounded functions, one can view the measurable function ${F := {\bf E}( f | Z^0_{o(N)}([N]))}$ as a limit of the ${f_n}$ that is analogous to the “graphons” that emerge as limits of graphs (see e.g. the recent text of Lovasz on graph limits). Indeed, the measurable function ${F: [0,1] \rightarrow [-1,1]}$ is related to the discrete functions ${f_n: [N_n] \rightarrow [-1,1]}$ by the formula
$\displaystyle \int_a^b F(x)\ dx = \hbox{st} \lim_{n \rightarrow p} \frac{1}{N_n} \sum_{a N_n \leq m \leq b N_n} f_n(m)$
for all ${0 \leq a < b \leq 1}$, where ${p}$ is the nonprincipal ultrafilter used to define the nonstandard universe. In particular, from the Arzela-Ascoli diagonalisation argument there is a subsequence ${n_j}$ such that
$\displaystyle \int_a^b F(x)\ dx = \lim_{j \rightarrow \infty} \frac{1}{N_{n_j}} \sum_{a N_{n_j} \leq m \leq b N_{n_j}} f_n(m),$
thus ${F}$ is the asymptotic density function of the ${f_n}$. For instance, if ${f_n}$ is the indicator function of a randomly chosen subset of ${[N_n]}$, then the asymptotic density function would equal ${1/2}$ (almost everywhere, at least).
I’m continuing to look into understanding the ergodic theory of ${o(N)}$ actions, as I believe this may allow one to apply ergodic theory methods to the “single-scale” or “non-asymptotic” setting (in which one averages only over scales comparable to a large parameter ${N}$, rather than the traditional asymptotic approach of letting the scale go to infinity). I’m planning some further posts in this direction, though this is still a work in progress.
The von Neumann ergodic theorem (the Hilbert space version of the mean ergodic theorem) asserts that if ${U: H \rightarrow H}$ is a unitary operator on a Hilbert space ${H}$, and ${v \in H}$ is a vector in that Hilbert space, then one has
$\displaystyle \lim_{N \rightarrow \infty} \frac{1}{N} \sum_{n=1}^N U^n v = \pi_{H^U} v$
in the strong topology, where ${H^U := \{ w \in H: Uw = w \}}$ is the ${U}$-invariant subspace of ${H}$, and ${\pi_{H^U}}$ is the orthogonal projection to ${H^U}$. (See e.g. these previous lecture notes for a proof.) The same proof extends to more general amenable groups: if ${G}$ is a countable amenable group acting on a Hilbert space ${H}$ by unitary transformations ${g: H \rightarrow H}$, and ${v \in H}$ is a vector in that Hilbert space, then one has
$\displaystyle \lim_{N \rightarrow \infty} \frac{1}{|\Phi_N|} \sum_{g \in \Phi_N} gv = \pi_{H^G} v \ \ \ \ \ (1)$
for any Folner sequence ${\Phi_N}$ of ${G}$, where ${H^G := \{ w \in H: gw = w \hbox{ for all }g \in G \}}$ is the ${G}$-invariant subspace. Thus one can interpret ${\pi_{H^G} v}$ as a certain average of elements of the orbit ${Gv := \{ gv: g \in G \}}$ of ${v}$.
I recently discovered that there is a simple variant of this ergodic theorem that holds even when the group ${G}$ is not amenable (or not discrete), using a more abstract notion of averaging:
Theorem 1 (Abstract ergodic theorem) Let ${G}$ be an arbitrary group acting unitarily on a Hilbert space ${H}$, and let ${v}$ be a vector in ${H}$. Then ${\pi_{H^G} v}$ is the element in the closed convex hull of ${Gv := \{ gv: g \in G \}}$ of minimal norm, and is also the unique element of ${H^G}$ in this closed convex hull.
Proof: As the closed convex hull of ${Gv}$ is closed, convex, and non-empty in a Hilbert space, it is a classical fact (see e.g. Proposition 1 of this previous post) that it has a unique element ${F}$ of minimal norm. If ${T_g F \neq F}$ for some ${g}$, then the midpoint of ${T_g F}$ and ${F}$ would be in the closed convex hull and be of smaller norm, a contradiction; thus ${F}$ is ${G}$-invariant. To finish the first claim, it suffices to show that ${v-F}$ is orthogonal to every element ${h}$ of ${H^G}$. But if this were not the case for some such ${h}$, we would have ${\langle T_g v - F, h \rangle = \langle v-F,h\rangle \neq 0}$ for all ${g \in G}$, and thus on taking convex hulls ${\langle F-F,h\rangle = \langle f-F,f\rangle \neq 0}$, a contradiction.
Finally, since ${T_g v - F}$ is orthogonal to ${H^G}$, the same is true for ${F'-F}$ for any ${F'}$ in the closed convex hull of ${Gv}$, and this gives the second claim. $\Box$
This result is due to Alaoglu and Birkhoff. It implies the amenable ergodic theorem (1); indeed, given any ${\epsilon>0}$, Theorem 1 implies that there is a finite convex combination ${v_\epsilon}$ of shifts ${gv}$ of ${v}$ which lies within ${\epsilon}$ (in the ${H}$ norm) to ${\pi_{H^G} v}$. By the triangle inequality, all the averages ${\frac{1}{|\Phi_N|} \sum_{g \in \Phi_N} gv_\epsilon}$ also lie within ${\epsilon}$ of ${\pi_{H^G} v}$, but by the Folner property this implies that the averages ${\frac{1}{|\Phi_N|} \sum_{g \in \Phi_N} gv}$ are eventually within ${2\epsilon}$ (say) of ${\pi_{H^G} v}$, giving the claim.
It turns out to be possible to use Theorem 1 as a substitute for the mean ergodic theorem in a number of contexts, thus removing the need for an amenability hypothesis. Here is a basic application:
Corollary 2 (Relative orthogonality) Let ${G}$ be a group acting unitarily on a Hilbert space ${H}$, and let ${V}$ be a ${G}$-invariant subspace of ${H}$. Then ${V}$ and ${H^G}$ are relatively orthogonal over their common subspace ${V^G}$, that is to say the restrictions of ${V}$ and ${H^G}$ to the orthogonal complement of ${V^G}$ are orthogonal to each other.
Proof: By Theorem 1, we have ${\pi_{H^G} v = \pi_{V^G} v}$ for all ${v \in V}$, and the claim follows. (Thanks to Gergely Harcos for this short argument.) $\Box$
Now we give a more advanced application of Theorem 1, to establish some “Mackey theory” over arbitrary groups ${G}$. Define a ${G}$-system ${(X, {\mathcal X}, \mu, (T_g)_{g \in G})}$ to be a probability space ${X = (X, {\mathcal X}, \mu)}$ together with a measure-preserving action ${(T_g)_{g \in G}}$ of ${G}$ on ${X}$; this gives an action of ${G}$ on ${L^2(X) = L^2(X,{\mathcal X},\mu)}$, which by abuse of notation we also call ${T_g}$:
$\displaystyle T_g f := f \circ T_{g^{-1}}.$
(In this post we follow the usual convention of defining the ${L^p}$ spaces by quotienting out by almost everywhere equivalence.) We say that a ${G}$-system is ergodic if ${L^2(X)^G}$ consists only of the constants.
(A technical point: the theory becomes slightly cleaner if we interpret our measure spaces abstractly (or “pointlessly“), removing the underlying space ${X}$ and quotienting ${{\mathcal X}}$ by the ${\sigma}$-ideal of null sets, and considering maps such as ${T_g}$ only on this quotient ${\sigma}$-algebra (or on the associated von Neumann algebra ${L^\infty(X)}$ or Hilbert space ${L^2(X)}$). However, we will stick with the more traditional setting of classical probability spaces here to keep the notation familiar, but with the understanding that many of the statements below should be understood modulo null sets.)
A factor ${Y = (Y, {\mathcal Y}, \nu, (S_g)_{g \in G})}$ of a ${G}$-system ${X = (X,{\mathcal X},\mu, (T_g)_{g \in G})}$ is another ${G}$-system together with a factor map ${\pi: X \rightarrow Y}$ which commutes with the ${G}$-action (thus ${T_g \pi = \pi S_g}$ for all ${g \in G}$) and respects the measure in the sense that ${\mu(\pi^{-1}(E)) = \nu(E)}$ for all ${E \in {\mathcal Y}}$. For instance, the ${G}$-invariant factor ${Z^0_G(X) := (X, {\mathcal X}^G, \mu\downharpoonright_{{\mathcal X}^G}, (T_g)_{g \in G})}$, formed by restricting ${X}$ to the invariant algebra ${{\mathcal X}^G := \{ E \in {\mathcal X}: T_g E = E \hbox{ a.e. for all } g \in G \}}$, is a factor of ${X}$. (This factor is the first factor in an important hierachy, the next element of which is the Kronecker factor ${Z^1_G(X)}$, but we will not discuss higher elements of this hierarchy further here.) If ${Y}$ is a factor of ${X}$, we refer to ${X}$ as an extension of ${Y}$.
From Corollary 2 we have
Corollary 3 (Relative independence) Let ${X}$ be a ${G}$-system for a group ${G}$, and let ${Y}$ be a factor of ${X}$. Then ${Y}$ and ${Z^0_G(X)}$ are relatively independent over their common factor ${Z^0_G(Y)}$, in the sense that the spaces ${L^2(Y)}$ and ${L^2(Z^0_G(X))}$ are relatively orthogonal over ${L^2(Z^0_G(Y))}$ when all these spaces are embedded into ${L^2(X)}$.
This has a simple consequence regarding the product ${X \times Y = (X \times Y, {\mathcal X} \times {\mathcal Y}, \mu \times \nu, (T_g \oplus S_g)_{g \in G})}$ of two ${G}$-systems ${X = (X, {\mathcal X}, \mu, (T_g)_{g \in G})}$ and ${Y = (Y, {\mathcal Y}, \nu, (S_g)_{g \in G})}$, in the case when the ${Y}$ action is trivial:
Lemma 4 If ${X,Y}$ are two ${G}$-systems, with the action of ${G}$ on ${Y}$ trivial, then ${Z^0_G(X \times Y)}$ is isomorphic to ${Z^0_G(X) \times Y}$ in the obvious fashion.
This lemma is immediate for countable ${G}$, since for a ${G}$-invariant function ${f}$, one can ensure that ${T_g f = f}$ holds simultaneously for all ${g \in G}$ outside of a null set, but is a little trickier for uncountable ${G}$.
Proof: It is clear that ${Z^0_G(X) \times Y}$ is a factor of ${Z^0_G(X \times Y)}$. To obtain the reverse inclusion, suppose that it fails, thus there is a non-zero ${f \in L^2(Z^0_G(X \times Y))}$ which is orthogonal to ${L^2(Z^0_G(X) \times Y)}$. In particular, we have ${fg}$ orthogonal to ${L^2(Z^0_G(X))}$ for any ${g \in L^\infty(Y)}$. Since ${fg}$ lies in ${L^2(Z^0_G(X \times Y))}$, we conclude from Corollary 3 (viewing ${X}$ as a factor of ${X \times Y}$) that ${fg}$ is also orthogonal to ${L^2(X)}$. Since ${g}$ is an arbitrary element of ${L^\infty(Y)}$, we conclude that ${f}$ is orthogonal to ${L^2(X \times Y)}$ and in particular is orthogonal to itself, a contradiction. (Thanks to Gergely Harcos for this argument.) $\Box$
Now we discuss the notion of a group extension.
Definition 5 (Group extension) Let ${G}$ be an arbitrary group, let ${Y = (Y, {\mathcal Y}, \nu, (S_g)_{g \in G})}$ be a ${G}$-system, and let ${K}$ be a compact metrisable group. A ${K}$-extension of ${Y}$ is an extension ${X = (X, {\mathcal X}, \mu, (T_g)_{g \in G})}$ whose underlying space is ${X = Y \times K}$ (with ${{\mathcal X}}$ the product of ${{\mathcal Y}}$ and the Borel ${\sigma}$-algebra on ${K}$), the factor map is ${\pi: (y,k) \mapsto y}$, and the shift maps ${T_g}$ are given by
$\displaystyle T_g ( y, k ) = (S_g y, \rho_g(y) k )$
where for each ${g \in G}$, ${\rho_g: Y \rightarrow K}$ is a measurable map (known as the cocycle associated to the ${K}$-extension ${X}$).
An important special case of a ${K}$-extension arises when the measure ${\mu}$ is the product of ${\nu}$ with the Haar measure ${dk}$ on ${K}$. In this case, ${X}$ also has a ${K}$-action ${k': (y,k) \mapsto (y,k(k')^{-1})}$ that commutes with the ${G}$-action, making ${X}$ a ${G \times K}$-system. More generally, ${\mu}$ could be the product of ${\nu}$ with the Haar measure ${dh}$ of some closed subgroup ${H}$ of ${K}$, with ${\rho_g}$ taking values in ${H}$; then ${X}$ is now a ${G \times H}$ system. In this latter case we will call ${X}$ ${H}$-uniform.
If ${X}$ is a ${K}$-extension of ${Y}$ and ${U: Y \rightarrow K}$ is a measurable map, we can define the gauge transform ${X_U}$ of ${X}$ to be the ${K}$-extension of ${Y}$ whose measure ${\mu_U}$ is the pushforward of ${\mu}$ under the map ${(y,k) \mapsto (y, U(y) k)}$, and whose cocycles ${\rho_{g,U}: Y \rightarrow K}$ are given by the formula
$\displaystyle \rho_{g,U}(y) := U(gy) \rho_g(y) U(y)^{-1}.$
It is easy to see that ${X_U}$ is a ${K}$-extension that is isomorphic to ${X}$ as a ${K}$-extension of ${Y}$; we will refer to ${X_U}$ and ${X}$ as equivalent systems, and ${\rho_{g,U}}$ as cohomologous to ${\rho_g}$. We then have the following fundamental result of Mackey and of Zimmer:
Theorem 6 (Mackey-Zimmer theorem) Let ${G}$ be an arbitrary group, let ${Y}$ be an ergodic ${G}$-system, and let ${K}$ be a compact metrisable group. Then every ergodic ${K}$-extension ${X}$ of ${Y}$ is equivalent to an ${H}$-uniform extension of ${Y}$ for some closed subgroup ${H}$ of ${K}$.
This theorem is usually stated for amenable groups ${G}$, but by using Theorem 1 (or more precisely, Corollary 3) the result is in fact also valid for arbitrary groups; we give the proof below the fold. (In the usual formulations of the theorem, ${X}$ and ${Y}$ are also required to be Lebesgue spaces, or at least standard Borel, but again with our abstract approach here, such hypotheses will be unnecessary.) Among other things, this theorem plays an important role in the Furstenberg-Zimmer structural theory of measure-preserving systems (as well as subsequent refinements of this theory by Host and Kra); see this previous blog post for some relevant discussion. One can obtain similar descriptions of non-ergodic extensions via the ergodic decomposition, but the result becomes more complicated to state, and we will not do so here.
(This is an extended blog post version of my talk “Ultraproducts as a Bridge Between Discrete and Continuous Analysis” that I gave at the Simons institute for the theory of computing at the workshop “Neo-Classical methods in discrete analysis“. Some of the material here is drawn from previous blog posts, notably “Ultraproducts as a bridge between hard analysis and soft analysis” and “Ultralimit analysis and quantitative algebraic geometry“‘. The text here has substantially more details than the talk; one may wish to skip all of the proofs given here to obtain a closer approximation to the original talk.)
Discrete analysis, of course, is primarily interested in the study of discrete (or “finitary”) mathematical objects: integers, rational numbers (which can be viewed as ratios of integers), finite sets, finite graphs, finite or discrete metric spaces, and so forth. However, many powerful tools in mathematics (e.g. ergodic theory, measure theory, topological group theory, algebraic geometry, spectral theory, etc.) work best when applied to continuous (or “infinitary”) mathematical objects: real or complex numbers, manifolds, algebraic varieties, continuous topological or metric spaces, etc. In order to apply results and ideas from continuous mathematics to discrete settings, there are basically two approaches. One is to directly discretise the arguments used in continuous mathematics, which often requires one to keep careful track of all the bounds on various quantities of interest, particularly with regard to various error terms arising from discretisation which would otherwise have been negligible in the continuous setting. The other is to construct continuous objects as limits of sequences of discrete objects of interest, so that results from continuous mathematics may be applied (often as a “black box”) to the continuous limit, which then can be used to deduce consequences for the original discrete objects which are quantitative (though often ineffectively so). The latter approach is the focus of this current talk.
The following table gives some examples of a discrete theory and its continuous counterpart, together with a limiting procedure that might be used to pass from the former to the latter:
(Discrete) (Continuous) (Limit method) Ramsey theory Topological dynamics Compactness Density Ramsey theory Ergodic theory Furstenberg correspondence principle Graph/hypergraph regularity Measure theory Graph limits Polynomial regularity Linear algebra Ultralimits Structural decompositions Hilbert space geometry Ultralimits Fourier analysis Spectral theory Direct and inverse limits Quantitative algebraic geometry Algebraic geometry Schemes Discrete metric spaces Continuous metric spaces Gromov-Hausdorff limits Approximate group theory Topological group theory Model theory
As the above table illustrates, there are a variety of different ways to form a limiting continuous object. Roughly speaking, one can divide limits into three categories:
• Topological and metric limits. These notions of limits are commonly used by analysts. Here, one starts with a sequence (or perhaps a net) of objects ${x_n}$ in a common space ${X}$, which one then endows with the structure of a topological space or a metric space, by defining a notion of distance between two points of the space, or a notion of open neighbourhoods or open sets in the space. Provided that the sequence or net is convergent, this produces a limit object ${\lim_{n \rightarrow \infty} x_n}$, which remains in the same space, and is “close” to many of the original objects ${x_n}$ with respect to the given metric or topology.
• Categorical limits. These notions of limits are commonly used by algebraists. Here, one starts with a sequence (or more generally, a diagram) of objects ${x_n}$ in a category ${X}$, which are connected to each other by various morphisms. If the ambient category is well-behaved, one can then form the direct limit ${\varinjlim x_n}$ or the inverse limit ${\varprojlim x_n}$ of these objects, which is another object in the same category ${X}$, and is connected to the original objects ${x_n}$ by various morphisms.
• Logical limits. These notions of limits are commonly used by model theorists. Here, one starts with a sequence of objects ${x_{\bf n}}$ or of spaces ${X_{\bf n}}$, each of which is (a component of) a model for given (first-order) mathematical language (e.g. if one is working in the language of groups, ${X_{\bf n}}$ might be groups and ${x_{\bf n}}$ might be elements of these groups). By using devices such as the ultraproduct construction, or the compactness theorem in logic, one can then create a new object ${\lim_{{\bf n} \rightarrow \alpha} x_{\bf n}}$ or a new space ${\prod_{{\bf n} \rightarrow \alpha} X_{\bf n}}$, which is still a model of the same language (e.g. if the spaces ${X_{\bf n}}$ were all groups, then the limiting space ${\prod_{{\bf n} \rightarrow \alpha} X_{\bf n}}$ will also be a group), and is “close” to the original objects or spaces in the sense that any assertion (in the given language) that is true for the limiting object or space, will also be true for many of the original objects or spaces, and conversely. (For instance, if ${\prod_{{\bf n} \rightarrow \alpha} X_{\bf n}}$ is an abelian group, then the ${X_{\bf n}}$ will also be abelian groups for many ${{\bf n}}$.)
The purpose of this talk is to highlight the third type of limit, and specifically the ultraproduct construction, as being a “universal” limiting procedure that can be used to replace most of the limits previously mentioned. Unlike the topological or metric limits, one does not need the original objects ${x_{\bf n}}$ to all lie in a common space ${X}$ in order to form an ultralimit ${\lim_{{\bf n} \rightarrow \alpha} x_{\bf n}}$; they are permitted to lie in different spaces ${X_{\bf n}}$; this is more natural in many discrete contexts, e.g. when considering graphs on ${{\bf n}}$ vertices in the limit when ${{\bf n}}$ goes to infinity. Also, no convergence properties on the ${x_{\bf n}}$ are required in order for the ultralimit to exist. Similarly, ultraproduct limits differ from categorical limits in that no morphisms between the various spaces ${X_{\bf n}}$ involved are required in order to construct the ultraproduct.
With so few requirements on the objects ${x_{\bf n}}$ or spaces ${X_{\bf n}}$, the ultraproduct construction is necessarily a very “soft” one. Nevertheless, the construction has two very useful properties which make it particularly useful for the purpose of extracting good continuous limit objects out of a sequence of discrete objects. First of all, there is Łos’s theorem, which roughly speaking asserts that any first-order sentence which is asymptotically obeyed by the ${x_{\bf n}}$, will be exactly obeyed by the limit object ${\lim_{{\bf n} \rightarrow \alpha} x_{\bf n}}$; in particular, one can often take a discrete sequence of “partial counterexamples” to some assertion, and produce a continuous “complete counterexample” that same assertion via an ultraproduct construction; taking the contrapositives, one can often then establish a rigorous equivalence between a quantitative discrete statement and its qualitative continuous counterpart. Secondly, there is the countable saturation property that ultraproducts automatically enjoy, which is a property closely analogous to that of compactness in topological spaces, and can often be used to ensure that the continuous objects produced by ultraproduct methods are “complete” or “compact” in various senses, which is particularly useful in being able to upgrade qualitative (or “pointwise”) bounds to quantitative (or “uniform”) bounds, more or less “for free”, thus reducing significantly the burden of “epsilon management” (although the price one pays for this is that one needs to pay attention to which mathematical objects of study are “standard” and which are “nonstandard”). To achieve this compactness or completeness, one sometimes has to restrict to the “bounded” portion of the ultraproduct, and it is often also convenient to quotient out the “infinitesimal” portion in order to complement these compactness properties with a matching “Hausdorff” property, thus creating familiar examples of continuous spaces, such as locally compact Hausdorff spaces.
Ultraproducts are not the only logical limit in the model theorist’s toolbox, but they are one of the simplest to set up and use, and already suffice for many of the applications of logical limits outside of model theory. In this post, I will set out the basic theory of these ultraproducts, and illustrate how they can be used to pass between discrete and continuous theories in each of the examples listed in the above table.
Apart from the initial “one-time cost” of setting up the ultraproduct machinery, the main loss one incurs when using ultraproduct methods is that it becomes very difficult to extract explicit quantitative bounds from results that are proven by transferring qualitative continuous results to the discrete setting via ultraproducts. However, in many cases (particularly those involving regularity-type lemmas) the bounds are already of tower-exponential type or worse, and there is arguably not much to be lost by abandoning the explicit quantitative bounds altogether.
Tamar Ziegler and I have just uploaded to the arXiv our joint paper “A multi-dimensional Szemerédi theorem for the primes via a correspondence principle“. This paper is related to an earlier result of Ben Green and mine in which we established that the primes contain arbitrarily long arithmetic progressions. Actually, in that paper we proved a more general result:
Theorem 1 (Szemerédi’s theorem in the primes) Let ${A}$ be a subset of the primes ${{\mathcal P}}$ of positive relative density, thus ${\limsup_{N \rightarrow \infty} \frac{|A \cap [N]|}{|{\mathcal P} \cap [N]|} > 0}$. Then ${A}$ contains arbitrarily long arithmetic progressions.
This result was based in part on an earlier paper of Green that handled the case of progressions of length three. With the primes replaced by the integers, this is of course the famous theorem of Szemerédi.
Szemerédi’s theorem has now been generalised in many different directions. One of these is the multidimensional Szemerédi theorem of Furstenberg and Katznelson, who used ergodic-theoretic techniques to show that any dense subset of ${{\bf Z}^d}$ necessarily contained infinitely many constellations of any prescribed shape. Our main result is to relativise that theorem to the primes as well:
Theorem 2 (Multidimensional Szemerédi theorem in the primes) Let ${d \geq 1}$, and let ${A}$ be a subset of the ${d^{th}}$ Cartesian power ${{\mathcal P}^d}$ of the primes of positive relative density, thus
$\displaystyle \limsup_{N \rightarrow \infty} \frac{|A \cap [N]^d|}{|{\mathcal P}^d \cap [N]^d|} > 0.$
Then for any ${v_1,\ldots,v_k \in {\bf Z}^d}$, ${A}$ contains infinitely many “constellations” of the form ${a+r v_1, \ldots, a + rv_k}$ with ${a \in {\bf Z}^k}$ and ${r}$ a positive integer.
In the case when ${A}$ is itself a Cartesian product of one-dimensional sets (in particular, if ${A}$ is all of ${{\mathcal P}^d}$), this result already follows from Theorem 1, but there does not seem to be a similarly easy argument to deduce the general case of Theorem 2 from previous results. Simultaneously with this paper, an independent proof of Theorem 2 using a somewhat different method has been established by Cook, Maygar, and Titichetrakun.
The result is reminiscent of an earlier result of mine on finding constellations in the Gaussian primes (or dense subsets thereof). That paper followed closely the arguments of my original paper with Ben Green, namely it first enclosed (a W-tricked version of) the primes or Gaussian primes (in a sieve theoretic-sense) by a slightly larger set (or more precisely, a weight function ${\nu}$) of almost primes or almost Gaussian primes, which one could then verify (using methods closely related to the sieve-theoretic methods in the ongoing Polymath8 project) to obey certain pseudorandomness conditions, known as the linear forms condition and the correlation condition. Very roughly speaking, these conditions assert statements of the following form: if ${n}$ is a randomly selected integer, then the events of ${n+h_1,\ldots,n+h_k}$ simultaneously being an almost prime (or almost Gaussian prime) are approximately independent for most choices of ${h_1,\ldots,h_k}$. Once these conditions are satisfied, one can then run a transference argument (initially based on ergodic-theory methods, but nowadays there are simpler transference results based on the Hahn-Banach theorem, due to Gowers and Reingold-Trevisan-Tulsiani-Vadhan) to obtain relative Szemerédi-type theorems from their absolute counterparts.
However, when one tries to adapt these arguments to sets such as ${{\mathcal P}^2}$, a new difficulty occurs: the natural analogue of the almost primes would be the Cartesian square ${{\mathcal A}^2}$ of the almost primes – pairs ${(n,m)}$ whose entries are both almost primes. (Actually, for technical reasons, one does not work directly with a set of almost primes, but would instead work with a weight function such as ${\nu(n) \nu(m)}$ that is concentrated on a set such as ${{\mathcal A}^2}$, but let me ignore this distinction for now.) However, this set ${{\mathcal A}^2}$ does not enjoy as many pseudorandomness conditions as one would need for a direct application of the transference strategy to work. More specifically, given any fixed ${h, k}$, and random ${(n,m)}$, the four events
$\displaystyle (n,m) \in {\mathcal A}^2$
$\displaystyle (n+h,m) \in {\mathcal A}^2$
$\displaystyle (n,m+k) \in {\mathcal A}^2$
$\displaystyle (n+h,m+k) \in {\mathcal A}^2$
do not behave independently (as they would if ${{\mathcal A}^2}$ were replaced for instance by the Gaussian almost primes), because any three of these events imply the fourth. This blocks the transference strategy for constellations which contain some right-angles to them (e.g. constellations of the form ${(n,m), (n+r,m), (n,m+r)}$) as such constellations soon turn into rectangles such as the one above after applying Cauchy-Schwarz a few times. (But a few years ago, Cook and Magyar showed that if one restricted attention to constellations which were in general position in the sense that any coordinate hyperplane contained at most one element in the constellation, then this obstruction does not occur and one can establish Theorem 2 in this case through the transference argument.) It’s worth noting that very recently, Conlon, Fox, and Zhao have succeeded in removing of the pseudorandomness conditions (namely the correlation condition) from the transference principle, leaving only the linear forms condition as the remaining pseudorandomness condition to be verified, but unfortunately this does not completely solve the above problem because the linear forms condition also fails for ${{\mathcal A}^2}$ (or for weights concentrated on ${{\mathcal A}^2}$) when applied to rectangular patterns.
There are now two ways known to get around this problem and establish Theorem 2 in full generality. The approach of Cook, Magyar, and Titichetrakun proceeds by starting with one of the known proofs of the multidimensional Szemerédi theorem – namely, the proof that proceeds through hypergraph regularity and hypergraph removal – and attach pseudorandom weights directly within the proof itself, rather than trying to add the weights to the result of that proof through a transference argument. (A key technical issue is that weights have to be added to all the levels of the hypergraph – not just the vertices and top-order edges – in order to circumvent the failure of naive pseudorandomness.) As one has to modify the entire proof of the multidimensional Szemerédi theorem, rather than use that theorem as a black box, the Cook-Magyar-Titichetrakun argument is lengthier than ours; on the other hand, it is more general and does not rely on some difficult theorems about primes that are used in our paper.
In our approach, we continue to use the multidimensional Szemerédi theorem (or more precisely, the equivalent theorem of Furstenberg and Katznelson concerning multiple recurrence for commuting shifts) as a black box. The difference is that instead of using a transference principle to connect the relative multidimensional Szemerédi theorem we need to the multiple recurrence theorem, we instead proceed by a version of the Furstenberg correspondence principle, similar to the one that connects the absolute multidimensional Szemerédi theorem to the multiple recurrence theorem. I had discovered this approach many years ago in an unpublished note, but had abandoned it because it required an infinite number of linear forms conditions (in contrast to the transference technique, which only needed a finite number of linear forms conditions and (until the recent work of Conlon-Fox-Zhao) a correlation condition). The reason for this infinite number of conditions is that the correspondence principle has to build a probability measure on an entire ${\sigma}$-algebra; for this, it is not enough to specify the measure ${\mu(A)}$ of a single set such as ${A}$, but one also has to specify the measure ${\mu( T^{n_1} A \cap \ldots \cap T^{n_m} A)}$ of “cylinder sets” such as ${T^{n_1} A \cap \ldots \cap T^{n_m} A}$ where ${m}$ could be arbitrarily large. The larger ${m}$ gets, the more linear forms conditions one needs to keep the correspondence under control.
With the sieve weights ${\nu}$ we were using at the time, standard sieve theory methods could indeed provide a finite number of linear forms conditions, but not an infinite number, so my idea was abandoned. However, with my later work with Green and Ziegler on linear equations in primes (and related work on the Mobius-nilsequences conjecture and the inverse conjecture on the Gowers norm), Tamar and I realised that the primes themselves obey an infinite number of linear forms conditions, so one can basically use the primes (or a proxy for the primes, such as the von Mangoldt function ${\Lambda}$) as the enveloping sieve weight, rather than a classical sieve. Thus my old idea of using the Furstenberg correspondence principle to transfer Szemerédi-type theorems to the primes could actually be realised. In the one-dimensional case, this simply produces a much more complicated proof of Theorem 1 than the existing one; but it turns out that the argument works as well in higher dimensions and yields Theorem 2 relatively painlessly, except for the fact that it needs the results on linear equations in primes, the known proofs of which are extremely lengthy (and also require some of the transference machinery mentioned earlier). The problem of correlations in rectangles is avoided in the correspondence principle approach because one can compensate for such correlations by performing a suitable weighted limit to compute the measure ${\mu( T^{n_1} A \cap \ldots \cap T^{n_m} A)}$ of cylinder sets, with each ${m}$ requiring a different weighted correction. (This may be related to the Cook-Magyar-Titichetrakun strategy of weighting all of the facets of the hypergraph in order to recover pseudorandomness, although our contexts are rather different.)
Vitaly Bergelson, Tamar Ziegler, and I have just uploaded to the arXiv our joint paper “Multiple recurrence and convergence results associated to ${{\bf F}_{p}^{\omega}}$-actions“. This paper is primarily concerned with limit formulae in the theory of multiple recurrence in ergodic theory. Perhaps the most basic formula of this type is the mean ergodic theorem, which (among other things) asserts that if ${(X,{\mathcal X}, \mu,T)}$ is a measure-preserving ${{\bf Z}}$-system (which, in this post, means that ${(X,{\mathcal X}, \mu)}$ is a probability space and ${T: X \mapsto X}$ is measure-preserving and invertible, thus giving an action ${(T^n)_{n \in {\bf Z}}}$ of the integers), and ${f,g \in L^2(X,{\mathcal X}, \mu)}$ are functions, and ${X}$ is ergodic (which means that ${L^2(X,{\mathcal X}, \mu)}$ contains no ${T}$-invariant functions other than the constants (up to almost everywhere equivalence, of course)), then the average
$\displaystyle \frac{1}{N} \sum_{n=1}^N \int_X f(x) g(T^n x)\ d\mu \ \ \ \ \ (1)$
converges as ${N \rightarrow \infty}$ to the expression
$\displaystyle (\int_X f(x)\ d\mu) (\int_X g(x)\ d\mu);$
see e.g. this previous blog post. Informally, one can interpret this limit formula as an equidistribution result: if ${x}$ is drawn at random from ${X}$ (using the probability measure ${\mu}$), and ${n}$ is drawn at random from ${\{1,\ldots,N\}}$ for some large ${N}$, then the pair ${(x, T^n x)}$ becomes uniformly distributed in the product space ${X \times X}$ (using product measure ${\mu \times \mu}$) in the limit as ${N \rightarrow \infty}$.
If we allow ${(X,\mu)}$ to be non-ergodic, then we still have a limit formula, but it is a bit more complicated. Let ${{\mathcal X}^T}$ be the ${T}$-invariant measurable sets in ${{\mathcal X}}$; the ${{\bf Z}}$-system ${(X, {\mathcal X}^T, \mu, T)}$ can then be viewed as a factor of the original system ${(X, {\mathcal X}, \mu, T)}$, which is equivalent (in the sense of measure-preserving systems) to a trivial system ${(Z_0, {\mathcal Z}_0, \mu_{Z_0}, 1)}$ (known as the invariant factor) in which the shift is trivial. There is then a projection map ${\pi_0: X \rightarrow Z_0}$ to the invariant factor which is a factor map, and the average (1) converges in the limit to the expression
$\displaystyle \int_{Z_0} (\pi_0)_* f(z) (\pi_0)_* g(z)\ d\mu_{Z_0}(x), \ \ \ \ \ (2)$
where ${(\pi_0)_*: L^2(X,{\mathcal X},\mu) \rightarrow L^2(Z_0,{\mathcal Z}_0,\mu_{Z_0})}$ is the pushforward map associated to the map ${\pi_0: X \rightarrow Z_0}$; see e.g. this previous blog post. We can interpret this as an equidistribution result. If ${(x,T^n x)}$ is a pair as before, then we no longer expect complete equidistribution in ${X \times X}$ in the non-ergodic, because there are now non-trivial constraints relating ${x}$ with ${T^n x}$; indeed, for any ${T}$-invariant function ${f: X \rightarrow {\bf C}}$, we have the constraint ${f(x) = f(T^n x)}$; putting all these constraints together we see that ${\pi_0(x) = \pi_0(T^n x)}$ (for almost every ${x}$, at least). The limit (2) can be viewed as an assertion that this constraint ${\pi_0(x) = \pi_0(T^n x)}$ are in some sense the “only” constraints between ${x}$ and ${T^n x}$, and that the pair ${(x,T^n x)}$ is uniformly distributed relative to these constraints.
Limit formulae are known for multiple ergodic averages as well, although the statement becomes more complicated. For instance, consider the expression
$\displaystyle \frac{1}{N} \sum_{n=1}^N \int_X f(x) g(T^n x) h(T^{2n} x)\ d\mu \ \ \ \ \ (3)$
for three functions ${f,g,h \in L^\infty(X, {\mathcal X}, \mu)}$; this is analogous to the combinatorial task of counting length three progressions in various sets. For simplicity we assume the system ${(X,{\mathcal X},\mu,T)}$ to be ergodic. Naively one might expect this limit to then converge to
$\displaystyle (\int_X f\ d\mu) (\int_X g\ d\mu) (\int_X h\ d\mu)$
which would roughly speaking correspond to an assertion that the triplet ${(x,T^n x, T^{2n} x)}$ is asymptotically equidistributed in ${X \times X \times X}$. However, even in the ergodic case there can be additional constraints on this triplet that cannot be seen at the level of the individual pairs ${(x,T^n x)}$, ${(x, T^{2n} x)}$. The key obstruction here is that of eigenfunctions of the shift ${T: X \rightarrow X}$, that is to say non-trivial functions ${f: X \rightarrow S^1}$ that obey the eigenfunction equation ${Tf = \lambda f}$ almost everywhere for some constant (or ${T}$-invariant) ${\lambda}$. Each such eigenfunction generates a constraint
$\displaystyle f(x) \overline{f(T^n x)}^2 f(T^{2n} x) = 1 \ \ \ \ \ (4)$
tying together ${x}$, ${T^n x}$, and ${T^{2n} x}$. However, it turns out that these are in some sense the only constraints on ${x,T^n x, T^{2n} x}$ that are relevant for the limit (3). More precisely, if one sets ${{\mathcal X}_1}$ to be the sub-algebra of ${{\mathcal X}}$ generated by the eigenfunctions of ${T}$, then it turns out that the factor ${(X, {\mathcal X}_1, \mu, T)}$ is isomorphic to a shift system ${(Z_1, {\mathcal Z}_1, \mu_{Z_1}, x \mapsto x+\alpha)}$ known as the Kronecker factor, for some compact abelian group ${Z_1 = (Z_1,+)}$ and some (irrational) shift ${\alpha \in Z_1}$; the factor map ${\pi_1: X \rightarrow Z_1}$ pushes eigenfunctions forward to (affine) characters on ${Z_1}$. It is then known that the limit of (3) is
$\displaystyle \int_\Sigma (\pi_1)_* f(x_0) (\pi_1)_* g(x_1) (\pi_1)_* h(x_2)\ d\mu_\Sigma$
where ${\Sigma \subset Z_1^3}$ is the closed subgroup
$\displaystyle \Sigma = \{ (x_1,x_2,x_3) \in Z_1^3: x_1-2x_2+x_3=0 \}$
and ${\mu_\Sigma}$ is the Haar probability measure on ${\Sigma}$; see this previous blog post. The equation ${x_1-2x_2+x_3=0}$ defining ${\Sigma}$ corresponds to the constraint (4) mentioned earlier. Among other things, this limit formula implies Roth’s theorem, which in the context of ergodic theory is the assertion that the limit (or at least the limit inferior) of (3) is positive when ${f=g=h}$ is non-negative and not identically vanishing.
If one considers a quadruple average
$\displaystyle \frac{1}{N} \sum_{n=1}^N \int_X f(x) g(T^n x) h(T^{2n} x) k(T^{3n} x)\ d\mu \ \ \ \ \ (5)$
(analogous to counting length four progressions) then the situation becomes more complicated still, even in the ergodic case. In addition to the (linear) eigenfunctions that already showed up in the computation of the triple average (3), a new type of constraint also arises from quadratic eigenfunctions ${f: X \rightarrow S^1}$, which obey an eigenfunction equation ${Tf = \lambda f}$ in which ${\lambda}$ is no longer constant, but is now a linear eigenfunction. For such functions, ${f(T^n x)}$ behaves quadratically in ${n}$, and one can compute the existence of a constraint
$\displaystyle f(x) \overline{f(T^n x)}^3 f(T^{2n} x)^3 \overline{f(T^{3n} x)} = 1 \ \ \ \ \ (6)$
between ${x}$, ${T^n x}$, ${T^{2n} x}$, and ${T^{3n} x}$ that is not detected at the triple average level. As it turns out, this is not the only type of constraint relevant for (5); there is a more general class of constraint involving two-step nilsystems which we will not detail here, but see e.g. this previous blog post for more discussion. Nevertheless there is still a similar limit formula to previous examples, involving a special factor ${(Z_2, {\mathcal Z}_2, \mu_{Z_2}, S)}$ which turns out to be an inverse limit of two-step nilsystems; this limit theorem can be extracted from the structural theory in this paper of Host and Kra combined with a limit formula for nilsystems obtained by Lesigne, but will not be reproduced here. The pattern continues to higher averages (and higher step nilsystems); this was first done explicitly by Ziegler, and can also in principle be extracted from the structural theory of Host-Kra combined with nilsystem equidistribution results of Leibman. These sorts of limit formulae can lead to various recurrence results refining Roth’s theorem in various ways; see this paper of Bergelson, Host, and Kra for some examples of this.
The above discussion was concerned with ${{\bf Z}}$-systems, but one can adapt much of the theory to measure-preserving ${G}$-systems for other discrete countable abelian groups ${G}$, in which one now has a family ${(T_g)_{g \in G}}$ of shifts indexed by ${G}$ rather than a single shift, obeying the compatibility relation ${T_{g+h}=T_g T_h}$. The role of the intervals ${\{1,\ldots,N\}}$ in this more general setting is replaced by that of Folner sequences. For arbitrary countable abelian ${G}$, the theory for double averages (1) and triple limits (3) is essentially identical to the ${{\bf Z}}$-system case. But when one turns to quadruple and higher limits, the situation becomes more complicated (and, for arbitrary ${G}$, still not fully understood). However one model case which is now well understood is the finite field case when ${G = {\bf F}_p^\omega = \bigcup_{n=1}^\infty {\bf F}_p^n}$ is an infinite-dimensional vector space over a finite field ${{\bf F}_p}$ (with the finite subspaces ${{\bf F}_p^n}$ then being a good choice for the Folner sequence). Here, the analogue of the structural theory of Host and Kra was worked out by Vitaly, Tamar, and myself in these previous papers (treating the high characteristic and low characteristic cases respectively). In the finite field setting, it turns out that nilsystems no longer appear, and one only needs to deal with linear, quadratic, and higher order eigenfunctions (known collectively as phase polynomials). It is then natural to look for a limit formula that asserts, roughly speaking, that if ${x}$ is drawn at random from a ${{\bf F}_p^\omega}$-system and ${n}$ drawn randomly from a large subspace of ${{\bf F}_p^\omega}$, then the only constraints between ${x, T^n x, \ldots, T^{(p-1)n} x}$ are those that arise from phase polynomials. The main theorem of this paper is to establish this limit formula (which, again, is a little complicated to state explicitly and will not be done here). In particular, we establish for the first time that the limit actually exists (a result which, for ${{\bf Z}}$-systems, was one of the main results of this paper of Host and Kra).
As a consequence, we can recover finite field analogues of most of the results of Bergelson-Host-Kra, though interestingly some of the counterexamples demonstrating sharpness of their results for ${{\bf Z}}$-systems (based on Behrend set constructions) do not seem to be present in the finite field setting (cf. this previous blog post on the cap set problem). In particular, we are able to largely settle the question of when one has a Khintchine-type theorem that asserts that for any measurable set ${A}$ in an ergodic ${{\bf F}_p^\omega}$-system and any ${\epsilon>0}$, one has
$\displaystyle \mu( T_{c_1 n} A \cap \ldots \cap T_{c_k n} A ) > \mu(A)^k - \epsilon$
for a syndetic set of ${n}$, where ${c_1,\ldots,c_k \in {\bf F}_p}$ are distinct residue classes. It turns out that Khintchine-type theorems always hold for ${k=1,2,3}$ (and for ${k=1,2}$ ergodicity is not required), and for ${k=4}$ it holds whenever ${c_1,c_2,c_3,c_4}$ form a parallelogram, but not otherwise (though the counterexample here was such a painful computation that we ended up removing it from the paper, and may end up putting it online somewhere instead), and for larger ${k}$ we could show that the Khintchine property failed for generic choices of ${c_1,\ldots,c_k}$, though the problem of determining exactly the tuples for which the Khintchine property failed looked to be rather messy and we did not completely settle it. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 1534, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9775468111038208, "perplexity": 164.1633558049971}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948615810.90/warc/CC-MAIN-20171218102808-20171218124808-00008.warc.gz"} |
https://brilliant.org/problems/fiery-factorials/ | # Fiery Factorials!
$\large \dfrac{1! \times2!\times 3!\times \cdots \times 10!}{(1!)^2(3!)^2(5!)^2(7!)^2(9!)^2}=15\times2^n$
What is the value of $$n$$ such that it satisfy the equation above?
× | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9928529262542725, "perplexity": 1541.262592017758}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549424575.75/warc/CC-MAIN-20170723162614-20170723182614-00580.warc.gz"} |
https://brainmass.com/physics/atmosphere/air-pressure-32682 | Explore BrainMass
# Air pressure
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This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
(a) assuming an average sea-level pressure of 1000hPa, what is the mass of the atmosphere? assume that the earth is as sphere with radius 6400km.
(b) what is the mass of the atmosphere above 500hPa? you can again assume a sphere of radius 6400km for this calculation. what fraction of the mass of the earth's atmosphere lies above 500 hPa?
(c) repeat (b) for the 250hPa surface. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9472998976707458, "perplexity": 1326.7567025320386}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652663016949.77/warc/CC-MAIN-20220528154416-20220528184416-00655.warc.gz"} |
https://aas.org/archives/BAAS/v34n2/aas200/36.htm | AAS 200th meeting, Albuquerque, NM, June 2002
Session 22. Extra Galactic Magnetic Fields: Their Origin and Manifestation through Structure of Quasars, Radio Lobes and within Clusters
Special Session Oral, Monday, June 3, 2002, 10:00-11:30am, La Cienega
## [22.01] Source of the Largest Extragalactic Magnetic Field Energies
P.P. Kronberg (LANL), Q.W. Dufton (University of Toronto), H. Li, S.A. Colgate (LANL)
The recently implied ubiquity of ~\,108\, M\odot central black holes (GBH) in large elliptical galaxies raises the question How much of the black hole energy that is released can we independently and quantitively measure in the surrounding IGM?''. A recent source-by-source analysis by us to answer this question determined that the best IGM calorimeters for this purpose are the largest (~0.5 to 5\,Mpc), not the most radio luminous extragalactic radio sources. These occur in relatively rarified IGM environments away from cluster cores. The upper bound of the energy content of the synchrotron-emitting lobes, much of which is magnetic, is ~8 \times 1060ergs, -- a significant fraction of the GBHs' gravitational binding energy. There are reasons why this global total energy estimate may even be conservative. The proximity of this number to the AGNs' total (gravitational) energy reservoir, and hence with the total radiated energy over all bands (which escapes at c) has many implications that are not yet understood in detail. The captured, i.e. magnetic, energy which propagates away much more slowly, constitutes a very large amount of retained energy in the locality of the host galaxy. For a cluster member AGN, this magnetic energy will be deposited within the host galaxy cluster. For a field'' radio galaxy, it will remain captured within ~ a Mpc.
We briefly discuss the wideranging implications that this hitherto hidden, intergalactic magnetic energy has for subsequent galaxy formation on these scales, and some implications for the IGM energetics of galaxy clusters.
Bulletin of the American Astronomical Society, 34
© 2002. The American Astronomical Soceity. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8183358311653137, "perplexity": 4802.723080870918}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721008.78/warc/CC-MAIN-20161020183841-00019-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://tex.stackexchange.com/questions/237442/start-inparaenum-numeration-with-another-letter-than-a/454237 | Start inparaenum-numeration with another letter than “a)”
How can I start \begin{inparaenum} with an arbitrary letter (other than a)) after a break in the inparaenum-environment. I would like to let a new inparaenum-environment be related to the end of the earlier one.
\usepackage{amsmath,amssymb,stmaryrd}
\usepackage{paralist}
\usepackage{tabto}
\begin{document}
\begin{enumerate}
\item \textbf{Working with logarithms}:
\NumTabs{3}
\begin{inparaenum}[a)]
Calculate the following
\item $\ln 1 =$
\tab \item $\ln e =$
\tab \item $\ln \frac{1}{e} =$
\end{inparaenum}
Express the following in terms of $\ln 2$:
\NumTabs{3}
\begin{inparaenum}[d)]
\item $\ln 4$
\tab \item $\ln \sqrt[3]{2^5}$
\tab \item $\ln \frac{1}{16}$
\end{inparaenum}
\end{enumerate}
\end{document}
Is there a way to let the second row start with d instead of using d in all positions? Thank you very much :)
• The \begin{inparamenum}[d)] is wrong and the compiler complains about it ;-) – user31729 Apr 8 '15 at 8:24
You can set the number by hand; since it's a second level list, you have to set enumii.
\usepackage{amsmath,amssymb,stmaryrd}
\usepackage{paralist}
\usepackage{tabto}
\begin{document}
\begin{enumerate}
\item \textbf{Working with logarithms}:
\NumTabs{3}
\begin{inparaenum}[a)]
Calculate the following
\item $\ln 1 =$
\tab \item $\ln e =$
\tab \item $\ln \frac{1}{e} =$
\end{inparaenum}
Express the following in terms of $\ln 2$:
\NumTabs{3}
\begin{inparaenum}[a)]\setcounter{enumii}{3}
\item $\ln 4$
\tab \item $\ln \sqrt[3]{2^5}$
\tab \item $\ln \frac{1}{16}$
\end{inparaenum}
\end{enumerate}
\end{document}
\usepackage{amsmath,amssymb,stmaryrd}
\usepackage[inline]{enumitem}
\usepackage{tabto}
\begin{document}
\begin{enumerate}
\NumTabs{3}
\item \textbf{Working with logarithms}:
Calculate the following\\[\medskipamount]
\begin{enumerate*}[label=\alph*),itemjoin=\tab,before={}]
\item $\ln 1 =$
\item $\ln e =$
\item $\ln \frac{1}{e} =$
\end{enumerate*}
Express the following in terms of $\ln 2$:\\[\medskipamount]
\begin{enumerate*}[label=\alph*),itemjoin=\tab,resume]
\item $\ln 4$
\item $\ln \sqrt[3]{2^5}$
\item $\ln \frac{1}{16}$
\end{enumerate*}
\end{enumerate}
\end{document}
• The \setcounter{enumii}{3} works fine as long as I stay in the upper enumerate-environment. Doesn't work in a new enumerate-inparaenum-environment. But that's ok for the moment. Thank you – Mac Apr 8 '15 at 8:43
• @Mac: This deliberately depends on the nesting of the environments – user31729 Apr 8 '15 at 10:10
• @Mac I suggest you to switch to enumitem, that's much more powerful than paralist. – egreg Apr 8 '15 at 10:40
Another version, using my assoccnt package, without the need of remembering the counter value:
The contenumii counter is used to store the total value of the enumii counter automatically and stored back after inparaenum.
Caveat: It will be incremented each time when another second level enumeration is used
Use the \SuspendCounters feature from the assoccnt package then
\usepackage{amsmath,amssymb,stmaryrd}
\usepackage{paralist}
\usepackage{tabto}
\usepackage{assoccnt}
\newcounter{contenumii}
\DeclareAssociatedCounters{enumii}{contenumii}
\begin{document}
\begin{enumerate}
\item \textbf{Working with logarithms}:
\NumTabs{3}
\begin{inparaenum}[a)]
Calculate the following
\item $\ln 1 =$
\tab \item $\ln e =$
\tab \item $\ln \frac{1}{e} =$
\end{inparaenum}
Express the following in terms of $\ln 2$:
\NumTabs{3}
\begin{inparaenum}[a)]\setcounter{enumii}{\number\value{contenumii}}
\item $\ln 4$
\tab \item $\ln \sqrt[3]{2^5}$
\tab \item $\ln \frac{1}{16}$
\end{inparaenum}
\end{enumerate}
\end{document}
Other options: enumitem package, with the resume option, but this is not as compact as paralist (initially, can be changed, perhaps)
• I cannot compile this.. it says assoccnt.sty not found. Somehow not loading the new package. I'm not sure what to do about it. But thanks for the idea :) – Mac Apr 8 '15 at 8:46
• @Mac: No problem at all... but it's on TeXLive and CTAN of course – user31729 Apr 8 '15 at 8:48
• That means I have to load it manually? I'm used to let this be done by TexnicCenter. I'm still looking for how to force my editor do this – Mac Apr 8 '15 at 8:58
• @Mac: I don't use those editor stuff like TexnicCenter etc. I update manually all the time (daily!) by invoking tlmgr on a Linux console – user31729 Apr 8 '15 at 9:11
• @ChristianHupfer Daily?! Why? – cfr Apr 9 '15 at 2:54
I know it's a very late answer, but I found this to work for me:
% Counter used in continuing lists.
\newcounter{contlist}
% ...
\begin{inparaenum}[(a)]
\item line-item 1
\item line-item 2
\end{inparaenum}
\setcounter{contlist}{\value{enumi}}
% ...
\begin{inparaenum}[(a)]
\setcounter{enumi}{\value{contlist}}
\item line-item 3
\item line-item 4
\end{inparaenum}
This works across different styles of enumerations [1, (b), (iii), etc].
• Welcome to TeX.SX! The other answers are also applicable to other styles, so how does this add anything new? – TeXnician Oct 7 '18 at 13:44
• It answers the original question, and doesn't need hand-coding. Nor does it need other styles/environments. – sn00p Oct 10 '18 at 10:58 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9744435548782349, "perplexity": 4658.078689278763}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250597458.22/warc/CC-MAIN-20200120052454-20200120080454-00079.warc.gz"} |
http://arzyabexpert.com/guild-wars-jnoyivf/what-is-the-following-sum-3-sqrt-125x%5E10y%5E13-96c4ab | ##### بلاگ
Let's look at some Excel SQRT function examples and explore how to use the SQRT function as a worksheet function in Microsoft Excel: Based on the Excel spreadsheet above, the following SQRT examples would return: =SQRT(A1) Result: 5 =SQRT(A2) Result: 5.796550698 =SQRT(A3) Result: #NUM! Thus, for calculating the product of the following square roots sqrt(33)*sqrt(6), enter simplify_surd(sqrt(33)*sqrt(6)), the result 3*sqrt(22) is returned. Ans. If the argument is NaN or negative, then the result is NaN. Let's look at some Oracle SQRT function examples and explore how to use the SQRT function in Oracle/PLSQL. So the answer would be 2. The second term abs(x)^N/(N!) So, you can take a 3 out of the sqrt., because 3^2 is 9. Conic Sections In mathematics, a square root of a number x is a number y such that y 2 = x; in other words, a number y whose square (the result of multiplying the number by itself, or y ⋅ y) is x. Description. If so, then #b = 11# and we would have #f(5+6/n) = sqrt(7+(6/n)^2)#, so we would get #f(x) = sqrt(7+(x-5)^2)#.) If x^2=-1=i^2, x=sqrt(-1)=(-1)^(1/2)=+-i sqrt(-2)=sqrt(-1)sqrt2=+-1.4142i and, likewise, sqrt(-18)=+-3(1.4142)i. Question: Is the following sum rational or irrational? Boolean has an integer value of 1, thus sum becomes 1 + 50 + 5 + 10 = 66. Now you can add the two sqrts. d) [True, 50, 5, 10] Sum is: 66 . For example: SQRT(9) Result: 3 SQRT(37) Result: 6.08276253029822 SQRT(5.617) Result: 2.37002109695251 You then have: 3 sqrt-2. For negative and complex numbers z = u + i*w, the complex square root sqrt(z) returns. It is also the area of the unit circle. If the argument passed is positive zero or negative zero then the result will be same as that of the argument. 2. Putting these two facts together gives the following, $A \approx \sum\limits_{n = 1}^\infty {\frac{1}{n}} > \int_{{\,1}}^{{\,\infty }}{{\frac{1}{x}\,dx}} = \infty$ Notice that this tells us that we must have, $\sum\limits_{n = 1}^\infty {\frac{1}{n}} > \infty \hspace{0.5in} \Rightarrow \hspace{0.5in}\sum\limits_{n = 1}^\infty {\frac{1}{n}} = \infty$ Since we can’t really be larger than (\sqrt(8))/(3)+\sqrt(16) TutorsOnSpot.com. Compared to other sites, www.OnSolver.com has a huge advantage, because you can find the sum of not only numerical but also functional series, which will determine the convergence domain of the original series, using the most known methods. For example, the following is a valid expression: (-1)^(2n+pi/3) Summation formula and Sigma (Σ) notation. ; S(i) refers to sum of Fibonacci numbers till F(i). Well, it's a right Riemann sum, so we're using the value of the function right over there, write it two plus five over N. So, this value right over here. A proof of the Alternating Series Test is also given. In this section we will discuss using the Alternating Series Test to determine if an infinite series converges or diverges. is bounded, that is that sum_(n=0)^oo x^n/(n!) So, be careful to not make this very common mistake! Now we could keep going. The sqrt() function in C++ returns the square root of a number. Correct answer to the question Ineed ! 0. The following table contains some important mathematical constants: Name Symbol Value Meaning Pi, Archimedes' constant or Ludoph's number: π ≈3.141592653589793 A transcendental number that is the ratio of the length of a circle's circumference to its diameter. $5 = \sqrt {25} = \sqrt {9 + 16} \ne \sqrt 9 + \sqrt {16} = 3 + 4 = 7$ If we “break up” the root into the sum of the two pieces we clearly get different answers! Derivatives Derivative Applications Limits Integrals Integral Applications Riemann Sum Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Functions Line Equations Functions Arithmetic & Comp. We can rewrite the relation F(n + 1) = F(n) + F(n – 1) as below: Relationship Deduction. 2) What is the output of the following program? This is the natural log, the natural log of two plus five over N, and since this is the first rectangle times one, times one. If the argument is positive infinity, then the result is positive infinity. sqrt: square root: tan-tangent: tanh -hyperbolic tangent: In the expression you can enter minus as a negation (sign), and also use implicit multiplication (2n will be interpreted as 2 * n). what is the following sum? Example. We have shown that for any x in (-oo,oo), sum_(n=0)^oo abs(x)^n/(n!) Riemann sums help us approximate definite integrals, but they also help us formally define definite integrals. Free series convergence calculator - test infinite series for convergence step-by-step This is a mistake. Precalculus . 5x(3 sqrt(x^2 y)+2(3 sqrt^5y) a- 7x(^6 square root of x^2y) b-7x^2(^6 square root of xy^2) c-7x^2(^3 square root of xy^2) d-7x(^3square root of x^2y) - e-eduanswers.com To use the comparison test to determine the convergence or divergence of a series $$\sum_{n=1}^∞a_n$$, it is necessary to find a suitable series with which to compare it. sum_(n=0)^oo abs(x)^n/(N^n) is the sum of a geometric series with positive common ratio abs(x)/N < 1, so converges. The graph of $x^2+(y-\sqrt[3]{x^2})^2=1$ is very interesting and is show below using desmos. For the elements of X that are negative or complex, sqrt(X) produces complex results. If the series is convergent, determine whether it is absolutely or conditionally convergent. Learn how this is achieved and how we can move between the representation of area as a definite integral and as a Riemann sum. Part A: In complete sentences, explain the relationships between all pairs of special angles 1, 2, 3 and 4 created by transversal line b and parallel lines d and e. Part B: for the given diagram, use the measure of The following special angle relationships are created by transversal line b and parallel lines d and e : 1 to find the measures of ∠2, ∠3, and ∠4. Efficient approach: The idea is to find the relationship between the sum of Fibonacci numbers and n th Fibonacci number and use Binet’s Formula to calculate its value. 5 (3 sqrt) + 9 (3 sqrt) Answers (1) Unique 29 December, 11:51. First you must simplify the sqrt-18. (d) Explanation: The List is initially has 3 elements. Hence it is also convergent. The Alternating Series Test can be used only if the terms of the series alternate in sign. Find a perfect square that is a multiple of 18: in this case it would be 9, because 9 x 2 = 18. Homework Writing Market. Now, #sqrt(7+(6/n)^2) = f(b+ Deltax) = f(-1+6/n)# Which will be true if #f(x) = sqrt(7+(x+1)^2)# This #f(x)# works for the other two given terms, so it must be the correct #f(x)#. The insert() adds element 5 at index 2, moving element 10 at index 3 and the List becomes [True, 50, 5, 10]. import math math.sqrt( x ) Note − This function is not accessible directly, so we need to import the math module and then we need to call this function using the math static object.. Parameters. B = sqrt(X) returns the square root of each element of the array X. Surds fraction calculator (square root quotient) The online square root calculator can symplify surds root quotients in exact form. The sqrt function’s domain includes negative and complex numbers, which can lead to unexpected results if used unintentionally. Following is the syntax for sqrt() method −. What is the following sum? (I wonder if it should be #int_5^b f(x) dx#? =SQRT(82.6) Result: 9.088454214 Our Services. The sqrt() method returns the square root of x for x > 0.. Syntax. Start studying MIS 207. Learn vocabulary, terms, and more with flashcards, games, and other study tools. 3 xx sqrt125 = 15sqrt5 and root(3)125 = 5 I have heard many students read root(3)n as "the third square root of n". is absolutely convergent. Answers Mine. Truly, each term has two values and the sum has four values, in Mathematical Exactitude. The square root is root(2)n (usually denoted sqrtx), the third (or cube) root is root(3)n, the fourth root is root(4)n and so on. 1 sqrt.- 2 + 3 sqrt.- 2 = 4 sqrt. What is the radius of convergence of the series #sum_(n=0)^oo(n*(x+2)^n)/3^(n+1)#? Whichever was meant the first step for simplifying is the same. This one right over here the width is the same, five over N but what's the height? The java.lang.Math.sqrt() returns the square root of a value of type double passed to it as argument. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.. Visit Stack Exchange Don't curse me feeling that I am making a mole appear as mountain. x − This is a numeric expression.. Return Value I am sorry. F(i) refers to the i th Fibonacci number. This is the concept of arithmetic, we are required to calculate the following; 5sqrt (3) + 9sqrt (3) Here we shall take the two terms to be like terms; thus; 5sqrt (3) + 9sqrt (3) =14sqrt (3) Thus the answer is: 14sqrt (3) Comment; Complaint; Link ; Know the Answer? See all questions in Determining the Radius and Interval of Convergence for a Power Series Impact of this question The sum and difference formulas can be used to find the exact values of the sine, cosine, or tangent of an angle. Determine whether the following series is convergent or divergent. This is useful for analysis when the sum of a series online must be presented and found as a solution of limits of partial sums of series. If you aren’t sure that you believe this consider the following quick number example. So, the sum is +-1.4142(1+-3)i=+-5.657i and +-2.8281, i=sqrt(-1), nearly.. Let us compile and run the above program that will produce the following result − Square root of 4.000000 is 2.000000 Square root of 5.000000 is 2.236068 math_h.htm Answer. This section we will discuss using the Alternating series Test is also.... A numeric expression.. Return value Start studying MIS 207 the elements of x x! I am making a mole appear as mountain, i=sqrt ( -1 ), nearly the (! If used unintentionally has an integer value of 1, thus sum 1. S ( i ) refers to the i th Fibonacci number convergent, determine whether it is or! ) +\sqrt ( 16 ) TutorsOnSpot.com = 4 sqrt to what is the following sum 3 sqrt 125x^10y^13 if an infinite series converges diverges. B = sqrt ( x ) dx # -1 ), nearly is 9 an integer value of double! Definite integrals representation of area as a riemann sum root quotient ) the online root... This consider the following series is convergent or divergent 3^2 is 9 we will using. The java.lang.Math.sqrt ( ) function in Oracle/PLSQL the width is the output of the series is,... Boolean has an integer value of type double passed to it as argument same as that of the unit.. 1+-3 ) i=+-5.657i and +-2.8281, i=sqrt ( -1 ), nearly term has two values and the is! More with flashcards, games, and more with flashcards, games, and study... In Mathematical Exactitude x for x > 0.. Syntax infinity, then the result is zero. Look at some Oracle sqrt function ’ s domain includes negative and complex numbers, which can to... Section we will discuss using the Alternating series Test is also given 50 + 5 + 10 = 66 height. Find the exact values of the sqrt., because 3^2 is 9 same. Was meant the first step for simplifying is the same, five n... + 5 + 10 = 66, which can lead to unexpected results if unintentionally! The Alternating series Test to determine if an infinite series converges or diverges 's the height = 66 ) in! You believe this consider the following program 10 = 66 Test is also given 4 sqrt use... A number negative and complex numbers z = u + i *,. That sum_ ( n=0 ) ^oo x^n/ ( n! ( 1+-3 ) i=+-5.657i and +-2.8281, (. You believe this consider the following program same, five over n What. The argument passed is positive zero or negative zero then the result be... Unit circle s ( i wonder if it should be # int_5^b (. Is achieved and how we can move between the representation of area as a definite and... ) i=+-5.657i and +-2.8281, i=sqrt ( -1 ), nearly, terms, and more flashcards. ), nearly learn vocabulary, terms, and more with flashcards, games, and other study tools unit... The square root sqrt ( x ) dx # negative zero then the result is NaN or negative, the., cosine, or tangent of an angle sum and difference formulas can be used to find the exact of... Common mistake 29 December, 11:51 if an infinite series converges or diverges at some Oracle sqrt function and... Is a numeric expression.. Return value Start studying MIS 207 Oracle sqrt function and. / ( 3 sqrt ) + 9 ( 3 ) +\sqrt ( 16 ) TutorsOnSpot.com other study tools method the... Can move between the representation of area as a riemann sum 3 sqrt.- 2 = 4.. Wonder if it should be # int_5^b f ( i ) quick number example be as... Following is the Syntax for sqrt ( ) method returns the square root calculator can symplify surds root quotients exact!, but they also help us approximate definite integrals tangent of an angle whether it also. And other study tools s ( i ), in Mathematical Exactitude here the width is the same five... If the argument is NaN or negative zero then the result is positive.. Some Oracle sqrt function in C++ returns the square root of each element of the following quick number example the! > 0.. Syntax numbers, which can lead to unexpected results if used unintentionally but they also us. A value of 1, thus sum becomes 1 + 50 + +... An infinite series converges what is the following sum 3 sqrt 125x^10y^13 diverges as a definite integral and as a definite integral and as riemann. Is positive infinity ( 1 ) Unique 29 December, 11:51 − this is a expression. Has an integer value of type double passed to it as argument argument passed is positive,... Syntax for sqrt ( z ) returns ( 3 sqrt ) + (... ) TutorsOnSpot.com is 9 's look at some Oracle sqrt function ’ s domain includes and... Also help us formally define definite integrals, but they also help us definite. Java.Lang.Math.Sqrt ( ) method − same as that of the Alternating series Test to determine if an series. Right over here the width is the same x for x > 0.. Syntax tangent of an angle is. − this is achieved and how we can move between the representation of area as a definite and! Same as that of the array x as argument and as a definite integral and as definite! T sure that you believe this consider the following series is convergent, determine whether it is or! Look at some Oracle sqrt function examples and explore how to use the sqrt ( ) method − as... 50 + 5 + 10 = 66 the square root of x for x > 0 Syntax! Achieved and how we can move between the representation of area as a riemann sum a numeric..! S domain includes negative and complex numbers z = u + i * w, the complex square of! 3 sqrt ) + 9 ( 3 sqrt ) + 9 ( 3 )! If used unintentionally if you aren ’ t sure that you believe this the... 5 + 10 = 66 how to use the sqrt function examples and explore how use... 10 ] sum is: 66 symplify surds root quotients in exact form ) and! Z = u + i * w, the sum has four,. Will discuss using the Alternating series Test can be used to find the exact values of the sqrt. because!, each term has two values and the sum is +-1.4142 ( 1+-3 ) i=+-5.657i and +-2.8281 i=sqrt.: 66 vocabulary, terms, and more with flashcards, games, and other tools... Numbers till f ( x ) produces complex results also given with flashcards, games and. 5 + 10 = 66 the terms of the array x ) / 3. Us formally define definite integrals formally define definite integrals w, the complex root. True, 50, 5, 10 ] sum is: 66 Return value Start MIS. Negative, then the result will what is the following sum 3 sqrt 125x^10y^13 same as that of the unit circle series... Root sqrt ( z ) returns the square root of x for x > 0.. Syntax we! [ True, 50, 5, 10 ] sum is: 66 examples... ) Explanation: the List is initially has 3 elements ( 16 ) TutorsOnSpot.com 9 ( 3 ) +\sqrt 16. 2 = 4 sqrt of an angle 2 ) What is the Syntax for (... Calculator can symplify surds root quotients in exact form / what is the following sum 3 sqrt 125x^10y^13 3 ) +\sqrt 16... Array x or complex, sqrt ( z ) returns of x that are negative complex... B = sqrt ( x ) produces complex results ^oo x^n/ ( n )... In C++ returns the square root of a value of 1, thus sum becomes 1 + +... Of the argument is NaN as that of the sine, cosine, or tangent of an angle the! 2 + 3 sqrt.- 2 + 3 sqrt.- 2 = 4 sqrt = (! List is initially has 3 elements becomes 1 + 50 + 5 + =! 'S the height, five over n but What 's the height width is the of... Boolean has an integer value of 1, thus sum becomes 1 + +! Answers ( 1 ) Unique 29 December, 11:51, sqrt ( x produces. Is a numeric expression.. Return value Start studying MIS 207 passed is positive or. December, 11:51 surds root quotients in exact form so, the what is the following sum 3 sqrt 125x^10y^13... Are negative or complex, sqrt ( z ) returns approximate definite integrals, but they help... Following is the same because 3^2 is 9 Syntax for sqrt ( x ) dx # results if unintentionally... Used only if the argument is NaN but they also help us approximate definite integrals, but they help. Initially has 3 elements it should be # int_5^b f ( x ) returns the square root a! 'S look at some Oracle sqrt function examples and explore how to use the sqrt function examples explore. Of an angle in C++ returns the square root of a number +\sqrt 16... For sqrt ( x ) returns the square root calculator can symplify surds quotients! 5 + 10 = 66 this is a numeric expression.. Return value Start studying 207. Following series is convergent, determine whether it is absolutely or conditionally convergent between the representation of as! Achieved and how we can move between the representation of area as a definite and... Of a value of type double passed to it as argument using the Alternating series Test to determine an... ) +\sqrt ( 16 ) TutorsOnSpot.com integer value of 1, thus sum becomes 1 + 50 5. 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https://www.flexiprep.com/Important-Topics/Mathematics/Functions.html | # Functions and Types of Functions: What Are Functions in Mathematics? (For CBSE, ICSE, IAS, NET, NRA 2022)
Glide to success with Doorsteptutor material for competitive exams : get questions, notes, tests, video lectures and more- for all subjects of your exam.
# Title: Functions and Types of Functions
• Functions are relations where each input has a particular output.
• We can define a function as a special relation which maps each element of set A with one and only one element of set B. Both the sets A and B must be non-empty.
• In this lesson, the concepts of functions in mathematics and the different types of functions are covered using various examples for better understanding.
## What Are Functions in Mathematics?
• A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output.
• A function defines a particular output for a particular input. Hence, is a function such that for a ∈ A there is a unique element b ∈ B such that
• Let A & B be any two non-empty sets, mapping from A to B will be a function only when every element in set A has one end only one image in set B.
## Example
• Another definition of functions is that it is a relation f in which each element of set A is mapped with only one element belonging to set B.
• Also, in a function, there cannot be two pairs with the same first element.
## A Condition for a Function
• Set A and Set B should be non-empty.
• In a function, a particular input is given to get a particular output. So, A function denotes that f is a function from A to B, where A is a domain and B is a co-domain.
• Many widely used mathematical formulas are expressions of known functions.
• For an element, a, which belongs to A, a unique element b, is there such that
• For example, the formula for the area of a circle, , gives the dependent variable A (the area) as a function of the independent variable r (the radius) .
• The unique element b to which f relates a, is denoted by f (a) , and is called f of a, or the value of f at a, or the image of a under f.
• The range of f (image of a under f)
• It is the set of all values of f (x) taken together.
• Range of f =
• A real-valued function has either P or any one of its subsets as its range.
• Further, if its domain is also either P or a subset of P, it is called a real function.
## Representation of Functions
• Functions are generally represented as f (x)
• Let,
• It is said as f of x is equal to x cube.
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https://brilliant.org/discussions/thread/no-conjectures/ | # No conjectures...
This is an inequality problem I found in a book, but I wonder if it can be done in simple way. Try this:
If $$a_1,a_2, \cdots ,a_n$$ are positive reals less than one and $$S_n = a_1 + \cdots + a_n$$, then show that
${1-S_n < (1-a_1)(1-a_2) \cdots (1-a_n) < \displaystyle \frac{1}{1+S_n}}$.
Note by A Brilliant Member
7 years ago
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2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
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Also, ai < 1 for all i = 1, 2...n 1 - (a1)^2 < 1 or, (1 - a1)(1+a1) < 1 or, (1- a1) < 1/(1+a1) Multiplying over all terms, (1 - a1)(1- a2)...(1 - an) < 1/(1+a1)(1+a2)...(1+an) But (1+a1)(1+a2)...(1+an) > 1 + Sn This immediately proves it.
- 7 years ago
Again not correct, $a_i^2 < 1$ doesn't imply $1- a_i^2 < 1$. You need to specify proofs more. Anyway for latex, you may use latex editor for Chrome.
- 7 years ago
Actually, Since $1 > a_i^2 > 0$, we have $1 > 1 - a_i^2 > 0$
Right, but he wrote "or", implying that the statement followed from $1- a_i^2 >0$. The proof was correct though.
- 7 years ago
The OR I wrote was just a continuation :P
- 7 years ago
Like "IMPLIES"
- 7 years ago
And I really don't get you. ai < 1 for all i. So, ai^2 < 1 for all i. This gives, 0 < 1 - ai^2 < 1. Now you just factor out (1- ai^2) into (1 - ai)(1+ai).
- 7 years ago
(1 - a1) (1 - a2) = 1 - a1 - a2 + a1a2 > 1 - (a1 + a2) So, for n such brackets, you can write (1- a1)(1- a2)...(1-an) > 1 - (a1+a2+a3...+an) = 1 - Sn
- 7 years ago
Not rigorous. :( Also there is no chance to justify your claim, in $(1-a_1)(1-a_2) \cdots (1-a_n)$, there are many more terms other than $(-1)^n a_1a_2a_3 \cdots a_n$. Not correct.
- 7 years ago
I think you just missed what I meant to say. I avoided writing entire proofs because I personally don't like answers without LATEX and here I am, without knowing LATEX. Anyway, I have already proved that (1-a1)(1-a2) > 1 - (a1+a2). Then, (1-a1)(1-a2)(1-a3) > (1-a3)(1-a1-a2)>1-(a1+a2+a3). Got it?
- 7 years ago
Now, this can, similarly be extended to n terms.
- 7 years ago
I know it would come, my friend, but I thought of getting a rigorous proof. Anyway no problem! :)
- 7 years ago
And guys, it would be really really helpful if you can tell me any site where I can write LATEX directly and when I paste it here in comments box, the entire writing with LATEX will come.
- 7 years ago
I use the latex editor Latexian, as it displays the equations as I am typing them.
An online free site would be WriteLatex, which I have used in collaboration with others internationally.
Staff - 7 years ago | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 16, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.961830735206604, "perplexity": 2847.7965864588678}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243990419.12/warc/CC-MAIN-20210511214444-20210512004444-00403.warc.gz"} |
http://mathhelpforum.com/algebra/156829-translating-english-phrases-into-algebraic-expressions-print.html | # translating English phrases into algebraic expressions
• Sep 20th 2010, 02:15 PM
fc1123
translating English phrases into algebraic expressions
okay I have another problem a number, half of that number, and one-third of that number are added. The result is 22. So far I have
x+1/2x+1/3x=22.
• Sep 20th 2010, 02:19 PM
yeKciM
Quote:
Originally Posted by fc1123
okay I have another problem a number, half of that number, and one-third of that number are added. The result is 22. So far I have
x+1/2x+1/3x=22.
write it like this ..... x+ x/2 + x/3 = 22 or
$x+\frac {x}{2} +\frac {x}{3} = 22$
that way someone can think you wrote ... x+ 1/(2x) + 1 /(3x) ....
it's the same as the another one that you post ... just this one multiply with 6
$6x +.....$
• Sep 20th 2010, 02:25 PM
fc1123
thank you so much for help. I am enjoying this forum because it is helping me a lot to understand the problems. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9392825961112976, "perplexity": 1843.0785137455784}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917121216.64/warc/CC-MAIN-20170423031201-00112-ip-10-145-167-34.ec2.internal.warc.gz"} |
http://en.wikipedia.org/wiki/Chebotaryov_density_theorem | # Chebotarev's density theorem
(Redirected from Chebotaryov density theorem)
Chebotarev's density theorem in algebraic number theory describes statistically the splitting of primes in a given Galois extension K of the field Q of rational numbers. Generally speaking, a prime integer will factor into several ideal primes in the ring of algebraic integers of K. There are only finitely many patterns of splitting that may occur. Although the full description of the splitting of every prime p in a general Galois extension is a major unsolved problem, the Chebotarev density theorem says that the frequency of the occurrence of a given pattern, for all primes p less than a large integer N, tends to a certain limit as N goes to infinity. It was proved by Nikolai Chebotaryov in his thesis in 1922, published in (Tschebotareff 1926).
A special case that is easier to state says that if K is an algebraic number field which is a Galois extension of Q of degree n, then the prime numbers that completely split in K have density
1/n
among all primes. More generally, splitting behavior can be specified by assigning to (almost) every prime number an invariant, its Frobenius element, which strictly is a representative of a well-defined conjugacy class in the Galois group
Gal(K/Q).
Then the theorem says that the asymptotic distribution of these invariants is uniform over the group, so that a conjugacy class with k elements occurs with frequency asymptotic to
k/n.
## History and motivation
When Carl Friedrich Gauss first introduced the notion of complex integers Z[i], he observed that the ordinary prime numbers may factor further in this new set of integers. In fact, if a prime p is congruent to 1 mod 4, then it factors into a product of two distinct prime gaussian integers, or "splits completely"; if p is congruent to 3 mod 4, then it remains prime, or is "inert"; and if p is 2 then it becomes a product of the square of the prime (1+i) and the invertible gaussian integer -i; we say that 2 "ramifies". For instance,
$5 = (1 + 2i)(1-2i)$ splits completely;
$3$ is inert;
$2 = -i(1+i)^2$ ramifies.
From this description, it appears that as one considers larger and larger primes, the frequency of a prime splitting completely approaches 1/2, and likewise for the primes that remain primes in Z[i]. Dirichlet's theorem on arithmetic progressions demonstrates that this is indeed the case. Even though the prime numbers themselves appear rather erratically, splitting of the primes in the extension
$\Bbb{Z}\subset \Bbb{Z}[i]$
follows a simple statistical law.
Similar statistical laws also hold for splitting of primes in the cyclotomic extensions, obtained from the field of rational numbers by adjoining a primitive root of unity of a given order. For example, the ordinary integer primes group into four classes, each with probability 1/4, according to their pattern of splitting in the ring of integers corresponding to the 8th roots of unity. In this case, the field extension has degree 4 and is abelian, with the Galois group isomorphic to the Klein four-group. It turned out that the Galois group of the extension plays a key role in the pattern of splitting of primes. Georg Frobenius established the framework for investigating this pattern and proved a special case of the theorem. The general statement was proved by Nikolai Grigoryevich Chebotaryov in 1922.
## Relation with Dirichlet's theorem
The Chebotarev density theorem may be viewed as a generalisation of Dirichlet's theorem on arithmetic progressions. A quantitative form of Dirichlet's theorem states that if N2 is an integer and a is coprime to N, then the proportion of the primes p congruent to a mod N is asymptotic to 1/n, where n=φ(N) is the Euler totient function. This is a special case of the Chebotarev density theorem for the Nth cyclotomic field K. Indeed, the Galois group of K/Q is abelian and can be canonically identified with the group of invertible residue classes mod N. The splitting invariant of a prime p not dividing N is simply its residue class because the number of distinct primes into which p splits is φ(N)/m, where m is multiplicative order of p modulo N; hence by the Chebotarev density theorem, primes are asymptotically uniformly distributed among different residue classes coprime to N.
## Formulation
Lenstra & Stevenhagen (1996) give an earlier result of Frobenius in this area. Suppose K is a Galois extension of the rational number field Q, and P(t) a monic integer polynomial such that K is a splitting field of P. It makes sense to factorise P modulo a prime number p. Its 'splitting type' is the list of degrees of irreducible factors of P mod p, i.e. P factorizes in some fashion over the prime field Fp. If n is the degree of P, then the splitting type is a partition Π of n. Considering also the Galois group G of K over Q, each g in G is a permutation of the roots of P in K; in other words by choosing an ordering of α and its algebraic conjugates, G is faithfully represented as a subgroup of the symmetric group Sn. We can write g by means of its cycle representation, which gives a 'cycle type' c(g), again a partition of n.
The theorem of Frobenius states that for any given choice of Π the primes p for which the splitting type of P mod p is Π has a natural density δ, with δ equal to the proportion of g in G that have cycle type Π.
The statement of the more general Chebotarev theorem is in terms of the Frobenius element of a prime (ideal), which is in fact an associated conjugacy class C of elements of the Galois group G. If we fix C then the theorem says that asymptotically a proportion |C|/|G| of primes have associated Frobenius element as C. When G is abelian the classes of course each have size 1. For the case of a non-abelian group of order 6 they have size 1, 2 and 3, and there are correspondingly (for example) 50% of primes p that have an order 2 element as their Frobenius. So these primes have residue degree 2, so they split into exactly three prime ideals in a degree 6 extension of Q with it as Galois group.[1]
## Statement
Let L be a finite Galois extension of a number field K with Galois group G. Let X be a subset of G that is stable under conjugation. The set of primes v of K that are unramified in L and whose associated Frobenius conjugacy class Fv is contained in X has density
$\frac{\#X}{\#G}.$[2]
### Infinite extensions
The statement of the Chebotarev density theorem can be generalized to the case of an infinite Galois extension L / K that is unramified outside a finite set S of primes of K (i.e. if there is a finite set S of primes of K such that any prime of K not in S is unramified in the extension L / K). In this case, the Galois group G of L / K is a profinite group equipped with the Krull topology. Since G is compact in this topology, there is a unique Haar measure μ on G. For every prime v of K not in S there is an associated Frobenius conjugacy class Fv. The Chebotarev density theorem in this situation can be stated as follows:[2]
Let X be a subset of G that is stable under conjugation and whose boundary has Haar measure zero. Then, the set of primes v of K not in S such that Fv ⊆ X has density
$\frac{\mu(X)}{\mu(G)}.$
This reduces to the finite case when L / K is finite (the Haar measure is then just the counting measure).
A consequence of this version of the theorem is that the Frobenius elements of the unramified primes of L are dense in G.
## Important consequences
The Chebotarev density theorem reduces the problem of classifying Galois extensions of a number field to that of describing the splitting of primes in extensions. Specifically, it implies that as a Galois extension of K, L is uniquely determined by the set of primes of K that split completely in it.[3] A related corollary is that if almost all prime ideals of K split completely in L, then in fact L = K.[4]
## Notes
1. ^ This particular example already follows from the Frobenius result, because G is a symmetric group. In general, conjugacy in G is more demanding than having the same cycle type.
2. ^ a b Section I.2.2 of Serre
3. ^ Corollary VII.13.10 of Neukirch
4. ^ Corollary VII.13.7 of Neukirch | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9695205092430115, "perplexity": 208.89931649350828}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386163800358/warc/CC-MAIN-20131204133000-00005-ip-10-33-133-15.ec2.internal.warc.gz"} |
http://www.ck12.org/book/CK-12-Algebra-I---Second-Edition/r1/section/3.2/ | <meta http-equiv="refresh" content="1; url=/nojavascript/"> Two-Step Equations | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Algebra I - Second Edition Go to the latest version.
# 3.2: Two-Step Equations
Created by: CK-12
## Learning Objectives
• Solve a two-step equation using addition, subtraction, multiplication, and division.
• Solve a two-step equation by combining like terms.
• Solve real-world problems using two-step equations.
## Solve a Two-Step Equation
We’ve seen how to solve for an unknown by isolating it on one side of an equation and then evaluating the other side. Now we’ll see how to solve equations where the variable takes more than one step to isolate.
Example 1
Rebecca has three bags containing the same number of marbles, plus two marbles left over. She places them on one side of a balance. Chris, who has more marbles than Rebecca, adds marbles to the other side of the balance. He finds that with 29 marbles, the scales balance. How many marbles are in each bag? Assume the bags weigh nothing.
Solution
We know that the system balances, so the weights on each side must be equal. If we use $x$ to represent the number of marbles in each bag, then we can see that on the left side of the scale we have three bags (each containing $x$ marbles) plus two extra marbles, and on the right side of the scale we have 29 marbles. The balancing of the scales is similar to the balancing of the following equation.
$3x + 2 = 29$
“Three bags plus two marbles equals 29 marbles”
To solve for $x$, we need to first get all the variables (terms containing an $x$) alone on one side of the equation. We’ve already got all the $x$’s on one side; now we just need to isolate them.
$3x + 2 &= 29\\3x + 2 - 2 &= 29 - 2 \qquad \text{Get rid of the 2 on the left by subtracting it from both sides.}\\3x &= 27\\\frac{3x}{3} &= \frac{27}{3} \qquad \quad \ \text{Divide both sides by 3.}\\x &= 9$
There are nine marbles in each bag.
We can do the same with the real objects as we did with the equation. Just as we subtracted 2 from both sides of the equals sign, we could remove two marbles from each side of the scale. Because we removed the same number of marbles from each side, we know the scales will still balance.
Then, because there are three bags of marbles on the left-hand side of the scale, we can divide the marbles on the right-hand side into three equal piles. You can see that there are nine marbles in each.
Three bags of marbles balances three piles of nine marbles.
So each bag of marbles balances nine marbles, meaning that each bag contains nine marbles.
Check out http://www.mste.uiuc.edu/pavel/java/balance/ for more interactive balance beam activities!
Example 2
Solve $6(x + 4) = 12$.
This equation has the $x$ buried in parentheses. To dig it out, we can proceed in one of two ways: we can either distribute the six on the left, or divide both sides by six to remove it from the left. Since the right-hand side of the equation is a multiple of six, it makes sense to divide. That gives us $x + 4 = 2$. Then we can subtract 4 from both sides to get $x = -2$.
Example 3
Solve $\frac{x - 3}{5} = 7$.
It’s always a good idea to get rid of fractions first. Multiplying both sides by 5 gives us $x - 3 = 35$, and then we can add 3 to both sides to get $x = 38$.
Example 4
Solve $\frac{5}{9}(x + 1) =\frac{2}{7}$.
First, we’ll cancel the fraction on the left by multiplying by the reciprocal (the multiplicative inverse).
$\frac{9}{5} \cdot \frac{5}{9}(x + 1) &= \frac{9}{5} \cdot \frac{2}{7}\\(x + 1) &= \frac{18}{35}$
Then we subtract 1 from both sides. ($\frac{35}{35}$ is equivalent to 1.)
$x + 1 &= \frac{18}{35}\\x + 1 - 1 &= \frac{18}{35} - \frac{35}{35}\\x &= \frac{18 - 35}{35} \\x &= \frac{-17}{35}$
These examples are called two-step equations, because we need to perform two separate operations on the equation to isolate the variable.
## Solve a Two-Step Equation by Combining Like Terms
When we look at a linear equation we see two kinds of terms: those that contain the unknown variable, and those that don’t. When we look at an equation that has an $x$ on both sides, we know that in order to solve it, we need to get all the $x-$terms on one side of the equation. This is called combining like terms. The terms with an $x$ in them are like terms because they contain the same variable (or, as you will see in later chapters, the same combination of variables).
Like Terms Unlike Terms
$4x, 10x, -3.5x,$ and $\frac{x}{12}$ $3x$ and $3y$
$3y, 0.000001y,$ and $y$ $4xy$ and $4x$
$xy, 6xy,$ and $2.39xy$ $0.5x$ and $0.5$
To add or subtract like terms, we can use the Distributive Property of Multiplication.
$3x + 4x &= (3 + 4)x = 7x \\0.03xy - 0.01xy &= (0.03 - 0.01)xy = 0.02xy\\-y + 16y + 5y &= (-1 + 16 + 5)y = 10y\\5z + 2z - 7z &= (5 + 2 - 7)z = 0z = 0$
To solve an equation with two or more like terms, we need to combine the terms first.
Example 5
Solve $(x + 5) - (2x - 3)=6$.
There are two like terms: the $x$ and the $-2x$ (don’t forget that the negative sign applies to everything in the parentheses). So we need to get those terms together. The associative and distributive properties let us rewrite the equation as $x + 5 - 2x + 3 = 6$, and then the commutative property lets us switch around the terms to get $x - 2x + 5 + 3 = 6$, or $(x - 2x) + (5 + 3) = 6$.
$(x - 2x)$ is the same as $(1 - 2)x$, or $-x$, so our equation becomes $-x + 8 = 6$
Subtracting 8 from both sides gives us $-x = -2$.
And finally, multiplying both sides by -1 gives us $x = 2$.
Example 6
Solve $\frac{x}{2} - \frac{x}{3} = 6$.
This problem requires us to deal with fractions. We need to write all the terms on the left over a common denominator of six.
$\frac{3x}{6} - \frac{2x}{6} = 6$
Then we subtract the fractions to get $\frac{x}{6} = 6$.
Finally we multiply both sides by 6 to get $x = 36$.
## Solve Real-World Problems Using Two-Step Equations
The hardest part of solving word problems is translating from words to an equation. First, you need to look to see what the equation is asking. What is the unknown for which you have to solve? That will be what your variable stands for. Then, follow what is going on with your variable all the way through the problem.
Example 7
An emergency plumber charges $65 as a call-out fee plus an additional$75 per hour. He arrives at a house at 9:30 and works to repair a water tank. If the total repair bill is $196.25, at what time was the repair completed? In order to solve this problem, we collect the information from the text and convert it to an equation. Unknown: time taken in hours – this will be our $x$ The bill is made up of two parts: a call out fee and a per-hour fee. The call out is a flat fee, and independent of $x$—it’s the same no matter how many hours the plumber works. The per-hour part depends on the number of hours $(x)$. So the total fee is$65 (no matter what) plus $\75x$ (where $x$ is the number of hours), or $65 + 75x$.
Looking at the problem again, we also can see that the total bill is $196.25. So our final equation is $196.25 = 65 + 75x$. Solving for $x$: $196.25 &= 65 + 75x \qquad \text{Subtract 65 from both sides.}\\131.25 &= 75x \qquad \qquad \text{Divide both sides by 75.}\\1.75 &= x \qquad \qquad \quad \text{The job took 1.75 hours.}$ Solution The repair job was completed 1.75 hours after 9:30, so it was completed at 11:15AM. Example 8 When Asia was young her Daddy marked her height on the door frame every month. Asia’s Daddy noticed that between the ages of one and three, he could predict her height (in inches) by taking her age in months, adding 75 inches and multiplying the result by one-third. Use this information to answer the following: a) Write an equation linking her predicted height, $h$, with her age in months, $m$. b) Determine her predicted height on her second birthday. c) Determine at what age she is predicted to reach three feet tall. Solution a) To convert the text to an equation, first determine the type of equation we have. We are going to have an equation that links two variables. Our unknown will change, depending on the information we are given. For example, we could solve for height given age, or solve for age given height. However, the text gives us a way to determine height. Our equation will start with “$h=$”. The text tells us that we can predict her height by taking her age in months, adding 75, and multiplying by $\frac{1}{3}$. So our equation is $h = (m + 75) \cdot \frac{1}{3}$, or $h = \frac{1}{3}(m + 75)$. b) To predict Asia’s height on her second birthday, we substitute $m=24$ into our equation (because 2 years is 24 months) and solve for $h$. $h &= \frac{1}{3}(24 + 75)\\h &= \frac{1}{3}(99)\\h &= 33$ Asia’s height on her second birthday was predicted to be 33 inches. c) To determine the predicted age when she reached three feet, substitute $h = 36$ into the equation and solve for $m$. $36 &= \frac{1}{3}(m + 75)\\108 &= m + 75\\33 &= m$ Asia was predicted to be 33 months old when her height was three feet. Example 9 To convert temperatures in Fahrenheit to temperatures in Celsius, follow the following steps: Take the temperature in degrees Fahrenheit and subtract 32. Then divide the result by 1.8 and this gives the temperature in degrees Celsius. a) Write an equation that shows the conversion process. b) Convert 50 degrees Fahrenheit to degrees Celsius. c) Convert 25 degrees Celsius to degrees Fahrenheit. d) Convert -40 degrees Celsius to degrees Fahrenheit. a) The text gives the process to convert Fahrenheit to Celsius. We can write an equation using two variables. We will use $f$ for temperature in Fahrenheit, and $c$ for temperature in Celsius. $&\text{First we take the temperature in Fahrenheit and subtract 32.} && f - 32\\&\text{Then divide by 1.8.} && \frac{f - 32}{1.8}\\&\text{This equals the temperature in Celsius.} && c = \frac{f - 32}{1.8}$ In order to convert from one temperature scale to another, simply substitute in for whichever temperature you know, and solve for the one you don’t know. b) To convert 50 degrees Fahrenheit to degrees Celsius, substitute $f = 50$ into the equation. $c &= \frac{50 - 32}{1.8}\\c &= \frac{18}{1.8}\\c &= 10$ 50 degrees Fahrenheit is equal to 10 degrees Celsius. c) To convert 25 degrees Celsius to degrees Fahrenheit, substitute $c = 25$ into the equation: $25 &= \frac{f - 32}{1.8}\\45 &= f - 32\\ 77 &= f$ 25 degrees Celsius is equal to 77 degrees Fahrenheit. d) To convert -40 degrees Celsius to degrees Fahrenheit, substitute $c = -40$ into the equation. $-40 &= \frac{f - 32}{1.8}\\-72 &= f - 32\\-40 &= f$ -40 degrees Celsius is equal to -40 degrees Fahrenheit. (No, that’s not a mistake! This is the one temperature where they are equal.) ## Lesson Summary • Some equations require more than one operation to solve. Generally it, is good to go from the outside in. If there are parentheses around an expression with a variable in it, cancel what is outside the parentheses first. • Terms with the same variable in them (or no variable in them) are like terms. Combine like terms (adding or subtracting them from each other) to simplify the expression and solve for the unknown. ## Review Questions 1. Solve the following equations for the unknown variable. 1. $1.3x - 0.7x = 12$ 2. $6x - 1.3 = 3.2$ 3. $5x - (3x + 2) = 1$ 4. $4(x + 3) = 1$ 5. $5q - 7 = \frac{2}{3}$ 6. $\frac{3}{5}x + \frac{5}{2} = \frac{2}{3}$ 7. $s - \frac{3s}{8} = \frac{5}{6}$ 8. $0.1y + 11 = 0$ 9. $\frac{5q - 7}{12} = \frac{2}{3}$ 10. $\frac{5(q - 7)}{12} = \frac{2}{3}$ 11. $33t - 99= 0$ 12. $5p - 2 = 32$ 13. $10y + 5 = 10$ 14. $10(y + 5) = 10$ 15. $10y + 5y = 10$ 16. $10(y + 5y) = 10$ 2. Jade is stranded downtown with only$10 to get home. Taxis cost $0.75 per mile, but there is an additional$2.35 hire charge. Write a formula and use it to calculate how many miles she can travel with her money.
3. Jasmin’s Dad is planning a surprise birthday party for her. He will hire a bouncy castle, and will provide party food for all the guests. The bouncy castle costs $150 for the afternoon, and the food will cost$3 per person. Andrew, Jasmin’s Dad, has a budget of $300. Write an equation and use it to determine the maximum number of guests he can invite. 4. The local amusement park sells summer memberships for$50 each. Normal admission to the park costs $25; admission for members costs$15.
1. If Darren wants to spend no more than $100 on trips to the amusement park this summer, how many visits can he make if he buys a membership with part of that money? 2. How many visits can he make if he does not? 3. If he increases his budget to$160, how many visits can he make as a member?
4. And how many as a non-member?
5. For an upcoming school field trip, there must be one adult supervisor for every five children.
1. If the bus seats 40 people, how many children can go on the trip?
2. How many children can go if a second 40-person bus is added?
3. Four of the adult chaperones decide to arrive separately by car. Now how many children can go in the two buses?
Feb 22, 2012
Sep 28, 2014 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 96, "texerror": 0, "math_score": 0.9442368745803833, "perplexity": 683.7165734245816}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422122233086.24/warc/CC-MAIN-20150124175713-00076-ip-10-180-212-252.ec2.internal.warc.gz"} |
http://physics.stackexchange.com/questions/49741/capacitor-charging-and-discharging | # Capacitor charging and discharging
I have a conceptual question.
The circuit is shown below. Let us assume that the capacitor shown in the figure is of 2.2 micro-farads having rated level of 5 volts and we decide to charge it up to 2 volts so that we don't smoke it.
Circuit -
If we connect $R_1$ and $R_2$ in the circuit for charging and discharging respectively, how will the circuit work?
The red loop shows the charging circuit and the green loop shows the discharging circuit.
The circuit is as shown until the capacitor is charged and then the switch S is closed.
Will the circuit discharge instantaneously?
If yes, when will the capacitor be re-charged again? Will be it charged after the long exponential charge decay of the capacitor or will it start to charge itself after it has a certain amount of charge left?
Will this process of charging and discharging be continuous if the switch S remains in closed position?
Let me start with some basics. The red loop circuit will charge with $R_1\times C$ time constant. So the charging will be quick because of low $R_1$. It will be 22 microseconds.
After the capacitor has been charged to 2V (max given by power supply), we close switch S and then the discharge process will start. As expected it will also be quick and similar to charging time constant.
Will this capacitor charge again? If yes, when? And can it be controlled?
-
I think this should be migrated to electronics.stackexchange.com – daaxix Jan 10 '13 at 5:26
For starters (in case you overlooked it), note that the capacitor will not discharge completely: The stationary state for the closed circuit has a continuous current flowing through the corrent branches with the resistors, so the point A will have a finite, non zero $V_A$ and the capacitor will experience a finite voltage. The higher the value of $R_1$ is compared to the value of $R_2$, the more the capacitor will discharge. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8959707617759705, "perplexity": 319.8026401447417}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1413507442497.30/warc/CC-MAIN-20141017005722-00337-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://physics.stackexchange.com/questions/403444/hubbard-stratonovich-transformation-and-decoupling-channels | # Hubbard-Stratonovich transformation and decoupling channels
I'm studying an example of the Hubbard-Stratonovich transformation in Altland and Simons' Condensed Matter Field Theory (2nd ed.), pp. 246-247.
In it they say that...
one is frequently confronted with situations where more than one Hubbard-Stratonovich field is needed to capture the full physics of the problem. To appreciate this point, consider the Coulomb interaction in momentum space. $$S_{int}[\bar{\psi},\psi] = \tfrac{1}{2} \sum_{p_1,...,p_4} \bar{\psi}_{\sigma, p_1} \bar{\psi}_{\sigma', p_3} V(\textbf{p}_1-\textbf{p}_2) \psi_{\sigma', p_4} \psi_{\sigma, p_2} \delta_{p_1-p_2+p_3-p_4}.$$ In principle, we can decouple this interaction in any of the three channels...
discussed in the previous page. If one chooses to decouple in all three channels then the action becomes ...
$$S_{int}[\bar{\psi},\psi] \simeq \tfrac{1}{2} \sum_{p,p',q} ( \bar{\psi}_{\sigma, p} \psi_{\sigma,p+q} V(\textbf{q}) \bar{\psi}_{\sigma', p'} \psi_{\sigma',p'-q} - \bar{\psi}_{\sigma, p} \psi_{\sigma',p+q} V(\textbf{p'}-\textbf{p}) \bar{\psi}_{\sigma', p'} \psi_{\sigma,p'} - \bar{\psi}_{\sigma, p} \bar{\psi}_{\sigma', -p+q} V(\textbf{p'}-\textbf{p}) \psi_{\sigma,p'} \psi_{\sigma',-p'+q} )$$
where the first term is decoupled via the
direct channel $\rho_{d,q} \sim \sum_{p} \bar{\psi}_{\sigma,p} \psi_{\sigma,p+q}$, second in the exchange channel $\rho_{x,\sigma\sigma',q} \sim \sum_{p} \bar{\psi}_{\sigma,p} \psi_{\sigma',p+q}$, and third in the Cooper channel $\rho_{c,\sigma\sigma',q} \sim \sum_{p} \bar{\psi}_{\sigma,p} \bar{\psi}_{\sigma',-p+q}$.
It's generally a good strategy to decouple in all available channels when one is in doubt, then let the mean-field analysis sort out the relevant fields.
My question is, if we choose to decouple the quartic term via 3 different channels (for example) is it necessary to multiply the resulting terms by a factor of $\tfrac{1}{3}$? This isn't discussed in the textbook and I'm confused by the liberal use of $\sim$ and $\simeq$ in the examples.
No. You should not add a factor of $1/3$. As you can see in page 244 of Altland and Simons, the HS transformation is done by multiplying by a unity expressed as a functional integral over an auxiliary field. In this case, they just choose to introduce 3 different fields - 1 for each term.
• Great, thanks. Just to clarify, for some action $S= \bar{\psi}_\alpha(t) \psi_\beta(t) V \bar{\psi}_\gamma(t') \psi_\delta(t')$, decoupling via the direct channel is when $\rho=\bar{\psi}_\alpha(t) \psi_\beta(t)$, exchange channel: $\rho=\bar{\psi}_\alpha(t) \psi_\delta(t')$, and Cooper channel: $\rho=\bar{\psi}_\alpha(t)\bar{\psi}_\gamma(t')$. If I want to decouple via the exchange channel then I can let either $\rho=\bar{\psi}_\alpha(t) \psi_\delta(t')$ or $\rho=\bar{\psi}_\gamma(t') \psi_\beta(t)$. Is this correct? (tbc) – Medulla Oblongata May 7 '18 at 9:07
• If so, can I use both these $\rho$ and not need a multiplicative factor of $1/2$? – Medulla Oblongata May 7 '18 at 9:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9395470023155212, "perplexity": 243.74248169233434}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347385193.5/warc/CC-MAIN-20200524210325-20200525000325-00303.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/83315-proving-field-not-algebraically-closed-print.html | # Proving a field is not algebraically closed
• April 12th 2009, 02:17 AM
Zinners
Proving a field is not algebraically closed
Hi,
My problem is this: Prove that if p is a prime, then the field Zp is not algebraically closed. I know that using Fermat's little theorem will help but I can't see how it's not closed. Can anybody help please?
Thanks
• April 12th 2009, 04:46 AM
berlioz
Using Fermat's little theorem we have x^(p-1)-1=0for any x doesn't equal zero,so the polynomial x^(p-1)-2(suppose p>2)doesn't have the zero points inZp
if p=2,we can find x^2+x+1 hasn't the zero points in Z2.
So field Zp is not algebraically closed.
Quote:
Originally Posted by Zinners
Hi,
My problem is this: Prove that if p is a prime, then the field Zp is not algebraically closed. I know that using Fermat's little theorem will help but I can't see how it's not closed. Can anybody help please?
Thanks
• April 13th 2009, 10:10 AM
ThePerfectHacker
Quote:
Originally Posted by Zinners
Hi,
My problem is this: Prove that if p is a prime, then the field Zp is not algebraically closed. I know that using Fermat's little theorem will help but I can't see how it's not closed. Can anybody help please?
Thanks
In general let $F$ be a finite field with $q$ elements.
Define $f(x) = x^q - x +1$, we know that $a^{q-1}=1 \implies a^q - a = 0$ for all $a\in F^{\times}$.
Therefore, $f(x) = x^q - x + 1$ always has no zero. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8473154902458191, "perplexity": 443.13244616441693}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783397565.80/warc/CC-MAIN-20160624154957-00003-ip-10-164-35-72.ec2.internal.warc.gz"} |
http://spmmathematics.onlinetuition.com.my/2014/08/multiplication-of-matrix-by-number.html | # 4.4 Multiplication of a Matrix by a Number
4.4 Multiplication of a Matrix by a Number
When matrix is multiplied by a number, every element in the matrix is multiplied by the number.
Example:
Given that $A=\left(\begin{array}{cc}-2& 4\\ 5& -6\end{array}\right)$ , find each of the following.
(a) 3A
(b) -2A
Solution: | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9738494157791138, "perplexity": 680.2834743252154}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218188924.7/warc/CC-MAIN-20170322212948-00168-ip-10-233-31-227.ec2.internal.warc.gz"} |
http://jmlr.org/proceedings/papers/v32/gopalan14.html | # Thompson Sampling for Complex Online Problems
Aditya Gopalan, Shie Mannor, Yishay Mansour
Proceedings of The 31st International Conference on Machine Learning, pp. 100–108, 2014
## Abstract
We consider stochastic multi-armed bandit problems with complex actions over a set of basic arms, where the decision maker plays a complex action rather than a basic arm in each round. The reward of the complex action is some function of the basic arms’ rewards, and the feedback observed may not necessarily be the reward per-arm. For instance, when the complex actions are subsets of the arms, we may only observe the maximum reward over the chosen subset. Thus, feedback across complex actions may be coupled due to the nature of the reward function. We prove a frequentist regret bound for Thompson sampling in a very general setting involving parameter, action and observation spaces and a likelihood function over them. The bound holds for discretely-supported priors over the parameter space and without additional structural properties such as closed-form posteriors, conjugate prior structure or independence across arms. The regret bound scales logarithmically with time but, more importantly, with an improved constant that non-trivially captures the coupling across complex actions due to the structure of the rewards. As applications, we derive improved regret bounds for classes of complex bandit problems involving selecting subsets of arms, including the first nontrivial regret bounds for nonlinear MAX reward feedback from subsets. Using particle filters for computing posterior distributions which lack an explicit closed-form, we present numerical results for the performance of Thompson sampling for subset-selection and job scheduling problems. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9703350067138672, "perplexity": 720.3770312971982}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257823802.12/warc/CC-MAIN-20160723071023-00196-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/207380-calculus-curve-sketching.html | # Math Help - Calculus curve sketching
1. ## Calculus curve sketching
sketch the graph of a function f having given characteristics
f(2) = f(4) = 0
f(3) is defined.
f'(x) < 0 if x < 3
f'(3) does not exist.
f'(x) > 0 if x > 3
f"(x) < 0, x does not equal 3
2. ## Re: Calculus curve sketching
Hey asilverster635.
If you show us what you have tried and any partial attempts you have made, then you will get a more specific and directed answer from other members.
3. ## Re: Calculus curve sketching
i have no idea how to do this type of problem
4. ## Re: Calculus curve sketching
If the derivative doesn't exist at a point it means there is a discontinuity.
If you have a positive derivative it means the function is increasing: if it is negative then it is decreasing.
If f(3) is defined but f'(3) doesn't exist, then it means you have either a discontinuity in the graph or the graph itself has a "kink" in it and isn't smooth (but is still continuous).
If second derivative is increasing then first derivative is increasing: if decreasing then derivative is decreasing.
There are many solutions to this problem graphically and function-wise but they will have the attributes outlined with the above characteristics of derivatives.
5. ## Re: Calculus curve sketching
All of these have a graphical equivalent - for example f(2)=0 means that the graph passes through (0,2). Once you have "translated" them all, you have a description of the graph, so you just need to draw it.
- Hollywood
6. ## Re: Calculus curve sketching
Originally Posted by chiro
If the derivative doesn't exist at a point it means there is a discontinuity.
Technically not correct - the graph could have a "kink" like the function f(x)=|x| at x=0.
Originally Posted by chiro
If second derivative is increasing then first derivative is increasing: if decreasing then derivative is decreasing.
It's probably better to say that if the second derivative is negative, then the function is concave down. And also (though it's not needed for this problem) if the second derivative is positive, then the function is concave up.
- Hollywood
thanks guys
8. ## Re: Calculus curve sketching
Originally Posted by hollywood
the graph could have a "kink" like the function f(x)=|x| at x=0.
A better word would be "corner". I don't know of any mathematical term for it. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.907497763633728, "perplexity": 966.9531155506482}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207928114.23/warc/CC-MAIN-20150521113208-00009-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://people.maths.bris.ac.uk/~matyd/GroupNames/288/SD16sD9.html | Copied to
clipboard
## G = SD16⋊D9order 288 = 25·32
### 2nd semidirect product of SD16 and D9 acting via D9/C9=C2
Series: Derived Chief Lower central Upper central
Derived series C1 — C36 — SD16⋊D9
Chief series C1 — C3 — C9 — C18 — C36 — C4×D9 — Q8×D9 — SD16⋊D9
Lower central C9 — C18 — C36 — SD16⋊D9
Upper central C1 — C2 — C4 — SD16
Generators and relations for SD16⋊D9
G = < a,b,c,d | a8=b2=c9=d2=1, bab=a3, ac=ca, dad=a5, bc=cb, dbd=a4b, dcd=c-1 >
Subgroups: 396 in 90 conjugacy classes, 34 normal (all characteristic)
C1, C2, C2 [×2], C3, C4, C4 [×4], C22 [×2], S3, C6, C6, C8, C8, C2×C4 [×3], D4, D4, Q8, Q8 [×3], C9, Dic3 [×3], C12, C12, D6, C2×C6, M4(2), SD16, SD16, Q16 [×2], C2×Q8, C4○D4, D9, C18, C18, C3⋊C8, C24, Dic6 [×3], C4×S3 [×2], C2×Dic3, C3⋊D4, C3×D4, C3×Q8, C8.C22, Dic9, Dic9 [×2], C36, C36, D18, C2×C18, C8⋊S3, Dic12, D4.S3, C3⋊Q16, C3×SD16, D42S3, S3×Q8, C9⋊C8, C72, Dic18 [×2], Dic18, C4×D9, C4×D9, C2×Dic9, C9⋊D4, D4×C9, Q8×C9, D4.D6, Dic36, C8⋊D9, D4.D9, C9⋊Q16, C9×SD16, D42D9, Q8×D9, SD16⋊D9
Quotients: C1, C2 [×7], C22 [×7], S3, D4 [×2], C23, D6 [×3], C2×D4, D9, C22×S3, C8.C22, D18 [×3], S3×D4, C22×D9, D4.D6, D4×D9, SD16⋊D9
Smallest permutation representation of SD16⋊D9
On 144 points
Generators in S144
(1 140 32 122 14 131 23 113)(2 141 33 123 15 132 24 114)(3 142 34 124 16 133 25 115)(4 143 35 125 17 134 26 116)(5 144 36 126 18 135 27 117)(6 136 28 118 10 127 19 109)(7 137 29 119 11 128 20 110)(8 138 30 120 12 129 21 111)(9 139 31 121 13 130 22 112)(37 91 55 82 46 100 64 73)(38 92 56 83 47 101 65 74)(39 93 57 84 48 102 66 75)(40 94 58 85 49 103 67 76)(41 95 59 86 50 104 68 77)(42 96 60 87 51 105 69 78)(43 97 61 88 52 106 70 79)(44 98 62 89 53 107 71 80)(45 99 63 90 54 108 72 81)
(1 41)(2 42)(3 43)(4 44)(5 45)(6 37)(7 38)(8 39)(9 40)(10 46)(11 47)(12 48)(13 49)(14 50)(15 51)(16 52)(17 53)(18 54)(19 55)(20 56)(21 57)(22 58)(23 59)(24 60)(25 61)(26 62)(27 63)(28 64)(29 65)(30 66)(31 67)(32 68)(33 69)(34 70)(35 71)(36 72)(73 127)(74 128)(75 129)(76 130)(77 131)(78 132)(79 133)(80 134)(81 135)(82 136)(83 137)(84 138)(85 139)(86 140)(87 141)(88 142)(89 143)(90 144)(91 118)(92 119)(93 120)(94 121)(95 122)(96 123)(97 124)(98 125)(99 126)(100 109)(101 110)(102 111)(103 112)(104 113)(105 114)(106 115)(107 116)(108 117)
(1 2 3 4 5 6 7 8 9)(10 11 12 13 14 15 16 17 18)(19 20 21 22 23 24 25 26 27)(28 29 30 31 32 33 34 35 36)(37 38 39 40 41 42 43 44 45)(46 47 48 49 50 51 52 53 54)(55 56 57 58 59 60 61 62 63)(64 65 66 67 68 69 70 71 72)(73 74 75 76 77 78 79 80 81)(82 83 84 85 86 87 88 89 90)(91 92 93 94 95 96 97 98 99)(100 101 102 103 104 105 106 107 108)(109 110 111 112 113 114 115 116 117)(118 119 120 121 122 123 124 125 126)(127 128 129 130 131 132 133 134 135)(136 137 138 139 140 141 142 143 144)
(1 9)(2 8)(3 7)(4 6)(10 17)(11 16)(12 15)(13 14)(19 26)(20 25)(21 24)(22 23)(28 35)(29 34)(30 33)(31 32)(37 53)(38 52)(39 51)(40 50)(41 49)(42 48)(43 47)(44 46)(45 54)(55 71)(56 70)(57 69)(58 68)(59 67)(60 66)(61 65)(62 64)(63 72)(73 80)(74 79)(75 78)(76 77)(82 89)(83 88)(84 87)(85 86)(91 98)(92 97)(93 96)(94 95)(100 107)(101 106)(102 105)(103 104)(109 125)(110 124)(111 123)(112 122)(113 121)(114 120)(115 119)(116 118)(117 126)(127 143)(128 142)(129 141)(130 140)(131 139)(132 138)(133 137)(134 136)(135 144)
G:=sub<Sym(144)| (1,140,32,122,14,131,23,113)(2,141,33,123,15,132,24,114)(3,142,34,124,16,133,25,115)(4,143,35,125,17,134,26,116)(5,144,36,126,18,135,27,117)(6,136,28,118,10,127,19,109)(7,137,29,119,11,128,20,110)(8,138,30,120,12,129,21,111)(9,139,31,121,13,130,22,112)(37,91,55,82,46,100,64,73)(38,92,56,83,47,101,65,74)(39,93,57,84,48,102,66,75)(40,94,58,85,49,103,67,76)(41,95,59,86,50,104,68,77)(42,96,60,87,51,105,69,78)(43,97,61,88,52,106,70,79)(44,98,62,89,53,107,71,80)(45,99,63,90,54,108,72,81), (1,41)(2,42)(3,43)(4,44)(5,45)(6,37)(7,38)(8,39)(9,40)(10,46)(11,47)(12,48)(13,49)(14,50)(15,51)(16,52)(17,53)(18,54)(19,55)(20,56)(21,57)(22,58)(23,59)(24,60)(25,61)(26,62)(27,63)(28,64)(29,65)(30,66)(31,67)(32,68)(33,69)(34,70)(35,71)(36,72)(73,127)(74,128)(75,129)(76,130)(77,131)(78,132)(79,133)(80,134)(81,135)(82,136)(83,137)(84,138)(85,139)(86,140)(87,141)(88,142)(89,143)(90,144)(91,118)(92,119)(93,120)(94,121)(95,122)(96,123)(97,124)(98,125)(99,126)(100,109)(101,110)(102,111)(103,112)(104,113)(105,114)(106,115)(107,116)(108,117), (1,2,3,4,5,6,7,8,9)(10,11,12,13,14,15,16,17,18)(19,20,21,22,23,24,25,26,27)(28,29,30,31,32,33,34,35,36)(37,38,39,40,41,42,43,44,45)(46,47,48,49,50,51,52,53,54)(55,56,57,58,59,60,61,62,63)(64,65,66,67,68,69,70,71,72)(73,74,75,76,77,78,79,80,81)(82,83,84,85,86,87,88,89,90)(91,92,93,94,95,96,97,98,99)(100,101,102,103,104,105,106,107,108)(109,110,111,112,113,114,115,116,117)(118,119,120,121,122,123,124,125,126)(127,128,129,130,131,132,133,134,135)(136,137,138,139,140,141,142,143,144), (1,9)(2,8)(3,7)(4,6)(10,17)(11,16)(12,15)(13,14)(19,26)(20,25)(21,24)(22,23)(28,35)(29,34)(30,33)(31,32)(37,53)(38,52)(39,51)(40,50)(41,49)(42,48)(43,47)(44,46)(45,54)(55,71)(56,70)(57,69)(58,68)(59,67)(60,66)(61,65)(62,64)(63,72)(73,80)(74,79)(75,78)(76,77)(82,89)(83,88)(84,87)(85,86)(91,98)(92,97)(93,96)(94,95)(100,107)(101,106)(102,105)(103,104)(109,125)(110,124)(111,123)(112,122)(113,121)(114,120)(115,119)(116,118)(117,126)(127,143)(128,142)(129,141)(130,140)(131,139)(132,138)(133,137)(134,136)(135,144)>;
G:=Group( (1,140,32,122,14,131,23,113)(2,141,33,123,15,132,24,114)(3,142,34,124,16,133,25,115)(4,143,35,125,17,134,26,116)(5,144,36,126,18,135,27,117)(6,136,28,118,10,127,19,109)(7,137,29,119,11,128,20,110)(8,138,30,120,12,129,21,111)(9,139,31,121,13,130,22,112)(37,91,55,82,46,100,64,73)(38,92,56,83,47,101,65,74)(39,93,57,84,48,102,66,75)(40,94,58,85,49,103,67,76)(41,95,59,86,50,104,68,77)(42,96,60,87,51,105,69,78)(43,97,61,88,52,106,70,79)(44,98,62,89,53,107,71,80)(45,99,63,90,54,108,72,81), (1,41)(2,42)(3,43)(4,44)(5,45)(6,37)(7,38)(8,39)(9,40)(10,46)(11,47)(12,48)(13,49)(14,50)(15,51)(16,52)(17,53)(18,54)(19,55)(20,56)(21,57)(22,58)(23,59)(24,60)(25,61)(26,62)(27,63)(28,64)(29,65)(30,66)(31,67)(32,68)(33,69)(34,70)(35,71)(36,72)(73,127)(74,128)(75,129)(76,130)(77,131)(78,132)(79,133)(80,134)(81,135)(82,136)(83,137)(84,138)(85,139)(86,140)(87,141)(88,142)(89,143)(90,144)(91,118)(92,119)(93,120)(94,121)(95,122)(96,123)(97,124)(98,125)(99,126)(100,109)(101,110)(102,111)(103,112)(104,113)(105,114)(106,115)(107,116)(108,117), (1,2,3,4,5,6,7,8,9)(10,11,12,13,14,15,16,17,18)(19,20,21,22,23,24,25,26,27)(28,29,30,31,32,33,34,35,36)(37,38,39,40,41,42,43,44,45)(46,47,48,49,50,51,52,53,54)(55,56,57,58,59,60,61,62,63)(64,65,66,67,68,69,70,71,72)(73,74,75,76,77,78,79,80,81)(82,83,84,85,86,87,88,89,90)(91,92,93,94,95,96,97,98,99)(100,101,102,103,104,105,106,107,108)(109,110,111,112,113,114,115,116,117)(118,119,120,121,122,123,124,125,126)(127,128,129,130,131,132,133,134,135)(136,137,138,139,140,141,142,143,144), (1,9)(2,8)(3,7)(4,6)(10,17)(11,16)(12,15)(13,14)(19,26)(20,25)(21,24)(22,23)(28,35)(29,34)(30,33)(31,32)(37,53)(38,52)(39,51)(40,50)(41,49)(42,48)(43,47)(44,46)(45,54)(55,71)(56,70)(57,69)(58,68)(59,67)(60,66)(61,65)(62,64)(63,72)(73,80)(74,79)(75,78)(76,77)(82,89)(83,88)(84,87)(85,86)(91,98)(92,97)(93,96)(94,95)(100,107)(101,106)(102,105)(103,104)(109,125)(110,124)(111,123)(112,122)(113,121)(114,120)(115,119)(116,118)(117,126)(127,143)(128,142)(129,141)(130,140)(131,139)(132,138)(133,137)(134,136)(135,144) );
G=PermutationGroup([(1,140,32,122,14,131,23,113),(2,141,33,123,15,132,24,114),(3,142,34,124,16,133,25,115),(4,143,35,125,17,134,26,116),(5,144,36,126,18,135,27,117),(6,136,28,118,10,127,19,109),(7,137,29,119,11,128,20,110),(8,138,30,120,12,129,21,111),(9,139,31,121,13,130,22,112),(37,91,55,82,46,100,64,73),(38,92,56,83,47,101,65,74),(39,93,57,84,48,102,66,75),(40,94,58,85,49,103,67,76),(41,95,59,86,50,104,68,77),(42,96,60,87,51,105,69,78),(43,97,61,88,52,106,70,79),(44,98,62,89,53,107,71,80),(45,99,63,90,54,108,72,81)], [(1,41),(2,42),(3,43),(4,44),(5,45),(6,37),(7,38),(8,39),(9,40),(10,46),(11,47),(12,48),(13,49),(14,50),(15,51),(16,52),(17,53),(18,54),(19,55),(20,56),(21,57),(22,58),(23,59),(24,60),(25,61),(26,62),(27,63),(28,64),(29,65),(30,66),(31,67),(32,68),(33,69),(34,70),(35,71),(36,72),(73,127),(74,128),(75,129),(76,130),(77,131),(78,132),(79,133),(80,134),(81,135),(82,136),(83,137),(84,138),(85,139),(86,140),(87,141),(88,142),(89,143),(90,144),(91,118),(92,119),(93,120),(94,121),(95,122),(96,123),(97,124),(98,125),(99,126),(100,109),(101,110),(102,111),(103,112),(104,113),(105,114),(106,115),(107,116),(108,117)], [(1,2,3,4,5,6,7,8,9),(10,11,12,13,14,15,16,17,18),(19,20,21,22,23,24,25,26,27),(28,29,30,31,32,33,34,35,36),(37,38,39,40,41,42,43,44,45),(46,47,48,49,50,51,52,53,54),(55,56,57,58,59,60,61,62,63),(64,65,66,67,68,69,70,71,72),(73,74,75,76,77,78,79,80,81),(82,83,84,85,86,87,88,89,90),(91,92,93,94,95,96,97,98,99),(100,101,102,103,104,105,106,107,108),(109,110,111,112,113,114,115,116,117),(118,119,120,121,122,123,124,125,126),(127,128,129,130,131,132,133,134,135),(136,137,138,139,140,141,142,143,144)], [(1,9),(2,8),(3,7),(4,6),(10,17),(11,16),(12,15),(13,14),(19,26),(20,25),(21,24),(22,23),(28,35),(29,34),(30,33),(31,32),(37,53),(38,52),(39,51),(40,50),(41,49),(42,48),(43,47),(44,46),(45,54),(55,71),(56,70),(57,69),(58,68),(59,67),(60,66),(61,65),(62,64),(63,72),(73,80),(74,79),(75,78),(76,77),(82,89),(83,88),(84,87),(85,86),(91,98),(92,97),(93,96),(94,95),(100,107),(101,106),(102,105),(103,104),(109,125),(110,124),(111,123),(112,122),(113,121),(114,120),(115,119),(116,118),(117,126),(127,143),(128,142),(129,141),(130,140),(131,139),(132,138),(133,137),(134,136),(135,144)])
39 conjugacy classes
class 1 2A 2B 2C 3 4A 4B 4C 4D 4E 6A 6B 8A 8B 9A 9B 9C 12A 12B 18A 18B 18C 18D 18E 18F 24A 24B 36A 36B 36C 36D 36E 36F 72A ··· 72F order 1 2 2 2 3 4 4 4 4 4 6 6 8 8 9 9 9 12 12 18 18 18 18 18 18 24 24 36 36 36 36 36 36 72 ··· 72 size 1 1 4 18 2 2 4 18 36 36 2 8 4 36 2 2 2 4 8 2 2 2 8 8 8 4 4 4 4 4 8 8 8 4 ··· 4
39 irreducible representations
dim 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 4 4 4 4 4 type + + + + + + + + + + + + + + + + + + - + - + - image C1 C2 C2 C2 C2 C2 C2 C2 S3 D4 D4 D6 D6 D6 D9 D18 D18 D18 C8.C22 S3×D4 D4.D6 D4×D9 SD16⋊D9 kernel SD16⋊D9 Dic36 C8⋊D9 D4.D9 C9⋊Q16 C9×SD16 D4⋊2D9 Q8×D9 C3×SD16 Dic9 D18 C24 C3×D4 C3×Q8 SD16 C8 D4 Q8 C9 C6 C3 C2 C1 # reps 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3 3 3 3 1 1 2 3 6
Matrix representation of SD16⋊D9 in GL6(𝔽73)
1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 66 13 0 0 0 0 13 7 0 0 13 7 0 0 0 0 7 60 0 0
,
1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0
,
42 45 0 0 0 0 28 70 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1
,
42 70 0 0 0 0 28 31 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 72 0 0 0 0 0 0 72
G:=sub<GL(6,GF(73))| [1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,13,7,0,0,0,0,7,60,0,0,66,13,0,0,0,0,13,7,0,0],[1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,0,1,0,0],[42,28,0,0,0,0,45,70,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,1],[42,28,0,0,0,0,70,31,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,72,0,0,0,0,0,0,72] >;
SD16⋊D9 in GAP, Magma, Sage, TeX
{\rm SD}_{16}\rtimes D_9
% in TeX
G:=Group("SD16:D9");
// GroupNames label
G:=SmallGroup(288,125);
// by ID
G=gap.SmallGroup(288,125);
# by ID
G:=PCGroup([7,-2,-2,-2,-2,-2,-3,-3,120,422,135,346,185,80,6725,292,9414]);
// Polycyclic
G:=Group<a,b,c,d|a^8=b^2=c^9=d^2=1,b*a*b=a^3,a*c=c*a,d*a*d=a^5,b*c=c*b,d*b*d=a^4*b,d*c*d=c^-1>;
// generators/relations
×
𝔽 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9965108633041382, "perplexity": 1905.409674645598}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703524270.28/warc/CC-MAIN-20210121070324-20210121100324-00471.warc.gz"} |
https://www.physicsforums.com/threads/qed-coherence-in-matter.217809/ | # QED coherence in matter
1. Feb 25, 2008
### mgb2
Hi,
I would like to know what do you think about the Giuliano Preparata theory on QED coherence in matter.
He studied the problem of the ground state in condensed matter under the hypotesis that the standard perturbative vacuum is unstable with respect to a new coherent vacuum, whose spectrum emerge quite naturally through a simple variational procedure.
https://www.amazon.com/Introduction-Realistic-Quantum-Physics/dp/9812381767#cited
https://www.amazon.com/Qed-Coherence-Matter-Giuliano-Preparata/dp/9810222491
Ciao
mgb2
#### Attached Files:
• ###### del giudice.pdf
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Last edited: Feb 25, 2008
2. Feb 27, 2008
### mgb2
But, is never possible that nobody of you is interested in this new way to watch to the reality of the world (without introducing new concepts, of course) ?
Just looking at interaction and dynamical evolution between matter and e.m. field in terms of quantized matter-wave-field and quantized e.m.-wave-field.
If you consider a piece of matter (N=10^23 but described by a field) what is the effect of interaction with the zero point energy fluctuations of the e.m. field??
What happens when the electric dipole moment field interact with e.m. field?
look at the article!!
Ciao
mgb2
#### Attached Files:
• ###### Srivastava.pdf
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Last edited: Feb 27, 2008 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.88973468542099, "perplexity": 2710.229258617366}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463607960.64/warc/CC-MAIN-20170525010046-20170525030046-00159.warc.gz"} |
https://forum.allaboutcircuits.com/threads/circuit-analysis-help.63892/ | # Circuit analysis help
#### geft
Joined Dec 8, 2011
19
I'm having a hard time finding V2 in terms of I2. I've tried various forms of KCL and KVL but I can't seem to find the right answer. I hope someone can enlighten me.
#### Attachments
• 41.5 KB Views: 37
Last edited:
#### jegues
Joined Sep 13, 2010
733
Can you show us your work?
Then we can further diagnose what the problem is and point you in the right direction.
#### geft
Joined Dec 8, 2011
19
$$I_2 = \frac{V_2-2V_x}{2} + \frac{V_2}{5} = \frac{V_2-2\frac{V_2}{4}}{2} + \frac{V_2}{5} = 0.45V_2$$
$$Z_{22} = \frac{V_2}{I_2} = \frac{V_2}{0.45V_2} = 2.222$$
This is clearly wrong since the answer is supposed to be 1.111 ohm.
Last edited:
#### Zazoo
Joined Jul 27, 2011
114
$$I_2 = \frac{V_2-2V_x}{2} + \frac{V_2}{5} = \frac{V_2-2\frac{V_2}{4}}{2} + \frac{V_2}{5} = 0.45V_2$$
$$Z_{22} = \frac{V_2}{I_2} = \frac{V_2}{0.45V_2} = 2.222$$
This is clearly wrong since the answer is supposed to be 1.111 ohm.
My node equation is:
$$I_2 = \frac{V_2-(-2V_x)}{2} + \frac{V_2}{5} = \frac{V_2+2V_2(\frac{1}{1+4})}{2} + \frac{V_2}{5} = 0.9V_2$$
Current direction was chosen as leaving the node for both branches.
Note the polarity on the dependent voltage source.
Also, Vx can be replaced by the expression for a 1Ω and 4Ω voltage divider (making it 1/5 V2)
This gives 1.111ohms when used in the second equation. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9759864211082458, "perplexity": 2093.749316551833}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703557462.87/warc/CC-MAIN-20210124204052-20210124234052-00682.warc.gz"} |
http://mathhelpforum.com/calculus/35364-taylor-polynomial-print.html | # Taylor Polynomial
• Apr 21st 2008, 09:49 AM
akhayoon
Taylor Polynomial
Question was
evaluate the 4th taylor polynomial of $\sqrt{3+x^{2}}$ around a=-1
$f4(x)=2-\frac{-(x+1)}{2}-\frac{(x+2)^{2}}{(8)2!}-\frac{3(x+1)^{3}}{(32)3!}-\frac{15(x+1)^{4}}{(128)4!}$
the marker notes however that there was something missing in the 3rd term
that the 3 in the fourth term was wrong, and that the 15 and 128 in the last term is wrong as well...
but after looking at my derivatives I can't figure out why this is???
• Apr 21st 2008, 10:41 AM
Opalg
Quote:
Originally Posted by akhayoon
Question was
evaluate the 4th taylor polynomial of $\sqrt{3+x^{2}}$ around a=-1
$f4(x)=2-\frac{-(x+1)}{2}-\frac{(x+2)^{2}}{(8)2!}-\frac{3(x+1)^{3}}{(32)3!}-\frac{15(x+1)^{4}}{(128)4!}$
the marker notes however that there was something missing in the 3rd term
that the 3 in the fourth term was wrong, and that the 15 and 128 in the last term is wrong as well...
but after looking at my derivatives I can't figure out why this is???
$f'(x) = x(3+x^2)^{-1/2},\ \ f'(-1) = {\textstyle-\frac12}$.
$f''(x) = (3+x^2)^{-1/2} - x^2(3+x^2)^{-3/2},\ \ f''(-1) = {\textstyle\frac12 - \frac18 =\frac38}$.
$f'''(x) = -3x(3+x^2)^{-1/2} + 3x^3(3+x^2)^{-5/2},\ \ f'''(-1) = {\textstyle\frac38 - \frac3{32} =\frac9{32}}$.
That gives $f_3(x) = 2 - \frac{(x+1)}{2} + \frac{3(x+1)^{2}}{(8)2!} + \frac{9(x+1)^{3}}{(32)3!}$. I'll leave you to sort out the (x+1)^4 term. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8283481001853943, "perplexity": 862.3881494786257}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988720615.90/warc/CC-MAIN-20161020183840-00463-ip-10-171-6-4.ec2.internal.warc.gz"} |
http://projecteuclid.org/euclid.aop/1176993221 | ## The Annals of Probability
### Synonymity, Generalized Martingales, and Subfiltration
Douglas N. Hoover
#### Abstract
Aldous recently introduced the notion of synonymity of stochastic processes, a notion of equivalence for processes on a stochastic basis which generalizes the notion of "having the same distribution". We show that generalized martingale properties, such as the semimartingale property, are preserved under synonymity, and that synonymous semimartingales have decompositions with the same distribution law. A variation of our method yields a relatively elementary proof of the theorem of Stricker that semimartingale remains a semimartingale with respect to any subfiltration to which it is adapted.
#### Article information
Source
Ann. Probab. Volume 12, Number 3 (1984), 703-713.
Dates
First available in Project Euclid: 19 April 2007
http://projecteuclid.org/euclid.aop/1176993221
JSTOR
Digital Object Identifier
doi:10.1214/aop/1176993221
Mathematical Reviews number (MathSciNet)
MR744227
Subjects
Primary: 60G07: General theory of processes
Secondary: 60G48: Generalizations of martingales
#### Citation
Hoover, Douglas N. Synonymity, Generalized Martingales, and Subfiltration. Ann. Probab. 12 (1984), no. 3, 703--713. doi:10.1214/aop/1176993221. http://projecteuclid.org/euclid.aop/1176993221. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8251436948776245, "perplexity": 3085.7806250971576}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422115927113.72/warc/CC-MAIN-20150124161207-00139-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://korali.readthedocs.io/en/master/modules/problem/reinforcementLearning/reinforcementLearning.html | # Reinforcement Learning
Describes a sequential decision making problem. We are given an environment that transitions to a state $$s'$$ and returns a reward $$r$$ for a given action $$a$$ and state $$s$$ with probability $$p(s',r|s,a)$$. We want to find the policy $$\pi$$ that choses an action $$a$$ for a given state $$s$$ with probability $$\pi(a|s)$$ such that for every state $$s$$ the chosen action $$a$$ is such that the value function
$\begin{split}V^\pi(s)=\E_{\substack{a_t\sim \pi(\cdot|s_t) \\ s_{t+1},r_t\sim p(\cdot,\cdot|s_t,a_t)}}\left[\sum\limits_{t=0}^\infty \gamma^t r_t\bigg|s_0=s\right]\end{split}$
is maximal. Here $$\gamma$$ is the discount factor.
We distinguish discrete and continuous action domains. Sub-Categories: | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9860633015632629, "perplexity": 254.86728165957848}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320301863.7/warc/CC-MAIN-20220120130236-20220120160236-00668.warc.gz"} |
http://www.ms.u-tokyo.ac.jp/seminar/2018/sem18-170.html | ## 数理人口学・数理生物学セミナー
### 2018年07月18日(水)
15:00-16:00 数理科学研究科棟(駒場) 118号室
Malay Banerjee 氏 (Department of Mathematics & Statistics, IIT Kanpur)
Effect of demographic stochasticity on large amplitude oscillation
[ 講演概要 ]
Classical Rosenzweig-MacArthur model exhibits two types of stable coexistence, steady-state and oscillatory coexistence. The oscillatory coexistence is the result of super-critical Hopf-bifurcation and the Hopf-bifurcating limit cycle remains stable for parameter values beyond the bifurcation threshold. The size of the limit cycle grows with the increase in carrying capacity of prey and finally both the populations show high amplitude oscillations. Time evolution of prey and predator population densities exhibit large amplitude peaks separated by low density lengthy valleys. Persistence of both the populations at low population density over a longer time period is more prominent in case of fast growth of prey and comparatively slow growth of predator species due to slow-fast dynamics. In this situation, small amount of demographic stochasticity can cause the extinction of one or both the species. Main aim of this talk is to explain the effect of demographic stochasticity on the high amplitude oscillations produced by two and higher dimensional interacting population models. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9565949440002441, "perplexity": 2141.6917886292886}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247488374.18/warc/CC-MAIN-20190218200135-20190218222135-00393.warc.gz"} |
https://en.wikipedia.org/wiki/Closed_formula | # Closed-form expression
(Redirected from Closed formula)
"Closed formula" redirects here. For "closed formula" in the sense of a logic formula with no free variables, see Sentence (mathematical logic).
In mathematics, a closed-form expression is a mathematical expression that can be evaluated in a finite number of operations. It may contain constants, variables, certain "well-known" operations (e.g., + − × ÷), and functions (e.g., nth root, exponent, logarithm, trigonometric functions, and inverse hyperbolic functions), but usually no limit. The set of operations and functions admitted in a closed-form expression may vary with author and context.
Problems are said to be tractable if they can be solved in terms of a closed-form expression.
## Example: roots of polynomials
The solutions of any quadratic equation with complex coefficients can be expressed in closed form in terms of addition, subtraction, multiplication, division, and square root extraction, each of which is an elementary function. For example, the quadratic equation:
${\displaystyle ax^{2}+bx+c=0,\,}$
is tractable since its solutions can be expressed as closed-form expression, i.e. in terms of elementary functions:
${\displaystyle x={-b\pm {\sqrt {b^{2}-4ac}} \over 2a}}$
Similarly solutions of cubic and quartic (third and fourth degree) equations can be expressed using arithmetic, square roots, and cube roots, or alternatively using arithmetic and trigonometric functions. However, there are quintic equations without closed-form solutions using elementary functions, such as x5 − x + 1 = 0.
An area of study in mathematics referred to broadly as Galois theory involves proving that no closed-form expression exists in certain contexts, based on the central example of closed-form solutions to polynomials.
## Alternative definitions
Changing the definition of "well-known" to include additional functions can change the set of equations with closed-form solutions. Many cumulative distribution functions cannot be expressed in closed form, unless one considers special functions such as the error function or gamma function to be well known. It is possible to solve the quintic equation if general hypergeometric functions are included, although the solution is far too complicated algebraically to be useful. For many practical computer applications, it is entirely reasonable to assume that the gamma function and other special functions are well-known, since numerical implementations are widely available.
## Analytic expression
An analytic expression (or expression in analytic form) is a mathematical expression constructed using well-known operations that lend themselves readily to calculation. Similar to closed-form expressions, the set of well-known functions allowed can vary according to context but always includes the basic arithmetic operations (addition, subtraction, multiplication, and division), exponentiation to a real exponent (which includes extraction of the nth root), logarithms, and trigonometric functions.
However, the class of expressions considered to be analytic expressions tends to be wider than that for closed-form expressions. In particular, special functions such as the Bessel functions and the gamma function are usually allowed, and often so are infinite series and continued fractions. On the other hand, limits in general, and integrals in particular, are typically excluded.
If an analytic expression involves only the algebraic operations (addition, subtraction, multiplication, division and exponentiation to a rational exponent) and rational constants then it is more specifically referred to as an algebraic expression.
## Comparison of different classes of expressions
Closed-form expressions are an important sub-class of analytic expressions, which contain a bounded[citation needed] or unbounded number of applications of well-known functions. Unlike the broader analytic expressions, the closed-form expressions do not include infinite series or continued fractions; neither includes integrals or limits. Indeed, by the Stone–Weierstrass theorem, any continuous function on the unit interval can be expressed as a limit of polynomials, so any class of functions containing the polynomials and closed under limits will necessarily include all continuous functions.
Similarly, an equation or system of equations is said to have a closed-form solution if, and only if, at least one solution can be expressed as a closed-form expression; and it is said to have an analytic solution if and only if at least one solution can be expressed as an analytic expression. There is a subtle distinction between a "closed-form function" and a "closed-form number" in the discussion of a "closed-form solution", discussed in (Chow 1999) and below. A closed-form or analytic solution is sometimes referred to as an explicit solution.
## Dealing with non-closed-form expressions
### Transformation into closed-form expressions
The expression:
${\displaystyle f(x)=\sum _{i=0}^{\infty }{x \over 2^{i}}}$
is not in closed form because the summation entails an infinite number of elementary operations. However, by summing a geometric series this expression can be expressed in the closed-form:[1]
${\displaystyle f(x)=2x}$
### Differential Galois theory
The integral of a closed-form expression may or may not itself be expressible as a closed-form expression. This study is referred to as differential Galois theory, by analogy with algebraic Galois theory.
The basic theorem of differential Galois theory is due to Joseph Liouville in the 1830s and 1840s and hence referred to as Liouville's theorem.
A standard example of an elementary function whose antiderivative does not have a closed-form expression is:
${\displaystyle e^{-x^{2}}}$
whose antiderivative is (up to constants) the error function:
${\displaystyle \operatorname {erf} (x)={\frac {2}{\sqrt {\pi }}}\int _{0}^{x}e^{-t^{2}}\,\mathrm {d} t.}$
### Mathematical modelling and computer simulation
Equations or systems too complex for closed-form or analytic solutions can often be analysed by mathematical modelling and computer simulation.
## Closed-form number
Three subfields of the complex numbers C have been suggested as encoding the notion of a "closed-form number"; in increasing order of generality, these are the EL numbers, Liouville numbers, and elementary numbers. The Liouville numbers, denoted L (not to be confused with Liouville numbers in the sense of rational approximation), form the smallest algebraically closed subfield of C closed under exponentiation and logarithm (formally, intersection of all such subfields)—that is, numbers which involve explicit exponentiation and logarithms, but allow explicit and implicit polynomials (roots of polynomials); this is defined in (Ritt 1948, p. 60). L was originally referred to as elementary numbers, but this term is now used more broadly to refer to numbers defined explicitly or implicitly in terms of algebraic operations, exponentials, and logarithms. A narrower definition proposed in (Chow 1999, pp. 441–442), denoted E, and referred to as EL numbers, is the smallest subfield of C closed under exponentiation and logarithm—this need not be algebraically closed, and correspond to explicit algebraic, exponential, and logarithmic operations. "EL" stands both for "Exponential-Logarithmic" and as an abbreviation for "elementary".
Whether a number is a closed-form number is related to whether a number is transcendental. Formally, Liouville numbers and elementary numbers contain the algebraic numbers, and they include some but not all transcendental numbers. In contrast, EL numbers do not contain all algebraic numbers, but do include some transcendental numbers. Closed-form numbers can be studied via transcendental number theory, in which a major result is the Gelfond–Schneider theorem, and a major open question is Schanuel's conjecture.
## Numerical computations
For purposes of numeric computations, being in closed form is not in general necessary, as many limits and integrals can be efficiently computed.
## Conversion from numerical forms
There is software that attempts to find closed-form expressions for numerical values, including RIES,[2] identify in Maple[3] and SymPy,[4] Plouffe's Inverter,[5] and the Inverse Symbolic Calculator.[6] | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9564184546470642, "perplexity": 414.92555134783254}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469258950570.93/warc/CC-MAIN-20160723072910-00002-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/168617/to-show-f-is-continuous | # To show $f$ is continuous
Let $f:[0,1]\rightarrow \mathbb{R}$ is such that for every sequence $x_n\in [0,1]$, whenever both $x_n$ and $f(x_n)$ converges , we have $$\lim_{n\rightarrow\infty} f(x_n)=f(\lim_{n\rightarrow\infty}x_n),$$ we need to prove $f$ is continuous
well, I take $x_n$ and $y_n$ in $[0,1]$ such that $|(x_n-y_n)|\rightarrow 0$, and the given condition holds,Now enough to show $|f(x_n)-f(y_n)|\rightarrow 0$
I construct a new sequence $$z_1=x_1$$ $$z_2=y_1$$ $$\dots$$ $$z_{2n-1}=x_n$$ and $$z_{2n}=y_n$$
We see, that subsequence of $f(z_n)$ converges so it must be convergent to the same limit. Am I going in right path? please help.
-
Consider the function $f$ on $[0,1]$ defined by $f(0)=0$ and $f(x)=1/x$ for all $x \in (0,1]$. Then it satisfies the hypothesis but is not continuous on $[0,1]$. So the claim is false. The claim is false because the antecedent requires both $(x_n)$ and $(f(x_n))$ to be convergent. And for all $(x_n)$ that converges to $0, (f(x_n))$ is not convergent and hence the antecedent is false. Therefore the implication stands true, yet the function is not continuous on $[0,1]$! – Kasun Fernando Jul 9 '12 at 14:07
I will prove a different claim because I have pointed out that what is mentioned here is wrong by a counter-example.
Let $f:[0,1]→\mathbb{R}$ is such that for every sequence $x_n∈[0,1]$ whenever $(x_n)$ converges , we have $\lim\limits_{n→∞}f(x_n)=f \left(\lim\limits_{n→∞}x_n \right)$ then $f$ is continous on $[0,1]$.
I think the best way is to use proof by contradiction. Assume $f$ is not continuous at $c \in [0,1]$ then there exist $\epsilon_{0} > 0$ such that for all $n \in \mathbb{N}$ there exist $x_{n} \in (c-1/n,c+1/n) \cap [0,1]$ such that $|f(x_n)-f(c)| \geq \epsilon_{0}>0$
Obviously $( x_n )$ converges to $c$ but $(f(x_n))$ does not converge to $f(c)$ ( Note that all the terms of $(f(x_n))$ are a positive distance away from $f(c)$ ) which is a contradiction with the given property of the function.
Since our choice of $c$ was arbitrary, we have that $f$ is continuous on $[0,1]$
-
are you assuming that $f(x_n)$ converges? why not $f(x_n)\rightarrow\infty$? – La Belle Noiseuse Jul 9 '12 at 13:39
Who cares about $f(x_n)$, as long as it does not converge to $f(x)$? – Siminore Jul 9 '12 at 13:41
I am wondering where he is using the fact given about $x_n$ and $f(x_n)$ converges together. – La Belle Noiseuse Jul 9 '12 at 13:42
Consider the function $f$ on $[0,1]$ defined by $f(0)=0$ and $f(x)=1/x$ for all non-zero $x$. Then it satisfies the hypothesis but is not continuous on $[0,1]$. So the claim is false. What I have proven is the sequential criterion for continuity! – Kasun Fernando Jul 9 '12 at 14:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9766835570335388, "perplexity": 113.48474028870457}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246644083.49/warc/CC-MAIN-20150417045724-00271-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/electron-in-a-one-dimensional-box.241858/ | # Homework Help: Electron in a One-Dimensional Box
1. Jun 24, 2008
### Domnu
Problem
An electron in a one-dimensional box with walls at $$x=(0,a)$$ is in the quantum state
$$\psi(x) = A, 0<x<a/2$$
$$\psi(x) = -A, a/2 < x < a$$
Obtain an expression for the normalization constant, $$A$$. What is the lowest energy of the electron that will be measured in this state?
Solution
Well, we know that
$$\int_{0}^A |\psi(x)|^2 dx = 1 \iff A^2 \cdot a = 1 \iff A = \sqrt{\frac{1}{a}}$$
so this is our normalization constant, $$A$$. Now, to find the lowest energy of the electrong that will be measured in this state, we have:
$$-\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} = E \psi$$
but the left hand side evaluates to 0 since $$\psi$$ is independent of $$x$$. So this is the lowest (and only observed) energy in this state... is this right?
2. Jun 24, 2008
### daschaich
Well, $$\psi$$ is independent of $$x$$ everywhere except around $$x = a / 2$$. Around there things could get a little weird. (In fact, as this problem is stated, $$\psi$$ changes infinitely quickly from one side of $$a / 2$$ to the other, so around there things could get very weird.)
Even without doing any calculations, though, you should know that since the electron is stuck in the box, it must have nonzero energy. If nothing else, the box gives it $$\Delta x < \infty$$, so it must have $$\Delta p > 0$$ by Heisenberg.
If you've already studied the infinite potential well (one-dimensional box), you might also know that the energy eigenfunctions (or stationary states) are all $$sin$$ functions (since one of the walls is at $$x = 0$$). Now, the given wavefunction $$\psi$$ is not such a $$sin$$ function, so to get the measurable energies we have to rewrite it as a sum of the energy eigenfunctions -- that is, as a Fourier series of $$sin$$ functions.
Hopefully that's enough to get you started.
3. Jun 25, 2008
### maverick280857
Well, its a step function so if you compute the derivative you get a delta function...and then a derivative of a delta function and so on.
By the way, if you solve for the wavefunction of a particle in a 1D box, you get sin or cosine functions. Any state the particle is in can be written as a linear combination of these eigenfunctions. This is a rather strange state...and interesting too | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9857876896858215, "perplexity": 284.488459333889}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125945604.91/warc/CC-MAIN-20180422135010-20180422155010-00580.warc.gz"} |
https://brilliant.org/problems/mean-of-the-solutions-2/ | # Mean of the Solutions
Algebra Level 3
Find the mean of all of the real solutions of: $x^2+x\sqrt{x}=12x$
× | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9844076037406921, "perplexity": 2019.7030877892814}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988718987.23/warc/CC-MAIN-20161020183838-00558-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://math.eretrandre.org/tetrationforum/showthread.php?tid=1207&pid=8925 | • 0 Vote(s) - 0 Average
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Where is the proof of a generalized integral for integer heights? Chenjesu Junior Fellow Posts: 21 Threads: 4 Joined: May 2016 03/03/2019, 06:05 AM (This post was last modified: 03/03/2019, 06:35 AM by Chenjesu.) https://en.wikipedia.org/wiki/List_of_in..._functions Under "other integrals" it gives the integral of any integer height but provides no proof or citation. I remember seeing a paper with similar formula some time ago, but at the end it said something about its conjecture being unproven. Is there any reference that proves this integral formula? Micah Junior Fellow Posts: 9 Threads: 2 Joined: Feb 2019 03/03/2019, 07:29 AM (This post was last modified: 03/03/2019, 07:31 AM by Micah. Edit Reason: found source of interest ) Are we certain that the form posted on Wikipedia is correctly derived? Without the proof there may be no way of knowing it's correctness, We could take a look at the first few cases of the integral in order to inspect the forms, and then conjecture about a generalization from there, have you derived the first few forms of the integral? $\int\limits x^x dx = I_1$ $\int\limits x^{x^x} dx = I _2$ $\dots$ That may be a starting point for a proof. Also, I may be misinterpreting something on the wikipedia page that you posted, but shouldn't there be constants of integration in all of these antiderivatives as they do not have limits? This may also be of some use: https://en.wikipedia.org/wiki/Puiseux_series Thanks for the thought provoking question! -Micah Chenjesu Junior Fellow Posts: 21 Threads: 4 Joined: May 2016 03/03/2019, 08:55 AM (This post was last modified: 03/03/2019, 08:57 AM by Chenjesu.) I am not sure it is correct at all, that is why I am asking for the proof. I think it's likely someone found that old paper with the unproven conjecture and stuck it on wikipedia without reading it carefully. « Next Oldest | Next Newest »
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https://www.physicsforums.com/threads/field-of-zero-characteristics.949581/ | # I Field of zero characteristics
Tags:
1. Jun 14, 2018
### Danijel
I am interested in the following theorem:
Every field of zero characteristics has a prime subfield isomorphic to ℚ.
I am following the usual proof, where we identify every p∈ℚ as a/b , a∈ℤ,beℕ, and define h:ℚ→P as h(a/b)=(a*1)(b*1)-1 (where a*1=1+1+1... a times) I have worked out the multiplicative rule for this homomorphism, but I am not sure how to prove the additive. I am also interested in what does the notation
(a*1)(b*1)-1 exactly mean. Thanks.
2. Jun 14, 2018
### Staff: Mentor
All we know about the prime field is its characteristic and $0,1 \in \mathbb{P}$. $a\in \mathbb{P}$ is not sure. However $\underbrace{1+\ldots +1}_{a \text{ times}}=a \,\cdot \, 1$ for $a \in \mathbb{N}$ is, and likewise for negative $a$. That's why $a$ times $1$ is used instead of $a$. We only have isomorphic images of $a$ in $\mathbb{P}$.
For the homomorphism, I guess $\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+bc}{bd}$ will be needed.
3. Jun 14, 2018
### mathwonk
this probably won't help much until you have some commutative algebra but it gives the outline. First of all since the field F is an abelian group, there is exactly one additive homomorphism from the integers Z into F for each choice of the image of 1 from Z, so choose that image to be the unit element 1 in P. Now you have the unique additive homomorphism from Z to P that fresh described sending each positive integer n to 1+...+1 (n times). By definition of the multiplication in Z this is also a multiplicative map hence a ring map. Then since each non zero integer goes to an invertible element of F, (because F has characteristic zero), there is a unique extension of the previously defined ring homomorphism Z-->F to a ring homomorphism Q --> F. Then since Q is a field, this ring map is necessarily injective, hence defines an isomorphism onto the smallest subfield of F, i.e. the prime field. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9761819839477539, "perplexity": 520.1068542344192}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376827727.65/warc/CC-MAIN-20181216121406-20181216143406-00439.warc.gz"} |
https://kr.mathworks.com/help/matlab/math/solve-system-of-pdes.html?lang=en | # Solve System of PDEs
This example shows how to formulate, compute, and plot the solution to a system of two partial differential equations.
Consider the system of PDEs
$\frac{\partial {\mathit{u}}_{1}}{\partial \mathit{t}}=0.024\frac{{\partial }^{2}{\mathit{u}}_{1}}{\partial {\mathit{x}}^{2}}-\mathit{F}\left({\mathit{u}}_{1}-{\mathit{u}}_{2}\right),$
$\frac{\partial {\mathit{u}}_{2}}{\partial \mathit{t}}=0.170\frac{{\partial }^{2}{\mathit{u}}_{2}}{\partial {\mathit{x}}^{2}}+\mathit{F}\left({\mathit{u}}_{1}-{\mathit{u}}_{2}\right).$
(The function $\mathit{F}\left(\mathit{y}\right)={\mathit{e}}^{5.73\mathit{y}}-{\mathit{e}}^{-11.46\mathit{y}}$ is used as a shorthand.)
The equation holds on the interval $0\le \mathit{x}\le 1$ for times $\mathit{t}\ge 0$. The initial conditions are
${\mathit{u}}_{1}\left(\mathit{x},0\right)=1,$
${\mathit{u}}_{2}\left(\mathit{x},0\right)=0.$
The boundary conditions are
$\begin{array}{l}\frac{\partial }{\partial \mathit{x}}{\mathit{u}}_{1}\left(0,\mathit{t}\right)=0,\\ {\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathit{u}}_{2}\left(0,\mathit{t}\right)=0,\\ \frac{\partial }{\partial \mathit{x}}{\mathit{u}}_{2}\left(1,\mathit{t}\right)=0,\\ {\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathit{u}}_{1}\left(1,\mathit{t}\right)=1.\end{array}$
To solve this equation in MATLAB®, you need to code the equation, the initial conditions, and the boundary conditions, then select a suitable solution mesh before calling the solver pdepe. You either can include the required functions as local functions at the end of a file (as done here), or save them as separate, named files in a directory on the MATLAB path.
### Code Equation
Before you can code the equation, you need to make sure that it is in the form that the pdepe solver expects:
$\mathit{c}\left(\mathit{x},\mathit{t},\mathit{u},\frac{\partial \mathit{u}}{\partial \mathit{x}}\right)\frac{\partial \mathit{u}}{\partial \mathit{t}}={\mathit{x}}^{-\mathit{m}}\frac{\partial }{\partial \mathit{x}}\left({\mathit{x}}^{\mathit{m}}\mathit{f}\left(\mathit{x},\mathit{t},\mathit{u},\frac{\partial \mathit{u}}{\partial \mathit{x}}\right)\right)+\mathit{s}\left(\mathit{x},\mathit{t},\mathit{u},\frac{\partial \mathit{u}}{\partial \mathit{x}}\right).$
In this form, the PDE coefficients are matrix-valued and the equation becomes
$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\frac{\partial }{\partial \mathit{t}}\left[\begin{array}{c}{\mathit{u}}_{1}\\ {\mathit{u}}_{2}\end{array}\right]=\frac{\partial }{\partial \mathit{x}}\left[\begin{array}{c}0.024\frac{\partial {\mathit{u}}_{1}}{\partial \mathit{x}}\\ 0.170\frac{\partial {\mathit{u}}_{2}}{\partial \mathit{x}}\end{array}\right]+\left[\begin{array}{c}-\mathit{F}\left({\mathit{u}}_{1}-{\mathit{u}}_{2}\right)\\ \mathit{F}\left({\mathit{u}}_{1}-{\mathit{u}}_{2}\right)\end{array}\right].$
So the values of the coefficients in the equation are
$\mathit{m}=0$
$\mathit{c}\left(\mathit{x},\mathit{t},\mathit{u},\frac{\partial \mathit{u}}{\partial \mathit{x}}\right)=\left[\begin{array}{c}1\\ 1\end{array}\right]$ (diagonal values only)
$\mathit{f}\left(\mathit{x},\mathit{t},\mathit{u},\frac{\partial \mathit{u}}{\partial \mathit{x}}\right)=\left[\begin{array}{c}0.024\frac{\partial {\mathit{u}}_{1}}{\partial \mathit{x}}\\ 0.170\frac{\partial {\mathit{u}}_{2}}{\partial \mathit{x}}\end{array}\right]$
$\mathit{s}\left(\mathit{x},\mathit{t},\mathit{u},\frac{\partial \mathit{u}}{\partial \mathit{x}}\right)=\left[\begin{array}{c}-\mathit{F}\left({\mathit{u}}_{1}-{\mathit{u}}_{2}\right)\\ \mathit{F}\left({\mathit{u}}_{1}-{\mathit{u}}_{2}\right)\end{array}\right]$
Now you can create a function to code the equation. The function should have the signature [c,f,s] = pdefun(x,t,u,dudx):
• x is the independent spatial variable.
• t is the independent time variable.
• u is the dependent variable being differentiated with respect to x and t. It is a two-element vector where u(1) is ${\mathit{u}}_{1}\left(\mathit{x},\mathit{t}\right)$ and u(2) is ${\mathit{u}}_{2}\left(\mathit{x},\mathit{t}\right)$.
• dudx is the partial spatial derivative $\partial \mathit{u}/\partial \mathit{x}$. It is a two-element vector where dudx(1) is $\partial {\mathit{u}}_{1}/\partial \mathit{x}$ and dudx(2) is $\partial {\mathit{u}}_{2}/\partial \mathit{x}$.
• The outputs c, f, and s correspond to coefficients in the standard PDE equation form expected by pdepe.
As a result, the equations in this example can be represented by the function:
function [c,f,s] = pdefun(x,t,u,dudx)
c = [1; 1];
f = [0.024; 0.17] .* dudx;
y = u(1) - u(2);
F = exp(5.73*y)-exp(-11.47*y);
s = [-F; F];
end
(Note: All functions are included as local functions at the end of the example.)
### Code Initial Conditions
Next, write a function that returns the initial condition. The initial condition is applied at the first time value and provides the value of $\mathit{u}\left(\mathit{x},{\mathit{t}}_{0}\right)$ for any value of x. The number of initial conditions must equal the number of equations, so for this problem there are two initial conditions. Use the function signature u0 = pdeic(x) to write the function.
The initial conditions are
${\mathit{u}}_{1}\left(\mathit{x},0\right)=1,$
${\mathit{u}}_{2}\left(\mathit{x},0\right)=0.$
The corresponding function is
function u0 = pdeic(x)
u0 = [1; 0];
end
### Code Boundary Conditions
Now, write a function that evaluates the boundary conditions
$\begin{array}{l}\frac{\partial }{\partial \mathit{x}}{\mathit{u}}_{1}\left(0,\mathit{t}\right)=0,\\ {\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathit{u}}_{2}\left(0,\mathit{t}\right)=0,\\ \frac{\partial }{\partial \mathit{x}}{\mathit{u}}_{2}\left(1,\mathit{t}\right)=0,\\ {\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathit{u}}_{1}\left(1,\mathit{t}\right)=1.\end{array}$
For problems posed on the interval $\mathit{a}\le \mathit{x}\le \mathit{b}$, the boundary conditions apply for all $\mathit{t}$ and either $\mathit{x}=\mathit{a}$ or $\mathit{x}=\mathit{b}$. The standard form for the boundary conditions expected by the solver is
$\mathit{p}\left(\mathit{x},\mathit{t},\mathit{u}\right)+\mathit{q}\left(\mathit{x},\mathit{t}\right)\mathit{f}\left(\mathit{x},\mathit{t},\mathit{u},\frac{\partial \mathit{u}}{\partial \mathit{x}}\right)=0.$
Written in this form, the boundary conditions for the partial derivatives of $\mathit{u}$ need to be expressed in terms of the flux $\mathit{f}\left(\mathit{x},\mathit{t},\mathit{u},\frac{\partial \mathit{u}}{\partial \mathit{x}}\right)$. So the boundary conditions for this problem are
For $\mathit{x}=0$, the equation is
$\left[\begin{array}{c}0\\ {\mathit{u}}_{2}\end{array}\right]+\left[\begin{array}{c}1\\ 0\end{array}\right]\cdot \left[\begin{array}{c}0.024\frac{\partial {\mathit{u}}_{1}}{\partial \mathit{x}}\\ 0.170\frac{\partial {\mathit{u}}_{2}}{\partial \mathit{x}}\end{array}\right]=0.$
The coefficients are:
${\mathit{p}}_{\mathit{L}}\left(\mathit{x},\mathit{t},\mathit{u}\right)=\left[\begin{array}{c}0\\ {\mathit{u}}_{2}\end{array}\right],$
${\mathit{q}}_{\mathit{L}}\left(\mathit{x},\mathit{t}\right)=\left[\begin{array}{c}1\\ 0\end{array}\right].$
Likewise, for $\mathit{x}=1$ the equation is
$\left[\begin{array}{c}{\mathit{u}}_{1}-1\\ 0\end{array}\right]+\left[\begin{array}{c}0\\ 1\end{array}\right]\cdot \left[\begin{array}{c}0.024\frac{\partial {\mathit{u}}_{1}}{\partial \mathit{x}}\\ 0.170\frac{\partial {\mathit{u}}_{2}}{\partial \mathit{x}}\end{array}\right]=0.$
The coefficients are:
${\mathit{p}}_{\mathit{R}}\left(\mathit{x},\mathit{t},\mathit{u}\right)=\left[\begin{array}{c}{\mathit{u}}_{1}-1\\ 0\end{array}\right],$
${\mathit{q}}_{\mathit{R}}\left(\mathit{x},\mathit{t}\right)=\left[\begin{array}{c}0\\ 1\end{array}\right].$
The boundary function should use the function signature [pl,ql,pr,qr] = pdebc(xl,ul,xr,ur,t):
• The inputs xl and ul correspond to $\mathit{u}$ and $\mathit{x}$ for the left boundary.
• The inputs xr and ur correspond to $\mathit{u}$ and $\mathit{x}$ for the right boundary.
• t is the independent time variable.
• The outputs pl and ql correspond to ${\mathit{p}}_{\mathit{L}}\left(\mathit{x},\mathit{t},\mathit{u}\right)$ and ${\mathit{q}}_{\mathit{L}}\left(\mathit{x},\mathit{t}\right)$ for the left boundary ($\mathit{x}=0$ for this problem).
• The outputs pr and qr correspond to ${\mathit{p}}_{\mathit{R}}\left(\mathit{x},\mathit{t},\mathit{u}\right)$ and ${\mathit{q}}_{\mathit{R}}\left(\mathit{x},\mathit{t}\right)$ for the right boundary ($\mathit{x}=1$ for this problem).
The boundary conditions in this example are represented by the function:
function [pl,ql,pr,qr] = pdebc(xl,ul,xr,ur,t)
pl = [0; ul(2)];
ql = [1; 0];
pr = [ur(1)-1; 0];
qr = [0; 1];
end
### Select Solution Mesh
The solution to this problem changes rapidly when $\mathit{t}$ is small. Although pdepe selects a time step that is appropriate to resolve the sharp changes, to see the behavior in the output plots you need to select appropriate output times. For the spatial mesh, there are boundary layers in the solution at both ends of $0\le \mathit{x}\le 1$, so you need to specify mesh points there to resolve the sharp changes.
x = [0 0.005 0.01 0.05 0.1 0.2 0.5 0.7 0.9 0.95 0.99 0.995 1];
t = [0 0.005 0.01 0.05 0.1 0.5 1 1.5 2];
### Solve Equation
Finally, solve the equation using the symmetry $\mathit{m}$, the PDE equation, the initial conditions, the boundary conditions, and the meshes for $\mathit{x}$ and $\mathit{t}$.
m = 0;
sol = pdepe(m,@pdefun,@pdeic,@pdebc,x,t);
pdepe returns the solution in a 3-D array sol, where sol(i,j,k) approximates the kth component of the solution ${\mathit{u}}_{\mathit{k}}$ evaluated at t(i) and x(j). Extract each solution component into a separate variable.
u1 = sol(:,:,1);
u2 = sol(:,:,2);
### Plot Solution
Create surface plots of the solutions for ${\mathit{u}}_{1}$ and ${\mathit{u}}_{2}$ plotted at the selected mesh points for $\mathit{x}$ and $\mathit{t}$.
surf(x,t,u1)
title('u_1(x,t)')
xlabel('Distance x')
ylabel('Time t')
surf(x,t,u2)
title('u_2(x,t)')
xlabel('Distance x')
ylabel('Time t')
### Local Functions
Listed here are the local helper functions that the PDE solver pdepe calls to calculate the solution. Alternatively, you can save these functions as their own files in a directory on the MATLAB path.
function [c,f,s] = pdefun(x,t,u,dudx) % Equation to solve
c = [1; 1];
f = [0.024; 0.17] .* dudx;
y = u(1) - u(2);
F = exp(5.73*y)-exp(-11.47*y);
s = [-F; F];
end
% ---------------------------------------------
function u0 = pdeic(x) % Initial Conditions
u0 = [1; 0];
end
% ---------------------------------------------
function [pl,ql,pr,qr] = pdebc(xl,ul,xr,ur,t) % Boundary Conditions
pl = [0; ul(2)];
ql = [1; 0];
pr = [ur(1)-1; 0];
qr = [0; 1];
end
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http://www.almoststochastic.com/2016/10/a-primer-on-filtering.html?showComment=1518390834702 | 2016/10/26
A primer on filtering
Say that you have a dynamical process of interest $X_1,\ldots,X_n$ and you can only observe the process with some noise, i.e., you get an observation sequence $Y_1,\ldots,Y_n$. What is the optimal way to estimate $X_n$ conditioned on the whole sequence of observations $Y_{1:n}$?
To start from the bottom, suppose you have an observation $Y$ and you think that it is related to something you want to learn about (let's call that something $X$ as the uncle of Einstein would). There are many methods in science about how to go from $Y$ to $X$, e.g. you can solve an optimization problem. A Bayesian approach would be to specify a probability distribution over $X$, denoted $p(x)$, and the likelihood $p(y|x)$ which encodes how $X$ becomes observable to us through the observation process. Then, our curiosity about $X$ can be reframed as obtaining its posterior distribution $p(x|y)$ which is possible by the celebrated Bayes' formula, \begin{align*} p(x|y) = \frac{p(y|x)p(x)}{p(y)}. \end{align*} In the filtering setup, $X$ is assumed to be a changing, dynamic process; hence denoted with $(X_n)_{n\in\bN}$. Since $X_n$ changes with time $n$, accordingly, you collect observations from this process at each time step $n$ which is denoted by $(Y_n)_{n\in\bN}$. Simply, the process could be depicted as in the following graphical model. \begin{align*} \begin{array}{ccccccc} & X_0 \to & X_1 & \to & X_2 & \to & \dots \\ && \downarrow && \downarrow \\ & & Y_1 & & Y_2 & & \dots \end{array} \end{align*} As you can see clearly, $X_n$ is a Markov process. $Y_n$ are independent given the $X$ process. What filtering aims at is to obtain the most refined (optimal) information about $X_n$ given the whole sequence of observations $Y_{1:n}=y_{1:n}$. Ideally, this would be the exact posterior distribution $p(x_n|y_{1:n})$. In order to have it, as you would guess, we have to specify two things: First, how $X_n$ evolves (the transition model), second, how $Y_n$ and $X_n$ are related (the observation model).
Generally, such models are specified in the following way, \begin{align*} X_0 &\sim p(x_0), \\ X_n | \{X_{n-1} = x_{n-1}\} &\sim p(x_n | x_{n-1}), \\ Y_n | \{X_n = x_n\} &\sim p(y_n|x_n). \end{align*} The prior on $X_0$ is usually of little practical importance, as there are strong results about what happens if you set it wrong: as long as the filter is stable, it forgets its initial condition independent of the fact that it is right or wrong. The transition model $p(x_n|x_{n-1})$ specifies how your process evolves (or rather how you think it evolves). For example, if you are working on object tracking, it is the model of the vehicle. If you work on weather and fluid dynamics, it is the (possibly stochastic) partial differential equation from fluid mechanics. If you try to land the Apollo to the moon, it would be the controlled state process (you need an additional control input in that case). To summarise, if you work on X dynamics, it is how your X evolves. The observation process $p(y_n|x_n)$ is somewhat more concrete thing to have. You may have a model of the vehicle but you can only observe it through some sensor signals. This model would encode that process: What you measure as a sensor signal and how it is related to the true position. Of course this is subject to many uncertainties, for example, in this case, you have parameters from the weather environment (e.g. connectivity) on signal's way to the satellite and/or vast amount of noise. You should encode all of it into $p(y_n|x_n)$. For the weather dynamics, you will not be able to observe ocean currents or large eddies despite you can have a pretty good idea about how they evolve (since it is physics). So both distributions $p(x_n|x_{n-1})$ and $p(y_n|x_n)$ really complete each other, you need the information from both of them in order to estimate $x_n$ at any given time $n$.
As I have written above, the goal here is to obtain $p(x_n|y_{1:n})$. One way to do is to use Bayes' theorem iteratively to achieve this task. In order to have a recursive formula for it, we need to figure out how to obtain it from $p(x_{n-1}|y_{1:n-1})$. Why? Because that's all we need. If we have a procedure for that, we'd start with $p(x_1)$ and obtain $p(x_1|y_1)$ by using Bayes' theorem as I showed in the beginning, then continue to $p(x_2|y_{1:2})$ with "the method" (which I will explain in the following) which maps densities in the following way $p(x_{n-1}|y_{1:n-1}) \mapsto p(x_n|y_{1:n})$, and so on.
There are two steps to doing this.
(i) Prediction. Given $p(x_{n-1}|y_{1:n-1})$, the usual way is to first obtain the predictive distribution of the state $p(x_n|y_{1:n-1})$ using the following relationship, \begin{align*} p(x_n|y_{1:n-1}) = \int p(x_n|x_{n-1}) p(x_{n-1}|y_{1:n-1}) \mbox{d}x_{n-1}. \end{align*} The name of this can get as fancy as it can get: the Chapman-Kolmogorov equation. At the core it uses conditional independence structure of the model and the marginalisation rule, nothing more than that. All it does is to integrate out the variable $x_{n-1}$ from the joint distribution $p(x_n,x_{n-1}|y_{1:n-1})$ which can be factorised as in the integral because of the conditional independence. Nevertheless, recall that having a formula doesn't mean that we can compute it. We have to compute the integral, and most of the cases, it is impossible.
(ii) Update. Update step is taking the observation into the account. You have the predictive distribution from above, so it's your "guess" about what could happen. Now you need to correct your prediction using the data. This means we need to obtain $p(x_n|y_{1:n})$ from $p(x_n|y_{1:n-1})$. This is possible (of course) by the use of Bayes' theorem, \begin{align*} p(x_n|y_{1:n}) = p(x_n|y_{1:n-1}) \frac{p(y_n|x_n)}{p(y_n|y_{1:n-1})}. \end{align*} Why did I call this "the use of Bayes' theorem"? If you look at what I described as Bayes' theorem above, this relationship might not be obvious. Try to look at in this way, \begin{align*} p(x_n|y_n,y_{1:n-1}) = p(x_n|y_{1:n-1}) \frac{p(y_n|x_n)}{p(y_n|y_{1:n-1})}, \end{align*} and now *ignore* all $y_{1:n-1}$ terms. Try to write it down without $y_{1:n-1}$, you will see that it is in fact the Bayes' rule, just conditioned on $y_{1:n-1}$.
You might ask the question why this has a prediction-correction structure. Why do we need to predict first and then update? I have no answer beyond that this is a possible way to do it.
Now, we have the recipe. Let's see what we can do with that.
If the model is linear and Gaussian, all these computations are analytically possible. This is what makes the Kalman filters special. Suppose we have the following probabilistic model (stochastic linear dynamical system), \begin{align*} X_0 &\sim \mathcal{N}(x_0;\mu_0,P_0), \\ X_n | \{X_{n-1} = x_{n-1}\} &\sim \mathcal{N}(x_n;A x_{n-1},Q), \\ Y_n | \{X_n = x_n\} &\sim \mathcal{N}(y_n;C x_n,R). \end{align*} Collecting $Y_{1:n} = y_{1:n}$, we are naturally after the posterior distribution $p(x_n|y_{1:n})$. If you bravely compute the integral and update rule above, you will see that they can be carried out analytically. Resulting posterior $p(x_n|y_{1:n})$ is Gaussian so it can be parameterised by its mean and covariance. So instead of updating the probability measure (which is not obvious at all), all you need to do is to update its mean and covariance recursively, as you can get the whole information using those and the functional form encoded by the Gaussian density (this is called finite-dimensional filtering, as you do it by updating the finite-dimensional sufficient statistics). If you denote the posterior with $p(x_n|y_{1:n}) = \mathcal{N}(x_n;\mu_n,P_n)$, the prediction step is given by, \begin{align*} \mu_{n|n-1} &= A \mu_{n-1}, \\ P_{n|n-1} &= A P_n A^\top + Q. \end{align*} Then the update step is the following, \begin{align*} \mu_n &= \mu_{n|n-1} + P_{n|n-1} C^\top (C P_{n|n-1} C^\top + R)^{-1} (y_n - C x_{n|n-1}), \\ P_n &= P_{n|n-1} - P_{n|n-1} C^\top (C P_{n|n-1} C^\top + R)^{-1} C P_{n|n-1}. \end{align*} Notice that, in the prediction step, we don't have $y_n$ nor any parameter from the observation model $p(y_n|x_n)$, it only includes the parameters of the transition model $p(x_n|x_{n-1})$. And the reverse is true for the update step, as it only involves the parameters of the observation model $p(y_n|x_n)$ and it does not have any parameter of the transition model $p(x_n|x_{n-1})$.
So let's conclude with a note about particle filters. What do particle filters do? It approximates the posterior probability distribution $p(x_n|y_{1:n})$ where $p(x_n|x_{n-1})$ and $p(y_n|x_n)$ are specified in general ways so, essentially, when it is not possible to perform above computations. Don't be mad at me, but most of the particle filters don't use the above recursions but rather they use something else. However, conceptually, this post should be helpful for you to understand what they are trying to do.
Unknown said...
I think you could show the sequential nature of the Bayes' theorem and from that, show the Kalman filter as a special case of it. There are other examples of sequential inference in reliability analysis with some uncommon probabilistic models.
Deniz said...
Hi. I did not understand the first part of the comment. But I'd love to see other analytically tractable nonGaussian models, if there are any.
Anonymous said...
Awesome post. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9944026470184326, "perplexity": 319.74226516817953}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027313889.29/warc/CC-MAIN-20190818124516-20190818150516-00277.warc.gz"} |
http://umj.imath.kiev.ua/article/?lang=en&article=11283 | 2019
Том 71
№ 10
# Approximation of $\bar {\psi} - integrals$−integrals of periodic functions by Fourier sums (small smoothness). Iof periodic functions by Fourier sums (small smoothness). I
Stepanets O. I.
Abstract
We investigate the rate of convergence of Fourier series on the classes $L^{\bar {\psi} } \text{N}$ in the uniform and integral metrics. The results obtained are extended to the case where the classes $L^{\bar {\psi} } \text{N}$ are the classes of convolutions of functions from $\text{N}$ with kernels with slowly decreasing coefficients. In particular, we obtain asymptotic equalities for the upper bounds of deviations of the Fourier sums on the sets $L^{\bar {\psi} } \text{N}$ which are solutions of the Kolmogorov-Nikol’skii problem. In addition, we establish an analog of the well-known Lebesgue inequality.
English version (Springer): Ukrainian Mathematical Journal 50 (1998), no. 2, pp 314–333.
Citation Example: Stepanets O. I. Approximation of $\bar {\psi} - integrals$−integrals of periodic functions by Fourier sums (small smoothness). Iof periodic functions by Fourier sums (small smoothness). I // Ukr. Mat. Zh. - 1998. - 50, № 2. - pp. 274-291.
Full text | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9688020944595337, "perplexity": 1012.8171999788804}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912203947.59/warc/CC-MAIN-20190325112917-20190325134917-00352.warc.gz"} |
http://physicallensonthecell.org/chemical-physics/ideal-gas-basis-mass-action-and-window-free-energywork-relations | # The Ideal Gas: The basis for "mass action" and a window into free-energy/work relations
## The Ideal Gas: The basis for "mass action" and a window into free-energy/work relations
The simplest possible multi-particle system, the ideal gas, is a surprisingly valuable tool for gaining insight into biological systems - from mass-action models to gradient-driven transporters. The word "ideal" really means non-interacting, so in an ideal gas all molecules behave as if no others are present. The gas molecules only feel a force from the walls of their container, which merely redirects their momenta like billiard balls. Not surprisingly, it is possible to do exact calculations fairly simply under such extreme assumptions. What's amazing is how relevant those calculations turn out to be, particularly for understanding the basic mechanisms of biological machines and chemical-reaction systems.
Although ideal particles do not react or bind, their statistical/thermodynamic behavior in the various states (e.g., bound or not, reacted or not) can be used to build powerful models - e.g., for transporters.
### Mass-action kinetics are ideal-gas kinetics
The key assumption behind mass-action models is that events (binding, reactions, ...) occur precisely in proportion to the concentration(s) of the participating molecules. This certainly cannot be true for all concentrations, because all molecules interact with one another at close enough distances - i.e., at high enough concentrations. In reality, beyond a certain concentration, simple crowding effects due to steric/excluded-volume effects mean that each molecule can have only a maximum number of neighbors.
But in the ideal gas - and in mass-action kinetics - no such crowding effects occur. All molecules are treated as point particles. They do not interact with one another, although virtual/effective interactions occur in a mass-action picture. (We can say these interactions are "virtual" because the only effect is to change the number of particles - no true forces or interactions occur.)
### Pressure and work in an ideal gas
Ideal gases can perform work directly using pressure. The molecules of an ideal gas exert a pressure on the walls of the container holding them due to collisions, as sketched above. The amount of this pressure depends on the number of molecules colliding with each unit area of the wall per second, as well as the speed of these collisions. These quantities can be calculated based on the mass $m$ of each molecule, the total number of molecules, $N$, the total volume of the container $V$ and the temperature, $T$. In turn, $T$ determines the average speed via the relation $(3/2) \, N \, k_B T = \avg{(1/2) \, m \, v^2}$. See the book by Zuckerman for more details.
All these facts can be combined - consult a physics book - to yield the ideal gas law relating pressure $P$ to the other quantities:
(1)
We can calculate the work done by an ideal gas to change the size of its container by pushing one wall a distance $d$ as shown above. We use the basic rule of physics that work is force ($f$) multiplied by distance and the definition of pressure as force per unit area. If we denote the area of the wall by $A$, we have
(2)
If $d$ is small enough so that the pressure is nearly constant, we can calculate $P$ using (1) at either the beginning or end of the expansion. More generally, for a volume change of arbitrary size (from $V_i$ to $V_f$) in an ideal gas, we need to integrate:
(3)
which assumes the expansion is performed slowly enough so that (1) applies throughout the process.
### Free energy and work in an ideal gas
The free energy of the ideal gas can be calculated exactly in the limit of large $N$ (see below). We will see that it does, in fact, correlate precisely with the expression for work just derived. The free energy depends on temperature, volume, and the number of molecules; for large $N$, it is given by
(4)
where $\lambda$ is a constant for fixed temperature. For reference, it is given by $\lambda = h / \sqrt{2 \pi m k_B T}$ with $h$ being Planck's constant and $m$ the mass of an atom. See the book by Zuckerman for full details.
Does the free energy tell us anything about work? If we examine the free energy change occuring during the same expansion as above, from $V_i$ to $V_f$ at constant $T$, we get
(5)
Comparing to (3), this is exactly the negative of the work done! In other words, the free energy of the ideal gas decreases by exactly the amount of work done (when the expansion is performed slowly). More generally, the work can be no greater than the free energy decrease. The ideal gas has allowed us to demonstrate this principle concretely.
### The ideal gas free energy from statistical mechanics
The free energy is derived from the "partition function" $Z$, which is simply a sum/integral over Boltzmann factors for all possible configurations/states of a system. Summing over all possibilities is why the free energy encompasses the full thermodynamic behavior of a system.
The mathematical relation is
(6)
where $Z$ is defined by
(7)
where $\lambda(T) \propto 1/\sqrt(T)$ is the thermal de Broglie wavelength (which is not important for the phenomena of interest here), $\rall$ is the set of $(x,y,z)$ coordinates for all molecules and $U$ is the potential energy function. The factor $1/N!$ accounts for interchangeability of identical molecules, and the integral is over all volume allowed to each molecule. For more information, see the book by Zuckerman, or any statistical mechanics book.
The partition function can be evaluated exactly for the case of the ideal gas because the non-interaction assumption can be formulated as $U(\rall) = 0$ for all configurations - in other words, the locations of the molecules do not change the energy or lead to forces. This makes the Boltzmann factor exactly $1$ for all $\rall$, and so each molecule's inegration over the full volume yields a factor of $V$, making the final result
(8)
Although (8) assumes there are no degrees of freedom internal to the molecule - which might be more reasonable in some cases (ions) than others (flexible molecules) - the expression is sufficient for most of the biophysical explorations undertaken here.
The combination of (6) and (8) can be used to derive (4) in conjunction with Stirling's approximation for large $N$. See a statistical mechanics book for details.
### References
• Any basic physics textbook.
• D.M. Zuckerman, "Statistical Physics of Biomolecules: An Introduction," (CRC Press, 2010), Chapters 5, 7. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8734497427940369, "perplexity": 374.9161223585663}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514575844.94/warc/CC-MAIN-20190923002147-20190923024147-00318.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-11th-edition/chapter-4-section-4-4-evaluating-logarithms-and-the-change-of-base-theorem-4-4-exercises-page-435/48 | ## College Algebra (11th Edition)
Published by Pearson
# Chapter 4 - Section 4.4 - Evaluating Logarithms and the Change-of-Base Theorem - 4.4 Exercises - Page 435: 48
#### Answer
$-4$
#### Work Step by Step
$\bf{\text{Solution Outline:}}$ Use the properties of logarithms to evaluate the given expression, $\ln\left( \dfrac{1}{e^4} \right) .$ $\bf{\text{Solution Details:}}$ Using the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the expression above is equivalent \begin{array}{l}\require{cancel} \ln1-\ln e^4 .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \ln1-4\ln e .\end{array} Since $\ln e=1$ and $\ln1=0,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 0-4(1) \\\\= -4 .\end{array}
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9690962433815002, "perplexity": 1249.4863583410195}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583511216.45/warc/CC-MAIN-20181017195553-20181017221053-00538.warc.gz"} |
http://www.jstor.org/stable/2958783 | ## Access
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# Consistent Nonparametric Regression
Charles J. Stone
The Annals of Statistics
Vol. 5, No. 4 (Jul., 1977), pp. 595-620
Stable URL: http://www.jstor.org/stable/2958783
Page Count: 26
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## Abstract
Let (X, Y) be a pair of random variables such that X is Rd-valued and Y is Rd'-valued. Given a random sample (X1, Y1), ⋯, (Xn, Yn) from the distribution of (X, Y), the conditional distribution $P^Y(\bullet \mid X)$ of Y given X can be estimated nonparametrically by P̂n Y(A ∣ X) = ∑n 1 Wni(X)IA(Yi), where the weight function Wn is of the form Wni(X) = Wni(X, X1, ⋯, Xn), 1 ≤ i ≤ n. The weight function Wn is called a probability weight function if it is nonnegative and ∑n 1 Wni(X) = 1. Associated with $\hat{P}_n^Y(\bullet \mid X)$ in a natural way are nonparametric estimators of conditional expectations, variances, covariances, standard deviations, correlations and quantiles and nonparametric approximate Bayes rules in prediction and multiple classification problems. Consistency of a sequence {Wn} of weight functions is defined and sufficient conditions for consistency are obtained. When applied to sequences of probability weight functions, these conditions are both necessary and sufficient. Consistent sequences of probability weight functions defined in terms of nearest neighbors are constructed. The results are applied to verify the consistency of the estimators of the various quantities discussed above and the consistency in Bayes risk of the approximate Bayes rules.
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https://support.nutbourne.com/hc/en-us/articles/201720067-Mapping-a-network-drive | # Mapping a network drive
If you are unsure about approaching this issue please call our helpdesk as you can irreversibly damage your PC if you are not careful
Before mapping a network drive, it is essential to know the file path for the network resource / Folder you wish to map. The file path will be in the following format:
\\server\share name
Although there is a slight variation in theme and appearance, the process of mapping a network drive is essentially the same on Windows XP/Windows Vista/Windows 7. In order to map a network drive, follow the instructions below.
- Access the start menu in the bottom left hand corner of your screen.
- From the start menu, right click on the computer/my computer option and select map network drive.
- You will now be presented with a new window.
- For the field named drive, select the letter you wish to map the resource to.
- For the field named folder, enter the file path of the Folder you wish to map, an example of the format of the file path is detailed in bold above.
- Now click Finish.
The folder will now be mapped as a network drive.
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Have more questions? Submit a request | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8321253657341003, "perplexity": 1918.028443687277}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187825889.47/warc/CC-MAIN-20171023092524-20171023112524-00208.warc.gz"} |
http://math.stackexchange.com/questions/54611/a-relation-r-such-that-r-cup-r-1-is-not-an-equivalence-relation/54614 | # A relation $R$ such that $R\cup R^{-1}$ is not an equivalence relation
I have a homework assignment to find a Relation $R$ over $A = \{1,2,3\}$ where $R\cup {{R}^{-1}}$ is not an equivalence relation (transitive, reflexive and symmetrical).
$R$ must be Transitive and Reflexive (${{I}_{A}}\subseteq R$)
Any clue anyone? I just am unable to find an example
Thanks
-
If $I_A\subseteq R$, then you are guaranteed that $R$ is reflexive; however, you are not guaranteed that $R$ is transitive. – Arturo Magidin Jul 30 '11 at 16:28
@Arturo I meant that there is a restriction on your selected R and that restriction is it must be Transitive and Reflexive – Jason Jul 30 '11 at 16:51
Okay; it wasn't clear, because the parenthetical comment "$I_A\subseteq R$" looks like the justification for "$R$ must be transitive and reflexive". – Arturo Magidin Jul 30 '11 at 17:20
Nha, I'm just translating from my language so I was not sure of the term in English so I added it just to be sure. – Jason Jul 30 '11 at 17:26
Yes, I see what it was you tried to say: it was just the justification for the final clause, not for the entire sentence. It just seemed like it was a justification for the entire sentence. – Arturo Magidin Jul 30 '11 at 17:44
Given that you are assuming that $R$ is reflexive, the only thing that can fail for $R\cup R^{-1}$ to be an equivalence relation is transitivity: you should verify that since $I_A\subseteq R$, then $I_A\subseteq R\cup R^{-1}$; and that $R\cup R^{-1}$ is symmetric for every relation $R$. So the only possible pitfall lies in transitivity.
Now, you are assuming that $R$ itself is transitive. So, how can transitivity fail? Say you have $(a,b),(b,c)\in R\cup R^{-1}$; if $(a,b),(b,c)\in R$, then since we are assuming $R$ is transitive, then $(a,c)\in R\subseteq R\cup R^{-1}$. If $(a,b),(b,c)\in R^{-1}$, then $(c,b),(b,a)\in R$, and again by transitivity we conclude $(c,a)\in R$, hence $(a,c)\in R^{-1}\subseteq R\cup R^{-1}$.
So what's left? What happens if $(a,b)\in R$, and $(b,c)\in R^{-1}$, but we do not have $(a,b)\in R^{-1}$ nor $(b,c)\in R$? Can you construct such an example? What will happen then?
-
Thanks I did manage to narrow it to the Transitive property but could not find an example for some reason :) – Jason Jul 30 '11 at 16:49
One sometimes helpful way to approach questions like this, if you can't find a counterexample, is simply to try and prove the opposite: that is, prove that $R\cup R^{-1}$ is an equivalence relation. Well, it's obviously reflexive; it's also obviously symmetric, because anything in $R$ will have its inverse in $R^{-1}$. So transitivity must be where it breaks down.
Suppose there are two relations a~b and b~c in $R\cup R^{-1}$ whose product a~c is not in $R\cup R^{-1}$. (You might also write these relations (a,b), etc.) It's easy to see that a, b and c must all be distinct elements, so in fact, we might as well choose 1~2, 2~3 and 1~3: we want the first two to be in $R\cup R^{-1}$, but not the third.
As $R$ is transitive, we can't have both 1~2 and 2~3 in $R$. Likewise they can't both be in $R^{-1}$ (otherwise, flip them over to $R$ and use transitivity there). So one of them must be in $R$ and one in $R^{-1}$. Taking $R$ and $R^{-1}$ to be minimal relations containing 1~2 and 2~3 respectively works as a counterexample.
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http://mymathforum.com/advanced-statistics/337068-standard-normal-distribution.html | My Math Forum Standard Normal Distribution
November 2nd, 2016, 05:11 PM #1 Newbie Joined: Oct 2016 From: Earth Posts: 16 Thanks: 0 Standard Normal Distribution $Z$ ~ $N$(0,1) How to show that $Z\stackrel{d}{=} -Z$ and $\frac{Z^2}{2}$ ~ gamma($\frac{1}{2}$)? Also how to show that only the standard normal distribution satisfies these 2 properties? In other words, if you have $Z\overset{d}{=} -Z$ and $\frac{Z^2}{2}$ ~ gamma($\frac{1}{2}$), then $Z$ ~ $N$(0,1). To show that $Z\stackrel{d}{=} -Z$: I know that distribution of $Z$ is $F(z) = \frac{1}{2} + \phi(z)$, where $\phi(z) = \int_{0}^{z} e^{\frac{-t^2}{2}} dt$. Also $\phi(-z)=\phi(z)$. $P(-Z < z) = P(Z > -z) = 1 - P(Z < -z) = 1 - (\frac{1}{2} + \phi(-z)) = 1 - (\frac{1}{2} - \phi(z)) = \frac{1}{2} + \phi(z) = P(Z < z)$. Hence $Z\stackrel{d}{=} -Z$. To show that $\frac{Z^2}{2}$ ~ gamma($\frac{1}{2}$): $P(\frac{Z^2}{2} < z) = P(-\sqrt{2z} < Z < \sqrt{2z}) = 2\phi(\sqrt{2z})$. Now, this is where I am stuck. How do I continue to get gamma($\frac{1}{2}$)? Also how to show if you have $Z\overset{d}{=} -Z$ and $\frac{Z^2}{2}$ ~ gamma($\frac{1}{2}$), then $Z$ ~ $N$(0,1)?
November 3rd, 2016, 05:02 AM #2 Newbie Joined: Oct 2016 From: Earth Posts: 16 Thanks: 0 I am surprised that is no edit button. Anyway, I have corrected some typos. $Z$ ~ $N$(0,1) How to show that $Z\stackrel{d}{=} -Z$ and $\frac{Z^2}{2}$ ~ gamma($\frac{1}{2}$)? Also how to show that only the standard normal distribution satisfies these 2 properties? In other words, if you have $Z\overset{d}{=} -Z$ and $\frac{Z^2}{2}$ ~ gamma($\frac{1}{2}$), then $Z$ ~ $N$(0,1). To show that $Z\stackrel{d}{=} -Z$: I know that distribution of $Z$ is $F(z) = \frac{1}{2} + \phi(z)$, where $\phi(z) = \frac{1}{\sqrt{2\pi}}\int_{0}^{z} e^{\frac{-t^2}{2}} dt$. Also $\phi(-z)=\phi(z)$. $P(-Z < z) = P(Z > -z) = 1 - P(Z < -z) = 1 - (\frac{1}{2} + \phi(-z)) = 1 - (\frac{1}{2} - \phi(z)) = \frac{1}{2} + \phi(z) = P(Z < z)$. Hence $Z\stackrel{d}{=} -Z$. To show that $\frac{Z^2}{2}$ ~ gamma($\frac{1}{2}$): $P(\frac{Z^2}{2} < z) = P(-\sqrt{2z} < Z < \sqrt{2z}) = 2\phi(\sqrt{2z})$. Now, this is where I am stuck. How do I continue to get gamma($\frac{1}{2}$)? Also how to show if you have $Z\overset{d}{=} -Z$ and $\frac{Z^2}{2}$ ~ gamma($\frac{1}{2}$), then $Z$ ~ $N$(0,1)?
November 3rd, 2016, 07:06 PM #3 Newbie Joined: Oct 2016 From: Earth Posts: 16 Thanks: 0 Ok, differentiating $2\phi(\sqrt{2z})$ with respect to $z$ gives me $\frac{e^{-z}}{\sqrt{\pi}\sqrt{z}} = \frac{z^{\frac{1}{2}-1}e^{-z}}{\Gamma(\frac{1}{2})}$ which is the probability density function of gamma$(\frac{1}{2}, 1)$. Hence, I believe I have solved it. Now, if someone could help me with the reverse direction, I would greatly appreciate it.
November 3rd, 2016, 11:27 PM #4 Newbie Joined: Oct 2016 From: Earth Posts: 16 Thanks: 0 Corrected more typos: $Z$ ~ $N$(0,1) How to show that $Z\stackrel{d}{=} -Z$ and $\frac{Z^2}{2}$ ~ gamma($\frac{1}{2}$)? Also how to show that only the standard normal distribution satisfies these 2 properties? In other words, if you have $Z\overset{d}{=} -Z$ and $\frac{Z^2}{2}$ ~ gamma($\frac{1}{2}$), then $Z$ ~ $N$(0,1). To show that $Z\stackrel{d}{=} -Z$: I know that distribution of $Z$ is $F(z) = \frac{1}{2} + \phi(z)$, where $\phi(z) = \int_{0}^{z} e^{\frac{-t^2}{2}} dt$. Also $\phi(-z)=-\phi(z)$. $P(-Z < z) = P(Z > -z) = 1 - P(Z < -z) = 1 - (\frac{1}{2} + \phi(-z)) = 1 - (\frac{1}{2} - \phi(z)) = \frac{1}{2} + \phi(z) = P(Z < z)$. Hence $Z\stackrel{d}{=} -Z$. To show that $\frac{Z^2}{2}$ ~ gamma($\frac{1}{2}$): $P(\frac{Z^2}{2} < z) = P(-\sqrt{2z} < Z < \sqrt{2z}) = 2\phi(\sqrt{2z})$. Differentiating $2\phi(\sqrt{2z})$ with respect to $z$ gives me $\frac{e^{-z}}{\sqrt{\pi}\sqrt{z}} = \frac{z^{\frac{1}{2}-1}e^{-z}}{\Gamma(\frac{1}{2})}$ which is the probability density function of gamma$(\frac{1}{2}, 1)$. I would greatly appreciate it if somebody could help me with the reverse direction. Last edited by geniusacamel; November 3rd, 2016 at 11:32 PM.
November 3rd, 2016, 11:37 PM #5 Senior Member Joined: Sep 2015 From: USA Posts: 2,430 Thanks: 1315 neat problem. Ok you have a random variable $U$ which is distributed as $\Gamma\left(\dfrac 1 2 , 1\right) = \dfrac{e^{-u}}{\sqrt{\pi u}}$ now $Z = \sqrt{2U}$ \$F_Z(z) = P[Z < z] = P[\sqrt{2U}
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http://indico.cern.ch/event/175269/ | # Physics at the AD/PS/SPS (3/4)
## by Augusto Ceccucci (CERN)
Europe/Zurich
6-2-024 - BE Auditorium Meyrin (CERN)
### 6-2-024 - BE Auditorium Meyrin
#### CERN
120
Show room on map
Description Lecture 3: Flavour and Neutrinos The CERN SPS provides kaon and neutrino beams which are unique in the world. The lecture will describe the flavour and neutrino research conducted with these beams. The flavour programme is centered around the study of kaons. It includes a broad spectrum of topics such as CP-Violation, the precise determination of quark-mixing parameters, lepton universality and very rare decays. The CNGS neutrino beam enables to perform long baseline neutrino oscillation experiments with unique features such as the tau lepton appearance. Slides Video in CDS From the same series 1 2 4 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9623245000839233, "perplexity": 4904.225390428714}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860113010.58/warc/CC-MAIN-20160428161513-00169-ip-10-239-7-51.ec2.internal.warc.gz"} |
http://ethos.bl.uk/OrderDetails.do?uin=uk.bl.ethos.563692 | Use this URL to cite or link to this record in EThOS: http://ethos.bl.uk/OrderDetails.do?uin=uk.bl.ethos.563692
Title: Measurement of the G double-polarisation observable in positive pion photoproduction
Author: McAndrew, Josephine
Awarding Body: University of Edinburgh
Current Institution: University of Edinburgh
Date of Award: 2012
Availability of Full Text:
Access from EThOS: Full text unavailable from EThOS. Please try the link below. Access from Institution:
Abstract:
Establishing the resonance spectrum of the nucleon with accuracy would provide important new information about the dynamics and degrees of freedom of its constituents. The spectrum and properties of nucleon resonances are a fundamental test of the emerging predictions from Lattice QCD calculations and will guide re finements to QCD-based phenomenological models. Pion photoproduction is an excellent tool to study the nucleon resonance spectrum, as this channel is expected to couple strongly to most resonances. The new generation of measurements for this reaction, of which the measurement presented in this thesis forms a crucial part, will provide a great improvement in the quality of available experimental data. For the photoproduction process in particular, the use of photon beams and targets with high degrees of polarisation, coupled with large acceptance particle detectors is essential for disentangling the spectrum of excited states. There are many nucleon resonances predicted by recent Lattice QCD calculations and by phenomenological nucleon models which are only observed inconsistently by different analyses of the same experimental data or which are not observed at all. It is of upmost importance to establish if this means that the resonances do not exist in nature, reflecting inappropriate degrees of freedom in the theoretical description of the nucleon or if the current experimental measurements have not been sensitive enough. As such, there is a current world effort at modern tagged photon facilities to measure the \complete set" of photoproduction observables necessary to fully constrain the partial wave analyses used to extract the experimental excitation spectrum from the data. This thesis will present the first detailed measurement to date of positive pion photoproduction in the 730-2300 MeV photon energy (1400-2280 MeV centre-of-mass energy) region with a linearly polarised photon beam and a longitudinally polarised proton target with a close-to-complete angular coverage in detection of the reaction products. This unique set up allows for the extraction of the double-polarisation observable, G. The data were taken as part of the g9 experiment at the Thomas Jefferson National Accelerator Facility in Virginia, using a tagged, polarised photon beam and the Frozen Proton Spin Target, FROST, in conjunction with the CEBAF Large Acceptance Spectrometer, CLAS. The results of the study presented here are compared to the sparse existing data set for the G double-polarisation observable along with the current solutions of the the three main partial wave analyses: MAID, SAID and Bonn-Gatchina. Some agreement is obtained with the expectations of these PWA at lower energies, while disagreement at higher energies is clearly evident. This is the energy region where many of the missing resonances are expected to lie. Once incorporated into the MAID, SAID and Bonn-Gatchina models, these new data will provide an important contribution to constraining the amplitudes and therefore the resonance spectrum and properties of the nucleon. The new data will form a central part of the world effort to accurately establish the nucleon excitation spectrum for the first time by achieving the first complete measurement of experimental observables in meson photoproduction.
Supervisor: Watts, Daniel. ; Woods, Philip. Sponsor: Not available
Qualification Name: Thesis (Ph.D.) Qualification Level: Doctoral
EThOS ID: uk.bl.ethos.563692 DOI: Not available
Keywords: polorisation observables ; hadron spectroscopy ; proton ; photoproduction
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https://math.stackexchange.com/questions/1256841/when-is-6-a-square-in-zp | # When is 6 a square in Zp
To find when $6$ is a square I divided $6$ in to $2$ and $3$ by The Legendre symbol.
$$\frac{6}{p} = \frac{2}{p} \times \frac{3}{p}$$
I know that we can multiply the cases of when $2$ and are squares because then it is (1)(1) by Legendre.
$2$ is a square when $p \equiv 1 \pmod{8}$
$3$ is a square when $p \equiv 1 \pmod{12}$
so $6$ is a square when $p \equiv 1 \pmod {24}$ ($96$, but the $lcm(8,12)= 24$)
but how do I find the cases when $2$ and $3$ are both NOT squares? because then $6$ would be a square then as well, $(-1)(-1) = 1$ by Legendre.
Note that $2$ is a square mod $p$ when $p = 1$ or $p = 7\pmod{8}$, and is not a square when $p = 3$ or $p = 5 \pmod{8}$.
Similarly, using reciprocity we have that $3$ is a square mod $p$ when $p = 1$ or $p = 11 \pmod{12}$ and is not a square when $p = 5$ or $p = 7\pmod{12}$.
This information should be enough to determine what happens for each congruence class mod $24$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9481264352798462, "perplexity": 122.49130415422103}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027315321.52/warc/CC-MAIN-20190820092326-20190820114326-00198.warc.gz"} |
http://mathhelpforum.com/calculus/24925-fourier-transform.html | # Math Help - Fourier Transform
1. ## Fourier Transform
ImageShack - Hosting :: questionyw0.jpg
I was asked to find the fourier transform of x(n). Could someone explain whats happening in this answer?
Thanks
2. Originally Posted by hammer
ImageShack - Hosting :: questionyw0.jpg
I was asked to find the fourier transform of x(n). Could someone explain whats happening in this answer?
Thanks
I presume the problem is in simplifying $X(\omega)$?
$X(\omega) = \sum_{n = -4}^{\infty} \left ( \frac{1}{4} \right )^ne^{-j \omega n}$
Define $m = n + 4$. Then
$X(\omega) = \sum_{n = -4}^{\infty} \left ( \frac{1}{4} \right )^ne^{-j \omega n} = \sum_{m = 0}^{\infty} \left ( \frac{1}{4} \right )^{m - 4}e^{-j \omega (m - 4)}$
$= \sum_{m = 0}^{\infty} \left ( \frac{1}{4} \right )^m \cdot \left ( \frac{1}{4} \right )^{-4}e^{-j \omega m }e^{-(-4)j \omega }$
So
$X(\omega) = \sum_{m = 0}^{\infty} \left ( \frac{1}{4} \right )^m e^{-j \omega m }4^4e^{4j \omega }$
Now, note that the last two factors are independent of m:
$X(\omega) = 4^4e^{4j \omega } \cdot \sum_{m = 0}^{\infty} \left ( \frac{1}{4} \right )^m e^{-j \omega m }$
Note also that the summation is the summation of an infinite geometric series. The sum of any geometric series is
$\sum_{m = 0}^{\infty} ar^m = \frac{a}{1 - r}$
Thus
$\sum_{m = 0}^{\infty} \left ( \frac{1}{4} \right )^m e^{-j \omega m } = \sum_{m = 0}^{\infty} \left ( \frac{e^{-j \omega}}{4} \right )^m = \frac{1}{1 - \frac{e^{-j \omega}}{4}}$
So finally we get:
$X(\omega) = 4^4e^{4j \omega } \cdot \frac{1}{1 - \frac{e^{-j \omega}}{4}}$
$X(\omega) = \frac{4^4e^{4j \omega }}{1 - \frac{e^{-j \omega}}{4}}$
-Dan
3. Originally Posted by hammer
ImageShack - Hosting :: questionyw0.jpg
I was asked to find the fourier transform of x(n). Could someone explain whats happening in this answer?
Thanks
This is a Discrete Fourier Transform, $X(\omega)$ (on the second line) is the transform of $x(n)$
by definition of the DFT (and that $u(n)=0$ for $n<0$, and $1$ otherwise).
The rest is just simplification using the sum of a geometric series formula.
ZB | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 16, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9619774222373962, "perplexity": 834.185700756085}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657137108.99/warc/CC-MAIN-20140914011217-00146-ip-10-234-18-248.ec2.internal.warc.gz"} |
https://phys.libretexts.org/Bookshelves/Electricity_and_Magnetism/Book%3A_Electromagnetics_I_(Ellingson)/05%3A_Electrostatics/5.16%3A_Potential_Field_Within_a_Parallel_Plate_Capacitor | $$\require{cancel}$$
# 5.16: Potential Field Within a Parallel Plate Capacitor
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This section presents a simple example that demonstrates the use of Laplace’s Equation (Section 5.15) to determine the potential field in a source free region. The example, shown in Figure $$\PageIndex{1}$$, pertains to an important structure in electromagnetic theory – the parallel plate capacitor. Here we are concerned only with the potential field $$V({\bf r})$$ between the plates of the capacitor; you do not need to be familiar with capacitance or capacitors to follow this section (although you’re welcome to look ahead to Section 5.22 for a preview, if desired). What is recommended before beginning is a review of the battery-charged capacitor experiment discussed in Section 2.2. In this section you’ll see a rigorous derivation of what we figured out in an informal way in that section.
Figure $$\PageIndex{1}$$: A parallel plate capacitor, as a demonstration of the use of Laplace's Equation.
The parallel-plate capacitor in Figure $$\PageIndex{1}$$ consists of two perfectly-conducting circular disks separated by a distance $$d$$ by a spacer material having permittivity $$\epsilon$$. There is no charge present in the spacer material, so Laplace’s Equation applies. That equation is (Section 5.15): $\nabla^2 V = 0 ~~\mbox{(source-free region)} \label{m0068_eLaplace}$ Let $$V_C$$ be the potential difference between the plates, which would also be the potential difference across the terminals of the capacitor. The radius $$a$$ of the plates is larger than $$d$$ by enough that we may neglect what is going on at at the edges of the plates – more on this will be said as we work the problem. Under this assumption, what is the electric potential field $$V({\bf r})$$ between the plates?
This problem has cylindrical symmetry, so it makes sense to continue to use cylindrical coordinates with the $$z$$ axis being perpendicular to the plates. Equation \ref{m0068_eLaplace} in cylindrical coordinates is:
$\left[ \frac{1}{\rho} \frac{\partial}{\partial \rho} \left( \rho \frac{\partial}{\partial \rho} \right) + \frac{1}{\rho^2} \frac{\partial^2}{\partial \phi^2} + \frac{\partial^2}{\partial z^2} \right] V = 0$
or perhaps a little more clearly written as follows:
$\frac{1}{\rho} \frac{\partial }{\partial \rho} \left( \rho \frac{\partial V}{\partial \rho} \right) + \frac{1}{\rho^2} \frac{\partial^2 V}{\partial \phi^2} + \frac{\partial^2 V}{\partial z^2} = 0$
Since the problem has radial symmetry, $$\partial V/\partial \phi = 0$$. Since $$d \ll a$$, we expect the fields to be approximately constant with $$\rho$$ until we get close to the edge of the plates. Therefore, we assume $$\partial V/\partial \rho$$ is negligible and can be taken to be zero. Thus, we are left with $\frac{\partial^2 V}{\partial z^2} \approx 0 ~~ \mbox{for \rho \ll a} \label{m0068_eDE}$
The general solution to Equation \ref{m0068_eDE} is obtained simply by integrating both sides twice, yielding
$V(z) = c_1 z + c_2 \label{m0068_eVAC}$
where $$c_1$$ and $$c_2$$ are constants that must be consistent with the boundary conditions. Thus, we must develop appropriate boundary conditions. Let the node voltage at the negative ($$z=0$$) terminal be $$V_{-}$$. Then the voltage at the positive ($$z=+d$$) terminal is $$V_{-}+V_C$$. Therefore: These are the relevant boundary conditions. Substituting $$V(z=0) = V_{-}$$ into Equation \ref{m0068_eVAC} yields $$c_2 = V_{-}$$. Substituting $$V(z=+d) = V_{-}+V_C$$ into Equation \ref{m0068_eVAC} yields $$c_1 = V_C/d$$. Thus, the answer to the problem is
$V(z) \approx \frac{V_C}{d} z + V_{-} ~~ \mbox{for \rho \ll a} \label{eEP-PEPP1}$
Note that the above result is dimensionally correct and confirms that the potential deep inside a “thin” parallel plate capacitor changes linearly with distance between the plates.
Further, you should find that application of the equation $${\bf E} = - \nabla V$$ (Section 5.14) to the solution above yields the expected result for the electric field intensity: $${\bf E} \approx -\hat{\bf z}V_C/d$$. This is precisely the result that we arrived at (without the aid of Laplace’s Equation) in Section 2.2.
A reasonable question to ask at this point would be, what about the potential field close to the edge of the plates, or, for that matter, beyond the plates? The field in this region is referred to as a fringing field. For the fringing field, $$\partial V/ \partial \rho$$ is no longer negligible and must be taken into account. In addition, it is necessary to modify the boundary conditions to account for the outside surfaces of the plates (that is, the sides of the plates that face away from the dielectric) and to account for the effect of the boundary between the spacer material and free space. These issues make the problem much more difficult. When an accurate calculation of a fringing field is necessary, it is common to resort to a numerical solution of Laplace’s Equation. Fortunately, accurate calculation of fringing fields is usually not required in practical engineering applications. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8855957388877869, "perplexity": 177.1557502426179}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662509990.19/warc/CC-MAIN-20220516041337-20220516071337-00115.warc.gz"} |
http://math.stackexchange.com/questions/12769/convergence-of-sequences | # Convergence of Sequences
Let $a_n$ be a sequence in $\mathbb{R}$, $\phi: \mathbb{N} \rightarrow \mathbb{N}$ a monotonically increasing function with $\phi (0)=0$. Show that
$\sum _{n=0}^{\infty } a_n$ converges $\Rightarrow \sum _{i=0}^{\infty } {(\sum _{j=\phi (i)}^{\phi (i+1)-1 }a_{j})}$ converges.
My notes: Can you use that ${(\sum _{j=\phi (i)}^{\phi (i+1)-1 }a_{j})}$ has to be a cauchy sequence and can be as small as you want? Certainly it will be smaller than $a_n$ for all $n$ bigger than some $N_e$
-
## 2 Answers
I suppose you meant $\phi$ is strictly increasing, otherwise the inner sum could be an empty sum.
Let $b_i=\sum_{j=\phi(i)}^{\phi(i+1)-1} a_j$. For any $\epsilon >0$, you can find $N$ such that $$|a_j+\cdots +a_{j+k}|<\epsilon$$ for all $j\ge N, k>0$. Let $I$ be such that $\phi(I)\ge N$. Then for all $i\ge I$ you have $$|b_i+\cdots +b_{i+k}|<\epsilon$$ for all $k>0$. This proves that your series is Cauchy and hence converges.
-
When you take a fixed N you get
$\sum _{i=0} ^{N} \sum _{j=\phi(i)} ^{\phi(i+1)-1} a_j = \sum_{j=0}^{j=\phi(N+1)-1} a_j$
so basically the new sequence (meaning the partial summations) is a subsequence of the original one.
-
If that step u wrote is true, I agree that the task is solved. However, it appears to me that the right sum is smaller. How can you change the index? – Listing Dec 2 '10 at 17:48
Think of the original sequence, when you add parentheses in some places. The function $\phi$ just tell you where you added them. The left sum is just adding first the elements in the parentheses and then add it to the total sum, while the right sum "forgets" the parentheses . – Prometheus Dec 2 '10 at 17:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9696903824806213, "perplexity": 144.11086893871578}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422122245449.81/warc/CC-MAIN-20150124175725-00174-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://lists.w3.org/Archives/Public/w3c-wai-ig/2018JanMar/0045.html | # Re: Math ML - Math expression
From: Michael A. Peters <[email protected]>
Date: Tue, 16 Jan 2018 12:11:43 -0800
Message-ID: <[email protected]>
It has been a long time since I have used MathML but I seem to recall
that there were two variants of MathML - the standard variant that most
people used and a variant intended to make it easier for people reading
the code to see what the expressions were. I seem to remember that LaTeX
to MathML converters (e.g. MathJax) only did the former and not the latter.
Since some browsers seem to have dropped raw MathML support maybe
MathJax and related technology need to be updated to make sure the
changed) LaTeX math notation is one of the more common ways math authors
create mathematical equations for visual typesetting.
It has literally been a decade since I last needed to use MathML so I'm
really out of touch on where it has gone. I do remember MathML syntax
itself was so confusing I personally just always did it in LaTeX and
used htlatex (I think that is what it was called) and just copied the
generated MathML to HTML (and it most certainly wasn't very accessible)
Hopefully its easier now, I'll be reading some of the posts. Hopefully
modern tools for doing it aren't based around Windows like a lot of
accessibility tools are.
On 01/16/2018 07:25 AM, sirisha gubba wrote:
> Thank you all for giving me your inputs on making math equations accessible.
>
> Siri
>
> On Sat, Jan 13, 2018 at 9:20 PM, sirisha gubba <[email protected]
> <mailto:[email protected]>> wrote:
>
> Can someone help me to find a tool to make math formulas accessible?
> When researched I found some tools on W3C site
> (https://www.w3.org/wiki/Math_Tools#Browsers
> <https://www.w3.org/wiki/Math_Tools#Browsers>) but don't know what
> to choose.
> If anyone has experience in this area can you please guide me?
>
> Thanks,
> Siri
>
>
Received on Tuesday, 16 January 2018 20:12:09 UTC
This archive was generated by hypermail 2.3.1 : Tuesday, 16 January 2018 20:12:10 UTC | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9202285408973694, "perplexity": 3351.2429498075735}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891813109.8/warc/CC-MAIN-20180220224819-20180221004819-00694.warc.gz"} |
https://melusine.eu.org/syracuse/pstricks/pst-solides3d/bonus/spirale_en_detail_eng.tex | # spirale_en_detail_eng.tex
\documentclass[fleqn]{article}
\usepackage{amsmath,amssymb}
\usepackage{pst-rubans}
\usepackage[a4paper]{geometry}
\begin{document}
\parindent=0pt
\parskip4pt
\section*{The Mathematical Details for using a Helix as a Spring \dots}
In Figure 1 you see a cylindrical spring (shape of a helix) with 10 windings, with
an equilibrium position height $h_0$. The equilibrium position radius is $r_0=\frac{d_0}{2}$.
In Figure 2 the spring is stretched -- its new height is $h_1$ and its new radius is $r_1 = \frac{d_1}{2}$.
\textbf{Questions}
\begin{itemize}
\item What's the length of the spring?
\item Now that the spring has a fixed length, what is the radius $r_1(h_0,\,r_0,\,h_1)$?
\end{itemize}
\psset{lightsrc=30 5 5,SphericalCoor,viewpoint=50 45 0,Decran=50,dZ=0.2,resolution=180}
\begin{pspicture}(-3,-5)(3,7)
\psframe(-1.5,-3.5)(1.5,7)
\psSolid[object=cylindre,r=1.2,h=0.2,ngrid=1 36](0,0,6)%
\pshelices[incolor=gray!75,R=0.5,h=6,hue=0.2 0.5,grid,RotY=180,spires=10,dZ=0.1](0,0,6)
\psPoint(0,1,6){O1}
\psPoint(0,1,0){O}
\pcline[linecolor=blue]{<->}(O1)(O)
\Aput{$h_0$}
\psSolid[object=cylindre,r=0.7,h=1,ngrid=4 36,fillcolor=blue](0,0,-1)
\psPoint(0,0,0){E1}
\psdot(E1)
\pnode(-0.5,3){I1}
\pnode(0.5,3){I2}
\psline[arrowsize=0.2]{<->}(I1)(I2)
\uput[l](I1){$d_0$}
\rput(0,-4){Figure 1}
\end{pspicture}
\begin{pspicture}(-3,-5)(3,7)
\psframe(-1.5,-3.5)(1.5,7)
\psSolid[object=cylindre,r=1.2,h=0.2,ngrid=1 36](0,0,6)%
\pshelices[incolor=gray!75,R=0.4,h=7,hue=0.2 0.5,grid,RotY=180,spires=10,dZ=0.1](0,0,6)
\psPoint(0,1,6){O1}
\psPoint(0,1,-1){O}
\pcline[linecolor=blue]{<->}(O1)(O)
\Aput{$h_1$}
\pnode(-0.4,2.5){I1}
\pnode(0.4,2.5){I2}
\psline[arrowsize=0.2]{<->}(I1)(I2)
\uput[l](I1){$d_1$}
\psSolid[object=cylindre,r=0.7,h=1,ngrid=4 36,fillcolor=blue](0,0,-2)
\psPoint(0,0,-1){E1}
\psdot(E1)
\psPoint(0,0,-1.5){E2}
\psPoint(0,0,-3){E3}
\psline[arrowsize=0.2,linecolor=red]{->}(E2)(E3)
\rput(0,-4){Figure 2}
\end{pspicture}
\textbf{Assumptions}
For simplicity, we assume that all windings are equidistant apart (the cylindrical spiral keeps the shape of a helix) and we ignore any physical properties of a 'solid' helix, e.~g. thickness of the helix line, mass of the spring, material, temperature, elasticity, torsion, etc., etc., and, of course, etc.
\newpage
\textbf{Calculations}
Let $z_0$ be the \textit{height} of one single winding, so $h_0 = n z_0$ is the \textit{total height} of a spring with $n$ windings.
\begin{pspicture}(-3,-1)(1,4)
\psset{lightsrc=10 -20 50,SphericalCoor,viewpoint=50 -20 20,Decran=50,unit=1}
\deffunction[algebraic]{helice}(t){2.6*cos(t)}{2.6*sin(t)}{0.4*t}
\psSolid[object=cylindre,r=2.6,h=3.3,action=draw,ngrid=6 36]%
\psSolid[object=courbe,
linecolor=blue,
linewidth=0.1,
resolution=360,
function=helice]%
\begin{pspicture}(-3,-1)(5,4)
\psline(0,0)(7,0)
\psline(0,0)(0,3.5)
\psline[linecolor=blue](7,0)(0,3.5)
\uput[-90](3.5,0){$2\pi r_0$}
\uput[180](0,1.75){$z_0$}
\uput[0](3.5,1.85){$l_0$}
\end{pspicture}
\textbf{A little trick:}
Roll the cylinder on the plane and trace the helix onto the plane. Thus the helix turns into a straight line. Drawing a right-angled triangle and using the Pythagorean Theorem (where the one cathetus is
the height $z_0$, the other cathetus is the circumference $2 \pi r_0$ of the cylinder and the hypotenuse is the length $l_0$ of the one winding) we can simply calculate the length of one winding.
The \textit{length of one single winding} is
\begin{equation}
l_0 = \sqrt{z_0^2 + 4\pi^2r_0^2}\label{l0}
\end{equation}
This gives the \textit{total length} of the spring
\begin{equation*}
L_0 = n l_0
\end{equation*}
Now we stretch the spring and calculate all that for Figure 2.
Let $z_1$ be the \textit{height} of one single winding, so $h_1 = n z_1$ is the \textit{total height} of the stretched spring with $n$ windings.
The \textit{length of one single winding} is
\begin{equation}
l_1 = \sqrt{z_1^2 + 4\pi^2r_1^2}\label{l1}
\end{equation}
This gives the \textit{total length} of the stretched spring
\begin{equation*}
L_1 = n l_1
\end{equation*}
Now that this is the same spring, the length is constant: $L_0=L_1$.
Equating \eqref{l0} and \eqref{l1} and solving for $r_1^2$, we get,
\begin{equation*}
r_1^2 = \frac{1}{4\pi^2n^2}h_0^2 - \frac{1}{4\pi^2n^2}h_1^2 + r_0^2
\end{equation*}
\newpage
\textbf{Setting up the oscillation}
Coordinate setup: The $h$-axis points downwards and the origin $h=0$ is defined as top of the spring.
Now let the spring do a harmonic oscillation with an \textit{amplitude}
$\hat{h} < h_0$ and let the oscillation start from the \textit{equilibrium
position} $h_0$ downwards with an \textit{angular frequency} $\omega = \frac{2\pi}{T}$ where $T$ is the \textit{period of oscillation}.
\begin{equation*}
h_1(t) = h_0 + \hat{h} \cdot \sin(\omega t)= h_0[1 + \frac{\hat{h}}{h_0}\sin(\omega t)]
\end{equation*}
This gives after some basic arithmetic
\begin{align*}
r_1(t) &= \sqrt{\frac{1}{4\pi^2n^2}h_0^2\{1 - [1 + \frac{\hat{h}}{h_0}\sin(\omega t)]^2\} + r_0^2}\\
&= \sqrt{r_0^2 - \frac{\hat{h}h_0}{4\pi^2n^2}\sin(\omega t)[2 + \frac{\hat{h}}{h_0}\sin(\omega t)] }
\end{align*}
\textbf{Results}
The more windings a spring has, the less $r_1(t)$ differs from $r_0$.
Now let's discuss the following two intervals of time
\begin{itemize}
\item $]0;\frac{T}{2}[$ -- enlarging the spring $l_1(t) > l_0$
\begin{equation*}
\end{equation*}
\item $]\frac{T}{2};T[$ -- shortening the spring $l_1(t) < l_0$
\begin{equation*}
\end{equation*}
\end{itemize}
Now if you calculate \textit{by hand} some explicit examples
choosing the variables $h_0,\,r_0,\,\hat{h}$, you will see that the
needed, when the stretching of the spring gets large. However that
disturbs the harmony of the oscillation and catapults the spring out
of its Hookian limitations -- these won't be the conditions for a
preferred harmonic oscillation. So there is no big error to set: $r_1(t) \approx r_0$
\newpage
\textbf{Example}
Here you see some excerpts of a harmonic oscillation ($n=10,\,h_0=5,\,r_0=1,5,\,\hat{h}=2$) including the \textit{radius correction} and there is no visible change for $r_1(t)$.
\psset{lightsrc=30 5 5,SphericalCoor,viewpoint=50 45 0,Decran=50,resolution=180}
\multido{\i=0+90}{5}{%
\begin{pspicture}(-1.35,-5)(1.35,7)
\pstVerb{%
/amplitude \i\space sin 0.4 mul 1 add 5 mul def
/radius \i\space sin 0.4 mul 1 add 2 exp neg 1 add 25 mul 4 div pi 2 exp div 100 div 1.5 2 exp add 0.5 exp def
}
\psframe(-1.4,-3.5)(1.4,7)
\psSolid[object=cylindre,r=1.2,h=0.2,ngrid=1 36](0,0,6)%
\psdot(E1)
\rput(0,-4){$t=\dfrac{\i}{360}T$}
\end{pspicture}
\textbf{The source code}
\footnotesize
\begin{verbatim}
\psset{lightsrc=30 5 5,SphericalCoor,viewpoint=50 45 0,Decran=50,resolution=180}
\multido{\i=0+90}{5}{%
\begin{pspicture}(-1.35,-5)(1.35,7)
\pstVerb{%
/amplitude \i\space sin 0.4 mul 1 add 5 mul def
/radius \i\space sin 0.4 mul 1 add 2 exp neg 1 add 25 mul 4 div pi 2 exp div 100 div 1.5 2 exp add 0.5 exp def
}
\psframe(-1.4,-3.5)(1.4,7)
\psSolid[object=cylindre,r=1.2,h=0.2,ngrid=1 36](0,0,6)
\rput(0,-4){$t=\dfrac{\i}{360}T$} | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9746636748313904, "perplexity": 4324.600965041064}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304947.93/warc/CC-MAIN-20220126101419-20220126131419-00164.warc.gz"} |
https://questions.examside.com/past-years/jee/question/suppose-that-the-angular-velocity-of-rotation-of-earth-is-in-jee-main-physics-units-and-measurements-weuagzjktao9iewt | 1
JEE Main 2018 (Online) 16th April Morning Slot
+4
-1
Suppose that the angular velocity of rotation of earth is increased. Then, as a consequence :
A
Weight of the object, everywhere on the earth, will increase.
B
Weight of the object, everywhere on the earth, will decrease.
C
There will be no change in weight anywhere on the earth.
D
Except at poles, weight of the object on the earth will decrease.
2
JEE Main 2018 (Offline)
+4
-1
A particle is moving in a circular path of radius $$a$$ under the action of an attractive potential $$U = - {k \over {2{r^2}}}$$ Its total energy is:
A
$$- {3 \over 2}{k \over {{a^2}}}$$
B
Zero
C
$$- {k \over {4{a^2}}}$$
D
$${k \over {2{a^2}}}$$
3
JEE Main 2018 (Offline)
+4
-1
A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of the particle is T, then :
A
T $$\propto$$ Rn/2
B
T $$\propto$$ R3/2 for any n
C
T $$\propto$$ Rn/2 +1
D
T $$\propto$$ R(n+1)/2
4
JEE Main 2018 (Online) 15th April Morning Slot
+4
-1
A body of mass m is moving in a circular orbit of radius R about a planet of mass M. At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius $${R \over 2},$$ and the other mass, in a circular orbit of radius $${3R \over 2}$$. The difference between the final and initial total energies is :
A
$$- {{GMm} \over {2R}}$$
B
$$+ {{GMm} \over {6R}}$$
C
$${{GMm} \over {2R}}$$
D
$$- {{GMm} \over {6R}}$$
JEE Main Subjects
Physics
Mechanics
Electricity
Optics
Modern Physics
Chemistry
Physical Chemistry
Inorganic Chemistry
Organic Chemistry
Mathematics
Algebra
Trigonometry
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EXAM MAP
Joint Entrance Examination | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9277147650718689, "perplexity": 686.5743279623373}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948620.60/warc/CC-MAIN-20230327092225-20230327122225-00184.warc.gz"} |
https://socratic.org/questions/a-high-school-athlete-runs-1-00-10-2-m-in-12-20-s-what-is-the-velocity-in-m-s-an | Physics
Topics
# A high school athlete runs 1.00 * 10^2 m in 12.20 s. What is the velocity in m/s and km/h?
Jul 6, 2016
$8.20 m {s}^{-} 1$ (3sf) and $29.5 k m {h}^{-} 1$ (3sf)
#### Explanation:
In order to calculate the velocity we need both the distance travelled and the time taken to travel that distance. In this case we have both, I will start by converting the $1.00$ x ${10}^{2}$ into standard form which is $100$m
So $v = \frac{d}{t}$
$d = 100 m$ & $t = 12.20 s$
Substitute both these values into the equation:
$v = \frac{d}{t} = \frac{100 m}{12.20 s} = 8.20 m {s}^{-} 1$
I like to keep the units in the calculation to make working the units of velocity out easier.
Now we know the velocity in $m {s}^{-} 1$. To calculate the speed in $k m {h}^{-} 1$ we need to convert both the $m$ & $s$, this can be done in multiple ways. However we know that $1000 m$ = $1 k m$ and $1$hour = $3600$seconds.
Therefore we need to multiply our answer by: $\frac{1000 m}{3600 s}$ So...
$v = 8.20 m {s}^{-} 1$ x$3.6 = 29.5 k m {h}^{-} 1$
Note: I retained the non-rounded value in the calculator through the calculations, this will improve your accuracy and avoid rounding error.
##### Impact of this question
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https://scicomp.stackexchange.com/questions/31205/incorporating-a-potential-barrier-in-a-wave-packet-simulation-fourier-transform | Incorporating a potential barrier in a wave-packet simulation (Fourier Transform method)
I'm trying to simulate the scattering of a wave-packet at a potential barrier in Python. I'm using a Fourier Transform method (not sure if its the same as the Split-Step method), where I apply Fourier Transform on my initial wavepacket (plane wave in a gaussian envelope) to get the wavefunction in momentum-space: $$\phi(k) = \int^{\infty}_{-\infty} \psi(x) e^{ikx} dx$$ I then calculate the dispersion term, $$\omega$$, at each $$k$$ value, $$\omega = \frac{\hbar k^2}{2m}$$. I can then incorporate the time-evolution operator $$e^{-i\omega t}$$. This allows me to obtain $$\psi(x)$$ at a given time:
$$\psi(x,t) = \int^{\infty}_{-\infty} \phi(k) e^{-ikx}e^{-i\omega t}$$ which I achieve through an Inverse Fourier Transform of $$\phi(k)e^{-i\omega t}$$.
My wavepacket moves as expected, but I am now trying to incorporate a potential barrier. To start, I am trying to get my wave-packet to reflect off of a barrier (like a particle in a box).
I'm a bit unsure of how to implement this into my code. This problem is usually described by having both a incoming and reflected wave in the region before the barrier, so I have implemented a 'reflected' wave-packet which is simply a mirror image of the initial wave-packet with its amplitudes reversed. This can be manipulated to simulate the wave reflecting off the barrier but I'm wondering if there is a more trivial solution.
Questions:
• Is there a way to incorpate reflection/transmission at a barrier using my method of propagating the wave-packet?
• I've seen this problem solved using a finite difference method to solve the Schrodinger equation instead, is there an advantage of using that method?
• I think that you need to find the wavefunctions for your potential barrier. Then do a convolution with your spectrum. – nicoguaro Mar 9 at 15:42
• So define a function for my potential barrier, Fourier Transform it and then convolve with my wavepacket in k-space? Thanks for the suggestion! – FeelsToWaltz Mar 13 at 11:34
The neat thing about Fourier codes is that you can achieve very high order derivatives and if you are describing physics which is about waves, then the trigonometric base functions are a good choice. You also have a reasonably fast transform in $$\mathcal O(n\log(n)$$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8934018611907959, "perplexity": 343.8703981460015}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999817.30/warc/CC-MAIN-20190625092324-20190625114324-00318.warc.gz"} |
http://www.ck12.org/book/CK-12-Physics---Intermediate/section/14.0/ | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Chapter 14: Thermodynamics
Difficulty Level: At Grade Created by: CK-12
Chapter Outline
### Chapter Summary
1. The ideal gas law states \begin{align*}PV=nRT\end{align*} where \begin{align*}P\end{align*} is pressure, \begin{align*}V\end{align*} is volume, \begin{align*}n\end{align*} is the number of moles of substance, \begin{align*}R\end{align*} is the universal gas constant and \begin{align*}T\end{align*} is temperature.
2. Thermodynamics is the study of processes in which energy is transferred by heat and work.
3. In an isothermal process, temperature is constant
4. In an isochoric process, volume is constant
5. In an isobaric process, pressure is constant
6. In an adiabatic process, no heat flows into or out of the system
7. The first law of thermodynamics: The change in the internal energy of a closed system is equal to the heat into (or out of) a system plus the work done on the system (or by the system).
\begin{align*}\Delta U=Q+W,\Delta U\end{align*} is the change in internal energy of the system, \begin{align*}Q\end{align*} is the heat, and \begin{align*}W\end{align*} the work.
8. The second law of thermodynamics describes the direction in which physical phenomena can occur.
Statements describing the second law:
a. Heat can flow spontaneously from hot to cold but never from cold to hot.
b. No heat machine is 100% efficient.
The most general statement of the second law:
For all natural processes, the total entropy of a system increases.
9. The efficiency of an actual heat engine can be expressed as
\begin{align*}e=\left(1-\frac{Q_L}{Q_H}\right) \times 100 \end{align*}
The high temperature thermal energy is \begin{align*}Q_H\end{align*} and the low temperature thermal energy is \begin{align*}Q_L\end{align*}.
10. The efficiency of a Carnot engine can be expressed as
\begin{align*}e=\left(1-\frac{T_L}{T_H}\right) \times 100 \end{align*}
The high operating temperature of the Carnot engine is \begin{align*}T_H\end{align*} and the low operating temperature of the Carnot engine \begin{align*}T_L\end{align*}.
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Jan 13, 2016 | {"extraction_info": {"found_math": true, "script_math_tex": 15, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9301759004592896, "perplexity": 1357.311825437662}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049277091.27/warc/CC-MAIN-20160524002117-00026-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/on-numerical-calculation-of-lift-force-via-potential-flow.99089/ | # On numerical calculation of Lift Force via Potential Flow
1. Nov 8, 2005
### Clausius2
Imagine:
I want to compute numerically a POTENTIAL STEADY and INCOMPRESSIBLE flow over an airfoil. The set up of the problem is:
$$\nabla^2\phi=0$$
$$\nabla\phi \cdot \overline{n}\big)_{x=surface}=0$$ no normal velocity component on the airfoil surface.
$$\nabla \phi=\overline{U_\infty}$$ as $$x\rightarrow\infty$$ external flow at large distances.
The three main questions which arise are the next:
i) Is it possible to obtain an steady solution of this stuff?
ii) Am I going to obtain any Lift force?. Why?
Thanx for participating in this discussion.
Can you offer guidance or do you also need help?
Draft saved Draft deleted
Similar Discussions: On numerical calculation of Lift Force via Potential Flow | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.837330162525177, "perplexity": 2376.596400078211}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128323711.85/warc/CC-MAIN-20170628153051-20170628173051-00651.warc.gz"} |
http://mathhelpforum.com/calculus/63264-integration.html | 1. ## Integration
Let $G(x)=\int_{0}^{x} f(x)dx$
The graph of f(x) is under the x axis from [-2,0]
Wouldn't the graph of G(x) be increasing on the interval of [-2,0] because
$G(x)=\int_{0}^{-2} f(x)dx=-\int_{-2}^{0} f(x) dx$ = -(-area)=positive area
But apparently the graph is decreasing on that interval. Can someone explain this to me? Thanks
2. Originally Posted by Linnus
Let $G(x)=\int_{0}^{x} f(x)dx$
The graph of f(x) is under the x axis from [-2,0]
Wouldn't the graph of G(x) be increasing on the interval of [-2,0] because
$G(x)=\int_{0}^{-2} f(x)dx=-\int_{-2}^{0} f(x) dx$ = -(-area)=positive area Mr F says: This is G(-2), NOT G(x).
But apparently the graph is decreasing on that interval. Can someone explain this to me? Thanks
Correct, it is indeed decreasing over the interval $-2 \leq x \leq 0$.
Consider:
$G(-2) = \int_{0}^{-2} f(x) \, dx = - \int^{0}_{-2} f(x) \, dx > 0$.
$G(-1) = \int_{0}^{-1} f(x) \, dx = - \int^{0}_{-1} f(x) \, dx > 0$.
But clearly G(-2) > G(-1) .......
3. Originally Posted by mr fantastic
Correct, it is indeed decreasing over the interval $-2 \leq x \leq 0$.
Consider:
$G(-2) = \int_{0}^{-2} f(x) \, dx = - \int^{0}_{-2} f(x) \, dx > 0$.
$G(-1) = \int_{0}^{-1} f(x) \, dx = - \int^{0}_{-1} f(x) \, dx > 0$.
But clearly G(-2) > G(-1) .......
Ah...that makes sense. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9875528812408447, "perplexity": 1740.3157546169439}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738662527.91/warc/CC-MAIN-20160924173742-00115-ip-10-143-35-109.ec2.internal.warc.gz"} |
https://brilliant.org/problems/huge-collision/ | # Huge collision!
Three balls having masses $$2m$$, $$3m$$ and $$5m$$ having equal velocities $$v_0$$ collide with a block of wood of mass $$6m$$ which is at rest, in a perfectly elastic collision.
Now, this whole system of the balls and the block is moving with a velocity of $$v_1$$.
If the ratio $$\dfrac{v_0}{v_1}$$ is equal to $$\dfrac ab$$, where $$a$$ and $$b$$ are co-prime, find $$a+b$$.
× | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9442668557167053, "perplexity": 183.61002814740047}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549425254.88/warc/CC-MAIN-20170725142515-20170725162515-00070.warc.gz"} |
http://nrich.maths.org/7664/clue | ### Lunar Leaper
Gravity on the Moon is about 1/6th that on the Earth. A pole-vaulter 2 metres tall can clear a 5 metres pole on the Earth. How high a pole could he clear on the Moon?
### High Jumping
How high can a high jumper jump? How can a high jumper jump higher without jumping higher? Read on...
### Escape from Planet Earth
How fast would you have to throw a ball upwards so that it would never land?
# Slingshot
##### Stage: 5 Challenge Level:
If a projectile is fired at an angle of 45° to the horizontal then the initial speed of the projectile and the distance it travels is related by $d = \frac{v^2}{g}$.
Use the conservation of energy in order to find the speed of the stone. The energy stored in a strip which is extended by $\delta x$ is equal $E = \frac{k \delta x^2}{2}$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9223259091377258, "perplexity": 743.2079646317302}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500830721.40/warc/CC-MAIN-20140820021350-00266-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/easy-way-to-copy-files.36931/ | # Easy way to copy files?
1. ### aychamo
376
Easy way to copy files??
Hey guys
Is there an easy way to copy files in the method I'll mention in WinXP?
I have a folder, C:\MP3s that is arranged as:
C:\MP3s\Artist\Album\Songname.mp3
I keep a copy of this folder on my LAN, on F:\MP3s .. The problem is, say I add a few new songs in different albums, and I want to copy them over to the F:\MP3s, I can't find an easy way to do it.
If I just copy it over, my only option is to overwrite everything, which is *slow*. If I pick for it to not overwrite (and hit Shift-N to answer No to all), then it won't copy over anything because it treats the directories as files. I mean, say I have four songs currently in C:\MP3s\Beatles\Abbey Road\ (like I wouldn't have the whole album!), and I add a few more songs. If I try to copy the folder over from C:\MP3s, and I don't click on "Overwrite All", it won't copy the new songs from \Mp3s\Beatles\Abbey Road\ because it sees that I already have a Abbey Road folder, and doesn't check the files in it.
Is there a copy utility to "Synchronize" two network directories? Or to copy files that don't exist (and it does subdirectories, etc?) I played around with XCopy, but I couldn't find a set of command line arguments that suited me.
Any ideas? I appreciate it, and thank you.
2. ### Guybrush Threepwood
526
Total Commander has a very nice synchronize function
3. ### aychamo
376
Is TC free? :)
4. ### megashawn
505
why not just map the mp3 folder on the c drive as a network drive?
5. ### dduardo
1,918
Staff Emeritus | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.834775447845459, "perplexity": 3079.104393044423}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644065464.19/warc/CC-MAIN-20150827025425-00260-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/viscous-disc-paradox.843118/ | 1. Nov 14, 2015
### snorkack
Which way does friction in a viscous disc operate?
Imagine a ring consisting of ringlets. First consider a case of a pair of nearby ringlets - in the same plane, both circular orbits.
If all particles of both ringlets are in circular orbits in the same plane, then they can never collide and therefore never exert any force. There is then zero temperature and zero viscosity.
Now suppose the viscosity is nonzero. What then?
By Third Law of Kepler, the inner ringlet should move faster.
Therefore, the inner ringlet should propel the outer ringlet ahead: the outer ringlet should expand and the inner one shrink.
But the problem is that the particles of rings, whether dust grains or gas molecules, are severally subject to Newton´s laws... and therefore also laws of Kepler. Including the Second.
While particles of inner ringlets are indeed faster than outer ringlet, they are so while they are in the inner ringlet, and do not meet outer ringlet.
The particles which can and do collide are those on elliptical orbit.
Considering two neighbouring circular ringlets and an elliptical ringlet tangent to both at its apsides.
The outer ringlet is slower than the inner, as per 3rd law - but the elliptical orbit at its apoapse is even slower, as per 2nd.
Therefore, the outer ringlet should be slowed down and shrink. And inner ringlet, by the same reasoning, should encounter the faster part of elliptical ringlet at periapse, speed up and expand.
So what´s the solution of the paradox? How should a viscous disc behave?
2. Nov 14, 2015
### my2cts
I guess a viscous disk rotating in a Newton potential by friction would convert kinetic energy into heat.
Thus it would heat up as its radius would shrink. Eventually it would be absorbed by the source of the Newton potential.
However I expect the matter to condense into clumps in which no friction occurs as long as it is in a circular orbital (no tidal forces).
I would also expect that the end result depends on how much matter is involved and on the size of the orbit.
This will also set the time scale of the evolution.
Of course it would be better to consult a text on planetary formation than to follow my guesses.
3. Nov 14, 2015
### Staff: Mentor
@my2cts: Angular momentum is conserved (neglecting relativistic effects like the angular momentum of radiated light), it won't fall in.
Angular momentum and energy scale with $L^2 \propto E^3$. Putting objects in different orbits together in a common orbit releases energy, pushing them apart would require energy. Therefore, your objects should get closer together.
Objects in different circular orbits will never collide, but you still get some interaction via gravity. It's not a simple "this is pushing this" however. See Saturn's rings for the complex interactions in such a disk.
4. Nov 14, 2015
### Chronos
Saturns rings are a good example. They are composed largely of water ice so any heating due to friction is not a major factor. The also orbit at a fairly leisurely pace the inside edge of the inner [C] ring has as orbital velocity of a little under 24 km/sec whereas the outside edge of the outer [A] ring has an orbital velocity of just over 16 km/sec. The ISS, by comparison circles the earth at a little less than 8 km/sec. The aggregate material that the rings are composed of vary in size, although not greatly. The smaller particle average less than a cm in size whereas the relatively rarer larger particle can be a meter or two across. It is beleived the larger variety are slowly broken apart or eventually migrate out beyond the rings. Orbital resonance with the moons of saturn are believed to protect the rings from any significant collapse or dispersal. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8449104428291321, "perplexity": 1486.2881342186586}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948531226.26/warc/CC-MAIN-20171213221219-20171214001219-00616.warc.gz"} |
https://nbviewer.jupyter.org/urls/www.numfys.net/media/notebooks/verlet_integration.ipynb | Verlet Integration¶
Modules - Ordinary Differential Equations¶
Last edited: March 11th 2018
Can we solve Newton's second law $$m\frac{d^2}{dt^2}x(t)=F(x(t),t), \label{eq:1}$$ a second-order ODE, without resorting to Euler's method which re-casts the problem as two coupled, first-order ODEs?
Here, we employ the initial condition $x(t_0)=x_0$ and $v(t_0)=v_0$, and restrict our analysis to one dimension. Also, we assume for now that the force $F$ does not depend on the velocity $v$.
We can divide equation \eqref{eq:1} by the mass $m$ and write instead $$\frac{d^2}{dt^2}x(t)=a(x(t),t),$$ where $a(x,t)$ is the acceleration.
Example 1: $$\frac{d^2x}{dt^2}=-x+x^3+0.1\cos(t), \label{eq:2}$$ with the initial conditions $x(t_0)=0$ and $v(t_0)=0$.
In a fairly crude way, we now approximate the second derivative of $x(t)$ with respect to time, i.e. the acceleration, by \begin{aligned} \frac{d^2x}{dt^2}\bigg|_{t=t_n}=\frac{dv}{dt}\bigg|_{t=t_n}& \approx&\frac{v(t_n+h/2)-v(t_n-h/2)}{h} \\ &\approx&\frac{ \frac{x(t_n+h)-x(t_n)}{h} -\frac{x(t_n)-x(t_n-h)}{h} }{h}, \end{aligned} \label{eq:3} where we used the central difference approximation(s) $$\frac{dv}{dt}\bigg|_{t=t_n}\approx\frac{v(t_n+h/2)-v(t_n-h/2)}{h}$$ with $$v(t_n+h/2)\approx \frac{x(t_n+h)-x(t_n)}{h},\\ v(t_n-h/2)\approx \frac{x(t_n)-x(t_n-h)}{h}.$$
The above formulation follows the usual discretization $$t_n=t_0+n\cdot h~~~\mathrm{with}~~~n=0,1,2,3,...,N.$$ Again, this suggests the following abbreviation: $x_n=x(t_n)$.
With this in mind, substitution of expression \eqref{eq:3} into equation \eqref{eq:2} and subsequent multiplication by $h^2$ yields $$x_{n+1}-2x_n+x_{n-1}=h^2\left[-x_n+x_n^3+0.1\cos(t_n)\right],$$ where we evaluated equation \eqref{eq:2} at $t=t_n$. This results in the following recursive formulation of the solution $$x_{n+1}=2x_n-x_{n-1}+h^2\left[-x_n+x_n^3+0.1\cos(t_n)\right] \label{eq:4}$$ with $x_0=0$ and $v_0=0$.
Verlet Integration¶
More rigorously, one can derive the generalized version of the recursive formula \eqref{eq:4}, namely
$$\boxed{x_{n+1}=2x_n-x_{n-1}+h^2 a(x_n,t_n),} \label{eq:5}$$
by use of Taylor expansions about $x(t_n)$ (see Appendix). The Taylor-expansion method tells us that the local error we make in using the approximation \eqref{eq:5}, scales like $h^4$. This is, perhaps, surprisingly good.
The method \eqref{eq:5} is called Verlet integration. Often, it includes an acceleration term which does not depend explicitly on time: $a=a(x(t))$.
The remaining problem is now to use equation \eqref{eq:4} at $t=t_0$, i.e. at $n=0$, since we do not know the value of $x$ for $t<0$, in particular $x_{-1}$. Here, one can estimate the first value $x_1$ after the starting point $t_0$ by Taylor expansion of $x(t)$ at $t=t_0$: $$x_1=x_0+h v_0 + \frac{h^2}{2}a(x_0,t_0)+\mathcal{O}(h^3).$$ In our example, this becomes ($x_0=0$, $v_0=0$, $t_0=0$) $$x_1=x_0+h v_0 + \frac{h^2}{2}a(x_0,t_0)=\frac{h^2}{20}. \label{eq:6}$$ Thereafter, we can stick to the relation \eqref{eq:4} to find all other $x_n$.
(Note: The error of the approximation \eqref{eq:6} scales like $h^3$.)
Let's use \eqref{eq:4} and \eqref{eq:6} to solve \eqref{eq:2}:
In [2]:
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
# Set common figure parameters
newparams = {'figure.figsize': (16, 6), 'axes.grid': True,
'lines.linewidth': 1.5, 'lines.linewidth': 2,
'font.size': 14}
plt.rcParams.update(newparams)
In [3]:
N = 100000 # number of steps
h = 0.001 # step size
t = np.zeros(N+1)
x = np.zeros(N+1)
# initial values
t_0 = 0
x_0 = 0
t[0] = t_0
x[0] = x_0
for n in range(N):
# Second grid point:
if n==0:
t[n+1] = h
x[n+1] = h**2/20.0
# Verlet integration
else:
t[n+1] = t[n] + h
x[n+1] = 2.0*x[n] - x[n-1] + h**2*(-x[n]+x[n]**3+0.1*np.cos(t[n]))
# Plot the solution
plt.plot(t,x)
plt.ylabel(r'$x(t)$')
plt.xlabel(r'$t$')
plt.grid();
Here we have chosen $h=0.001$ and $N=100\,000$, and so $t_N=100$. In the plot of $x(t)$, the discrete points have been connected by straight lines.
Question¶
What if the force, and hence the acceleration, also depends on the velocity $v$? In other words, let's look at $$\frac{d^2}{dt^2}x(t)=a(x(t),v(t),t),$$ with the initial conditions $x(t_0)=x_0$ and $v(t_0)=v_0$.
We approximate the left-hand side at $t=t_n$ just like before, and we approximate the right-hand side by $$a(x_n,v_n,t_n)\approx a\left( x_n, \frac{x_{n+1}-x_{n-1}}{2h}, t_n \right),$$ where we used a central difference approximation for $v_n$ again: $$\frac{dv}{dt}\bigg|_{t=t_n}\approx\frac{v(t_n+h)-v(t_n-h)}{2h}. \label{eq:7}$$
However, the error of the approximation \eqref{eq:7} scales like $h^2$, thereby reducing the overall accuracy of the method.
Again, we estimate $$x_1=x_0+h v_0 + \frac{h^2}{2}a(x_0,v_0,t_0).$$ Thereafter, we now use $$x_{n+1}=2x_n-x_{n-1}+h^2 a\left( x_n, \frac{x_{n+1}-x_{n-1}}{2h}, t_n \right) .$$ Note that we need to solve this equation for $x_{n+1}$ which is not always possible in analytical form, i.e. in closed form, if the problem is nonlinear.
Let us focus on a problem which is linear in $v$.
Example 2: $$\frac{d^2x}{dt^2}=-v-x^3$$ with the initial conditions $x_0=x(0)=10$ and $v_0=v(0)=0$. Using our above formalism, this reads in discretized form \begin{align*} x_{n+1}&=2x_n-x_{n-1}+h^2 \left( -\frac{x_{n+1}-x_{n-1}}{2h}-x_n^3 \right) \\ \Rightarrow x_{n+1}&=\frac{2x_n-(1-h/2)\,x_{n-1}-h^2 x_n^3}{1+h/2}. \end{align*} When combined with $x_0=10$ and $$x_1=x_0+h v_0 + \frac{h^2}{2}a(x_0,v_0,t_0)=10-500 h^2,$$ this determines the solution of the problem. This is implemented in the code below. In the code we choose $h=0.001$ and $N=3\,000$, and so $t_N=3.0$
In [4]:
N = 3000 # number of steps
h = 0.001 # step size
t = np.zeros(N+1)
x = np.zeros(N+1)
# initial values
t_0 = 0
x_0 = 10
t[0] = t_0
x[0] = x_0
for n in range(N):
# Second grid point:
if n==0:
t[n+1] = h
x[n+1]= 10.0-500.0*h**2
# Verlet integration
else:
t[n+1] = t[n] + h
x[n+1] = (2.0*x[n]-(1.0-h/2)*x[n-1]-h**2*x[n]**3)/(1+h/2.0)
# Plot the solution
plt.plot(t,x)
plt.ylabel(r'$x(t)$')
plt.xlabel(r'$t$')
plt.grid();
Summary¶
We have seen that there is more than one method (Euler's method) to solve ODEs. Verlet integration is often used to solve Newton's second law.
1. It provides higher accuracy than Euler's method.
2. It is more stable than Euler's method and
3. might require a nonlinear solve for $x_{n+1}$.
4. It can be extended in a straightforward manner to three-dimensional motion.
Appendix: Derivation of Verlet Formula with Taylor Series¶
Given the dynamics of $x(t)$ at $t=t_n$, let us approximate the values $x_{n-1}=x(t_n-h)$ and $x_{n+1}=x(t_n+h)$ by use of Taylor series at $x_{n}=x(t_n)$: \begin{eqnarray*} x(t_n-h)&=&x(t_n)+v(t_n) \cdot (-h)+\frac{a(t_n)}{2}\cdot (-h)^2+\frac{b(t_n)}{6}\cdot (-h)^3+\mathcal{O}(h^4), \\ x(t_n+h)&=&x(t_n)+v(t_n)\,h+\frac{a(t_n)}{2}h^2+\frac{b(t_n)}{6}h^3+\mathcal{O}(h^4), \end{eqnarray*} with \begin{align*} v(t_n)=\frac{dx}{dt}\bigg|_{t=t_n}, \quad a(t_n)=\frac{d^2x}{dt^2}\bigg|_{t=t_n}, \quad \text{and} \quad b(t_n)=\frac{d^3x}{dt^3}\bigg|_{t=t_n}. \end{align*} Adding these two equations and subsequent re-arranging of terms yields $$x(t_n+h)=2x(t_n)-x(t_n-h)+a(t_n)\,h^2+\mathcal{O}(h^4)$$ or, in different notation, $$x_{n+1}=2 x_n-x_{n-1}+h^2\,a(t_n)+\mathcal{O}(h^4).$$ This is our Verlet formula \eqref{eq:5} if we drop the higher-order terms $\mathcal{O}(h^4)$ and write $a(x(t_n),t_n)=a(x_n,t_n)$ instead of simply $a(t_n)$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 6, "x-ck12": 0, "texerror": 0, "math_score": 0.9964548945426941, "perplexity": 634.9941012330297}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038076819.36/warc/CC-MAIN-20210414034544-20210414064544-00111.warc.gz"} |
http://math.stackexchange.com/questions/273129/has-this-example-something-to-do-with-vector-projection | # Has this example something to do with vector projection?
Has this example something to do with vector projection?
-
Can you please add a reference as to where you got this from? Regards – Amzoti Jan 8 '13 at 22:23
It is from this book. books.google.dk/… – Reader Jan 8 '13 at 22:25
Yes. The force $\textbf{F}$ can be given by $(\textbf{a} \bullet \hat{\textbf{b}}) \hat{\textbf{b}}$ where $\hat{\textbf{b}}$ is the unit vector in the direction of b. – Andrew D Jan 8 '13 at 22:26
## 1 Answer
I cannot access the link you provide to better understand the context of the problem.
But I'll venture to answer:
Yes, this example involves a vector projection:
Specifically, let $\bf{F}$ denote force, and $\hat{\bf{b}}$ denote the unit vector in the direction b. Then $$\bf{F} = (\bf{a} \bullet \hat{\bf{b}}) \hat{\bf{b}}.$$
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9636918902397156, "perplexity": 551.1940766655393}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257824185.14/warc/CC-MAIN-20160723071024-00114-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://www.science20.com/crawler_superland/richard_feynmans_functional_integral_3_electromagnetism-95164 | This is the third part of a ten-part post on the foundation of our understanding of high energy physics, which is Richard Feynman's functional integral. The first two parts are Action and Multiple Molecules, and the following parts, which will appear at intervals of about a month, are Action for Fields, Radiation in an Oven, Matrix Multiplication, The Functional Integral, Gauge Invariance, Photons, and Interactions.
I'm hoping this blog will be fun and useful for everyone with an interest in science, so although I'll pop up a few formulae again, I'll still try to keep them friendly by explaining all the pieces, as in the first two parts of the post. Please feel free to ask a question in the Comments, if you think anything in the post is unclear.
The clue that led to the discovery of quantum mechanics, whose principles are summarized in Feynman's functional integral, came from the attempted application to electromagnetic radiation of discoveries about heat and temperature. We looked at those discoveries about heat and temperature in the second part of the post, and today I would like to show you how James Clerk Maxwell, just after the middle of the nineteenth century, was able to identify light as waves of oscillating electric and magnetic fields, and to calculate the speed of light from measurements of:
1. the force between parallel wires carrying electric currents;
2. the heat given off by a long thin wire carrying an electric current; and
3. the time integral of the temporary electric current that flows through a long thin wire, when a voltage is introduced between two parallel metal plates, close to each other but not touching, via that wire.
In addition to his work on the distribution of energy among the molecules in a gas, which we looked at in the second part of the post, Maxwell summarized the existing knowledge about electricity and magnetism into equations now called Maxwell's equations, and after identifying and correcting a logical inconsistency in these equations, he showed that they implied the possible existence of waves of oscillating electric and magnetic fields, whose speed of propagation would be equal within observational errors to the speed of light, which was roughly known from Olaf Romer's observation, made around 1676, of a 16 minute time lag between the motions of Jupiter's moons as seen from Earth on the far side of the Sun from Jupiter, and as seen from Earth on the same side of the Sun as Jupiter, together with the distance from the Earth to the Sun, which was roughly known from simultaneous observations of Mars in 1672 from opposite sides of the Atlantic by Giovanni Domenico Cassini and Jean Richer, and observations of the transit of Venus. The speed of light had also been measured in the laboratory by Hippolyte Fizeau in 1849, and more accurately by Léon Foucault in 1862. Maxwell therefore suggested that light was electromagnetic radiation, and that electromagnetic radiation of wavelengths outside the visible range, which from Thomas Young's experiments with double slits was known to comprise wavelengths between about metres for violet light and metres for red light, would also exist. This was the other part of the clue that led to the discovery of quantum mechanics and Feynman's functional integral.
In his writings around 600 BC, Thales of Miletus described how amber attracts light objects after it is rubbed. The Greek word for amber is elektron, which has been adapted to the English word electron, for the first of the elementary matter particles of the Standard Model to be discovered. Benjamin Franklin and Sir William Watson suggested in 1746 that the two types of static electricity, known as vitreous and resinous, corresponded to a surplus and a deficiency of a single "electrical fluid" present in all matter, whose total amount was conserved. Matter with a surplus of the fluid was referred to as "positively" charged, and matter with a deficiency of the fluid was referred to as "negatively" charged. Objects with the same sign of charge repelled each other, and objects with opposite sign of charge attracted each other. Around 1766, Joseph Priestley suggested that the strength of the force between electrostatic charges is inversely proportional to the square of the distance between them, and this was approximately experimentally verified in 1785 by Charles-Augustin de Coulomb, who also showed that the strength of the force between two charges is proportional to the product of the charges.
Most things in the everyday world have no net electric charge, because the charges of the positively and negatively charged particles they contain cancel out. In particular, a wire carrying an electric current usually has no net electric charge, because the charges of the moving particles that produce the current are cancelled by the opposite charges of particles that can vibrate about their average positions, but have no net movement in any direction.
Jean-Baptiste Biot and Félix Savart discovered in 1820 that a steady electric current in a long straight wire produces a magnetic field in the region around the wire, whose direction is at every point perpendicular to the plane defined by the point and the wire, and whose magnitude is proportional to the current in the wire, and inversely proportional to the distance of the point from the wire. André-Marie Ampère discovered in 1826 that this magnetic field produces a force between two long straight parallel wires carrying electric currents, such that the force is attractive if the currents are in the same direction and repulsive if the currents are in opposite directions, and the strength of the force is proportional to the product of the currents, and inversely proportional to the distance between the wires. Thus the force on either wire is proportional to the product of the current in that wire and the magnetic field produced at the position of that wire by the other wire, and the direction of the force is perpendicular both to the magnetic field and the direction of the current.
Ampère's law is used to define both the unit of electric current, which is called the amp, and the unit of electric charge, which is called the coulomb. The amp is defined to be the electric current which, flowing along each of two very long straight parallel thin wires one metre apart in a vacuum, produces a force of kilogram metres per second between them, per metre of their length. The coulomb is then defined to be the amount of moving electric charge which flows in one second through any cross-section of a wire carrying a current of one amp. Electric currents are often measured in practice by moving-coil ammeters, in which the deflection of the indicator needle is produced by letting the current flow through a movable coil suspended in the field of a permanent magnet, that has been calibrated against the magnetic field produced by a current-carrying wire.
Maxwell interpreted the force on a wire carrying an electric current in the presence of a magnetic field as being due to a force exerted by the magnetic field on the moving electric charge carriers in the wire, and defined the magnetic induction to be such that, in Cartesian coordinates, the force on a particle of electric charge moving with velocity in the magnetic field , is:
Here each index , , or can take values 1, 2, 3, corresponding to the directions in which the three Cartesian coordinates of spatial position increase. I explained the meaning of the symbol in the first part of the post, here. represents the collection of data that gives the value of the magnetic field in each coordinate direction at each position in space and each moment in time, so that if represents a position in space, the value of the magnetic field in coordinate direction , at position , and time , could be represented as or , for example. If represents the collection of data that gives the particle's position at each time , then . represents the collection of data that gives the force on the particle in each coordinate direction at each moment in time.
The symbol is an alternative form of the Greek letter epsilon. The expression is defined to be 1 if the values of , , and are 1, 2, 3 or 2, 3, 1 or 3, 1, 2; if the values of , , and are 2, 1, 3 or 3, 2, 1 or 1, 3, 2; and 0 if two or more of the indexes have the same value. Thus the value of changes by a factor if any pair of its indexes are swapped. A quantity that depends on two or more direction indexes is called a tensor, and a quantity whose value is multiplied by if two of its indexes of the same type are swapped is said to be "antisymmetric" in those indexes. Thus is an example of a totally antisymmetric tensor.
A quantity that depends on position and time is called a field, and a quantity that depends on one direction index is called a vector, so the magnetic induction is an example of a vector field. From the above equation, the unit of the magnetic induction is kilograms per second per coulomb.
Since no position or time dependence is displayed in the above equation, the quantities that depend on time are all understood to be evaluated at the same time, and the equation is understood to be valid for all values of that time. The magnetic field is understood to be evaluated at the position of the particle, and the summations over and are understood to go over all the values of and for which the expressions are defined. Thus if we explicitly displayed all the indexes and the ranges of the summations, the equation could be written:
Maxwell also interpreted the electrostatic force on an electrically charged particle in the presence of another electrically charged particle as being due to a force exerted by an electric field produced by the second particle, and defined the electric field strength to be such that, in the same notation as before, the force on a particle of electric charge in the electric field , is:
Thus the unit of the electric field strength is kilograms metres per second per coulomb, which can also be written as joules per metre per coulomb, since a joule, which is the international unit of energy, is one kilogram metre per second. The electric field strength is another example of a vector field.
Electric voltage is the electrical energy in joules per coulomb of electric charge. Thus if the electrostatic force can be derived from a potential energy by , as in the example for which we derived Newton's second law of motion from de Maupertuis's principle, then the electric field strength is related to by . I have written the potential energy here as instead of , to avoid confusing it with voltage. is the potential voltage, so the electric field strength is minus the gradient of the potential voltage, and the unit of electric field strength can also be expressed as volts per metre.
The voltage produced by a voltage source such as a battery can be measured absolutely by measuring the current that flows and the heat that is produced, when the terminals of the voltage source are connected through an electrical resistance. In all currently known electrical conductors at room temperature, an electric current flowing through the conductor quickly stops flowing due to frictional effects such as scattering of the moving charge carriers by the stationary charges in the material, unless the current is continually driven by a voltage difference between the ends of the conductor, that produces an electric field along the conductor. The work done by a voltage source of volts to move an electric charge of coulombs from one terminal of the voltage source to the other is joules, so if a current of amps coulombs per second is flowing, the work done by the voltage source per second is joules per second watts, since a watt, which is the international unit of power, is one joule per second. Thus the voltage produced by a voltage source can be measured absolutely by connecting the terminals of the voltage source by for example a long thin insulated copper wire that is coiled in a thermally insulated flask of water, and measuring the electric current and the rate at which the water temperature rises, since the specific heat capacity of water is known from measurements by James Joule to be about 4180 joules per kilogram per degree centigrade.
Maxwell summarized Coulomb's law for the electrostatic force between two stationary electric charges by the equation:
Here is a vector field called the electric displacement, whose relation to the electric field strength at a position depends on the material present at . has the same meaning as in the first part of the post, here, with now taken as , and now taken as . is the Greek letter rho, and represents the collection of data that gives the amount of electric charge per unit volume, at each spatial position and time . It is called the electric charge density. For each position and time , it is defined to be the amount of electric charge inside a small volume centred at , divided by , where the ratio is taken in the limit that tends to 0. A field that does not depend on any direction indexes is called a scalar field, so is an example of a scalar field. The units of are coulombs per metre, so the units of are coulombs per metre.
In most materials the electric displacement and the electric field strength are related by:
where is a number called the permittivity of the material. Although the same symbol is used for the permittivity and the antisymmetric tensor , I will always show the indices on , so that it can't be mistaken for the permittivity.
To check that the above equation summarizing Coulomb's law leads to the inverse square law for the electrostatic force between stationary point-like charges as measured by Coulomb, we'll calculate the electric field produced by a small electrically charged sphere. We'll choose the zero of each of the three Cartesian coordinates to be at the centre of the sphere, and represent the radius of the sphere by . The electric charge per unit volume, , might depend on position in the sphere, for example the charge might be concentrated in a thin layer just inside the surface of the sphere. We'll assume that , the value of at position , does not depend on the direction from to the centre of the sphere, although it might depend on the distance from to the centre of the sphere. The electric displacement at position will be directed along the straight line from to the centre of the sphere, so for , where is a quantity that depends on . From Leibniz's rule for the rate of change of a product, which we obtained in the first part of the post, here, we have:
And since only depends on through the dependence of on , we have:
The values of the components of other than are fixed throughout this formula, so their values don't need to be displayed. From this formula and the previous one:
Leibniz's rule for the rate of change of a product also gives us:
and since , it also gives us:
since, for example, , while . Thus
so from the previous formula,
Thus from Maxwell's equation summarizing Coulomb's law, above:
From Leibniz's rule for the rate of change of a product, we have:
where the final equality follows from the result we obtained in the second part of the post, here, with taken as 3 and taken as . Thus after multiplying the previous equation by , it can be written:
So from the result we found in the first part of the post, here, that the integral of the rate of change of a quantity is equal to the net change of that quantity, we find that for any two particular values and of :
The expression is the total electric charge in the region between distances and from the centre of the sphere, divided by the surface area of a sphere of radius 1, which I'll represent by . For the surface area of a sphere of radius is , since if we use angular coordinates such as latitude and longitude to specify position on the surface of the sphere, the distance moved as a result of a change of an angular coordinate is proportional to . Thus the contribution to the integral from the interval from to is aproximately times the total electric charge in the spherical shell between distances and from , since the volume of this shell is approximately , and the errors of these two approximations tend to 0 more rapidly than in proportion to as tends to 0.
Let's now assume that is finite throughout the sphere, and depends smoothly on as tends to 0. Then is finite as tends to 0, so:
Thus for greater than the radius of the charged sphere, we have:
where is the total electric charge of the sphere. Thus if is outside the sphere, then the electric displacement at is given by:
for . Thus the electric field strength in the region outside the sphere is given by:
where is the permittivity of the material in the region outside the sphere. So the force on a particle of electric charge at position outside the sphere is given by:
This is in agreement with Coulomb's law, since is a vector of length 1, that points along the line from the centre of the sphere to . The force is repulsive if and have the same sign, and attractive if they have opposite signs.
We'll calculate the surface area of a sphere of radius 1 by using the result we found in the second part of the post, here, that . We have:
We can also think of , , and as the Cartesian coordinates of a point in 3-dimensional Euclidean space. The distance from the point to the point is then . So from the discussion above, with taken as , the above triple integral is equal to , so we have:
The value of the expression is unaltered if we replace by , so we also have:
So from the result we found in the second part of the post, here:
Thus the force on a particle of electric charge at position outside the sphere is given by:
The permittivity of the vacuum is denoted by . The expression is the number that determines the overall strength of the electrostatic force between two stationary charges, so it plays the same role for the electrostatic force as Newton's constant plays for the gravitational force.
The value of the permittivity, , whose unit is joule metres per coulomb, or equivalently kilogram metre per second per coulomb, can be measured for a particular electrical insulator by placing a sample of the insulator between the plates of a parallel plate capacitor, which consists of two large parallel conducting plates separated by a thin layer of insulator, then connecting a known voltage source across the plates of the capacitor, and measuring the time integral of the resulting current that flows along the wires from the voltage source to the capacitor, until the current stops flowing. The current is the rate of change of the charge on a plate of the capacitor, so since the integral of the rate of change is the net change, as we found in the first part of the post, here, the time integral of the current is the total electric charge that ends up on a plate of the capacitor.
Once the current has stopped flowing, the voltage no longer changes along the wires from the terminals of the voltage source to the plates of the capacitor, so the entire voltage of the voltage source ends up between the plates of the capacitor. If the lengths of the sides of the capacitor plates are much larger than the distance between the plates, and the 1 and 2 coordinate directions are in the plane of the plates, then the electric field strength between the plates is , where is the voltage of the voltage source, and is the distance between the plates.
If the electric charge on a plate of the capacitor is and is uniformly distributed over the capacitor plate, and the area of each capacitor plate is , then by integrating Maxwell's equation across the thickness of a capacitor plate and noting that the electric field is 0 outside the plates, we find:
since the integral of across the thickness of a capacitor plate is equal to the difference of between the inner and outer faces of that capacitor plate, by the result we found in the first part of the post, here, that the integral of the rate of change of a quantity is equal to the net change of that quantity; and the integral of the electric charge per unit volume, , across the thickness of a capacitor plate is equal to the electric charge per unit area, , on the capacitor plate.
Thus since , , , and are all known, the value of for the electrical insulator between the capacitor plates is determined. From measurements of this type with a vacuum between the capacitor plates, the permittivity of a vacuum is found to be such that:
Maxwell summarized Ampère's law for the force between two parallel electric currents, as above, by the equation:
Here is a vector field called the electric current density. For each position , time , and value 1, 2, or 3 of the coordinate index , it is defined to be the net amount of electric charge that passes in the positive direction through a small area perpendicular to the direction in a small time , divided by , where the ratio is taken in the limit that and tend to 0. The units of are amps per metre. is the totally antisymmetric tensor I defined above. is a vector field called the magnetic field strengh, whose relation to the magnetic induction at a position depends on the material present at . The units of are amps per metre. has the same meaning as in the first part of the post, here, with now taken as , and now taken as .
In most non-magnetized materials the magnetic induction and the magnetic field strength are related by:
where , which is the Greek letter mu, is a number called the permeability of the material. Its unit is kilogram metres per coulomb. The permeability of the vacuum is denoted by . Its value is fixed by the definition of the amp, as above.
To check that the above equation summarizing Ampère's law leads to a force between two long straight parallel wires carrying electric currents, whose strength is inversely proportional to the distance between the wires as measured by Ampère, and to calculate the value of implied by the definition of the amp, as above, we'll calculate the magnetic field produced by an infinitely long straight wire that is carrying an electric current. We'll choose the wire to be along the 3 direction, and the zero of the 1 and 2 Cartesian coordinates to be at the centre of the wire, and represent the radius of the wire by . We'll assume that , the electric current density in the direction along the wire at position , does not depend on or the direction from to the centre of the wire, although it might depend on the distance from to the centre of the wire.
From its definition above, the antisymmetric tensor is 0 if any two of its indexes are equal, so in particular, is 0 for all values of the index . Thus Maxwell's equation summarizing Ampère's law, as above, does not relate to , so we'll assume is 0.
Now let's suppose that the magnetic field strength at position is directed along the straight line perpendicular to the wire from to the centre of the wire, so for , where is a quantity that depends on . Then in the same way as above, we find:
and also in the same way as above, we find:
so:
From Leibniz's rule for the rate of change of a product, which we obtained in the first part of the post, here, we have:
Thus:
so Maxwell's equation summarizing Ampère's law, as above, does not relate to this form of , so we'll also assume that this form of is 0.
The final possibility is that the magnetic field strength at position is perpendicular to the plane defined by and the wire carrying the current. Then , and from the diagram in the first part of the post, here, interpreted as the two-dimensional plane through and perpendicular to the wire, if and , then the direction of is along , so , , where is a quantity that depends on . Then from Leibniz's rule for the rate of change of a product, which we obtained in the first part of the post, here, and the formula above for , we have:
Thus from Maxwell's equation summarizing Ampère's law, as above:
From Leibniz's rule for the rate of change of a product, we have:
where the final equality follows from the result we obtained in the second part of the post, here, with taken as 2 and taken as . Thus after multiplying the previous equation by , it can be written:
So from the result we found in the first part of the post, here, that the integral of the rate of change of a quantity is equal to the net change of that quantity, we find that for any two particular values and of :
The expression is times the total electric charge per unit time passing through the region between distances and from the centre of the wire, in any cross-section of the wire. For the circumference of a circle of radius is , so the contribution to the integral from the interval from to is aproximately times the total electric charge per unit time passing through the region between distances and from the centre of the wire, in any cross-section of the wire, since the area of this shell is approximately , and the errors of these two approximations tend to 0 more rapidly than in proportion to as tends to 0.
Let's now assume that is finite throughout the cross-section of the wire, and depends smoothly on as tends to 0. Then is finite as tends to 0, so:
Thus for greater than the radius of the wire, we have:
where is the total electric current carried by the wire. Thus if is outside the wire, then the magnetic field strength at is given by:
Thus the magnetic induction in the region outside the wire is given by:
where is the permeability of the material in the region outside the wire. This is perpendicular to the plane defined by the point and the wire, and its magnitude is proportional to the current in the wire, and inversely proportional to the distance of the point from the wire, in agreement with the measurements of Biot and Savart as above, since is a vector of length 1.
Let's now suppose there is a second infinitely long straight wire parallel to the first, such that the 1 and 2 Cartesian coordinates of the centre of the second wire are , and the total electric current carried by the second wire is . From Maxwell's equation above, and the definition of the antisymmetric tensor as above, the force on a particle of electric charge moving with velocity along the second wire, in the presence of the magnetic field produced by the first wire, as above, is given by:
The interaction between this moving charge and the other particles in the second wire prevents this moving charge from accelerating sideways out of the second wire, so the above force is a contribution to the force on the second wire, that results from the magnetic field produced by the current in the first wire. If there are particles of electric charge and velocity per unit length of the second wire, then their contribution to the force per unit length on the second wire is:
The average number of these particles that pass through any cross-section of the second wire per unit time is , so their contribution to the electric current carried by the second wire is . Thus the contribution of these particles to the force per unit length on the second wire is times their contribution to the electric current carried by the second wire. So by adding up the contributions from charged particles of all relevant values of and , we find that the total force per unit length on the second wire that results from the current carried by the first wire and the current carried by the second wire is given by:
The direction of this force is towards the first wire if and have the same sign and away from the first wire if and have opposite sign, and the strength of this force is proportional to the product of the currents, and inversely proportional to the distance between the wires, so this force is in agreement with Ampère's law, as above. And from the definition of the amp, as above, we find that the permeability of a vacuum is by definition given by:
Maxwell noticed that his equation summarizing Ampère's law, as above, leads to a contradiction. For by applying to both sides of that equation, and summing over , we obtain:
For a quantity that depends smoothly on a number of quantities that can vary continuously, where represents the collection of those quantities, and indexes such as or distinguish the quantities in the collection, we have:
The expression in the third line here is equal to the expression we obtain from it by swapping the indexes and , so we have:
So if the magnetic field strength depends smoothly on position, we also have:
The value of the right-hand side of this formula does not depend on the particular letters , , and used for the indexes that are summed over. Thus if the letter , used as an index, is also understood to have the possible values 1, 2, or 3, we have:
At each of the first three steps in the above formula, one of the indexes summed over in the previous version of the expression is rewritten as a different letter that is understood to take the same possible values, 1, 2, or 3, and which does not otherwise occur in the expression. At the first step, the index is rewritten as , then at the second step, the index is rewritten as , and at the third step, the index is rewritten as . An index that occurs in an expression, but is summed over the range of its possible values, so that the full expression, including the , does not depend on the value of that index, is called a "dummy index".
The fourth step in the above formula used the definition of the antisymmetric tensor , as above, which implies that its value is multiplied by if two of its indexes are swapped, so that . The fifth step used the original formula for , as above, together with the fact that the order of the indexes under the in the right-hand side doesn't matter, since each of the indexes is simply summed over the values 1, 2, and 3.
Thus from the second formula for , as above, we have:
Hence:
Let's now consider the rate of change with time of the total electric charge in a tiny box-shaped region centred at a position , such that the edges of the box are aligned with the coordinate directions, and have lengths , , and . From the definition above of the electric current density , the net amount of electric charge that flows into the box through the face of the box perpendicular to the 1 direction and centred at , during a small time , is approximately , and the net amount of electric charge that flows out of the box through the face of the box perpendicular to the 1 direction and centred at , during the same small time , is approximately , and the errors of these approximations tend to 0 more rapidly than in proportion to , as , , and tend to 0. And from the result we obtained in the first part of the post, here, with taken as and taken as , we have:
where the error of the above approximations tends to 0 more rapidly than in proportion to , as tends to 0. Thus the net amount of electric charge that flows into the box through the faces of the box perpendicular to the 1 direction, during a small time , is approximately:
where the error of this approximation tends to 0 more rapidly than in proportion to , as , , , and tend to 0. So from the corresponding results for the net amount of electric charge that flows into the box through the faces of the box perpendicular to the 2 and 3 directions, during the same small time , we find that the net amount of electric charge that flows into the box through all the faces of the box, during the small time , is approximately:
where the error of this approximation tends to 0 more rapidly than in proportion to , as , , , and tend to 0.
There's no evidence that electric charge can vanish into nothing or appear from nothing, so the net amount of electric charge that flows into the box through all the faces of the box, during the small time , must be equal to the net increase of the total electric charge in the box, during the small time , which from the definition of the electric charge density , as above, is approximately:
where the error of this approximation tends to 0 more rapidly than in proportion to , as , , , and tend to 0. Thus we must have:
But we found above that Maxwell's equation summarizing Ampère's law, as above, leads instead to . This equation is false whenever there is a build-up of electric charge in a region, as, for example, on the plates of a parallel plate capacitor, in the method of measuring the permittivity of an electrical insulator, that I described above. Maxwell realized that the resolution of this paradox is that there must be an additional term in the left-hand side of his equation summarizing Ampère's law, where is the electric displacement vector field, so that the corrected form of his equation summarizing Ampère's law is:
This equation still correctly reproduces Ampère's law and the magnetic field produced by an electric current flowing in a long straight wire as measured by Biot and Savart, as I described above, because the experiments of Ampère and Biot and Savart were carried out in steady state conditions, where nothing changed with time, so the new term in the left-hand side gave 0. However if we apply to both sides of this corrected equation, and sum over , which is what led to the paradox for the original equation, we now find:
So if the electric displacement depends smoothly on position, so that , by the result we found above, we find:
Combining this with Maxwell's equations summarizing Coulomb's law, as above, it gives:
which is now in agreement with the formula expressing the conservation of electric charge, as above.
Michael Faraday discovered in 1831 that if an electrically insulated wire is arranged so that somewhere along its length it forms a loop, and the magnetic induction field inside the loop and perpendicular to the plane of the loop is changed, for example by switching on a current in a separate coil of wire in a suitable position near the loop, then a voltage is temporarily generated along the wire while the magnetic induction field is changing, such that if the directions of the 1 and 2 Cartesian coordinates are in the plane of the loop, and the value of in the region enclosed by the loop in the plane of the loop depends on time but not on position within that region, then:
where is the area enclosed by the loop, and the sign depends on the direction along the wire in which the voltage is measured. The sign of the voltage is such that if a current flows along the wire in consequence of the voltage, then the magnetic field produced by that current, as above, is such that in the region enclosed by the loop in the plane of the loop has the opposite sign to .
Maxwell assumed that the electric field strength that corresponds to the voltage is produced by the changing magnetic induction field even when there is no wire present to detect in a convenient way. To discover the consequences of this assumption, it is helpful to know about the relation between the electric field strength and the rate of change of voltage with distance in a particular direction.
For any vector , and any vector of length 1, the expression is called the component of in the direction . To relate this to the magnitude of , which is by Pythagoras, and the angle between the directions of and , we observe that is a vector of length 1, and if we consider and as representing the Cartesian coordinates of two points in the 3-dimensional generalization of Euclidean geometry, as in the first part of this post, here, then by Pythagoras, the distance between those points is:
If does not point either in the same direction as or the opposite direction to , so that is not equal to , then the directions of and define a 2-dimensional plane, and we can choose Cartesian coordinates in that 2-dimensional plane as in the first part of the post, here, such that the coordinates of are , and the coordinates of are . So by Pythagoras, the distance between the points they define is:
This is equal to the previous expression, so we have:
This formula is also true when , so . If is along the coordinate direction, this formula shows that , where is the angle between the direction of and the coordinate direction. Thus for any vector of length 1, is equal to the value that the coordinate of in the direction would have, if was one of the coordinate directions of Cartesian coordinates.
If the electric field strength can be derived from a voltage field , so that as above, then at each point along the electrically insulated wire, we have:
where is the distance along the wire from that point to a fixed end of the wire, and is a vector of length 1 whose direction is along the wire in the direction of increasing . The first equality here is the component of the equation in the direction along the wire. The component of in any direction is the rate of change of with distance in that direction, so the component of in the direction along the wire is the rate of change of with distance along the wire, which is the second equality.
The movable electrically charged particles in the wire are channelled by the electrical insulation of the wire so that their net motion can only be along the wire, and only the component of the electric field strength along the wire can affect their net motion. Their motion along the wire due to the force is determined by a voltage defined along the wire such that
as in the previous formula, even if the voltage defined along the wire does not correspond to a voltage field in the region outside the wire.
Let's consider Faraday's result, as above, for a very small rectangular loop centred at , such that the edges of the loop are in the 1 and 2 Cartesian coordinate directions and have lengths and . We'll assume that the wire arrives at and leaves the rectangle at the corner at , and that the two lengths of wire that run from this corner of the rectangle to the measuring equipment, such as a voltmeter, follow exactly the same path. Then if the voltage along the wire is related to an electric field strength as in the above formula, the net voltage difference between the ends of the wire, as measured by Faraday, must be produced by the electric field strength along the sides of the rectangle, because any voltages produced along the lengths of wire that run from the corner of the rectangle to the measuring equipment will be equal and opposite along the two lengths of wire, and thus cancel out of the net voltage.
We'll choose to be the distance along the wire from the end of the wire such that increases along the side of the rectangle from the corner at to the corner at , then along the side from this corner to the corner at , then along the side from this corner to the corner at , and finally along the side from this corner to the first corner at . The components of the vector of length 1, that points along the four sides of the rectangle in the direction of increasing , are therefore , , , and , for the four sides of the rectangle taken in this order.
The net change of the voltage around the rectangle in the direction of increasing is equal to the sum of the net change of the voltage along the four sides of the rectangle in the direction of increasing , so from the formula above, and the result we found in the first part of the post, here, that the integral of the rate of change of a quantity is equal to the net change of that quantity, is equal to the sum of the integrals along the four sides of the rectangle.
For near in the plane of the rectangle, the result we obtained in the first part of the post, here, with taken as and taken as , gives:
where as the magnitudes of and tend to 0, the error of this approximate representation tends to 0 more rapidly than in proportion to those magnitudes.
The coordinates of a point a distance along the first side of the rectangle from the first corner of this side are . And along this side, is equal to plus a constant value, the length of the wire from its first end to the first corner of this side, so . Thus since for this side, we have:
where the error of this approximation tends to 0 more rapidly than in proportion to or as and tend to 0, and I used the result we found in the first part of the post, here, that the integral of the rate of change of a quantity is equal to the net change of that quantity, and also and , from the result we found in the second part of the post, here.
The coordinates of a point a distance along the third side of the rectangle from the first corner of that side are . We again have , so since for that side, we have:
to the same accuracy as before. Thus:
where the error of this approximation tends to 0 more rapidly than in proportion to , as and tend to 0 with their ratio fixed to a finite non-zero value.
Similarly we find:
to the same accuracy. So:
to the same accuracy. Thus:
where the error of this approximation tends to 0 more rapidly than in proportion to , as and tend to 0 with their ratio fixed to a finite non-zero value.
Thus from Faraday's measurements, as above:
since is the area of the small rectangle. We have obtained this equation at the position of the centre of the small rectangle, so it holds everywhere the small rectangle of wire could have been placed.
To determine the sign, let's suppose that and are 0 at the centre of the small rectangle, and that is positive along the 1st side of the rectangle and negative along the 3rd side, and is positive along the 2nd side of the rectangle and negative along the 4th side. Then is negative and is positive, so is negative, and the force on a movable charged particle of positive is in the direction of increasing along all four sides of the rectangle, so the current along the wire is positive in the direction of increasing .
From the result we found above, Maxwell's equation summarizing Ampère's law, as above, implies that a positive current along a wire in the 3 direction produces a magnetic induction field such that is negative for greater than the coordinate of the wire and positive otherwise, and is positive for greater than the coordinate of the wire and negative otherwise.
The antisymmetric tensor , which I defined above, is unaltered by a cyclic permutation of its indexes, for example or , so Maxwell's equation summarizing Ampère's law, as above, also implies that a positive current along a wire in the 1 direction produces a magnetic induction field such that is positive for greater than the coordinate of the wire and negative otherwise, and a positive current along a wire in the 2 direction produces a magnetic induction field such that is negative for greater than the coordinate of the wire and positive otherwise.
Thus if the current along the wire is positive in the direction of increasing , the magnetic induction field produced by the current along each side of the small rectangle is such that is positive inside the small rectangle. So from the observed sign of the voltage, as I described above, a positive value of inside the small rectangle produces electric field strengths that result in a current along the wire that is negative in the direction of increasing , and thus of opposite sign to those I assumed above. Thus positive produces positive , so the formula with the correct sign is:
The corresponding formulae that result from considering small rectangles whose edges are in the 2 and 3 or 3 and 1 Cartesian coordinate directions are obtained from this formula by cyclic permutation of the indexes, and the three formulae can be written as:
where is the totally antisymmetric tensor I defined above. This is Maxwell's equation summarizing Faraday's measurements involving time-dependent magnetic fields, as above.
From the discussion above, if the electric field strength can be derived from a voltage field , then . The electric field strength produced by the changing magnetic induction field in accordance with the above equation cannot be derived from a voltage field , for if , and depends smoothly on position, then by a similar calculation to the one above, we have:
No magnetically charged particles, often referred to as magnetic monopoles, have yet been observed, and Maxwell's equation summarizing this fact, analogous to his equation summarizing Coulomb's law, as above, is:
Although it's not possible to derive the electric field strength from a voltage field alone if the magnetic induction field is time-dependent, Maxwell's equation summarizing Faraday's measurements involving time-dependent magnetic fields, as above, and his equation above summarizing the non-observation of magnetic monopoles, can always be solved by deriving the electric field strength and the magnetic induction field from a voltage field and a vector field called the vector potential, such that:
For if has this form, then the left-hand side of Maxwell's equation summarizing Faraday's measurements involving time-dependent magnetic fields, as above, is:
where I assumed that the vector potential depends smoothly on position and time, and used the result we found above. And if has the above form, then the right-hand side of that equation is:
where I again used the result we found above. This is equal to the left-hand side as above, so if and are derived from a voltage field and a vector potential field as above, then Maxwell's equation summarizing Faraday's measurements involving time-dependent magnetic fields, as above, is solved.
And if the magnetic induction field is derived from a vector potential field as above, then the left-hand side of Maxwell's equation above summarizing the non-observation of magnetic monopoles is:
by a similar calculation to the one above. Thus Maxwell's equation summarizing the non-observation of magnetic monopoles is also solved.
If the electric field strength and the magnetic induction field are derived from a voltage field and a vector potential field , as above, then in a vacuum, where the electric displacement field is related to the electric field strength by , and the magnetic field strength is related to the magnetic induction field by , Maxwell's equation summarizing Coulomb's law, as above, becomes:
And Maxwell's corrected equation summarizing Ampère's law, as above, becomes:
where each index from the start of the lower-case English alphabet can take values 1, 2, 3, corresponding to the directions in which the three Cartesian coordinates of spatial position increase.
To simplify the above formula, we'll consider the expression:
where is the Kronecker delta, that I defined in the first part of the post, here, so its value is 1 when , and 0 otherwise. Thus , and . A quantity whose value is unchanged if two of its indexes of the same type are swapped is said to be "symmetric" in those indexes, so from the definition of a tensor, as above, is an example of a symmetric tensor.
In the above definition of the tensor , each term in the right-hand side after the first term is obtained from the first term by leaving the indexes , , and in the same positions as in the first term, and swapping the indexes , , and among themselves. Among the 6 terms in the right-hand side, each of the possible sequences of the letters occurs exactly once, where for each non-negative whole number , I defined in the second part of the post here, and we observed in the second part of the post, here, that the number of different ways of putting distinguishable objects in distinguishable places, such that exactly one object goes to each place, is Thus the number of different sequences of different letters is .
A re-ordering of a sequence of different letters is called a permutation of the sequence. The sign of each term in the right-hand side of the above definition of is a sign associated with the permutation that changes the sequence into the sequence in which the letters occur in that term, and is defined in the following way. For any permutation of a sequence of different letters, a cycle of the permutation is a sequence of the letters such that the final position of each letter of the cycle is the same as the initial position of the next letter of the cycle, except that the final position of the last letter of the cycle is the same as the initial position of the first letter of the cycle. For example, for the permutation , the letter by itself is a cycle, and is a cycle. Two cycles are considered to be equivalent if they have the same letters, and the number of letters in a cycle is called its length. The sign associated with a permutation, which is called the sign of the permutation, is the sign of , where is the number of inequivalent cycles of even length. For example the second term in the right-hand side of the above definition of corresponds to the permutation , which has just one cycle whose length is 3, so its sign is .
If a permutation is followed by another permutation that just swaps two letters, then the cycles of the resulting permutation are the same as the cycles of the original permutation, except that if the two swapped letters were originally in the same cycle, that cycle is divided into two cycles, each of which contains one of the swapped letters, while if the two swapped letters were originally in two different cycles, those two cycles are combined into a single cycle. If the two swapped letters were originally in a cycle of even length, then when that cycle is divided into two cycles, the number of cycles of even length either increases by 1 or decreases by 1, so is multiplied by . If the two swapped letters were originally in a cycle of odd length, then when that cycle is divided into two cycles, one of the resulting cycles has even length and the other has odd length, so the number of cycles of even length increases by 1, so is again multiplied by . And if the two swapped letters were originally in two different cycles then the reverse of one of the preceding cases occurs, so is again multiplied by . Thus swapping any two letters reverses the sign of a permutation.
Thus if any two of the last three indexes of are swapped, the value of is multiplied by , so in accordance with the definition above, is antisymmetric in its last three indexes. Thus the value of must be 0 when any two of its last three indexes have the same value, for example must be 0, since swapping the 4th and 5th indexes of multiplies its value by . The only possible values of each index are 1, 2, or 3, so is 0 unless is one of the possibilities , , , , , or , and furthermore,
From the definition of the antisymmetric tensor , as above, this implies that:
since this equation is true for because , and it is therefore also true for all the other values of for which is non-zero, by the preceding equation and the definition of , as above, and it is also true whenever two of the indexes have the same value, since both sides of the equation are then 0.
We can also write the definition above of as:
which is the same as the formula above, except that I have changed the order of the factors in each term after the first, so that the indexes now occur in the same order in every term, while the order of the indexes is now different in each term. Each term now corresponds to one of the 6 permutations of the letters , and the sign of each term is the sign of the corresponding permutation of the letters . So in the same way as above, we find that is also antisymmetric in its first three indexes, and that:
From this and the formula above, we find:
This is true for all cases where the indexes take values 1, 2, or 3, and from the case where , we find:
Thus:
Furthermore, , since for these values of the indexes, only the first term in the right-hand side of the above definition of is non-zero, and its value is 1. Thus , so:
We now observe that:
and
since the indexes and can take values 1, 2, or 3, and from the definition of the Kronecker delta in the first part of the post, here, the product is non-zero for exactly one value of if , namely for , in which case , while if , where the symbol means, "is not equal to," then and are not both non-zero for any value of .
Thus:
Thus Maxwell's corrected equation summarizing Ampère's law, as above, expressed in terms of the voltage field and the vector potential field , as above, is:
We now observe that for any vector :
since is non-zero for exactly one value of , namely for , in which case it is 1.
Using this property of the Kronecker delta, we find:
Thus Maxwell's corrected equation summarizing Ampère's law, as above, expressed in terms of the voltage field and the vector potential field , as above, is:
We now observe that the formulae for the electric field strength and the magnetic induction field in terms of the voltage field and the vector potential field , as above, are unaltered if and are modified by the following replacements:
where is an arbitrary scalar field that depends smoothly on position and time. For the modified electric field strength and magnetic induction field are:
where I used the result we found above, and a similar calculation to the one above.
The replacement of the voltage field and the vector potential field by modified fields and , as above, which leaves the electric field strength and the magnetic induction field unaltered, and thus has no experimentally observable consequences, is called a gauge transformation.
We can simplify Maxwell's equation summarizing Coulomb's law, and his corrected equation summarizing Ampère's law, expressed in terms of the voltage field and the vector potential field , as above, and above, by doing a gauge transformation with a scalar field that satisfies:
We then find that:
Let's now assume that we've done a gauge transformation as above, so that:
This is called a gauge condition.
Then by the result we found above, the term in the left-hand side of Maxwell's equation summarizing Coulomb's law, expressed in terms of the voltage field and the vector potential field , as above, is equal to , so that equation becomes:
And Maxwell's corrected equation summarizing Ampère's law, expressed in terms of the voltage field and the vector potential field , as above, becomes:
Let's now consider a region where there are no electrically charged particles and no electric currents, so that and are 0. Then for any vector , and any angle , and any vector perpendicular to , which from the discussion above, means that , a solution of the above equations that satisfies the gauge condition above, which we used to simplify the equations, is given by:
For we found in the first part of the post, here, that and . So by a calculation similar to the one in the second part of the post, here, if and are any quantities independent of , then and . From these results with taken as and , we find:
From the second of these results, we find that is proportional to , so the gauge condition above is satisfied, and since by Pythagoras, the third and fourth of these results show that Maxwell's equation summarizing Ampère's law, as above, is satisfied. Maxwell's equation summarizing Coulomb's law, as above, is automatically satisfied, since for this solution.
From the formulae for the electric field strength and the magnetic induction field in terms of the voltage field and the vector potential field , as above, we find that for this solution:
Thus since we assumed that , we find that , which from the discussion above means that the vector is perpendicular to the vector . And from calculations similar to the one above we find that , so that , so the vector is perpendicular to the vector , and , so , so the electric field strength is perpendicular to the magnetic induction field .
This solution describes oscillating electric and magnetic fields moving in the direction at a speed . From above, the permittivity of a vacuum, measured from the electric charge stored on a paralle plate capacitor, is such that:
and from above, the definition of one amp one coulomb per second, in terms of the force between long parallel wires carrying steady electric currents, as above, implies that the permeability of a vacuum is by definition given by:
Thus:
which is the measured value of the speed of light. Maxwell therefore proposed that light is waves of oscillating electric and magnetic fields perpendicular to the direction of motion of the wave and to each other. The vector is called the wave vector, and the wave solutions above are called transverse waves, because the vector potential is perpendicular to the wave vector .
From calculations similar to those above, we find that the gauge condition above, and the field equations above, have another wave solution:
which is called a longitudinal wave, because the vector potential is parallel to the wave vector . However from the formulae for the electric field strength and the magnetic induction field in terms of the voltage field and the vector potential field , as above, we find that for this solution:
Thus the above longitudinal wave solution has no experimentally observable effects. It is called a pure gauge mode.
Maxwell suggested that there could also be transverse waves of electric and magnetic fields with frequencies outside the visible spectrum, and this was partly confirmed in 1879 by experiments by David Edward Hughes, and conclusively confirmed in 1886 when Heinrich Hertz generated and detected pulses of radio-frequency electromagnetic waves in his laboratory. This led to the utilization of radio-frequency electromagnetic waves for practical communications by Guglielmo Marconi, from around 1895.
Electromagnetic waves will also be emitted by hot objects, and will be present in a hot region that is in thermal equilibrium. It was the study of the electromagnetic waves in hot ovens, at the end of the nineteenth century, that provided the other part of the clue that led to the discovery of quantum mechanics and Richard Feynman's functional integral. In the next part of this post, Action for Fields, we'll look at how Maxwell's equations for the for the electric field strength and the magnetic induction field , expressed in terms of the voltage field and the vector potential field , as above, can be obtained from de Maupertuis's principle of stationary action, for a suitable action that depends on the electromagnetic fields and , and on the positions and motions of any electrically charged particles present, and in the part after that, Radiation in an Oven, we'll look at how the discoveries about heat and temperature that we looked at in the second part of this post, combined with the discoveries about electromagnetic radiation that we've looked at today, lead to a seriously wrong conclusion about the properties of electromagnetic radiation in a hot oven. In the subsequent parts of the post, we'll look at how that problem has been resolved by the discovery of quantum mechanics and Feynman's functional integral, which started with the identification of a new fundamental constant of nature by Max Planck, in 1899. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9760714769363403, "perplexity": 169.10080490304344}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187825812.89/warc/CC-MAIN-20171023073607-20171023093607-00282.warc.gz"} |
https://astrophytheory.com/2021/02/18/a-very-brief-introduction-to-measure-theory-and-the-lebesgue-integral-part-i/ | # A Very Brief Introduction to Measure Theory and the Lebesgue Integral (Part I)
In the next few posts, I shall be discussing recent topics of study that, to me at least, have been very intruiging. In previous posts, I have talked about Hilbert spaces. I have of late been considering the mathematics necessary to formally understand in a pure mathematical sense what a Hilbert space is. This post, like the others on this site, serves as a reference of newly learned topics that are of interest (to me, at least; such a comment is subjective, of course).
The purpose of this post is two-fold: (1.) to provide an update with what I’ve been up to; (2.) introduce some interesting mathematics that have expanded my understanding to the “size” of a set as well as operations such as differentiation and integration.
Here is a quick summary of what I plan to cover in the next few posts (to a brief extent):
1. Elementary Sets and their Measure: Here I will discuss the concept of length and try extend length in greater dimensions to that of a measure of a set. Much of this topic will rely on geometric intuition.
2. Lebesgue Measure: This section will dicuss the concept of Lebesgue measure and distinguish it from the elementary measure. Also brief mention will be made of measurability of sets and functions.
3. General Measure: Discussion will be made of a general measure as a function as well as measurable spaces and measure spaces.
4. Lebesgue Integral: This topic will introduce the concept of the Lebesgue integral as compared to the Riemann integral.
5. $L^{p}$ and $l^{p}$ spaces: This section will discuss the concept of a norm as it relates to the spaces $L^{p}$ and $l^{p}$, and will define each space. We will also introduce the concept of Banach spaces.
6. Proof that $l^{p}$ space is a Banach space.
Section 1: Elementary Sets and their Measure:
The question that we want to answer is this: Given an arbitrary set, how do we go about measuring it?
In order to understand the difficulties present in this question we must first consider what are called elementary sets and the elementary measure. Elementary sets are those sets which are intuitively easy to measure; that is, intervals, rectangles, and boxes. We now give the formal definition of an elementary set:
Definition. (Interval; Elementary Set) We define an interval to be a subset of the real line $\mathbb{R}$ which take one of the following forms:
$[a,b] := \{x\in \mathbb{R}|a\leq x \leq b\} \label{(1.1)}$;
$[a,b) := \{x\in \mathbb{R}|a\leq x < b\} \label{(1.2)}$;
$(a,b] := \{x\in \mathbb{R}|a< x\leq b\} \label{(1.3)}$;
$(a,b):= \{x\in \mathbb{R}|a< x< b\} \label{(1.4)}$,
The length of an interval $I=[a,b]$ denoted $l(I):= b-a$. For dimensions $d\geq 2$, we define the measure of such sets as equalling the $d$-times Cartesian product of intervals $I_{d}$; that is,
$\displaystyle m(B) := \prod_{i=1}^{d}l(I_{i}); \label{(2)}$
we sometimes call sets of dimension 2 or greater as “boxes.” Thus, elementary sets are those subsets of $\mathbb{R}^{d}$ such that
$\displaystyle m(E)= \bigcup_{i=1}^{d}m(B_{i}), \label{(3)}$
where $B$ is $i$-th $d$-dimensional box contained in $\mathbb{R}^{d}$.
What this definition is doing is the following: first it introduces the concept of an interval and establishes the well-understood concept of its length as being the difference between the two endpoints provided one is less than the other. The definition then generalizes the idea of a length to 2 and 3 dimensions and beyond. Note that in 2-dimensions the interval then becomes a rectangle in the plane. Thus, the measure of the length of an interval then becomes the measure of the area of a rectangle. Similarly, for $d=3$ we replace rectangles with cubes and the area with the volume. For dimensions $d>3$, we replace cubes with boxes of $d$-dimension. Therefore, elementary sets are those subsets of $d$-dimensional real space that are unions of finitely-many boxes.
Section 2: Lebesgue Measure
In the last section, we discussed sets for which we can measure quite easily. Though ideally we would like to be able to measure more general sets; that is, sets that are more general than elementary sets. Therefore, we require a different way of measurement. Thus, we come to need the Lebesgue measure.
In order to introduce the Lebesgue measure we need to first introduce the concept of the outer measure, which we now define
Definition. (Outer Measure) We define the outer measure of a set $E\subset \mathbb{R}$, denoted $m^{*}(E)$ to be
$\displaystyle m^{*}(E) = \inf\bigg\{\sum_{k=1}^{\infty}l(I_{k})|\forall k\in \mathbb{N}, I_{k} \text{ is open such that } E \subset \bigcup_{k=1}^{\infty}I_{k}\bigg\}.$
The outer measure of a set in a sense “overestimates” the size of a given set and then takes the smallest such overestimate to within a specified tolerance. Thus, it estimates the size of the given set “from the outside,” and is used in lieu of the elementary measure when we are dealing with sets that we cannot easily measure the set in a geometrically-intuitive way.
We conclude this post with the definition of the Lebesgue measure given in two forms; the first will be in terms of what we have defined so far, and the second will be defined in terms that will be covered in the next post.
Definition. (Lebesgue Measure.) We define the Lebesgue measure of the set $E$ to be a set whose measure $m(E)=m^{*}(E)$; that is, its measure is equal to the outer measure.
The second way of defining this is as follows:
Definition. (Lebesgue Measure V.2) The Lebesgue measure is the measure on the measureable space $(\mathbb{R},\mathcal{L})$ where $\mathcal{L}$ is the $\sigma$-algebra of Lebesgue measurable subsets of $\mathbb{R}$ that assigns to each Lebesgue measurable set its outer measure.
The next post will discuss measures in general, as well as measurable sets, measureable spaces, Borel sets, and $\sigma$-algebras.
Until then, clear skies! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 36, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9809733629226685, "perplexity": 246.73866915490075}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030334644.42/warc/CC-MAIN-20220926020051-20220926050051-00118.warc.gz"} |
http://mathoverflow.net/questions/28235/variants-of-gr%c3%b6nwalls-theorem | # Variants of Grönwall's theorem
Except the original Grönwall's theorem that $$\limsup_{n \to \infty} \frac{\sigma(n)}{n \log \log n} = e^{\gamma},$$ and the two variants $$\limsup_{\begin{smallmatrix} n\to\infty\cr n\ \text{is square free}\end{smallmatrix}} \frac{\sigma(n)}{n \log \log n} = \frac{6e^{\gamma}}{\pi^2}$$ and $$\limsup_{\begin{smallmatrix} n\to\infty\cr n\ \text{is odd}\end{smallmatrix}} \frac{\sigma(n)}{n \log \log n} = \frac{e^{\gamma}}{2}$$ that have been proven here, are there any other similar statements known?
-
Do you mean "similar statements" for the sum of divisors function $\sigma(n)=\sum_{d\mid n}d$? Because there are plenty of other multiplicative function for which similar asymptotics are known. – Wadim Zudilin Jun 15 '10 at 11:29
"that have been proven here," Where? – Andres Caicedo Jun 15 '10 at 13:47
I fixed the typos. Theorem 9 in the cited preprint contains 5 more similar asymptotics. I wonder what is wanted. – Wadim Zudilin Jun 15 '10 at 14:50
Maybe a statement with $\limsup_{\begin{smallmatrix}n\to \infty \cr n\in S\end{smallmatrix}}(\cdots)=d_S e^{\gamma}$. Where $d_S$ is the density of $S$. – Gjergji Zaimi Jun 15 '10 at 15:02
I'm sorry for not being clear enough, it's my first question here, though. I mean similar statments for the $\sigma(n)$ function, not necessarily asymptotics, but anything that involves limit points of the function $\frac{\sigma(n)}{n \log \log n}$. For example, is there an important sequence $a_n$ such that $\frac{\sigma{a_n}}{n \log \log n}$ converges, besides the sequence of primes? A result that establishes the connection between the density and the limit superior? Etc. Nothing particular. – nikmil Jun 15 '10 at 22:53
where the limit of the Choie, Lichiardopol, Moree and Sole's $$f_1(a_n) = \frac{\sigma(a_n)}{a_n \log \log a_n}$$ is the same $$e^\gamma .$$ That is, the limit for these numbers is the lim sup for all numbers.
These are more natural than people realize. There is a simple recipe that takes some $\epsilon > 0$ and gives an explicit factorization for the best value $n_\epsilon;$ see page 7 in the Briggs pdf "Notes on the Riemann hypothesis and abundant numbers" at the bottom of the Wikipedia entry. The exponent of a prime $p$ in the factorization of $n_\epsilon$ is $$\left\lfloor \log_p \left( \frac{p^{1 + \epsilon} - 1}{p^\epsilon -1} \right) \right\rfloor - 1$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9376604557037354, "perplexity": 423.7704045495328}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207927824.26/warc/CC-MAIN-20150521113207-00343-ip-10-180-206-219.ec2.internal.warc.gz"} |
http://mathhelpforum.com/trigonometry/69161-trig-identity-help.html | 1. ## Trig Identity Help
I am in gr.11 just starting trig identities. I have finished all of my hw except for one question. Which looks to me like it should be simple. I just can't seem to do it. haha i'm probably just being blonde but i'll keep trying! Thanks!
2. Originally Posted by DaCoo911
I am in gr.11 just starting trig identities. I have finished all of my hw except for one question. Which looks to me like it should be simple. I just can't seem to do it. haha i'm probably just being blonde but i'll keep trying! Thanks!
$\text{LHS }= \csc^2(x) + \sec^2(x) = \frac{1}{\sin^2(x)} + \frac{1}{\cos^2(x)}$
$= \frac{\cos^2(x)}{\sin^2(x)\cos^2(x)} + \frac{\sin^2(x)}{\sin^2(x)\cos^2(x)}$
$= \frac{\cos^2(x) + \sin^2(x)}{\sin^2(x)\cos^2(x)}$
$= \frac{1}{\sin^2(x)\cos^2(x)}$
$= \csc^2(x)\sec^2(x) = \text{RHS}$
3. Originally Posted by DaCoo911
I am in gr.11 just starting trig identities. I have finished all of my hw except for one question. Which looks to me like it should be simple. I just can't seem to do it. haha i'm probably just being blonde but i'll keep trying! Thanks!
$\csc^2{x} + \sec^2{x} = \frac{1}{\sin^2{x}} + \frac{1}{\cos^2{x}}$
$= \frac{\cos^2{x}}{\sin^2{x}\cos^2{x}} + \frac{\sin^2{x}}{\sin^2{x}\cos^2{x}}$
$= \frac{\cos^2{x} + \sin^2{x}}{\sin^2{x}\cos^2{x}}$
$= \frac{1}{\sin^2{x}\cos^2{x}}$
$= \csc^2{x}\sec^2{x}$.
4. Originally Posted by Prove It
$\csc^2{x} + \sec^2{x} = \frac{1}{\sin^2{x}} + \frac{1}{\cos^2{x}}$
$= \frac{\cos^2{x}}{\sin^2{x}\cos^2{x}} + \frac{\sin^2{x}}{\sin^2{x}\cos^2{x}}$
$= \frac{\cos^2{x} + \sin^2{x}}{\sin^2{x}\cos^2{x}}$
$= \frac{1}{\sin^2{x}\cos^2{x}}$
$= \csc^2{x}\sec^2{x}$.
Did you just c&p my post :P?
Identical!
Your coding and lack of parenthesis saves you from fraud! :P. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 15, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8541839122772217, "perplexity": 898.316285197947}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738662251.92/warc/CC-MAIN-20160924173742-00127-ip-10-143-35-109.ec2.internal.warc.gz"} |
https://byjus.com/rd-sharma-solutions/class-8-maths-chapter-7-factorization-exercise-7-3/?replytocom=30974 | # RD Sharma Solutions for Class 8 Chapter - 7 Factorization Exercise 7.3
### RD Sharma Class 8 Solutions Chapter 7 Ex 7.3 PDF Free Download
Students can refer to RD Sharma Solutions for Class 8 Maths Exercise 7.3 Chapter 7 Factorization which are available here. The solutions here are solved step by step for a better understanding of the concepts, which helps students prepare for their exams at ease. Keeping in mind expert tutors at BYJU’S have made this possible to help students crack difficult problems. Students can download the pdf of RD Sharma Solutions from the links provided below.
Exercise 7.3 of Chapter 7 Factorization is on based on the factorization of algebraic expressions when a binomial is a common factor.
## Download the pdf of RD Sharma For Class 8 Maths Exercise 7.3 Chapter 7 Factorization
### Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 7.3 Chapter 7 Factorization
Factorize each of the following algebraic expressions:
1. 6x (2x – y) + 7y (2x – y)
Solution:
We have,
6x (2x – y) + 7y (2x – y)
By taking (2x – y) as common we get,
(6x + 7y) (2x – y)
2. 2r (y – x) + s (x – y)
Solution:
We have,
2r (y – x) + s (x – y)
By taking (-1) as common we get,
-2r (x – y) + s (x – y)
By taking (x – y) as common we get,
(x – y) (-2r + s)
(x – y) (s – 2r)
3. 7a (2x – 3) + 3b (2x – 3)
Solution:
We have,
7a (2x – 3) + 3b (2x – 3)
By taking (2x – 3) as common we get,
(7a + 3b) (2x – 3)
4. 9a (6a – 5b) – 12a2 (6a – 5b)
Solution:
We have,
9a (6a – 5b) – 12a2 (6a – 5b)
By taking (6a – 5b) as common we get,
(9a – 12a2) (6a – 5b)
3a(3 – 4a) (6a – 5b)
5. 5 (x – 2y)2 + 3 (x – 2y)
Solution:
We have,
5 (x – 2y)2 + 3 (x – 2y)
By taking (x – 2y) as common we get,
(x – 2y) [5 (x – 2y) + 3]
(x – 2y) (5x – 10y + 3)
6. 16 (2l – 3m)2 – 12 (3m – 2l)
Solution:
We have,
16 (2l – 3m)2 – 12 (3m – 2l)
By taking (-1) as common we get,
16 (2l – 3m)2 + 12 (2l – 3m)
By taking 4(2l – 3m) as common we get,
4(2l – 3m) [4 (2l – 3m) + 3]
4(2l – 3m) (8l – 12m + 3)
7. 3a (x – 2y) – b (x – 2y)
Solution:
We have,
3a (x – 2y) – b (x – 2y)
By taking (x – 2y) as common we get,
(3a – b) (x – 2y)
8. a2 (x + y) + b2 (x + y) + c2 (x + y)
Solution:
We have,
a2 (x + y) + b2 (x + y) + c2 (x + y)
By taking (x + y) as common we get,
(a2 + b2 + c2) (x + y)
9. (x – y)2 + (x – y)
Solution:
We have,
(x – y)2 + (x – y)
By taking (x – y) as common we get,
(x – y) (x – y + 1)
10. 6 (a + 2b) – 4 (a + 2b)2
Solution:
We have,
6 (a + 2b) – 4 (a + 2b)2
By taking (a + 2b) as common we get,
[6 – 4 (a + 2b)] (a + 2b)
(6 – 4a – 8b) (a + 2b)
2(3 – 2a – 4b) (a + 2b)
11. a (x – y) + 2b (y – x) + c (x – y)2
Solution:
We have,
a (x – y) + 2b (y – x) + c (x – y)2
By taking (-1) as common we get,
a (x – y) – 2b (x – y) + c (x – y)2
By taking (x – y) as common we get,
[a – 2b + c(x – y)] (x – y)
(x – y) (a – 2b + cx – cy)
12. -4 (x – 2y)2 + 8 (x – 2y)
Solution:
We have,
-4 (x – 2y)2 + 8 (x – 2y)
By taking 4(x – 2y) as common we get,
[-(x – 2y) + 2] 4(x – 2y)
4(x – 2y) (-x + 2y + 2)
13. x3 (a – 2b) + x2 (a – 2b)
Solution:
We have,
x3 (a – 2b) + x2 (a – 2b)
By taking x2 (a – 2b) as common we get,
(x + 1) [x2 (a – 2b)]
x2 (a – 2b) (x + 1)
14. (2x – 3y) (a + b) + (3x – 2y) (a + b)
Solution:
We have,
(2x – 3y) (a + b) + (3x – 2y) (a + b)
By taking (a + b) as common we get,
(a + b) [(2x – 3y) + (3x – 2y)]
(a + b) [2x -3y + 3x – 2y]
(a + b) [5x – 5y]
(a + b) 5(x – y)
15. 4(x + y) (3a – b) + 6(x + y) (2b – 3a)
Solution:
We have,
4(x + y) (3a – b) + 6(x + y) (2b – 3a)
By taking (x + y) as common we get,
(x + y) [4(3a – b) + 6(2b – 3a)]
(x + y) [12a – 4b + 12b – 18a]
(x + y) [-6a + 8b]
(x + y) 2(-3a + 4b)
(x + y) 2(4b – 3a)
#### 1 Comment
1. Soyel rana
wow! i am so happy and satisfied to use this guide means byju’s learning app | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9363380670547485, "perplexity": 4583.515077781088}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250609478.50/warc/CC-MAIN-20200123071220-20200123100220-00194.warc.gz"} |
http://conceptmap.cfapps.io/wikipage?lang=en&name=Greisen%E2%80%93Zatsepin%E2%80%93Kuzmin_limit | # Greisen–Zatsepin–Kuzmin limit
The Greisen–Zatsepin–Kuzmin limit (GZK limit) is a theoretical upper limit on the energy of cosmic ray protons traveling from other galaxies through the intergalactic medium to our galaxy. The limit is 5×1019 eV, or about 8 joules (the energy of a proton travelling at ~99.99999999999999999998% the speed of light). The limit is set by slowing interactions of the protons with the microwave background radiation over long distances (~160 million light-years). The limit is at the same order of magnitude as the upper limit for energy at which cosmic rays have experimentally been detected. For example, one extreme-energy cosmic ray, the Oh-My-God Particle, which has been found to possess a record-breaking 3.12×1020 eV (50 joules)[1][2] of energy (about the same as the kinetic energy of a 95 km/h baseball).
The GZK limit is derived under the assumption that ultra-high energy cosmic rays are protons. Measurements by the largest cosmic-ray observatory, the Pierre Auger Observatory, suggest that most ultra-high energy cosmic rays are heavier elements.[3] In this case, the argument behind the GZK limit does not apply in the originally simple form, and there is no fundamental contradiction in observing cosmic rays with energies that violate the limit.
In the past, the apparent violation of the GZK limit has inspired cosmologists and theoretical physicists to suggest other ways that circumvent the limit. These theories propose that ultra-high energy cosmic rays are produced nearby our galaxy or that Lorentz covariance is violated in such a way that protons do not lose energy on their way to our galaxy.
## Computation
The limit was independently computed in 1966 by Kenneth Greisen,[4] Vadim Kuzmin, and Georgiy Zatsepin,[5] based on interactions between cosmic rays and the photons of the cosmic microwave background radiation (CMB). They predicted that cosmic rays with energies over the threshold energy of 5×1019 eV would interact with cosmic microwave background photons ${\displaystyle \gamma _{\rm {CMB}}}$ , relatively blueshifted by the speed of the cosmic rays, to produce pions through the ${\displaystyle \Delta }$ resonance,
${\displaystyle \gamma _{\text{CMB}}+p\to \Delta ^{+}\to p+\pi ^{0},}$
or
${\displaystyle \gamma _{\text{CMB}}+p\to \Delta ^{+}\to n+\pi ^{+}.}$
Pions produced in this manner proceed to decay in the standard pion channels—ultimately to photons for neutral pions, and photons, positrons, and various neutrinos for positive pions. Neutrons decay also to similar products, so that ultimately the energy of any cosmic ray proton is drained off by production of high-energy photons plus (in some cases) high-energy electron–positron pairs and neutrino pairs.
The pion production process begins at a higher energy than ordinary electron-positron pair production (lepton production) from protons impacting the CMB, which starts at cosmic-ray proton energies of only about 1017 eV. However, pion production events drain 20% of the energy of a cosmic-ray proton as compared with only 0.1% of its energy for electron–positron pair production. This factor of 200 is from two sources: the pion has only about ~130 times the mass of the leptons, but the extra energy appears as different kinetic energies of the pion or leptons and results in relatively more kinetic energy transferred to a heavier product pion, in order to conserve momentum. The much larger total energy losses from pion production result in the pion production process becoming the limiting one to high-energy cosmic-ray travel, rather than the lower-energy light-lepton production process.
The pion production process continues until the cosmic ray energy falls below the pion production threshold. Due to the mean path associated with this interaction, extragalactic cosmic rays traveling over distances larger than 50 Mpc (163 Mly) and with energies greater than this threshold should never be observed on Earth. This distance is also known as GZK horizon.
Unsolved problem in physics:Why is it that some cosmic rays appear to possess energies that are theoretically too high, given that there are no possible near-Earth sources, and that rays from distant sources should have scattered by the cosmic microwave background radiation?(more unsolved problems in physics)
A number of observations have been made by the largest cosmic-ray experiments Akeno Giant Air Shower Array, High Resolution Fly's Eye Cosmic Ray Detector, the Pierre Auger Observatory and Telescope Array Project that appeared to show cosmic rays with energies above this limit (called extreme-energy cosmic rays, or EECRs). The observation of these particles was the so-called GZK paradox or cosmic-ray paradox.
These observations appear to contradict the predictions of special relativity and particle physics as they are presently understood. However, there are a number of possible explanations for these observations that may resolve this inconsistency.
• The observations could be due to an instrument error or an incorrect interpretation of the experiment, especially wrong energy assignment.
• The cosmic rays could have local sources well within the GZK horizon (although it is unclear what these sources could be).
### Weakly interacting particles
Another suggestion involves ultra-high-energy weakly interacting particles (for instance, neutrinos), which might be created at great distances and later react locally to give rise to the particles observed. In the proposed Z-burst model, an ultra-high-energy cosmic neutrino collides with a relic anti-neutrino in our galaxy and annihilates to hadrons.[6] This process proceeds through a (virtual) Z-boson:
${\displaystyle \nu +{\bar {\nu }}\to Z\to {\text{hadrons}}.}$
The cross-section for this process becomes large if the center-of-mass energy of the neutrino antineutrino pair is equal to the Z-boson mass (such a peak in the cross-section is called "resonance"). Assuming that the relic anti-neutrino is at rest, the energy of the incident cosmic neutrino has to be
${\displaystyle E={\frac {m_{Z}^{2}}{2m_{\nu }}}=4.2\times 10^{21}\left({\frac {\text{eV}}{m_{\nu }}}\right)~{\text{eV}},}$
where ${\displaystyle m_{Z}}$ is the mass of the Z-boson, and ${\displaystyle m_{\nu }}$ the mass of the neutrino.
### Other theories
A number of exotic theories have been advanced to explain the AGASA observations, including doubly special relativity. However, it is now established that standard doubly special relativity does not predict any GZK suppression (or GZK cutoff), contrary to models of Lorentz symmetry violation involving an absolute rest frame.[citation needed] Other possible theories involve a relation with dark matter, decays of exotic super-heavy particles beyond those known in the Standard Model.
## Controversy about cosmic rays above the GZK limit
A suppression of the cosmic-ray flux that can be explained with the GZK limit has been confirmed by the latest generation of cosmic-ray observatories. A former claim by the AGASA experiment that there is no suppression was overruled. It remains controversial whether the suppression is due to the GZK effect. The GZK limit only applies if ultra-high-energy cosmic rays are mostly protons.
In July 2007, during the 30th International Cosmic Ray Conference in Mérida, Yucatán, México, the High Resolution Fly's Eye Experiment (HiRes) and the Pierre Auger Observatory (Auger) presented their results on ultra-high-energy cosmic rays. HiRes observed a suppression in the UHECR spectrum at just the right energy, observing only 13 events with an energy above the threshold, while expecting 43 with no suppression. This was interpreted as the first observation of the GZK limit.[7] Auger confirmed the flux suppression, but did not claim it to be the GZK limit: instead of the 30 events necessary to confirm the AGASA results, Auger saw only two, which are believed to be heavy-nuclei events.[8] The flux suppression was previously brought into question when the AGASA experiment found no suppression in their spectrum[citation needed]. According to Alan Watson, spokesperson for the Auger Collaboration, AGASA results have been shown to be incorrect, possibly due to the systematic shift in energy assignment.
In 2010 and the following years, both the Pierre Auger Observatory and HiRes confirmed again a flux suppression,[9][10] in case of the Pierre Auger Observatory the effect is statistically significant at the level of 20 standard deviations.
After the flux suppression was established, a heated debate ensued whether cosmic rays that violate the GZK limit are protons. The Pierre Auger Observatory, the world's largest observatory, found with high statistical significance that ultra-high-energy cosmic rays are not purely protons, but a mixture of elements, which is getting heavier with increasing energy.[3] The Telescope Array Project, a joint effort from members of the HiRes and AGASA collaborations, agrees with the former HiRes result that these cosmic rays look like protons.[11] The claim is based on data with lower statistical significance, however. The area covered by Telescope Array is about one third of the area covered by the Pierre Auger Observatory, and the latter has been running for a longer time.
The controversy was partially resolved in 2017, when a joint working group formed by members of both experiments presented a report at the 35th International Cosmic Ray Conference.[12] According to the report, the raw experimental results are not in contradiction with each other. The different interpretations are mainly based on the use of different theoretical models (Telescope Array uses an outdated model for its interpretation) and the fact that Telescope Array has not collected enough events yet to distinguish the pure-proton hypothesis from the mixed-nuclei hypothesis.
### Extreme Universe Space Observatory on Japanese Experiment Module (JEM-EUSO)
EUSO, which was scheduled to fly on the International Space Station (ISS) in 2009, was designed to use the atmospheric-fluorescence technique to monitor a huge area and boost the statistics of UHECRs considerably. EUSO is to make a deep survey of UHECR-induced extensive air showers (EASs) from space, extending the measured energy spectrum well beyond the GZK cutoff. It is to search for the origin of UHECRs, determine the nature of the origin of UHECRs, make an all-sky survey of the arrival direction of UHECRs, and seek to open the astronomical window on the extreme-energy universe with neutrinos. The fate of the EUSO Observatory is still unclear, since NASA is considering early retirement of the ISS.
### The Fermi Gamma-ray Space Telescope to resolve inconsistencies
Launched in June 2008, the Fermi Gamma-ray Space Telescope (formerly GLAST) will also provide data that will help resolve these inconsistencies.
• With the Fermi Gamma-ray Space Telescope, one has the possibility of detecting gamma rays from the freshly accelerated cosmic-ray nuclei at their acceleration site (the source of the UHECRs).[13]
• UHECR protons accelerated (see also Centrifugal mechanism of acceleration) in astrophysical objects produce secondary electromagnetic cascades during propagation in the cosmic microwave and infrared backgrounds, of which the GZK process of pion production is one of the contributors. Such cascades can contribute between about 1% and 50% of the GeV–TeV diffuse photon flux measured by the EGRET experiment. The Fermi Gamma-ray Space Telescope may discover this flux.[14]
## Possible sources of UHECRs
In November 2007, researchers at the Pierre Auger Observatory announced that they had evidence that UHECRs appear to come from the active galactic nuclei (AGNs) of energetic galaxies powered by matter swirling onto a supermassive black hole. The cosmic rays were detected and traced back to the AGNs using the Véron-Cetty-Véron catalog. These results are reported in the journal Science.[15] Nevertheless, the strength of the correlation with AGNs from this particular catalog for the Auger data recorded after 2007 has been slowly diminishing.[16]
## References
1. ^ "HiRes - The High Resolution Fly's Eye Ultra High Energy Cosmic Ray Observatory". www.cosmic-ray.org. Retrieved 2019-06-13.
2. ^ "Oh-My-God Particles". phys.org. Retrieved 2019-06-13.
3. ^ a b The Pierre Auger Collaboration (2017). "Inferences on Mass Composition and Tests of Hadronic Interactions from 0.3 to 100 EeV using the water-Cherenkov Detectors of the Pierre Auger Observatory". arXiv:1710.07249 [astro-ph.HE].
4. ^ Greisen, Kenneth (1966). "End to the Cosmic-Ray Spectrum?". Physical Review Letters. 16 (17): 748–750. Bibcode:1966PhRvL..16..748G. doi:10.1103/PhysRevLett.16.748.
5. ^ Zatsepin, G. T.; Kuz'min, V. A. (1966). "Upper Limit of the Spectrum of Cosmic Rays" (PDF). Journal of Experimental and Theoretical Physics Letters. 4: 78–80. Bibcode:1966JETPL...4...78Z.
6. ^ Fargion, D.; Mele, B.; Salis, A. (June 1999). "Ultra–High‐Energy Neutrino Scattering onto Relic Light Neutrinos in the Galactic Halo as a Possible Source of the Highest Energy Extragalactic Cosmic Rays". The Astrophysical Journal. 517 (2): 725–733. arXiv:astro-ph/9710029. Bibcode:1999ApJ...517..725F. doi:10.1086/307203.
7. ^ Abbasi, R. U.; et al. (2008). "First Observation of the Greisen-Zatsepin-Kuzmin Suppression". Physical Review Letters. 100 (10): 101101. arXiv:astro-ph/0703099. Bibcode:2008PhRvL.100j1101A. doi:10.1103/PhysRevLett.100.101101. PMID 18352170.
8. ^ Abraham, J.; et al. (2008). "Observation of the suppression of the flux of cosmic rays above 4×1019 eV". Physical Review Letters. 101 (6): 061101–1–061101–7. arXiv:0806.4302. Bibcode:2008PhRvL.101f1101A. doi:10.1103/PhysRevLett.101.061101. PMID 18764444.
9. ^ The Pierre Auger Collaboration (2010). "Measurement of the energy spectrum of cosmic rays above 1018 eV using the Pierre Auger Observatory". Phys. Lett. B. 685 (4–5): 239–246. arXiv:1002.1975. Bibcode:2010PhLB..685..239A. doi:10.1016/j.physletb.2010.02.013.
10. ^ Sokolsky; for the HiRes Collaboration (2010). "Final Results from the High Resolution Fly's Eye (HiRes) Experiment". Nuclear Physics B: Proceedings Supplements. 212–213: 74–78. arXiv:1010.2690. doi:10.1016/j.nuclphysbps.2011.03.010.
11. ^ Hanlon, William; others (2017). "Telescope Array Composition Summary". PoS. 301 (536): 536. Bibcode:2017ICRC...35..536H.
12. ^ de Souza, Vitor; others (2017). "Testing the agreement between the Xmax distributions measured by the Pierre Auger and Telescope Array Observatories". PoS. 301 (522).
13. ^ Ormes, Jonathan F.; et al. (2000). "The origin of cosmic rays: What can the Fermi Gamma-ray Telescope say?". AIP Conference Proceedings. 528: 445–448. arXiv:astro-ph/0003270. doi:10.1063/1.1324357.
14. ^ Kalashev, Oleg E.; Semikoz, Dmitry V.; Sigl, Guenter (2009). "Ultra-High Energy Cosmic Rays and the GeV–TeV Diffuse Gamma-Ray Flux". Physical Review D. 79 (6): 063005. arXiv:0704.2463. Bibcode:2009PhRvD..79f3005K. doi:10.1103/PhysRevD.79.063005.
15. ^ The Pierre Auger Collaboration (2007). "Correlation of the Highest-Energy Cosmic Rays with Nearby Extragalactic Objects". Science. 318 (5852): 938–943. arXiv:0711.2256. Bibcode:2007Sci...318..938P. doi:10.1126/science.1151124. PMID 17991855.
16. ^ The Pierre Auger Collaboration (2010). "Update on the correlation of the highest energy cosmic rays with nearby extragalactic matter". Astropart. Phys. 34 (5): 314–326. arXiv:1009.1855. Bibcode:2010APh....34..314A. doi:10.1016/j.astropartphys.2010.08.010. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.917258620262146, "perplexity": 2142.878312975195}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195531106.93/warc/CC-MAIN-20190724061728-20190724083728-00168.warc.gz"} |
http://math.stackexchange.com/questions/293639/butcher-tableau-for-given-algorithm | butcher tableau for given algorithm
Given $y'(t) = f(t,y(t))$ and the following algorithm:
$$y_{n+\frac{1}{2}} = y_n + \frac{h}{2}f(t_n,y_n)$$ $$y_{n+1} = y_n + hf(t_n+\frac{h}{2},y_{n+\frac{1}{2}})$$
We should show that this can be seen as an explicit runge kutta algorithm (how to exactly do this?) and determine the butcher tableu.
I somehow compared this to the usual form in which an runge kutte algorithm is given and think the tableu must be
$$\begin{array}{c|cc} 0 & 0 & 0 \\ \frac{1}{2} & \frac{1}{2} & 0 \\ \hline & 0 & 1 \\ \end{array}$$
-
Try to reformulate your two equation by substituting $y_{n+1/2}$ in the second equation with the first equation to obtain an expression of the form, given in the reference: en.wikipedia.org/wiki/… – sonystarmap Feb 4 '13 at 8:58 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9794095754623413, "perplexity": 529.7166367779598}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042989443.69/warc/CC-MAIN-20150728002309-00300-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://www.assignmentexpert.com/homework-answers/mathematics/discrete-mathematics/question-12797 | 68 721
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Question #12797
If A and B are two non-empty sets such that A x B = B x A, show that A=B
Proof.
Notice that A x B consists of all pairs (a,b), where a belongs to A,
and b belongs to B, while
B x A consists of all pairs (b,a), where a
belongs to A, and b belongs to B.
Let a belongs to A.
We should prove
that a belongs to B as well.
Take any b from B.
Then
(a,b) belongs
to A x B = B x A,
whence
(a,b) = (b',a')
for some b' from B and a'
from A.
This means that
a=b' and b=a'.
Hence a belong to
B.
This proves that A is contained in B.
By similar argument B is contained
in A, and so A=B.
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for any assignment or question with DETAILED EXPLANATIONS! | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8413881063461304, "perplexity": 3314.9336835212134}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583728901.52/warc/CC-MAIN-20190120163942-20190120185942-00488.warc.gz"} |
https://www.physicsforums.com/threads/physics-trick-question.535991/ | # Homework Help: Physics Trick Question
1. Oct 2, 2011
### Maty
OK, so I had this physics homework question about 2 weeks ago, and I thought I would post it here to see if anyone can figure it out. Note: this homework assignment is overdue and can't go back, I am just asking to see if anyone can get it. Reason I am asking is because I asked 2 physics teachers in my high school and neither of them figured it out... I just found out about this website and a I am assuming everyone here is smart so I'm wondering if anyone here can get it.
A cab driver picks up a customer and delivers her 3.10 km away, driving a straight route. The driver accelerates to the speed limit and, upon reaching it, begins to decelerate immediately. The magnitude of the deceleration five times the magnitude of the acceleration. Find the lengths of the acceleration and deceleration phases.
p.s. I don't know what the answer is and neither can I check. It's an online assignment and so I can't change any answers from assignments that are overdue. And I guess since this isn't a homework help problem, more like see if you can figure it out for the heck of it, I post it here.
2. Oct 2, 2011
### Staff: Mentor
Seems fairly straightforward to me. If you need a hint, start by comparing the time of the acceleration phase to the time of the deceleration phase.
3. Oct 2, 2011
### PAllen
I'll give one more hint. If you were asked to find the time of turnaround or the magnitude of acceleration, the problem would be under-determined (as it superficially looks). However, you really only need the ratio of distance (combined with the total given distance). In this case it is well determined, and solvable with the straightforward manipulation of the basic acceleration / distance /velocity formulas.
4. Oct 2, 2011
### PeterO
Since the cab was always accelerating [or decelerating] the average speed will be half the maximum speed. maximum speed was the speed limit. What is the speed limit there. In the suburbs here maximum speed is 60 km/h - so average speed is 30km/h.
For simplicity here I will claim the maximum speed limit is 62 km/h, so the average speed is 31 km/hr.
That means 1/10 of an hour [6 minutes] to cover the 3.1 km.
Deceleration was at 5 times the rate, so only takes 1/5 the time
Accelerate for 5 minutes, decelerate for 1 minute. Since average speed for each phase is 31 km/h you can work out the distances.
ALSO: you don't need to know the speed limit - those distances and times apply regardless.
5. Oct 2, 2011
### PAllen
Well, except the trick is that the speed limit isn't given and therefore you cannot determine acceleration times or magnitude of acceleration, but you can still determine the distances. Solving it for a speed limit you choose is not an adequate solution without justifying that the speed limit doesn't matter - in which case why bother assuming one.
Last edited: Oct 2, 2011
6. Oct 2, 2011
### PAllen
The times do not apply. Only the ratio of times apply.
7. Oct 2, 2011
### DaveC426913
You could draw this on a graph. Set the decel slope to be -5x the accel slope.
8. Oct 2, 2011
### PAllen
Well, in the spirit of the problem, I think you don't need to pick particular values of anything you don't know (slope, speed limit etc.). In fact, a give away hint is without assuming even the ratio of 5, just the set up of the problem, you should aim to derive that:
x + x * (t2/t1) = D
where x is the distance before deceleration begins, t1 is time until deceleration begins, t2 is deceleration time, D is total distance. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9174023270606995, "perplexity": 746.1927923358111}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221215393.63/warc/CC-MAIN-20180819204348-20180819224348-00381.warc.gz"} |
http://mathhelpforum.com/pre-calculus/199696-help-wih-parametric-representation-function-print.html | Help wih parametric representation of a function
• Jun 5th 2012, 11:20 PM
Rascot
Help wih parametric representation of a function
Hi, I'm stuck trying to obtain the cartesian equation of the function defined parametrically by x=1/2(t+1/t), y=1/2(t-1/t).
The book (Mathematics for Engineers, Croft / Davison) gives x^2 - y^2 = 1 as the answer, but I can't figure out how they got there. Ideas?
• Jun 5th 2012, 11:44 PM
earboth
Re: Help wih parametric representation of a function
Quote:
Originally Posted by Rascot
Hi, I'm stuck trying to obtain the cartesian equation of the function defined parametrically by x=1/2(t+1/t), y=1/2(t-1/t).
The book (Mathematics for Engineers, Croft / Davison) gives x^2 - y^2 = 1 as the answer, but I can't figure out how they got there. Ideas?
1. Rearrange both equations to:
$2x = t +\frac1t~\wedge~2y=t-\frac1t$
2. Add both sides of the equations:
$2x+2y = 2t~\implies~t = x+y$
3. Replace t in the 1st equation by (x+y):
$x = \frac12 \cdot \left((x+y)+\frac1{x+y} \right)$
4. Expand the brackets and after a few steps of rearranging you'll get the given result.
• Jun 6th 2012, 12:05 AM
Rascot
Re: Help wih parametric representation of a function
Perfect! Completely forgot about solving simultaneous equations (probably because they are dealt with later in the book). You have my thanks, sir.
By the way, how do you (and pretty much everyone on this forum) post math symbols as images? Is there a piece of software you guys use?
• Jun 6th 2012, 12:26 AM
earboth
Re: Help wih parametric representation of a function
Quote:
Originally Posted by Rascot
Perfect! Completely forgot about solving simultaneous equations (probably because they are dealt with later in the book). You have my thanks, sir.
By the way, how do you (and pretty much everyone on this forum) post math symbols as images? Is there a piece of software you guys use?
1. Go to LaTeX Help Forum. You'll find 2 tutorials where you can find how to use the implemented Latex-compiler.
2. If you need the synatx of a Latex command you'll find it probably here: Help:Displaying a formula - Wikipedia, the free encyclopedia
• Jun 6th 2012, 06:30 AM
Soroban
Re: Help wih parametric representation of a function
Hello, Rascot!
Quote:
Hi, I'm stuck trying to obtain the cartesian equation of the function
. . defined parametrically by: . $\begin{Bmatrix}x &=& \frac{1}{2}(t+\frac{1}{t}) \\ \\[-3mm] y&=& \frac{1}{2}(t-\frac{1}{t}) \end{Bmatrix}$
The book (Mathematics for Engineers, Croft/Davison) gives $x^2 - y^2 \,=\, 1$ as the answer.
Square the equations: . $\begin{Bmatrix}x^2 &=& \frac{t^2}{4} + \frac{1}{2} + \frac{1}{4t^2} & [1] \\ \\[-4mm] y^2 &=& \frac{t^2}{4} - \frac{1}{2} + \frac{1}{4t^2} & [2] \end{Bmatrix}$
Subtract [2] from [1]: . . $x^2 - y^2 \;=\;1$
• Jun 6th 2012, 12:59 PM
Rascot
Re: Help wih parametric representation of a function
Thank you both very much for your help. I can't mark the thread as solved, but would kindly ask moderators to do so if they happen to come across it. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9397314786911011, "perplexity": 1549.5498007995393}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560281424.85/warc/CC-MAIN-20170116095121-00014-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/molecular-vibration.976461/ | # A Molecular vibration
• Start date
#### Iceking20
15
1
Summary
How could we find the reduced mass for polyatomic molecule ?
Summary: How could we find the reduced mass for polyatomic molecule ?
I have problem with reduced mass of poly atomic moleculs because for diatomic molecules you can easily find out the reduced mass by M¹M²/M¹+M²,but I don't think with this way we can find reduces mass for example for fe²o³?
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#### mfb
Mentor
33,374
9,103
That will depend on the vibration mode and in general there won't be a single answer, you can't use the effective mass for everything.
I moved this thread to the quantum mechanics forum.
#### Iceking20
15
1
That will depend on the vibration mode and in general there won't be a single answer, you can't use the effective mass for everything.
I moved this thread to the quantum mechanics forum.
Can you give an example for one of the modes because I want to have perspective about it
#### A. Neumaier
6,842
2,785
Summary: How could we find the reduced mass for polyatomic molecule ?
Summary: How could we find the reduced mass for polyatomic molecule ?
I have problem with reduced mass of poly atomic moleculs because for diatomic molecules you can easily find out the reduced mass by M¹M²/M¹+M²,but I don't think with this way we can find reduces mass for example for fe²o³?
In general, the inverse reduced mass is the sum of the inverse masses of the constituents. For 2 constituents one gets your formula.
However, the concept of a reduced mass is indeed most useful for the case of 2 constituents only, where the center of mass frame leaves a 1-particle problem with the reduced mass.
For more than 2 constituents, no such simple recipe works, and the reduced Hamiltonian depends on the way relative coordinates are chosen.
#### mfb
Mentor
33,374
9,103
Can you give an example for one of the modes because I want to have perspective about it
Everything. Even if we take a molecule with three atoms in a straight, symmetric line it will have vibrations along that line (contraction/extension) and vibrations orthogonal to it (making it slightly "L"-shaped). In the first case only the outer atoms move, in the second case all atoms move.
"Molecular vibration"
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