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https://math.stackexchange.com/questions/4279729/limit-of-sequence-involving-arcsin
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# limit of sequence involving arcsin
Suppose a sequence $$y_n\to y > 0$$ and let $$a$$ be a real number. Determine whether the limit $$\lim\limits_{n\to\infty} 2^n \arcsin (\frac{a}{2^n y_n})$$ exists (is finite) and if so, determine its value.
I think the answer is $$\frac{a}{y},$$ but I'm not sure which limit properties to use. Would the Squeeze theorem be useful, for instance? Intuitively, as $$n\to\infty, \frac{a}{2^n y_n}\to 0,$$ so $$\arcsin (\frac{a}{2^n y_n} )\approx \frac{a}{2^ny_n}.$$ I'm not sure if I can really apply the product property of limits; i.e. $$\lim\limits_{n\to\infty} a_n b_n = \lim\limits_{n\to\infty} a_n\cdot \lim\limits_{n\to\infty} b_n$$ provided the RHS value is finite. However, I'm not really sure how to formalize this using the definition of limits. Let $$\epsilon > 0.$$ I want to show that there exists $$N\in\mathbb{N}$$ so that for $$n\ge N, |2^n \arcsin (\frac{a}{2^n y_n}) -L| < \epsilon,$$ where $$L$$ is the limit of the sequence.
• Are you allowed to assume knowledge of the limit $\lim_{x\to 0}\frac{\arcsin(x)}{x}=1$ in your solution? Oct 18, 2021 at 3:11
Take $$2^{n}y_{n}=ax_{n}$$
Then you have $$\lim_{n\to\infty}\frac{ax_{n}}{y_{n}}\arcsin(\frac{1}{x^{n}})$$.
(such a substitution is justified as your question assumes $$y_{n}\neq 0$$ as it was in the denominator)
Which you can write as $$\lim_{n\to\infty}\frac{a}{y_{n}}x_{n}\arcsin(\frac{1}{x_{n}})$$.
Now you take $$a_{n}=\frac{a}{y_{n}}$$ and $$b_{n}=x_{n}\arcsin(\frac{1}{x_{n}})$$.
Now you can use product as both these sequences converge separately. Now $$x\cdot\arcsin(\frac{1}{x})$$ is a continuous bounded function in $$[1,\infty)$$. And differentiable in $$(1,\infty)$$. So you can use that to justify $$\lim_{n\to\infty}x_{n}\arcsin(\frac{1}{x_{n}})=\lim_{x\to\infty}x\cdot\arcsin(\frac{1}{x})=\lim_{n\to\infty}\frac{\arcsin(\frac{1}{x})}{\frac{1}{x}}$$. Which would justify the use of L'Hospital (as both $$\arcsin(\frac{1}{x})$$ and $$\frac{1}{x}$$ are continuous in $$[1,\infty)$$ and differentiable in $$(1,\infty)$$.) and concluding that this limit is indeed $$1$$.
And $$\lim_{n\to\infty}a_{n}=\frac{a}{y}$$.
So you have your answer as $$\frac{a}{y}$$
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We've updated our
TEXT
# D1.02: Examples 2–5
### Example 2
If we invest $750 at 6% annual interest, compounded quarterly, the formula for the amount A that the investment is worth after t years is $A=750\cdot{{\left(1+\frac{0.06}{4}\right)}^{4t}}$. Find the amount the investment is worth after five years. Then evaluate this for several values of t between 0 and 30 years, and sketch a graph of this formula. Answer: First, recall how to get an exponent in your calculator. On most calculators, it’s a ^ key or a ${{y}^{x}}$ key. Find that key and practice using it to evaluate ${{2}^{3}}$. When you can do that correctly, then evaluate the following expression, writing these intermediate values as indicated below.$A=750\cdot{{\left(1+\frac{0.06}{4}\right)}^{4t}}=750\cdot{{\left(1+\frac{0.06}{4}\right)}^{4\cdot5}}=750\cdot{{\left(1.015\right)}^{20}}=750\cdot1.346855007=1010.141255$ So the investment is worth$1010.14 after 5 years. It is more convenient to be able to put all of these values into the calculator at the same time rather than writing intermediate steps. We will learn to do that in this course, but that’s not the point of this lesson. The point here is to practice making graphs. This will be useful to you in checking your work when you evaluate formulas. If one of the values you compute doesn’t seem to fit the pattern of the others, that warns you to check your computation again.
Formula input and output
t A
5 1010.141
0 750.000
10 1360.514
20 2467.997
30 4476.992
We can use the graph to estimate at what time the amount of the investment will be about $4000. Look at the graph to find 4000 on the vertical axis and then notice that the corresponding value on the horizontal axis is about 27.5. So at approximately 27.5 years, the amount of the investment will be$4000. Check that by plugging it into the formula. $A=750\cdot{{\left(1+\frac{0.06}{4}\right)}^{4\cdot27.5}}=\,\,\,\text{etc.}\,\,=3857.68$ This is as close as we could reasonably expect from using a graph to approximate the input value.
### Example 3
Consider our approximation from Example 2. We wanted to find the number of years to leave the money in so that the amount of the investment would be $4000. The graph suggested that $t=27.5$ years. But then we found that after 27.5 years, the amount was only$3857.68. Clearly we should leave the money in somewhat longer. How much longer?
Answer: If we know how to do some complicated algebra (using logarithms) we can quickly obtain an answer to “how long should we leave the money to earn $4000?” We will not do that particular type of algebra in this course. Instead we will use numerical work to get refine our estimate from the graph. Here, when we look at the graph and the numbers, it is clear that 30 years is too long and 27.5 years is too short. So we try some value in between. Let’s try 28 years. $A=750\cdot{{\left(1+\frac{0.06}{4}\right)}^{4\cdot28}}=\,\,\text{etc.}\,\,=\,\,3974.28$. This is much closer to the$4000 which was our goal. If we wanted to try to get it even more accurately, we would use a value larger than 28 years, but only very slightly larger. Since the amount is compounded quarterly, the next time the amount will increase is at 28.25 years, so let’s try that. $A=750\cdot{{\left(1+\frac{0.06}{4}\right)}^{4\cdot28.25}}=4033.89$. That is further away from $4000 than the value for 28 years, so the best answer for this question is that the amount of the investment will be at about$4000 at 28 years.
### Example 4
Following Example 2, we’d like to have a formula that allows us to vary the interest rate and number of times per year it is compounded as well as varying the number of years of the investment. So let r be the interest rate, converted to a decimal and n be the number of times per year it is compounded, and, as before t is the number of years and A is the amount of the investment. Then the formula is $A=750\cdot{{\left(1+\frac{r}{n}\right)}^{n\cdot{t}}}$. This is an example of a formula with several input values. Use this formula to find the amount of the investment after 5 years if the interest rate is 8% and it is compounded monthly.
$A=750\cdot{{\left(1+\frac{r}{n}\right)}^{n\cdot{t}}}=750\cdot{{\left(1+\frac{0.08}{12}\right)}^{12\cdot5}}=750\cdot{{\left(1.006666667\right)}^{60}}=750\cdot1.4898457=1117.384281$
So after 5 years, this investment is worth $1117.38. We can’t graph this easily, because we would need more than a two-dimensional graph. We need one dimension for the output value and one dimension for every input value and investigating four-dimensional graphs is beyond the scope of this course! However, in most practical applications, technicians and scientists isolate one or two input variables that they are most interested in and then analyze the problem with a two-dimensional graph (for one input variable) or a three-dimensional graph (for two input variables.) We will look at some three-dimensional graphs later in the course. ### Example 5 Following Example 3, suppose we want to allow the initial amount of the investment to change. So we need a variable for that. Since it is an amount, we’d like to call it A, but we already have an A in this formula that means something else. We could use a different letter, but in applications problems we often choose to call both values A and distinguish between them by a subscript. In this problem, we would usually call the original amount of money ${{A}_{0}}$ and the final amount of money after t years ${{A}_{t}}$. So the formula is ${{A}_{t}}={{A}_{0}}\cdot{{\left(1+\frac{r}{n}\right)}^{n\cdot{t}}}$. Use this formula to find the amount of an investment after 6 years if the initial amount is$900, the annual rate is 0.07 and it is compounded twice a year.
\begin{align}&{{A}_{t}}={{A}_{0}}\cdot{{\left(1+\frac{r}{n}\right)}^{n\cdot{t}}}\\&{{A}_{6}}=900\cdot{{\left(1+\frac{0.07}{2}\right)}^{2\cdot6}}=900\cdot{{(1.035)}^{12}}=900\cdot1.511068657=1359.961792\\\end{align}
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
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## #1 2006-10-23 07:50:30
fusilli_jerry89
Member
Registered: 2006-06-23
Posts: 86
### Fireman Problem
A fireman has a hose and is standing on top of a 20 m high building. His hose shoots water at 12 m/s and he wishes to hit the top of another 20 m hig building 21 m away. what angle should he aim his hose at?
I calculated that in the x direction, the velocities are both 12cosθ, the acceleration is zero, distance is 21.0 and te time is unknown. In the y direction, te initial velocity is 12sinθ and the final is -12sinθ, the acceleration is -9.8, the distance is zero, and the time is also unknown.
I then set up a system:
x: t=21/12cosθ
y: t=24sinθ/9.8
0=21/12cosθ-24sinθ/9.8
(21/12)secθ=(24/9.8)sinθ
(21/12)/(24/9.8)secθ=sinθ
sinθ/secθ=205.8/288
sinθcosθ=205.8/288
I'm stuck now...
Last edited by fusilli_jerry89 (2006-10-23 07:52:29)
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## #2 2006-10-23 10:20:46
luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470
### Re: Fireman Problem
from what youve posted, there is no answer
because it would result in taking the inverse sine of a number round about the size of 14 which is outside its range
Last edited by luca-deltodesco (2006-10-23 10:29:40)
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The End Of All Things To Come.
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## #3 2006-10-23 10:30:17
luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470
### Re: Fireman Problem
fusilli_jerry89 wrote:
A fireman has a hose and is standing on top of a 20 m high building. His hose shoots water at 12 m/s and he wishes to hit the top of another 20 m hig building 21 m away. what angle should he aim his hose at?
I calculated that in the x direction, the velocities are both 12cosθ, the acceleration is zero, distance is 21.0 and te time is unknown. In the y direction, te initial velocity is 12sinθ and the final is -12sinθ, the acceleration is -9.8, the distance is zero, and the time is also unknown.
if you take the angle to be from the x axis counter clockwise
you require that s_x = 21m. so you can calculate t for theta by
then you also require that for this value of theta, s_y will be 0, since buildings are at same height, i.e.
but again, its outside of the range?
anyone tell me where im going wrong?
Last edited by luca-deltodesco (2006-10-23 10:39:24)
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The End Of All Things To Come.
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## #4 2006-10-24 02:00:36
mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900
### Re: Fireman Problem
You're not going wrong, it's just that the hose is too weak. The optimum angle of projection is 45° (when the start and finish are at the same height), and that has a range of v²/g, where v is the initial velocity of the projection and v is the gravitational force.
So in this situation, the maximum range of the water from the hose would be 144/9.8, which is around 14.7m, falling considerably short of the 21m target.
Why did the vector cross the road?
It wanted to be normal.
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## #5 2006-10-24 06:12:27
luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470
### Re: Fireman Problem
mathsyperson wrote:
You're not going wrong, it's just that the hose is too weak.
good, because i went over that twice, and i was getting worried.
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# BASIC LINEAR ALGEBRA by fdh56iuoui
VIEWS: 17 PAGES: 50
• pg 1
``` BASIC LINEAR ALGEBRA
AN EXERCISE APPROACH
Gabriel Nagy
Kansas State University
c Gabriel Nagy
CHAPTER 1
Vector spaces and linear maps
In this chapter we introduce the basic algebraic notions of vector spaces and
linear maps.
1. Vector spaces
Suppose k is a field. (Although the theory works for arbitrary fields, we will
eventually focus only on the choices k = R, C.)
Definition. A k-vector space is an abelian group (V, +), equipped with an
external operation1
k × V (λ, v) −→ λv ∈ V,
called scalar multiplication, with the following properties:
• λ · (v + w) = (λ · v) + (λ · w), for all λ ∈ k, v, w ∈ V .
• (λ + µ) · v = (λv) + (µv), for all λ, µ ∈ k, v ∈ V .
• (λ · µ)v = λ · (µ · v), for all λ, µ ∈ k, v ∈ V .
• 1 · v = v, for all v ∈ V .
The elements of a vector space are sometimes called vectors.
Examples. The field k itself is a k-vector space, with its own multiplication
as scalar multiplication.
A trivial group (with one element) is always a k-vector space (with the only
possible scalar multiplication).
1 Suppose V is a k-vector space. Prove that
0 · v = 0, for all v ∈ V.
(The zero in the left-hand side is the field k. The zero in the right-hand side is the
neutral element in the abelian group V .)
Use the above fact to conclude that for any v ∈ V , the vector −v (the inverse
of v in the abelian group V ) can also be described by
−v = (−1) · v.
2 Fix a field k, a non-empty set I, and a family (Vi )i∈I of k-vector spaces.
Consider the product i∈I Vk , equipped with the operations:
• (vi )i∈I + (wi )i∈I = (vi + wi )i∈I ;
• λ · (vi )i∈I = (λvi )i∈I .
1 When convenient, the symbol · may be omitted.
1
2 1. VECTOR SPACES AND LINEAR MAPS
Prove that i∈I Vk is a k-vector space. This structure is called the k-vector space
direct product of the family (Vi )i∈I .
Definition. Suppose V is a k-vector space. A subset X ⊂ V is called a
k-linear subspace, if
• Whenever x, y ∈ X, we have x + y ∈ X.
• Whenever x, y ∈ Xand λ ∈ k, we have λx ∈ X.
3 If X is a k-linear subspace of the k-vector space V , then X itself is a k-vector
space, when equipped with the operations “inherited” from V . Prove than any
linear subspace of V contains the zero vector 0 ∈ V
4 Let (Vi )i∈I be a family of k-vector spaces (indexed by a non-empty set I). For
an element v = (vi )i∈I ∈ i∈I Vi let us define the set
v = {i ∈ I : vi = 0}.
Prove that the set
{v ∈ Vi : v is finite }
i∈I
is a linear subspace of i∈I Vi . This space is called the k-vector space direct sum
of the family (Vi )i∈I , and is denoted by i∈I Vi .
Definition. Suppose we have a family (Vi )i∈I of vector spaces. For a fixed
index j ∈ I, define the map εj : Vj → i∈I Vi as follows. For a vector v ∈ Vj we
construct εj (v) = (wi )i∈I , where
v if i = j
wi =
0 if i = j.
We call the maps εj : Vj → i∈I Vi , j ∈ I, the standard inclusions. The maps
πj : Vi (vi )i∈I −→ vj ∈ Vj
i∈I
are called the coordinate maps.
5 Let (Vi )i∈I be a family of vector spaces. Prove that the standard inclusions
εi , i ∈ I. are injective. In fact prove that πi ◦ εi = IdVi . Prove that any element
v ∈ i∈I Vi is given as
v= (εi ◦ πi )(v).
i∈ v
In other words, if v = (vi )i∈I , then v = i∈ v εi (vi ).
6 Suppose (Xj )j∈J is a family of k-linear subspaces of V . Prove that the inter-
section j∈J Xj is again a k-linear subspace of V .
1. VECTOR SPACES 3
Definition. Let V be a k-vector space, and let M ⊂ V be an arbitrary subset
of V . Consider the family
F = {X : X k-linear subspace of V , and X ⊃ M }.
The set
Spank (M ) = X,
X∈F
which is a linear subspace of V by the preceding exercise, is called the k-linear span
of M in V .
Convention. Spank (∅) = {0}.
Example. The linear span of a singleton is described as
Spank ({v}) = kv(= {λv : λ ∈ k}).
7 Prove that if M and N are subsets of a k-vector space V , with M ⊂ N , then we
also have the inclusion Spank (M ) ⊂ Spank (N ). Give an example where M N,
but their spans coincide.
8 Let V be a k-vector space, and M be a subset of V . For an element v ∈ V ,
prove that the following are equivalent:
(i) v ∈ Spank (M );
(ii) there exists an integer n ≥ 1, elements x1 , . . . , xn ∈ M , and scalars
λ1 , . . . , λn ∈ k such that2 v = λ1 x1 + · · · + λn xn .
Hint: First prove that the set of elements satisfying property (ii) is a linear subspace. Second,
prove that the linear span of M contains all elements satisfying (ii).
Notation. Suppose V is a vector space, and A1 , . . . , An are subsets of V . We
define
A1 + · · · + A2 = {a1 + · · · + an : ak ∈ Ak , k = 1, . . . , n}.
9 Let V be a k-vector space. Suppose A1 , . . . , An are k-homogeneous, in the
sense that for every k = 1, . . . , n we have the equality:
Ak = {λx : λ ∈ k, x ∈ Ak }.
Prove the equality
Span(A1 ∪ · · · ∪ An ) = Span(A1 ) + · · · + Span(An ).
10 Let V be a k-vector space, and (Xj )j∈J be a family of linear subspaces of V .
For an element v ∈ V , prove that the following are equivalent:
(i) v ∈ Spank j∈J Xj );
(ii) there exists an integer n ≥ 1 and x1 , . . . , xn ∈ j∈J Xj , such that v =
x1 + · · · + xn .
2 From now on we will use the usual convention which gives the scalar multiplication prece-
4 1. VECTOR SPACES AND LINEAR MAPS
Comment. If X1 , . . . , Xn are linear subspaces of the vector space V , then using
the notation preceding Exercise ??, and the above result, we get
Span(X1 ∪ · · · ∪ Xn ) = X1 + · · · + Xn .
11 In general, a union of linear subspaces is not a linear subspace. Give an
example of two linear subspaces X1 , X2 ⊂ R2 , such that X1 ∪ X2 is not a linear
subspace.
12 Prove that the union of a directed family of linear subspaces is a linear
subspace. That is, if (Xj )j∈J is a family of k-linear subspaces of the k-vector space
V , with the property
• For any j, k ∈ J there exists some ∈ J such that Xj ⊂ X ⊃ Xk ,
then j∈J Xj is again a k-linear subspace of V .
Hint: Use the preceding exercise.
Definition. Suppose V is a k-vector space. A set M ⊂ V is said to be k-
linearly independent, if (compare with Exercise ??) for every strict subset P M ,
one has the strict inclusion Spank (P ) Spank (M ).
13 Let M be a subset in the k-vector space V . Prove that the following are
equivalent:
(i) M is linearly independent.
(ii) If n ≥ 1 is an integer, if x1 , . . . , xn ∈ M are different elements, and if
λ1 , . . . , λn ∈ k satisfy
λ1 x1 + · · · + λn xn = 0,
then λ1 = · · · = λn = 0.
Hint: To prove (i) ⇒ (ii) show that if one has a relation as in (ii), then one of the x’s can be
eliminated, without changing the linear Span.
14 Prove that a linearly independent set M cannot contain the zero element.
Prove that a subset of a linearly independent set is again linearly independent.
15 Prove that the union of a directed family of linearly independent sets is again
a linearly independent set. That is, if (Xj )j∈J is a family of k-linearly independent
subsets of the k-vector space V , with the property
• For any j, k ∈ J there exists some ∈ J such that Xj ⊂ X ⊃ Xk ,
then j∈J Xj is again a k-linearly independent subset of V .
16 Suppose V is a vector space, and x1 , x2 , . . . is a sequence (finite or infinite)
of different non-zero vectors in V . Prove that the following are equivalent:
(i) The set M = {x1 , x2 , . . . } is linearly independent.
(ii) The sequence of susbspaces Wk = Span({x1 , . . . , xk }) is strictly increasing,
in the sense that we have strict inclusions
W1 W2 ....
1. VECTOR SPACES 5
Hint: The implication (i) ⇒ (ii) is clear from the definition.
Conversely, if M were not linearly independent, there exist scalars λ1 , . . . , λn ∈ k such that
λ1 x1 + · · · + λn xn = 0, and at least one of the λ’s non-zero. If we take k = max{j : 1 ≤ j ≤
n and λj = 0}, then we get xk ∈ Span{x1 , . . . , xk−1 , which proves that Wk = Wk−1 .
Definition. Let V be a k-vector space. A subset B ⊂ V is called a k-linear
basis for V , if:
• Spank (B) = V ;
• B is k-linearly independent.
17 Prove the following:
Theorem 1.1.1. Let V be a k-vector space, and let P ⊂ M ⊂ V be subsets,
with P linearly independent, and Spank (M ) = V . Then there exists a linear basis
B such that P ⊂ B ⊂ M .
Sketch of proof: Consider the set
B = {B ⊂ V : B linearly independent, and P ⊂ B ⊂ M },
equipped with the inclusion as the order relation. Use Zorn’s Lemma to prove that B has a
maximal element, and then prove that such a maximal element must span the whole space. (One
key step is the checking of the hypothesis of Zorn’s Lemma. Use Exercise 13??.)
18 Suppose B is a linear basis for the vector space V , and P and M are subsets
of V , such that one has strict inclusions P B M . Prove that P and M are no
longer bases (although P is linearly independent and Span(M ) = V ).
19 Let W be a linear subspace of the vector space V , and let A be a linear basis
for W . Prove that there exists a linear basis B for V , with B ⊃ A.
Hint: Use Theorem 1.1.1.
Definition. A vector space is said to be finite dimensional, if it has a finite
basis.
20 Prove the following
Lemma 1.1.1. (Exchange Lemma) Suppose A and B are linear bases for the
vector space V . Prove that for every a ∈ A, there exists some b ∈ B, such that the
set (A {a}) ∪ {b} is again a linear basis.
Hint: Write a = β1 b1 + · · · + βn bn , for some b1 , . . . , bn ∈ B, and some β1 , . . . , βn ∈ k {0}.
At least one of the b’s does not belong to Span (A {a}). Choose this as the exchange for a.
We have the inclusion (A {a}) ∪ {b} ⊂ A ∪ {b}, with the first set linearly independent, but the
second one not. Using the fact that Span (A ∪ {b}) = V , Theorem 1.1.1 will force (A {a}) ∪ {b}
to be the basis.
21 Prove that any two linear bases, in a finite dimensional vector space, have the
same number of elements.
Hint: Fix a finite basis A = {a1 , . . . , an }, and let B be an arbitrary basis. Use the Exchange
Lemma 1.1.1 to construct inductively a sequence of elements b1 , . . . , bn ∈ B, such that
• For every k = 1, . . . , n, the set {b1 , . . . , bk−1 , ak , . . . , an } is a linear basis.
6 1. VECTOR SPACES AND LINEAR MAPS
Note that all the b’s must be different. We end up with a linear basis B0 = {b1 , . . . , bn }, with
|B0 | = n. Use Exercise ?? to conclude that B0 = B.
22* Prove the following generalization of the above result.
Theorem 1.1.2. Any two linear bases in a vector space have the same car-
dinality. (In cardinal arithmetic two sets have the same cardinality if there is a
bijection between them.)
Sketch of proof: Suppose A and B are two bases for V . If either A or B is finite, we apply
the previous exercise. So we can assume that both A and B are infinite.
For every a ∈ A we write it (uniquely) as a = β1 b1 + · · · + βn bn , for some b1 , . . . , bn ∈ B,
and some β1 , . . . , βn ∈ k {0}, and we set PB (a) = {b1 , . . . , bn }, so that PB (a) is the smallest
subset of P ⊂ B, with Span(P ) a. If we denote by F in(B) the set of all finite subsets of B, we
have now a map Π : A a −→ PB (a) ∈ F in(B). This map is not injective. However, for every
P ∈ F in(B), we have Π−1 (P ) ⊂ Span(P ), which means that Π−1 (P ) is a linearly independent set
in the finite dimensional vector space Span(P ). By Theorem 1.1.1, combined with the previous
exercise, this forces each of the sets Π−1 (P ), P ∈ F in(B), to be finite.
Then A is a disjoint union of preimages A = P ∈F in(B) Π−1 (P ), each of the sets in this
union being finite. Since F in(B) is infinite, we get
Card(A) = Card Π−1 (P ) ≤ Card F in(B) = Card(B).
P ∈F in(B)
By symmetry, we also have Card(B) ≤ Card(A), and we are done.
Definition. Given a k-vector space V , the above Theorem states that the
cardinality of a linear basis for V is independent of the choice of the basis. This
“number” is denoted by dimk V , and is called the dimension of V . In the finite
dimensional case, the dimension is a non-negative integer. If V is the trivial (zero)
space, we define its dimension to be zero.
23 Let n ≥ 1 be an integer. Define, for each j ∈ {1, . . . , n}, the element ej =
(δij )n ∈ kn . (Here δij stands for the Kronecker symbol, defined to be 1, if i = j,
i=1
and 0 if i = j.) Prove that the set {e1 , . . . , en } is a linear basis for the vector space
kn , therefore we have dim kn = n.
24 Generalize the above result, by proving that
dim k = Card(I).
i∈I
25 Suppose W is a linear subspace of the vector space V . Prove that dim W ≤
dim V . (This means that if A is a linear basis for W , and B is a linear basis for V ,
then Card(A) ≤ Card(B), which in turn means that there exists an injective map
f : A → B.)
Hint: Find another basis B for V , with B ⊃ A. Then use Theorem 1.1.2.
26 Suppose V is a vector space. Prove that the following are equivalent:
(i) V is finite dimensional.
(ii) Whenever W is a linear subspace of V , with dim W = dim V , it follows
that W = V .
1. VECTOR SPACES 7
(iii) Every infinite increasing sequence W1 ⊂ W2 ⊂ W3 ⊂ · · · ⊂ V is an infinite
of linear subspaces is stationary, in the sense that there exists some k ≥ 1,
such that Wn = Wk , for all n ≥ k.
Definition. Suppose W = (Wj )j∈J is a family of non-zero linear subspaces
of a vector space V . We say that W is linearly independent, if for every k ∈ J one
has
Wk ∩ Span Wj = {0}.
j∈J {k}
27 Let V be a vector space, and X be a subset of V {0}. The following are
equivalent:
(i) The set X is linearly independent (as a subset of V ).
(ii) The family (kx)x∈X is a linearly independent family of linear subspaces
of V .
28 Let W = (Wi )i∈I be a linearly independent family of linear subspaces of V .
Suppose X = (Xi )i∈I is another family of linear subspaces, such that Xi ⊂ Wi , for
all i ∈ I. Define J = j ∈ I : Xj = {0} . Prove that XJ = (Xj )j∈J is also a
linearly independent family.
29 Suppose V is a vector space, and W1 , W2 , . . . is a sequence (finite or infinite)
of non-zero linear subspaces of V . Prove that the following are equivalent:
(i) The sequence (W1 , W2 , . . . ) is linearly independent family.
(ii) For every k we have
Wk ∩ (W1 + · · · + Wk ) = {0}.
Hint: Argue along the same lines of the hint given to exercise ??.
30 Let V be a vector space, and let W1 and W2 be two non-zero linear subspaces.
Prove that (W1 , W2 ) is a linearly independent pair of linear subspaces, if and only
if W1 ∩ W2 = {0}.
31 Let W be a linear subspace of the vector space V . Prove that there exists a
linear subspace X of V , such that W ∩ X = {0}, and W + X = V .
Hint: Start with some linear basis A for W . Find a linear basis B for V , with B ⊃ A. Take
X = Span(B A).
32 Suppose W = (Wj )j∈J is a family of linear subspaces of a vector space V .
Prove that the following are equivalent
(i) W is linearly independent.
(ii) If n ≥ 1 is an integer, if j1 , . . . , jn ∈ J are different indices, and if w1 ∈
Wj1 , . . . , wn ∈ Wjn are elements such that w1 + · · · + wn = 0, it follows
that w1 = · · · = wn = 0.
(iii) There exists a choice, for each j ∈ J, of a linear basis Bj for Wj , such
that
(α) the sets (Bj )j∈J are mutually disjoint, i.e. for any j, k ∈ I with j = k,
we have Bj ∩ Bk = ∅;
(β) the set j∈J Bj is linearly independent.
8 1. VECTOR SPACES AND LINEAR MAPS
(iii’) For any choice, for each j ∈ J, of a linear basis Bj for Wj , we have the
properties (α) and (β) above.
33 Let (Wj )j∈J be a linearly independent family of linear subspaces of V . If we
choose, for each j ∈ J, a linear basis Bj for Wj , then j∈J Bj is linear basis for
the vector space Span j∈J Wj .
34 Let W1 , . . . , Wn be a finite linearly independent family of linear subspaces of
V . Prove that
dim (W1 + · · · + Wn ) = dim W1 + · · · + dim Wn .
35 Let V be a finite dimensional vector space. Suppose W1 , . . . , Wn are linear
subspaces of V , with
• W1 + · · · + Wn = V ;
• dim W1 + · · · + dim Wn .
Prove that (Wk )n is a linearly independent family.
k=1
Hint: For each k ∈ {1, . . . , n}, let Bk be a linear basis for Wk . Put B = B1 ∪ · · · ∪ Bn . On the
one hand, we have
(1) |B| ≤ |B1 | + · · · + |Bn | = dim W1 + · · · + dim Wn = dim V.
On the other hand, we clearly have Span(B) = V , so we must have |B| ≥ dim V . This means that
we must have equality in (1), so in particular the sets B1 , . . . , Bn are mutually disjoint, and their
union is a linear basis for V . Then the result follows from exercise ??
36 Let (Vi )i∈I be a family of vector spaces, and let εj : Vj → i∈I Vi , j ∈ I, be
the standard inclusions.
(i) εj (Vj ) j∈I is a linearly independent family of linear subspaces of i∈I Vi .
(ii) Suppose for each j ∈ I, we choose a linear basis Bj for Vj . Then εj (Bj )
is a linear basis for Vj .
Using the fact that εj are injective, conclude that dim εj (Vj ) = dim Vj , for all j ∈ I
As an application of the above, prove (use exercise ??) that the dimension of a
finite direct sum is given as
n n
dim Vi = dim Vi .
i=1 i=1
Definition. Suppose X is a k-linear subspace of the k-vector space V . In
particular X is a subgroup of the abelian group (V, +) We can then define the
quotient group V /X, which is again an abelian group. Formally, the quotient is
defined as the set of equivalence classes modulo X:
v≡w (mod X), if and only if v − w ∈ X.
The addition is defined as [v]X + [w]X = [v + w]X . (Here [v]X stands for the
equivalence class of v.) We can also define the scalar multiplication by λ · [v]X =
[λv]X . With these operations V /X becomes a vector space, called the quotient
vector space. For a subset A ⊂ V , we denote by [A]X the set {[a]X : a ∈ A} ⊂ V /X.
1. VECTOR SPACES 9
37 Verify the statements made in the above definition.
Definitions. Suppose X is a k-linear subspace of the k-vector space V .
• For a set M ⊂ V we define its k-linear span relative to X to be the linear
subspace
Spank (M ; X) = Spank (M ∪ X) = X + Spank (M ).
• A set P ⊂ V is said to be k-linearly independent relative to X, if the
map P p −→ [p]X ∈ V /X is injective, and the set [P ]X is linearly
independent in the quotient vector space V /X.
• A set B ⊂ V is called a k-linear X-basis for V ,
- Span(B; X) = V ;
- B is linearly independent relative to X.
38 Let X be a linear subspace of the vector space V .
A. If P ⊂ V is linearly independent relative to X, then P is linearly inde-
pendent.
B. For a set P ⊂ V , the following are equivalent:
(i) P is linearly independent relative to X.
(ii) If n ≥ 1 is an integer, if p1 , . . . , pn ∈ P are different elements, and if
λ1 , . . . , λn ∈ k satisfy
λ1 p1 + · · · + λn pn ∈ X,
then λ1 = · · · = λn = 0.
(iii) There exists a linear basis B for X, such that P ∪ B is linearly
independent.
(iii’) If B is any linear basis for X, then P ∪ B is linearly independent.
(iv) P is linearly independent, and X ∩ Span(P ) = {0}.
39 Let X and Y be linear subspaces of the vector space V , such that X ⊂ Y .
A. Prove that for every set M ⊂ V , we have the inclusion Span(M ; X) ⊂
Span(M ; Y ).
B. Prove that if a set P is linearly independent relative to Y , then P is also
linearly independent relative to X.
40 Let X be a linear subspace of the vector space V . For a set B ⊂ V prove that
the following are equivalent:
(i) B is a linear X-basis for V .
(ii) The map B b −→ [b]X ∈ V /X is injective, and [B]X is a linear basis for
V /X.
In this situation, prove the equality
dim V /X = Card(B).
41 Let X be a linear subspace of the vector space V , and let A be a linear basis
for X. Suppose B ⊂ V has the property A ∩ B = ∅. Prove that the following are
equivalent:
(i) B is a linear X-basis for V .
10 1. VECTOR SPACES AND LINEAR MAPS
(ii) A ∪ B is a linear basis for V .
Use this fact, in conjunction with Exercise ??, to prove the equality
(2) dim X + dim V /X = dim V.
42 Let W1 and W2 be linear subspaces of the vector space V . Let B ⊂ W1 be
an arbitrary subset. Prove that the following are equivalent:
(i) B is a linear (W1 ∩ W2 )-basis for W1 .
(i) B is a linear W2 -basis for W1 + W2 .
Use this fact to prove the following “Paralellogram Law”
(3) dim W1 + dim W2 = dim (W1 + W2 ) + dim (W1 ∩ W2 ).
Hint: Assume B satisfies condition (i). Prove first that Span(B; W2 ) = W1 + W2 . It suffices to
prove that W1 ⊂ Span(B; W2 ), which is pretty easy. Second, prove that B is linearly independent
relative to W2 . Argue that, if v = β1 b1 + · · · + βn bn ∈ W2 , then in fact we have v ∈ W1 ∩ W2 ,
which using (i) forces β1 = · · · = βn = 0. Use Exercise ?? to conclude that B is a linear W2 -basis
for W1 + W2 .
If B satisfies condition (ii), then it is clear that B is also linearly independent relative to
W1 ∩ W2 . To prove the equality Span(B; W1 ∩ W2 ) = W1 , it suffices to prove only inclusion “⊃.”
Start with some element w1 ∈ W1 . Using (ii) there exist b ∈ Span(B) and w2 ∈ W2 , such that
w1 = b + w2 . This forces w2 = w1 − b to belong to W1 ∩ W2 , so w1 ⊂ Span B ∪ (W1 ∩ W2 ) , thus
proving the desired inclusion.
To prove the equality (3), we use the above equivalence, combined with exercise ?? to get
dim W1 /(W1 ∩ W2 ) = Card(B) = dim (W1 + W2 )/W2 .
By adding dim(W1 ∩ W2 ) + dim W2 to both sides, and using exercise ??, the result follows.
2. Linear maps
Definition. Suppose V and W are k-vector spaces. A map T : V → W is
said to be k-linear, if:
• T is a group homomorphism, i.e. T (x + y) = T (x) + T (y), for all x, y ∈ V ;
• T is compatible with the scalar multiplication, i.e. T (λx) = λT (x), for all
λ ∈ k, x ∈ V .
43 Suppose V is a k-vector space. For a scalar µ ∈ k, we define the multiplication
map
Mµ : V v −→ µv ∈ V.
Prove that Mµ is k-linear. If we take µ = 1, then Mµ = IdV , the identity map on
V.
44 Let V and W be vector spaces, and let T : V → W be a linear map. For any
subset M ⊂ V , prove the equality
Span T (M ) = T Span(M ) .
45 Prove that the composition of two linear maps is again linear. Prove that the
inverse of a bijective linear map is again linear. A bijective linear map is called a
linear isomorphism.
2. LINEAR MAPS 11
46 Let X be a linear subspace of the vector space V , and let T : V → W be a
linear map. Prove that the restriction T X : X → W is again a linear map.
47 Let X be a k-linear subspace of a k-vector space V . Prove that the quotient
map (defined on page 8) V v −→ [v]X ∈ V /X is k-linear. Also prove that the
inclusion ι : X → V is linear.
48 Let T : V1 → V2 be a linear map, let X1 be a linear subspace of V1 , and let
X2 be a linear subspace of V2 . Prove the
Lemma 1.2.1. (Factorization Lemma) The following are equivalent:
(i) There exists a linear map S : V1 /X1 → V2 /X2 such that the diagram
quotient map
−−−−
V1 − − − − → V1 /X1
T S
quotient map
−−−−
V2 − − − − → V2 /X2
is commutative. (This means that, if we denote by qk : Vk → Vk /Xk ,
k = 1, 2 the quotient maps, then we have the equality q2 ◦ T = S ◦ q1 .)
(ii) T (X1 ) ⊂ X2 .
Moreover, in this case the map S, described implicitly in (i), is unique.
49 Let (Vi )i∈I be a family of vector spaces. For each j ∈ I, denote by εj :
Vj → i∈I Vi the standard inclusion, and by πj : i∈I Vi → Vj the coordinate
projection. Prove that εj and πj are linear maps.
50 Prove the following:
Theorem 1.2.1. (Universal Property of the direct sum) Let (Vi )i∈I be a family
of vector spaces. For each j ∈ I, denote by εj : Vj → i∈I Vi the standard
inclusion. Let W be a vector space, and let (Tj : Vj → W )j∈I be a collection of
linear maps. Then there exists a unique linear map T : i∈I Vi → W , such that
T ◦ εj = Tj , for all j ∈ J.
Hint: We know that any element v = (vi )i∈I ∈ i∈I Vi is represented as
v= (εj ◦ πj )(v) = εj (vj ).
j∈ v j∈ v
Define T (v) = j∈ v (Tj ◦ πj )(v) = j∈ v Tj (vj ).
Corollary 1.2.1. Let V be a k-vector space and let W = (Wi )j∈J be a family
of linear subspaces of V . Then there exists a unique linear map ΓW : i∈I Wi → V ,
such that, for every i ∈ I, we have
(ΓW ◦ εi )(w) = w, for all w ∈ Wi .
For any w = (wi )i∈I ∈ i∈I Wi , we have
ΓW (w) = wi .
i∈ w
12 1. VECTOR SPACES AND LINEAR MAPS
Comment. A particular case of the above result is when all the Wj ’s have
dimension one (or zero). In this case the family W is represented by an element
b = (bj )j∈J ∈ j∈J V , by Wj = kbj . Following the above construction, we have
a linear map denoted ΓV :
b j∈J k → V , which is defined as follows. For every
λ = (λj )j∈J ∈ j∈J k, we have ΓV (λ) = j∈ λ λj bj ∈ V .
b
51 Let V be a k-vector space and let W = (Wi )j∈J be a family of linear subspaces
of V . Let ΓW : j∈J Wj → V be the linear map defined in exercise ??.
A. Prove that ΓW is injective, if and only if W is a linearly independent
family.
B. Prove that ΓW is surjective, if and only if V = Span j∈J Wj .
Conclude that ΓW is an isomorphism, if and only if both condition below are true:
(i) W is a linearly independent family;
(ii) V = Span j∈J Wj .
Definition. A family W = (Wj )j∈J of linear subspaces of V , satisfying the
conditions (i) and (ii) above, is called a direct sum decomposition of V .
Comment. A particular case of the above result can be derived, along the
same lines as in the comment following exercise ??. Specifically, if V is a vector
space, and if we have a system b = (bj )j∈J ∈ j∈J V , then
A. ΓV : j∈J k → V is surjective, if and only if V = Span {bj : j ∈ J} .
b
B. ΓV : j∈J k → V is injective, if and only if all the bj ’s are different and
b
the set {bj : j ∈ J} is linearly independent.
In particular, ΓV : j∈J k → V is a linear isomorphism, if and only if all the bj ’s
b
are different and the set {bj : j ∈ J} is a linear basis for V .
52 Let (Vj )j∈J be a direct sum decomposition for a vector space V . Let W
be a vector space, and assume that, for each j ∈ J, we are given a linear map
Tj : Vj ∈ W . Prove that there exists a unique linear map T : V → W , such that
T Vj
= Tj , for all j ∈ J.
53 Let V and W be vector spaces, and let T : V → W be a linear map. Since T is
a group homomorphism, we know that Ker T ⊂ V and Ran T ⊂ W are subgroups.
Prove that they are in fact linear subspaces.
Comment. We know from group theory that the triviality of the kernel is
equivalent to the injectivity of the homomorphism. As a particular case, we get the
following:
• A linear map is injective, if and only if its kernel is the zero subspace.
54 Generalize the above fact as follows. Let V and W be vector spaces, let
T : V → W be a linear map, and let Z be a linear subspace of V . Prove that the
following are equivalent:
(i) The restriction T Z : Z → W is injective.
(ii) Z ∩ Ker T = {0}.
2. LINEAR MAPS 13
55 Let X be a linear subspace of the vector space V , and let q : V → V /X denote
the quotient map.
A. Prove that Ker q = X. Use this to show that, given a linear subspace W of
V , the restriction q W : W → V /X is injective, if and only if W ∩X = {0}.
B. Given a linear subspace W of V , the restriction q W : W → V /X is
surjective, if and only if W + X = V .
56* Prove the following technical result.
Lemma 1.2.2. (Descent Lemma) Let V be a vector space, and let T : V → V
be a linear map. For every Define the subspaces W0 = {0} and
Wk = Ker(T ◦ · · · ◦ T ),
k factors
for all k ≥ 1.
(i) One has the inclusions W0 ⊂ W1 ⊂ W2 ⊂ . . . .
(ii) For every k ≥ 1, one has Wk = T −1 (Wk−1 )
(iii) For every k ≥ 1, there exists a unique linear map Sk : Wk+1 /Wk →
Wk /Wk−1 such that the diagram
quotient map
−−−−
Wk+1 − − − − → Wk+1 /Wk
S
T k
Wk −−−−
− − − − → Wk /Wk−1
quotient map
is commutative.
(iv) All linear maps Sk : Wk+1 /Wk → Wk /Wk−1 are injective.
(v) Suppose there exists some , such that W = W +1 . Then Wk = W , for
all k > .
Sketch of proof: Parts (i) and (ii) are obvious.
From (ii) we get T (Wk ) ⊂ Wk−1 , and then part (iii) follows from the Factorization Lemma
??.
To prove part (iv), start with some x ∈ Ker Sk . This means that x = [w]Wk , for some
w ∈ Wk+1 , and T (w) ∈ Wk−1 . This forces w ∈ Wk , so x = 0 in the quotient space Wk+1 /Wk .
This means that Ker Sk = {0}, and we are done.
To prove part (v), observe first that the given condition forces W +1 /W = {0}. Use then
induction, based on (iv).
57 Let V and Z be vector spaces, let W be a linear subspace of V , and let
T : W → Z be a linear map. Prove that there exists a linear map S : V → Z, such
that S W = T .
Hint: It suffices to prove this in the particular case when Z = W and T = Id. By exercise
??,we can choose a linear subspace X of V , with X ∩ W = {0} and X + W = V . If we take
q : V → V /X the quotient map, then its restriction q W : W → V /X is a linear isomorphism.
Put S = (q W
)−1 ◦ q.
58 Let V be a non-zero k-vector space, and let v ∈ V {0}. Prove the existence
of a k-linear map φ : V → k with φ(v) = 1.
14 1. VECTOR SPACES AND LINEAR MAPS
Hint: Take W = Span(v) = kv, and define the map ψ : k λ −→ λv ∈ W . Prove that
ψ : k → W is a linear isomorphism. Take φ0 : W → k to be the inverse of ψ, and apply the
previous exercise.
59 Let V and W be vector spaces, and let T : V → W be a linear map. For a
subset M ⊂ V , prove the equality
T −1 Span T (M ) = Span(M ; Ker T ).
60 Let V and W be vector spaces, and let T : V → W be a linear map. For a
subset M ⊂ V , prove that the following are equivalent:
(i) M is linearly independent relative to Ker T .
(ii) The restriction T M : M → W is injective and the subset T (M ) ⊂ W is
linearly independent.
61 Let V and W be vector spaces, and let T : V → W be a linear map. Prove
that the following are equivalent:
(i) T is injective.
(ii) There exists a linear basis B for V , such that the restriction T B : B → W
is injective, and T (B) is linearly independent.
(ii’) If B is any linear basis for V , then the restriction T B : B → W is
injective, and T (B) is linearly independent.
62 Let V and W be vector spaces, and let T : V → W be a linear map. Prove
that the following are equivalent:
(i) T is surjective.
(ii) There exists a linear basis B for V , such that the restriction Span T (B) =
W . independent.
(ii’) If B is any linear basis for V , Span T (B) = W .
63 Let T : V → W be a linear isomorphism from the vector space V into the
vector space W . Prove that a set B ⊂ V is linear basis for V , if and only if T (B)
is a linear basis for W . Conclude that dim V = dim W .
64 Let V and W be vector spaces, and let T : V → W be a linear map. Prove
that the following are equivalent:
(i) T is a linear isomorphism.
(ii) There exists a linear basis B for V , such that the restriction T B : B → W
is injective, and T (B) is a linear basis for W .
(ii’) If B is any linear basis for V , then the restriction T B : B → W is
injective, and T (B) is a linear basis for W .
65 Let V and W be vector spaces, and let T : V → W be a linear map. Prove
the following.
2. LINEAR MAPS 15
Theorem 1.2.2. (Isomorphism Theorem) There exists a unique bijective linear
map T : V /Ker T → Ran T such that the following diagram is commutative
quotient map
−−−−
V − − − − → V /Ker T
T T
W −−−
←−− Ran T
inclusion
ˆ
(this means that T = ι ◦ T ◦ q, where ι is the inclusion, and q is the quotient map).
Use exercise ?? to conclude that
(4) dim(Ker T ) + dim(Ran T ) = dim V.
Hint: ˆ
To get the existence, and uniqueness of T , use the Factorization Lemma ??, with the
subspaces Ker T ⊂ V and {0} ⊂ W .
66 Let V and W be finite dimensional vector spaces, with dim V = dim W , and
let T : V → W be a linear map.
A. Prove the “Fredholm alternative”
dim(Ker T ) = dim (W/Ran T ).
B. Prove that the following are equivalent:
(i) T is a linear isomorphism.
(ii) T is surjective.
(iii) T is injective.
Hint: Part A follows immediately from (4) and (2). To prove the equivalence in part B, use the
fact that for a linear subspace Y ⊂ W , the condition dim(W/Y ) = 0 is equivalent to the fact that
Y = W.
67* Give an infinite dimensional counterexample to the “Fredholm alternative.”
∞
Specifically, consider the infinite dimensional vetor space V = n=1 k and con-
struct a non-invertible surjective linear map T : V → V .
Notation. Suppose V and W are k-vector spaces. We denote by Link (V, W )
the set of all k-linear maps V → W .
68 Let V and W be k-vector spaces. For T, S ∈ Lin(V, W ) we define the map
T + S : V → W by
(T + S)(v) = T (v) + S(v), for all v ∈ V.
Prove that T + S is again a linear map. Prove that Lin(V, W ) is an abelian group,
when equipped with the addition operation. The neutral element is the null map
0(v) = 0.
69 Let V and W be k-vector spaces. For T ∈ Lin(V, W ) and λ ∈ k, we define
the map λT : V → W by
(λT )(v) = λT (v), for all v ∈ V.
Prove that λT is a linear map. Prove that Lin(V, W ), when equipped with this
multiplication, and the addition defined above, is k-vector space.
16 1. VECTOR SPACES AND LINEAR MAPS
70 The above construction is a special case of a more general one. Start with a
vector space W and a set X. Consider the set M ap(X, W ) of all functions X → W .
For f, g ∈ M ap(X, W ) we define f + g ∈ M ap(X, W ) by
(f + g)(x) = f (x) + g(x), for all x ∈ X,
and for f ∈ M ap(X, W ) and λ ∈ k, we define λf ∈ M ap(X, W ) by
(λf )(x) = λf (x), for all x ∈ X.
The M ap(X, W ) becomes a k-vector space. Prove that if V is a vector space, then
Lin(V, W ) is a linear subspace in M ap(V, W ).
Comment. The set M ap(X, W ) is precisely the product x∈X W . The above
vector space structure on M ap(X, W ) is precisely the direct product vector space
structure.
71 Suppose V , W , X and Y are vector spaces, and T : X → V and S : W → Y
are linear maps. Prove that the map
Lin(V, W ) R −→ S ◦ R ◦ T ∈ Lin(X, Y )
is linear.
72 Let V be a k-vector space. Prove that the map
Link (k, V ) T −→ T (1) ∈ V
is a k-linear isomorphism.
73 Let V and W be vector spaces. Suppose Z V is a proper linear subspace.
Assume we have elements v ∈ V Z and w ∈ W . Prove that there exists a linear
map T ∈ Lin(V, W ), such that T Z = 0, and T (v) = w.
Hint: By assumption, the element v = [v]Z is non-zero in the quotient space V /Z. Use exercise
?? to find a linear map φ : V /Z → k, with φ(v) = 1. Take σ : K → W to be unique linear map
with σ(1) = w. Now we have a composition σ ◦ φ ∈ Lin(V /Z, W ), with (σ ◦ φ)(v) = w. Finally,
compose this map with the quotien map V → V /Z.
74 Suppose V is a vector space, and X is a finite set. Prove that
dim M ap(X, V ) = Card(X) · dim V.
Hint: Use exercise ??, plus the identification M ap(X, V ) = x∈X V.
75 Let V and W be vector spaces. Fix a subset X ⊂ V , and consider the map
ΣX : Lin(V, W ) T −→ T X
∈ M ap(X, W ).
Prove that ΣX is a linear map.
76* Use same notations as in the previous exercise. Assume both V and W are
non-zero.
A. Prove that ΣX is injective, if and only if V = Span(X).
B. Prove that ΣX is surjective, if and only if X is linearly independent.
2. LINEAR MAPS 17
Conclude that ΣX is an isomorphism, if and only if X is a linear basis for V .
Hint: A. If Span(X) V , choose an element v ∈ V Span(X), and an element w ∈ W {0}.
Use exercise ?? to produce a linear map T : V → W , such that T (v) = w and T X = 0. Such an
element is non-zero, but belongs to Ker ΣX , so ΣX is not injective. Conversely, any T ∈ Ker ΣX
will have the property that X ⊂ Ker T . In particular, if Span(X) = V , this will force Ker T = V ,
hence T = 0.
B. Assume X is not linearly independent. This means that there exist different elements
x1 , . . . , xn ∈ X and scalars α1 , . . . , αn ∈ k {0}, such that α1 x1 + · · · + αn xn = 0. Choose a an
element w ∈ W {0} and define the map f : X → W by
w if x = x1
f (x) =
0 if x = x1
Prove that f does not belong to Ran ΣX , so ΣX is not surjective. Conversely, assume X is linearly
independent, and let f : X → W , thought as an element b = f (x) x∈X ∈ x∈X W . Define the
linear map ΓW : x∈X k → W , as in exercise ??. Likewise, if we put Z = Span(X), we can use
b
the element a = (x)x∈X ∈ x∈X Z, to define a linear map ΓZ : a x∈X k → Z. This time, Γa
Z
W ◦ (ΓZ )−1 ∈ Lin(Z, W ), we have
is a linear isomorphism. If we consider the composition T = Γb a
managed to produce a linear map with T (x) = f (x), for all x ∈ X. We now extend T to a linear
map S : V → W , which will satisfy ΣX (S) = f .
77 Suppose V and W are vector spaces.
A. Prove that the following are equivalent:
(i) dim V ≤ dim W ;
(ii) there exists an injective linear map T : V → W ;
(iii) there exists a surjective linear map S : W → V .
B. Prove that the following are equivalent:
(i) dim V = dim W ;
(ii) there exists a linear isomorphism T : V → W .
Hint: Fix A a linear basis for V , and B a linear basis for W , so that ΣA : Lin(V, W ) →
M ap(A, W ) and ΣB : Lin(W, V ) → M ap(B, V ) are linear isomorphisms. To prove A (i) ⇒ (ii),
we use the definition of cardinal inequality
Card(A) ≤ Card(B) ⇔ ∃ f : A → B injective .
For f as above, if we take T ∈ Lin(V, W ) with the property that ΣA (T ) = f , it follows that T is
injective. To prove the implication (ii) ⇒ (iii), start with some injective linear map T : V → W ,
and consider S : Ran T → V to be the inverse of the isomorphism T : V → Ran T . Use exercise ??
to extend S to a linear map R : W → V . Since S is surjective, so is R. The implication (iii) ⇒ (i)
follows from the (2).
B. The implication (i) ⇒ (ii) has already been discussed. To prove the converse, use the
equivalence
Card(A) ≤ Card(B) ⇔ ∃ f : A → B bijective .
For f as above, if we take T ∈ Lin(V, W ) with the property that ΣA (T ) = f , it follows that T is
an isomorphism.
78 Assume V and W are vector spaces, with V finite dimensional. Prove that
(5) dim Lin(V, W ) = dim V · dim W.
As a particular case, we have
(6) dim Link (V, k) = dim V
Definition. For a k-vector space V , the vector space Link (V, k) is called the
dual of V .
18 1. VECTOR SPACES AND LINEAR MAPS
79 Let V be a k-vector space. Suppose B is a linear basis for V . For every b ∈ B,
define the map fb : B → k by
1 if x = b
fb =
0 if x = b
Use exercise ?? to find, for each b ∈ B, a unique element b∗ ∈ Link (V, k), with
ΣB (b∗ ) = fb . The linear map b∗ is uniquely characterized by b∗ (b) = 1, and
b∗ (x) = 0, for all x ∈ B {b}.
(i) Prove that set B ∗ = {b∗ : b ∈ B} is linearly independent in Link (V, k).
(ii) Prove that the map B b −→ b∗ ∈ B ∗ is injective, so we have Card(B) =
Card(B ∗ ).
80 Let V be a k-vector space, and let B be a linear basis for V . Using the
notations from the preceding exercise, prove that the following are equivalent
(i) B ∗ is a linear basis for Link (V, k).
(ii) V is finite dimensional.
Hint: Use the isomorphism ΣB : Link (V, k) → b∈B k, we have
Span ΣB (B ∗ ) = Span({fb : b ∈ B}).
But when we identify M ap(B, k) with b∈B k, we get
Span({fb : b ∈ B}) = k.
b∈B
Therefore the condition Span(B ∗ ) = Link (V, k) is equivalent to the condition b∈B k= b∈B k,
which is equivalent to the condition that B is finite.
Definition. Suppose V is a finite dimensional k-vector space, and B is a linear
basis for V . Then (see the above result) the linear basis B ∗ for Link (V, k) is called
the dual basis to B. In fact the notion of the dual basis includes also the bijection
B b −→ b∗ ∈ B ∗ as part of the data.
3. The minimal polynomial
Notations. From now on, the composition of linear maps will be written
without the ◦ symbol. In particular, if V is a k-vector space, if n ≥ 2 is and
integer, and if T ∈ Link (V, V ), we write T n instead of T ◦ T ◦ · · · ◦ T (n factors).
Instead of Link (V, V ) we shall simply write Link (V ).
81 Suppose V is a k-vector space. The composition operation in Link (V, V ) is
obviously associative. Prove that, for every T ∈ Link (V ), the maps
Link (V ) S −→ ST ∈ Link (V )
Link (V ) S −→ T S ∈ Link (V )
are linear. (In particular, the composition product is distributive with respect to
Comment. The above result states that Link (V ) is a unital k-algebra. The
term “unital” refers to the existence of a multiplicative identity element. In our
case, this is the identity IdV , which from now on will be denoted by I (or IV when
we need to specify the space).
3. THE MINIMAL POLYNOMIAL 19
82 If dim V ≥ 2, prove that the algebra Lin(V ) is non-commutative, in the sense
that there exist elements T, S ∈ Lin(V ) with T S = ST . (Give an example when
T S = 0, but ST = 0.
83 Let V be a vector space, and let S, T : V → V be linear maps. Suppose S
and T commute, i.e. ST = T S. Prove
(i) For any integers m, n ≥ 1, the linear maps S m and T n commute: S m Tn =
T nSm.
(ii) The Newton Binomial Formula holds:
n
n n
(S + T ) = k S n−k T k .
k=0
Hints: For (i) it Suffices to analyze the case n = 1. Use induction on m
For (ii) use (i) and induction.
The following is a well-known result from algebra
Theorem 1.3.1. Suppose L is a unital k-algebra. Then for any element X ∈ L,
there exists a unique unital k-algebra homomorphism ΦX : k[t] → L, such that
ΦX (t) = X.
Here k[t] stands for the algebra of polynomials in a (formal) variable t, with
coefficients in k. The unital algebra homomorphism condition means that
• ΦX is linear;
• ΦX is multiplicative, in the sense that φx (P Q) = ΦX (P )ΦX (Q), for all
p, q ∈ k[t];
• Φx (1) = I.
To be more precise, the homomorphism ΦX is constructed as follows. Start with
some polynomial P (t) = α0 + α1 t + · · · + αn tn . Then
ΦX (P ) = α0 I + α1 X + · · · + αn X n .
Definition. Suppose V is a k-vector space, and X : V → V is a linear
map. If we apply the above Theorem for the algebra Link (V ), and the element X,
the corresponding map ΦX : k[t] → Link (V ) is called the polynomial functional
calculus of X. For a polynimial P ∈ k[t], the element Φx (P ) will simply be denoted
by P (X).
Remark 1.3.1. If P and Q are polynomials, then P (X)Q(X) = Q(X)P (X),
simply because both are equal to (P Q)(X). This means that, although Link (V ) is
not commutative, the sub-algebra
{P (X) : P ∈ k[t]} ⊂ Link (V )
is commutative.
84 Let V be a vector space, and let S, T : V → V be linear maps. Assume S
and T commute. Prove that, for any two polynomials P, Q ∈ k[t], the linear maps
P (S) and Q(T ) commute.
85 Prove that the algebra k[t] is infinite dimensional, as a k-vector space. More
explicitly, the set {1} ∪ {tn : n ≥ 1} is a linear basis for k[t].
20 1. VECTOR SPACES AND LINEAR MAPS
86 Let V be a finite dimensional k-vector space. For every X ∈ Link (V ), prove
that there exists a non-zero polynomial P ∈ k[t], such that P (X) = 0.
Hint: Consider the polynomial functional calculus ΦX : k[t] → Link (V ). On the one hand, we
notice that dim Link (V ) = (dim V )2 < ∞ (by exercise ??). This will force dim(Ran ΦX ) < ∞.
On the other hand, we know that
dim(Ker ΦX ) + dim(Ran ΦX ) = dim(k[t]).
But since k[t] is infinite dimensional, this forces Ker φX to be infinite dimensional. In particular
Ker ΦX = {0}.
87 Suppose V is a finite dimensional k-vector space, and X ∈ Link (V ). Prove
that there exists a unique polynomial M ∈ k[t] with the following properties:
(i) M (X) = 0.
(ii) M is monic, in the sense that the leading coefficient is 1.
(iii) If P ∈ k[t] is any polynomial with P (X) = 0, then P is divisible by M
Hint: One (conceptual) method of proving this result is to quote the property of k[t] of being a
principal ideal domain. Then Ker ΦX , being an ideal, must be presented as Ker ΦX = M · k[t],
for a unique monic polynomial M .
Here is the sketch of a “direct” proof (which in fact traces the steps used in proving that k[t]
is a pid, in this particular case). Choose a (non-zero) polynomial N ∈ k[t], of minimal degree,
such that N (X) = 0. Replace N with M = λN , so that M is monic (we will still have M (X) = 0).
Suppose now P is a polynomial with P (x) = 0. Divide P by M with remainder, so that we have
P = M Q+R, for some polynomials Q and R, with deg R < deg M . Using the fact that M (X) = 0,
we get
R(X) = M (X)Q(X) + R(X) = P (X) = 0.
By minimality, this forces R = 0, thus proving property (iii). The uniqueness is a direct application
of (iii).
Definition. Suppose V is a finite dimensional k-vector space, and X : V → V
is a linear map. The polynomial M , described in the preceding exercise, is called
the minimal polynomial of X, and will be denoted by MX .
88* Give an infinite dimensional counter example to exercise ??.
Hint: Take V = k[t] and define the linear map X : k[t] P (t) −→ tP (t) ∈ k[t]. Prove that
there is no polynomial M such that M (X) = 0.
89 Let V be a finite dimensional vector space, and fix some X ∈ Lin(V ). Prove
that the following are equivalent:
• X is a scalar multiple of the identity, i.e. there exists some α ∈ k such
that X = αI.
• The minimal polynomial MX (t) is of degree one.
Definition. Suppose V is a vector space. A linear map X : V → V is said to
be nilpotent, if there exists some integer n ≥ 1, such that X n = 0.
90 Let V be a finite dimensional vector space, and let X ∈ Lin(V ) be nilpotent.
Prove that the minimal polynomial is of the form MX (t) = tp , for some p ≤ dim V .
In particular, we have X dim V = 0.
Hint: Fix n such that X n = 0 Consider the linear subspaces W0 = {0} and Wp = Ker(X k ),
k ≥ 1. It is clear that MX (t) = tp , where p is the smallest index for which Wp = V . Prove that
3. THE MINIMAL POLYNOMIAL 21
for every k ∈ {1, . . . , p} we have the strict inclusion Wk−1 Wk . (Use the Descent Lemma ??)
Use then (recursively) the dimension formula for quotient spaces (2), to get
p
dim V = dim(Wk /Wk−1 ).
k=1
Since each term in the sum is ≥ 1, this will force dim V ≥ p.
91 Let V be a vector space, and let M, N : V → V be nilpotent linear maps. If
M and N commute, then prove that M + N is again nilpotent. If X : V → V is an
arbitrary linear map, which commutes with N , then prove that XN is nilpotent.
92* Suppose V is finite dimensional, X ∈ Lin(V ), and W is an invariant linear
subspace, i.e. with the property T (W ) ⊂ W . Let T : V /X → V /X be the unique
linear map which makes the digram
quotient map
−−−−
V − − − − → V /W
X T
quotient map
−−−−
V − − − − → V /W
commutative (see exercise ??). Let us also consider the linear map S = X W :
W → W . Prove that
(i) The minimal polynomial MX (t) of X divides the product of minimal poly-
nomials MS (t)MT (t).
(ii) The polynomial MX (t) is divisble by both minimal polynomials MS (t) and
MT (t), hence MX (t) is divisible by the lowest common multiple (lcm) of
MS (t) and MT (t).
Conclude that, if MS (t) and MT (t) are relatively prime, then
MX (t) = MS (t)MT (t).
Hint: Observe first that, for any polynomial P ∈ k[t], one has a commutative diagram
quotient map
− − − −→
V − − − − − V /W
P (X)
P (T )
quotient map
− − − −→
V − − − − − V /W
and the equality P (S) = P (X) W . In particular, since MX (X) = 0, we immediately get MX (S) =
0 and MX (T ) = 0, thus proving (ii). To prove (i), we only need to show that MS (X)MT (X) = 0.
Start with an arbitrary element v ∈ V . Since MT (T ) = 0, using the above diagram we get the
equality
q [MT (X)](v) = 0, in V /W,
which means that the element w = [MT (X)](v) belongs to W . But then, using MS (S) = 0, we
get
0 = [MS (S)](w) = [MS (X)](w) = [MS (X)] [MT (X)](v) = [MS (X)MT (X)](v).
93 The following example shows that, in general the minimal polynomial MX (t)
can differ from both the product MS (t)MT (t) and lcm(MS (t), MT (t)). Consider
the space V = k4 and the linear map
X : k4 (α1 , α2 , α3 , α4 ) −→ (0, α1 , 0, α2 ) ∈ k4 .
22 1. VECTOR SPACES AND LINEAR MAPS
Take the linear subspace W = {(0, 0, λ, µ) : λ, µ ∈ k} ⊂ V . Check the equalities:
MS (t) = t, MT (t) = t2 , and MX (t) = t4 .
Hint: Prove that S = 0, T 2 = 0, but T = 0, and X 4 = 0, but X 3 = 0.
94* Suppose V is finite dimensional, X ∈ Lin(V ), and W1 , . . . , Wn are invariant
subspaces, such that W1 + · · · + Wn = V . For each k = 1, . . . , n we take Mk (t) to
be the minimal polynomial of X Wk . Prove that the minimal polynomial of X is
given as
MX (t) = lcm(M1 (t), . . . , Mk (t)).
Hint: Use induction on k. In fact, it suffices to prove the case k = 2. As in the previ-
ous exercise, prove that MX (t) is divisible by lcm(M1 (t), . . . , Mk (t)). Conversely, if we de-
note lcm(M1 (t), . . . , Mk (t)) simply by P (t), we know that for every k, we have a factorization
P (t) = Qk (t)Mk (t). In particular, for every w ∈ Wk , we get
[P (X)](w) = [Qk (X)] [Mk (X)](w) = [Qk (X)] [Mk (X Wk
)](w) = 0.
Now we have P (X) Wk
= 0, for all k, which gives P (X) = 0, thus proving that P (t) is divisible
by MX (t).
95* Let V be a finite dimensional k-vector space, and let X : V → V be a linear
map. Prove the following
Theorem 1.3.2. (Abstract Spectral Decomposition) Assume the minimal poly-
nomial is decomposed as
(7) MX (t) = M1 (t) · M2 (t) · · · · · Mn (t),
such that for any j, k ∈ {1, . . . , n} with j = k, we have gcd(Mj , Mk ) = 1. Define
the polynomials
MX (t)
Pj (t) = , j = 1, . . . , n.
Mj (t)
Since gcd(P1 , P2 , . . . , Pn ) = 1, we know there exist polynomials Q1 , . . . , Qn ∈ k[t],
such that
(8) Q1 (t)P1 (t) + Q2 (t)P2 (t) + · · · + Qn (t)Pn (t) = 1.
Define the linear maps
Ej = Qj (X)Pj (X) ∈ Link (V ), j = 1, 2, . . . , n.
(i) The linear maps Ej , j = 1, . . . , n are idempotents, i.e. Ej 2 = Ej .
(ii) For any j, k ∈ {1, . . . , n} with j = k, we have Ej Ek = 0.
(iii) E1 + E2 + · · · + En = I.
(iv) For each j ∈ {1, . . . , n} we have the equality Ran Ej = Ker[Mj (X)].
(v) For every j, the minimal polynomial of the restriction X Ran Ej is precisely
Mj (t).
Sketch of proof: By construction, we have (iii), so in fact conditions (i) and (ii) are equivalent.
To prove (ii) start with j = k. Using the fact that Pj Pk is divisible by all the Mi ’s it follows that
Pj Pk is divisible by MX , so we get Pj (X)Pk (X) = 0. Using polynomial functional calculus, we
get
Ej Ek = Qj (X)Pj (X)Qk (X)Pk (X) = Qj (X)Qk (X)Pj (X)Pk (X) = 0.
To prove (iv) we first observe that since Mj (t)Pj (t) = MX (t), we have (again by functional
calculus)
Mj (X)Ej = Mj (X)Qj (X)Pj (X) = Qj (X)Mj (X)Pj (X) = Qj (X)MX (X) = 0,
3. THE MINIMAL POLYNOMIAL 23
which proves the inclusion Ran Ej ⊂ Ker[Mj (X)]. Conversely, since Pk (t) is divisible by Mj (t)
for all k = j, we see that Pk (X) Ker[M (X)] = 0, which forces Ek Ker[M (X)] = 0, for all k = j.
j j
Using (iii) this gives Ej Ker[Mj (X)]
= Id Ker[Mj (X)]
, thus proving the other inclusion.
To prove (v), denote for simplicity the subspaces Ran Ej by Wj . By (iii) we know that
W1 + · · · + Wn = V . By exercise ?? we know that
0 0
(9) MX = lcm(M1 , . . . , Mn ),
0
where Mj is the minimal polynomial of X . By (iii) we know that Mj (X ) = 0, hence Mj (t)
Wj Wj
0 0 0
is divisble by Mj (t). In particular we will also have gcd(Mj , Mk ) = 1, for all j = k, so by (9) we
will get
0 0 0
MX (t) = M1 (t)M2 (t) · · · · · Mn (t).
0
This forces Mj = Mj , for all j
Definition. Use the hypothesis of the above Theorem. The idempotents
E1 , . . . , En are called the spectral idempotents of X, associated with the decom-
position (7). The subspaces Ran Ej = Ker[Mj (X)], j = 1, . . . , n are called the
spectral subspaces of X, associated with the decomposition (7). Notice that
(i) V = Ker[M1 (X)] + Ker[M2 (X)] + · · · + Ker[Mn (X)].
n
(ii) Ker[Mj (X)] j=1 is a linearly independent family of linear subspaces.
n
In other words, Ker[Mj (X)] j=1
is a direct sum decomposition of V .
96 Prove the properties (i) and (ii) above.
97 The above definition to carries a slight ambiguity, in the sense that the
spectral idempotents Ej are defined using a particular choice of the polynomials
Qj for which (8) holds. Prove that if we choose another sequence of polynomials,
Q1 , . . . , Qn , such that
Q1 (t)P1 (t) + Q2 (t)P2 (t) + · · · + Qn (t)Pn (t) = 1,
then we have the equalities
Qj (X)Pj (X) = Ej , for all j = 1, . . . , n.
Therefore the spectral idempotents are un-ambiguously defined.
Hint: Define E j = Qj (X)Pj (X), j = 1, . . . , n. Use the Theorem to show Ran E j = Ran Ej , and
then use the Theorem again to show that E j = Ej , for all j.
Another approach is to look at the spectral decomposition
V = Ker[M1 (X)] + Ker[M2 (X)] + · · · + Ker[Mn (X)],
which depends only on the factorization (7). For every element v ∈ V , there are unique elements
wj ∈ Ker[Mj (X)], j = 1, . . . , n, such that v = w1 + · · · + wn . Then Ej (v) = wj .
98 Prove the following
Theorem 1.3.3. (Uniqueness of Spectral Decomposition) Let V be a finite
dimensional vector space, let X : V → V be a linear map, and let W1 , . . . , Wn be a
collection of invariant linear subspaces. For each j = 1, . . . , n, define Mj (t) to be
the minimal polynomial of X W ∈ Lin(Wj ). Assume that:
j
(i) W1 + · · · + Wn = V .
(ii) For every j, k ∈ {1, . . . , n} with j = k, we have gcd(Mj , Mk ) = 1.
24 1. VECTOR SPACES AND LINEAR MAPS
Then the minimal polynomial decomposes as MX (t) = M1 (t) · · · · · Mn (t), and
moreover, the Wj ’s are precisely the spectral subspaces associated with this decom-
position, i.e. Wj = Ker[Mj (X)], for all j = 1, . . . , n.
Hint: The decomposition of MX (t) has already been discussed (see exercise ??). It is clear
that Wj ⊂ Ker[Mj (X)]. On the one hand, (Wj )n j=1 is a linearly independent family of linear
subspaces, so
dim V = dim(W1 + · · · + Wn ) = dim W1 + · · · + dim Wn .
On the other hand, we have dim Wj ≤ dim(Ker[Mj (X)]), and we also have
dim V = dim(Ker[M1 (X)]) + · · · + dim(Ker[Mn (X)]).
By finiteness, this forces dim Wj = dim(Ker[Mj (X)]), so we must have Wj = Ker[Mj (X)].
99* Work again under the hypothesis of the Abstract Spectral Decomposition
Theorem. The following is an alternative construction of the spectral idempotents.
Suppose R1 , . . . , Rn are polynomials such that, for each j ∈ {1, . . . , n}, we have:
(i) Rj is divisible by Mk , for all k ∈ {1, . . . , n}, k = j;
(ii) 1 − Rj is divisible by Mj .
(See the hint on why such polynomials exist.) Prove that Rj (X) = Ej , for all j.
Hint: The simplest example of the existence of R1 , . . . , Rn is produced by taking Rj (t) =
Qj (t)Pj (t), j = 1, . . . , n.
Define Fj = Rj (X). On the one hand, j, k ∈ {1, . . . , m} with j = k, the polynomial Rj Rk is
divisible by both Pj and Pk , hence it is divisible by gcd(Pj , Pk ) = MX . In particular we get
Fj Fk = Rj (X)Rk (X) = (Rj Rk )(X) = 0.
On the other hand, for a fixed j, it is clear that Rk is divisible by Mj , for all k = j. Therefore,
if we consider the polynomial R(t) = R1 (t) + · · · + Rn (t), we see that R − Rj is divisible by Mj .
But so is 1 − Rj . This means that 1 − R is divisble by Mj . Since this is true for all j, it follows
that 1 − R is divisible by gcd(M1 , . . . , Mn ) = MX . This forces R(X) = 1, which means that
F1 + · · · + Fn = I, so we must have Fj 2 = Fj .
n
Now we have a linearly independent family of linear subspaces Ran Fj j=1 , with
(10) Ran Fn + · · · + Ran Fn = V.
Notice that Mj Rj is divisible by MX , so Mj (X)Fj = Mj (X)Rj (X) = 0. This means that
Ran Fj ⊂ Ker[Mj (X)]. Because of (10), this forces Ran Fj = Ker[Mj (X)] = Ran Ej , which in
turn forces Fj = Ej .
100 Suppose V and W are k-vector spaces, and S : V → W is a linear isomor-
phism. Prove that the map
Link (V ) X −→ SXS −1 ∈ Link (W )
is an isomorphism of k-algebras. Use this fact tow prove that, for a linear map
X ∈ Link (V ) one has
P (SXS −1 ) = SP (X)S −1 , for all P ∈ k[t].
101 Suppose V , W and S are as above, and V (hence also W ) finite dimensional.
Prove that for any X ∈ Lin(V ) we have the equality of minimal polynomials
MX (t) = MSXS −1 (t).
4. SPECTRUM 25
4. Spectrum
Convention. Throughout this section we restrict ourselves to a particular
choice of the field k = C, the field of complex numbers. All vector spaces are
assumed to be over C.
This choice of the field is justified by the fact that the arithmetic in C[t] is
particularly nice, as shown by the following:
Theorem 1.4.1. (Fundamental Theorem of Algebra) For every non-zero poly-
nomial P ∈ C[t] there exists a collection of different numbers λ1 , . . . , λn ∈ C, and
integers m1 , . . . , mk ≥ 1, such that
P (t) = (t − λ1 )m1 · · · · · (t − λn )mn .
The numbers λ1 , . . . , λn are called the roots of P . They are precisely the
soultions (in C) of the equation P (λ) = 0. The numbers m1 , . . . , mn are called the
multiplicities. More precisely, mk is called the multiplicity of the root λk . Remark
that
deg P = m1 + · · · + mn .
Definitions. Let V be a vector space, and X : V → V be a linear map. The
spectrum of X is defined to be the set
Spec(X) = {λ ∈ C : MX (λ) = 0}.
Suppose λ ∈ Spec(X). The multiplicity of λ, as a root of the minimal polynomial, is
called the spectral multiplicity of λ, and is denoted by mX (λ). The linear subspace
N (X, λ) = Ker[(X − λI)mX (λ) ,
is called the spectral subspace of λ. The number
dX (λ) = dim N (X, λ)
is called the spectral dimension of λ.
According to the Fundamental Theorem of Algebra, if we list the spectrum as
Spec(X) = {λ1 , . . . , λn }, with the λ’s different, we have a decomposition
(11) MX (t) = (t − λ1 )mX (λ1 ) · · · · · (t − λn )mX (λn ) .
When we apply the Abstract Spectral Decomposition Theorem for the polynomials
Mj (t) = (t − λj )mX (λj ) , j = 1, . . . , n, we get the following
Theorem 1.4.2. (Spectral Decomposition Theorem) Suppose V is a finite
dimensional vector space, and X : V → V is a linear map. List the spectrum as
Spec(X) = {λ1 , . . . , λn }, with all λ’s different.
(i) The spectral subspaces N (X, λj ), j = 1, . . . , n, form a linearly independent
family of linear subspaces.
(ii) N (X, λ1 ) + · · · + N (X, λn ) = V .
If for every j ∈ {1, . . . , n} we define Xj = X N (X,λj )
∈ Lin N (X, λj ) , then
(iii) the minimal polynomial of Xj is (t − λj )mX (λj ) ; in particular the spectrum
is a singleton Spec(Xj ) = {λj }.
26 1. VECTOR SPACES AND LINEAR MAPS
Definitions. Suppose V is a finite dimensional vector space, and X : V → V
is a linear map. The family N (X, λ) λ∈Spec(X) , which by Theorem ?? is a direct
sum decomposition of V , is called the spectral decomposition of X. The spectral
idempotents of X are the ones described in the Abstract Spectral Decomposition
Theorem (exercise ??), corresponding to (11). They are be denoted by EX (λ),
λ ∈ Spec(X). These idempotents have the following properties (which uniquely
characterize them):
(i) EX (λ)EX (µ) = 0, for all λ, µ ∈ Spec(X) with λ = µ;
(ii) Ran EX (λ) = N (X, λ), for all λ ∈ Spec(X).
As a consequence, we also get λ∈Spec(X) EX (λ) = I.
102 Let V be a finite dimensional vector space, and let X : V → V be a linear
map. Prove that, for every λ ∈ Spec(X), we have the inequality mX (λ) ≤ dX (λ).
Hint: Consider the linear map Xλ = X N (X,λ)
∈ Lin N (X, λ) . We know that the minimal
polynomial of Xλ is Mλ (t) = (t−λ)mX (λ) .Then the linear map T = Xλ −λI : N (X, λ) → N (X, λ)
is nilpotent. Use exercise ?? to conclude that (Xλ − λI)dX (λ) = 0, hence obtaining the fact that
(t − λ)dX (λ) is divisible by Mλ (t).
Definition. Let V be a finite dimensional vector space, let X : V → V be a
linear map, and let Spec(X) = {λ1 , . . . , λn } be a listing if its spectrum (with all
λ’s different). The polynomial
HX (t) = (t − λ1 )dX (λ1 ) · · · · · (t − λn )dX (λn )
is called the chracteristic polynomial of X.
Remark 1.4.1. The above result gives the fact that the minimal polynomial
MX divides the characteristic polynomial HX .
103 Prove that deg HX = dim V . As a consequence of the above remark, we
have.
Proposition 1.4.1. The degree of the minimal polynomial does not exceed
dim V .
Hint: Use the Spectral Decomposition to derive the fact that deg HX = dim V .
104 Suppose V and W are finite dimensional vector spaces and S : V → W
is a linear isomorphism. Prove that for any X ∈ Lin(V ) we have the equality of
characteristic polynomials HX (t) = HSXS −1 (t).
Hint: Using the equality of minimal polynomials MX (t) = MSXS −1 (t), we know that Spec(X) =
Spec(SXS −1 ). Prove that, for every λ ∈ Spec(X) we have
S N (X, λ) = N (SXS −1 , λ),
hence dX (λ) = dSXS −1 (λ).
105 Prove the following characterization of the spectrum.
Proposition 1.4.2. Let V be a finite dimensional vector space, and X : V →
V be a linear map. For a complex number λ, the following are equivalent:
(i) λ ∈ Spec(X);
4. SPECTRUM 27
(ii) the linear map X − λI is not injective;
(iii) the linear map X − λI is not surjective;
(iv) the linear map X − λI is not bijective;
Hint: The equivalence of (ii)-(iv) follows from the Fredholm alternative. If λ ∈ Spec(X), upon
writing MX (t) = P (t)(t − λ), we see that 0 = MX (X) = P (X)(X − λI), so X − λI cannot be
invertible. Conversely, if λ ∈ Spec(X), then when we divide MX (t) by t − λ we get a non-zero
remainder. In other words, we get MX (t) = (t − λ)Q(t) + µ, for some polynomial Q and some
µ = 0. But then, using functional calculus, we get
0 = MX (X) = (X − λI)Q(X) + µI,
which proves that the linear map Y = −µ−1 Q(X) satisfies (X − λ)Y = Y (X − λI) = I, so X − λI
is invertible.
The above characterization, specifically condition (ii) brings up an important
concept.
Definition. A number λ is said to be an eigenvalue for the linear map X :
V → V , if there exists a non-zero element v ∈ V , with X(v) = λv. Such a v is then
called an eigenvector for X, corresponding to λ.
With this terminiology, we have λ ∈ Spec(X), if and only if λ is an eigenvalue
for X. (This condition is equivalent to the fact that Ker(X − λI) = {0}, which is
the same as the non-injectivity of X − λI.)
106 Prove the following:
Theorem 1.4.3. (Spectral Mapping Theorem) If V is a finite dimensional
vector space, and X : V → V is linear, then for any polynomial P ∈ C[t] one has
the equality
Spec P (X) = P Spec(X) .
Sketch of proof: To prove the inclusion “⊃” start with some λ ∈ Spec(X), and use exercise
?? to get a λ-eigenvector v for X. Then show that v is a P (λ)-eigenvector for P (X), hence
P (λ) ∈ Spec P (X) .
To prove the inclusion “⊂” start with some µ ∈ Spec P (X) , and Use the Fundamental
Theorem of Algebra to get a factorization
P (t) − µ = (t − λ1 ) · (t − λn ),
for some λ − 1, . . . , λn ∈ C (not necessarily different). Use functional calculus to obtain
P (X) − µI = (X − λ1 I) · (X − λn I).
Argue that at least one of the linear maps X − λj I, j = 1, . . . , n is non-invertible, so that at least
one of the λj ’s belongs to Spec(X). For such a λj we then have µ = P (λj ) ∈ P Spec(X) .
The Spectral Mapping Theorem has numerous applications. One of them is
the following.
107 Let V be a finite dimensional vector space, and let X : V → V be a linear
map. Consider the linear map S : V → V defined by
S= λEX (λ),
λ∈Spec(X)
where EX (λ), λ ∈ Spec(X), are the spectral idempotents of X. Prove that
28 1. VECTOR SPACES AND LINEAR MAPS
(i) The minimal polynomial of S is
MS (t) = (t − λ).
λ∈Spec(X)
In particular MS (t) has simple roots.
(ii) The linear map N = X − S is nilpotent.
Hints: Let us recall (see exercise ??) that the spectral idempotents can be constructed as
EX (λ) = Rλ (X), where Rλ (t), λ ∈ Spec(X) is a family of polynomials with the following prop-
erties:
(α) Rλ (t) is divisible by (t − µ)mX (µ) , for all λ, µ ∈ Spec(X) with λ = µ;
(β) 1 − Rλ (t) is divisible by (t − λ)mX (λ) , for all λ ∈ Spec(X).
In particular, we get
(12) Rλ (µ) = δλµ , for all λ, µ ∈ Spec(X).
Now, by functional calculus, the linear map S is given by S = P (X), where
P (t) = λRλ (t),
λ∈Spec(X)
and then (12) will give
(13) P (λ) = λ, for all λ ∈ Spec(X).
Using the Spectral Mapping Theorem, we get the equality Spec(S) = Spec(X). Notice that, by
construction, the subspaces Wλ = Ran EX (λ) = Ker[(X − λI)mX (λ) ], λ ∈ Spec(X) = Spec(S),
for a direct sum decomposition of V . It is pretty obvious that S W = λIWλ , for all λ ∈ Spec(S),
λ
which means that the minimal polynomial of S W is Mλ (t) = t − λ. Using exercise ?? we then
λ
have the desired form for MS (t).
If we look at N = X − S, we see that we can write, N = Q(X), where Q(t) = t − P (t). Then
using (13) we have
Q(λ) = 0, for all λ ∈ Spec(X),
so by the Spectral Mapping Theorem we get Spec(N ) = {0}, so N is nilpotent.
Definition. Let V be a finite dimensional vector space. A linear map S :
V → V is said to be semisimple, if the minimal polynomial MS (t) has simple roots.
The above result then states that any linear X : V → V map has a “Semisimple
+ Nilpotent” decomposition X = S + N , with both S and N representable using
functional calculus of X.
108 Let V be a finite dimensional vector space, and let X : V → V be a linear
map. Prove that the following are equivalent:
(i) X is semisimple.
(ii) There exist idempotents E1 , . . . , En , with
• Ei Ej = 0, for all i, j ∈ {1, . . . , n}, with i = j,
• E1 + · · · + En = I,
and there exist complex numbers λ1 , . . . , λn ∈ C, such that
X = λ 1 E 1 + · · · + λ n En .
(iii) There exists a linear basis for V consisting of eigenvectors for X.
Hint: Suppose X is semisimple. Use the construction given in exercise ?? to produce the
semisimple map
S= λEX (λ).
λ∈Spec(X)
We know that S can be written as S = P (X), where P ∈ C[t] is a certain polynomial, with
P (λ) = λ, for all λ ∈ Spec(X). In particular, the polynomial Q(t) = t − P (t) is divisible by
4. SPECTRUM 29
λ∈spec(X) (t − λ) = MX (t), hence Q(X) = 0, which by functional calculus gives X = P (X) = S,
thus proving (ii).
Assume now X satisfies (ii). In particular the subspaces Wj = Ran Ej , j = 1, . . . , n form a
direct sum decomposition for V . If, for every j ∈ {1, . . . , n}, we choose a basis Bj for Wj , then
B = n Bλ is a desired linear basis (each b ∈ Bj is a λj -eigenvector for X).
j=1
Finally, assume B is a linear basis for V consisting of eigenvectors for S. This means that
we have a map B b −→ θ(b) ∈ C, such that
S(b) = θ(b)b, for all b ∈ B.
Put Λ = {θ(b) : b ∈ B}. For each λ ∈ Λ, we define the set Bλ = θ−1 ({λ}), and we put
Wλ = Span(Bλ ). Then it is clear that S W = λIWλ , for all λ ∈ Λ, so that the minimal
λ
polynomial of S Wλ
is Mλ (t) = t − λ. Since B = λ∈Λ Bλ , we have λ∈Λ Wλ = V , so by
exercise ?? we conclude that the minimal polynomial of S is MS (t) = λ∈Λ (t − λ).
109 Let V be a finite dimensional vector space, and let S, T : V → V be
semisimple linear maps, which commute, i.e. ST = T S. Prove that S + T and ST
are again semisimple.
Hints: Use the spectral decompositions to write S = λ∈Spec(S) λES (λ), with the ES (λ)’s
obtained from functional calculus of S. Likewise, we have the spectral decomposition T =
µ∈Spec(T ) λET (µ), with the ET (µ)’s obtained from functional calculus of T . In particular,
the ES (λ)’s commute with the ET (µ)’s. Using this fact we see that, if we define the linear maps
Fλµ = ES (λ)ET (µ), then the Fλµ ’s will be idempotents. Moreover, we will have
• Fλ1 µ1 Fλ2 µ2 = 0, if (λ1 , µ1 ) = (λ2 , µ2 );
• Fλµ = I.
λ∈Spec(S)
µ∈Spec(T )
Now we have
T +S = (λ + µ)Fλµ , TS = (λµ)Fλµ ,
λ∈Spec(S) λ∈Spec(S)
µ∈Spec(T ) µ∈Spec(T )
and we can apply exercise ??.
110 The result from exercise ?? can be strengthened. Prove the following.
Theorem 1.4.4. (“Semisimple + Niplotent” Theorem) Let V be a finite di-
mensional vector space, and let X : V → V be a linear map. Then there exists a
unique decomposition X = S + N , with
(i) S semisimple;
(ii) N niplotent;
(iii) S and N commute, i.e. SN = N S.
Skecth of proof: The existence is clear from exercise ??.
To prove uniqueness, assume X = S + N is another decomposition, with properties (i)-
(iii). Notice that both S and N commute with X, hence they also commute with the spectral
idempotents EX (λ), λ ∈ Spec(X). In particular, S and S commutive, and also N and N
commute. Now we have
S −S =N −N ,
with the lefthand side semisimple (exercise ??) and the righthand nilpotent (exercise ??). But the
only linear map, which is both semisimple and nilpotent, is the null map. This forces S = S and
S = N.
30 1. VECTOR SPACES AND LINEAR MAPS
5. The similarity problem and the spectral invariant
Definition. Let V be a vector space, and let X, Y : V → V be linear maps.
We say that X is similar to Y , if there exists a linear isomorphism S : V → V , such
that Y = SXS −1 .
111 Prove that “is similar to” is an equivalence relation.
The following is a fundamental question in linear algebra.
Similarity Problem. Suppose V is a finite dimensional vector space, and
X, Y : V → V are linear maps. When is X similar to Y ?
An additional problem is to construct at least one linear isomorphism S : V →
V , such that Y = SXS −1 .
The approach to the Similarity Problem is by constructing various (computable)
invariants. More precisely, for every linear map X : V → V , we want to construct
an “object” ΘX , such that
• if X is similar to Y , then ΘX = ΘY .
If we have such a construction, we call the correspondence Lin(V ) X −→ ΘX a
similarity invariant. Then our main goal will be to construct a “strong” similarity
invariant, in the sense that
• If ΘX = ΘY , then X is similar to Y .
In this section we will eventually construct such a strong invariant for complex
vector spaces.
Convention. All vector spaces in this section are assumed to be over C.
Examples. The following are examples of similarity invariants
(i) the spectrum Lin(V ) X −→ Spec(X);
(ii) the minimal polynomial Lin(V ) X −→ MX ∈ C[t];
(iii) the characteristic polynomial Lin(V ) X −→ HX ∈ C[t].
Note that (ii) and (iii) are stronger than (i).
112 The mininmal and the charecteristic polynomial are not strong enough.
Give examples of linear maps T, X ∈ Lin(C3 ), and Y, Z ∈ Lin(C2 ) such that
MT (t) = MX (t), HY (t) = HZ (t), but HT (t) = HX (t) and MY (t) = MZ (t). This
shows that T is not similar to X, although MT (t) = MX (t), and Y is not similar
to Z, although HY (t) = HZ (t).
Hint: Take T (α1 , α2 , α3 ) = (0, α2 , α3 ), and X(α1 , α2 , α3 ) = (0, 0, α3 ).
Take Y (α1 , α2 ) = (0, α1 ), Z(α1 , α2 ) = (0, 0).
113 Even when the minimal and characteristic polynomial are combined, i.e.
when we consider the invariant ΘX = (MX , HX ), we still do not get a strong
invariant. Consider the linear maps X, Y ∈ Lin(C4 ), defined by
X(α1 , α2 , α3 , α4 ) = (0, α1 , 0, α3 )
Y (α1 , α2 , α3 , α4 ) = (0, α1 , 0, 0)
Later on we shall see why X and Y are not similar. For the moment prove that
MX (t) = MY (t) = t2 , and HX (t) = HY (t) = t4 .
5. THE SIMILARITY PROBLEM AND THE SPECTRAL INVARIANT 31
Definition. Let V be a finite dimensional vector space, and let X : V → V
is a linear map. For λ ∈ Spec(X), we define the linear subspaces Nk (X, λ) =
Ker[(X − λI)k ], k ≥ 1. These subspaces are called the generalized eigenspaces of
X, corresponding to λ. More specifically, Nk (X, λ) will be called the generalized
eigenspace of order k. It will be convenient to use the convention N0 (X, λ) = {0}.
114 Suppose V is a finite dimensional vector space, and X : V → V is a linear
map. Fix λ ∈ Spec(X). Prove that for any integers p > k ≥ 1 one has the inclusion
X k Np (X, λ) ⊂ Np−k (X, λ).
115 Suppose V is a finite dimensional vector space, and X : V → V is a linear
map. Fix λ ∈ Spec(X). Prove that the sequence of generalized λ-eigenspaces
exhibits the following inclusion pattern:
(14) {0} N1 (X, λ) ··· NmX (λ) (X, λ) = NmX (λ)+1 (X, λ) = · · · = N (X, λ).
Hint: It is clear that we have the inclusions
{0} N1 (X, λ) ⊂ N2 (X, λ) ⊂ . . . ,
Define the maps Yk = X N (X,λ) ∈ Lin Nk (X, λ) . On the one hand, it is clear that (Yk −λI)k =
k
0, so that if we denote by Mk the minimal polynomial of Yk , we know that
(α) The polynomial Mk (t) divides the polynomial (t − λ)k .
On the other hand, by exercise ?? we also know that the minimal polynomial Mk of Yk also
divides the minimal polynomial MX of X. Combined with (α) this gives
(β) The polynomial Mk (t) divides the polynomial (t − λ)min{k,mX (λ)} .
Because Yk = Yk+1 Nk (X,λ)
, again by exercise ?? we get
(γ) Mk divides Mk+1 , for all k ≥ 1.
Suppose now k ≥ mX (λ). Combining (β) and (γ) yields Mk (t) = MmX (λ) (t) = (t − λ)mX (λ) .
In particular, we get (Yk − λ)mX (λ) = 0, i.e. (X − λI)mX (λ) N (X,λ) = 0, forcing the inclusion
k
Nk (X, λ) ⊂ NmX (λ) (X, λ). This then gives in fact the equality Nk (X, λ) = NmX (λ) (X, λ) =
N (X, λ).
The next step would be to prove that, if k ∈ {1, . . . , mX (λ)}, then Nk−1 (X, λ) Nk (X, λ).
By the Descent Lemma (exercise ??), it suffices to check this only for k = mX (λ). Suppose
NmX (λ)−1 (X, λ) = N (X, λ). Using (β) this will force (X − λI)mX (λ)−1 N (X,λ) = 0. This is
however impossible, because the minimal polynomial of X N (X,λ)
is (t − λ)mX (λ) .
Definition. Let V be a finite dimensional vector space, and let X : V → V
be a linear map. For each λ ∈ Spec(X) we define the sequence of integers
∆X (λ) = [dim N1 (X, λ), dim N2 (X, λ), . . . , dim NmX (λ) (X, λ)].
The family D X = ∆X (λ) λ∈Spec(X)
is called the spectral invariant of X.
Comments. It is understood that the spectral invariant ∆X has the following
data built in it:
• The spectrum Spec(X) as the indexing set.
• The spectral multiplicities. For each λ ∈ Λ, the spectral multiplicty
mX (λ) is the length of the sequence ∆X (λ).
So, given X, Y ∈ Lin(V ), the equality ∆X = ∆Y means:
32 1. VECTOR SPACES AND LINEAR MAPS
(i) Spec(X) = Spec(Y );
(ii) mX (λ) = mY (λ), for all λ ∈ Spec(X) = Spec(Y );
(iii) dim Nk (X, λ) = dim Nk (Y, λ), for all λ ∈ Spec(X) = Spec(Y ) and all
k with 1 ≤ k ≤ mX (λ) = mY (λ).
116 Prove that the spectral invariant is a similarity invariant.
Hint: Assume Y = SXS −1 . We know already that the spectra and the multiplicities are equal.
Prove that for any λ ∈ Spec(X) = Spec(Y ) and any integer k ≥ 1 we have S Nk (X, λ) =
Nk (Y, λ), so (rememeber that linear isomorphisms preserve dimensions) we get dim Nk (X, λ) =
dim Nk (Y, λ).
117 Prove that the linear maps X, Y ∈ Lin(C4 ) defined in exercise ?? have
different spectral invariants. In particular X is not similar to Y .
The remainder of this section is devoted to the converse of exercise ??:
Theorem 1.5.1. (Spectral Invariant Theorem) If V is a finite dimensional
vector space, and if X, Y : V → V are linear maps with D X = ∆Y , then X is
similar to Y .
The proof of this result is quite ellaborate, so we will break it into several steps.
The first result we are concerned with is a set of restrictions that occur on the
spectral invariant.
118 Let V be a finite dimensional vector space, and let X : V → V be a linear
map. Prove that, for each λ ∈ Spec(X), the sequence
mX (λ)
dim[Nk (X, λ)/Nk−1 (X, λ)] k=1
is non-decreasing.
Hint: Use the Descent Lemma (exercise ??) for the linear map T = X − λI.
Comments. Since we have strict inclusions Nk−1 (X, λ) Nk (X, λ), k =
1, . . . , mX (λ), by exercise ??, it follows that the terms in the above sequence also
satsifies:
dim[Nk (X, λ)/Nk−1 (X, λ)] ≥ 1, for all k = 1, . . . , mX (λ).
We also know that, for each k ≥ 1, we have
dim[Nk (X, λ)/Nk−1 (X, λ)] = dim Nk (X, λ) − dim Nk−1 (X, λ).
So the spectral invariant exhibits the following pattern
dim N1 (X, λ) ≥ dim N2 (X, λ) − dim N1 (X, λ) ≥ . . .
(15)
· · · ≥ dim NmX (λ) (X, λ) − dim NmX (λ)−1 (X, λ) ≥ 1.
119 Let V be a finite dimensional vector space, and let X : V → V be
a linear map. Fix some λ ∈ Spec(X). Prove there exists a unique sequence
[c1 , c2 , . . . , cmX (λ) ] of non-negative integers, with cmX (λ) ≥ 1, such that for every
k = 1, . . . , mX (λ), we have:
(16) dim Nk (X, λ) = c1 + 2c2 c + · · · + kck + kck+1 + · · · + kcmX (λ) .
5. THE SIMILARITY PROBLEM AND THE SPECTRAL INVARIANT 33
Hint: Define first the numbers
pk = dim[Nk (X, λ)/Nk−1 (X, λ)] = dim Nk (X, λ) − dim Nk−1 (X, λ), k = 1, . . . , mX (λ),
so that by exercies ?? we have
p1 ≥ p2 ≥ · · · ≥ pmX (λ) ≥ 1,
and
(17) dim Nk (X, λ) = p1 + p2 + · · · + pk , for all k = 1, . . . , mX (λ).
Define cmX (λ) = pmX (λ) , and ck = pk − pk+1 for all k with 1 ≤ k < mX (λ). Then we will have
pk = ck + ck+1 + · · · + cmX (λ) , for all k = 1, . . . , mX (λ),
and the conclusion follows from (17)
Definitions. Let V be a finite dimensional vector space, and let X : V → V
be a linear map. For every λ ∈ Spec(X), the numbers defined in the above exercise
will be called the λ-cell numbers of X. They will be denoted by ck (X, λ). From the
above discussion, it is clear that if one defines a system ΥX = ΥX (λ) λ∈Spec(X) ,
where
ΥX (λ) = [c1 (X, λ), c2 (X, λ), . . . , cmX (λ) (X, λ)],
then we have constructed a new similarity invariant, called the Jordan invariant.
For each λ ∈ Spec(X), the set
ΥX (λ) = k ∈ {1, . . . , mX (λ)} : ck (X, λ) = 0
will be called the λ-cell support of X.
Comment. It is clear that the spectral invariant and the Jordan invariant are
equivalent, in the sense that
• ΥX = ΥX , if and only if ∆X = ∆Y .
The proof of the Spectral Invariant Theorem will use the Jordan invariant in an
essential way.
120 Let V be a finite dimensional vector space, and let T : V → V be a nilpotent
linear map, so that Spec(T ) = {0}, and the minimal polynomial is MT (t) = tm for
some m ≥ 1. Prove that there eixsts a sequence of linear subspaces
Z1 ⊂ Nk (T, 0), Z2 ⊂ N2 (T, 0), . . . , Zm ⊂ Nm (T, 0),
such that, for every p ∈ {1, . . . , m} we have:
(i) Zp + T (Zp+1 ) + · · · + T m−p (Zm ) + Np−1 (T, 0) = Np (T, 0);
(ii) Zp ∩ T (Zp+1 ) + · · · + T m−p (Zm ) + Np−1 (T, 0) = {0}.
Hint: Denote, for simplicity, the generalized eigenspaces Nk (T, 0) by Nk , k = 1, . . . , m. Do
reverse induction on p. Start off by choosing a linear subspace Zm ⊂ Nm such that
• Zm + Nm−1 = Nm ;
• Zm ∩ Nm−1 = {0}.
The existence of such a subspace follows from exercise ??. Suppose the spaces Zk+1 , Zk+2 , . . . , Zm
have been chosen, such that properties (i) and (ii) hold for p ∈ {k + 1, k + 2, . . . , m}. Consider
the linear subspace
Wk = T (Zk+1 ) + T 2 (Zk+2 ) + · · · + T m−k (Zm ) + Nk−1 ⊂ Nk ,
and use again exercise ?? to find a linear subspace Zk , such that
• Zk + Wk = Nk ;
• Zk ∩ Wk = {0}.
34 1. VECTOR SPACES AND LINEAR MAPS
Definition. With the above notations, the sequence (Z1 , Z2 , . . . , Zm ) will be
called a cell skeleton for T .
121 Use the setting and notations from exercise ??. Prove that if (Z1 , . . . , Zm )
is a cell skeleton for T , then for every p ∈ {1, . . . , m} we have the equality
p m
Np (T, 0) = T j− (Zj ).
=1 j=
0
(Here we use the convention: T (Zj ) = Zj .)
Hint: Use property (i) and induction on p.
In exercises ??-?? below, we fix a finite dimensional vector space V , a nilpotent
linear map T : V → V . Denote by m the degree of the minimal polynomial, i.e.
we have MT (t) = tm . For simplicity, the generalized eigenspaces Nk (T, 0) will be
denoted by Nk , k = 1, . . . , m.
122 Suppose p ∈ {1, . . . , m}. If W ⊂ Np is a linear subspace, with W ∩ Np−1 =
{0}. Suppose k is an integer with 1 ≤ k < p. Prove that
• The linear map T p−k W : W → V is injective.
• The space T p−k (W ) is a linear subspace of Nk , with T p−k (W ) ∩ Nk−1 =
{0}.
Hint: Notice that, by the construction of the generalized eigenspaces, as Nj = Ker(T j ), we have
the inclusions
(18) T i (Nj ) ⊂ Nj−i , for all i, j with 1 ≤ i < j ≤ m.
In particular, we have the inclusions
T p−k (Np ) ⊂ Nk and T p−k (Np−1 ) ⊂ Nk−1 .
From the first inclusion, we immediately get the inclusion T p−k (W ) ⊂ Nk . Using the Factorization
Lemma (exercise ??), there exists a linear map S : Np /Np−1 → Nk /Nk−1 , such that the diagram
quotient map
− − − −→
Np − − − − − Np /Np−1
p−k
T S
quotient map
− − − −→
Nk − − − − − Nk /Nk−1
is commutative. So, if we denote by qj : Nj → Nj /Nj−1 , j = 1, . . . , m, the quotient maps, we
have
(19) S ◦ qp = qk ◦ T p−k .
It turns out that S is injective. (Indeed, if we start with some element x ∈ Np /Np−1 with
S(x) = 0, then if we write x = qp (v) for some v ∈ Np , then we get 0 = S(x) = S qp (x) =
qk T p−k (v) , which means that T p−k (v) belongs to Nk−1 = Ker(T k−1 ). This will then give
T p−1 (v) = T k−1 T p−k (v) = 0, which means that v belongs to Ker(T p−1 ) = Np−1 . Finally this
forces x = qp (v) = 0.)
Now, since W ∩ Ker qp = W ∩ Np−1 = {0}, by exercise ?? we know that the restriction qp W
is injective. This gives the fact that the map (S ◦ qp ) W is injective.
When the equality (19) to W we see that the composition (qk ◦ T p−k ) W is injective, so this
will force T p−k W to be injective. Moreover, the injectivity of (qk ◦ T p − k) W will also force the
equality T p−k (W ) ∩ Ker qk = {0}, which means precisely that T p−k (W ) ∩ Nk−1 = {0}.
123* Let (Z1 , Z2 , . . . , Zm ) be a cell skeleton for T . Fix some integer p with
1 ≤ p < m. Define the linear subspace
Vp = T (Zp+1 ) + T 2 (Zp+2 ) + · · · + T m−p (Zm ).
5. THE SIMILARITY PROBLEM AND THE SPECTRAL INVARIANT 35
Let Xp : Zp+1 ⊕ Zp+2 ⊕ · · · ⊕ Zm → Vp be the unique linear map with the property
that Xp Z = T k Z , for all k = 1, 2, . . . , m − p. Prove that
p+k p+k
(i) Xp is a linear isomorphism.
(ii) Vp ∩ Np−1 = {0}.
Hint: Use reverse induction on p. Start with p = m − 1. In this case Vm−1 = T (Zm ), and
everything is obvious from exercise ??. Assume now the properties (i) (ii) are true for p = k + 1,
and let us prove them for p = k. By construction, the linear subspace Zp+1 ⊂ Np+1 is constructed
in such a way that
(α) Zk+1 + Vk+1 + Np = Nk+1 ;
(β) Zk+1 ∩ [Vk+1 + Nk ] = {0}.
Using the inductive hypothesis, we know that Vk+1 ∩ Nk = {0}, which means that Vk+1 and
Nk form a linearly independent pair of subspaces. In particular, using exercise ?? and condition
(β), we get the fact that the triple (Zk+1 , Vk+1 , Nk ) is also a linearly independent family. In
particular, the subspace Yk+1 = Zk+1 + Vk+1 has the properties:
(γ) Yk+1 ∩ Nk = {0};
(δ) the linear map
Y : Zk+1 ⊕ Vk+1 (z, v) −→ z + v ∈ Yk+1
is a linear isomorphism.
Using (γ) and exercise ??, we know that T Yk+1
is injective, and T (Yk+1 ) ∩ Nk = {0}, thus giving
property (i) for p = k. Observe that
T (Yk+1 ) = T (Zk+1 ) + T (Vk+1 ) = T (Zk+1 ) + T 2 (Zk+2 ) + · · · + T m−k (Zk ) = Vk ,
so we get a linear isomorphism T Yk+1
: Yk+1 → Vp . Using (δ), and the inductive hypothesis, we
immediately obtain property (ii) for p = k.
124 Let (Z1 , Z2 , . . . , Zm ) be a cell skeleton for T . Prove the equalities:
dim Zp = cp (T, 0), for all p = 1, . . . , m,
where c1 (T, 0), . . . , cm (T, 0) are the cell numbers of T .
Hint: Define, Vp = T (Zp+1 ) + · · · + T m−p (Zm ), and Wp = Vp + Np−1 , for all p = 1, . . . , m − 1.
Also put Wm = Zm . Denote dim Zk simply by ck . On the one hand, by the preceding exercise,
we know that
(20) dim Vp = dim Zp+1 + dim Zp+2 + · · · + dim Zm = cp+1 + cp+2 + · · · + cm ,
for all p = 1, . . . , m − 1. On the other hand, again by the preceding exercise, we know that
(21) dim Wp = dim Vp + dim Np−1 .
Recall that, by construction, the subspace Zp ⊂ Np is chosen such that
• Zp ∩ Wp = {0};
• Zp + Wp = Np .
This will give dim Np = dim Zp + dim Wp = cp + dim Wp . Using (21) and (20) we get
dim Np = cp + dim Vp + dim Np−1 = cp + cp+1 + · · · + cm + dim Np−1 .
By the construction of the cell numbers (see exercise ??) this gives the desired result.
125 Let us denote by J the cell support of T , i.e.
J = j ∈ {1, . . . , m} : cj (X, 0) = 0 .
Let (Z1 , . . . , Zm ) be a cell skeleton for T . Prove that the family
T k (Zj ) j∈J
0≤k<j
is a direct sum decomposition of V . (Here we use the convention: T 0 (Zj ) = Zj .)
36 1. VECTOR SPACES AND LINEAR MAPS
Hint: On the one hand, we know that
dim T k (Zj ) = dim Zj = cj (X, 0), for all j, k with 0 ≤ k < j ≤ m.
Of course, if j ∈ J, these dimensions are zero. In particular, using the properties of cell numbers
(exercise ??), we get we have
m j−1 m j−1 m
dim T k (Zj ) = dim T k (Zj ) = cj (X, 0) = jcj (X, 0) =
(22) j∈J j=1 k=0 j=1 k=0 j=1
0≤k<j
= dm (X, 0) = dim Nm = dim V.
On the other hand, by exercise ??
m m
V = Nm = T j− (Zj ),
=1 j=
and after making the change of indices k = j − , we get
m j−1
V = T k (Zj ) = T k (Zj ),
j=1 k=0 j∈J
0≤k<j
and the result then follows from (22) and exercise ??.
126* Prove the following
Theorem 1.5.2. (Spectral Invariant Theorem for Nilpotents) Let V and W be
finite dimensional vector spaces, and let T : V → V and X : W → W be nilpotent
linear maps. Assume
dim[Ker(T k )] = dim[Ker(X k )], for all k ≥ 1.
Then there exists a linear isomorphism S : V → W such that X = ST S −1 .
Sketch of proof: Denote for simplicity the numbers dim[Ker(T k )] = dim[Ker(X k )] by dk . We
know that we have 0 < d1 < · · · < dm = dm+1 = . . . , where m is the degree of the minimal
polynomial. In particular, we get:
• the minimal polynomials of T and X coincide: MT (t) = MX (t) = tm ;
• dim V = dim W = dm .
As a consequence of the hypothesis, we also get the equality of the cell numbers:
ck (T, 0) = ck (X, 0), for all k ∈ {1, . . . , m}.
Denote, for simplicity these cell numbers by ck . Define the set
J = j ∈ {1, . . . , m} : cj = 0 .
Let (Z1 , . . . , Zm ) be a cell skeleton for T , and let (G1 , . . . , Gm ) be a cell skeleton for X. For each
j ∈ J, define the subspaces
(23) Vj = Zj + T (Zj ) + · · · + T j−1 (Zj );
(24) Wj = Gj + X(Gj ) + · · · + X j−1 (Gj ).
Fix for the moment j ∈ J. We know that
(i) dim Zj = dim Gj = cj (> 0);
j−1 j−1
(ii) the families T k (Zj ) k=0 and X k (Gj ) k=0 are linearly independent;
(iii) the linear maps T k : Zj → T k (Zj ) and X k : Gj → X k (Gj ), k = 1, . . . , j − 1 are linear
isomorphisms.
5. THE SIMILARITY PROBLEM AND THE SPECTRAL INVARIANT 37
Using (i), there exists a linear isomorphism Sj0 : Zj → Gj . Using (iii), for each k = 1, . . . , j − 1
there exist linear isomorphisms Sjk : T k (Zj ) → X k (Gj ) such that the diagram
Tk
Zj − − − T k (Zj )
− −→
Sj0
S
jk
Xk
Gj − − − X k (Zj )
− −→
is commutative. (Simply define Sjk = X k ◦ Sj0 ◦ (T k Zj
)−1 .) By (ii), we know that (23) and (24)
are in fact direct sum decompositions. In particular, there exists a (unique) linear isomorphism
Sj : Vj → Wj such that
Sj T k (Zj )
= Sjk , for all k = 0, . . . , j − 1.
It is not hard to prove that Vj is invariant for T , and Wj is invariant for X. Moreover, by
construction we will now have a commutative diagram:
T
− −→
Vj − − − Vj )
(25) Sj
S
j
X
− −→
Wj − − − W j
Let now j vary in J. Using exercise ?? we have direct sum decompositions
V = Vj and W = Wj ,
j∈J j∈J
so there exists a unique linear isomorphism S : V → W such that
S Vj
= Sj , for all j ∈ J.
Using (25) we then get ST = XS, and we are done.
127* Prove the Spectral Invariant Theorem.
Hints: Assume X, Y : V → V have the same spectral invariants. Denote the set Spec(X) =
Spec(Y ) simply by Λ. We know that the minimal polynomials of X and Y coincide
MX (t) = MY (t) = (t − λ)m(λ) .
λ∈Λ
Put Vλ = N (X, λ) and Wλ = N (Y, λ). Use the spectral decomposition, to write direct sum
decompositions
(26) V = Vλ = Wλ .
λ∈Λ λ∈Λ
Fix for the moment λ ∈ Λ. We know that Vλ is invariant for X, and Wλ is invariant for Y .
Moreover, if we define Xλ = X V ∈ Lin(Vλ ) and Yλ = Y W ∈ Lin(Wλ ), then these linear maps
λ λ
will have the same spectral invariants. We have Spec(Xλ ) = Spec(Yλ ) = {λ}, and
dk (Xλ , λ) = dk (X, λ) = dk (Y, λ) = dk (Yλ , λ).
It follows that the nilpotents Xλ − λI and Yλ − λI have the same spectral invariants. By the
Spectral Theorem for Nilpotents, there exists a linear isomorphism Sλ : Vλ → Wλ such that
Sλ (Xλ − λI) = (Yλ − λI)Sλ . This clearly gives
(27) Sλ Xλ = Yλ Sλ .
Let now λ vary. Using the direct sum decompositions (26), there exists a (unique) linear isomor-
phism S : V → V , such that
S V = Sλ , for all λ ∈ Λ.
λ
Using (27), we immediatel get SX = Y S, and we are done.
CHAPTER 2
Matrices
1. Arithmetic of matrices
Definitions. Suppose k is a field, and m, n ≥ 1 are integers. A rectangular
table:
a11 a12 . . . a1n
a21 a22 . . . a2n
(28) . ,
. .
.. .
. ... .
.
am1 am2 . . . amn
filled with elements of k, is called an m × n matrix with coefficients in k. The set
of all such matrices will be denoted by M atm×n (k).
A square matrix is one with m = n. For the set of all square n × n matrices,
we will use the simplified notation M atn (k).
The indexing convention for writing the coefficients will be that the first index
represents the row number, and the second index represents the column number.
So the matrix (28) can also be written as
aij 1≤i≤m .
1≤j≤n
128 Prove that M atm×n (k) is a k-vector space, when equipped with the following
operations:
a11 a12 . . . a1n b11 b12 . . . b1n
a21 a22 . . . a2n b21 b22 . . . b2n
. + . . =
. . .
.. .
. ... . .
. . .
. ... .
.
am1 am2 . . . amn bm1 bm2 . . . bmn
a11 + b11 a12 + b12 . . . a1n + b1n
a21 + b21 a22 + b22 . . . a2n + b2n
= ,
.
. .
. .
.
. . ... .
am1 + bm1 am2 + bm2 . . . amn + bmn
a11 a12 . . . a1n λa11 λa12 . . . λa1n
a21 a22 . . . a2n λa21 λa22 . . . λa2n
λ· . . = .
. . .
.. .
. ... . .
. . .
. ... .
.
am1 am2 . . . amn λam1 λam2 . . . λamn
The zero matrix in 0m×n ∈ M atm×n (k) is the matrix filled with zeros. Prove that
dim M atm×n (k) = mn.
39
40 2. MATRICES
Matrices arise naturally in the study of linear maps between finite dimensional
vector spaces.
Definition. Suppose m, n ≥ 1 are integers. Let V be a vector space of di-
mension n, and let W be a vector space of dimension m. Suppose we have two
ordered 1 linear bases A = (v1 , . . . , vn ) for V , and B = (w1 , . . . , wm ) for W . Sup-
pose a linear map T : V → W is given. For each j ∈ {1, . . . , n}, there are unique
scalars α1j , α2j , . . . , αmj such that
T (vj ) = α1j w1 + α2j w2 + · · · + αmj wm .
The m × n matrix
α11 α12 ... α1n
α21 α22 ... α2n
A,B
MT = .
. .
.. .
. ... .
.
αm1 αm2 ... αmn
is called the matrix of T , with respect to the pair (A, B). The map
ΩAB : Lin(V, W ) T −→ M atm×n (k)
is called the matrix representation associated with the pair (A, B).
129 With the above notations, prove that the matrix representation
ΩAB : Lin(V, W ) → M atm×n (k)
is a linear isomorphism.
Example. The simplest example of the above construction is the following.
Assume V = k, with ordered basis A = (1), and W = k, with the standard ordered
linear basis B = (e1 , . . . , em ). We have the following bijective correspondences:
(29) Lin(k, km ) T −→ AB
MT ∈ M atm×1 (k)
(30) Lin(k, km ) T −→ T (1) ∈ km .
Composing the correspondence (29) with the inverse of the correspondence (30),
we get a bijective correspondence km → M atm×1 (k).
130 Prove that the bijective correspondence described above is given as
α1
α2
km (α1 , . . . , αm ) −→ . .
.
.
αm
Convention. From now on we will think km as identified with the collection
of m × 1 matrices. Such matrices are called column matrices.
1 An ordered linear basis is a prefered listing of a linear basis (by means of specifying which
is the first vector, the second vector, etc.)
1. ARITHMETIC OF MATRICES 41
Comment. With the above identification, the standard ordered linear basis
(e1 , . . . , em ) of km is described by
0
.
.
.
ei = 1 ← i , i = 1, . . . , m.
.
.
.
0
131 Let m, n ≥ 1 be integers, and let T : kn → km be a linear map. Let A =
(n) (n) (m) (m)
(e1 , . . . , en ) be the standard orderd basis for kn , and let B = (e1 , . . . , en )
be the standard orderd basis for km . Prove that the matrix of T with respect to
(A, B) is given by
(n) (n)
AB
MT = T (e1 ) T (e2 ) . . . T (e(n) ) .
n
AB (n) (n)
That is, the columns of MT are precisely the column matrices T (e1 ), . . . T (en ).
Definition. Suppose m, n, p ≥ 1 are integers, and we have matrices X ∈
M atm×n (k), and Y ∈ M atn×p (k), say
X = xij 1≤i≤m and Y = yk 1≤k≤n .
1≤j≤n 1≤ ≤p
Construct the matrix
Z = zij 1≤i≤m ∈ M atm×p (k)
1≤j≤p
by defining
zij = xi1 y1j + xi2 y2j + · · · + xin ynj , for all i ∈ {1, . . . , m}, j ∈ {1, . . . , p}.
The matrix Z is called the product of X and Y , and is simply denoted by XY .
The following exercise explains why the product is defined in this way.
132 Let m, n, p ≥ 1 be integers, and let U, V, W be vector spaces with dim U = p,
dim V = n and dim W = m. Fix an ordered linear bases A for U , B for V , and
C for W . Suppose we are given linear maps S : U → V and T : V → W . Let
X ∈ M atm×n (k) be the matrix of T with respect to (B, C), and let Y ∈ M atn×p (k)
be the matrix of S with respect to (A, B). Prove that the matrix of T ◦ S : U → W ,
with respect to (A, C) is equal to the product XY .
Comments. The above result says that we have a commutative diagram
composition
Lin(V, W ) × Lin(U, V ) − − − → Lin(U, W )
−−−−
(31) Ω ×Ω
Ω
BC AB AC
M atm×n (k) × M atn×p (k) − −→
−− − M atm×p (k)
product
133 Use (31) to derive the following properties of the product:
42 2. MATRICES
A. The product is associative. That is, given matrices X ∈ M atm×n (k),
Y ∈ M atn×p (k), and Z ∈ M atp×r (k), one has the equality
(XY )Z = X(Y Z).
B. Given matrices M ∈ M atm×n (k) and P ∈ M atp×r (k), the maps
M atn×p (k) X −→ M X ∈ M atm×p (k)
M atn×p (k) X −→ XP ∈ M atn×r (k)
are linear
Definition. Given an integer n ≥ 1, we define the identity matrix
1 0 ... 0
0 1 . . . 0
In = . . .
. .
. . .. .
.
.
0 0 ... 1
134 Let V be a vector space of dimension n, and let B an ordered linear basis
for V . Prove that ΩBB (IdV ) = In .
In particular, using (31), obtain the identities
Im X = XIn = X, for all X ∈ M atm×n (k).
Notation. Let n ≥ 1 be an integer, let V be a vector space of dimension n, and
let B be an ordered linear basis for V . The linear map ΩBB : Lin(V ) → M atn (k)
will be denoted by ΩB .
Comments. The above results, when applied to square matrices, give the fact
that M atn (k) is a unital k-algebra. In particular (see Section 3 in Chapter 1), for
any matrix M ∈ M atn (k) one has a polynomial functional calculus
k[t] P −→ P (M ) ∈ M atn (k).
If V is a vector space of dimension n, and if B is an ordered linear basis for V ,
then
ΩB : Lin(V ) → M atn (k)
is an isomorphism of k-algebra. It is clear that, for a linear map T : V → V , we
have
ΩB P (T ) = P ΩB (T ) , for all P ∈ k[t].
135 Let n ≥ 1 be an integer, and let T : kn → kn be a linear map. Let A
136 Let m, n ≥ 1 be integers, and let T : kn → km be a linear map. Let
M ∈ M atm×n (k) be the matrix of T with respect to the standard ordered linear
bases. Prove that
(32) T (X) = M X, for all X ∈ kn .
(Recall that we identify kn with M atn×1 (k).)
2. GAUSS-JORDAN ELIMINATION 43
Definition. Let m, n ≥ 1 be integers, and let T : kn → km be a linear map.
An m × n matrix M , satisfying (32) is said to represent T .
137 Let m, n ≥ 1 be integers, and let T : kn → km be a linear map. Prove that
the matrix M connstructed in exercise ?? is the unique one that represents T .
138 Let n ≥ 1 be an integer, and let X be an n × n matrix. Prove that there
exists a unique non-zero polynomial M ∈ k[t] with the following with the following
properties:
(i) M (X) = 0.
(ii) M is monic, in the sense that the leading coefficient is 1.
(iii) If P ∈ k[t] is any polynomial with P (X) = 0, then P is divisible by M
Hint: Use the same arguments as in exercise ??.
Definition. The polynomial described above is called the minimal polynomial
of the matrix X, and is denoted by MX .
139 Let n ≥ 1 be an integer, let V be a vector space of dimension n, and let B
be an ordered linear basis for V . Suppose a linear map T : V → V is given. Define
the n × n matrix X = ΩB (T ). Prove that the minimal polynomials of T and X
coincide:
MX (t) = MT (t).
Hint: Use the fact that ΩB is an isomorphism of k-algebras.
140 Let n ≥ 1 be an integer, and let A ∈ M atn (k). Define the linear map
TA : k n X −→ AX ∈ kn ,
represented by A. Prove that the minimal polynomials of T and A coincide:
MT (t) = MA (t).
Hint: Show that the correspondence
M atn (k) A −→ TA ∈ Lin(kn )
is an isomorphism of k-algebras.
Using this fact, one can derive the The matrix representations of linear maps
The matrix operations
Rank
Systems of linear equations
2. Gauss-Jordan elimination
Row transformations
Gauss-Jordan reduction
invariance of rank; rank test for consistency
Applications to finding the kernel
Aplications to completing basis
44 2. MATRICES
3. Similarity
The similarity problem
Upper triangular forms
4. Determinants
Def. of determinant
det(AB)=det(A)det(B)
row/column expansion
linear independence via determinants
inverse of a matrix via determinants
5. A conceptual approach to determinants*
Tensor product
Tensor algebra
Exterior algebra
Meaning of determinant and its basic properties
CHAPTER 3
The Jordan canonical form
1. Characteristic polynomial
Def. of char polyn
Roots of char polyn = eigenvalues
Cayley-Hamilton thm
Multiplicity in char polyn = dim of spectral subspaces
2. Jordan basis
The algorithm
The Jordan canonical form
Spectral multiplicity diagram vs. Jordan canonical form
Conclusion: spectral multiplicity diagram is a complete invariant for the simi-
larity problem
Jordan dec: SS+NILP
3. Applications
Entire functions
Exponentials and trigonometric functions
Holomorphic functional calculus*
Spectral mapping thm
4. Real Jordan form*
Complexification
Minimal/char real polynomial, vs. minimal/char complex polynomial
Real Jordan form
Applications
45
CHAPTER 4
Linear algebra on finite dimensional Hilbert spaces
1. Hilbert spaces
Inner products and norms
CBS ineq.
Orthon. basis
2. Linear maps on Hilbert spaces
Duality ker X*, Ran X*, etc.
The norm of a linear map
Spectral properties of the adjoint
3. Normal operators
Normal and self-adjoint maps
Positivity
The unitary group
Polar decomposition*
Generalized eig spaces for normal operators = eigensppace
The spectral thm
4. Applications
Spectral thm for self-adjoint
Functional calculus depends only on the values on the spectrum
Square roots
Positivity via diagonal minors
47
Appendices
A. Results from group theory
B. The symmetric group
49
```
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Math
posted by .
How do I solve:
-9x + 2x to the third power +8x -12x to the third power
• Math -
Isn't there suppsed to be an "=" sign in there somewhere? If so, where?
You can't "solve" it if it isn't an equation.
• Math -
-9x+2x=-7x
(-7x)^3=(-7)^3 *x^3=-343x^3
8x-12x=-4x
(-4x)^3=(-4)^3 *x^3=-64x^3
-343x^3 +(-64x^3)=
-343x^3-64x^3= -407x^3
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# How do you calculate expected value?
Nov 15, 2015
Same thing as the mean
#### Explanation:
Expected value is another way of stating the mean. It can be calculated one of two ways:
For a set of numbers, take the sum divided by the number of data points. So 4,6,11: (4+6+11)/3 = 21/3 = 7
If it's a probability distribution then it's the sum of probabilities times each possible outcome
X=1 with probability 0.5
X=2 with probability 0.2
X=3 with probability 0.3
Mean = 1(.5) + 2(.2) + 3(.3) = .5 + .4 + .9 = 1.8
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# Calculate Probabilities Using Addition and Multiplication Rules
The Addition Rule of Probability is a rule for determining is used to find the probability that event A or event B happens. It’s associated with the use of the conjunction “or.” For example,
• We may want to find the probability of rolling a 3 or a 6 on a fair die
• An insurance company might want to determine the probability that a claim comes from a make policyholder or the claim is greater than \$5,000.
There are two versions of the Addition Rule depending on whether the events involved are mutually exclusive (do not share elements) or non-mutually exclusive (share elements). Let’s see how each version looks like.
1.1 The Addition Rule for Non-Mutually Exclusive Events
Non-mutually exclusive events are those that can happen at the same time. In the language of probabilities, such events are said to have some elements in common. Examples include losing and scoring in a game of soccer, driving and listening to music, or going into labor during the day and giving birth to a girl. More practical examples within the field of finance and insurance include owning a home and having an active insurance policy, having a credit card while also servicing a car loan, or defaulting on premium payment and getting into an accident.
When two events, A and B, are non-mutually exclusive, the Addition Rule tells us that the probability that A or B will occur is the sum of the probability that either event will happen minus the probability that both will happen:
$$P(A \cup B) = P(A) + P(B) − P(A \cap B)$$
The last term, $$P(A \cap B)$$, enters the equation twice, i.e., once in P(A) and once in P(B). As such, we must subtract it once so that it is not double-counted.
A Venn diagram perfectly illustrates the idea of double counting. The intersection between A and B has to be subtracted when considering the probability of A or B.
\begin{align*} \textbf{Figure 1.1.1 – }& \textbf{The Addition rule for Non-Mutually Exclusive} \\ & \textbf{Events} \end{align*}
Example 1.1.1
Let’s say you’ve taken out a single card from a regular pack of cards. What would be the probability that the card is either an ace or heart?
Let A be the event of picking an ace and B be the event of picking a heart.
$$P(A) = \frac {4}{52}$$ and $$P(B) = \frac {13}{52}$$
$$P(A \text{ and } B) = \frac {1}{52}$$
The two events are non-mutually exclusive because there’s always a chance that the card will be both an ace and heart.
Thus,
\begin{align*} P(A \cup B) & = P(A) + P(B) − P(A \cap B) \\ & = \frac {4}{52} + \frac {13}{52} – \frac {1}{52} \\ & = \frac {16}{52} = \frac {4}{13} \end{align*}
Note: In a standard deck of cards, there are four suites: hearts, clubs, spades, and diamonds. There are thirteen cards in each suite: an ace, a 2, a 3, a 4, a 5, a 6, an 8, a 9, a 10, a jack, a queen, and a king. Thus, the entire deck has 52 cards in total.
Example 1.1.2
The gender breakdown in a math class of 40 students is 25 boys to 15 girls. During the end of year exam, 12 boys and 5 girls made an A grade. If we were to pick a student at random from the class, what would be the probability of choosing a boy or an A student?
Let G be the event of choosing a girl and A be the event of choosing an A student.
$$P(G) =\frac {15}{40}$$ and $$P(A) = \frac {17}{40}$$
The probability of choosing a girl who has made grade A is $$\frac {5}{40}$$, i.e.,
$$P(G \text{ and } A) = \frac {5}{40}$$
The two events are non-mutually exclusive because there’s always a chance that the student chosen will be both be a girl and an A student.
Thus,
\begin{align*} P(G \cup A) & = P(G) + P(A) − P(G \cap A) \\ & = \frac {15}{40} + \frac {17}{40} – \frac {5}{40} \\ & = \frac {27}{40} \end{align*}
Example 1.1.3
A consulting firm employs 100 people. 51 of these have degrees in finance, 63 have degrees in economics, and 28 have degrees in both. What is the probability that an employee chosen at random has a degree in either finance or economics?
Let F and E be the events that an employee has a degree in finance and economics, respectively.
Our interest is $$P(F \cup E)$$
\begin{align*} P(F \cup E) & = P(F) + P(E) − P(F \cap E) \\ & = 0.51 + 0.63 – 0.28 \\ & = 0.86 = 86\% \end{align*}
Note: Given three non-mutually exclusive events A, B, and C, the addition rule appears as follows:
\begin{align*} P(A \cup B \cup C) & = {P(A) + P(B) + P(C) – P(A \cap B) – P(B \cap C)} \\ & {- P(A \cap C) + P(A \cap B \cap C)} \end{align*}
Question 1.1.1
A life insurance company subjects potential policyholders to varying levels of medical evaluation depending on certain factors such as age and smoking status of an individual. The probability that a visit to the Life Office results in neither lab work nor referral to a specialist is 50%. Of those coming to the Life Office, 25% are referred to specialists and 35% require lab work.
Calculate the probability that a visit to the Life Office results in both lab work and referral to a specialist.
Solution
Let:
L = event that a visit results in lab work
S = event a visit results in referral to a specialist
Thus, we’re interested in $$P(L \cap S)$$
But what do we have?
From the question, $$P(L) = 0.35$$, $$P(S) = 0.25$$, and $$P(L^C \cap S^C) = 0.50$$
From the addition rule, we know that $$P(L \cup S) = P(L) + P(S) − P(L \cap S)$$
Thus, $$P(L \cap S) = P(L) + P(S) – P(L \cup S)$$
But we also know that $$P(L \cup S) = 1 – P(L \cup S)^C$$
Thus, $$P(L \cap S) = P(L) + P(S) – (1 – P(L \cup S)^C) = P(L) + P(S) – 1 + P(L \cup S)^C$$
Also, recall that $$P(L \cup S)^C = P(L^C \cap S^C)$$
Thus, $$P(L) + P(S) – 1 + P(L \cup S)^C = P(L) + P(S) – 1 + P(L^C \cap S^C)$$
$$P(L \cap S) = 0.35 + 0.25 – 1 + 0.50 = 0.10$$
1.2 The Addition Rule for Mutually Exclusive Events
Mutually exclusive is a term that describes two or more events that cannot occur at the same time. In most cases, it signifies a situation where one outcome supersedes another. The rolling of dice is a basic example. It is not possible to roll both a four and a six at the same time. Likewise, getting a four on the initial roll has no influence on whether or not a subsequent roll will produce a six. All throws of a die are independent.
When two events, A and B, are mutually exclusive, the Addition Rule tells us that the probability that A or B will occur is the sum of the probability that either event will happen.
$$P(A \cup B) = P(A) + P(B)$$
Example 1.2.1
There are four equal sections in a spinner, each colored yellow, blue, orange, and black. How likely is it that the spinner will land on yellow or blue after a single spin?
Let:
Y = event that the spinner lands on yellow
B = event that the spinner lands on red
$$P(Y) =\frac {1}{4}$$
$$P(B) = \frac {1}{4}$$
All four sections cannot occur simultaneously, and therefore,
$$P(A \cup B) = P(A) + P(B) = \frac {1}{4} + \frac {1}{4} = \frac {1}{2}$$
Example 1.2.2
From a deck of 52 playing cards, what is the probability of drawing either a jack or a queen?
Let J represent the event that a jack is drawn and B represent the event that a queen is drawn.
$$P(J \cup B) = P(J) + P(B) = \frac {4}{52} + \frac {4}{52} = \frac {8}{52}$$
2. Multiplication Rule
The multiplication rule in probability allows us to calculate the probability of multiple events occurring together as long as we know the probabilities of those events individually. There are two versions for the multiplication rule: The general multiplication rule and the specific multiplication rule.
2.1 The General Multiplication rule
The general multiplication rule applies for events that are dependent (not independent). According to the rule, the probability that both events A and B will occur simultaneously is equal to the product of the probability of B occurring and the conditional probability that event A will occur given that B occurs.
$$P(A \cap B) = P(B) . P(A|B)$$
Example 2.1.1
Imagine you are interested in the probability of drawing queens in two consecutive draws, without replacement. Initially, the deck contains 4 queens out of 52 cards, thus the probability of a queen on the first draw is 4/52. Once you draw a queen (event Q1), the probability of drawing yet another queen changes. The new “dependent” probability of drawing that second queen (event Q2) is now 3/51.
In notation form:
\begin{align*} P(Q1 \cap Q2) & = P(Q1) . P(Q2|Q1) \\ & = \frac {4}{52} . \frac {3}{51} = 0.4525\% \end{align*}
2.2 The Specific Multiplication rule
The specific multiplication rule of probability applies for events that are independent. According to the rule, the probability that both events A and B will occur simultaneously is equal to the product of their individual probabilities.
$$P(A \cap B) = P(A) . P(B)$$
Example 2.2.1
How likely is it that you will get “tails” twice in a row when you flip a fair coin? To put it differently, what is the probability of getting tails on the first flip and tails on the second flip?
The probability of getting two “tails” in a row would be:
$$P(T \cap T) = \frac {1}{2} \times \frac {1}{2} = \frac {1}{4}$$
Learning Outcome
1.b Topic: General Probability – Calculate probabilities using the addition and multiplication rules.
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# Thread: Prove continuity at a point using epsilon/delta
1. ## Prove continuity at a point using epsilon/delta
Hi,
I need to prove that f(x) = cuberoot[x+3] is continuous at x = -2 (using: u^3 - v^3 = (u-v)(u^2 + uv + v^2). I know that I need to manipulate mod(f(x) - f(-2)) to be less than epsilon for some delta but I'm struggling with the algebra.
Any help would be greatly appreciated!
2. Originally Posted by s0791264
Hi,
I need to prove that f(x) = cuberoot[x+3] is continuous at x = -2 (using: u^3 - v^3 = (u-v)(u^2 + uv + v^2). I know that I need to manipulate mod(f(x) - f(-2)) to be less than epsilon for some delta but I'm struggling with the algebra.
Any help would be greatly appreciated!
$f(x) = \sqrt[3](x+3)$ at x = -2
what you want to show is the following:
for $\epsilon > 0$, there exists $\delta > 0$ such that:
$|f(x) - f(c)| < \epsilon$ for all x that belongs to dom f\{c} s.t. $|x-c| < \delta$
i remember working backward in problem like these. Start out with something like this...
let $\epsilon > 0$,
for $|x+2| < \delta$
$|\sqrt[3](x+3) - \sqrt[3](-2+3)| = |\sqrt[3](x+3) - 1|$
... (you'll prolly need to use some algebraic formula for cube/cluberoots and perhaps triangle inequality as well
...
...
$< \epsilon$
P.S. I'll try this later on.
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# Critical point
by cateater2000
Tags: critical, point
P: 35 let f(x)=sin(1/x)*x^2 for x not 0, and f(0)=0. show that x=0 is a critical point for f which is neither a local minimum, a local maximum, nor an inflection point. well I tried differentiating this, and got f'=-cos(1/x) +2xsin(1/x). to find a critical point i make f'=0. Not sure how to do this. Do I take the limx->0? Any hints or tips would be great
Sci Advisor HW Helper PF Gold P: 12,016 What you've done here, is to find the derivative of f at all points EXCEPT at x=0! But you are to find f'(0)... Use the definition of the derivative.
P: 35 k thanks i'll try that
Sci Advisor HW Helper PF Gold P: 12,016 Critical point If you're a bit unsure what I mean, the definition of f'(0) is: $$f'(0)=\lim_{h\to{0}}\frac{f(0+h)-f(0)}{h}$$
P: 35 yeh i got that to work, now how do I show that it's not a local min,max or inflection. Would I look at the second derivative? If that's not defined it's not anything?
HW Helper
PF Gold
P: 12,016
Quote by cateater2000 yeh i got that to work, now how do I show that it's not a local min,max or inflection. Would I look at the second derivative? If that's not defined it's not anything?
The standard second-derivative fails, since the first derivative is discontinuous at x=0 (the 2.derivative is not defined).
It remains to be shown that f(0) is not a local maximum/minimum.
This should be fairly easy to show..
Use, for example, the following definition of local maximum:
We say that a function f has a local maximum at $$x_{0}$$, iff there exists a $$\delta>0$$ so that for all $$x\in{D}(x_{0},\delta),f(x)\leq{f}(x_{0})$$
I've assumed that the x's in the open $$\delta$$-disk are in the domain of f, as is the case in your problem.
Note that this definition makes no assumption of differentiability or continuity of f.
P: 35 ok i finished that part of the question.( this is a 4 part question) I can't figure out these 2 parts. Any tips would be fantastic f(x)=x^2*sin(1/x) 1.let g(x)=2x^2 +f(x) (f from the first question i asked) Show g has a global minimum at x=0 but g'(x) changes sign infinitely often on both (0,e) and (-e,0) for any e>0. For this question I can easily show 0 is a critical point. But when I show it's a minimum is what's difficult, when I differentiate twice I cannot see that f''(0)>0 2. Let h(x)=x+2f(x). Show h'(0)>0, but h is not monotone increasing on any interval that includes 0. I know how to show h'(0)>0 but have no idea how to show it's monotone increasing. Again any help would be fantastic thanks in advance
Sci Advisor HW Helper PF Gold P: 12,016 Show that for 1., g(x)>=x^2 for ALL x. How can that help you in showing that x=0 must be a global minimum?
Related Discussions Atomic, Solid State, Comp. Physics 1 Calculus & Beyond Homework 3 Calculus 5 Calculus 2 Calculus 1
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# Eureka Math Geometry Module 5 Lesson 1 Answer Key
## Engage NY Eureka Math Geometry Module 5 Lesson 1 Answer Key
### Eureka Math Geometry Module 5 Lesson 1 Exploratory Challenge Answer Key
Exploratory Challenge
Choose one of the colored points (C,D, …) that you marked. Draw the right triangle formed by the line segment connecting the original two points A and B and that colored point. Take a copy of the triangle, and rotate it 180˚ about the midpoint of $$\overline{A B}$$.
Label the acute angles in the original triangle as x and y, and label the corresponding angles in the rotated triangle the same.
Todd says ACBC’ is a rectangle. Maryam says ACBC’ is a quadrilateral, but she is not sure it is a rectangle. Todd is right but does not know how to explain himself to Maryam. Can you help him out?
a. What composite figure is formed by the two triangles? How would you prove it?
A rectangle is formed. We need to show that all four angles measure 90°.
i. What is the sum of the measures of x and y? Why?
90°; the sum of the measures of the acute angles in any right triangle is 90°.
ii. How do we know that the figure whose vertices are the colored points (C,D, …) and points A and B is a rectangle?
All four angles measure 90°. The colored points (C,D, …) are constructed as right angles, and the angles at points A and B measure x + y, which is 90°.
b. Draw the two diagonals of the rectangle. Where is the midpoint of the segment connecting the two original points A and B? Why?
The midpoint of the segment connecting points A and B is the intersection of the diagonals of the rectangle because the diagonals of a rectangle are congruent and bisect each other.
c. Label the intersection of the diagonals as point P. How does the distance from point P to a colored point (C,D, …) compare to the distance from P to points A and B?
The distances from P to each of the points are equal.
d. Choose another colored point, and construct a rectangle using the same process you followed before. Draw the two diagonals of the new rectangle. How do the diagonals of the new and old rectangle compare? How do you know?
One diagonal is the same (the one between points A and B), but the other is different since it is between the new colored point and its image under a rotation. The new diagonals intersect at the same point P because diagonals of a rectangle intersect at their midpoints, and the midpoint of the segment connecting points A and B has not changed. The distance from P to each colored point equals the distance from P to each original point A and B. By transitivity, the distance from P to the first colored point, C, equals the distance from P to the second colored point, D.
e. How does your drawing demonstrate that all the colored points you marked do indeed lie on a circle?
For any colored point, we can construct a rectangle with that colored point and the two original points, A and B, as vertices. The diagonals of this rectangle intersect at the same point P because diagonals intersect at their midpoints, and the midpoint of the diagonal between points A and B is P. The distance from P to that colored point equals the distance from P to points A and B. By transitivity, the distance from P to the first colored point, C, equals the distance from P to any other colored point.
By definition, a circle is the set of all points in the plane that are the same distance from a given center point. Therefore, each colored point on the drawing lies on the circle with center P and a radius equal to half the length of the original line segment joining points A and B
### Eureka Math Geometry Module 5 Lesson 1 Example Answer Key
Example
In the Exploratory Challenge, you proved the converse of a famous theorem in geometry. Thales’ theorem states the following: If A, B, and C are three distinct points on a circle, and $$\overline{A B}$$ is a diameter of the circle, then ∠ACB is right.
Notice that, in the proof in the Exploratory Challenge, you started with a right angle (the corner of the colored paper) and created a circle. With Thales’ theorem, you must start with the circle and then create a right angle.
Prove Thales’ theorem.
a. Draw circle P with distinct points A, B, and C on the circle and diameter (AB) ̅. Prove that ∠ACB is a right angle.
Sample image shown to the right.
b. Draw a third radius ($$\overline{P C}$$). What types of triangles are △APC and △BPC? How do you know?
They are isosceles triangles. Both sides of each triangle are radii of circle P and are, therefore, of equal length.
c. Using the diagram that you just created, develop a strategy to prove Thales’ theorem.
Look at each of the angle measures of the triangles, and see if we can prove m∠ACB is 90°.
d. Label the base angles of △APC as b° and the base angles of △BPC as a°. Express the measure of ∠ACB in terms of a° and b°.
The measure of ∠ACB is a° + b°.
e. How can the previous conclusion be used to prove that ∠ACB is a right angle?
2a + 2b = 180 because the sum of the angle measures in a triangle is 180°. Then, a + b = 90, so ∠ACB is a right angle.
### Eureka Math Geometry Module 5 Lesson 1 Exercise Answer Key
Exercises
Exercise 1.
$$\overline{A B}$$ is a diameter of the circle shown. The radius is 12.5 cm, and AC = 7 cm.
a. Find m∠C.
90°
b. Find AB.
25 cm
c. Find BC.
24 cm
Exercise 2.
In the circle shown, $$\overline{B C}$$ is a diameter with center A.
a. Find m∠DAB.
144°
b. Find m∠BAE.
128°
c. Find m∠DAE.
88°
### Eureka Math Geometry Module 5 Lesson 1 Problem Set Answer Key
Question 1.
A, B, and C are three points on a circle, and angle ABC is a right angle. What is wrong with the picture below? Explain your reasoning.
Draw in three radii (from O to each of the three triangle vertices), and label congruent base angles of each of the three resulting isosceles triangles. See diagram to see angle measures. In the big triangle (△ABC), we get 2a + 2b + 2c = 180. Using the distributive property and division, we obtain 2(a + b + c) = 180, and a + b + c = 90. But we also have 90 = m∠B = b + c. Substitution results in a + b + c = b + c, giving a a value of 0-a contradiction.
Question 2.
Show that there is something mathematically wrong with the picture below.
Draw three radii ($$\overline{O A}$$, $$\overline{O B}$$, and $$\overline{O C}$$). Label m∠BAC as a° and m∠BCA as c°. Also label m∠OAC as x˚ and m∠OCA as x° since △AOC is isosceles (both sides are radii). If m∠ABC is a right angle (as indicated on the drawing), then a° + c° = 90°. Since △AOB is isosceles, m∠ABO = a° + x°. Similarly, m∠CBO = c° + x°. Now adding the measures of the angles of △ABC results in a° + a° + x° + c° + x° + c° = 180°. Using the distributive property and division, we obtain a° + c° + x° = 90°. Substitution takes us to a°s + c° = a° + c° + x°, which is a contradiction. Therefore, the figure above is mathematically impossible.
Question 3.
In the figure below, $$\overline{A B}$$ is the diameter of a circle of radius 17 miles. If BC = 30 miles, what is AC?
16 miles
Question 4.
In the figure below, O is the center of the circle, and $$\overline{A D}$$ is a diameter.
a. Find m∠AOB.
48°
b. If m∠AOB∶m∠COD = 3∶4, what is m∠BOC?
68°
Question 5.
$$\overline{P Q}$$ is a diameter of a circle, and M is another point on the circle. The point R lies on $$\overleftrightarrow{\boldsymbol{M Q}}$$ such that RM = MQ. Show that m∠PRM = m∠PQM. (Hint: Draw a picture to help you explain your thinking.)
Since RM = MQ (given), m∠RMP = m∠QMP (both are right angles, ∠QMP by Thales’ theorem and ∠RMP by the angle addition postulate), and MP = MP (reflexive property), then △PRM ≅ △PQM by SAS. It follows that ∠PRM ≅ ∠PQM (corresponding sides of congruent triangles) and that m∠PRM = m∠PQM (by definition of congruent angles).
Question 6.
Inscribe △ABC in a circle of diameter 1 such that $$\overline{A C}$$ is a diameter. Explain why:
a. sin(∠A) = BC.
$$\overline{A C}$$ is the hypotenuse, and AC = 1. Since sine is the ratio of the opposite side to the hypotenuse, sin(∠A) necessarily equals the length of the opposite side, that is, the length of $$\overline{B C}$$.
b. cos(∠A) = AB.
$$\overline{A C}$$ is the hypotenuse, and AC = 1. Since cosine is the ratio of the adjacent side to the hypotenuse, cos(∠A) necessarily equals the length of the adjacent side, that is, the length of $$\overline{A B}$$.
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Homework 12 Solutions
# Homework 12 Solutions - lim(kl9356 – Homework 12 –...
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Unformatted text preview: lim (kl9356) – Homework 12 – Weathers – (17104) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points m 1 1 m 2 before after v v v 2 v m 1 m 2 3 4 α β An m 2 = 1 . 4 kg can of soup is thrown upward with a velocity of v 2 = 5 . 9 m / s. It is immediately struck from the side by an m 1 = 0 . 5 kg rock traveling at v 1 = 8 . 8 m / s. The rock ricochets off at an angle of α = 52 ◦ with a velocity of v 3 = 6 . 1 m / s. What is the angle of the can’s motion after the collision? Correct answer: 66 . 7 ◦ . Explanation: Basic Concepts: Conservation of Mo- mentum p before = p after . Solution: Horizontally m 1 v 1 + m 2 (0) = m 1 v 3 cos α + m 2 v 4 cos β , so that m 2 v 4 cos β = m 1 v 1- m 1 v 3 cos α (1) = (0 . 5 kg) (8 . 8 m / s)- (0 . 5 kg) (6 . 1 m / s) cos52 ◦ = 2 . 52223 kg m / s . Vertically m 1 (0) + m 2 v 2 = m 1 v 3 sin α + m 2 v 4 sin β , so that m 2 v 4 sin β = m 2 v 2- m 1 v 3 sin α (2) = (1 . 4 kg) (5 . 9 m / s)- (0 . 5 kg) (6 . 1 m / s) sin52 ◦ = 5 . 85657 kg m / s . Thus tan β = m 2 v 4 sin β m 2 v 4 cos β = (5 . 85657 kg m / s) (2 . 52223 kg m / s) = 2 . 32198 , and β = arctan(2 . 32198) = 66 . 7 ◦ . 002 (part 2 of 2) 10.0 points With what speed does the can move immedi- ately after the collision? Correct answer: 4 . 55472 m / s. Explanation: Using equation (1) above, v 4 = m 1 v 1- m 1 v 3 cos α m 2 cos β = (0 . 5 kg) (8 . 8 m / s) (1 . 4 kg) cos66 . 7 ◦- (0 . 5 kg) (6 . 1 m / s) cos(52 ◦ ) (1 . 4 kg) cos(66 . 7 ◦ ) = 4 . 55472 m / s or using equation (2) above, v 4 = m 2 v 2- m 1 v 3 sin α m 2 sin β = (5 . 9 m / s) sin(66 . 7 ◦ )- (0 . 5 kg) (6 . 1 m / s) sin(52 ◦ ) (1 . 4 kg) sin(66 . 7 ◦ ) = 4 . 55472 m / s . 003 10.0 points Two ice skaters approach each other at right angles. Skater A has a mass of 75 kg and travels in the + x direction at 2 . 58 m / s. Skater B has a mass of 38 . 2 kg and is moving in the + y direction at 1 . 68 m / s. They collide and cling together. Find the final speed of the couple. Correct answer: 1 . 80092 m / s. lim (kl9356) – Homework 12 – Weathers – (17104) 2 Explanation: From conservation of momentum ∆ p = 0 m A v A ˆ ı + m B v B ˆ = ( m A + m B ) v f Therefore v f = radicalbig ( m A v A ) 2 + ( m B v B ) 2 m A + m B = radicalbig (193 . 5 kg m / s) 2 + (64 . 176 kg m / s) 2 75 kg + 38 . 2 kg = 1 . 80092 m / s 004 10.0 points A uniform flat plate of metal with a circular hole is situated in the reference frame shown in the figure below.- 10- 8- 6- 4- 2 0 2 4 6 8 10- 10- 8- 6- 4- 2 2 4 6 8 10 Calculate the x-coordinate of the center of mass x cm of the metal plate....
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# Expected Value (EV) in Poker: A Crash Course for Beginners
If you want to consistently win at poker, understanding expected value (EV) is mandatory. EV is one of the most important concept for making optimal decisions that maximize your win rate. In this crash course for poker beginners, we’ll cover everything you need to know about EV including what it is, how to calculate it, real hand examples, and its limitations.
## What is Expected Value in Poker?
Expected value is a mathematical concept that determines whether a particular play will generate profit or incur losses over many repetitions. EV analyzes whether the potential reward of the play sufficiently outweighs the risk taken.
Positive EV means the play is profitable and worth making. Negative EV means the play will lose money over time and should be avoided. Understanding whether your potential poker plays have positive or negative expected value is the key to making winning choices.
## Calculating Expected Value
Here is the formula for calculating EV:
EV = (Chance of Winning) x (Size of Win) – (Chance of Losing) x (Size of Loss)
Let’s use a simple example to demonstrate. Your opponent has moved all in on the river for \$200 into \$300, so you now have to call \$200 for a pot of \$500 which makes the pot \$700, factoring in your call. You estimate they are bluffing 50% of the time. Lets run the math:
1. 0.50 (percentage chance of winning as decimal) X 700 (pot size) = \$350
2. .50 (chance of losing as decimal) x 200 (amount to call/lose) = \$100
3. \$350 less \$100 = \$250
As you can see, making the call is a positive EV move as it returns \$250 in the long run. If we change the estimation of our opponents bluff to 20%, we are losing \$20 in the long run.
## Making +EV Plays
When you consistently make +EV plays over thousands of hands, your overall win rate improves substantially. Seeking +EV situations and avoiding unprofitable -EV spots must guide your poker decisions. This is how you crush the game over the long-term.
## Examples of Using Expected Value
You can leverage EV to determine:
• If calling a flop bet with a flush draw has adequate pot odds
• Whether to call down a river bet with top pair weak kicker
• If 4-bet bluffing with a middling hand has good risk/reward against a nit
• Which hands are profitable to open raise preflop from each position
EV applies to every poker scenario and decision throughout a hand.
## EV Changes Dynamically Based on Many Factors
It’s crucial to realize EV is fluid, not static. It fluctuates hand to hand based on your position, the opponent’s tendencies, pot odds, board texture, and game flow. No plays always have automatic +EV or -EV. You must estimate EV in the moment based on all factors and information available.
## EV Estimations Have Limitations
While EV is very powerful, it does have limitations. Firstly, your EV calculations are only as good as your estimates of the key variables. The percentages you use for win probability, pot size assumptions, and risk/reward ratios impact EV accuracy significantly.
Secondly, EV tells you which plays are +EV over thousands of repetitions. It cannot predict the outcome of a single occurrence. Sometimes you make the mathematically +EV play and still lose in the moment due to variance.
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# Homework Help: Find the mass of a cat as it walks across a plank
1. Jun 21, 2013
### agentlee
1. The problem statement, all variables and given/known data
A cat walks along a uniform plank that is 4.00m long and has a mass of 7.00kg. The plank is supported buy two sawhorses, one .440m from the left end of the board and the other 1.50m from its right end. When the cat reaches the right ened, the plank just begins to tip. What is the mass of the cat?
2. Relevant equations
ƩFx = 0
ƩFy = 0
Ʃτ = 0
τ=I∂=Fr
3. The attempt at a solution
This is a solved example from my book, but I don't understand how they do it.
They're saying to calculate the torque about the right sawhorse, so they do
Mg(.500m) - mg(1.50)m = 0
This is probably an obvious one but, I have no idea where the .500m came from. Can someone explain this to me?
2. Jun 21, 2013
### TSny
What point do you think they're choosing for the axis of rotation?
3. Jun 21, 2013
### agentlee
I forgot to mention they labeled F1=0, so that's the axis of rotation right?
4. Jun 21, 2013
### barryj
Where is the center of mass of the plank?
5. Jun 21, 2013
### TSny
No. The reason F1 = 0 is because the plank is about to tip and is therefore losing contact with the sawhorse on the left. But they are not choosing the sawhorse on the left as the axis of rotation.
As the plank begins to tip, about what point does it rotate?
6. Jun 21, 2013
### agentlee
Where the cat is positioned?
7. Jun 21, 2013
### TSny
No, that's not it either.
8. Jun 21, 2013
### agentlee
I see what you mean now, okay. So through a valiant use of process of elimination, the second sawhorse
9. Jun 21, 2013
### TSny
So, when you want to find the torque due to the weight of the board, at what point of the plank do you put the weight of the board?
10. Jun 21, 2013
### agentlee
Oh wow...The weight's at halfway and the plank is at 1.5m so 2-1.5...wow lol
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# What's My Function?: Question Preview (ID: 9547)
### Below is a preview of the questions contained within the game titled WHAT'S MY FUNCTION?: What's My Function Is A Mathematical Game Where You Answer Questions About Relations And Functions And If You Get The Correct Answer The Ball Falls Into The Bucket. To play games using this data set, follow the directions below. Good luck and have fun. Enjoy! [print these questions]
Which expression best describes the equation y=3x+4?
a) this equation is an indirect proportion.
b) this equation is neither an indirect nor a direct proportion.
c) this equation is a direct proportion.
d) this equation is both an indirect and direct equation.
Which expression best describes the equation y=5x-6
a) this is a direct proportion.
b) this an indirect proportion.
c) this is both an indirect and direct proportion.
d) this is both neither an indirect nor a direct proportion.
Which coordinate is true in the expression y=5x+2 when x=6?
a) (13,6)
b) (6,3)
c) (3,6)
d) (6,13)
Which coordinate is true in the expression y=4x-5 when x=5?
a) (20,5)
b) (2,5)
c) (5,20)
d) (2,20)
Which expression best describes the equation y=7x-2?
a) this is an indirect proportion.
b) this is a direct proportion.
c) this is neither a direct nor an indirect proportion.
d) this is both a direct and an indirect proportion.
Which coordinate is true in the expression y=6x+7 when x=8?
a) (56,8)
b) (5,8)
c) (8,6)
d) (8,56)
Which coordinate is true in the expression y=9x-4 when x=2?
a) (2,14)
b) (14,2)
c) (4,2)
d) (2,4)
Which best describes the equation y=1x+6?
a) this is an indirect proportion.
b) this is both an indirect and a direct proportion.
c) this a direct proportion.
d) this is neither an indirect nor a direct proportion.
Which expression best describes the equation y=3x-9?
a) this is a direct proportion.
b) this is both a direct proportion and an indirect proportion.
c) this is neither a direct proportion nor an indirect proportion.
d) this is an indirect proportion.
Which coordinate is true in the expression y=4x-5 when x=9?
a) (31,9)
b) (9,31)
c) (3,9)
d) (9,3)
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TEACHERS / EDUCATORS
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# Fractional inches 3/64, 1/8, 1/4, 3/8, 1/2, 3/16, 3/4 to centimeter (cm)
Enter a whole number of inches (in, “):
+
Select a fractional part of an inch:
Equals:
0.00in
Conversion result in cm (centimeter):
0.00cm
## How to convert from fractional inches (1/8, 1/4, 3/8, 1/2, 3/4) to cm?
You need to multiply the fraction in inches by 2.54 and then you get the result in cm.
Since 1 inch contains 2.54 cm
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×
Search anything:
Bitwise Operators + tricks
bitwise algorithm
Get this book -> Problems on Array: For Interviews and Competitive Programming
In this article at OpenGenus, we have covered the basics of Bit manipulations, conversions between number systems and clever tricks with bitwise operators.
Table of content
1. Bit Manipulation
• And
• OR
• XOR
• Complement
2. Number System
• Decimal
• Binary
• Octal
3. Conversions
• Decimal to base b
• Base to decimal
4. Left Shift operator
5. Right shift operator
6. Odd and Even
7. How to convert positive number to negative number
8. Power of 0
Bit Manipulation
Operators
1. AND β> In logical operations, the AND operator compares two conditions and returns true only if both conditions are true. It evaluates whether both operands are truthy or falsey.
a b a & b
0 0 0
0 1 0
1 0 0
1 1 1
`````` When you & with 1 the digits remain the same
``````
EX. 110010001 & 1111111111 = 110010001
1. OR β> The OR operator compares two conditions and returns true if at least one of the conditions is true. It evaluates whether either operand is truthy.
a b a or b
0 0 0
0 1 1
1 0 1
1 1 1
2. XOR The XOR operator compares two conditions and returns true if exactly one of the conditions is true, but not both. It evaluates the exclusive combination of the operands' truth values.
( If and only if ) β> Only one statement should be true.
a b a ^ b
o 0 0
o 1 1
1 0 1
1 1 0
``````Observations: a ^ 1 = {like a ^ b -> 1 ^ 1 = 0. i.e. complement of a that is opposite of 1 i.e. 0
a ^ 0 = a
a ^ a = 0
``````
3. Complement ( ~ )
The complement operator, also known as the negation operator, is used to invert the truth value of a condition. It is typically represented by the exclamation mark (!) symbol. When applied to a true condition, it evaluates to false, and vice versa. It flips the logical state of the operand, transforming true to false and false to true.
a = 1010010
a (complement) = 0101101
Number System
1. Decimal β> 0,1,2....9 Base: 10
Decimal is a number system based on the power of 10. It uses ten symbols (0-9) to represent numbers. Each digit's value depends on its position, with the rightmost digit representing ones, the next representing tens, and so on.
2. Binary β> 0 & 1 Base: 2
Binary is a number system based on the power of 2. It uses only two symbols (0 and 1) to represent numbers. Each digit's value depends on its position, with the rightmost digit representing ones, the next representing twos, and so on.
3. Octal β> 0,1,2....7 Base: 8
Octal is a number system based on the power of 8. It uses eight symbols (0-7) to represent numbers. Each digit's value depends on its position, with the rightmost digit representing ones, the next representing eights, and so on.
``````0,1,2,3,4,5,6,7,10,11,12,13,14,15,16,17,20
``````
we do not count numbers which contain 8 and 9
Hexadecimal is a number system based on the power of 16. It uses sixteen symbols (0-9 and A-F) to represent numbers. Each digit's value depends on its position, with the rightmost digit representing ones, the next representing sixteens, and so on.
``````0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F\$
``````
Conversions
1. Decimal to base b
The process of converting a decimal number to base-b involves representing the decimal number using digits from 0 to b-1. It includes dividing the decimal number by b, noting the remainders, and concatenating them in reverse order to form the base-b representation.
Keep dividing by base, take remainders and write in reverse order.
2. Base to decimal
The process of converting a decimal number to base-b involves representing the decimal number using digits from 0 to b-1. It includes dividing the decimal number by b, noting the remainders, and concatenating them in reverse order to form the base-b representation.
Continuing the operator..
5. Left Shift operator ( << )
The left shift operator (<<) is a bitwise operator that shifts the bits of a number to the left by a specified number of positions. It effectively multiplies the number by 2 raised to the power of the shift amount. The leftmost bits are discarded, and zeros are filled in from the right. This operation is equivalent to multiplying the number by 2 for each shift.
now using << 1010 = 10100 | 10100 in decimal is 20
When left shift operator is uded the nubmer is doubled
``````a <<1 = 2a
``````
``````a << b = a*2^b
``````
6. Right shift operator ( >> )
The right shift operator (>>) is a bitwise operator that shifts the bits of a number to the right by a specified number of positions. It effectively divides the number by 2 raised to the power of the shift amount. The rightmost bits are discarded, and depending on the sign of the number, either zeros or ones are filled in from the left. This operation is equivalent to dividing the number by 2 for each shift.
Q: 0011001 >> 1 => 001100 => 1100
Two left zeros are ignored in binary numbers.
Odd and Even
10011 --> Leaving this every other is a power of 2 (the last bit is know as Least Significant Bit)
rest of the left bits will always be even as all the numbers multiply with power of 2
Henc if 2 to the powert 0 plane == 1 that will ODD
ohterwise it will be EVEN
``````Mask: -> !(1<<(n-1))
``````
How to convert Positive number to Negative Number
Step 1: Take the complement of the positive number. In binary representation, this involves flipping all the bits (0s become 1s and 1s become 0s).
Step 2: Add 1 to the complement obtained in Step 1.
The result of this process is the negative representation of the original positive number. This method utilizes the two's complement representation, which is commonly used in computer systems to represent negative numbers. By applying these steps, you effectively invert the sign of the number while maintaining the integrity of the binary representation.
``````EX. (10)_10 = (00001010)_2
Step - 1; 11110101
Step - 2: 1110101 + 1 = (11110110)_2 = (-10)
``````
These steps also known as 2βs complement method.
Some Shortcut formulas.
Power of 0
IF n & (n - 1) = 0 // It is power of 2.
In power of 2 number all the numbers are 0 except the most significant bit
The expression "n & (n - 1) = 0" is used to determine if a number is a power of 2. When this expression evaluates to zero, it indicates that the number is a power of 2. In a power of 2 number, all the bits are 0 except for the most significant bit, which is set to 1. By subtracting 1 from a power of 2 number, all the bits to the right of the most significant bit become 1. Therefore, when performing a bitwise AND operation between the number and its decrement, the result will be 0 only if the number is a power of 2.
Ex. 10000000
if we & n with 1111111 ( n - 1 ) it gives 0
``````IF 0 β> a
a % 4 = 0 a
a % 4 = 1 1
a % 4 = 2 a + 1
a % 4 = 3 0
``````
Manish Kumar
I am first year student at KIIT. I am currently learning Web Development, started writing blogs as to record my learning journey, it gives me deeper understanding when I learn and write it later.
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https://math.stackexchange.com/questions/1414437/4-sin2-frac-theta2-s-n1-sin-n-theta-n-sin-n1-theta-and-4-sin2-f
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# $4\sin^2\frac{\theta}{2}.S=(n+1)\sin n\theta-n \sin (n+1)\theta$, and $4\sin^2\frac{\theta}{2}.C=-1+(n+1)\cos n\theta-n \cos (n+1)\theta$
If $S\equiv \sin\theta+2\sin2\theta+3\sin3\theta+......+n\sin n\theta$ and $C\equiv \cos\theta+2\cos2\theta+3\cos3\theta+......+n\cos n\theta$,prove that
$4\sin^2\frac{\theta}{2}.S=(n+1)\sin n\theta-n \sin (n+1)\theta$,
and $4\sin^2\frac{\theta}{2}.C=-1+(n+1)\cos n\theta-n \cos (n+1)\theta$
My try:$4\sin^2\frac{\theta}{2}.S$
$=2(1-\cos \theta)[\sin\theta+2\sin2\theta+3\sin3\theta+......+n\sin n\theta]$
$=2[\sin\theta+2\sin2\theta+3\sin3\theta+......+n\sin n\theta-\cos\theta\sin\theta-2\cos\theta\sin2\theta+.....-n\cos\theta\sin n\theta]$
but it is not getting further simplified.Please help me.
## 2 Answers
HINT:
$$2r(1-\cos y)\cdot\sin ry=2r\sin ry-r(2\sin ry\cos y)$$ $$=2r\sin ry-r\{\sin(r+1)y-\sin(r-1)y\}$$
$$=-r\{\sin(r+1)y-\sin ry\}+r\{\sin ry-\sin(r-1)y\}$$
Do you know about Telescoping Series?
Set $r=1,2,3,\cdots,n-1,n$ and add
Similarly, $$2r(1-\cos y)\cdot\cos ry=2r\cos ry-r(2\cos ry\cos y)=?$$
HINT:
Using Euler's formula, $$S_n=C+iS=\sum_{r=1}^n re^{ir\theta}$$
Now using Arithmetic-Geometric Series, $$S_n=\dfrac{1-(n+1)e^{in\theta}+ne^{i(n+1)\theta}}{(1-e^{i\theta})^2}$$
Simplify and equate the real & the imaginary parts.
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http://www.livestrong.com/article/525801-how-to-calculate-an-elevation-gain-for-a-treadmill/
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0
• You're all caught up!
# How to Calculate an Elevation Gain for a Treadmill
by
John Woloch
John Woloch writes professionally for various websites. He has published in the Dutch journal "Crux" and writes frequently on oil painting, classical languages and topics involving math and biochemistry. Woloch holds a Master of Arts in English from the University of Chicago, a Master of Arts in classics from Ohio State University and a postbaccalaureate pre-medical degree from Georgetown University.
Feet running on a treadmill. Photo Credit Ancika/iStock/Getty Images
You don't need crampons and an ice-ax to be a mountain killer. The incline settings on your treadmill let you simulate the effort required for a change in vertical height, or elevation gain. You can build your endurance and stamina by using and calculating elevation gain on your treadmill. As you prepare for an upcoming race, compare your treadmill elevation work with the race map. If the race route contains steeper elevation changes, you can adjust your training accordingly.
### Step 2
Divide the percent grade you have written by 100 using a calculator. For example, 7/100 = 0.07.
### Step 3
Multiply your answer by the number of miles you have run on your treadmill. For example, you have run three miles: 0.07 x 3 = 0.21. You have completed an elevation gain of 0.21 miles.
### Step 4
Multiply your answer by 5,280. For example, 0.21 x 5280 = 1108.8. You have completed an elevation gain of approximately 1,108 feet.
### Step 5
Divide your answer by 3.281. For example, 1108.8/3.281 = 337.945. You have completed an elevation gain of approximately 338 meters.
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http://jwilson.coe.uga.edu/EMAT6680Fa10/Plummer/Assignment6/Assignment6.html
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Assignment 6
Given three points A, B, and C, we will draw a line that intersects segment AC at point X and also intersects segment BC at point Y such that
AX = XY = YB
Let’s create an angle ABC as follows
For now we’ve left off point B and have a dashed line that intersects point C. Point B will be on that line, but we do not know exactly where yet. We will find it as our last step.
To start off, we can place an arbitrary point on segment AC and call that point X.
Next we need to take the distance from X to A and translate that distance from X to an intersection with BC. We can do that by creating a circle with its center at point X and point A on its circumference.
The distance from X to A is the radius of the circle and is the same as the distance from X to any point on that circle. To translate the length of AX to BC we look at the intersection point between the circle and segment BC and call that point Y. We now construct the segment XY. Since point Y is on the circle, and all radii of a circle are congruent, we can conclude that AX = XY.
The final step is to translate the length of XY (which is the same length of AX) along the line from point C so that we can locate our point B. We do this by creating another circle with Y as its center and point X on its circumference.
Since XY is the radius of our first circle and also a radius of our second circle, we can conclude that both circles are congruent. We can also conclude that any radius from our first circle will be congruent with any radius from our second circle. To locate our point B, we look at the intersection point between our second circle and the dashed line from point C. We can call this intersection point B.
So now we have a line that intersects segment AC at point X and also intersects segment BC at point Y such that
AX = XY = YB
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https://whatisconvert.com/22-knots-in-meters-second
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# What is 22 Knots in Meters/Second?
## Convert 22 Knots to Meters/Second
To calculate 22 Knots to the corresponding value in Meters/Second, multiply the quantity in Knots by 0.514444444444 (conversion factor). In this case we should multiply 22 Knots by 0.514444444444 to get the equivalent result in Meters/Second:
22 Knots x 0.514444444444 = 11.317777777768 Meters/Second
22 Knots is equivalent to 11.317777777768 Meters/Second.
## How to convert from Knots to Meters/Second
The conversion factor from Knots to Meters/Second is 0.514444444444. To find out how many Knots in Meters/Second, multiply by the conversion factor or use the Velocity converter above. Twenty-two Knots is equivalent to eleven point three one eight Meters/Second.
## Definition of Knot
The knot is a unit of speed equal to one nautical mile (1.852 km) per hour, approximately 1.151 mph. The ISO Standard symbol for the knot is kn. The same symbol is preferred by the IEEE; kt is also common. The knot is a non-SI unit that is "accepted for use with the SI". Worldwide, the knot is used in meteorology, and in maritime and air navigation—for example, a vessel travelling at 1 knot along a meridian travels approximately one minute of geographic latitude in one hour. Etymologically, the term derives from counting the number of knots in the line that unspooled from the reel of a chip log in a specific time.
## Definition of Meter/Second
Metre per second (American English: meter per second) is an SI derived unit of both speed (scalar) and velocity (vector quantity which specifies both magnitude and a specific direction), defined by distance in metres divided by time in seconds. The SI unit symbols are m·s−1, m s−1 or m/s sometimes (unofficially) abbreviated as "mps". Where metres per second are several orders of magnitude too slow to be convenient, such as in astronomical measurements, velocities may be given in kilometres per second, where 1 km/s is 1000 metres per second, sometimes unofficially abbreviated as "kps".
## Using the Knots to Meters/Second converter you can get answers to questions like the following:
• How many Meters/Second are in 22 Knots?
• 22 Knots is equal to how many Meters/Second?
• How to convert 22 Knots to Meters/Second?
• How many is 22 Knots in Meters/Second?
• What is 22 Knots in Meters/Second?
• How much is 22 Knots in Meters/Second?
• How many m/s are in 22 kt?
• 22 kt is equal to how many m/s?
• How to convert 22 kt to m/s?
• How many is 22 kt in m/s?
• What is 22 kt in m/s?
• How much is 22 kt in m/s?
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## Thursday, October 31, 2013
### Happy Halloween & Math activity Thursday: 3.NBT.A.3
Happy Halloween everyone!
Today for MaTh I thought I would do something with candy : )
So CCSS 3.NBT.A.3 states: Multiply one-digit whole numbers by multiples of 10 in the range 10–90 (e.g., 9 × 80, 5 × 60) using strategies based on place value and properties of operations.
So, hopefully your third graders know their times tables for this one because they will need to know the one digit 9 x 8 easily to do 9 x 80. If you kids need a little help with those you might consider purchasing my conceptual multiplication cards. They can help your visual learners
To visually show a problem, kids need to understand that multiplication is really about groups. In my card above you see that 3 x 6 is really 3 groups of 6 or you can show it as 6 groups of 3 also.
So, for my activity to do with this standard I put together a place value system with candy corn. In the ones place we have the individual candy corn, the tens place we have them in bundles of ten, then for the hundreds place we have a bowl of ten bundles. I could go on and make a thousands place which would probably be a basket to hold ten bowls.
So, I did a sample problem 4 x 20. How many ways can we approach this? Some children might know how to do it on paper with 4 x 0 is 0 and 4 x 2 in the same column is 8, so it is 80 (but do they understand it conceptually?). Do they know they can represent it as 4 groups of 20 or 20 groups of 4.
If the kids show it they will see...
(sorry, the other picture got sucked into cyber space and was accidentally deleted!)
Of those two ways, which way is easier to show? Why?
Most likely you will agree that 4 groups of 20 is easier to show. With the 20 groups of 4 it is there adds an additional setp to figure it out. Multiply the array 4 x 5 and then multiply that by the pieces of candy. (4 x 5) x 4 still is 80. They just found the factors of 20.
BTW, you can do this with other things if you don't have or want kids having candy: grapes, raisins, beans, beads, or something else. Do you ever use non-standard manipulative to do math in your class?
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# How to distribute points evenly on the surface of hyperspheres in higher dimensions?
I am interested in evenly distributing N points on the surface of spheres in dimensions 3 and higher.
To be more specific:
• Given a number of points N and number of dimensions D (where D > 1, N > 1)
• The distance of every point to the origin must be 1
• The minimum distance between any two points should be as large as possible
• The distance of each point to it's closest neighbour doesn't necessarily have to be the same for every point (indeed it's not possible for it to be the same unless the number of points forms the vertices of a platonic solid or if N <= D).
I am not interested in:
• Creating a uniform random distribution on a hypersphere, because I want the minimum distance between any two points to be as large as possible instead of being randomly distributed.
• Particle repulsion simulation type methods, because they are hard to implement and take an extremely long time to run for large N (Ideally the method should be deterministic and in O(n)).
One method that satisfies these criteria is called the fibonacci lattice, but I have only been able to find code implementations for that in 2d and 3d.
The method behind the fibonacci lattice (also known as the fibonacci spiral) is to generate a 1d line that spirals around the surface of the sphere such that the surface area covered by the line is roughly the same at every turn. You can then drop N points equally distributed on the spiral and they will roughly be evenly distributed on the surface of the sphere.
In this answer there is a python implementation for 3 dimensions that generates the following:
I wanted to know whether the fibonacci spiral could be extended to dimensions higher than 3 and posted a question on the maths stack exchange. To my surprise I received two amazing answers which as far as I can tell (because I don't fully understand the maths shown) shows it's indeed possible to extend this method to N dimensions.
Unfortunately I don't understand enough of the maths shown to be able to turn either answer into (pseudo)code. I am an experienced computer programmer, but my maths background only goes so far.
I will copy in what I believe to be the most important part of one of the answers below (unfortunately SO doesn't support mathjax so I had to copy as an image)
Difficulties presented by the above that I struggle with:
• How to resolve the inverse function used for ψn?
• The example given is for d = 3. How do I generate formulae for arbitrary d?
Would anyone here who understands the maths involved be able to make progress towards a pseudo code implementation of either answer to the linked fibonacci lattice question? I understand a full implementation may be quite difficult so I'd be happy with a part implementation that leads me far enough to be able to complete the rest myself.
To make it easier, I've already coded a function that takes spherical coordinates in N dimensions and turns them into cartesian coordinates, so the implementation can output either one as I can easily convert.
Additionally I see that one answer uses the next prime number for each additional dimension. I can easily code a function that outputs each successive prime, so you can assume that's already implemented.
Failing an implementation of the fibonacci lattice in N dimensions, I'd be happy to accept a different method that satisfies the above constraints.
• I understand that the question is essentially "Take the equations from this other answer and turn it into pseudo code". I hope that's an appropriate type of question to ask on here but let me know if it's not. Additionally, let me know if I should copy any information from that answer into this question so that it's less of a "link only" type question. Jul 20, 2019 at 8:59
• Can you edit your question and briefly define the base concepts here? For instance I might be able to implement an n-dimensional Fibonacci lattice if I knew what a Fibonacci lattice is, but not knowing it I unfortunately will skip this question, being low on spare time. Jul 22, 2019 at 10:11
• @LajosArpad I hope I have now added some more detail that will help. Jul 22, 2019 at 10:50
• Thank you for the further information, but I still do not know what a Fibonacci lattice is. You have given some attributes regarding it, but did not define the concept. I will see whether I have time to look into it, but it's inprobable, unfortunately. Jul 22, 2019 at 12:14
• Thanks for the effort. I understand it's quite a complicated concept, and unless you have prior knowledge it probably requires reading the linked question in full at math.stackexchange.com/a/3297830/688579 for a proper understanding. I know that requires quite a bit of effort which is why I've offered all of my rep as a bounty, if I could offer more, then I would. Unfortunately Stack Overflow doesn't support math jax, which limits the amount I can copy from that question into this one without it getting tedious. Jul 22, 2019 at 12:23
Very interesting question. I wanted to implement this into mine 4D rendering engine as I was curious how would it look like but I was too lazy and incompetent to handle ND transcendent problems from the math side.
Instead I come up with different solution to this problem. Its not a Fibonaci Latice !!! Instead I expand the parametrical equation of a hypersphere or n-sphere into hyperspiral and then just fit the spiral parameters so the points are more or less equidistant.
It sounds horrible I know but its not that hard and the results look correct to me (finally :) after solving some silly typos copy/paste bugs)
The main idea is to use the n-dimensional parametrical equations for hypersphere to compute its surface points from angles and radius. Here implementation:
see the [edit2]. Now the problem boils down to 2 main problems:
1. compute number of screws
so if we want that our points are equidistant so they must lye on the spiral path in equidistances (see bullet #2) but also the screws itself should have the same distance between each other. For that we can exploit geometrical properties of the hypersphere. Let start with 2D:
so simply screws = r/d. The number of points can be also inferred as points = area/d^2 = PI*r^2/d^2.
so we can simply write 2D spiral as:
t = <0.0,1.0>
a = 2.0*M_PI*screws*t;
x = r*t*cos(a);
y = r*t*sin(a);
To be more simple we can assume r=1.0 so d=d/r (and just scale the points later). Then the expansions (each dimension just adds angle parameter) look like this:
2D:
screws=1.0/d; // radius/d
points=M_PI/(d*d); // surface_area/d^2
a = 2.0*M_PI*t*screws;
x = t*cos(a);
y = t*sin(a);
3D:
screws=M_PI/d; // half_circumference/d
points=4.0*M_PI/(d*d); // surface_area/d^2
a= M_PI*t;
b=2.0*M_PI*t*screws;
x=cos(a) ;
y=sin(a)*cos(b);
z=sin(a)*sin(b);
4D:
screws = M_PI/d;
points = 3.0*M_PI*M_PI*M_PI/(4.0*d*d*d);
a= M_PI*t;
b= M_PI*t*screws;
c=2.0*M_PI*t*screws*screws;
x=cos(a) ;
y=sin(a)*cos(b) ;
z=sin(a)*sin(b)*cos(c);
w=sin(a)*sin(b)*sin(c);
Now beware points for 4D are just my assumption. I empirically found out that they relate to constant/d^3 but not exactly. The screws are different for each angle. Mine assumption is that there is no other scale than screws^i but it might need some constant tweaking (did not do analysis of the resulting point-cloud as the result look ok to me)
Now we can generate any point on spiral from single parameter t=<0.0,1.0>.
Note if you reverse the equation so d=f(points) you can have points as input value but beware its just approximate number of points not exact !!!
2. generate step on spirals so points are equidistant
This is the part I skip the algebraic mess and use fitting instead. I simply binary search delta t so the resulting point is d distant to previous point. So simply generate point t=0 and then binary search t near estimated position until is d distant to the start point. Then repeat this until t<=1.0 ...
You can use binary search or what ever. I know its not as fast as O(1) algebraic approach but no need to derive the stuff for each dimension... Looks 10 iterations are enough for fitting so its not that slow either.
Here implementation from my 4D engine C++/GL/VCL:
void ND_mesh::set_HyperSpiral(int N,double r,double d)
{
int i,j;
reset(N);
d/=r; // unit hyper-sphere
double dd=d*d; // d^2
if (n==2)
{
// r=1,d=!,screws=?
// S = PI*r^2
// screws = r/d
// points = S/d^2
int i0,i;
double a,da,t,dt,dtt;
double x,y,x0,y0;
double screws=1.0/d;
double points=M_PI/(d*d);
dbg=points;
da=2.0*M_PI*screws;
dt=0.1*(1.0/points);
for (t=0.0,i0=0,i=1;;i0=i,i++)
{
for (dtt=dt,j=0;j<10;j++,dtt*=0.5)
{
t+=dtt;
a=da*t;
x=(t*cos(a))-x0; x*=x;
y=(t*sin(a))-y0; y*=y;
if ((!j)&&(x+y<dd)){ j--; t-=dtt; dtt*=4.0; continue; }
if (x+y>dd) t-=dtt;
}
if (t>1.0) break;
a=da*t;
as2(i0,i);
}
}
if (n==3)
{
// r=1,d=!,screws=?
// S = 4*PI*r^2
// screws = 2*PI*r/(2*d)
// points = S/d^2
int i0,i;
double a,b,da,db,t,dt,dtt;
double x,y,z,x0,y0,z0;
double screws=M_PI/d;
double points=4.0*M_PI/(d*d);
dbg=points;
da= M_PI;
db=2.0*M_PI*screws;
dt=0.1*(1.0/points);
for (t=0.0,i0=0,i=1;;i0=i,i++)
{
for (dtt=dt,j=0;j<10;j++,dtt*=0.5)
{
t+=dtt;
a=da*t;
b=db*t;
x=cos(a) -x0; x*=x;
y=sin(a)*cos(b)-y0; y*=y;
z=sin(a)*sin(b)-z0; z*=z;
if ((!j)&&(x+y+z<dd)){ j--; t-=dtt; dtt*=4.0; continue; }
if (x+y+z>dd) t-=dtt;
}
if (t>1.0) break;
a=da*t;
b=db*t;
as2(i0,i);
}
}
if (n==4)
{
// r=1,d=!,screws=?
// S = 2*PI^2*r^3
// screws = 2*PI*r/(2*d)
// points = 3*PI^3/(4*d^3);
int i0,i;
double a,b,c,da,db,dc,t,dt,dtt;
double x,y,z,w,x0,y0,z0,w0;
double screws = M_PI/d;
double points=3.0*M_PI*M_PI*M_PI/(4.0*d*d*d);
dbg=points;
da= M_PI;
db= M_PI*screws;
dc=2.0*M_PI*screws*screws;
dt=0.1*(1.0/points);
for (t=0.0,i0=0,i=1;;i0=i,i++)
{
for (dtt=dt,j=0;j<10;j++,dtt*=0.5)
{
t+=dtt;
a=da*t;
b=db*t;
c=dc*t;
x=cos(a) -x0; x*=x;
y=sin(a)*cos(b) -y0; y*=y;
z=sin(a)*sin(b)*cos(c)-z0; z*=z;
w=sin(a)*sin(b)*sin(c)-w0; w*=w;
if ((!j)&&(x+y+z+w<dd)){ j--; t-=dtt; dtt*=4.0; continue; }
if (x+y+z+w>dd) t-=dtt;
} dt=dtt;
if (t>1.0) break;
a=da*t;
b=db*t;
c=dc*t;
as2(i0,i);
}
}
for (i=0;i<pnt.num;i++) pnt.dat[i]*=r;
for (i=0;i<s1.num;i++) s1.dat[i]*=n;
for (i=0;i<s2.num;i++) s2.dat[i]*=n;
for (i=0;i<s3.num;i++) s3.dat[i]*=n;
for (i=0;i<s4.num;i++) s4.dat[i]*=n;
}
Where n=N are set dimensionality, r is radius and d is desired distance between points. I am using a lot of stuff not declared here but what isimportant is just that pnt[] list the list of points of the object and as2(i0,i1) adds line from points at indexes i0,i1 to the mesh.
Here few screenshots...
3D perspective:
4D perspective:
4D cross-section with hyperplane w=0.0:
and the same with more points and bigger radius:
the shape changes with rotations in which its animated ...
[Edit1] more code/info
This is how my engine mesh class look like:
//---------------------------------------------------------------------------
//--- ND Mesh: ver 1.001 ----------------------------------------------------
//---------------------------------------------------------------------------
#ifndef _ND_mesh_h
#define _ND_mesh_h
//---------------------------------------------------------------------------
#include "list.h" // my dynamic list you can use std::vector<> instead
#include "nd_reper.h" // this is just 5x5 transform matrix
//---------------------------------------------------------------------------
enum _render_enum
{
_render_Wireframe=0,
_render_Polygon,
_render_enums
};
const AnsiString _render_txt[]=
{
"Wireframe",
"Polygon"
};
enum _view_enum
{
_view_Orthographic=0,
_view_Perspective,
_view_CrossSection,
_view_enums
};
const AnsiString _view_txt[]=
{
"Orthographic",
"Perspective",
"Cross section"
};
struct dim_reduction
{
int view; // _view_enum
double coordinate; // cross section hyperplane coordinate or camera focal point looking in W+ direction
double focal_length;
dim_reduction() { view=_view_Perspective; coordinate=-3.5; focal_length=2.0; }
dim_reduction(dim_reduction& a) { *this=a; }
~dim_reduction() {}
dim_reduction* operator = (const dim_reduction *a) { *this=*a; return this; }
//dim_reduction* operator = (const dim_reduction &a) { ...copy... return this; }
};
//---------------------------------------------------------------------------
class ND_mesh
{
public:
int n; // dimensions
List<double> pnt; // ND points (x0,x1,x2,x3,...x(n-1))
List<int> s1; // ND points (i0)
List<int> s2; // ND wireframe (i0,i1)
List<int> s3; // ND triangles (i0,i1,i2,)
List<int> s4; // ND tetrahedrons (i0,i1,i2,i3)
DWORD col; // object color 0x00BBGGRR
int dbg; // debug/test variable
ND_mesh() { reset(0); }
ND_mesh(ND_mesh& a) { *this=a; }
~ND_mesh() {}
ND_mesh* operator = (const ND_mesh *a) { *this=*a; return this; }
//ND_mesh* operator = (const ND_mesh &a) { ...copy... return this; }
void as1(int a0) { s1.add(a0); }
// init ND mesh
void reset(int N);
void set_HyperTetrahedron(int N,double a); // dimensions, side
void set_HyperCube (int N,double a); // dimensions, side
void set_HyperSphere (int N,double r,int points); // dimensions, radius, points per axis
void set_HyperSpiral (int N,double r,double d); // dimensions, radius, distance between points
// render
void glDraw(ND_reper &rep,dim_reduction *cfg,int render); // render mesh
};
//---------------------------------------------------------------------------
#define _cube(a0,a1,a2,a3,a4,a5,a6,a7) { as4(a1,a2,a4,a7); as4(a0,a1,a2,a4); as4(a2,a4,a6,a7); as4(a1,a2,a3,a7); as4(a1,a4,a5,a7); }
//---------------------------------------------------------------------------
void ND_mesh::reset(int N)
{
dbg=0;
if (N>=0) n=N;
pnt.num=0;
s1.num=0;
s2.num=0;
s3.num=0;
s4.num=0;
col=0x00AAAAAA;
}
//---------------------------------------------------------------------------
void ND_mesh::set_HyperSpiral(int N,double r,double d)
{
int i,j;
reset(N);
d/=r; // unit hyper-sphere
double dd=d*d; // d^2
if (n==2)
{
// r=1,d=!,screws=?
// S = PI*r^2
// screws = r/d
// points = S/d^2
int i0,i;
double a,da,t,dt,dtt;
double x,y,x0,y0;
double screws=1.0/d;
double points=M_PI/(d*d);
dbg=points;
da=2.0*M_PI*screws;
dt=0.1*(1.0/points);
for (t=0.0,i0=0,i=1;;i0=i,i++)
{
for (dtt=dt,j=0;j<10;j++,dtt*=0.5)
{
t+=dtt;
a=da*t;
x=(t*cos(a))-x0; x*=x;
y=(t*sin(a))-y0; y*=y;
if ((!j)&&(x+y<dd)){ j--; t-=dtt; dtt*=4.0; continue; }
if (x+y>dd) t-=dtt;
}
if (t>1.0) break;
a=da*t;
as2(i0,i);
}
}
if (n==3)
{
// r=1,d=!,screws=?
// S = 4*PI*r^2
// screws = 2*PI*r/(2*d)
// points = S/d^2
int i0,i;
double a,b,da,db,t,dt,dtt;
double x,y,z,x0,y0,z0;
double screws=M_PI/d;
double points=4.0*M_PI/(d*d);
dbg=points;
da= M_PI;
db=2.0*M_PI*screws;
dt=0.1*(1.0/points);
for (t=0.0,i0=0,i=1;;i0=i,i++)
{
for (dtt=dt,j=0;j<10;j++,dtt*=0.5)
{
t+=dtt;
a=da*t;
b=db*t;
x=cos(a) -x0; x*=x;
y=sin(a)*cos(b)-y0; y*=y;
z=sin(a)*sin(b)-z0; z*=z;
if ((!j)&&(x+y+z<dd)){ j--; t-=dtt; dtt*=4.0; continue; }
if (x+y+z>dd) t-=dtt;
}
if (t>1.0) break;
a=da*t;
b=db*t;
as2(i0,i);
}
}
if (n==4)
{
// r=1,d=!,screws=?
// S = 2*PI^2*r^3
// screws = 2*PI*r/(2*d)
// points = 3*PI^3/(4*d^3);
int i0,i;
double a,b,c,da,db,dc,t,dt,dtt;
double x,y,z,w,x0,y0,z0,w0;
double screws = M_PI/d;
double points=3.0*M_PI*M_PI*M_PI/(4.0*d*d*d);
dbg=points;
da= M_PI;
db= M_PI*screws;
dc=2.0*M_PI*screws*screws;
dt=0.1*(1.0/points);
for (t=0.0,i0=0,i=1;;i0=i,i++)
{
for (dtt=dt,j=0;j<10;j++,dtt*=0.5)
{
t+=dtt;
a=da*t;
b=db*t;
c=dc*t;
x=cos(a) -x0; x*=x;
y=sin(a)*cos(b) -y0; y*=y;
z=sin(a)*sin(b)*cos(c)-z0; z*=z;
w=sin(a)*sin(b)*sin(c)-w0; w*=w;
if ((!j)&&(x+y+z+w<dd)){ j--; t-=dtt; dtt*=4.0; continue; }
if (x+y+z+w>dd) t-=dtt;
} dt=dtt;
if (t>1.0) break;
a=da*t;
b=db*t;
c=dc*t;
as2(i0,i);
}
}
for (i=0;i<pnt.num;i++) pnt.dat[i]*=r;
for (i=0;i<s1.num;i++) s1.dat[i]*=n;
for (i=0;i<s2.num;i++) s2.dat[i]*=n;
for (i=0;i<s3.num;i++) s3.dat[i]*=n;
for (i=0;i<s4.num;i++) s4.dat[i]*=n;
}
//---------------------------------------------------------------------------
void ND_mesh::glDraw(ND_reper &rep,dim_reduction *cfg,int render)
{
int N,i,j,i0,i1,i2,i3;
const int n0=0,n1=n,n2=n+n,n3=n2+n,n4=n3+n;
double a,b,w,F,*p0,*p1,*p2,*p3,_zero=1e-6;
vector<4> v;
List<double> tmp,t0; // temp
List<double> S1,S2,S3,S4; // reduced simplexes
#define _swap(aa,bb) { double *p=aa.dat; aa.dat=bb.dat; bb.dat=p; int q=aa.siz; aa.siz=bb.siz; bb.siz=q; q=aa.num; aa.num=bb.num; bb.num=q; }
// apply transform matrix pnt -> tmp
tmp.allocate(pnt.num); tmp.num=pnt.num;
for (i=0;i<pnt.num;i+=n)
{
v.ld(0.0,0.0,0.0,0.0);
for (j=0;j<n;j++) v.a[j]=pnt.dat[i+j];
rep.l2g(v,v);
for (j=0;j<n;j++) tmp.dat[i+j]=v.a[j];
}
// copy simplexes and convert point indexes to points (only due to cross section)
S1.allocate(s1.num*n); S1.num=0; for (i=0;i<s1.num;i++) for (j=0;j<n;j++) S1.add(tmp.dat[s1.dat[i]+j]);
S2.allocate(s2.num*n); S2.num=0; for (i=0;i<s2.num;i++) for (j=0;j<n;j++) S2.add(tmp.dat[s2.dat[i]+j]);
S3.allocate(s3.num*n); S3.num=0; for (i=0;i<s3.num;i++) for (j=0;j<n;j++) S3.add(tmp.dat[s3.dat[i]+j]);
S4.allocate(s4.num*n); S4.num=0; for (i=0;i<s4.num;i++) for (j=0;j<n;j++) S4.add(tmp.dat[s4.dat[i]+j]);
// reduce dimensions
for (N=n;N>2;)
{
N--;
if (cfg[N].view==_view_Orthographic){} // no change
if (cfg[N].view==_view_Perspective)
{
w=cfg[N].coordinate;
F=cfg[N].focal_length;
for (i=0;i<S1.num;i+=n)
{
a=S1.dat[i+N]-w;
if (a>=F) a=F/a; else a=0.0;
for (j=0;j<n;j++) S1.dat[i+j]*=a;
}
for (i=0;i<S2.num;i+=n)
{
a=S2.dat[i+N]-w;
if (a>=F) a=F/a; else a=0.0;
for (j=0;j<n;j++) S2.dat[i+j]*=a;
}
for (i=0;i<S3.num;i+=n)
{
a=S3.dat[i+N]-w;
if (a>=F) a=F/a; else a=0.0;
for (j=0;j<n;j++) S3.dat[i+j]*=a;
}
for (i=0;i<S4.num;i+=n)
{
a=S4.dat[i+N]-w;
if (a>=F) a=F/a; else a=0.0;
for (j=0;j<n;j++) S4.dat[i+j]*=a;
}
}
if (cfg[N].view==_view_CrossSection)
{
w=cfg[N].coordinate;
_swap(S1,tmp); for (S1.num=0,i=0;i<tmp.num;i+=n1) // points
{
p0=tmp.dat+i+n0;
if (fabs(p0[N]-w)<=_zero)
{
}
}
_swap(S2,tmp); for (S2.num=0,i=0;i<tmp.num;i+=n2) // lines
{
p0=tmp.dat+i+n0; a=p0[N]; b=p0[N];// a=min,b=max
p1=tmp.dat+i+n1; if (a>p1[N]) a=p1[N]; if (b<p1[N]) b=p1[N];
if (fabs(a-w)+fabs(b-w)<=_zero) // fully inside
{
continue;
}
if ((a<=w)&&(b>=w)) // intersection -> points
{
a=(w-p0[N])/(p1[N]-p0[N]);
}
}
_swap(S3,tmp); for (S3.num=0,i=0;i<tmp.num;i+=n3) // triangles
{
p0=tmp.dat+i+n0; a=p0[N]; b=p0[N];// a=min,b=max
p1=tmp.dat+i+n1; if (a>p1[N]) a=p1[N]; if (b<p1[N]) b=p1[N];
p2=tmp.dat+i+n2; if (a>p2[N]) a=p2[N]; if (b<p2[N]) b=p2[N];
if (fabs(a-w)+fabs(b-w)<=_zero) // fully inside
{
continue;
}
if ((a<=w)&&(b>=w)) // cross section -> t0
{
t0.num=0;
i0=0; if (p0[N]<w-_zero) i0=1; if (p0[N]>w+_zero) i0=2;
i1=0; if (p1[N]<w-_zero) i1=1; if (p1[N]>w+_zero) i1=2;
i2=0; if (p2[N]<w-_zero) i2=1; if (p2[N]>w+_zero) i2=2;
if (i0+i1==3){ a=(w-p0[N])/(p1[N]-p0[N]); for (j=0;j<n;j++) t0.add(p0[j]+a*(p1[j]-p0[j])); }
if (i1+i2==3){ a=(w-p1[N])/(p2[N]-p1[N]); for (j=0;j<n;j++) t0.add(p1[j]+a*(p2[j]-p1[j])); }
if (i2+i0==3){ a=(w-p2[N])/(p0[N]-p2[N]); for (j=0;j<n;j++) t0.add(p2[j]+a*(p0[j]-p2[j])); }
if (t0.num==n1) for (j=0;j<t0.num;j++) S1.add(t0.dat[j]);// copy t0 to target simplex based on points count
}
}
_swap(S4,tmp); for (S4.num=0,i=0;i<tmp.num;i+=n4) // tetrahedrons
{
p0=tmp.dat+i+n0; a=p0[N]; b=p0[N];// a=min,b=max
p1=tmp.dat+i+n1; if (a>p1[N]) a=p1[N]; if (b<p1[N]) b=p1[N];
p2=tmp.dat+i+n2; if (a>p2[N]) a=p2[N]; if (b<p2[N]) b=p2[N];
p3=tmp.dat+i+n3; if (a>p3[N]) a=p3[N]; if (b<p3[N]) b=p3[N];
if (fabs(a-w)+fabs(b-w)<=_zero) // fully inside
{
continue;
}
if ((a<=w)&&(b>=w)) // cross section -> t0
{
t0.num=0;
i0=0; if (p0[N]<w-_zero) i0=1; if (p0[N]>w+_zero) i0=2;
i1=0; if (p1[N]<w-_zero) i1=1; if (p1[N]>w+_zero) i1=2;
i2=0; if (p2[N]<w-_zero) i2=1; if (p2[N]>w+_zero) i2=2;
i3=0; if (p3[N]<w-_zero) i3=1; if (p3[N]>w+_zero) i3=2;
if (i0+i1==3){ a=(w-p0[N])/(p1[N]-p0[N]); for (j=0;j<n;j++) t0.add(p0[j]+a*(p1[j]-p0[j])); }
if (i1+i2==3){ a=(w-p1[N])/(p2[N]-p1[N]); for (j=0;j<n;j++) t0.add(p1[j]+a*(p2[j]-p1[j])); }
if (i2+i0==3){ a=(w-p2[N])/(p0[N]-p2[N]); for (j=0;j<n;j++) t0.add(p2[j]+a*(p0[j]-p2[j])); }
if (i0+i3==3){ a=(w-p0[N])/(p3[N]-p0[N]); for (j=0;j<n;j++) t0.add(p0[j]+a*(p3[j]-p0[j])); }
if (i1+i3==3){ a=(w-p1[N])/(p3[N]-p1[N]); for (j=0;j<n;j++) t0.add(p1[j]+a*(p3[j]-p1[j])); }
if (i2+i3==3){ a=(w-p2[N])/(p3[N]-p2[N]); for (j=0;j<n;j++) t0.add(p2[j]+a*(p3[j]-p2[j])); }
if (t0.num==n1) for (j=0;j<t0.num;j++) S1.add(t0.dat[j]);// copy t0 to target simplex based on points count
}
}
}
}
glColor4ubv((BYTE*)(&col));
if (render==_render_Wireframe)
{
// add points from higher primitives
glPointSize(5.0);
glBegin(GL_POINTS);
glNormal3d(0.0,0.0,1.0);
if (n==2) for (i=0;i<S1.num;i+=n1) glVertex2dv(S1.dat+i);
if (n>=3) for (i=0;i<S1.num;i+=n1) glVertex3dv(S1.dat+i);
glEnd();
glPointSize(1.0);
glBegin(GL_LINES);
glNormal3d(0.0,0.0,1.0);
if (n==2)
{
for (i=0;i<S2.num;i+=n1) glVertex2dv(S2.dat+i);
for (i=0;i<S3.num;i+=n3)
{
glVertex2dv(S3.dat+i+n0); glVertex2dv(S3.dat+i+n1);
glVertex2dv(S3.dat+i+n1); glVertex2dv(S3.dat+i+n2);
glVertex2dv(S3.dat+i+n2); glVertex2dv(S3.dat+i+n0);
}
for (i=0;i<S4.num;i+=n4)
{
glVertex2dv(S4.dat+i+n0); glVertex2dv(S4.dat+i+n1);
glVertex2dv(S4.dat+i+n1); glVertex2dv(S4.dat+i+n2);
glVertex2dv(S4.dat+i+n2); glVertex2dv(S4.dat+i+n0);
glVertex2dv(S4.dat+i+n0); glVertex2dv(S4.dat+i+n3);
glVertex2dv(S4.dat+i+n1); glVertex2dv(S4.dat+i+n3);
glVertex2dv(S4.dat+i+n2); glVertex2dv(S4.dat+i+n3);
}
}
if (n>=3)
{
for (i=0;i<S2.num;i+=n1) glVertex3dv(S2.dat+i);
for (i=0;i<S3.num;i+=n3)
{
glVertex3dv(S3.dat+i+n0); glVertex3dv(S3.dat+i+n1);
glVertex3dv(S3.dat+i+n1); glVertex3dv(S3.dat+i+n2);
glVertex3dv(S3.dat+i+n2); glVertex3dv(S3.dat+i+n0);
}
for (i=0;i<S4.num;i+=n4)
{
glVertex3dv(S4.dat+i+n0); glVertex3dv(S4.dat+i+n1);
glVertex3dv(S4.dat+i+n1); glVertex3dv(S4.dat+i+n2);
glVertex3dv(S4.dat+i+n2); glVertex3dv(S4.dat+i+n0);
glVertex3dv(S4.dat+i+n0); glVertex3dv(S4.dat+i+n3);
glVertex3dv(S4.dat+i+n1); glVertex3dv(S4.dat+i+n3);
glVertex3dv(S4.dat+i+n2); glVertex3dv(S4.dat+i+n3);
}
}
glEnd();
}
if (render==_render_Polygon)
{
double nor[3],a[3],b[3],q;
#define _triangle2(ss,p0,p1,p2) \
{ \
glVertex2dv(ss.dat+i+p0); \
glVertex2dv(ss.dat+i+p1); \
glVertex2dv(ss.dat+i+p2); \
}
#define _triangle3(ss,p0,p1,p2) \
{ \
for(j=0;(j<3)&&(j<n);j++) \
{ \
a[j]=ss.dat[i+p1+j]-ss.dat[i+p0+j]; \
b[j]=ss.dat[i+p2+j]-ss.dat[i+p1+j]; \
} \
for(;j<3;j++){ a[j]=0.0; b[j]=0.0; } \
nor[0]=(a[1]*b[2])-(a[2]*b[1]); \
nor[1]=(a[2]*b[0])-(a[0]*b[2]); \
nor[2]=(a[0]*b[1])-(a[1]*b[0]); \
q=sqrt((nor[0]*nor[0])+(nor[1]*nor[1])+(nor[2]*nor[2])); \
if (q>1e-10) q=1.0/q; else q-0.0; \
for (j=0;j<3;j++) nor[j]*=q; \
glNormal3dv(nor); \
glVertex3dv(ss.dat+i+p0); \
glVertex3dv(ss.dat+i+p1); \
glVertex3dv(ss.dat+i+p2); \
}
#define _triangle3b(ss,p0,p1,p2) \
{ \
glNormal3dv(nor3.dat+(i/n)); \
glVertex3dv(ss.dat+i+p0); \
glVertex3dv(ss.dat+i+p1); \
glVertex3dv(ss.dat+i+p2); \
}
glBegin(GL_TRIANGLES);
if (n==2)
{
glNormal3d(0.0,0.0,1.0);
for (i=0;i<S3.num;i+=n3) _triangle2(S3,n0,n1,n2);
for (i=0;i<S4.num;i+=n4)
{
_triangle2(S4,n0,n1,n2);
_triangle2(S4,n3,n0,n1);
_triangle2(S4,n3,n1,n2);
_triangle2(S4,n3,n2,n0);
}
}
if (n>=3)
{
for (i=0;i<S3.num;i+=n3) _triangle3 (S3,n0,n1,n2);
for (i=0;i<S4.num;i+=n4)
{
_triangle3(S4,n0,n1,n2);
_triangle3(S4,n3,n0,n1);
_triangle3(S4,n3,n1,n2);
_triangle3(S4,n3,n2,n0);
}
glNormal3d(0.0,0.0,1.0);
}
glEnd();
#undef _triangle2
#undef _triangle3
}
#undef _swap
}
//---------------------------------------------------------------------------
#undef _cube
//---------------------------------------------------------------------------
#endif
//---------------------------------------------------------------------------
I use mine dynamic list template so:
List<double> xxx; is the same as double xxx[];
xxx.add(5); adds 5 to end of the list
xxx[7] access array element (safe)
xxx.dat[7] access array element (unsafe but fast direct access)
xxx.num is the actual used size of the array
xxx.reset() clears the array and set xxx.num=0
xxx.allocate(100) preallocate space for 100 items
so you need to port it to any list you have at disposal (like std:vector<>). I also use 5x5 transform matrix where
void ND_reper::g2l (vector<4> &l,vector<4> &g); // global xyzw -> local xyzw
void ND_reper::l2g (vector<4> &g,vector<4> &l); // global xyzw <- local xyzw
convert point either to global or local coordinates (by multiplying direct or inverse matrix by point). You can ignore it as its used just once in the rendering and you can copy the points instead (no rotation)... In the same header are also some constants:
const double pi = M_PI;
const double pi2 =2.0*M_PI;
const double pipol=0.5*M_PI;
const double deg=M_PI/180.0;
I got also vector and matrix math template integrated in the transform matrix header so vector<n> is n dimensional vector and matrix<n> is n*n square matrix but its used only for rendering so again you can ignore it. If youre interested here few links from whic all this was derived:
The enums and dimension reductions are used only for rendering. The cfg holds how should be each dimension reduced down to 2D.
AnsiString is a self relocating string from VCL so either use char* or string class you got in your environment. DWORD is just unsigned 32 bit int. Hope I did not forget something ...
• I've awarded the bounty because this looks like the most promising solution, and the bounty was about to expire. However I haven't been able to get this code to run due to the undeclared variables, would you be able to resolve that issue? Jul 28, 2019 at 21:21
• @Karl I added [edit1] with more (missing) code and descriptions. I did not post the other meshes code as I am afraid I am near 30K limit of answer (full code of the mesh has 23KB not counting helper files) if you are missing something else comment me. Jul 28, 2019 at 21:48
• @Karl I just updated the code a little (better screw and point ratios for 3D and 4D) Jul 28, 2019 at 22:22
All of the previous answer worked, but still lacked actual code. There were two real pieces missing, which this implements generally.
1. We need to compute the integral of sin^(d-2)(x). This has a closed form if you do recursive integration by parts. Here I implement it in the recursive fashion, although for dimension ~> 100 I found numeric integration of sin^d to be faster
2. We need to compute the inverse function of that integral, which for sin^d, d > 1 doesn't have a close form. Here I compute it using binary search, although there are likely better ways as stated in other answers.
These two combined with a way to generate primes results in the full algorithm:
from itertools import count, islice
from math import cos, gamma, pi, sin, sqrt
from typing import Callable, Iterator, List
def int_sin_m(x: float, m: int) -> float:
"""Computes the integral of sin^m(t) dt from 0 to x recursively"""
if m == 0:
return x
elif m == 1:
return 1 - cos(x)
else:
return (m - 1) / m * int_sin_m(x, m - 2) - cos(x) * sin(x) ** (
m - 1
) / m
def primes() -> Iterator[int]:
"""Returns an infinite generator of prime numbers"""
yield from (2, 3, 5, 7)
composites = {}
ps = primes()
next(ps)
p = next(ps)
assert p == 3
psq = p * p
for i in count(9, 2):
if i in composites: # composite
step = composites.pop(i)
elif i < psq: # prime
yield i
continue
else: # composite, = p*p
assert i == psq
step = 2 * p
p = next(ps)
psq = p * p
i += step
while i in composites:
i += step
composites[i] = step
def inverse_increasing(
func: Callable[[float], float],
target: float,
lower: float,
upper: float,
atol: float = 1e-10,
) -> float:
"""Returns func inverse of target between lower and upper
inverse is accurate to an absolute tolerance of atol, and
must be monotonically increasing over the interval lower
to upper
"""
mid = (lower + upper) / 2
approx = func(mid)
while abs(approx - target) > atol:
if approx > target:
upper = mid
else:
lower = mid
mid = (upper + lower) / 2
approx = func(mid)
return mid
def uniform_hypersphere(d: int, n: int) -> List[List[float]]:
"""Generate n points over the d dimensional hypersphere"""
assert d > 1
assert n > 0
points = [[1 for _ in range(d)] for _ in range(n)]
for i in range(n):
t = 2 * pi * i / n
points[i][0] *= sin(t)
points[i][1] *= cos(t)
for dim, prime in zip(range(2, d), primes()):
offset = sqrt(prime)
mult = gamma(dim / 2 + 0.5) / gamma(dim / 2) / sqrt(pi)
def dim_func(y):
return mult * int_sin_m(y, dim - 1)
for i in range(n):
deg = inverse_increasing(dim_func, i * offset % 1, 0, pi)
for j in range(dim):
points[i][j] *= sin(deg)
points[i][dim] *= cos(deg)
return points
Which produces the following image for 200 points on a sphere:
• Based on your answer, I optimized the code such that it can (approximately) generated a lot of points in high dimensional space (30000 points on 300 dims took around 23s). See pastebin.com/YPpVJMkc Dec 1, 2022 at 21:02
I got another insane idea on how to do this. Its entirely different than my previous approach hence new Answer...
Well one of the other answers suggest creating uniform distribution of points on hypercube surface and then normalize the points distance to center of hypercube to radius of hyperspace and use that for repulsion particle simulation. I did that in past for 3D with good results but in higher dimensions that would be insanely slow or complicated by BVH like structures.
But it got me thinking what about doing this backwards. So distribute the points on hypercube non linearly so after the normalization the points became linearly distributed on hypersphere...
So we simply step angle between +/-45 deg and compute the green points. The angle step da must divide 90 deg exactly and gives the point density. So all the 2D points will be a combination of +/-1.0 and tan(angle) for all faces.
When all points are done simply compute the size of each point to center and rescale it so it will be equal to hypersphere radius.
This can be easily expanded to any dimensionality
Each dimension above 2D just add one for cycle ang angle to iterate.
Here C++ example for 2D,3D,4D using my engine from previous answer of mine:
void ND_mesh::set_HyperSpherePCL(int N,double r,double da)
{
reset(N);
int na=floor(90.0*deg/da);
if (na<1) return;
da=90.0*deg/double(na-1);
if (n==2)
{
int i;
double a,x,y,l;
for (a=-45.0*deg,i=0;i<na;i++,a+=da)
{
x=tan(a); y=1.0;
l=sqrt((x*x)+(y*y));
x/=l; y/=l;
}
}
if (n==3)
{
int i,j;
double a,b,x,y,z,l;
for (a=-45.0*deg,i=0;i<na;i++,a+=da)
for (b=-45.0*deg,j=0;j<na;j++,b+=da)
{
x=tan(a); y=tan(b); z=1.0;
l=sqrt((x*x)+(y*y)+(z*z));
x/=l; y/=l; z/=l;
}
}
if (n==4)
{
int i,j,k;
double a,b,c,x,y,z,w,l;
for (a=-45.0*deg,i=0;i<na;i++,a+=da)
for (b=-45.0*deg,j=0;j<na;j++,b+=da)
for (c=-45.0*deg,k=0;k<na;k++,c+=da)
{
x=tan(a); y=tan(b); z=tan(c); w=1.0;
l=sqrt((x*x)+(y*y)+(z*z)+(w*w));
x/=l; y/=l; z/=l; w/=l;
}
}
for (int i=0;i<pnt.num/n;i++) as1(i);
rescale(r,n);
}
//---------------------------------------------------------------------------
The n=N is dimensionality r is radius and da is angualr step in [rad].
And perspective 2D/3D/4D previews:
And here more points and better size for 3D:
The cube pattern is slightly visible but the point distance looks OK to me. Its hard to see it on GIF as the backside points are merging with the front ones...
And this is the 2D square and 3D cube without normalization to sphere:
as you can see on edges is much smaller point density...
Preview is only using perspective projection as this does not generate mesh topology, just the points so cross section is not possible to do...
Also beware this produces some duplicate points on the edges (I think looping the angles one iteration less for some of the mirrors should remedy that but too lazy to implement that)
I strongly suggest to read this:
As a partial answer, you can use Newton's method to compute the inverse of f. Using x as the initial point in the Newton iteration is a good choice since f(x) is never more than 1 unit away from x. Here is a Python implementation:
import math
def f(x):
return x + 0.5*math.sin(2*x)
def f_inv(x,tol = 1e-8):
xn = x
y = f(xn)
while abs(y-x) > tol:
xn -= (y-x)/(1+math.cos(2*xn))
y = f(xn)
return xn
A nice fact about this application of Newton's method is that whenever cos(2*x) = -1 (where you would have division by 0) you automatically have sin(2*x) = 0 so that f(x) = x. In this case, the while loop is never entered and f_inv simply returns the original x.
• Good, this resolves the inverse function quite nicely. The only problem left is how to generate formulae for the angles in arbitrary d. Jul 22, 2019 at 11:42
• Nice and short implementation. Jul 23, 2019 at 19:35
We have n points, which are P1, ..., Pn. We have a dimension number d. Each (i = 1,n) point can be represented as:
Pi = (pi(x1), ..., pi(xd))
We know that
D(Pi, 0) = 1 <=>
sqrt((pi(x1) - pj(x1))^2 + ... + (pi(xd) - pj(xd))^2) = 1
and the minimal distance between any points, MD is
MD <= D(Pi, Pj)
A solution is acceptable if and only if MD could not be higher.
If d = 2, then we have a circle and put points on it. The circle is a polygon with the following properties:
• it has n angles
• n -> infinity
• each side is of similar length
So, a polygon of n angles, where n is a finite number and higher than 2, also, each side is of similar length is closer to a circle each time we increment n. Note that the firs polygon in d = 2 is the triangle. We have a single angle and our minimal angle unit is 360degrees / n.
Now, if we have a square and distribute the points evenly on it, then converting our square into circle via base transformation should be either the exact solution, or very close to it. If it is the exact solution, then this is a simple solution for the case when d = 2. If it is only very close, then with an approach of approximation we can determine what the solution is within a given precision of your choice.
I would use this idea for the case when d = 3. I would solve the problem for a cube, where the problem is much simpler and use base transformation to convert my cube points to my sphere points. And I would use this approach on d > 3, solving the problem for a hypercube and transform it to a hypersphere. Use the Manhattan distance when you evenly distribute your points on a hypercube of d dimensions.
Note that I do not know whether the solution for a hypercube transformed into a hypersphere is the exact solution or just close to it, but if it's not the exact solution, then we can increase precision with approximation.
So, this approach is a solution for the problem, which is not necessarily the best approach in terms of time complexity, so, if one has delved into the Fibonacci lattice area and knows how to generalize it for more dimensions, then his/her answer might be a better choice for acceptance than mine.
The invert of f(x) = x - 0.5sin2x can be determined if you defined the Taylor series of f(x). You will get a polynomial series of x which can be inverted.
• So we equidistribute points on the surface of a hypercube, and then to turn that into a hyphersphere we resize all of the point's vectors from the origin to have a length of 1. Unless I've misunderstood what you mean by the base transformation, that will have the result of points being more bunched up where the corners of the hypercube were. Jul 22, 2019 at 14:12
• @Karl I agree that the solution is unlikely to be acceptable as is (for the reason that you state), but perhaps it could be used to set up the initial state for the particle repulsion approach you allude to in the comments. If the initial state has a good distribution then convergence to the optimal might be faster. Jul 22, 2019 at 16:39
• @JohnColeman I have been researching particle repulsion methods for this problem for the last 4 years. One of the areas I researched was seeding a particle repulsion method using the technique described in this answer (if I understand the base transformation correctly). The results were not bad, but I wish to study deterministic methods now, which is why I'd like to focus on the fibonacci lattice. Jul 22, 2019 at 17:01
• @Karl we do not calculate the distance between the points using Euclidean geometry, but using Manhattan distance (adding the distances of different dimensions) according to my idea. This is of course only a starting point. If this happens to result in a deterministic equidistribution according to the specs, then this is a solution. If not, then this is a good starting point, but in that case it is undeterministic. It would be good to know if someone has the time to check whether the initial result matches the criteria and if not, how far is it from that. Jul 23, 2019 at 10:35
• @LajosArpad It does seem like a promising start Jul 23, 2019 at 10:41
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http://mathhelpforum.com/calculus/141011-help-local-max-min-concavity-increasing-decreasing.html
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# Thread: Help with local max/min, concavity, increasing decreasing?
1. ## Help with local max/min, concavity, increasing decreasing?
13)$\displaystyle f(x)=sin x + cos x 0\leq x \leq 2\pi$
$\displaystyle f'(x)=cosx-sinx = 0$
$\displaystyle x = \frac{\pi}{4} and \frac{5\pi}{4}$
how do I find where the function increase and decrease?
If I put these into a sign chart, I end up with all 0's
also how do I find the concavity?
I get that$\displaystyle f''(x) = -sin x - cos x$
so $\displaystyle x = \frac{7\pi}{4} and \frac{3\pi}{4}$
and for this also, if I plug it into f''(x) for a sign chart, the results will still be 0
is there another method for trig to do these problems?
2. Originally Posted by dorkymichelle
13)$\displaystyle f(x)=sin x + cos x 0\leq x \leq 2\pi$
$\displaystyle f'(x)=cosx-sinx = 0$
$\displaystyle x = \frac{\pi}{4} and \frac{5\pi}{4}$
how do I find where the function increase and decrease?
$\displaystyle x = \frac{\pi}{4} and \frac{5\pi}{4}$ are split points that split up the domain into three intervals
$\displaystyle 0\leq x\leq \frac{\pi}{4}$
$\displaystyle \frac{\pi}{4}\leq x\leq \frac{5\pi}{4}$ and
$\displaystyle \frac{5\pi}{4}\leq x\leq 2\pi$
Originally Posted by dorkymichelle
If I put these into a sign chart, I end up with all 0's
To determine the sign of $\displaystyle f'(x)$ in each of these intervals use a value of x in the interior of the interval
To determine the sign of $\displaystyle f'(x)$ in the interval, $\displaystyle 0\leq x\leq \frac{\pi}{4}$, for example, you could use $\displaystyle x=\frac{\pi}{6}$
$\displaystyle f'(\frac{\pi}{6})=\cos{\frac{\pi}{6}}-\sin{\frac{\pi}{6}}=\frac{\sqrt{3}}{2}-\frac{1}{2}>0$
If $\displaystyle f'(x)>0$, then f(x) is increasing in the interval
If $\displaystyle f'(x)<0$, then f(x) is decreasing in the interval
Determine the sign of $\displaystyle f'(x)$ in the other two intervals
Originally Posted by dorkymichelle
also how do I find the concavity?
I get that$\displaystyle f''(x) = -sin x - cos x$
so $\displaystyle x = \frac{7\pi}{4} and \frac{3\pi}{4}$
$\displaystyle x = \frac{7\pi}{4} and \frac{3\pi}{4}$ are split points for the second derivative
They split up the domain into the intervals
$\displaystyle 0\leq x\leq \frac{3\pi}{4}$
$\displaystyle \frac{3\pi}{4}\leq x\leq \frac{7\pi}{4}$ and
$\displaystyle \frac{7\pi}{4}\leq x\leq 2\pi$
Determine the sign of $\displaystyle f"(x)$ in each interval
If $\displaystyle f"(x)>0$, then f(x) is concave up
If $\displaystyle f"(x)<0$, then f(x) is concave down
3. Oh I see, I guess I just had a "trig scares me" moment and forgot to find a value between those intervals and used the crit. points instead.
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# Decimals to Fractions
• Jul 21st 2010, 11:43 AM
David Green
Decimals to Fractions
Hi All,
My basic maths are not the best on the planet in some areas, would somebody be able to advice how to change decimals to fractions and vica versa?
Example;
convert 13mm to a fraction?
Convert 13/15 to a decimal?
13/15 = 0.86, but what would I do to change the decimal to a fraction?
Any help much appreciated
• Jul 21st 2010, 12:44 PM
eumyang
Actually, 13/15 = 0.86666666.., a non-terminating, repeating decimal.
Here's a way to convert this decimal to a fraction. Let's assign a variable to this decimal:
$N = 0.866666666...$
Multiply both sides of this by 100 and 10:
\begin{aligned}
100N &= 86.666666666... \\
10N &= 8.666666666...
\end{aligned}
Subtract the 2nd line from the first:
$90N = 78$
Divide both sides by 90 to solve for N:
$N = \frac{78}{90} = \frac{13]{15}$
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| 4.125 | 4 |
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https://betterlesson.com/lesson/637800/exploring-orbits-where-the-centripetal-force-is-gravity?from=breadcrumb_lesson
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# Exploring Orbits Where the Centripetal Force is Gravity
2 teachers like this lesson
Print Lesson
## Objective
Students apply the circular motion concepts of centripetal force and tangential velocity to the orbit of the Earth around the Sun.
#### Big Idea
For a planet traveling in a circular orbit, the force of gravity is the centripetal force and the tangential velocity is required to keep the planet in orbit.
## Context and Supplies
In this lesson students apply the uniform circular motion concepts of centripetal force and tangential velocity to circular orbits of planets around the sun. It builds directly off the previous lesson of Playing "A Round" with Circular Motion with gravity as the centripetal force.
To do this, students access a PHET simulation on gravity. PHET is an excellent resource provided by the University of Colorado. Students work in groups of 2-3 and need access to a PC or Macintosh computer.
The NGSS engineering standard that is applicable here is HS-ETS1-4 which has students use a computer simulation to model complex real-world problems and interactions.
This activity sets up NGSS performance standard HS-PS2-4 where students use mathematical representations of Newton’s Law of Gravitation to describe and predict the gravitational forces between objects. In addition, this lesson has students applying CCSS Math Practice 7: Look for and make use of structure as well as Science Practice 2: Developing and using models and Science Practice 7: Engaging in argument from evidence.
## Opening - Gravitational Preview
5 minutes
As students enter the room, I have the first slide of the Gravity and Circular Motion power point which shows the moon orbiting the Earth. This power point has animations that don't appear in preview mode, so it is best to download it. I also have posted the day's learning goals on the whiteboard:
SWBAT
• Review the direction of the forces acting for a satellite orbiting a large body
• Review the direction of the velocity of a satellite orbiting a large body
• Review what happens to a satellite if you remove its tangential or orbital velocity.
But before students start the learning activity, we review circular motion concepts which were learned in the previous class on Playing Around with Circular Motion. The power point has slides which contain pictures and vectors from that lesson. It is important to remind students about centripetal motion, tangential velocity and the equations that model the motions of objects moving uniformly in a circle. The final slide shows what would happen if we eliminated the tangential velocity of the moon... the force of gravity would cause it to crash into the earth.
## Gravity Simulator
30 minutes
After I finish the review on tangential velocity, centripetal acceleration and the formulas associated with them, students apply those concepts to circular orbits using the PhET gravity simulation activity. The simulator listed on the sheet is provided by the University of Colorado and is one of many interactive science simulators present at PhET.
Students work in groups of 2-3, with one computer per group. Ideally, I would like to have all students complete this activity independently on their own computer as the questions and concepts are not rigorous but are good to lay a foundation for future learning. However, I do not have the resources to do this. I hand out the activity sheet and students choose their partners, collect their computer and navigate to the PhET Gravity and Orbits simulator. Students make observations on what they see on the simulator and sketch the force and velocity vector for the Earth orbiting the Sun (see picture below).
Next, they observe what happens if one removes the orbital velocity (the Earth is pulled into the sun) and if one removes gravity (the Earth travels on in a path tangent to the circle).
Lastly, the students complete a Venn diagram comparing the motion and forces of a ball on a string being swung in a circle and the moon going around the Earth. This compare and contrast activity is a great strategy to leads students to analyze their understanding on the important factors involved in circular motion.
## Review and Closure
15 minutes
As students finish the activity, we review the students answers on the gravity simulation activity. I ask for volunteers to share their opening observations on the simulator. Then I use the document camera and call up students whose answers are correct, clear and well explained. It takes about 10 minutes to review the students' answers. I include a demonstration of twirling a ball at the end of a string. As I run the demo, I call on students to supply their answers on how the ball is different from the moon and how it is similar.
I finish the class sharing Aristotle's view of the universe. This ancient Greek philosopher said that the shape of our world was a perfect sphere as are the sun and moon. Not only that, but all motion (orbits) in the "heavens" are perfect circles. We now know that planets do not travel in perfectly circular orbits, but rather ellipses. I show them what an elliptical orbit looks like on the PhET simulator projected on whiteboard. However, many of the orbits of the planets are so close to circular that our application of circular motion provides a good approximation.
In the next lesson, we will explore the concept of gravity in the historical context of the Renaissance and how Newton's discovery that the force that causes Earthly objects to fall IS THE SAME FORCE that causes the Moon to go around the Earth and the Earth to go around the Sun. Students will understand the significance of the first "Universal Law" and how that continues to impact our view of the universe today.
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0
# Is a obtuse triangle always always isosceles?
Updated: 12/11/2022
Wiki User
13y ago
no. obtuse simply means one of its angles in greater than 90 degrees, whereas isosceles is determined but the side lengths. they are not directly linked.
Wiki User
13y ago
Earn +20 pts
Q: Is a obtuse triangle always always isosceles?
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Related questions
No not always
No
### If a triangle is obtuse then is it isosceles?
An obtuse triangle is not necessarily isosceles, but it can be.
### What is an isosceles obtuse triangle?
An isosceles obtuse triangle is a triangle with 1 obtuse angle and only 2 congruent sides.
### What type of triangle has an angle that measures 120 degrees and an angle that measures thirty degrees for kids?
It is an obtuse angled isosceles triangle.It is an obtuse angled isosceles triangle.It is an obtuse angled isosceles triangle.It is an obtuse angled isosceles triangle.
### What is the differences between an isoceles triangle and an obtuse triangle?
-- An isosceles triangle can also be an obtuse triangle but it doesn't have to be. -- An obtuse triangle can also be an isosceles triangle but it doesn't have to be. So there's no consistent diferences betwen them.
### What is a triangle that has has 1 obtuse?
It can be an isosceles triangle or a scalene triangle if it has 1 obtuse angle
### How many angles do a acute isosceles triangle have?
It doesn't matter if a triangle is isosceles, scalene, obtuse, acute... There are always 3 angles in a triangle (which add up to 180).
### Does an obtuse triangle have to be scalene?
No. An obtuse triangle can be isosceles. It just can't be a right or equilateral triangle.
### Is every obtuse triangle a scalene triangle?
no- it can be an isosceles triangle
### What is a obtuse isosceles?
it i a triangle that only has one obtuse angle
### Different kinds of triangle?
there is equilateral triangle, right triangle, isosceles triangle, obtuse triangle, acute triangle, scalene triangle and oblique triangle
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# The MolePage 1
Slide 1
The Mole
Slide 2
CA Standards
Slide 3
## The Mole
1 dozen =
1 gross =
1 ream =
1 mole =
12
144
500
6.02 x 1023
There are exactly 12 grams of carbon-12 in one mole of carbon-12.
Slide 4
I didn’t discover it. Its just named after me!
Slide 5
## Calculating Formula Mass
Calculate the formula mass of carbon dioxide, CO2.
12.01 g + 2(16.00 g) =
44.01 g
One mole of CO2 (6.02 x 1023 molecules) has a mass of 44.01 grams
Slide 6
Mole Relationships
Mole
Atoms
or molecules
Liters
Grams
6.02 x 1023
Atomic
Mass
22. 4 L
22.4
L
Slide 7
## Calculations with Moles: Converting moles to grams
How many grams of lithium are in 3.50 moles of lithium?
3.50 mol Li
= g Li
1 mol Li
6.94 g Li
24.3
Slide 8
## Calculations with Moles: Converting grams to moles
How many moles of lithium are in 18.2 grams of lithium?
18.2 g Li
= mol Li
6.94 g Li
1 mol Li
2.62
Slide 9
## Calculations with Moles: Using Avogadro’s Number
How many atoms of lithium are in 3.50 moles of lithium?
3.50 mol
= atoms
1 mol
6.02 x 1023 atoms
2.07 x 1024
Slide 10
Calculations with Moles: Using Avogadro’s Number
How many atoms of lithium are in 18.2 g of
lithium?
18.2 g Li
= atoms Li
1 mol Li
6.022 x 1023 atoms Li
1.58 x 1024
6.94 g Li
1 mol Li
(18.2)(6.022 x 1023)/6.94
Slide 11
## Standard Molar Volume
Equal volumes of all gases at the same temperature and pressure contain the same number of molecules.
At STP (Standard Temperature and Pressure):
1 mole of a gas occupies 22.4 liters of volume
Slide 12
Go to page:
1 2
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## How do you calculate energy absorbed or released?
To calculate the amount of heat released in a chemical reaction, use the equation Q = mc ΔT, where Q is the heat energy transferred (in joules), m is the mass of the liquid being heated (in kilograms), c is the specific heat capacity of the liquid (joule per kilogram degrees Celsius), and ΔT is the change in …
## How do you calculate difference in time?
1. Convert both times to 24 hour format, adding 12 to any pm hours. 8:55am becomes 8:55 hours (start time)
2. If the start minutes are greater than the end minutes…
3. Subtract end time minutes from start time minutes…
4. Subtract the hours…
5. Put(not add) the hours and minutes together – 6:45 (6 hours and 45 minutes)
## What is the formula for elapsed time?
Calculate elapsed days/month/years To calculate elapsed days is so easy, you just need to apply this formula = B2-A2, A2 is the start date, B2 is the end date. Tip: To calculate elapsed month, you can use this formula =DATEDIF(A2,B2,”m”), A2 is the start date, B2 is the end date.
## How do you add time on a calculator?
For example, add the times 1:30:45 and 2:45:20 together.
1. Add hours, minutes, and seconds separately: hours = 1 + 2 = 3 hours.
2. Seconds are greater than 60, so add 1 minute and subtract 60 seconds: minutes = 75 + 1 = 76 minutes.
3. Minutes are greater than 60, so add 1 hour and subtract 60 minutes: hours = 3 + 1 = 4 hours.
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# Online Aptitude Test - Aptitude Test 3
Instruction:
• This is a FREE online test. DO NOT pay money to anyone to attend this test.
• Total number of questions : 20.
• Time alloted : 30 minutes.
• Each question carry 1 mark, no negative marks.
• DO NOT refresh the page.
• All the best :-).
1.
The average monthly income of P and Q is Rs. 5050. The average monthly income of Q and R is Rs. 6250 and the average monthly income of P and R is Rs. 5200. The monthly income of P is:
A.
3500
B.
4000
C.
4050
D.
5000
Correct Answer: Option B
Explanation:
Let P, Q and R represent their respective monthly incomes. Then, we have:
P + Q = (5050 x 2) = 10100 .... (i)
Q + R = (6250 x 2) = 12500 .... (ii)
P + R = (5200 x 2) = 10400 .... (iii)
Adding (i), (ii) and (iii), we get: 2(P + Q + R) = 33000 or P + Q + R = 16500 .... (iv)
Subtracting (ii) from (iv), we get P = 4000.
P's monthly income = Rs. 4000.
2.
A library has an average of 510 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is:
A.
250
B.
276
C.
280
D.
285
Correct Answer: Option D
Explanation:
Since the month begins with a Sunday, to there will be five Sundays in the month.
Required average
= 510 x 5 + 240 x 25 30
= 8550 30
= 285
3.
The sum of two number is 25 and their difference is 13. Find their product.
A.
104
B.
114
C.
315
D.
325
Correct Answer: Option B
Explanation:
Let the numbers be x and y.
Then, x + y = 25 and x - y = 13.
4xy = (x + y)2 - (x- y)2
= (25)2 - (13)2
= (625 - 169)
= 456
xy = 114.
Learn more problems on : Problems on Numbers
Direction (for Q.No. 4):
Each of the questions given below consists of a statement and / or a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statement(s) is / are sufficient to answer the given question. Read the both statements and
• Give answer (A) if the data in Statement I alone are sufficient to answer the question, while the data in Statement II alone are not sufficient to answer the question.
• Give answer (B) if the data in Statement II alone are sufficient to answer the question, while the data in Statement I alone are not sufficient to answer the question.
• Give answer (C) if the data either in Statement I or in Statement II alone are sufficient to answer the question.
• Give answer (D) if the data even in both Statements I and II together are not sufficient to answer the question.
• Give answer(E) if the data in both Statements I and II together are necessary to answer the question.
4.
What is the number? I. The sum of the two digits is 8. The ratio of the two digits is 1 : 3. II. The product of the two digit of a number is 12. The quotient of two digits is 3.
A.
I alone sufficient while II alone not sufficient to answer
B.
II alone sufficient while I alone not sufficient to answer
C.
Either I or II alone sufficient to answer
D.
Both I and II are not sufficient to answer
E.
Both I and II are necessary to answer
Correct Answer: Option C
Explanation:
Let the tens and units digit be x and y respectively. Then,
I. x + y = 8 and x = 1 y 3
I gives, 4y = 24 y = 6.
So, x + 6 = 8 x = 2.
II. xy = 12 and x = 3 y 1
II gives, x2 = 36 x = 6.
So, 3y = 6 y = 2.
Therefore, Either I or II alone sufficient to answer.
Learn more problems on : Problems on Numbers
5.
A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, then how old is B?
A.
7
B.
8
C.
9
D.
10
E.
11
Correct Answer: Option D
Explanation:
Let C's age be x years. Then, B's age = 2x years. A's age = (2x + 2) years.
(2x + 2) + 2x + x = 27
5x = 25
x = 5.
Hence, B's age = 2x = 10 years.
Learn more problems on : Problems on Ages
6.
(17)3.5 x (17)? = 178
A.
2.29
B.
2.75
C.
4.25
D.
4.5
Correct Answer: Option D
Explanation:
Let (17)3.5 x (17)x = 178.
Then, (17)3.5 + x = 178.
3.5 + x = 8
x = (8 - 3.5)
x = 4.5
Learn more problems on : Surds and Indices
7.
If x = 3 + 22, then the value of x - 1 is: x
A.
1
B.
2
C.
22
D.
33
Correct Answer: Option B
Explanation:
x - 1 2 = x + 1 - 2 x x
= (3 + 22) + 1 - 2 (3 + 22)
= (3 + 22) + 1 x (3 - 22) - 2 (3 + 22) (3 - 22)
= (3 + 22) + (3 - 22) - 2
= 4.
x - 1 = 2. x
Learn more problems on : Surds and Indices
8.
A and B together have Rs. 1210. If of A's amount is equal to of B's amount, how much amount does B have?
A.
Rs. 460
B.
Rs. 484
C.
Rs. 550
D.
Rs. 664
Correct Answer: Option B
Explanation:
4 A = 2 B 15 5
A = 2 x 15 B 5 4
A = 3 B 2
A = 3 B 2
A : B = 3 : 2.
B's share = Rs. 1210 x 2 = Rs. 484. 5
Learn more problems on : Ratio and Proportion
9.
Speed of a boat in standing water is 9 kmph and the speed of the stream is 1.5 kmph. A man rows to a place at a distance of 105 km and comes back to the starting point. The total time taken by him is:
A.
16 hours
B.
18 hours
C.
20 hours
D.
24 hours
Correct Answer: Option D
Explanation:
Speed upstream = 7.5 kmph.
Speed downstream = 10.5 kmph.
Total time taken = 105 + 105 hours = 24 hours. 7.5 10.5
Learn more problems on : Boats and Streams
10.
A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a. in 5 years. What is the sum?
A.
Rs. 4462.50
B.
Rs. 8032.50
C.
Rs. 8900
D.
Rs. 8925
E.
None of these
Correct Answer: Option D
Explanation:
Principal
= Rs. 100 x 4016.25 9 x 5
= Rs. 401625 45
= Rs. 8925.
Learn more problems on : Simple Interest
11.
A person takes a loan of Rs. 200 at 5% simple interest. He returns Rs. 100 at the end of 1 year. In order to clear his dues at the end of 2 years, he would pay:
A.
Rs. 105
B.
Rs. 110
C.
Rs. 115
D.
Rs. 115.50
Correct Answer: Option C
Explanation:
Amount to be paid
= Rs. 100 + 200 x 5 x 1 + 100 x 5 x 1 100 100
= Rs. 115.
Learn more problems on : Simple Interest
12.
The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:
A.
1520 m2
B.
2420 m2
C.
2480 m2
D.
2520 m2
Correct Answer: Option D
Explanation:
We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103.
Solving the two equations, we get: l = 63 and b = 40.
Area = (l x b) = (63 x 40) m2 = 2520 m2.
13.
What was the day of the week on 28th May, 2006?
A.
Thursday
B.
Friday
C.
Saturday
D.
Sunday
Correct Answer: Option D
Explanation:
28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)
Odd days in 1600 years = 0
Odd days in 400 years = 0
5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) 6 odd days
```Jan. Feb. March April May
(31 + 28 + 31 + 30 + 28 ) = 148 days
```
148 days = (21 weeks + 1 day) 1 odd day.
Total number of odd days = (0 + 0 + 6 + 1) = 7 0 odd day.
Given day is Sunday.
14.
The angle between the minute hand and the hour hand of a clock when the time is 4.20, is:
A.
0°
B.
10°
C.
5°
D.
20°
Correct Answer: Option B
Explanation:
Angle traced by hour hand in 13 hrs = 360 x 13 ° = 130°. 3 12 3
Angle traced by min. hand in 20 min. = 360 x 20 ° = 120°. 60
Required angle = (130 - 120)° = 10°.
15.
At what angle the hands of a clock are inclined at 15 minutes past 5?
A.
58 1 ° 2
B.
64°
C.
67 1 ° 2
D.
72 1 ° 2
Correct Answer: Option C
Explanation:
Angle traced by hour hand in 21 hrs = 360 x 21 ° = 157 1 ° 4 12 4 2
Angle traced by min. hand in 15 min. = 360 x 15 ° = 90°. 60
Required angle = 157 1 ° - 90° = 67 1 ° 2 2
16.
The angle between the minute hand and the hour hand of a clock when the time is 8.30, is:
A.
80°
B.
75°
C.
60°
D.
105°
Correct Answer: Option B
Explanation:
Angle traced by hour hand in 17 hrs = 360 x 17 ° = 255°. 2 12 2
Angle traced by min. hand in 30 min. = 360 x 30 ° = 180°. 60
Required angle = (255 - 180)° = 75°.
17.
At what time between 9 and 10 o'clock will the hands of a watch be together?
A.
45 min. past 9
B.
50 min. past 9
C.
49 1 min. past 9 11
D.
48 2 min. past 9 11
Correct Answer: Option C
Explanation:
To be together between 9 and 10 o'clock, the minute hand has to gain 45 min. spaces.
55 min. spaces gained in 60 min.
45 min. spaces are gained in 60 x 45 min or 49 1 min. 55 11
The hands are together at 49 1 min. past 9. 11
18.
In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
A.
63
B.
90
C.
126
D.
45
E.
135
Correct Answer: Option A
Explanation:
Required number of ways = (7C5 x 3C2) = (7C2 x 3C1) = 7 x 6 x 3 = 63. 2 x 1
Learn more problems on : Permutation and Combination
Direction (for Q.No. 19):
Find out the wrong number in the given sequence of numbers.
19.
8, 13, 21, 32, 47, 63, 83
A.
47
B.
63
C.
32
D.
83
Correct Answer: Option A
Explanation:
Go on adding 5, 8, 11, 14, 17, 20.
So, the number 47 is wrong and must be replaced by 46.
Learn more problems on : Odd Man Out and Series
Direction (for Q.No. 20):
Insert the missing number.
20.
1, 4, 9, 16, 25, 36, 49, (....)
A.
54
B.
56
C.
64
D.
81
Correct Answer: Option C
Explanation:
Numbers are 12, 22, 32, 42, 52, 62, 72.
So, the next number is 82 = 64.
Learn more problems on : Odd Man Out and Series
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21345200
21,345,200 (twenty-one million three hundred forty-five thousand two hundred) is an even eight-digits composite number following 21345199 and preceding 21345201. In scientific notation, it is written as 2.13452 × 107. The sum of its digits is 17. It has a total of 9 prime factors and 120 positive divisors. There are 7,741,440 positive integers (up to 21345200) that are relatively prime to 21345200.
Basic properties
• Is Prime? No
• Number parity Even
• Number length 8
• Sum of Digits 17
• Digital Root 8
Name
Short name 21 million 345 thousand 200 twenty-one million three hundred forty-five thousand two hundred
Notation
Scientific notation 2.13452 × 107 21.3452 × 106
Prime Factorization of 21345200
Prime Factorization 24 × 52 × 17 × 43 × 73
Composite number
Distinct Factors Total Factors Radical ω(n) 5 Total number of distinct prime factors Ω(n) 9 Total number of prime factors rad(n) 533630 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 21,345,200 is 24 × 52 × 17 × 43 × 73. Since it has a total of 9 prime factors, 21,345,200 is a composite number.
Divisors of 21345200
120 divisors
Even divisors 96 24 12 12
Total Divisors Sum of Divisors Aliquot Sum τ(n) 120 Total number of the positive divisors of n σ(n) 5.63223e+07 Sum of all the positive divisors of n s(n) 3.49771e+07 Sum of the proper positive divisors of n A(n) 469352 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 4620.09 Returns the nth root of the product of n divisors H(n) 45.478 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 21,345,200 can be divided by 120 positive divisors (out of which 96 are even, and 24 are odd). The sum of these divisors (counting 21,345,200) is 56,322,288, the average is 46,935,2.4.
Other Arithmetic Functions (n = 21345200)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 7741440 Total number of positive integers not greater than n that are coprime to n λ(n) 5040 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1348274 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 7,741,440 positive integers (less than 21,345,200) that are coprime with 21,345,200. And there are approximately 1,348,274 prime numbers less than or equal to 21,345,200.
Divisibility of 21345200
m n mod m 2 3 4 5 6 7 8 9 0 2 0 0 2 2 0 8
The number 21,345,200 is divisible by 2, 4, 5 and 8.
• Abundant
• Polite
• Practical
Base conversion (21345200)
Base System Value
2 Binary 1010001011011001110110000
3 Ternary 1111011110002222
4 Quaternary 1101123032300
5 Quinary 20431021300
6 Senary 2041300212
8 Octal 121331660
10 Decimal 21345200
12 Duodecimal 7194668
20 Vigesimal 6d8300
36 Base36 cpi28
Basic calculations (n = 21345200)
Multiplication
n×i
n×2 42690400 64035600 85380800 106726000
Division
ni
n⁄2 1.06726e+07 7.11507e+06 5.3363e+06 4.26904e+06
Exponentiation
ni
n2 455617563040000 9725248006601408000000 207587363750508374041600000000 4430993796727351345592760320000000000
Nth Root
i√n
2√n 4620.09 277.396 67.9712 29.2321
21345200 as geometric shapes
Circle
Diameter 4.26904e+07 1.34116e+08 1.43136e+15
Sphere
Volume 4.0737e+22 5.72546e+15 1.34116e+08
Square
Length = n
Perimeter 8.53808e+07 4.55618e+14 3.01867e+07
Cube
Length = n
Surface area 2.73371e+15 9.72525e+21 3.6971e+07
Equilateral Triangle
Length = n
Perimeter 6.40356e+07 1.97288e+14 1.84855e+07
Triangular Pyramid
Length = n
Surface area 7.89153e+14 1.14613e+21 1.74283e+07
Cryptographic Hash Functions
md5 6e4569e7c1c07b5a849c8b49304a7b54 aa97287272ba53f45c57903f3e78ea275846160e 0c6762982983f05e795e5044380ba68cce98ca0b61077199b5d33bc106d5a247 ed1c70369cd81811cbc4ce07ebcae065364f4ceb96fb8a9cbef412c4415e338dc2a9475b36420acd6bbb6f5801c5d4febfa2b6fef9b432161e7bb435dfccbf66 907161fd5dbadfac3b87f34d1c04ab3ef9d76f17
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# Decimal fractions
##### Active member
What decimal of an hour is a second ?
A. .0025
B. .0256
C. .00027
D. .000126
If 2994 ÷ 14.5 = 172, then 29.94 ÷ 1.45 = ?
A. 0.172
B. 1.72
C. 17.2
D. 172
The expression (11.98 x 11.98 + 11.98 x x + 0.02 x 0.02) will be a perfect square for x equal to:
A. 0.02
B. 0.2
C. 0.04
D. 0.4
##### Active member
3889 + 12.952 - ? = 3854.002
A. 47.095
B. 47.752
C. 47.932
D. 47.95
0.04 x 0.0162 is equal to:
A. 6.48 x 10^-3
B. 6.48 x 10^-4
C. 6.48 x 10^-5
D. 6.48 x 10^-6
The price of commodity X increases by 40 paise every year, while the price of commodity Y increases by 15 paise every year. If in 2001, the price of commodity X was Rs. 4.20 and that of Y was Rs. 6.30, in which year commodity X will cost 40 paise more than the commodity Y ?
A. 2010
B. 2011
C. 2012
D. 2013
##### Active member
617 + 6.017 + 0.617 + 6.0017 = ?
A. 6.2963
B. 62.965
C. 629.6357
D. None of these
0.002 x 0.5 = ?
A. 0.0001
B. 0.001
C. 0.01
D. 0.1
34.95 + 240.016 + 23.98 = ?
A. 298.0946
B. 298.111
C. 298.946
D. 299.09
##### Active member
Which of the following is equal to 3.14 x 10^6 ?
A. 314
B. 3140
C. 3140000
D. None of these
How many digits will be there to the right of the decimal point in the product of 95.75 and .02554 ?
A. 5
B. 6
C. 7
D. None of these
4.036 divided by 0.04 gives :
A. 1.009
B. 10.09
C. 100.9
D. None of these
##### Active member
What decimal of an hour is a second ?
A. .0025
B. .0256
C. .00027
D. .000126
If 2994 ÷ 14.5 = 172, then 29.94 ÷ 1.45 = ?
A. 0.172
B. 1.72
C. 17.2
D. 172
The expression (11.98 x 11.98 + 11.98 x x + 0.02 x 0.02) will be a perfect square for x equal to:
A. 0.02
B. 0.2
C. 0.04
D. 0.4
1. C, 2. C
##### Active member
3889 + 12.952 - ? = 3854.002
A. 47.095
B. 47.752
C. 47.932
D. 47.95
0.04 x 0.0162 is equal to:
A. 6.48 x 10^-3
B. 6.48 x 10^-4
C. 6.48 x 10^-5
D. 6.48 x 10^-6
The price of commodity X increases by 40 paise every year, while the price of commodity Y increases by 15 paise every year. If in 2001, the price of commodity X was Rs. 4.20 and that of Y was Rs. 6.30, in which year commodity X will cost 40 paise more than the commodity Y ?
A. 2010
B. 2011
C. 2012
D. 2013
1. D, 2. B
##### Active member
617 + 6.017 + 0.617 + 6.0017 = ?
A. 6.2963
B. 62.965
C. 629.6357
D. None of these
0.002 x 0.5 = ?
A. 0.0001
B. 0.001
C. 0.01
D. 0.1
34.95 + 240.016 + 23.98 = ?
A. 298.0946
B. 298.111
C. 298.946
D. 299.09
1. C, 2. B
##### Active member
Which of the following is equal to 3.14 x 10^6 ?
A. 314
B. 3140
C. 3140000
D. None of these
How many digits will be there to the right of the decimal point in the product of 95.75 and .02554 ?
A. 5
B. 6
C. 7
D. None of these
4.036 divided by 0.04 gives :
A. 1.009
B. 10.09
C. 100.9
D. None of these
1.c, 2.b
##### Active member
The least among the following is:
A.0.2
B.1 ÷ 0.2
C.0.2
D.(0.2)^2
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11.3 The ideal gas law
Page 1 / 11
• State the ideal gas law in terms of molecules and in terms of moles.
• Use the ideal gas law to calculate pressure change, temperature change, volume change, or the number of molecules or moles in a given volume.
• Use Avogadro’s number to convert between number of molecules and number of moles.
In this section, we continue to explore the thermal behavior of gases. In particular, we examine the characteristics of atoms and molecules that compose gases. (Most gases, for example nitrogen, ${\text{N}}_{2}$ , and oxygen, ${\text{O}}_{2}$ , are composed of two or more atoms. We will primarily use the term “molecule” in discussing a gas because the term can also be applied to monatomic gases, such as helium.)
Gases are easily compressed. We can see evidence of this in [link] , where you will note that gases have the largest coefficients of volume expansion. The large coefficients mean that gases expand and contract very rapidly with temperature changes. In addition, you will note that most gases expand at the same rate, or have the same $\beta$ . This raises the question as to why gases should all act in nearly the same way, when liquids and solids have widely varying expansion rates.
The answer lies in the large separation of atoms and molecules in gases, compared to their sizes, as illustrated in [link] . Because atoms and molecules have large separations, forces between them can be ignored, except when they collide with each other during collisions. The motion of atoms and molecules (at temperatures well above the boiling temperature) is fast, such that the gas occupies all of the accessible volume and the expansion of gases is rapid. In contrast, in liquids and solids, atoms and molecules are closer together and are quite sensitive to the forces between them.
To get some idea of how pressure, temperature, and volume of a gas are related to one another, consider what happens when you pump air into an initially deflated tire. The tire’s volume first increases in direct proportion to the amount of air injected, without much increase in the tire pressure. Once the tire has expanded to nearly its full size, the walls limit volume expansion. If we continue to pump air into it, the pressure increases. The pressure will further increase when the car is driven and the tires move. Most manufacturers specify optimal tire pressure for cold tires. (See [link] .)
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Berger describes sociologists as concerned with
4
Because I'm writing a report and I would like to be really precise for the references
where did you find the research and the first image (ECG and Blood pressure synchronized)? Thank you!!
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×
[–] 2 points3 points (0 children)
The launch of 63S next week will be the first attempt of a Soyuz performing a two orbit rendezvous, I believe. As others have said, this is the nominal flight path for Progress. The 63S crew is planned to dock with station 3:07 after launching. Pretty quick!
[–] 3 points4 points (0 children)
Good question! The reason that they land in Kazakhstan is because that is where the Baikonur Cosmodrome is, the launch facility of Roscosmos. Fun fact: the same launch pad used today is also where Yuri Gagarin launched from on the first human space flight ever!
The reason that Christina Koch, a NASA Astronaut, is landing there, is because we currently buy seats on Russian rockets to get to and from the ISS. Hopefully that will end this year with the commercial crew program.
[–] 1 point2 points (0 children)
Good question! In a circular orbit, orbital velocity depends only on the radius of the orbit (i.e. the height of the satellite). Higher satellites move slower than lower satellites. So, to answer your question, the velocity of a satellite is adjustable in the sense that you can change its orbit. The orbital velocity of a real life satellite in an elliptical orbit is dependent on both the radius and the semi-major axis of the orbit.
[–] 13 points14 points (0 children)
The force of gravity between two objects is defined by F = GMm/r2 where G is the gravitational constant, the M and m are the two masses, and r is the distance between the center of masses.
In your specific scenario the force between the two satellites would be = (6.67e-11 m3 kg-1 s-2 )(100 kg)(100 kg)/(100 m)2, which would give us 6.67e-11 Newtons. If we divide this by the mass of the satellite, we can find the acceleration equals 6.67 e-13 m/s2 (F=ma). This shows a very very small acceleration of the two satellites towards each other given that no other opposing forces are present. As to whether or not the resulting acceleration would be measurable, I am sure it can be with sensitive enough accelerometers.
For reference, if we use the same formula to calculate the force between the earth and one of these satellites (assuming 250km altitude) we get a force of 905 Newton’s. The satellite’s acceleration towards earth would be 9.05 m/s2.
I hope that sheds some more light on the situation for you.
[–] 1 point2 points (0 children)
Haha I live very far from Lubbock now and there is so much red dirt in the crevices of my car, it’s honestly amazing
[–] 4 points5 points (0 children)
It all came from goddamn Post
[–] 0 points1 point (0 children)
Pronounce “Chipotle”
[–] 1 point2 points (0 children)
2:30 would be awesome. Got an early flight out Sunday
[–] 1 point2 points (0 children)
Hey me too! Can't wait
[–] 0 points1 point (0 children)
Lol yeah what was that??
[–] 49 points50 points (0 children)
[–] 0 points1 point (0 children)
RAIDER!!!
[–] 0 points1 point (0 children)
Yeah I was thinking of doing that too! Fox sports go is working for now. Thanks for that link
[–] 0 points1 point (0 children)
I just graduated and moved to Florida too! How do you watch the games?
[–] 5 points6 points (0 children)
Hey man, it's not so baddd
[–] 1 point2 points (0 children)
No I'm just pleasantly surprised every time I don't get charged! It feeds me for like 2 full days I wouldn't mind paying a little more.
[–] 1 point2 points (0 children)
Thanks! I'll do that.
[–] 8 points9 points (0 children)
At my regular store and a few other stores I have been to, I ask for a 'Venn diagram' and they automatically know what I'm talking about. However, some other stores I have been to are really confused about it. Does corporate teach you anything about these? Have you ever heard of this?
It's when you overlap two tortillas, similar to an actual Venn diagram, and then roll them up together. It fits way more food and is the same price.
[–] 91 points92 points (0 children)
Thanks so much for doing this AMA! I am a mechanical engineer who is moving to Vero Beach soon. If I swing by, could I get shown around your facilities? I would love to learn more about how you guys do what you do.
[–] 2 points3 points (0 children)
[–] 4 points5 points (0 children)
Yessss, add a banana and a tad of hersheys syrup and blend it all up for a yummy smoothie
[–] 3 points4 points (0 children)
This is an incredible interview that gave me some inspiration during this hellish week, thanks for sharing
[–] 4 points5 points (0 children)
What cruise control module did you go with? I have a mazda 3 base and I miss cruise control so much
[–] 2 points3 points (0 children)
I am a senior in engineering right now, but when I was a freshman I took songwriting. It was pretty simple. I recommend that you know how to play an instrument because if not you will have to sing in front of everyone (which would have been super embarassing for me).
The teacher gives lectures which are pretty interesting and about 2 or 3 times a semester you must form groups and write a song. Nothing complicated. Pretty much everyone got A's I think.
[–] 0 points1 point (0 children)
Yeah engineering. Would you say there's a higher chance for STEM?
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## Elementary Statistics (12th Edition)
a) Fail to reject the null hypothesis. b)$\mu_1-\mu_2$ is between -23.62787 and 4.44929.
a) Null hypothesis:$\mu_1=\mu_2$, alternative hypothesis:$\mu_1$ is less than $\mu_2$. Hence the value of the test statistic: $t=\frac{(\overline{x_1}-\overline{x_2})-(\mu_1-\mu_2)}{\sqrt{s_1^2/n_1+s_2^2/n_2}}=\frac{(13.125-22.71429)-(0)}{\sqrt{8.96036^2/24+18.60285^2/14}}=-1.81.$ The degree of freedom: $min(n_1-1,n_2-1)=min(24-1,14-1)=13.$ The corresponding P-value by using the table: p is between 0.025 and 0.05. If the P-value is less than $\alpha$, which is the significance level, then this means the rejection of the null hypothesis. Hence:P is more than $\alpha=0.01$, because it is more than 0.0.025, hence we fail to reject the null hypothesis. b) The corresponding critical value using the table: $t_{\alpha/2}=t_{0.01}=2.65.$ The margin of error: $E=t_{\alpha/2}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=2.65\sqrt{\frac{8.96036^2}{24}+\frac{18.60285^2}{14}}=14.03858.$ Hence the confidence interval $\mu_1-\mu_2$ is between $\overline{x_1}-\overline{x_2}-E$=(13.125-22.71429)-14.03858=-23.62787 and $\overline{x_1}-\overline{x_2}+E$=(13.125-22.71429)+14.03858=4.44929.
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< Previous Section Home Next Section >
# Section 3.17Properties of Functions
## Basic Terminology
The table below summarizes basic terminology developed so far in the book. It also introduces new terms and meanings regarding properties of functions. The new meanings will be discussed more fully following the table.
Term Symbol Meaning
Interval of numbers from a to b $\,[a,b]$ if you include a and b
$\,(a,b)$ if you exclude a and b
All real numbers between a and b. The interval is called closed if you include a and b; the interval is called open if you exclude a and b.
Intervals can be half open or half closed, denoted $[a,b)$ when you include a and exclude b, or $(a,b]$ when you exclude a and include b.
Variable Conventionally, any letter at the upper end of the alphabet, such as x, y, or z But any notation can be used as a variable as long as you so state it. If you represent the value of a quantity whose measure varies within a situation, then you are using that letter as a variable.
Independent variable The variable whose values you control, or the variable whose values you take as determining the other variable's values.
Dependent variable
The variable whose values are determined by values of the independent variable.
Function Any letter, symbol, or word declared as representing a function A relationship between an independent variable and a dependent variable is called a function if each value of the independent variable is related to exactly one value of the dependent variable.
Functions can be defined by a rule for computing its values, such as $f(x)=3x^2-\cos(x)$ or conceptually, such as "Let $\text{dist}(x)$ be the number of meters from start run by Usain Bolt x seconds after the start in the final heat of the 2012 Summer Olympics."
In both cases, every value of x is associated with exactly one value $f(x)$. The rule $x^2-\cos(x)$ always produces just one number for any value of x. Usain Bolt is never two distances from start at a given moment in time during the race.
Domain of a function All values of the function's independent variable for which the function has a value. $f(x)=1/x$ is not defined when $x=0$, so 0 is not in the domain of f.
Range of a function
All values related by the function to values of the independent variable.
One-to-one function 1-1 A function f is said to be one-to-one when no two values of its independent variable are related to the same value of its dependent variable. In symbols, if a and b are in the domain of f and $a\ne b$, then $f(a)\ne f(b)$.
Discontinuous function A function f is said to be discontinuous at a value $x_0$ in its domain if there is a tolerance level $\epsilon > 0$ such that there is no interval around $x_0$ having all values of $f(x)$ within that tolerance.
Put another way, f is discontinuous at a value $x_0$ in its domain if there is a value of $\epsilon$ greater than 0 so that in any interval containing $x_0$, there are values $x_1$ and $x_2$ in the interval such that $|f(x_1)-f(x_2)|\gt \epsilon$.
Continuous function A function is said to be continous over an interval if it is not discontinuous at any value in the interval.
Increasing function A function f is said to be increasing over an interval if $f(a)\lt f(b)$ whenever $a\lt b$ for all values a and b in the interval.
Decreasing function A function f is said to be decreasing over an interval if $f(a)\gt f(b)$ whenever $a\lt b$ for all values a and b in the interval.
Even function
A function f is said to be even if $f(-x)=f(x)$ for all values of x in its domain. The graph of an even function in the Cartesian coordinate system is symmetric by a reflection about the y-axis.
Odd function
A function f is said to be odd if $f(-x)=-f(x)$ for all values of x in its domain. The graph of an odd function in the Cartesian coordinate system is symmetric by a rotation of 180° about the origin.
## Examples and Discussion
### Function
Do not equate "function" with "rule"!
A function definition might include an explicit rule of assignment, but it needn't. A function can also be defined conceptually, as in the example already given: "Let $\text{dist}(x)$ be the number of meters from start run by Usain Bolt x seconds after the start in the final heat of the 2012 Summer Olympics." In this race, Usain Bolt was exactly one distance from start at each moment during the race. We can say this even though we do not have an explicit rule to determine what number of meters he was from start at any moment during the race. The expression $\text{dist}(3.27)$ has a definite meaning and a definite value even though, at this moment, we have no way to determine the value.
There is a way to say "Any value of x is associated with exactly one value of y" formally:
Suppose you are told there is a relationship between values of x and y named g and that g is a function from x to y. If you are later told $g(a)\ne g(b)$, then you can conclude automatically that $a\ne b$.
How does this capture "... exactly one ..."? Because if $g(a)\ne g(b)$ while $a=b$, then this says $g(a)\ne g(a)$ (since $a=b$). One value of x would be associated with two values of y.
Take $g(x)=\pm \sqrt x$ for $x\ge 0$. We have, for example, $g(4)=-2$ and $g(4)=2$. By definition, g is not a function from x to y even though it relates values of x and y. There is at least one value of x (namely, 4) that is associated with two values of y, and therefore $x=4$ is not associated by g with exactly one value of y.
### One-to-one Function
Here is another way to say a function f is 1-1: f is 1-1 if every value in the range of f is associated with exactly one value in the domain of f.
### Discontinuous Functions
The idea of a function f being discontinuous at $x=a$ is that there is a "gap" in values of $f(x)$ for values of x around $x=a$.
Put another way, there is some level of tolerance for which you cannot approximate a value of $f(a)$ with the assurance that every value of $f(x)$ will be within this tolerance for all values of x in any interval containing $a$.
A simple example is $$f(x)= \left\{ \begin{array}{ll} x & \text{if } x \leq 1 \\[1ex] x+0.0000001 & \text{if } x \gt 1 \end{array} \right.$$ If we set our tolerance at 0.00000005 and let $x_0=1$, then no interval around $x_0=1$ will have all values of x in it having $f(x)$ within this tolerance of $f(1)$.
Figure 3.17.1 illustrates how graphs can be deceiving. The graph of $y=f(x)$ (f defined as above) appears to be continuous around $x=1$. The graph shows a vertical line at $x=1$ and two horizontal lines, one at $y=1.0000005$ and one at $y=0.99999995$. These blue lines show a tolerance of 0.0000005 around $f(1)=1$.
Figure 3.17.1. A function whose graph at one scale suggests it is continuous everywhere turns out to be discontinuous at $x=1$. No interval around $x=1$ has all values $f(x)$ within 0.0000005 of $f(1)$ for values of x in the interval.
### Continuous Functions
A function f is continuous over an interval if there is no value $x_0$ of x in the interval at which $f(x_0)$ is discontinuous. Put loosely, there are no "gaps" in the values of $f(x)$ over the interval.
Notice that we speak of discontinuity at a value of x, but we speak of continuity over an interval of values of x. The idea of discontinuity indeed involves intervals, but discontinuity happens at specific values in an interval.
If a function is continuous over an interval, we say it is continuous at each value in the interval.
### Increasing Functions
A more colloquial statement is, "A function is increasing over an interval if the value of the function increases as the value of its argument increases."
An equivalent, but less common, way to say a function increases over an interval is that the value of the function decreases as the value of its argument decreases.
### Decreasing Functions
A more colloquial statement is, "A function is decreasing over an interval if the value of the function decreases as the value of its argument increases."
An equivalent, but less common, way to say a function decreases over an interval is that the value of the function increases as the value of its argument decreases.
## Activities
In these activities you will see a video showing values of x related to values of y by a function named f. Each video is followed by questions for you to answer.
A note about the logic of explaining an answer of "No" in these activities: You must support a "No" answer to a question about a property by giving one specific example where the property is not satisfied.
For example, to justify "no" as your answer to the question, "Is f one-to-one over its domain?" you must state two specific values of x, say 2 and 4, and state that $f(2)=f(4)$ while $2\ne 4$. A property fails to hold for a relationship when it fails once.
### Activity 3.17.1
The animation below shows values of x being related to values of y by a relationship named f. Play the video, then answer the questions following the video using meanings for terms in the question. Click here to remind yourself of terms' meanings. You can replay the video and pause it, or scroll the play bar to examine parts of the video. (Move your pointer away from the video to remove the scroll bar.)
1. What, approximately, is the domain of f?
2. What, approximately, is the range of f?
3. Is the relationship named f a function from x to y? Explain.
4. Is f continuous over its domain? Explain.
5. Is f one-to-one over its domain? Explain.
6. Over what interval(s) in its domain, if any, is f increasing? Explain.
7. Over what interval(s) in its domain, if any, is f decreasing? Explain.
### Activity 3.17.2
The animation below shows values of x being related to values of y by a relationship named f. Play the video, then answer the questions following the video using meanings for terms in the question. Click here to remind yourself of terms' meanings. You can replay the video and pause it, or scroll the play bar to examine parts of the video. (Move your pointer away from the video to remove the scroll bar.)
1. What, approximately, is the domain of f?
2. What, approximately, is the range of f?
3. Is the relationship named f a function from x to y? Explain.
4. Is f continuous over its domain? Explain.
5. Is f one-to-one over its domain? Explain.
6. Over what interval(s) in its domain, if any, is f increasing? Explain.
7. Over what interval(s) in its domain, if any, is f decreasing? Explain.
### Activity 3.17.3
The animation below shows values of x being related to values of y by a relationship named f. Play the video, then answer the questions following the video using meanings for terms in the question. Click here to remind yourself of terms' meanings. You can replay the video and pause it, or scroll the play bar to examine parts of the video. (Move your pointer away from the video to remove the scroll bar.)
1. What, approximately, is the domain of f?
2. What, approximately, is the range of f?
3. Is the relationship named f a function from x to y? Explain.
4. Is f continuous over its domain? Explain.
5. Is f one-to-one over its domain? Explain.
6. Over what interval(s) in its domain, if any, is f increasing? Explain.
7. Over what interval(s) in its domain, if any, is f decreasing? Explain.
### Activity 3.17.4
The animation below shows values of x being related to values of y by a relationship named f. Play the video, then answer the questions following the video using meanings for terms in the question. Click here to remind yourself of terms' meanings. You can replay the video and pause it, or scroll the play bar to examine parts of the video. (Move your pointer away from the video to remove the scroll bar.)
1. What, approximately, is the domain of f?
2. What, approximately, is the range of f?
3. Is the relationship named f a function from x to y? Explain.
4. Is f continuous over its domain? Explain.
5. Is f one-to-one over its domain? Explain.
6. Over what interval(s) in its domain, if any, is f increasing? Explain.
7. Over what interval(s) in its domain, if any, is f decreasing? Explain.
## Exercise Set 3.17
1. Sketching graphs in the Cartesian coordinate system:
1. Sketch a graph of the relationship between values of x and values of y for the video in Activity 1.
2. Sketch a graph of the relationship between values of x and values of y for the video in Activity 2.
3. Sketch a graph of the relationship between values of x and values of y for the video in Activity 3.
4. Sketch a graph of the relationship between values of x and values of y for the video in Activity 4.
2. The animation below simulates a car driving on a race course. The car traverses the course 4 times. Quantities mentioned in a-c are in the order dependent quantity followed by independent quantity.
1. Consider the car's number of miles driven from start in relation to number of minutes elapsed since starting. Is this relationship a function? Is it continuous over its domain? Is it 1-1 over its domain?
2. Consider the car's position on the track in relation to number of minutes elapsed since starting. Is this relationship a function? Is it continuous over its domain? Is it 1-1 over its domain?
3. Consider the car's straight-line distance from start in relation to distance driven from start. Is this relationship a function? Is it continuous over its domain? Is it 1-1 over its domain?
3. You are told the graph of $y=h(x)$ is symmetric about the line $x=4$.
1. Define a function j so that $h(j(x))$ is even. Try your general solution with $h(x)=x^4-16x^3+96x^2-256x+260$. (Click here to get a GC file already set up for you.
2. Suppose the graph of $y=k(x)$ is symmetric about the line $x=c$. Define a function j so that $k(j(x))$ is even. Explain your answer.
4. Play the video shown below or scroll through it by moving the play bar.
1. What, approximately, is the domain of f?
2. What, approximately, is the range of f?
5. Let h be defined as $h(x)=1/(2-x)$.
1. What is the domain of h? The range of h?
2. Is h continuous over its domain? Remember, for h to be discontinuous at a value $x_0$ in its domain there must be a gap in y-values over an interval containing $x_0$ and the interval must be contained in h's domain.
6. State whether each function is even, odd, or neither even nor odd. Explain your answers. Graph $y=\_\_(x)$ and $y=\_\_(-x)$ [insert the function name] in GC if you are unsure of your answer.
1. $f(x)=\sin(x)$
2. $g(x)=\cos(x)$
3. $h(x)=\tan(x)$ Recall: $\tan(x)=\dfrac{\sin(x)}{\cos(x)},\, \cos(x)\ne 0$
4. $j(x)=x^2-4x+4$
5. $k(x)=x\sin(x)$
6. $m(x)=x\cos(x)$
7. $n(x)=3x+1$
7. The function f is even. The function g is odd. Both are defined for all real numbers. What can you conclude about:
1. $h(x)=f(x)+g(x)$
2. $k(x)=f(x)g(x)$
3. $l(x)=f(g(x))$
4. $m(x)=g(f(x))$
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# Areas of Parallelograms and Triangles
## Class 9 NCERT Maths
### NCERT
1 Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.
##### Solution :
(i)Yes.It can be observed that trapezium $ABCD$ and triangle $PCD$ have a common base $CD$ and these are lying between the same parallel lines $AB$ and $CD$.$\\$
(ii)No. It can be observed that parallelogram $PQRS$ and trapezium $MNRS$ have a common base RS. However, their vertices, (i.e., opposite to the common base) $P, Q$ of parallelogram and $M, N$ of trapezium, are not lying on the same line.$\\$
(iii)Yes. It can be observed that parallelogram $PQRS$ and triangle $TQR$ have a common base $QR$ and they are lying between the same parallel lines $PS$ and $QR.$$\\ (iv)No. It can be observed that parallelogram ABCD and triangle PQR are lying between same parallel lines AD and BC. However, these do not have any common base.\\ (v)Yes. It can be observed that parallelogram ABCD and parallelogram APQD have a common base AD and these are lying between the same parallel lines AD and BQ.\\ (vi)No. It can be observed that parallelogram PBCS and PQRS are lying on the same base PS. However, these do not lie between the same parallel lines. 2 In the given figure, ABCD is parallelogram, AE \perp DC and CF \perp AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD. ##### Solution : In parallelogram ABCD, CD = AB = 16 cm$$\\$ [Opposite sides of a parallelogram are equal] We know that$\\$ Area of a parallelogram = Base * Corresponding altitude$\\$ Area of parallelogram $ABCD = CD * AE = AD * CF$$\\ 16 cm * 8 cm = AD * 10 cm$$\\$ $AD=\dfrac{16*8}{10} cm =12.8 cm$$\\ Thus, the length of AD is 12.8 cm. 3 If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD show that ar (EFGH)=\dfrac{1}{2} ar (ABCD) ##### Solution : Let us join HF.$$\\$ In parallelogram $ABCD,$$\\ AD = BC and AD \parallel BC (Opposite sides of a parallelogram are equal and parallel)\\ AB = CD (Opposite sides of a parallelogram are equal)\\ \Rightarrow \dfrac{1}{2}AD=\dfrac{1}{2}BC and AH\parallel BF$$\\$ $\Rightarrow AH = BF$ and $AH \parallel BF$ ( $H$ and $F$ are the mid-points of $AD$ and $BC$)$\\$ Therefore, $ABFH$ is a parallelogram.$\\$ Since $\Delta HEF$ and parallelogram $ABFH$ are on the same base $HF$ and between the same parallel lines $AB$ and $HF,$$\\ \therefore Area (\Delta HEF) = \dfrac{1}{2} Area (ABFH)....(1)$$\\$ Similarly, it can be proved that$\\$ Area $(\Delta HGF) =\dfrac{1}{2}$ Area $(HDCF) ... (2)$$\\ On adding Equations (1) and (2), we obtain\\ Area (\Delta HEF) + Area (\Delta HGF)$$ \\$ $=\dfrac{1}{2}$Area $(ABFH) + \dfrac{1}{2}$Area $(HDCF)$$\\ =\dfrac{1}{2}[Area (ABFH) + Area (HDCF)]$$\\$ $\Rightarrow$ Area $(EFGH) =\dfrac{1}{2}$Area $(ABCD)$
4 $P$ and $Q$ are any two points lying on the sides $DC$ and $AD$ respectively of a parallelogram $ABCD$. Show that ar $(APB) = ar (BQC).$
##### Solution :
It can be observed that $\Delta BQC$ and parallelogram $ABCD$ lie on the same base $BC$ and these are between the same parallel lines $AD$ and $BC$.$\\$ $\therefore$ Area$(\Delta BQC)=\dfrac{1}{2}$Area$(ABCD)....(1)$$\\ Similarly, \Delta APB and parallelogram ABCD lie on the same base AB and between the same parallel lines AB and DC.\\ \therefore Area (\Delta APB)=\dfrac{1}{2}Area (ABCD)...(2)$$\\$ From Equations (1) and (2), we obtain$\\$ Area $(\Delta BQC)$ = Area $(\Delta APB)$
##### Solution :
Given: $A \Delta ABC, AD$ is the median and $E$ is the mid-point of median $AD.$
To prove:$ar(\Delta BED) = 1/4 ar(\Delta ABC)$ $\\$ Proof : In $\Delta ABC, AD$ is the median.$\\$ $\therefore ar (\Delta ABD) = ar (\Delta ADC)$ $\\$ [$therefore$ Median divides a $\Delta$ into two $\Delta s$ of equal area]$\\$ $ar(\Delta ABD) = \dfrac{1}{2} ar(ABC)$ .....(i)$\\$ In $\Delta ABD, BE$ is the median.$\\$ $ar (\Delta BED) = ar (\Delta BAE)$ $\\$ $\therefore ar (\Delta BED)= = \dfrac{1}{2}ar (\Delta BED)$ $\\$ $=ar(\Delta ABD) =\dfrac{1}{2}[ \dfrac{1}{2}ar (\Delta ABC)] = \dfrac{1}{2 }ar (\Delta ABC)$ $\\$
9 Show that the diagonals of a parallelogram divide it into four triangles of equal area.
##### Solution :
We know that diagonals of parallelogram bisect each other.$\\$ Therefore,$O$ is the mid-point of $AC$ and $BD.$ $\\$ $BO$ is the median in $\Delta ABC.$ Therefore, it will divide it into two triangles of equal areas.$\\$ $\therefore$ Area ($\Delta AOB) =$ Area ($\Delta BOC)$ ... (1)$\\$ In $\Delta BCD, CO$ is the median.$\\$ $\therefore$ Area ($\Delta BOC)$ = Area ($\Delta COD)$ ... (2)$\\$ Similarly, Area ($\Delta COD)$ = Area ($\Delta AOD)$ ... (3)$\\$ From Equations (1), (2), and (3), we obtain $\\$ Area ($\Delta AOB) =$ Area ($\Delta BOC)$ = Area ($\Delta COD)$ = Area ($\Delta AOD)$ $\\$ Therefore, it is evident that the diagonals of a parallelogram divide it into four triangles of equal area.
10 In the given figure, $ABC$ and $ABD$ are two triangles on the same base $AB$. If line-segment $CD$ is bisected by $AB$ at $O,$ show that $ar (ABC) = ar (ABD).$
##### Solution :
Consider $\Delta ACD.$ $\\$ Line-segment $CD$ is bisected by $AB$ at $O.$ Therefore, $AO$ is the median of $\Delta ACD.$ $\\$ $\therefore$ Area ($\Delta ACO) =$ Area ($\Delta ADO)$ ... (1)$\\$ Considering $\Delta BCD, BO$ is the median.$\\$ $\therefore$ Area ($\Delta BCO)$ = Area ($\Delta BDO)$ ... (2)$\\$ Adding Equations (1) and (2), we obtain$\\$ Area ($\Delta ACO)$+ Area $(\Delta BCO)$ = Area ($\Delta ADO)$ + Area ($\Delta BDO)$ $\\$ $\Rightarrow$ Area ($\Delta ABC)$ = Area ($\Delta ABD)$
11 $D, E$ and $F$ are respectively the mid-points of the sides $BC, CA$ and $AB$ of a $\Delta ABC$ . Show that: $\\$ (i)$BDEF$ is a parallelogram.$\\$ (ii) $ar (DEF) = \dfrac{1}{4}ar (ABC)$ $\\$ (iii) $ar (BDEF) = \dfrac{1}{2}ar (ABC)$
##### Solution :
(i) $F$ is the mid-point of $AB$ and $E$ is the mid-point of $AC.$ $\\$ $\therefore FE||BC$ and $FE =\dfrac{1}{2} BD$ $\\$ Line joining the mid-points of two sides of a triangle is parallel to the third and half of It$\\$ $\therefore FE||BD [BD$ is the part of $BC]$ $\\$ And $FE = BD$ $\\$ Also, $D$ is the mid-point of $BC.$$\\ B D = \dfrac{1}{2} B C \\ And FE||BC and FE = BD \\ Again E is the mid-point of AC and D is the mid-point of BC. \therefore DE||AB and DE = \dfrac{1}{2} AB \\ DE||AB [BF is the part of AB] And DE = BF \\ Again F is the mid-point of AB.$$\\$ $\therefore B F = \dfrac{1}{2} A B$ But$DE =\dfrac{ 1}{2} AB$ $\\$ $\therefore DE = BF$ $\\$ Now we have $FE||BD$ and $DE||BF$ And $FE = BD$ and $DE = BF$ $\\$ Therefore, $BDEF$ is a parallelogram.$\\$
(ii) $BDEF$ is a parallelogram.$\\$ $\therefore ar(\Delta BDF) = ar(\Delta DEF) ...........(i)$ $\\$ [diagonals of parallelogram divides it in two triangles of equal area] $DCEF$ is also parallelogram.$\\$ $\therefore ar (\Delta DEF) = ar (\Delta DEC) ..........(ii)$ $\\$ Also, $AEDF$ is also parallelogram.$\\$ $\therefore ar (\Delta AFE) = ar (\Delta DEF) ..........(iii)$ $\\$ From eq. (i), (ii) and (iii),$\\$ $ar (\dfrac DEF) = ar (\Delta BDF) = ar (\Delta DEC) = ar (\Delta AFE) ..........(iv)$ $\\$ Now, $ar (\delta ABC) = ar (\Delta DEF) + ar (\Delta BDF) + ar (\Delta DEC) + ar (\Delta AFE) ..........(v)$ $\\$ $ar (\Delta ABC) = ar (\Delta DEF) + ar (\Delta DEF) + ar (\Delta DEF) + ar (\Delta DEF)$ $\\$ [Using (iv) & (v)]$\\$ $ar (\Delta ABC) = 4 × ar (\Delta DEF)$ $\\$ $ar (\Delta DEF) = \dfrac{1}{4} ar (\Delta ABC)$ $\\$ (iii) $ar (||gm BDEF) = ar (\Delta BDF) + ar (\Delta DEF) = ar (\Delta DEF) + ar (\Delta DEF)$[Using (iv)]$\\$ $ar (||gm BDEF) = 2 ar (\Delta DEF)$ $\\$ $ar (||gm BDEF) = 2× \dfrac{1}{4} ar (\Delta ABC)$ $\\$ $ar (||gm BDEF) = \dfrac{1}{2} ar (\Delta ABC)$
12 In the given figure, diagonals $AC$ and $BD$ of quadrilateral $ABCD$ intersect at $O$ such that $OB = OD$. If $AB = CD$, then show that:$\\$ (i) $ar (DOC) = ar (AOB)$ $\\$ (ii) $ar (DCB) = ar (ACB)$ $\\$ (iii) $DA || CB$ or $ABCD$ is a parallelogram.$\\$ [Hint: From $D$ and $B$, draw perpendiculars to $AC.$]
15 The side $AB$ of a parallelogram $ABCD$ is produced to any point $P$. A line through $A$ and parallel to $CP$ meets $CB$produced at $Q$ and then parallelogram $PBQR$ is completed (see the following figure). Show that$\\$$ar (ABCD) = ar (PBQR). \\[Hint: Join AC and PQ. Now compare area (ACQ) and area (APQ)] ##### Solution : Let us join AC and PQ.$$\\$ $\Delta ACQ$ and $\Delta AQP$ are on the same base $AQ$ and between the same parallels $AQ$ and $CP.$ $\\$ Area $(\Delta ACQ)$ = Area $(\Delta APQ$)$\\$ Area$(\Delta ACQ) -$ Area $(\Delta ABQ)$ = Area $(\Delta APQ) -$ Area $(\Delta ABQ)$ Area $(\Delta ABC)$ = Area $(\Delta QBP)$... (1)$\\$ Since $AC$ and $PQ$ are diagonals of parallelograms $ABCD$ and $PBQR$ respectively, $\\$ Area$(\Delta ABC) = \dfrac{1}{2}$ Area $(ABCD)$... (2)$\\$ Area $(\Delta QBP) = \dfrac{1}{2}$ Area $(PBQR) ... (3)$ $\\$ From Equations (1), (2), and (3), we obtain$\\$ $\dfrac{1}{2}$ Area $(ABCD) = \dfrac{1}{2}$ Area $(PBQR)$ $\\$ Area$(ABCD)$= Area $(PBQR)$
16 Diagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB || DC$ intersect each other at $O.$ Prove that $ar (AOD) = ar (BOC).$
##### Solution :
It can be observed that $\Delta DAC$ and $\Delta DBC$ lie on the same base $DC$ and between the same parallels $AB$ and $CD.$ $\\$ Area $(\Delta DAC)$= Area $(\Delta DBC)$ $\\$ Area $(\Delta DAC) -$ Area $(\Delta DOC)$= Area $(\Delta DBC) -$ Area $(\delta DOC)$$\\ Area (\Delta AOD) = Area (\Delta BOC) 17 In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that\\ (i) ar (ACB) = ar (ACF)$$\\$ (ii) $ar (AEDF) = ar (ABCDE)$
(i) $\Delta ACB$ and $\Delta ACF$ lie on the same base $AC$ and are between The same parallels $AC$ and $BF.$$\\ Area (\Delta ACB) = Area (\Delta ACF)$$\\$ (ii) It can be observed that$\\$ Area $(\Delta ACB)$= Area $(\Delta ACF)$$\\ Area (\Delta ACB) + Area (ACDE) = Area (\Delta ACF) + Area (ACDE) \\ Area (ABCDE) = Area (AEDF) 18 A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented. ##### Solution : Let quadrilateral ABCD be the original shape of the field.\\ The proposal may be implemented as follows.\\ Join diagonal BD and draw a line parallel to BD through point A.$$\\$ Let it meet the extended side $CD$ of $ABCD$ at point $E.$$\\ Join BE and AD. Let them intersect each other at O.$$\\$ Then, portion $\Delta AOB$ can be cut from the original field so that the new shape of the field will be $\Delta BCE.$ (See figure).$\\$ We have to prove that the area of $\Delta AOB$ (portion that was cut so as to construct Health Centre) is equal to the area of $\Delta DEO$ (portion added to the field so as to make the area of the new field so formed equal to the area of the original field).
It can be observed that $\Delta DEB$ and $\Delta DAB$ lie on the same base $BD$ and are between the same parallels $BD$ and $AE.$$\\ Area (\Delta DEB) = Area (\Delta DAB) \\ Area (\Delta DEB) - Area (\Delta DOB) = Area (\Delta DAB) - Area (\Delta DOB) \\ Area (\Delta DEO) = Area (\Delta AOB) 19 ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).$$\\$ [Hint: Join $CX.$]
##### Solution :
It can be observed that $\Delta ADX$ and $\Delta ACX$ lie on the same base $AX$ and are between the same parallels $AB$ and $DC.$$\\ Area (\Delta ADX) = Area (\Delta ACX) ... (1)\\ \Delta ACY and \Delta ACX lie on the same base AC and are between the same parallels AC and XY.\\ Area (\Delta ACY) = Area (ACX) ... (2)$$\\$ From Equations (1) and (2), we obtain $\\$ Area $(\Delta ADX)$ = Area $(\Delta ACY)$
20 In the given figure, $AP || BQ || CR.$ Prove that $ar (AQC) = ar (PBR).$
##### Solution :
Since $\Delta ABQ$ and $\delta PBQ$ lie on the same base $BQ$ and are between the same parallels $AP$ and $BQ,$$\\ \therefore Area (\Delta ABQ) = Area (\Delta PBQ) ... (1)\\ Again, \Delta BCQ and \Delta BRQ lie on the same base BQ and are between the same parallels BQ and CR.$$\\$ $\therefore$ Area $(\Delta BCQ)$ = Area $(\Delta BRQ)$ ... (2)$\\$ On adding Equations (1) and (2), we obtain$\\$ Area $(\Delta ABQ)$ + Area $(\Delta BCQ)$ = Area $(\Delta PBQ)$ + Area $(\Delta BRQ)$ $\\$ $\therefore$ Area $(\Delta AQC)$ = Area $(\Delta PBR)$
21 Diagonals $AC$ and $BD$ of a quadrilateral $ABCD$ intersect at $O$ in such a way that $ar (AOD) = ar (BOC)$. Prove that $ABCD$ is a trapezium.
##### Solution :
It is given that$\\$ Area $(\Delta AOD)$= Area $(\Delta BOC)$ $\\$ Area $(\Delta AOD)$+ Area $(\Delta AOB)$= Area $(\Delta BOC)$ + Area $(\Delta AOB)$ $\\$ Area $(\Delta ADB)$= Area $(\Delta ACB)$ $\\$ We know that triangles on the same base having areas equal to each other lie between the same parallels.$\\$ Therefore, these triangles, $\Delta ADB$ and $\Delta ACB,$ are lying between the same parallels.$\\$ i.e., $AB || CD$ $\\$ Therefore, $ABCD$ is a trapezium.
22 In the given figure, $ar (DRC) = ar (DPC)$ and $ar (BDP) = ar (ARC)$. Show that both the quadrilaterals $ABCD$ and $DCPR$ are trapeziums.
##### Solution :
It is given that$\\$ Area $(\Delta DRC)$ = Area $(\Delta DPC)$ $\\$ As $\Delta DRC$ and $\delta DPC$ lie on the same base $DC$ and have equal areas, therefore, they must lie between the same parallel lines.$\\$ $\therefore DC || RP$ $\\$ Therefore, $DCPR$ is a trapezium.$\\$ It is also given that$\\$ Area $(\Delta BDP)$= Area $(\Delta ARC)$ $\\$ Area $(\Delta BDP) -$ Area $(\Delta DPC)$ = Area $(\Delta ARC) -$ Area $(\Delta DRC)$ $\\$ $\therefore$ Area $(\delta BDC)$ = Area $(\Delta ADC)$ $\\$ Since $\Delta BDC$ and $\Delta ADC$ are on the same base $CD$ and have equal areas, they must lie between the same parallel lines.$\\$ $\therefore AB || CD$ $\\$ Therefore, $ABCD$ is a trapezium.
23 Parallelogram $ABCD$ and rectangle $ABEF$ are on the same base $AB$ and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
##### Solution :
As the parallelogram and the rectangle have the same base and equal area, therefore, these will also lie between the same parallels. Consider the parallelogram $ABCD$ and rectangle $ABEF$ as follows.
Here, it can be observed that parallelogram $ABCD$ and rectangle $ABEF$ are between the same parallels $AB$ and $CF.$$\\ We know that opposite sides of a parallelogram or a rectangle are of equal lengths. Therefore,\\ AB = EF (For rectangle)\\ AB = CD (For parallelogram)\\ \therefore CD = EF \\ \therefore AB + CD = AB + EF ... (1)\\ Of all the line segments that can be drawn to a given line from a point not lying on it, the perpendicular line segment is the shortest.\\ \therefore AF < AD \\ And similarly, BE < BC$$\\$ $\therefore AF + BE < AD + BC$ ... (2)$\\$ From Equations (1) and (2), we obtain$\\$ $AB + EF + AF + BE < AD + BC + AB + CD$$\\ Perimeter of rectangle ABEF < Perimeter of parallelogram ABCD. 24 In the following figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC). \\ Can you now answer the question that you have left in the ’Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?\\ [Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide \Delta ABC into n triangles of equal areas.] ##### Solution : Let us draw a line segment AL \perp BC. Let us draw a line segment AL \perp BC. \\ We know that,\\ Area of a triangle = 12 × Base × Altitude\\ Area (\Delta ADE)= 12 ×DE×AL \\ Area (\Delta ABD)= 12 ×BD×AL \\ Area(\Delta AEC)= 12 ×EC×AL \\ It is given that DE = BD = EC \\ 12 ×DE×AL= 12 ×BD×AL= 12 ×EC×AL \\ Area (\Delta ADE) = Area (\Delta ABD) = Area (\Delta AEC) It can be observed that Budhia has divided her field into 3 equal parts. 25 In the following figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (\Delta ADE) = ar (\Delta BCF). ##### Solution : It is given that ABCD is a parallelogram. We know that opposite sides of a parallelogram are equal.\\ \therefore AD = BC ... (1)\\ Similarly, for parallelograms DCEF and ABFE, it can be proved that\\ DE = CF ... (2)$$\\$ And, $EA = FB ... (3)$$\\ In \Delta ADE and \Delta BCF, \\ AD = BC [Using equation (1)]\\ DE = CF [Using equation (2)]\\ EA = FB [Using equation (3)]\\ \therefore \Delta ADE \cong \Delta BCF (SSS congruence rule) \therefore Area (\Delta ADE) = Area (\Delta BCF) 26 In the following figure, ABCD is parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (\Delta BPC) = ar (\Delta DPQ). ##### Solution : It is given that ABCD is a parallelogram. AD || BC and AB || DC(Opposite sides of a parallelogram are parallel to each other) Join point A to point C. Consider \Delta APC and \Delta BPC$$\\$ $\Delta APC$ and $\Delta BPC$ are lying on the same base $PC$ and between the same parallels $PC$ and $AB$. Therefore,$\\$ Area $(\Delta APC)$ = Area $(\delta BPC)$ ... (1)$\\$ In quadrilateral $ACDQ,$ it is given that $AD = CQ$$\\ Since ABCD is a parallelogram,\\ AD || BC (Opposite sides of a parallelogram are parallel)\\ CQ is a line segment which is obtained when line segment BC is produced. \\ \therefore AD || CQ \\ We have,\\ AC = DQ and AC|| DQ \\ Hence, ACQD is a parallelogram.\\ Consider BDCQ and BACQ$$\\$ These are on the same base $CQ$ and between the same parallels $CQ$ and $AD.$ Therefore,$\\$ Area $(\Delta DCQ)$= Area $(\Delta ACQ)$$\\ \therefore Area (\Delta DCQ) - Area (\Delta PQC) = Area (\Delta ACQ) - Area (\Delta PQC)$$\\$ $\therefore$Area $(\Delta DPQ)$ = Area $(\Delta APC)$ ... (2)$\\$ From equations (1) and (2), we obtain$\\$ Area $(\Delta BPC)$ = Area$(\Delta DPQ)$
27 In the following figure, $ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC$. If $AE$ intersects $BC$ at $F,$ show that$\\$ (i) $ar (BDE) = \dfrac{1}{4} ar (ABC)$$\\ (ii) ar (BDE) = \dfrac{1}{2} ar (BAE)$$\\$ (iii) $ar (ABC) = 2 ar (BEC)$$\\ (iv) ar (BFE) = ar (AFD)$$\\$ (v) $ar (BFE) = 2 ar (FED)$$\\ (vi) ar (FED) = \dfrac{1}{8} ar (AFC)$$\\$ [Hint: Join $EC$ and $AD.$ Show that $BE || AC$ and $DE || AB$ etc.]
##### Solution :
(i) Let $G$ and $H$ be the mid-points of side $AB$ and $AC$ respectively. $\\$ Line segment $GH$ is joining the mid-points and is parallel to third side. $\\$ Therefore, $BC$will be half of the length of $BC$ (mid-point theorem).
$\therefore GH=\dfrac{1}{2} BC$ and $GH || BD$ $\\$ $\therefore GH = BD = DC$ and $GH || BD (D$ is the mid-point of $BC)$ $\\$ Similarly,$\\$ $\bullet GD=HC=HA$$\\ \bullet HD=AG=BG$$\\$ Therefore, clearly $\Delta ABC$ is divided into $4$ equal equilateral triangles viz $\Delta BGD, \Delta AGH, \Delta DHC$ and $\delta GHD$ $\\$ In other words, $\Delta BGD = \dfrac{1}{4} \Delta ABC$ $\\$ Now consider $\Delta BDG$ and $\Delta BDE$ $\\$ $BD = BD$(Common base)$\\$ As both triangles are equilateral triangle, we can say $BG = BE$$\\ DG = DE Therefore, \delta BDG \cong \Delta BDE [By SSS congruency]\\ Thus, area (\Delta BDG) = area (\Delta BDE) \\ ar (\Delta BDE) = \dfrac{1}{4} ar (\Delta ABC) Hence proved (ii) Area (\Delta BDE) = Area (\Delta AED) (Common base DE and DE || AB)$$\\$ Area $(\Delta BDE) -$ Area $(\Delta FED)$ $\\$ = Area $(\Delta AED) -$ Area $(\Delta FED)$ $\\$ Area $(\Delta BEF)$ = Area $(\Delta AFD)$ ... (1)$\\$ Now, Area $(\Delta ABD)$ = Area $(\Delta ABF)$ + Area $(\Delta AFD)$ $\\$ Area $(\Delta ABD)$ = Area $(\Delta ABF)$ + Area $(\Delta BEF)$ [From equation (1)]$\\$ Area$(\Delta ABD)$ = Area $(\Delta ABE) AD$ is the median in $\Delta ABC$
$ar (\delta ABD)=\dfrac{1}{2}ar(\Delta ABC)$ $\\$ $=\dfrac{4}{2}ar(\Delta BDE) \text{(As proved earlier)}$ $\\$ $ar (\Delta ABD)=2ar (\Delta BDE) \qquad (3)$ $\\$ From(2) and (3) , we obtain $\\$ $2 ar(\Delta BDE)=ar(\Delta ABE)$ $\\$ $ar(BDE)=\dfrac{1}{2}ar(BAE)$ $\\$ (iii)$ar(\Delta ABE)=ar(\Delta BEC)$ (Common base $BE$ and $BE||AC$)$\\$ $ar(\Delta ABF)+ar(\Delta BEF)=ar(\Delta BEC)$ $\\$ Using equation (1),we obtain $\\$ $ar(\Delta ABF)+ar(\Delta AFD)=ar(\Delta BEC)$ $\\$ $\dfrac{1}{2}ar(\Delta ABC)=ar(\Delta BEC)$ $\\$ $ar (\Delta ABC)=2 ar(\Delta BEC)$
(iv) It is seen that $\Delta BDE$ and $ar \Delta AED$ lie on the same base $(DE)$ and between the parallels $DE$ and $AB$ $\\$ $\therefore ar (\Delta BDE) = ar (\Delta AED)$ $\\$ $\therefore ar (\Delta BDE) - ar (\Delta FED)$ $\\$ $= ar (\Delta AED) - ar (\Delta FED)$ $\\$ $\therefore ar (\Delta BFE) = ar (\Delta AFD)$ $\\$ (v) Let $h$ be the height of vertex $E,$ corresponding to the side $BD$ in $\Delta BDE$. $\\$ Let $H$ be the height of vertex $A$, corresponding to the side $BC$ in $\Delta ABC.$ $\\$ In (i), it was shown that $ar (BDE) = \dfrac{1}{4} ar (ABC)$ $\\$ In (iv), it was shown that $ar (\Delta BFE) = ar (\Delta AFD).$ $\\$ $\therefore ar (\Delta BFE) = ar (\Delta AFD)$ $\\$ $= 2 ar (\Delta FED)$ Hence,$\\$ (vi) $ar( \Delta AFC)=ar(\Delta AFD)+ar( \Delta ADC)$ $\\$ $=2ar(\Delta FED)+ \dfrac{1}{2 }ar( \Delta ABC)$ [using(v) =$2ar( FED)+\dfrac{ 1}{2} [4 ar( \Delta BDE)]$ [Usingresultofpart(i)]$\\$ $=2ar( \Delta FED)+2ar(\Delta BDE)=2ar(\Delta FED)+2ar(\Delta AED)$ [ $\Delta BDE$ and $\Delta AED$ are on the same base and between same parallels] $\\$ $=2ar(\Delta FED)+2[ar(\Delta AFD)+ar(\Delta FED)]$ $\\$ $=2ar( \Delta FED)+2ar(\Delta AFD)+2ar(\Delta FED)$ [Using(viii)]$\\$ $=4ar( \Delta FED)+4ar(\Delta FED)$$\\ ar(\Delta AFC)=8ar(\Delta FED) \\ ar(\Delta FED)= \dfrac{1}{8} ar( AFC) 28 Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that \\[Hint: From A and C, draw perpendiculars to BD] ##### Solution : Given : A quadrilateral ABCD, in which diagonals AC and BD intersect each other at point E. To Prove : ar(\Delta AED)* ar(\Delta BEC) \\ =ar(\Delta ABE)* ar(\Delta CDE) \\ \text{Construction:} From A,draw AM\perp BD AM BD and fromC, draw CN\perp BD \\ Proof : ar(\Delta ABE)=\dfrac{1}{2}*BE*AM..........(1) \\ ar(\Delta AED )=\dfrac{1}{2}*DE * AM...........(2) \\ Dividing eq. (ii) by (i),we get , \\ \dfrac{ar(\Delta AED)}{ar(\Delta ABE)}=\dfrac{\dfrac{1}{2}*DE*AM}{\dfrac{1}{2}* BE * AM} \\ \dfrac{ar(\Delta AED)}{ar(\Delta ABE)}=\dfrac{DE}{BE}......(iii) \\ Similarly \dfrac{ar(\Delta AED)}{ar(\Delta ABE)}=\dfrac{DE}{BE}.........(iv) \\ From eq.(iii) and (iv) , we get \\ \dfrac{ar(\Delta AED)}{ar(\Delta ABE)}=\dfrac{ar(\Delta CDE)}{ar(\Delta BEC)} \\ \implies ar(\Delta AED)*ar(\Delta BEC)=ar(\Delta ABE)*ar(\Delta CDE)$$\\$ Hence proved.
29 $P$ and $Q$ are respectively the mid-points of sides $AB$ and $BC$ of a triangle $ABC$ and $R$ is the mid-point of $AP,$ show that$\\$ (i) $ar (PRQ) = \dfrac{1}{2} ar (ARC)$ $\\$ (ii)$ar(RQC)=\dfrac{3}{8}ar (ABC)$ $\\$ (iii)$ar(PBQ)=ar(ARC)$
##### Solution :
(i) $PC$ is the median of $\Delta ABC.$ $\\$ $ar (\Delta BPC) = ar (\Delta APC) ..........(i)$ $\\$ $RC$ is the median of $APC.$$\\ ar (\Delta ARC) = \dfrac{1}{2} ar ( \Delta APC) ..........(ii)$$\\$ [Median divides the triangle into two triangles of equal area] $\\$ $PQ$ is the median of $\Delta BPC$.$\\$
$\therefore ar (\Delta PQC) = \dfrac{1}{2} ar (\Delta BPC) ..........(iii)$ $\\$ From eq. (i) and (iii), we get,$\\$ $ar (\Delta PQC) = \dfrac{1}{2} ar ( \Delta APC) ..........(iv)$ $\\$ From eq. (ii) and (iv), we get,$\\$ $ar ( \Delta PQC) = ar (\Delta ARC) ..........(v)$ $\\$ We are given that $P$ and $Q$ are the mid-points of $AB$ and $BC$ respectively.$\\$ $PQ || AC$ and $PA=\dfrac{1}{2}AC$ $\\$ $ar ( \Delta APQ) = ar (\Delta PQC) ..........(vi)$ $\\$[triangles between same parallel are equal in area]$\\$ From eq. (v) and (vi), we get$\\$ $ar (\Delta APQ) = ar (\Delta ARC) ..........(vii)$ $\\$ $R$ is the mid-point of $AP.$ Therefore $RQ$ is the median of $\Delta APQ.$ $\\$ $ar ( \Delta PRQ) =\dfrac{ 1}{2} ar (\Delta APQ) ..........(viii)$ $\\$ From (vii) and (viii), we get,$\\$ $ar(\Delta PRQ)=\dfrac{ 1}{2} ar(\Delta ARC)$ $\\$
(ii) $PQ$ is the median of $\Delta BPC$ $\\$ $\therefore ar(\Delta PQC)= \dfrac{1}{2 } ar(\Delta BPC)$ $\\$ $= \delta{1}{2}×\dfrac{1}{2} ar(\Delta ABC)= \dfrac{1}{4} ar( ABC)..........(ix)$$\\ Also ar ( \Delta PRC) = \dfrac{1}{2} ar (\Delta APC) [Using (iv)]\\ ar(\Delta PRC)=\dfrac{ 1}{2}×\dfrac{1}{2}ar( \Delta ABC)= \dfrac{1}{4} ar(\Delta ABC)..........(x) \\ Adding eq. (ix) and (x), we get,\\ ar( \Delta PQC)+ar( \Delta PRC)= (\dfrac{1}{4}+\dfrac{1}{4})ar(\Delta ABC) \\ \implies ar (quad. PQCR) =\dfrac{ 1}{2} ar (\Delta ABC) ..........(xi) \\ Subtracting ar (\Delta PRQ) from the both sides,\\ ar(quad.PQCR)-ar(\Delta PRQ)=\dfrac{ 1}{2} ar(\Delta ABC)-ar(\Delta PRQ) \\ ar( \Delta RQC)=\dfrac{ 1}{2} ar( \Delta ABC)-\dfrac{ 1}{2} ar(\Delta ARC)[Usingresult(i)]\\ ar(\Delta ARC)=\dfrac{ 1}{2 }ar(\Delta ABC)-\dfrac{ 1}{2}×\dfrac{1}{2} ar(\Delta APC) \\ ar( \Delta RQC)=\dfrac{ 1}{2} ar(\Delta ABC)-\dfrac{ 1}{4} ar(\Delta APC) \\ ar(\Delta RQC)=\dfrac{ 1}{2 }ar(\Delta ABC)-\dfrac{ 1}{4}×\dfrac{1}{2} ar(\Delta ABC)[PC is median of ABC] \\ ar(\Delta RQC)=\dfrac{ 1}{2} ar( \Delta ABC)- \dfrac{1}{8} ar(\Delta ABC) \\ ar( \Delta RQC)=(\dfrac{1}{2}-\dfrac{1}{8})* ar(\Delta ABC) \\ ar( \Delta RQC)=\dfrac{ 8}{3} ar(\Delta ABC) \\ (iii)ar(\Delta PRQ)= \dfrac{1}{2} ar(\Delta ARC) [Using result (i) ]\\ 2ar(\Delta PRQ) = ar(\Delta ARC)..(xii) \\ ar (\Delta PRQ) = \dfrac{1}{2} ar (\Delta APQ) [RQ is the median of \Delta APQ] ..........(xiii) \\ But ar (\Delta APQ) = ar (\Delta PQC) [Using reason of eq. (vi)]\\ ..........(xiv) From eq. (xiii) and (xiv), we get,\\ ar (\Delta PRQ) = \dfrac{1}{2} ar (\Delta PQC) ..........(xv)\\ But ar (\Delta BPQ) = ar (\Delta PQC) [PQ is the median of BPC] ..........(xvi) \\ From eq. (xv) and (xvi), we get,\\ ar (\Delta PRQ) = \dfrac{1}{2} ar (\Delta BPQ) ..........(xvii) \\ Now from (xii) and (xvii), we get,\\ 2×\dfrac{1}{2}ar(\Delta BPQ)=ar(\Delta ARC) \\ ar(\Delta BPQ)=ar(\Delta ARC) 30 In the following figure, ABC is a right triangle right angled at A.$$ BCED, ACFG$ and $ABMN$ are squares on the sides $BC, CA$ and $AB$ respectively. Line segment $AX \perp DE$ meets $BC$ at $Y.$ Show that:$\\$ (i) $\Delta MBC \cong \Delta ABD$ $\\$ (ii) $ar(BYXD) = 2ar(MBC)$ $\\$ (iii) $ar(BYXD) = 2ar(ABMN)$ $\\$ (iv) $\Delta FCB \cong \Delta ACE$ $\\$ (v) $ar(CYXE) = 2ar(FCB)$ $\\$ (vi) $ar(CYXE) = ar(ACFG)$ $\\$ (vii) $ar(BCED) = ar(ABMN) + ar(ACFG)$ $\\$ Note: Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in class X.
##### Solution :
(i) We know that each angle of a square is $90^o.$ $\\$ Hence, $\angle ABM = \angle DBC = 90^o$ $\\$ $\therefore \angle ABM + \angle ABC = \angle DBC + \angle ABC$ $\\$ $\therefore \angle MBC = \angle ABD$ $\\$ In $\Delta MBC$ and $\Delta ABD,$ $\\$ $\angle MBC = \angle ABD$ (Proved above)$\\$ $MB = AB$ (Sides of square $ABMN$)$\\$ $BC = BD$ (Sides of square $BCED$)$\\$ $\therefore \Delta MBC \cong \Delta ABD$ ($SAS$ congruence rule)$\\$ (ii) We have $\Delta MBC \cong \Delta ABD$ $\\$ $\therefore ar (\Delta MBC) = ar (\Delta ABD) ... (1)$ $\\$ It is given that $AX \perp DE$ and $BD \perp DE$ (Adjacent sides of square $BDEC$)$\\$ $\therefore BD || AX$(Two lines perpendicular to same line are parallel to each other)$\\$ $\Delta ABD$ and parallelogram $BYXD$ are on the same base $BD$ and between the same parallels $BD$ and $AX.$ $\\$ Area $(\Delta YXD) = 2$ Area $(\Delta MBC)$ [Using equation (1)] ... (2)$\\$
(iii) $\Delta MBC$ and parallelogram $ABMN$ are lying on the same base $MB$ and between same parallels $MB$ and $NC.$$\\ 2 ar (\Delta MBC) = ar (ABMN)$$\\$ $ar (\Delta YXD) = ar (ABMN)$ [Using equation (2)] ... (3)$\\$ (iv) We know that each angle of a square is $90^o$ $\\$ $\therefore \angle FCA = \angle BCE = 90^o$ $\\$ $\therefore \angle FCA + \angle ACB =\angle BCE + \angle ACB$ $\\$ $\therefore \angle FCB = \angle ACE$ $\\$ In $\Delta FCB$ and $\Delta ACE,$ $\\$ $\angle FCB = \angle ACE$ $\\$ $FC = AC$(Sides of square $ACFG$)$\\$ $CB = CE$(Sides of square $BCED$)$\\$ $\Delta FCB \cong \Delta ACE (SAS$ congruence rule)$\\$
(v) It is given that $AX \perp DE$ and $CE \perp DE$ (Adjacent sides of square $BDEC)$$\\ Hence, CE || AX(Two lines perpendicular to the same line are parallel to each other)\\ Consider BACE and parallelogram CYXE$$\\$ $BACE$ and parallelogram $CYXE$ are on the same base $CE$ and between the same parallels $CE$ and $AX.$ $\\$ $\therefore ar (\Delta YXE) = 2 ar (\Delta ACE) ... (4)$ $\\$ We had proved that $\\$ $\therefore \Delta FCB \cong \Delta ACE$ $\\$ $ar (\Delta FCB) \cong ar (\Delta ACE) ... (5)$ $\\$ On comparing equations (4) and (5), we obtain$\\$ $ar (CYXE) = 2 ar (\Delta FCB) ... (6)$$\\ (vi) Consider BFCB and parallelogram ACFG \\ BFCB and parallelogram ACFG are lying on the same base CF and between the same parallels CF and BG.$$\\$ $\therefore ar (ACFG) = 2 ar (\Delta FCB)$ $\\$ $therefore ar (ACFG) = ar (CYXE)$ [Using equation (6)] ... (7)$\\$ (vii) From the figure, it is evident that$\\$ $ar (\Delta CED) = ar (\Delta YXD) + ar (CYXE)$ $\\$ $\therefore ar (\Delta CED) = ar (ABMN) + ar (ACFG)$ [Using equations (3) and (7)]
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# Evolutes
The locus of the centers of curvature of a given curve is called the evolute of that curve. Consider the circle of curvature corresponding to a point on a curve. If moves along the given curve, we may suppose the corresponding circle of curvature to roll along the curve with it, its radius varying so as to be always equal to the radius of curvature of the curve at the point . The curve in Figure 14.5 described by the center of the circle is the evolute of .
Figure 14.5: Geometric visualization of an evolute.
It is instructive to make an approximate construction of the evolute of a curve by estimating (from the shape of the curve) the lengths of the radii of curvature at different points on the curve and then drawing them in and drawing the locus of the centers of curvature.
Formula (14.7) gives the coordinates of any point on the evolute expressed in terms of the coordinates of the corresponding point of the given curve. But is a function of ; therefore
give us at once the parametric equations of the evolute in terms of the parameter x.
To find the ordinary rectangular equation of the evolute we eliminate between the two expressions. No general process of elimination can be given that will apply in all cases, the method to be adopted depending on the form of the given equation. In a large number of cases, however, the student can find the rectangular equation of the evolute by taking the following steps:
General directions for finding the equation of the evolute in rectangular coordinates.
• FIRST STEP. Find from (14.9).
• SECOND STEP. Solve the two resulting equations for and in terms of and .
• THIRD STEP. Substitute these values of and in the given equation. This gives a relation between the variables and which is the equation of the evolute.
Example 14.4.1 Find the equation of the evolute of the parabola .
Figure 14.6: Evolute of a parabola.
Solution. .
First step. , .
Second step. , .
Third step ; or, .
Remembering that denotes the abscissa and the ordinate of a rectangular system of coordinates, we see that the evolute of the parabola is the semi-cubical parabola ; the centers of curvature for , , , being at , , , respectively.
Example 14.4.2 Find the equation of the evolute of the parabola .
Figure 14.7: Evolute of an ellipse.
Solution. , .
First step. , .
Second step. , .
Third step. , the equation of the evolute of the ellipse , , , , are the centers of curvature corresponding to the points , , , , on the curve, and , , correspond to the points , , .
When the equations of the curve are given in parametric form, we proceed to find and , as in §12.5, from
(14.12)
and then substitute the results in formulas (14.9). This gives the parametric equations of the evolute in terms of the same parameter that occurs in the given equations.
Example 14.4.3 The parametric equations of a curve are
(14.13)
Find the equation of the evolute in parametric form, plot the curve and the evolute, find the radius of curvature at the point where , and draw the corresponding circle of curvature.
Solution. , , , . Substituting in above formulas (14.12) and then in (14.9), gives
(14.14)
the parametric equations of the evolute. Assuming values of the parameter , we calculate , ; from (14.13) and (14.14). Now plot the curve and its evolute.
Figure 14.8: Evolute of an parametric curve.
The point is common to the given curve and its evolute. The given curve (a semi-cubical parabola) lies entirely to the right and the evolute entirely to the left of .
The circle of curvature at , where , will have its center at on the evolute and radius . To verify our work, find radius of curvature at . From (12.5), we get
when . This should equal the distance
Example 14.4.4 Find the parametric equations of the evolute of the cycloid,
(14.15)
Figure 14.9: Evolute of a cycloid.
Solution. As in Example 12.5.2, we get
Substituting these results in formulas (14.9), we get the answer:
(14.16)
Remark 14.4.1 In the previous example, if we eliminate between equations (14.16), there results the rectangular equation of the evolute referred to the axes and . The coordinates of with respect to these axes are . Let us transform equations (14.16) to the new set of axes and . Then
Substituting in (14.16) and reducing, the equations of the evolute become
(14.17)
Since (14.17) and (14.14) are identical in form, we have: The evolute of a cycloid is itself a cycloid whose generating circle equals that of the given cycloid.
david joyner 2008-11-22
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Latest Banking jobs » Quantitative Aptitude Quiz For IBPS Clerk...
Quantitative Aptitude Quiz For IBPS Clerk Prelims 2021- 11th August
Directions (1-5): Study the bar chart given below and answer the following questions.
Bar chart shows the total votes (in ‘000) in 5 different cities (A, B, C, D & E) and percentage of valid votes out of total votes in these 5 cities.
Q1. In A, ratio of valid votes received by BJP, INC & SP is 11 : 3 : 5. If BJP win the election by 24000 votes, then find valid votes received by INC & SP together.
(a) 28000
(b) 40000
(c) 45000
(d) 32000
(e) 21000
Q2. If in D there are only two parties-BJP & INC and BJP got 70% of the total valid votes, then find by how many votes BJP won the election in D.
(a) 18000
(b) 25000
(c) 10000
(d) 16000
(e) 6000
Q3. If in E INC got 15000 more valid votes than AAP and BJP won the election by 15000 votes, then find the valid votes received by BJP. (There are only three parties in E – BJP, INC & AAP)
(a) 40000
(b) 38000
(c) 45000
(d) 35000
(e) 30000
Q4. In B there are only two parties – BJP & INC. If BJP got 60% of total votes in B and ratio of invalid votes received by BJP and INC is 2 : 1, then find valid votes received by INC in B.
(a) 24000
(b) 27000
(c) 22000
(d) 20000
(e) 25000
Q5. In C there are four parties – BJP, INC, SP & AAP. If ratio of valid votes received by BJP, INC, SP & AAP in C is 4 : 2 : 3 : 3 and ratio of invalid votes received by BJP, INC, SP & AAP in C is 1 : 3 : 4 : 2, then find difference between total votes received by INC & SP in C.
(a) 2000
(b) 3500
(c) 3400
(d) 2100
(e) 2700
Directions (6-10):- Given bar graph shows the details of number of students in a particular class of 3 different schools in 5 different years.
Q6.What is the difference between average number of students of school A across all the years and the average number of students of school B across all the years?
(a) 18
(b) 10
(c) 12
(d) 14
(e) 16
Q7.Find the respective ratio of the total number of students of school A in 2011 and 2012 together to the total number of students of school C in 2013 and 2014 together?
(a) 31:33
(b) 47:55
(c) 55:47
(d) 33:31
(e) 31:37
Q8.If in 2016, the total number of students in School A, School B and School C increases by 10%,20% and 15% respectively as compared to 2015, then find the total number of students in 2016 in all the schools together?
(a) 850
(b) 870
(c) 780
(d) 830
(e) 800
Q9.Total students of all the school together in 2013 is approximately what percentage more/less than the total students of school B in 2011 and 2015 together?
(a) 52%
(b) 59%
(c) 56%
(d) 63%
(e) 48%
Q10.Find the difference between the number of total students from all the schools in 2011 and 2013 together and the total number of students from all the schools in 2014 and 2015 together?
(a) 140
(b) 60
(c) 120
(d) 80
(e) 100
Directions (11-15): Given bar graph shows the percentage distribution of total number of students of each school (P, Q, R & S) who took admission in 3 different streams. Total students in P, Q, R & S are 700, 800, 400 & 900 respectively.
Q11.What is average number of students who have opted for MBBS in all the 4 colleges?
(a) 256
(b) 233
(c) 284
(d) 224
(e) 296
Q12.What is the ratio of the total number of student who have opted for both engg. and MBBS stream together in college Q to that of in same stream together in college R?
(a) 38:65
(b) 67:35
(c) 35:67
(d) 65:38
(e) 29:37
Q13. The number of student who have opted for MBBS in college P is what percent of the number of students who have opted for the engg. in college Q?
(a) 87.5%
(b) 50%
(c) 75%
(d) 100%
(e) 62.5%
Q14.What is the ratio of the no. of students who have opted for engg. in college R to that of those who have opted for same stream in college P?
(a) 14:11
(b) 17:13
(c) 11:14
(d) 13:17
(e) None of these
Q15.Which of the combination represents the colleges with maximum number of students, who have opted for pharmacy and those who have opted for engg. respectively?
(a) P & R
(b) Q & S
(c) Q & R
(d) R & S
(e) P & Q
Solutions
Congratulations!
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# Simple harmonic motion - positive or negative acceleration? watch
1. In questions like the following, how do you know when acceleration is positive or negative when you're only told that displacement is a certain value, since it could be in either direction of the equilibrium?
A simple pendulum is given a small displacement from its equilibrium position and performs simple harmonic motion.
Calculate the frequency of the oscillations of a simple pendulum of length 984 mm.
Calculate the acceleration of the bob of the simple pendulum when the displacement from the equilibrium position is 42 mm.
So I worked out the frequency and got 0.504 Hz which is correct, but in the second part I got -0.42, but the mark scheme says it's positive 0.42.
I used the equation a = -(2pif)^2(x)
If x is positive, it's impossible for a to be positive isn't it? Since the square of anything is positive?
2. The mark scheme is just giving the examiner the magnitude of the acceleration to check your numerical answer.
#You will not lose a mark over the minus sign. In fact, there is nothing wrong with your answer. Be careful of mark schemes, they are not model answers for students, they are notes for the examiner.
3. (Original post by Stonebridge)
The mark scheme is just giving the examiner the magnitude of the acceleration to check your numerical answer.
#You will not lose a mark over the minus sign. In fact, there is nothing wrong with your answer. Be careful of mark schemes, they are not model answers for students, they are notes for the examiner.
Ah ok, thanks!
4. (Original post by jf1994)
Ah ok, thanks!
One thing worth thinking about here, though, is if you understand the significance of that minus sign in this context.
The acceleration at that point is certainly "0.42 ms-2" but, for example, can it be determined from the information given whether that means the pendulum is speeding up or slowing down?
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How much mass will be converted to energy to accelerate the spaceship
1. May 1, 2013
AyooNisto
1. The problem statement, all variables and given/known data
i put my answers to the questions in bold
A spaceship and its occupants have a total mass of 1.8×105kg . The occupants would like to travel to a star that is 30 light-years away at a speed of 0.70c. To accelerate, the engine of the spaceship changes mass directly to energy.
How much mass will be converted to energy to accelerate the spaceship to this speed?
Assume the acceleration is rapid, so the speed for the entire trip can be taken to be 0.70c, and ignore decrease in total mass for the calculation. How long will the trip take according to the astronauts on board?
2. Relevant equations
[Broken]
3. The attempt at a solution
part a)
substituting the numebrs into the first equation
√(1-(0.70)2) = .7141428429
1/.7141428429 = 1.400280084
1.400280084 - 1 = .400280084moc2
.40(180000kg) = 72000kg
part b
t = d/v
(30y)c/0.70c = 42.9 y
42.9y = tΔ/√(1-(0.70)2) = tΔ0 = 30.6y
Last edited by a moderator: May 6, 2017
2. May 1, 2013
cepheid
Staff Emeritus
Please include units in your calculations, otherwise they are totally meaningless. (I'm not just being pedantic here: it's actually impossible to know what the calculations mean without units).
You have the right basic idea, but I think you need to be a little more careful. You have assumed that the mass of the ship is constant, when in fact it changes. The equation for the relativistic energy (rest energy PLUS kinetic) is indeed$$E = \gamma mc^2$$where $\gamma = (\sqrt{1-v^2/c^2})^{-1}$. So, the difference in energy between moving (gamma is [STRIKE]non zero[/STRIKE] greater than 1 ) and not moving (gamma is [STRIKE]zero[/STRIKE] 1) would indeed be $\Delta E = KE = (\gamma-1)mc^2$ IF the ship's mass were not changing. BUT the ship's mass is changing, so the difference in energy is actually $\Delta E = \gamma m_1c^2 - m_0c^2$ where m1 is the mass after reaching 0.7c and m0 is the original mass. So, the algebra to solve for m1 here is a little more involved.
Last edited: May 1, 2013
3. May 1, 2013
AyooNisto
i have included the units i hope that helps
4. May 1, 2013
cepheid
Staff Emeritus
Okay...but did you read any of the rest of my post? I pointed out that you were doing something wrong. You need to correct it.
Also, I edited my first post to correct a stupid mistake about the gammas. I did so transparently.
5. May 1, 2013
BruceW
You have calculated the final kinetic energy of the spaceship, given that the mass of the spaceship has not changed. But as cepheid says, they want you to do the calculation given that the rest mass of the spaceship is used up to provide the speed increase. I think the best way to do the problem is to think of what they want you to assume stays the same in this problem. (Usually it is the rest mass that stays the same, but this is not true in this problem).
6. May 1, 2013
haruspex
Isn't it quite a bit more involved? With conventional propellants, you have to consider how the mass changes over time, yielding a logarithmic relationship. Doesn't something similar apply here?
E.g., if you think about the momentum viewed from an inertial frame, and the acceleration comes from emitting radiation, as the craft accelerates the wavelength increases for the observer, reducing the momentum per photon.
7. May 1, 2013
BruceW
since the question doesn't give an exhaust velocity or specific impulse, I think we are meant to assume that all the energy is contained in either the rest mass of the ship or the KE of the ship. In other words, there is no propellant and momentum is not conserved at all. In this case, it is not very involved, but maybe kindof useless, since in real problems we will always have conservation of momentum.
edit: maybe he is meant to assume that the 'propellant' is just EM radiation. Then it would be possible to solve the problem (with conservation of momentum). But this would take some time to do from first principles... more time than I would expect from the way the question is worded. So I don't think this is what he is meant to do.
8. May 1, 2013
rcgldr
I was trying to edit my post, but it got fouled up and I deleted it to I create another post.
The first part of the post asks about the amount of mass required to accelerate the ship, ingoring mass loss. So it seems it's asking for how much mass if converted into energy equals the kinetic energy of the space ship. It's not clear if energy added to the "exhaust" plume is to be considered.
The second part of the post is asking about time, how long will the trip take at 0.70 C, apparently from a reference frame of the space ships initial velocity, which would make it's intial speed zero, before it accelerated to 0.70 C.
Last edited: May 1, 2013
9. May 1, 2013
BruceW
ah... that looks like what the OP did. But I interpreted the question differently. I essentially interpreted it as "what is the required change in rest mass of the spaceship, such that it can get to that velocity, given that total energy is constant (no energy is 'added'), and assuming no propellant, i.e. momentum not conserved"
I guess either interpretation is possible... so maybe I shouldn't have been so quick to disagree with the OP's original answer.
10. May 1, 2013
AkInfinity
mass doesnt increase, the correct phenomena is impedance aka relative mass. Just puting it out there since many people get confused with mass (the amount of matter in an object) and impedance (inertia; relative mass) which are both used interchengably as mass.
11. May 2, 2013
BruceW
Not necessarily. the rest-mass stays constant only if a pure four-force acts on the object. Otherwise, generally the rest-mass will change. And by the way part (b) in the question is worded, it looks like the rest-mass is supposed to change in part (a)
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# Math help from the Learning Centre
This guide provides useful resources for a wide variety of math topics. It is targeted at students enrolled in a math course or any other Centennial course that requires math knowledge and skills.
## What is Math?
Somehow it's okay for people to chuckle about not being good at math. Yet if I said "I never learned to read," they'd say I was an illiterate dolt. Neil Degrasse Tyson
Math is more than just solving variables or calculating numbers. It promotes skills you can apply to many contexts in every day living. Here are some skills inherent in learning mathematics.
You Sniff out Patterns while working on Mathematics
The analysis of current and historical events, all require one to be on the lookout for patterns. When you make decisions, you are behaving through a pattern, understanding those patterns develop your awareness.
In mathematics, patterns exist in every calculation. For example, understanding that multiplying a whole number sequentially starting from 1 (e.g., $$7\times 1=7$$, $$7\times 2=14$$, $$7\times 3=21$$, $$7\times 4=28$$) is the same as adding by 7 every time. Understanding these patterns help you connect multiplication with addition. There are hidden pattern such as square of integers between 1 and 100 (e.g., $$1^2$$,$$2^2$$,$$3^2$$,...$$100^2$$). Students should always be on the lookout for patterns. The search for regularity extend to daily lives.
You Learn to Experiment
When faced with a mathematical problem, a good habit is to immediately start playing with it. Simple ideas include recording results, keeping all but one variable fixed, and trying very small or large numbers. There are mental experiments, such as trying to perform operations without writing anything down.
One should also develop skepticism for experimental results, realizing the inaccuracies as one experiments.
You Learn to Describe
Describing precisely what you do is an important step in understanding. Mathematical sophistication comes from the ability to say what you mean. One way to see the utility and elegance of mathematical formulations is to struggle with problems which ordinary language descriptions are too cumbersome, forcing one to precisely describe in a way for someone else to understand. Making convincing arguments to your peers develops your reasoning abilities by articulating evidence of your convictions. Writing down thoughts, conjectures, arguments, questions, and opinions are useful in developing your skills to describe mathematics. Through description, you learn to collaborate with others.
You Learn to Tinker
Taking ideas apart and putting them back together allows one to see possibilities of how things can be put back in different ways. After experimenting with a rotation followed by a translation, you can wonder if the same applies to a translation followed by a rotation.
You Learn to Invent
Tinkering with ideas, apps, or machines leads to expertise in building new ones. Inventions range from rules for a game, steps to doing things, explanations of how things work, or mathematical theorems. Good inventions give the impression of being innovative. For example, good rules for a game make it intriguing for everyone and have internal consistencies that make sense. For example, if chess players had to do five jumping jacks before each move, the rule does not fit with the rest of the game and not many people would stand for it. Although it may seem that mathematics is full of rules and formulas, they emerge from the experiences of the inventors and arise from their attempts to bring clarity to situations. Experiencing the situations that gave rise to the rules and formulas allows one to develop meaning for algorithms.
You Learn to Visualize
Mathematics can be visualized. Algebra is a representation of what is inherently visual. Multiplication of two values can be visualized as an application of area. One purpose is to aid understanding of the process. Visual reasoning is applied, every time you drive, estimate how far an object is, and when you play games. It also helps you perform mental calculations.
You Learn to Conjecture and Guess
Making plausible conjectures takes time to develop. It is central not only in mathematics. Conjectures are driven by the experiences and evidence you gather from inventing, experimenting, and tinkering, and need to be tested out.
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FeedbackTank Volume Calculator - Inch Calculator
For example, lets find the volume of a cylinder tank that is 36 in diameter and 72 long. radius = 36 ÷ 2 radius = 18 tank volume = × 18 2 × 72 tank volume = 73,287 cu in. Thus, the capacity of this tank is 73,287 cubic inches. How to calculate the Vertical Tank?How to calculate the Vertical Tank?In the case of the vertical cylindrical tank, you need to perform the same type of measurement. However, since the tank is standing upright rather than lying on its side, you would replace the total length of the tank by the total height of the tank. Thus, the final calculation becomes the following V(tank) = r2h.Tank Volume Calculator What is the volume of a cylinder shaped tank?What is the volume of a cylinder shaped tank?Methods to calculate the volume of tanks and the volume of a liquid inside a tank. These calculations will give you cubic measures such as ft 3 or m 3 depending on your units of measure. Horizontal Cylinder Tank. Total volume of a cylinder shaped tank is the area, A, of the circular end times the length, l.Tank Volume Calculator
What is the volume of a vertical cylinder?What is the volume of a vertical cylinder?Total volume of a cylinder shaped tank is the area, A, of the circular end times the height, h. A = r 2 where r is the radius which is equal to d/2. Therefore The filled volume of a vertical cylinder tank is just a shorter cylinder with the same radius, r, and diameter, d, but height is now the fill height or f.Tank Volume Calculator - Online Calculator Resource* TankCalc
R increases the tank's size in the Y (vertical) and Z (depth) dimensions. A tank with a zero R entry will not have a volume, regardless of the other entries. The left and right end caps are fully (and separately) described with a dimensional variable r that describes the reach of the end cap along the X 01-02 Water flowing into cylindrical tank MATHalinoProblem 02 Water flows into a vertical cylindrical tank at 12 ft 3 /min, the surface rises 6 in/min. Find the radius of the tank.. Solution 02
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Example Inputting liquid level = 3, diameter = 24, tank length = 30, then clicking "Inches" will display the total tank volume in cubic inches and US Gallons and will also show the volume at the 3 inch level. In addition, a dipstick chart is automatically generated. The default dipstick chart is in increments of one but you can change that by clicking on one of the increment buttons.Tank Volume Calculator - ibec language instituteTank Body Height (restated from above) gallons liters in Liquid Volume in 3 ft 3 in gallons liters Overall Tank Height (calculated) Freeboard at this Liquid Level in in *Program does not calculate liquid volume into upper head on vertical tanks (mixing with liquid in head space is not good practice).Tank Volume Calculator - ibec language instituteTank Body Height (restated from above) gallons liters in Liquid Volume in 3 ft 3 in gallons liters Overall Tank Height (calculated) Freeboard at this Liquid Level in in *Program does not calculate liquid volume into upper head on vertical tanks (mixing with liquid in head space is not good practice).
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Entering information in the following forms create a tank chart for metal rectangular and flat-end cylindrical liquid storage tanks. Horizontal Liquid Storage Tank Charts. Ft. In. Length Diameter Tank Chart Details 1 Inch Increments 1/8th Inch Increments Number of Columns Vertical Tank Charts. Ft. In. Height Diameter Tank Chart Details japan vertical cylindrical tank building volumeTank Charts - Hall Tank CompanyUse this form to generate a chart of tank capacities. Hall Tank does not guarantee the capacity charts accuracy and in no way takes liability for loss due to its content. Calculating a chart will be considered acceptance of this agreement.Tank Charts - Hall Tank CompanyUse this form to generate a chart of tank capacities. Hall Tank does not guarantee the capacity charts accuracy and in no way takes liability for loss due to its content. Calculating a chart will be considered acceptance of this agreement.
Volume Calculator
This free volume calculator can compute the volumes of common shapes, including that of a sphere, cone, cube, cylinder, capsule, cap, conical frustum, ellipsoid, and square pyramid. Explore many other math calculators like the area and surface area calculators, as well as hundreds of other calculators related to finance, health, fitness, and more.Volume of a cylinder with calculator - Math Open ReferenceVolume of a partially filled cylinder. One practical application is where you have horizontal cylindrical tank partly filled with liquid. Using the formula above you can find the volume of the cylinder which gives it's maximum capacity, but you often need to know the volume of liquid in the tank API 650 ABOVEGROUND STORAGE TANKS, Part I Code japan vertical cylindrical tank building volumeThis standard covers the design, manufacture and installation of vertical aboveground cylindrical tanks, steel fabricated and welded, with nominal capacities ranging from 79.5 m3 to 1590 m3 (in standard sizes). 1.1.6) API 12F The requirements of this standard are similar to API 12D, in this case for tanks
Analysis of a draining tank - Olin
A tank of water is drained with siphon. The cylindrical tank has a constant cross sectional area, A, and volume, V. The draining tube has diameter, D, and length L. The end of the tube and the top of the water are separate in vertical distance by a height, h(t). The water that leaves the tank Chapter 2. Secondary Containment Facilitythrough 2.3 show how to calculate the volumes of horizontal, cylindrical, vertical, and cone-bottom tanks. ()() 2 0.5 21 2 1 2 2 Horizontal cylindrical tank fluid volume (center section of tank) D2hD h 1 L h n D V Di h s 84 D2 Spherical tank fluid volume (end sections of tank) Vh1.5Dh 3 Total tank Chapter 2. Secondary Containment Facilitythrough 2.3 show how to calculate the volumes of horizontal, cylindrical, vertical, and cone-bottom tanks. ()() 2 0.5 21 2 1 2 2 Horizontal cylindrical tank fluid volume (center section of tank) D2hD h 1 L h n D V Di h s 84 D2 Spherical tank fluid volume (end sections of tank) Vh1.5Dh 3 Total tank
Choosing the Right Water Storage for Your Community is an japan vertical cylindrical tank building volume
The purpose of water storage tanks is usually to meet peak demands, such as fire flows and times of the day when water use is high. There are a few items to consider when selecting a new water storage tank for your community or industry. Traditionally, it has been a common practice of many waterCircular Cylinder Rectangular PrismTank Volume Calculator - Vertical Cylindrical Tanks - MetricVertical cylindrical tank volume calculator diagram Fraction Precision Set 1/8 1/16 1/32 1/64 Decimal All Inch inputs and dimensions are actual physical finished sizes (unless otherwise noted)Effect of settlement of foundations on the failure risk of japan vertical cylindrical tank building volumeSep 30, 2019Hotaa E., Ignatowicz R. (2018). Failure risk of bottoms of cylindrical steel tanks supported on ring foundations. Building Materials 4/2018, p. 88-90, (in polish) [5] Wu T.Y., Liu G.R. (2000). Comparison of design methods for a tank-bottom annular plate and concrete ringwall. International Journal of Pressure Vessels and Piping 77, p. 511-517 japan vertical cylindrical tank building volume
Highland Tank - Gauge Charts
Please note that these charts are theoretical and are intended as a guide for estimating tank/vessel volumes. Required Choose Tank Style Horizontal Cylindrical Horizontal Cylindrical (Elliptical Heads) Horizontal Rectangular Vertical Flat Bottom Vertical Dished Bottom Vertical Coned BottomHighland Tank - Gauge ChartsPlease note that these charts are theoretical and are intended as a guide for estimating tank/vessel volumes. Required Choose Tank Style Horizontal Cylindrical Horizontal Cylindrical (Elliptical Heads) Horizontal Rectangular Vertical Flat Bottom Vertical Dished Bottom Vertical Coned BottomHorizontal Cylindrical Tank Volume and Level CalculatorVolume calculation on a partially filled cylindrical tank Some Theory. Using the theory. Use this calculator for computing the volume of partially-filled horizontal cylinder-shaped tanks.With horizontal cylinders, volume changes are not linear and in fact are rather complex as the theory above shows. Fortunately you have this tool to do the work for you.
Horizontal Cylindrical Tank Volume and Level Calculator
Volume calculation on a partially filled cylindrical tank Some Theory. Using the theory. Use this calculator for computing the volume of partially-filled horizontal cylinder-shaped tanks.With horizontal cylinders, volume changes are not linear and in fact are rather complex as the theory above shows. Fortunately you have this tool to do the work for you.Images of Japan vertical Cylindrical tank Building volume imagesTank Volume Calculator - Oil TanksThe tank size calculator on this page is designed for measuring the capacity of a variety of fuel tanks. Alternatively, you can use this tank volume calculator as a water volume calculator if you need to calculate some specific water volume. The functionality of this LPG Storage Bullet Tanks vs. LPG Storage Spheres japan vertical cylindrical tank building volumeBecause of this, the overall coating/painting cost per unit of volume is also lower for spherical vessels than for cylindrical tanks. Finally, the spherical shape also offers the most uniform stress resistance, which allows for a thinner wall. Thus materials cost per unit of volume is also lower for spherical vessels.
Mixing 101 Optimal Tank Design Dynamix Agitators
Vertical cylindrical tanks are the most common type of tank in use. A key consideration for cylindrical tanks is to ensure that they are either baffled or offset-mounted to prevent swirling from occurring. Refer to section 2 below (The Use of Baffling) for details. Generally baffles are not required for smaller tanks (<5,000 gallons in japan vertical cylindrical tank building volumeNumerical study of pressure build-up in vertical tanks for japan vertical cylindrical tank building volumeOct 01, 2019The tank CFD model is preliminary validated against small-scale experimental data obtained for cryogenic nitrogen and then extended to the simulation of an industrial cylindrical tank, whose volume is 100 m 3. The effect of fluid, i.e. ethylene and LNG (modelled as pure methane), filling level and possible insulation damage, on natural japan vertical cylindrical tank building volumeNumerical study of pressure build-up in vertical tanks for japan vertical cylindrical tank building volumeOct 01, 2019The tank CFD model is preliminary validated against small-scale experimental data obtained for cryogenic nitrogen and then extended to the simulation of an industrial cylindrical tank, whose volume is 100 m 3. The effect of fluid, i.e. ethylene and LNG (modelled as pure methane), filling level and possible insulation damage, on natural japan vertical cylindrical tank building volume
PROCESS FABRICATORS LUu)
containment type tanks are available in both vertical and horizontal configurations, with a maximum capacity of 30,000 gallons. Transport Tanks Process Fabricators, Inc. fabricates transport tanks in a wide variety of shapes and sizes. * Obround truck tanks * Cylindrical truck tanks * Rectangular truck tanks * Trailer tanks * Pallet tanksPeople also askWhat is horizontal cylinder tank volume?What is horizontal cylinder tank volume?Horizontal Cylinder Tank Total volume of a cylinder shaped tank is the area, A, of the circular end times the length, l. A = r 2 where r is the radius which is equal to 1/2 the diameter or d/2.Tank Volume CalculatorRelated searches for japan vertical cylindrical tank building vertical cylindrical tank volume calculatorvolume of cylindrical tankhorizontal cylindrical tank volume chartcylindrical tank volume charthorizontal cylindrical tank volume formulahorizontal cylindrical tank volume calculatorcalculate volume cylindrical tankcylindrical tank volume formulaSome results are removed in response to a notice of local law requirement. For more information, please see here.
Related searches for japan vertical cylindrical tank building
vertical cylindrical tank volume calculatorvolume of cylindrical tankhorizontal cylindrical tank volume chartcylindrical tank volume charthorizontal cylindrical tank volume formulahorizontal cylindrical tank volume calculatorcalculate volume cylindrical tankcylindrical tank volume formulaSome results are removed in response to a notice of local law requirement. For more information, please see here.Horizontal Cylindrical Tank Volume Calculator - MetricVertical Cylindrical Tank Volume Rectangular Tank Volume Directory Inch Inch Inch Inch Horizontal Cylindrical Tank Volume Calculator, Dip Chart and Fill Times - Metric. Hemispherical Ends Ellipsoidal Ends - adj Ellipsoidal Ends - 2:1 Flat Ends Straight LengthRelated searches for japan vertical cylindrical tank building vertical cylindrical tank volume calculatorvolume of cylindrical tankhorizontal cylindrical tank volume chartcylindrical tank volume charthorizontal cylindrical tank volume formulahorizontal cylindrical tank volume calculatorcalculate volume cylindrical tankcylindrical tank volume formulaSome results are removed in response to a notice of local law requirement. For more information, please see here.Horizontal Cylindrical Tank Volume Calculator - MetricVertical Cylindrical Tank Volume Rectangular Tank Volume Directory Inch Inch Inch Inch Horizontal Cylindrical Tank Volume Calculator, Dip Chart and Fill Times - Metric. Hemispherical Ends Ellipsoidal Ends - adj Ellipsoidal Ends - 2:1 Flat Ends Straight LengthSloshing of Cylindrical Tank due to Seismic AccelerationSloshing of Cylindrical Tank due to Seismic Acceleration . Yasumasa Shoji and Hidenori Munakata . Chiyoda Advanced Solutions Corporation . Technowave 100 Bldg. 1-25, Shin-Urashima-cho, 1-chome, Kanagawa-ku, Yokohama 221-0031, Japan . Abstract Cylindrical tanks are subjected to the seismic loads in certain countries, for example in Japan.
Sloshing of Cylindrical Tank due to Seismic Acceleration
Sloshing of Cylindrical Tank due to Seismic Acceleration . Yasumasa Shoji and Hidenori Munakata . Chiyoda Advanced Solutions Corporation . Technowave 100 Bldg. 1-25, Shin-Urashima-cho, 1-chome, Kanagawa-ku, Yokohama 221-0031, Japan . Abstract Cylindrical tanks are subjected to the seismic loads in certain countries, for example in Japan.Spill Prevention Control and Countermeasure (SPCC) PlanDisplacement Volume, DVTank 2 (ft 3) = x n (ft ) c (ft) c is the containment wall height used in Step 2 of A. = ft o Repeat to calculate the displacement of each additional horizontal cylindrical tank located with the largest tank in the dike or berm. 2. Calculate the total displacement volume from the additional vertical cylindrical tanks in theSpill Prevention Control and Countermeasure (SPCC) PlanV /V = ÷ = c is the secondary containment volume calculated in Step 1. d / e is the tank volume calculated in Step 2. c (ft 3) d or e (ft ) f % = x 100 = f g If percentage, g, is 100% or greater, the capacity of the secondary containment is sufficient to contain the shell capacity of the tank. If rain can collect in the dike or berm, continue to step 4.
Tank Volume Calculator
Total volume of a cylinder shaped tank is the area, A, of the circular end times the height, h. A = r 2 where r is the radius which is equal to d/2. Therefore V(tank) = r 2 h The filled volume of a vertical cylinder tank is just a shorter cylinder with the same radius, r, There is a cylindrical tank lying horizontally on the japan vertical cylindrical tank building volumeThe small cylindrical tank has a radius of 4 feet and a height of 6 feet; the large cylindrical tank has a radius of 8 feet and a height of 16 feet. The . asked by Anonymous on April 25, 2016; Calculus. Water is flowing into a vertical cylindrical tank at the rate of 24 cu. ft. per minute.User rating 4.6/5DESIGN RECOMMENDATION FOR STORAGE TANKS AND provide a recommendation of the seismic design for the various types of storage tanks in common use throughout Japan. The Sub-Committee first published Design Recommendation for Storage Tanks and Their Supports in 1984, and amended it in the 1990, 1996 and 2010 japan vertical cylindrical tank building volume Vertical, Cylindrical Storage Tanks ----- 154 Appendix B Assessment of japan vertical cylindrical tank building volume
Vessel Volume & Level Calculation
Vessel Volume & Level Calculation Estimates Volume filled in a Vessel with Ellipsoidal (2:1 Elliptical), Spherical (Hemispherical), Torispherical (ASME F&D, Standard F&D, 80:10 F&D) and Flat heads.
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Question
The sum of the digits of a two-digit number is 9. When we interchange the digits it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?
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Solution
Let the digits at tens place and ones place be x and (9−x) respectively.Therefore, original number =10x+(9−x)=9x+9On interchanging the digits, the digits at one's place and tens place will be x and (9−x) respectively.Therefore, new number after interchanging the digits =10(9−x)+x=90−10x+x=90−9xAccording to the given question,New number = Original number +2790−9x=9x+9+2790−9x=9x+36Transposing 9x to R.H.S and 36 to L.H.S, we obtain90−36=18x54=18xDividing both sides by 18, we obtain3=x and 9−x=6Hence, the digits at tens place and one's place of the number are 3 and 6 respectively.Therefore, the two-digit number is 9x+9=9×3+9=36
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# Metamath Proof Explorer
## Theorem divcl
Description: Closure law for division. (Contributed by NM, 21-Jul-2001) (Proof shortened by Mario Carneiro, 17-Feb-2014)
Ref Expression
Assertion divcl ${⊢}\left({A}\in ℂ\wedge {B}\in ℂ\wedge {B}\ne 0\right)\to \frac{{A}}{{B}}\in ℂ$
### Proof
Step Hyp Ref Expression
1 divval ${⊢}\left({A}\in ℂ\wedge {B}\in ℂ\wedge {B}\ne 0\right)\to \frac{{A}}{{B}}=\left(\iota {x}\in ℂ|{B}{x}={A}\right)$
2 receu ${⊢}\left({A}\in ℂ\wedge {B}\in ℂ\wedge {B}\ne 0\right)\to \exists !{x}\in ℂ\phantom{\rule{.4em}{0ex}}{B}{x}={A}$
3 riotacl ${⊢}\exists !{x}\in ℂ\phantom{\rule{.4em}{0ex}}{B}{x}={A}\to \left(\iota {x}\in ℂ|{B}{x}={A}\right)\in ℂ$
4 2 3 syl ${⊢}\left({A}\in ℂ\wedge {B}\in ℂ\wedge {B}\ne 0\right)\to \left(\iota {x}\in ℂ|{B}{x}={A}\right)\in ℂ$
5 1 4 eqeltrd ${⊢}\left({A}\in ℂ\wedge {B}\in ℂ\wedge {B}\ne 0\right)\to \frac{{A}}{{B}}\in ℂ$
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# Calculus
posted by .
Find the exact area of the region enclosed by the square root of (x) + the square root of (y) is = 1; x = 0 and y = 0.
I moved the first equation around to get:
y = [1- the square root of (x) ] ^2
Unfortunately, this gives me bounds from 1 to 1.
So I'm stuck with:
integral from 1-1 of [1- the square root of (x) ] ^2
I used FnInt on this equation and it returns an answer of 0.
Any help or suggestions would be greatly appreciated!
Thanks alot,
Mike
The limits of dx integration are from x=0 (the x=0 line) to x=1 (where the curve crosses the y=0 line). What you want to calculate in the integral od y dx from x=0 to x=1. That can be written
(Integral of) (1 - 2 sqrt x + x) dx
## Similar Questions
1. ### algebra
1.)correct 2.)wrong---> the first term is wrong 4(3a^2)^2 = 36a^4 3.) not sure on this one 4.) wrong---> - square root of six is not a solution Hope this Helps! 1)find p(-3)if p(x)=4x^3-5x^2+7x-10 =-184 2)If r(x)=4x^2-3x+7,find …
2. ### Algebra
Use the quadratic formula to solve the equation. Give exact answers: 2x^2 -1 = 6x. The choices are: a) -3 + square root(7)/2, -3 - square root(7)/2 b) 3 + square root(11)/2, 3 - square root(11)/2 c) 3 + square root(7)/2, 3 - square …
3. ### Algebra
Rationalize each expression by building perfect nth root factors for each denominator. Assume all variables represent positive quantities. I don’t understand how to compute these. Also I don’t have the square root sign so I typed …
4. ### College Algebra
I need some help with a handful of questions. I need to see the steps on how to get to the final answers. I would also appreciate if you can explain how you got to those steps. Rationalize each expression by building perfect nth root …
5. ### College Algebra
I need some help with a handful of questions. I need to see the steps on how to get to the final answers. I would also appreciate if you can explain how you got to those steps. Rationalize each expression by building perfect nth root …
6. ### Math simplifying mixed radicals
Please help me simplify these following mixed radicals:: and show me how you did it cause I need to learn : I don't know how to type a square root sign so I just wrote square root and the numbers infront of the square root are supposed …
7. ### Math Help!
Four students worked to find an estimate for square root 22. Who is closest to finding the true estimate?
8. ### college pre-calculus
simplify: square root (4/3)-square root (3/4) A. 2-sqaure root(3)/2 square root (3) B. square root (7/12) C. square root (3)/6 D. 2/square root (3) which answer?
9. ### Pre-Calculus
Solve 0 = 3x^2 + 5x -1 by completing the square. Express your answer as exact roots. 3(x^2 + 5/3 + 25/36) = -1 (x+ 5/6)^2 = 37/36 sq root( x + 5/6)^2 = sq root (37/36) x + 5/6 = +- sq root 37/6 x = -5/6 +- square root 37 /6 <--- …
10. ### Calculus
A rectangle is bounded by the x axis and the semicircle = square root 25-x^2. What length and width should the rectangle have so that its area is a maximum?
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## Soap Geometry
Sometimes, we look for relation between objects of different dimensionality, search for proportionality rules and try to factor away constants, or, at least, figure out what they are made of. A cute example of which came to me in the shower…
Knowing your weight, can you know how much soap you use?
Let us abstract the fact that weight is not mass; and consider mass only. So, let $M$ be the mass, $\bar{d}$ the average density, $h$ the average height, $w$ the average width, $t$ the average thickness (and assuming a suitable face separation).
Mass, density, and volume are related by:
$\displaystyle V = \frac{M}{\bar{d}}$
and indeed:
$\displaystyle V = \frac{M}{\bar{d}} = \frac{M}{\frac{M}{V}} = \frac{M}{M}V = V$
But also:
$V=h\:w\:t$
$V=h\:(c_w\:h)\:(c_t\:h)$
$V=c_w\:c_t\:h^3$
$V=c_v\:h^3$
where $c_w = w/h$, $c_t=t/h$. Surface is also related; of course. We have:
$S=2(h\:w+h\:t+w\:t)$
$S=2(c_w\:h^2+c_t\:h^2+c_w\:c_t\:h^2)$
$S=2(c_w+c_t+c_w\:c_t)h^2$
$S=c_s\:h^2$
Using $S$ with $V$:
$V=c_v\:h^3$
$\displaystyle{}V=\frac{c_v}{c_s}(c_s\:h^2)h$
$\displaystyle\frac{V}{h}=\frac{c_v}{c_s}(c_s\:h^2)$
$\displaystyle\frac{c_s}{c_v}\frac{V}{h}=(c_s\:h^2)$
$\displaystyle\frac{c_s}{c_v}\frac{V}{h}=S$
Therefore:
$S\propto{}V^{\frac{2}{3}}$
and, transitively:
$S\propto{}M^{\frac{2}{3}}$
So, the amount of soap you use is directly proportional to your surface, which is proportional to $M^\frac{2}{3}$.
*
* *
This reasonning holds for “ordinary” objects that are close to most objects we see in the world around us, but eventually breaks when considering exotic objects like, say, a sheet of paper. In the case of a sheet of paper, surface greatly exceeds volume as the volume of an ideal sheet of paper goes to zero as it approximates a region of the plane. In this case, the $S\propto{}V^{2/3}$ approximation doesn’t hold at all—it does, but with a degenerate proportionality constant.
### One Response to Soap Geometry
1. Nicolas A. Bérard-Nault says:
A quick empirical “rule of thumb” you can use to estimate your skin surface is:
$S = (71.84m^{0.425} \times (100l)^{0.725}) / 10^4$
http://wikiwix.com/cache/?url=http://formation.etud.u-psud.fr/physique/PCS0/grandeurs-phys/homme/homme.htm&title=%5B1%5D
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teaching resource
# Multiplication and Division Speed Drill Worksheets – Facts of 7
• Updated: 12 Apr 2021
A set of worksheets to help students develop fast and accurate recall of the seven times tables.
• Non-Editable: PDF
• Pages: 1 Page
• Years: 3 - 6
### Curriculum
• #### Key Stage 2 (KS2) - Lower
Key Stage 2 (KS2) - Lower covers students in Year 3 and Year 4.
• #### Mathematics
The principal focus of mathematics teaching in lower key stage 2 is to ensure that pupils become increasingly fluent with whole numbers and the four operations, including number facts and the concept of place value. This should ensure that pupils dev...
• #### Key Stage 2 (KS2) – Upper
Key Stage 2 (KS2) - Upper covers students in Year 5 and Year 6.
• #### Mathematics
The principal focus of mathematics teaching in upper key stage 2 is to ensure that pupils extend their understanding of the number system and place value to include larger integers. This should develop the connections that pupils make between multipl...
• #### Multiplication and division
teaching resource
# Multiplication and Division Speed Drill Worksheets – Facts of 7
• Updated: 12 Apr 2021
A set of worksheets to help students develop fast and accurate recall of the seven times tables.
• Non-Editable: PDF
• Pages: 1 Page
• Years: 3 - 6
A set of worksheets to help students develop fast and accurate recall of the seven times tables.
Use these multiplication and division worksheets to help your students become more familiar with the seven times table.
### Why Is Fast and Accurate Recall of the Times Tables Important for Students?
Once your students have gained an understanding of the underlying principles of multiplication (repeated addition, groups, arrays), is it time to focus on speedy and accurate recall of these facts. Having these facts at their fingertips will make mental computations more efficient and will significantly assistant their understanding of other mathematical concepts, such as fractions.
### What Does This Resource Contain?
This resource contains:
• one page of 80 multiplication questions for the seven times table
• one page of 80 multiplication and division questions for the seven times table
• answers for both sets of questions.
### How Could This Resource Be Used?
The key to fast and accurate recall of multiplication facts is practise, practise, practise!
Provide these worksheets to your students to work on at home during the week. At the end of the week, provide them with the same worksheet and time how long it takes them to complete the questions. Remember, the focus is to achieve both speed and accuracy. Remind your students that there is no point completing quickly if the answers are incorrect. Similarly, there is little point being accurate if it takes hours to do so!
For the purposes of differentiation, you must wish to vary the time at which you introduce the speed challenge to your students. Some students will need more time to consolidate their recall of these facts, and the time challenge may stress them or lead to a decrease in confidence if introduced too early.
### 0 Comments
Write a review to help other teachers and parents like yourself. If you'd like to request a change to this resource, or report an error, select the corresponding tab above.
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# CBA Review #2 Mechanics and Thermal Physics - PowerPoint PPT Presentation
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CBA Review #2 Mechanics and Thermal Physics. Work-Energy Momentum Conservation Impulse Thermodynamics. Work and Energy. Work = Fd Units = Joules. F = magnitude of the force d = magnitude of the displacement. A 100N force acts on a box 1.5m to the right.
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CBA Review #2 Mechanics and Thermal Physics
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### CBA Review #2 Mechanics and Thermal Physics
Work-Energy
Momentum Conservation
Impulse
Thermodynamics
### Work = Fd Units = Joules
F = magnitude of the force
d = magnitude of the displacement
A 100N force acts on a box 1.5m to the right.
How much work did the force do?
Work = 100(1.5) = 150 J
### Kinetic Energy – Energy of Motion
KE = (1/2)mv2
Example: Find the kinetic energy of a 20kg mass moving at 10 m/s.
KE = (1/2)mv2 = (1/2)(20)(10)2 = 1000J
Example: If you double your velocity, what happens to the KE?
KE (new) = (1/2)m(2v)2 = 4KE(original)
### Work Energy Theorem
Work Done = DKE
or
Fd = ½mvf2 – ½mvi2
### Example
A 10.0N force acts on a 4.00kg box initially at rest. If the force pushes the box 3.00 meters, what I the final velocity of the box?
Fd = ½mvf2 – ½mvi2
10(3) = (1/2)4vf2– 0
30 = (2)vf2
vf2 = 15
vf = 3.87 m/s
## Energy Types
Stored Energy – Potential Energy
Energy of Motion – Kinetic Energy
### Gravitational Potential Energy
The work you do to lift an object is
equal to its increase in gravitational PE.
Work to lift = PE = mgh
h = vertical distance from ground level.
### Example
How much work does a 100kg person do walking up a flight of stairs
that takes her 10m off of the ground?
Work = increase in PE = mgh = 100(9.8)10 = 9800 Joules.
### Total Mechanical Energy, E
E = KE + PE
If there is no friction, the total mechanical energy is conserved.
This means KE + PE is always the same number.
So if KE drops, PE must rise. If PE drops, KE must rise.
Another way to say this, is that KE is converted to PE, or PE is converted to KE.
### Example
A 1 kg rock is dropped from rest from a building 20m high. Fill in the table.
PE at the top is mgh =
1(9.8)(20) = 196J
0
196J
196J
KE at the top is zero.
98J
196J
98J
147J
49J
196J
E at the top (and everywhere)
is 0 + 196J = 196J
196J
196J
0
PE at the bottom is zero…………
## Momentum and Energy in Collisions
### Definition of Momentum
The symbol p stands for momentum.
Momentum is the product of mass and
velocity.
p = mv
### Examples of calculating momentum
A 2000kg car is moving at 30m/s. What is
the momentum of the car?
p = mv = (2000kg)(30 m/s) = 60,000 kg m/s
A .1 kg bullet has a momentum of 50 kg m/s. How fast is it moving?
v = p/m = 50/.1 = 500 m/s
### The Vector Nature of Momentum
Momentum is a vector – it points in the same
direction as the velocity.
In one dimension, momentum pointing to the right is positive.
Momentum pointing to the left is negative.
### Example:
Find the momentum of each ball. Be careful of the signs!
Answer: For the 3kg ball, p = 3(20) = 60 kg m/s
For the 10 kg ball, p = 2(-10) = -20 kg m/s
### Newton’s 2nd Law in terms of Momentum
Favg = maavg = mDv/Dt = (pf – pi)/ Dt = Dp/Dt
FavgDt = Dp
Impulse = Change in momentum
### Example
A 50N force is applied to a 20kg particle moving at 4m/s.
The force is applied for 4 seconds.
1. What is the impulse?
Impulse = FDt = 50(4) = 200 N.s
2. How fast is the particle moving after 4 seconds?
Impulse = Dp I = mvf – mvi 200 = 20vf - 20(4)
vf= (200 + 80) /20 = 14 m/s
### Conservation of Momentum
Momentum is Conserved for Collisions
Total momentum = Total momentum
before the collision after the collision
Pbefore = Pafter
### Completely Inelastic Collisions
• When two objects hit and stick together.
• Or, the reverse of this – when one object
breaks apart into two objects.
Momentum is Conserved
Total momentum = Total momentum
before the collision after the collision
Pbefore = Pafter
### Example
A cannon ( mass = 500kg) fires a cannon ball ( m = 50kg) at 40m/s.
How fast does the cannon move after it fires the cannon ball?
Before: Pi = 0
After: Pf = mballvball + mcannonvcannon
Pi = Pf
0 = mballvball + mcannonvcannon
(-mballvball )/mcannon =vcannon = (-50)(40)/500 = -4 m/s
### Example
A car mass = 1kg moving at 3m/s hits another 1kg car and they stick together.
How fast are they moving after they stick together?
Pi = mvi = 1(3) = 3 Pf = 2mv = 2v 2v = 3, v = 1.5 m/s
### Example
A car mass = 10kg moving at 2m/s hits another 15kg car moving to the left at
3m/s and they stick together. How fast are they moving after they stick together?
Pi = m1v1i + m2v2i = 10(2) + 15(-3) = -25
Pf = m1v1f + m2v2f = (m1 + m2 )vf = 25vf
-25 = 25vf vf = -25/25 = -1 m/s
### Elastic Forces - Hooke’s Law
F = kx PE = kx2
k =spring constant in N/m
Thermodynamics
• Name the three methods of heat transfer and give an example of each.
• Conduction: typically through solidsThe handle of a pot also gets hot
• Convection: hot gas/liquid risesConvection oven, or hot magma rises inside the core of the earth
• Radiation: transfer without mediumThe sun warms the earth through space
Thermodynamics
• What happens to metal when heated?
• Like most materials, metal expands when heated.
### Heat Energy, Q
Heat to raise the temperature of a
substance with no phase change:
Q = mcDT
Heat required for a change of state:
Q = mL
Thermodynamics
• What is entropy?
• Entropy is a measure of the disorder in a system. Without the input of work, the entropy of a system will always increase. The entropy of the universe is constantly increasing. For example:
• Ice :
• High order
• Low entropy
• Ice melted
• Less order
• Higher entropy
### Method of Mixtures
m1c1DT1 + m2c2DT2 = 0
400g of water at 50 degrees C is mixed with 600g
of water at 70 degrees C. What’s the final
temperature?
400(1)(Tf -50) + 600(1)(Tf -70) = 0
1000Tf - 62000 = 0 Tf = 62 degrees C.
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# Are there limits to cell size?
## Presentation on theme: "Are there limits to cell size?"— Presentation transcript:
Are there limits to cell size?
Demonstration Recap: Using 3 different sized agar blocks to represent cells (3 cm, 2 cm, and 1 cm), we observed the “nutrient uptake” of each when placed in solution. Although the nutrients traveled into the “cells” at the same speed, the “cell” that appeared to receive the most nourishment was the smallest one. This can be explained by the distance a particle from outside the cell would need to travel inwards to reach the center. In the small cell, this distance is the smallest.
Data: Surface Areas & Volumes
(6 x w x l) Volume = (w x l x h) Surface Area Volume 1 cm Cube 6 x 1 x 1 = 6 1 x 1 x 1 = 1 6/1 = 6 2 cm 6 x 2 x 2 = 24 2 x 2 x 2 = 8 24/8 = 3/1 = 3 3 cm 6 x 3 x 3 = 54 3 x 3 x 3 = 27 54/27 = 2/1 = 2
Data: Observations 1 cm Cube: (SA/V = 6) 2 cm Cube: (SA/V = 3)
Which “cell” was most efficient?
This is the cell in which the nutrients (pink) got closest to reaching the core of the “cell”. Answer: the smallest cell Based on this answer, can you tell me…
What is the relationship between cell size and the rate at which nutrients reach the core of the cell? The “cell” that received the most nutrition for all of its organelles was the smallest cell. It has a surface area to volume ratio of 6 to 1, which is the highest of all three “cells”. This means more of the cell was able to receive nutrients because it was directly exposed to the nutrient solution. The part of the cell that was not exposed did not receive nutrients until they traveled into the area. However, since the cell took up very little space (has a very small volume), the nutrients did not have far to travel once they got inside the cell.
Does a cell have to be round?
No, any shape is possible- however, the ratio of Surface Area to Volume may vary.
What is the best shape and size for a cell?
The best shape and size for a cell is one where the ratio of surface area to volume is high enough that the cell gets complete nourishment with minimal harm inflicted on itself in the process. HUH?! What?
For instance… IF you had a cell with a very small SA/V ratio (volume is very large), your cell might be so big that nutrients cannot reach their destination in time, and the cell will die.
But on the other hand… IF you have a cell with a very large SA/V ratio (surface area is very large), your cell will become very fragile and thin. It could die easily from heat exposure or snags.
The best cell shape… Would probably be a cell that is long and thin, tube-like. Good examples: nerve cells, muscle cells, xylem and phloem cells in plants.
Biggest & Smallest Cells
How big is the biggest cell? The biggest cell is an ostrich egg (about the size of a football). What are the smallest cells? Bacteria- Ever wonder why bacteria can produce a new generation in 20 – 30 minutes? And over ten million cells in less than a day?
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# Understanding Equivalence Relations
For $$A = \{(−4, −20), (−3, −9), (−2, −4), (−1, −11), (−1, −3), (1, 2), (1, 5), (2, 10), (2, 14), (3, 6), (4, 8), (4, 12)\}$$ define the relation $R$ on $A$ by $(a, b) R (c, d)$ if $ad = bc$.
• a) Verify that $R$ is an equivalence relation on $A$.
• b) Find the equivalence classes $[(2, 14)], [(−3, −9)],$ and $[(4, 8)]$.
• c) How many cells are there in the partition of $A$ induced by $R$?
-
What are your thoughts on the problem? What do you know about equivalence relations - do you know what a relation is, and what three properties a relation must satisfy to be an equivalence relations? The more you tell us about what work you have done on this question, the better we can help you. – Alex Wertheim Aug 13 '13 at 5:36
First of all you have to check the property of reflexivity for all elements of $A$. For instance: $-4 \cdot -20= -20 \cdot -4$.
Secondly you have to check all $aRb$ for symmetry. For instance $(-2,-4)R(3,6)$ because $-2 \cdot 6 = -4 \cdot 3$. Now we have to verify if $(3,6)R(-2,-4)$.
Thirdly you have to check the transitivity: if $aRb$ and $bRc$ then $aRc$. For instance: $(3,6)R(4,8)$. We already know that $(-2,-4)R(3,6)$. You now have to check if $(-2,-4)R(4,8)$.
Hints for parts $b$ en $c$ of your question. To get a good intuitive feel for equivalence classes and equivalence relations I suggest you take a piece of paper and draw some dots on it. Each dot will represent an element $a \in A$. Draw a line between two dots $a$ and $b$ if $aRb$. Also draw a line if it turns out $bRa$. Do you know what your drawing should look like when you are finished? Can you see the equivalence classes? How many are there?
Hope this helps
-
Leo's answer gives you all that you really need, but it might be helpful to see the computation of one equivalence class; I'll compute $[\langle 4,8\rangle]$. You know that $\langle 4,8\rangle\mathbin{R}\langle c,d\rangle$ if and only if $4d=8c$. This condition is equivalent to the condition $d=2c$, so we're looking for the pairs $\langle c,d\rangle\in A$ such that $d=2c$. The set consisting of precisely those pairs is
$$[\langle 4,8\rangle]=\{\langle -2,-4\rangle,\langle 1,2\rangle,\langle 3,6\rangle,\langle 4,8\rangle\}\;.$$
-
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Absolute Value To Piecewise Function Calculator
Introduction: The Absolute Value To Piecewise Function Calculator is a versatile online tool designed to calculate the absolute value of a given numerical input and integrate it into a piecewise function. This functionality is valuable in various mathematical applications, particularly in defining functions that change their behavior based on different input ranges.
Formula: The absolute value of a number is the distance of that number from zero on the number line. A piecewise function is a mathematical function defined by different rules for different input ranges.
How to Use:
1. Enter a numerical value into the input field.
2. Click the “Calculate” button to obtain the absolute value and the corresponding piecewise function.
3. The result will be displayed below the button.
Example: Suppose you want to integrate the absolute value of -5 into a piecewise function. Enter -5 into the calculator and click “Calculate.” The result will show the absolute value of -5 is 5, and the piecewise function could be defined as f(x) = { 5, x < 0; 0, x >= 0 }.
FAQs:
1. Q: Can I input decimal numbers? A: Yes, the calculator accepts both whole numbers and decimal numbers.
2. Q: How is the piecewise function defined? A: The piecewise function is defined based on the sign of the input value.
3. Q: Can I integrate other functions into a piecewise function? A: No, this calculator specifically integrates the absolute value into a piecewise function.
4. Q: What if I enter a non-numeric value? A: The calculator will prompt you to enter a valid number.
5. Q: Is there a limit to the size of the input number? A: The calculator can handle a wide range of numerical values.
Conclusion: The Absolute Value To Piecewise Function Calculator is a valuable tool for mathematicians, students, and professionals who need to integrate absolute values into piecewise functions. Whether you are working on mathematical modeling or analyzing functions with changing behavior, this calculator provides a quick and efficient solution.
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iSoul In the beginning is reality.
# Tag Archives: Space & Time
Matters relating to length and duration in physics and transportation
# 4D Formulations of Newtonian Mechanics
Four-Dimensional Formulations of Newtonian Mechanics
First we reproduce section 2 from Michael Friedman’s “Simultaneity in Newtonian Mechanics and Special Relativity” in Foundations of Space-Time Theories (ed. Earman et al., UMinn, 1977), p.405-407. Then we provide the dual.
According to the spatio-temporal point of view, the basic object of both our theories is a four-dimensional manifold. I shall use R4, the set of quadruples of real numbers, to represent the spatio-temporal manifold. Both theories agree that there is a natural system of straight lines defined on this manifold. If (a0, a1, a2, a3), (b0, b1, b2, b3) are two fixed points in R4, then a straight line is a subset of R4 consisting of elements (x0, x1, x2, x3) of the form
(1) x0 = a0r + b0
x1 = a1r + b1
x2 = a2r + b2
x3 = a3r + b3
where r ranges through the real numbers. A curve on R4 is a (suitably continuous and differentiable) map σ: R → R4. Such a curve σ(u) is a geodesic if and only if it satisfies
(2) x0 = a0u + b0
x1 = a1u + b1
x2 = a2u + b2
x3 = a3u + b3
where (x0, x1, x2, x3) = σ(u) and the ai and bi are constants. So if a curve is a geodesic its range is a straight line. Note that the geodesies are just the curves that satisfy
(3) d2xi/du2 = 0 i = 0, 1, 2, 3.
The importance of straight lines and geodesies is due to the fact that both theories agree that the trajectories of free particles are spatio-temporal straight lines. So we can represent such trajectories as geodesies in R4.
# Note
This blogger is focused on developing a new six-dimensional theory of time (an early version is posted here). The glossary is also being updated (see here). Good progress is being made, with the double frame and three-dimensional time aspect especially new. RG
# Temporo-spatial Newtonian mechanics
We follow the treatment by David Tong of Cambridge University in his Classical Dynamics.
A tempicle is defined as a moving object of insignificant time. The motion of a tempicle of vass n at the chronation t is governed by Newton’s Temporo-spatial Second Law, R = nb or, more precisely,
R(t; t′) = h′ (1.1)
where R is the release which, in general, can depend on both the chronation t as well as the lenticity t′, and h = nt′ is the fulmentum. Both R and h are 3-vectors which we denote by the bold font. A prime indicates differentiation with respect to stance x. Equation (1.1) reduces to R = nb if n′ = 0. But if n = n(x), then the form with h′ is correct.
General theorems governing differential equations guarantee that if we are given t and t′ at an initial stance x = x0, we can integrate equation (1.1) to determine t(x) for all x (as long as R remains finite). This is the goal of classical dynamics.
# Composition order
Written compositions organized by temporal order are narratives. Items such as descriptions of people, places, or objects are organized as they occur to the narrator, for example, as the narrator takes apart an object or walks through a building or meets various people. This is a common method of composition but there are others.
Spatial order is another method of composition. Items such as descriptions of people, places, or objects are organized by their physical or spatial positions or relationships, for example, starting at the top and proceeding downward. Explanations of a geopolitical matter might proceed in geographic order.
Travel can be described temporally or spatially. An itinerary is usually arranged temporally but telling about it afterwards might be more interesting if arranged spatially. There are other principles of organization such as climactic order (order of importance) and topical order.
In science the independent variable determines the type of organization. If the independent variable is time, the organization is temporal. If the independent variable is space or distance, the organization is spatial. The stance in spatial organization corresponds to the time in temporal organization.
The values of the independent variable are the index to the order of the composition. If the independent variable is time, then the times indicate the steps in the order. If the independent variable is space or distance, then the stances indicate the steps in the order. Once the step is indicated, the composition may be the same: whether it’s Tuesday, so the tour is in Paris or it’s Paris, so the tour is on Tuesday makes no difference.
# Equations of Motion Generalized
This is an update and expansion of the post here.
Here is a derivation of the spatio-temporal equations of motion, in which acceleration is constant. Let time = t, location = x, initial location = x(t0) = x0, velocity = v, initial velocity = v(t0) = v0, speed = v = |v|, and acceleration = a.
First equation of motion
v = ∫ a dt = v0 + at
Second equation of motion
x = ∫ (v0 + at) dt = x0 + v0t + ½at²
Third equation of motion
From v² = vv = (v0 + at) ∙ (v0 + at) = v0² + 2t(av0) + a²t², and
(2a) ∙ (xx0) = (2a) ∙ (v0t + ½at²) = 2t(av0) + a²t² = v² ‒ v0², it follows that
v² = v0² + 2(a ∙ (xx0)), or
v² − v0² = 2ax, with x0 = 0.
Here is a derivation of the temporo-spatial equations of motion, in which retardation is constant. Let stance = x, time (chronation) = t, initial time = t(x0) = t0, lenticity = w, initial lenticity = w(x0) = w0, pace w = |w|, and retardation = b.
# A theory of 6D space and time
Note: as the research develops this post will be updated.
Introduction
Experience shows motion takes place in three dimensions. There are two measures of the extent of motion: length and duration. The length of motion in three dimensions comprises three-dimensional space. The duration of motion in three dimensions comprises three-dimensional time. Length and duration are symmetric concepts, as will be shown below.
Introduction
An independent variable is specified prior to measuring any dependent variable, so an independent variable is the domain of a functionally-related dependent variable. The independent variable is commonly an interval of time. Distance is the independent variable of an inverse square law. In Hooke’s law the independent variable is mass.
A date-time or time-stamp is a combined time-of-day and date on the calendar, which is of interest in history and astronomy. A time interval or elapsed time is the difference between two date-times, which is of interest in science.
A linear reference is of interest in geography and transportation. The stance interval or distance is the difference between two linear references, which is of interest in science.
Distance is an equivalence relation between pairs of (spatial) points. Distime is an equivalence relation between pairs of instants.
An elapsed time or distime is the date-time that changes during an event or motion. A travel stance or distance is the change in linear reference during an event or motion.
Variables of time periods and distances are fixed. Variables of elapsed values are increasing from a starting point. Intervals are deltas of elapsed values. E.g., time periods are deltas of time. Distances are deltas of stances, that is, stations or points along a line or curve. Elapsed time and elapsed distance, or stance, are increasing variables.
Given that there are three dimensions of motion, and that every motion is measured by its length and duration, then motion requires three dimensions of length and three dimensions of duration. Three dimensions of length are called three-dimensional space. Three dimensions of duration are called three-dimensional time.
For example, motion on a two-dimensional surface can be presented as a two-dimensional map scaled in units of length or as a two-dimensional map scaled in units of duration. [The latter time maps are …]
# What is a clock?
What is a clock? it is a device that measures time, but what are the essentials of a clock? I submit these are the essentials of a clock:
(1) A clock requires a uniform motion. Because only the kinematics (not the dynamics) are significant, a uniform rotation is acceptable. But because the result will be represented as a line – a timeline or time axis – a linear uniform motion has a more direct connection with what is measured, so let us take the first essential as a uniform linear motion.
(2) In order for clocks to be measuring alike, it is necessary that there be a standard rate for all clocks. In addition, clocks should have a standardized beginning point, so that clocks are interchangeable.
(3) A clock requires a pointer which indicates the present time on a time scale as it moves at the standard uniform rate. This would be the hands and dial on a common analogue clock. On a linear clock it is a part whose position in motion is interpreted as the present value of time. The pointer and scale are essentials of a clock.
Furthermore, a clock must be interpreted as showing the present time of the observer’s rest frame.
Figure 1
All the essentials of a clock can be represented by a frame in standard uniform motion relative to the observer’s rest frame. In that case, a clock should be definable in terms of frames of reference: one rest frame and one frame in uniform motion relative to the observer’s rest frame, as in the following.
# Space, time, and dimension
The post continues the ones here, here, and here.
There are three dimensions of motion. The extent of motion in each dimension may be measured by either length or time (duration). There are three dimensions of length and three dimensions of time (duration) for a total of six dimensions.
But there is no six-dimensional metric. Why? Because a metric requires all dimensions to have the same units, which requires a ratio to convert one unit into the other unit. The denominator on a ratio is a one-dimensional quantity, which means either the length or time dimensions need to be reduced by two dimensions.
This ratio is a conversion factor that is either a speed, which multiplied by a time equals a length, or a pace, which multiplied by a length equals a time. In general a speed is the ratio Δdr²/|Δdt²| = Δdr²/(Δt1² + Δt2² + Δt3²)1/2, and a pace is the ratio Δdt²/|Δdr²| = Δdt²/(Δx² + Δy² + Δz²)1/2. The denominator is a distance or distime, which is a linear measure of length or time (duration).
The conversion factors required are the speed of light in a vacuum, c, or its inverse, the pace of light in a vacuum, k. The resulting four-dimensional metric is either c²dt² − dx² − dy² − dz² (with time reduced to one dimension) or dt1² − dt2² − dt3² − k²dr² (with space reduced to one dimension).
These metrics are often simplified by taking c = 1 and k = 1 so that symbolically they are the same. Their units are not the same, however.
Each metric may be further reduced by separating space and time, as in classical physics. Then the space metric is |Δdr²| = (Δx² + Δy² + Δz²)1/2 and the time metric is |Δdt²| = (Δt1² + Δt2² + Δt3²)1/2. In the classical (3+1) of three space dimensions and one time dimension, time is replaced by its metric, and in the classical (1+3) of one space dimension and three time dimensions, space is replaced by its metric.
# Space and time as frames
An observer is a body capable of use as a measurement apparatus. An inertial observer is an observer in inertial motion, i.e., one that is not accelerated with respect to an inertial system. An observer here shall mean an inertial observer.
An observer makes measurements relative to a frame of reference. A frame of reference is a physical system relative to which motion and rest may be measured. An inertial frame is a frame in which Newton’s first law holds (a body either remains at rest or moves in uniform motion, unless acted upon by a force). A frame of reference here shall mean an inertial frame.
A rest frame of observer P is a frame at rest relative to P. A motion frame of observer P is a frame in uniform motion relative to P. Each observer has at least one rest frame and at least one motion frame associated with it. An observer’s rest frame is three-dimensional, but their motion frame is effectively one-dimensional, that is, only one dimension is needed.
Space is the geometry of places and lengths in R3. A place point (or placepoint) is a (spatial) point. The space origin is a reference place point. The location of a place point is the space vector to it from the space origin. Chron (3D time) is the geometry of times and durations in R3. A time point (or timepoint) is an instant. The time origin is a reference time point. The chronation of a time point is the chron vector to it from the time origin.
A frame of reference is unmarked if there are no units specified for its coordinates. A frame of reference is marked by specifying (1) units of either length or duration for its coordinates and (2) an origin point. A space frame of observer P is a rest frame of P that is marked with units of length. A time frame of observer P is a motion frame of P that is marked with units of duration.
Speed, velocity, and acceleration require an independent motion frame. Pace, lenticity, and retardation require an independent rest frame. These independent frames are standardized as clocks or odologes so they are the same for all observers.
Let there be a frame K1 with axes a1, a2, and a3, that is a rest frame of observer P1, and let there be a motion frame K2 with axes 1, 2, and 3, that is a motion frame of P1 along the coincident a1-a´1 axis. See Figure 1.
Figure 1
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# measuring resistivity of copper wire
I have a power supply, multimeter, and a copper wire with cross sectional area (A) A= 2.14*10^-3 m^2, Length L = 0.13 m and the resistivity of copper is p=1.7 × 10-8. The goal is to find the resistance of my copper wire, but I'm not sure what other instruments I need to determine the resistance, or how to setup the circuit to find out the resistance. After using the resistance equation, I found that the resistance should be about 1*10^-6 (ohms), but I cant get close to that answer. How should the circuit be setup?
• How much current can your power supply provide? Do you have a second multimeter, or does the power supply have an internal ammeter? What is the lowest voltage your multimeter can accurately measure? – The Photon Oct 6 '17 at 4:29
• You are describing testing a piece of "wire" about the size of a soda can. Fix your figures first, or find a new piece of wire, more like 1 m long and 1 mm diameter, or 1x10^-6 m^2 area. It is possible to measure the resistivity of a piece of bulk copper, but you'd need hundreds of amps, and the calculations of the current flow are quite different to the simple formula. – tomnexus Oct 6 '17 at 5:53
• @tomnexus He does not need any power supply. I think it is just a homework, and "power supply" in the question is supposed to distract students. – Chupacabras Oct 6 '17 at 6:05
• Well, if the OP is trying to measaure 1 uOhm, it will be rather difficult, as tonnexus points out. Even 1000 A will only produce 1mV. – mkeith Oct 6 '17 at 6:28
• Measuring the resistance of very small resistances, like a copper wire with a multimeter is unreliable. I suggest using a Thomson bridge, which gives much better accuracy at low resistance values. – Bart Oct 6 '17 at 8:58
As MKEITH said, copper has a temperature coefficient; its 0.4%, or 4,000 ppm, per degree Centigrade.
Copper, as with other materials has a thermal timeconstant (or, when inverted, a parameter named thermal-diffusivity).
For a cubic meter of copper, the time constant is 9,600 seconds (about 3 hours).
For 10cm cube (4" on a side), the time constant is 100X faster or 96 seconds.
For 1cm cube, the time constant is another 100X faster, or 0.96 seconds.
For 1mm cube, Tau is another 100X faster, or 0.0096 seconds.
Thus measurement has be made extremely quickly, or the temperature rise must be very small.
In general, you use a power supply to run a known current through the wire, and measure the voltage drop using a voltmeter accurate to the mV level. Then you use Ohm's law to calculate the resistance. V = I * R. V is the voltage on the voltmeter, I is the current, and R is what you want to know. You probe with the voltmeter at the exact location where you want to know the resistance.
Note, copper is a good conductor, but its conductivity is fairly sensitive to temperature. If you run so much current through it that it warms up, the resistance will change. So best to measure reasonably quickly, then disconnect the power.
$R= ρ \times \frac{l}{A}$
$R= 1.7 \times 10^{-8} \Omega.m \times \frac{0.13 m}{0.00214 m^{2}} = 1.0327 \mu\Omega$
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# Quaternions & Rotations in 3D Space: What You Need to Know
• Registred
Only one is needed. No matter how many times you turn an object, you can always find a single axis and rotation angle that completely describes the orientation of that object in three dimensional space. Always. It's the Euler rotation theorem. Note: This theorem does not apply to four dimensional space and higher. It's special to two and three dimensional space.One way to represent rotations in three dimensional space is via that single axis orientation. You need a unit vector and an angle. Multiply each element of the unit vector by the angle and you have just what is needed, three parameters that fully describe the orientation of an object. Unfortunately, this single axis rotation vector is hard to use and even harder
#### Registred
Hey there,
I have a question regarding quaternions and rotations. It's related to 3D graphics programming, but nevertheless I'm sure the physics forum is the right place for my question.
As far as I've understood there are 3 primary ways we can store rotation: euler angles, quaternions and matrices. In 3D graphics programming, quaternions are often the preferred way because they need less memory than matrices but avoid gimbal lock.
To represent a 3D object with euler angles I just need 3 components: x, y and z, which each represent the angle around an axis.
Now, how many quaternions are needed to store a rotation? To my knowledge a quaternion stores an axis and the amount of rotation - so I'd need 3 quaternions to store a rotation - is that right? That would be 3 (quaternions) x 4 (values) = 12 values to be stored. To store a rotation in a matrix I would need 3 x 3 = 9 values, which is even less. So why not just use matrices? Why do quaternions need less values/memory than matrices? Do I really need 3 quaternions to represent a rotation?
Greetings!
For some reason, you're getting things mixed up.
Matrices ARE used to do coordinate rotations, even in 3D graphics programming. They are an efficient way not only of storing data, but computer hardware is specially designed to process them faster than regular data streams.
Quaternions have representations in matrix form, just like complex numbers.
To accommodate rotations and translations in 3D, 4x4 matrices are used, as illustrated in this article:
http://inside.mines.edu/~gmurray/ArbitraryAxisRotation/ [Broken]
Last edited by a moderator:
Quaternions have representations in matrix form, just like complex numbers.
How do you mean that? A quaternion can be represented by a matrix?
I know that OpenGL and other graphics APIs work with 4x4 matrices internally, however, frameworks built around these APIs often implement Quaternions for rotation and transform them to rotation matrices when needed. (At least this is the way I understood it, feel free to correct me if I'm wrong).
I'm really curious about how many quaternions I need to store for a rotation. I mean, i have 3 axes, but this doesn't mean that I need 3 quaternions, does it?
Registred said:
How do you mean that? A quaternion can be represented by a matrix?
A 2x2 matrix. Computer implementations? Probably not. Almost definitely not. Much more likely is as a 4 vector, or as a scalar (the real part) plus a three vector (the imaginary, or quaternionic part).
I know that OpenGL and other graphics APIs work with 4x4 matrices internally, however, frameworks built around these APIs often implement Quaternions for rotation and transform them to rotation matrices when needed. (At least this is the way I understood it, feel free to correct me if I'm wrong).
Those 4x4 matrices -- here are some appropriate adjectives. Ugly. Inefficient. Kludge. Don't go there.
I'm really curious about how many quaternions I need to store for a rotation. I mean, i have 3 axes, but this doesn't mean that I need 3 quaternions, does it?
Only one is needed. No matter how many times you turn an object, you can always find a single axis and rotation angle that completely describes the orientation of that object in three dimensional space. Always. It's the Euler rotation theorem. Note: This theorem does not apply to four dimensional space and higher. It's special to two and three dimensional space.
There are a lot of ways to represent rotations in three dimensional space. One approach is via 3x3 orthogonal matrices. This approach generalizes to any dimension. Rotations in four dimensional space are described by a 4x4 orthogonal matrix, in five dimensional space, a 5x5 orthogonal matrix, and so on. There's one big drawback to the matrix approach: There are a lot more numbers in the matrix than there are degrees of freedom. A 3x3 matrix contains nine numbers, but there are only three degrees of freedom in rotations in three dimensional space. Only three numbers are needed. A quaternion has four elements, so there's a slight bit of over-specification here. A 3x3 matrix has nine elements. That's not just over-specification. It's overkill.
One way to represent rotations in three dimensional space is via that single axis orientation. You need a unit vector and an angle. Multiply each element of the unit vector by the angle and you have just what is needed, three parameters that fully describe the orientation of an object. Unfortunately, this single axis rotation vector is hard to use and even harder to manipulate. (What happens if an object rotates about this axis by this number of radians, then about that axis by that number of radians? Good luck describing this with that three element single axis rotation representation. It can be done, but it is a mathematical nightmare.)
Unit quaternions are very closely allied with this single axis rotation, only now the mathematics are much better behaved. That problem of rotating about this axis, then that axis? This becomes a product of two quaternions. Any sequence of rotations is a product of unit quaternions.
Obviously quaternions can be represented by a 4 vector (not a matrix).
When SteamKing mentioned “Quaternions have representations in matrix form”, I believe he meant that a rotation matrix can be constructed using the elements of a quaternion.
One somewhat common representation of quaternions is as 2x2 complex matrices. Physicists use this scheme a lot; the Pauli spin matrices are isomorphic to the quaternions.
Thank you so much for this excellent explanation!
Only one is needed. No matter how many times you turn an object, you can always find a single axis and rotation angle that completely describes the orientation of that object in three dimensional space. Always. It's the Euler rotation theorem. Note: This theorem does not apply to four dimensional space and higher. It's special to two and three dimensional space.
This was the point I was unsure about - being able to represent every rotation with ONE quaternion is a reason to not use matrices. I'm already trying to implement a quaternion functionality and hope that I won't encounter any problems.
Registred said:
This was the point I was unsure about - being able to represent every rotation with ONE quaternion is a reason to not use matrices.
This is not an advantage of quaternions over matrices. The exact same thing applies to matrices. Every rotation can be represented by a single matrix.
There is no reason to implement that. Use an existing library. For that matter, do not sweat over matrices/quaternions - use whatever 3D rotation library works for you. As long as it works, leave it alone. Premature optimization is the root of all evil.
## 1. What are quaternions and how are they used in 3D space?
Quaternions are mathematical entities that are used to represent rotations in 3D space. They consist of four components - a scalar and three imaginary numbers, and are often denoted as w + xi + yj + zk. Quaternions are used to describe rotations because they have certain properties that make them more efficient and less prone to errors compared to other methods, such as Euler angles.
## 2. How do quaternions differ from other methods of representing rotations?
Unlike Euler angles, quaternions do not suffer from gimbal lock, which is a limitation that occurs when two of the three rotation axes align and restrict the movement of the object. Additionally, quaternions are more compact and have simpler mathematical operations, making them more efficient to use in calculations.
## 3. Can quaternions be used for both 3D and 2D rotations?
Yes, quaternions can be used for both 3D and 2D rotations. In 2D space, the z-axis component is not necessary and can be set to 0, effectively reducing the quaternion to a complex number with the real and imaginary components representing the rotation angle.
## 4. How are quaternions multiplied and how does it relate to rotations?
Quaternions are multiplied by multiplying their components based on the rules of quaternion algebra. This multiplication can be thought of as a combination of two rotations, where the first quaternion represents the current orientation of an object and the second quaternion represents the desired rotation. The resulting quaternion is then applied to the object, resulting in a new orientation.
## 5. Can quaternions be used for interpolating between two rotations?
Yes, quaternions can be used for interpolating between two rotations. This is often used in computer animation, where smoothly transitioning between two rotations is important. Quaternions can be used to find intermediate orientations between two given rotations, resulting in a smooth and natural-looking transition.
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Numbers and Operations - Easy Mathematics Step-by-Step
## Easy Mathematics Step-by-Step (2012)
### Chapter 1. Numbers and Operations
In this chapter, you learn about the various sets of numbers that make up the real numbers.
Natural Numbers and Whole Numbers
The natural numbers (or counting numbers) are the numbers
1, 2, 3, 4, 5, 6, 7, 8, …
The three dots indicate that the pattern continues without end.
You can represent the natural numbers as equally spaced points on a number line, increasing endlessly in the direction of the arrow, as shown in Figure 1.1.
Figure 1.1 Natural numbers
When you include the number 0 with the set of natural numbers, you have the whole numbers:
0, 1, 2, 3, 4, 5, 6, 7, 8, …
The number 0 is a whole number, but not a natural number.
Like the natural numbers, you can represent the whole numbers as equally spaced points on a number line, increasing endlessly in the direction of the arrow, as shown in Figure 1.2.
Figure 1.2 Whole numbers
The graph of a number is the point on the number line that corresponds to the number, and the number is the coordinate of the point. You graph a set of numbers by marking a large dot at each point corresponding to one of the numbers. The graph of the numbers 2, 3, and 7 is shown in Figure 1.3.
Figure 1.3 Graph of 2, 3, and 7
Integers
On the number line shown in Figure 1.4, the point 1 unit to the left of 0 corresponds to the number –1 (read as “negative one”), the point 2 units to the left of 0 corresponds to the number –2, the point 3 units to the left of 0 corresponds to the number –3, and so on. The number –1 is the opposite of 1, –2 is the opposite of 2, –3 is the opposite of 3, and so on. The number 0 is its own opposite.
Figure 1.4 Whole numbers and their opposites
A number and its opposite are exactly the same distance from 0. For instance, 3 and –3 are opposites, and each is 3 units from 0.
The number 0 is neither positive nor negative.
The whole numbers and their opposites make up the integers:
…, –3, –2, –1, 0, 1, 2, 3, …
The integers are either positive (1, 2, 3, …), negative (…, –3, –2, –1), or 0. Positive numbers are located to the right of 0 on the number line, and negative numbers are to the left of 0, as shown in Figure 1.5.
Figure 1.5 Integers
It is not necessary to write a + sign on positive numbers (although it’s not wrong to do so). If no sign is written, then you know the number is positive.
Problem Find the opposite of the given number.
a. 8
b. –4
Solution
a. 8
Step 1. Describe the location of 8 and its opposite on a number line.
8 is 8 units to the right of 0. The opposite of 8 is 8 units to the left of 0.
Step 2. State the opposite of 8.
The number that is 8 units to the left of 0 is –8.
b. –4
Step 1. Describe the location of –4 and its opposite on a number line.
–4 is 4 units to the left of 0. The opposite of –4 is 4 units to the right of 0.
Step 2. State the opposite of –4.
The number that is 4 units to the right of 0 is 4.
Problem Graph the integers –5, –2, 3, and 7.
Solution
Step 1. Draw a number line.
Step 2. Mark a large dot at each of the points corresponding to –5, –2, 3, and 7.
Rational, Irrational, and Real Numbers
The number is an example of a rational number. A rational number is a number that can be expressed as a quotient of an integer divided by an integer other than 0. That is, the rational numbers are all the numbers that can be expressed as
The number 0 is excluded as a denominator for because division by 0 is undefined, so has no meaning no matter what number you put in the place of p.
Fractions, decimals, and percents are rational numbers. All of the natural numbers, whole numbers, and integers are rational numbers as well because you can express each of these numbers, as shown here.
The decimal representations of rational numbers terminate or repeat. For instance, is a rational number whose decimal representation terminates, and is a rational number whose decimal representation repeats. You can show a repeating decimal by placing a line over the block of digits that repeats, like this: . You also might find it convenient to round the repeating decimal to a certain number of decimal places. For instance, rounded to two decimal places, .
The symbol ≈ means “is approximately equal to.”
Note: Fractions, decimals, and percents are discussed at length in Chapters 57.
The irrational numbers are the real numbers whose decimal representations neither terminate nor repeat. These numbers cannot be expressed as ratios of two integers. For instance, the positive number that multiplies by itself to give 2 is an irrational number called the positive square root of 2. You use the square root symbol to show the positive square root of 2 like this: . Every positive number has two square roots: a positive square root and a negative square root. The other square root of 2 is . It also is an irrational number. (See Chapter 10 for an additional discussion of square roots.)
You cannot express as the ratio of two integers, nor can you express it precisely in decimal form. Its decimal equivalent continues on and on without a pattern of any kind, so no matter how far you go with decimal places, you can only approximate . For instance, rounded to three decimal places, . Do not be misled, however, because even though you cannot determine an exact value for , it is a number that occurs frequently in the real world. For instance, designers and builders encounter as the length of the diagonal of a square that has sides with length of 1 unit, as shown in Figure 1.6.
Figure 1.6 Diagonal of unit square
Not all roots are irrational. For instance, .
There are infinitely many square roots and other roots as well that are irrational.
Be careful: Square roots of negative numbers are not real numbers.
Two famous irrational numbers are π and e. The number π is the ratio of the circumference of a circle to its diameter, and the number e is used extensively in calculus. Most scientific and graphing calculators have π and e keys. To nine decimal place accuracy, π ≈ 3.141592654 and e ≈ 2.718281828.
Although, in the past, you might have used 3.14 or for π, π does not equal either of these numbers. The numbers 3.14 and are rational numbers, but π is irrational.
The real numbers are all the rational and irrational numbers put together. They are all the numbers on the number line (see Figure 1.7). Every point on the number line corresponds to a real number, and every real number corresponds to a point on the number line.
Figure 1.7 Real number line
The relationship among the various sets of numbers included in the real numbers is shown in Figure 1.8.
Figure 1.8 Real numbers
Note: Hereafter in this book, all numbers are understood to be real numbers.
Problem Categorize the given number according to the various sets of the real numbers to which it belongs. (State all that apply.)
a. 0
b. 0.75
c. –25
d. 36
e. –0.35
Solution
Step 1. Recall the descriptions of the natural numbers, whole numbers, integers, rational numbers, irrational numbers, and real numbers.
Step 2. Categorize the number.
a. 0 is a whole number, an integer, a rational number, and a real number.
b. 0.75 is a rational number and a real number.
c. –25 is an integer, a rational number, and a real number.
d. 36 is a natural number, a whole number, an integer, a rational number, and a real number.
e. –0.35 is a rational number and a real number.
Problem Graph the numbers –4, –2.5, 0, , and 3.6.
Solution
Step 1. Draw a number line.
Step 2. Mark a large dot at each of the points corresponding to –4, –2.5, 0, , and 3.6.
Terminology for the Four Basic Operations
For the math you do in your everyday world, you work with the real numbers. Addition, subtraction, multiplication, and division are the four basic operations you use. Each of the operations has special symbolism and terminology associated with it. Table 1.1 shows the terminology and symbolism for the operations.
Table 1.1 The Four Basic Operations
As you can see from the examples in Table 1.1, addition and subtraction “undo” each other. Similarly, multiplication and division undo each other, as long as no division by 0 occurs.
Division Involving Zero
You must be very careful when you have zero in a division problem. The number 0 can be the dividend, provided the divisor is not 0—the quotient will be 0. But 0 can never be the divisor. The quotient of any number divided by 0 has no meaning; that is, division by 0 is undefined.
Problem State whether the quotient is 0 or undefined.
Solution
Step 1. Recall that division by 0 is undefined, so state that the quotient is undefined.
Step 1. Recall that the quotient is 0 when 0 is divided by a nonzero number, so state that the quotient is 0.
Step 1. Recall that the quotient is 0 when 0 is divided by a nonzero number, so state that the quotient is 0.
Step 1. Recall that division by 0 is undefined, so state that the quotient is undefined.
Step 1. Recall that division by 0 is undefined, so state that the quotient is undefined.
Properties of Real Numbers
The real numbers have the following 11 properties under the operations of addition and multiplication.
For all real numbers a, b, and c, you have:
1. Closure Property of Addition. is a real number. This property guarantees that the sum of any two real numbers is always a real number.
Examples
2. Closure Property of Multiplication. is a real number. This property guarantees that the product of any two real numbers is always a real number.
Examples
3. Commutative Property of Addition. . This property allows you to reverse the order of the numbers when you add, without changing the sum.
Examples
4. Commutative Property of Multiplication. . This property allows you to reverse the order of the numbers when you multiply, without changing the product.
Examples
5. Associative Property of Addition. . This property says that when you have three numbers to add together, the final sum will be the same regardless of the way you group the numbers (two at a time in the same order) to perform the addition.
Example
Suppose you want to compute . In the order given, you have two ways to group the numbers for addition:
Either way, 16 is the final sum.
6. Associative Property of Multiplication. . This property says that when you have three numbers to multiply together, the final product will be the same regardless of the way you group the numbers (two at a time in the same order) to perform the multiplication.
Example
Suppose you want to compute . In the order given, you have two ways to group the numbers for multiplication:
Either way, 7 is the final product.
The associative property is needed when you have to add or multiply more than two numbers because you can do addition or multiplication on only two numbers at a time. Thus, when you have three numbers, you must decide which two numbers you want to start with—the first two or the last two (assuming you keep the same order). Either way, your final answer is the same.
7. Additive Identity Property. There is a real number 0, called the additive identity, such that and . This property guarantees that you have the number 0 for which its sum with any real number is the number itself.
Examples
8. Multiplicative Identity Property. There is a real number 1, called the multiplicative identity, such that and . This property guarantees that you have the number 1 for which its product with any real number is the number itself.
Examples
9. Additive Inverse Property. Every real number a has an additive inverse, –a (its opposite), such that and . This property guarantees that every real number has an opposite whose sum with the number is 0.
Examples
10. Multiplicative Inverse Property. Every nonzero real number a has a multiplicative inverse, (its reciprocal), such that and . This property guarantees that every real number, except 0, has a reciprocal whose product with the number is 1.
Example
11. Distributive Property. and . This property says that when you have a number times a sum, you can either add first and then multiply or multiply first and then add. Either way, the final answer is the same.
Example
can be computed two ways:
add first to obtain or multiply first to obtain
The distributive property is the only property that involves both addition and multiplication at the same time. Another way to express the distributive property is to say that multiplication distributes over addition.
Either way, the answer is 45.
Subtraction and division are not mentioned in the properties listed because you can always turn subtraction into addition by “adding the opposite,” and you can turn division into multiplication by “multiplying by the reciprocal.” That is,
When you subtract a number, you get the same answer as you do when you add its opposite.
When you divide by a nonzero number, you get the same answer as you do when you multiply by its reciprocal.
Problem Identify the property illustrated.
Solution
Step 1. Recall the 11 properties: closure property of addition, closure property of multiplication, commutative property of addition, commutative property of multiplication, associative property of addition, associative property of multiplication, additive identity property, multiplicative identity property, additive inverse property, multiplicative inverse property, and distributive property.
Step 2. Identify the property illustrated.
additive identity property
Step 2. Identify the property illustrated.
closure property of addition
Step 2. Identify the property illustrated.
commutative property of multiplication
Besides the 11 properties given, the number 0 has the following unique characteristic.
Zero Factor Property
If a real number is multiplied by 0, then the product is ; and if the product of two numbers is 0, then at least one of the numbers is 0.
Problem Find the product.
a. –9 · 0
Solution
a. –9·0
Step 1. Given that 0 is a factor of the product, apply the zero factor property.
Step 1. Given that 0 is a factor of the product, apply the zero factor property.
This property explains why 0 does not have a reciprocal. There is no number that multiplies by 0 to give 1—because any number multiplied by 0 is 0.
Exercise 1
For 1–9, categorize the given number according to the various sets of the real numbers to which it belongs. (State all that apply.)
1. 10
2. –7.3
3. –74
4. –1000
5. 0.555 …
8. 0
For 10–12, state whether the quotient is 0 or undefined.
For 13–15, identify the property illustrated.
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# Math
1/2(56x-98)=147
X=7
1. So you are solving for x.
First multiply both sides by 2 to clear your fraction out : )
56x - 98 = 294
then add 98 to both sides and continue...
posted by MsPi_3.14159265
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# Arc length
The circumference of a circle = $$\pi d$$ or $$2\pi r$$.
Look at the sector of the circle shown below. To calculate the length of the arc, we need to know what fraction of the circle is shown. To do this, we use the angle and compare it with 360˚.
This angle is 144°.
That is $$\frac{{144^\circ }}{{360^\circ }} = \frac{2}{5}$$ of a full turn (360°).
So the arc is $$\frac{2}{5}$$ of the circumference.
$$c=\pi d=3.14\times 6$$ (Remember the diameter is double the radius.)
$=18.84cm$
Arc length = $$\frac{2}{5}\times 18.84 = 7.54cm$$
(There is so need to simplify $$\frac{144}{360}$$, you can use this in the arc calculation instead of $$\frac{2}{5}$$.)
The formula used to calculate the Arc Length is:
$Arc\,length = \frac{{Angle}}{{360^\circ }} \times \pi d$
Now try the example question below.
Question
Calculate the length of the arc shown in the diagram below.
$Arc\,length = \frac{{Angle}}{{360^\circ }} \times \pi d$
Remember also that $$d = 2 \times r$$
$= \frac{{150}}{{360}} \times \pi \times 8$
$= 10.47cm$
$= 7.5\,(to\,1\,d.p.)$
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# Math, Algebra
posted by on .
Q1. Find the length of the following segment created by these pairs of endpoints: (0, 1), (0, 7)
Q2. Find the length of the following segment created by these pairs of endpoints: (9, 6), (12, 6)
Q3. Find the length of the following segment created by these pairs of endpoints: (1, 2), (7, 2)
• Math, Algebra - ,
1. d^2=(0-0)^2+(1-7)^2 solve for d, it is not zero.
2. same formula, answer is correct.
3. The answer is not two.
Now on these, the answers are obvious: THINK. On the first, it is a line along the x axis going from 1 to 7, length..
On the second, it is a vertical line at x=6, going from y=9 to 12. Yep, that is a distance of 3
On the third, it is a horizontal line along y=2 from x=1 to seven, which is not a length of 2.
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# In the expansion of (x + y)^6, what is the coefficient of the x^3y^3
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Solution:
Trick is we have to eliminate terms with powers wherever we see them.
$$(x+y)^6 = (x+y)^2*(x+y)^2*(x+y)^2 = (x^2 + y^2 + 2xy)*(x^2 + y^2 + 2xy)*(x+y)^2$$
= $$(6 x^2 y^2 + 4 x^3 y + 4 x y^3)*(x^2 + y^2 + 2xy)$$ (eliminated higher power terms)
= $$4 x^3 y^3 + 4 x^3 y^3 + 12 x^3 y^3$$
= $$20 x^3 y^3$$
So, Option E
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Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3 [#permalink]
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(x+y)^6 = (x+y)^2 * (x+y)^2 * (x+y)^2
By expanding coefficient of x^3*y^3 = 20
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Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3 [#permalink]
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Always follow the triangle, things will be easier.
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Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3 [#permalink]
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Bunuel wrote:
Bunuel wrote:
In the expansion of $$(x + y)^6$$, what is the coefficient of the $$x^3y^3$$ term?
(A) 6
(B) 12
(C) 15
(D) 18
(E) 20
Kudos for a correct solution.
There is a long way to solve this problem, of course: write out (x + y) (x + y) (x + y) (x + y) (x + y) (x + y), expand mechanically, and get the coefficient of the x^3y^3 term. This would take too long on the GMAT, however. There are at least 2 shortcuts.
1) Start mechanically, but think about what you’re doing to make the x^3y^3 term. You might start with a much simpler case:
(x + y) (x + y) = x^2 + 2xy + y^2
Notice that you get a 2 on the xy term, because there are two xy products you can form as you expand:
(X + y) (x + Y) – you pick the x from the first (x + y) and the y from the second (x + y).
(x + Y) (X + y) – vice versa.
If you want to expand a much bigger product of (x + y)’s and find the coefficient of a particular term such as x^3y^3, then you need to think about all the different ways you can get three x’s and three y’s as you expand.
(X + y) (X + y) (X + y) (x + Y) (x + Y) (x + Y) – pick the three x’s first, then the three y’s.
(X + y) (x + Y) (X + y) (x + Y) (X + y) (x + Y) – pick x, y, x, y, x, y. etc.
So really what you’re asking is this: how many ways can you rearrange three x’s and three y’s! That’s a combinatorics problem (how many anagrams are there of the word “xxxyyy”?). The number of ways to rearrange these letters is 6!/(3!3!) = 20, and 20 is the coefficient on the x^3y^3 term.
2) Use Pascal’s Triangle. This is a handy little device to get the coefficients of (x + y)^n, when n is a relatively small integer. You build the triangle downwards with 1’s on the outside. All the interior numbers are sums of the two numbers above.
Each row gives you the coefficients of (x + y)^n for some n. Since the second row gives you 1 and 1 (the coefficients of (x + y)^1 = x + y, it’s actually the n+1’th row that gives you the coefficients of (x + y)^n. So, for instance, you can just read off the bottom row to get all the coefficients of (x + y)^6:
$$(x + y)^6 = x^6 + 6x^5y + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6$$.
The reason this works is because each number in Pascal’s triangle represents the number of legal “zigzags” that get you to that number from the 1 at the top. (A legal zigzag goes down one row and right or left just one number.) For instance, there are 20 legal zigzags that go from the 1 at the top of the triangle to the number 20 in the middle. Each of those zigzags has 3 left “zigs” and 3 right “zags” in it. For instance, you could go left-left-left-right-right-right and end up at 20, or you could go left-right-left-right-left-right and end up at 20, etc. This is exactly the same situation as counting the anagrams of a 6-letter word with two repeated letters (x and y, or L and R).
Attachment:
2015-09-13_2041.png
Is the methodology right to derive the answer. ?
I derived it this way
(x+y)^6
=(x+y)^3 * (x+y)^3
=(x^3+Y^3+3x^2*y+3y^2*x) (x^3+Y^3+3x^2*y+3y^2*x)
Now since we want the ans in x^3 Y^3 terms , we will only multiply those terms in left bracket to those terms in right bracket which yield x^3 Y^3.
For eg . 3x^2*y(from left bracket ) multiply to 3y^2*x ( from right bracket)= 9x^3*y^3
similarly 3y^2*x (from left bracket ) multiply to 3x^2*y ( from right bracket)= 9x^3*y^3
and (now single terms): x^3(from left bracket ) multiply to y^3 ( from right bracket) = x^3*y^3
similarly, y^3(from left bracket ) multiply to x^3( from right bracket)= x^3*y^3
Adding all these values we get = 20 *x^3 * Y^3 .. Thus ans (E).
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Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3 [#permalink]
Really good video that explains Pascal's triangle:
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Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3 [#permalink]
You can also use the formula mentioned by RenB in another similar question.
When given $$(x+y+z)^n$$, and required to find number of occurrences or sum of coefficients of $$x^p*y^q*z^r$$-
Sum of coefficients of $$x^p*y^q*z^r$$= $$\frac{n!}{p!*q!*r!}$$
In case given: $$(x-y-z)^n$$, and required to find number of occurrences or sum of coefficients of $$x^p*y^q*z^r$$
Sum of coefficients of $$x^p*y^q*z^r$$= $$\frac{n!}{p!*q!*r!}*(1)^p*(-1)^q*(-1)^r$$
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In the expansion of (x + y)^6, what is the coefficient of the x^3y^3 [#permalink]
Asked: In the expansion of $$(x + y)^6$$, what is the coefficient of the $$x^3y^3$$ term?
$$(x + y)^6 = ^6C_0x^6 + ^6C_1 x^5y + ^6C_2 x^4y^2 + ^6C_3 x^3y^3 + ^6C_4 x^2y^4+ ^6C_5 xy^5 + ^6C_6y^6$$
Coefficient of the $$x^3y^3$$ term = $$^6C_3 = 20$$
IMO E
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Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3 [#permalink]
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Re: In the expansion of (x + y)^6, what is the coefficient of the x^3y^3 [#permalink]
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# Surface Area of Cubes and Rectangular Prisms
#### How do you find the total surface area of a rectangular box?
A rectangular prism where all the sides are the same is a cube.
The surface area of a prism is the total sum of the areas of all its side faces. Representing a solid in two-dimensional form is called a net.
Surface area is the total area of all surfaces on a 3dimensional shape.
##### Cuboids or rectangular prisms have 6 rectangular faces.
We mark the length l, width w, and height h of the prism and use the formula,
Area of top and bottom face = 2 lw
Area of right and left side face = 2hw
Area of front and back face = 2hl
##### Example: Find the surface area of prism shown in the figure.
S = 2lw + 2lh + 2wh
= 2(3)(5) + 2(3)(6) + 2(5)(6)
= 30 + 36 + 60
= 126 square inches
##### Example: Find the surface area of an aquarium of dimensions 10 cm, 8 cm and 5 cm.
Formula for Surface area = 2lw + 2lh + 2wh
= 2.10.5 + 2.10.8 + 2.8.5
= 340 cm²
#### How do you find the surface area of a cube?
When all sides in a rectangular prism are equal, it is a cube. So, each face is a square.
A cube has 6 square faces.
In a cube all sides are equal so we calculate the surface area by 6 × side²
##### Example: Calculate the surface area of a cube of side 4 mm
Surface area of a cube = 6 × side²
= 6 × 4²
= 96 mm²
## Check point
1. Find the surface area of a rectangular wooden TV unit of length 18 cm, width 5 cm, height, 6 cm.
2. Find the surface area of this cube.
1. Aurora wants to calculate the surface area of a carton box which she needs to wrap. Length of the box is 7 cm, width is 5 cm, and 2 cm is the height.
1. Find the surface area of a rectangular tube if the length, width, and height are as follows:
1. Which of the following ice cubes will melt faster? “A” which measures 3 mm, 2 mm, 1 mm or “B” the other measures 2 mm, 1.5 mm, 1.3 mm. Calculate the surface area for both (Hint: the more is the surface area, the quicker it melts).
1. What will be the surface area of cube with side 32 inches?
1. 456 cm²
2. 726 mm²
3. 118 cm²
4. 76 square inches
5. A = 22 cm², B = 15.1 cm² A is the ice cube which will melt faster, due to larger surface area.
6. 6,144 square inches
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Competitions
Power of time
There are n processes to be completed by you. All the processes have a unique number assigned to them from 1 to n. You are given two things:
• The calling order in which all the processes are called.
• The ideal order in which all the processes should have been executed.
If the process in the calling order cannot be executed (does not fit the ideal order), then it is put to the end of the calling order queue for execution.
Now, let us demonstrate this by an example. Let's say that there are 3 processes, the calling order of the processes is: 3 - 2 - 1. The ideal order is: 1 - 3 - 2, i.e. process number 3 will only be executed after process number 1 has been completed; process number 2 will only be executed after process number 3 has been executed.
• Iteration #1: Since the ideal order has process #1 to be executed firstly, the calling ordered is changed, i.e., the first element must be pushed to the last place. Changing the position of the element takes 1 unit of time. The new calling order is: 2 - 1 - 3. Time taken in step #1: 1.
• Iteration #2: Since the ideal order has process #1 to be executed firstly, the calling ordered has to be changed again, i.e., the first element must be pushed to the last place. The new calling order is: 1 - 3 - 2. Time taken in step #2: 1.
• Iteration #3: Since the first element of the calling order is same as the ideal order, that process will be executed. And it will be thus popped out. Time taken in step #3: 1.
• Iteration #4: Since the new first element of the calling order is same as the ideal order, that process will be executed. Time taken in step #4: 1.
• Iteration #5: Since the last element of the calling order is same as the ideal order, that process will be executed. Time taken in step #5: 1.
Total time taken: 5 units.
Executing a process takes 1 unit of time. Changing the position takes 1 unit of time.
Input
First line contains the number of processes n (1n100). Second line contains the calling order of the processes. The third line contains the ideal order of the processes.
Output
Print the total time taken for the entire queue of processes to be executed.
Time limit 1 second
Memory limit 128 MiB
Input example #1
3
3 2 1
1 3 2
Output example #1
5
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http://www.instructables.com/id/BALLISTIC-PENDULUM-PHYSICS/
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# BALLISTIC PENDULUM PHYSICS!
video BALLISTIC PENDULUM PHYSICS!
LEARNING OBJECTIVE:
Conservation of momentum. m1 * v1 = m2 * v2 Mass of the projectile times speed of the projectile before impact = mass of the pendulum times speed of pendulum+projectile after impact. The ballistic pendulum was invented in 1742 by English mathematician Benjamin Robins. Sandbags have been used as pendulums for bullets. This was the first way to find out how fast a bullet was going.
The part of the video most pertinent to this experiment is from time marker 1:00 minute to time marker 3:00 minutes. The rest of the video shows an example of why conservation of momentum is relevant.
You can use a projectile as small as the airsoft pellet in the video, and a thimble for a pendulum, or you can use the guns detailed in my instructables "MONKEY HUNTER PHYSICS" and "MONKEY HUNTER PHYSICS EASIER GUN" These of course, being larger projectiles will need a larger, heavier pendulum.
1) weigh your projectile and pendulum.
2) shoot projectile at pendulum
3)measure how high the pendulum + projectile swing.
4) calculate velocity of the pendulum using the formula
v=(2GX)^0.5
Where v=velocity of pendulum, G=acceleration of Gravity, X= distance fallen (or in this case distance risen before it stops)
5) Calculate velocity of projectile using mV=Mv
Where m= small mass of projectile, V=big Velocity of projectile, M=big Mass of pendulum, and v= small velocity of pendulum.
6)If you have access to an electronic chronograph, you may want to compare results between the pendulum and the chronograph. Some pelletgun manufacturers also publish their muzzle velocities. Compare your results to published specs.
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http://www.oxfordmathcenter.com/drupal7/node/229
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# Example
Suppose $a_1 \equiv b_1 \pmod{m}$ and $a_2 \equiv b_2 \pmod{m}$. Prove the following:
1. $a_1 + a_2 \equiv b_1 + b_2 \pmod{m}$
2. $a_1 - a_2 \equiv b_1 - b_2 \pmod{m}$
3. $a_1 a_2 \equiv b_1 b_2 \pmod{m}$
First note, as a beginning to each of the following arguments, $a_1 \equiv b_1 \pmod{m}$ implies $b_1 = a_1 + mk_1$ for some integer $k_1$, while $a_2 \equiv b_2 \pmod{m}$ implies $b_2 = a_2 + mk_2$ for some integer $k_2$.
Now consider the following:
1. \displaystyle{\begin{align} b_1 + b_2 &= (a_1 + mk_1) + (a_2 + mk_2)\\ &= (a_1 + a_2) + m(k_1 + k_2)\\ \end{align}}
which implies $a_1 + a_2 \equiv b_1 + b_2 \pmod{m}$
2. \displaystyle{\begin{align} b_1 - b_2 &= (a_1 + mk_1) - (a_2 + mk_2)\\ &= (a_1 - a_2) + m(k_1 - k_2)\\ \end{align}}
which implies $a_1 - a_2 \equiv b_1 - b_2 \pmod{m}$
3. \displaystyle{\begin{align} b_1 b_2 &= (a_1 + mk_1) (a_2 + mk_2)\\ &= a_1 a_2 + a_1 m k_2 + a_2 m k_1 + m^2 k_1 k_2\\ &= a_1 a_2 + m(a_1 k_2 + a_2 k_1 + mk_1 k_2) \end{align}}
which implies $a_1 a_2 \equiv b_1 b_2 \pmod{m}$
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https://socratic.org/questions/how-do-you-factor-x-2-9x-18
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# How do you factor x^2+9x+18?
Apr 21, 2018
By finding what multiplies to give 18 and adds to give 9.
#### Explanation:
General form $a {x}^{2} + b x + c$
What does multiplies to give $a c$ and adds to give $b$
First is 18 positive or negative. If positive both factors will be positive or negative depending on the sign of $b$.
(Not this Question) If negative the factors will be opposite in sign.
List all the factors of $a c = 18$.
$1 , 18$
$2 , 9$
$3 , 6$
Now looking at these number intuitively what will add to give $b = 9$.
$3 , 6$
Now place into the form:
$\left(x + 6\right) \left(x + 3\right)$
This is the factorised version of ${x}^{2} + 9 x + 18$
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https://www.sangakoo.com/en/unit/the-pyramid-surface-area-and-volume
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# The pyramid: Surface area and volume
The pyramid is a polyhedron formed by any polygon (base) and triangular faces that coincide at the top point, called the apex.
The following figure is an example of quadrangular pyramid:
Calculate the area of a quadrangular pyramid with side of the basis $$10 \ m$$ and height $$5 \ m$$.
The area of the basis is $$A_{base} =100 \ m^2$$$To find the area of the laterals, first we have to find $$Ap$$, $$Ap^2=h^2+ap^2=5^2+\Big( \dfrac{10}{2}\Big)^2 \\ Ap=5\sqrt{2}$$$
So, the area of a side face is: $$A_{lateral}=\dfrac{10 \cdot 5\sqrt{2}}{2}=25\sqrt{2} \\ A_{total}=4 \cdot A_{lateral}+A_{basis}\\A_{total}=100\sqrt{2}+100=241,4 \ m²$$$Generalizing,$$A_{laterals}=\dfrac{perimetre_{basis} \cdot Ap}{2} \\ A_{total}=A_{laterals}+A_{basis}$$$
To find the volume of a pyramid, it is useful to remember that it is the third part of the volume of a prism of equal basis and height: $$V=\dfrac{A_{basis} \cdot height}{3}$$\$
In case of the previous example $$100$$ obtains a volume $$100 \cdot \dfrac{5}{3} = 166,67$$
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https://socratic.org/questions/what-is-the-domain-of-sqrt-3x-2-7x-6
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# What is the domain of sqrt(3x^2-7x-6)?
Dec 27, 2016
Factor under the square root.
$y = \sqrt{3 {x}^{2} - 9 x + 2 x - 6}$
$y = \sqrt{3 x \left(x - 3\right) + 2 \left(x - 3\right)}$
$y = \sqrt{\left(3 x + 2\right) \left(x - 3\right)}$
Set the relation to $0$ and solve for $x$.
$x = - \frac{2}{3} \mathmr{and} 3$
Use test points to check when the function will be negative and when it will be positive. You will find that the number under the square root will always be positive except over the interval $- \frac{2}{3} < x < 3$.
The domain is therefore $\left(- \infty , - \frac{2}{3}\right]$, $\left[3 , \infty\right)$.
Hopefully this helps!
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https://www.physicsforums.com/threads/deriving-the-frequency-in-a-beam-equation.740939/
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# Deriving the frequency in a beam equation.
• kezzstar
In summary, the frequency in a beam equation is derived by considering the properties of the beam, such as its length, mass, and stiffness, and using mathematical equations to solve for the natural frequency. This frequency represents the rate at which the beam can vibrate without any external forces, and is an important factor in understanding the behavior and stability of the beam. The frequency can also be affected by boundary conditions and loading conditions, making it a crucial aspect to consider in engineering and design applications.
kezzstar
1. Deriving frequency in a beam equation, E=Young's Modulus, I = Moment of Inertia, m=mass, L=length, f=frequency.
2. Deriving this equation: (∏/2)√(EI/mL3)
3.(EI(∂4y/∂x4 = -m(∂2y/∂x2) ≈ EI(∂4y/∂x4+ m(∂2y/∂x2 = 0
Don't know where to go from here...
Maybe assume a form of solution and use separation of variables?
The equation in (3) looks wrong. If it is supposed to be "force = mass x acceleration", where is the derivative with respect to time?
You haven't defined the problem clearly. If this is a beam of mass m, how is it fixed? Or is it a cantilever beam (assumed to have no mass) with a mass m at the end?
supported at each end. m in the equation is the mass of the beam itself. Yeah 3. is wrong, just found that too. Argh I'm stumped...
kezzstar said:
supported at each end. m in the equation is the mass of the beam itself. Yeah 3. is wrong, just found that too. Argh I'm stumped...
1 person
Chestermiller said:
Thanks for the reply. Checked that website and still no further forward.
kezzstar said:
Thanks for the reply. Checked that website and still no further forward.
The website gives you the exact equation you need to solve and the method of solution. What specifically are you not able to follow about the development in the reference. It's hard to help without knowing specifically what the stumbling block is.
chet
1 person
Chestermiller said:
The website gives you the exact equation you need to solve and the method of solution. What specifically are you not able to follow about the development in the reference. It's hard to help without knowing specifically what the stumbling block is.
chet
Where abouts in the text am I looking?
kezzstar said:
Where abouts in the text am I looking?
Pages 5.1 through 5.8. Eqn. 5.5 with w = 0 is the equation you are supposed to be solving.
Chet
1 person
## 1. What is the formula for calculating the frequency in a beam equation?
The formula for calculating the frequency in a beam equation is f = (1/2π) √(EI/mL^3), where f is the frequency, E is the modulus of elasticity, I is the moment of inertia, m is the mass per unit length, and L is the length of the beam.
## 2. How is the frequency related to the properties of the beam?
The frequency is inversely proportional to the length and square root of the modulus of elasticity and moment of inertia, and directly proportional to the square root of the mass per unit length. This means that as any of these properties increase, the frequency will decrease, and vice versa.
## 3. What is the significance of calculating the frequency in a beam equation?
Calculating the frequency in a beam equation is important because it helps determine the natural frequency of the beam, which is the frequency at which the beam will naturally vibrate when disturbed. This information is useful in designing and analyzing structures to ensure they can withstand vibrations and avoid resonance.
## 4. Can the frequency in a beam equation be changed?
Yes, the frequency in a beam equation can be changed by altering the properties of the beam such as its length, mass per unit length, or modulus of elasticity. It can also be changed by applying external forces or adding additional supports to the beam.
## 5. How does the frequency in a beam equation affect the stability of a structure?
The frequency in a beam equation is directly related to the stability of a structure. If the frequency of an external force applied to a structure matches the natural frequency of the structure, it can cause resonance and lead to instability or even failure of the structure. Therefore, it is important to consider the frequency in a beam equation when designing structures to ensure they are stable and can withstand any potential vibrations.
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https://decodingdatascience.com/measure-of-spread/
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When we think of data, we often think of it in terms of its central tendency – the average or median value that represents the “typical” value of a dataset. But central tendency only tells us part of the story. What about the other values in the dataset? How spread out are they? How much variability is there?
This is where measures of spread come in. Measures of spread are statistical tools that allow us to quantify the variability or dispersion of data. In this article, we will explore different types of measures of spread, how they are calculated, and how they can be used in data analysis.
What is Variability?
Before we dive into measures of spread, let’s first define what we mean by variability. In statistics, variability refers to the degree to which individual data points deviate from the central tendency. In other words, it measures how much the data “varies” or spreads out.
Variability is an important concept in statistics because it tells us how much we can trust the central tendency as a representation of the data. If there is a lot of variability, the central tendency may not be a good representation of the data.
There are several types of measures of spread, each with its own strengths and weaknesses. Here are some of the most common ones:
Range
The range is the simplest measure of spread and is calculated by subtracting the smallest value from the largest value in a dataset. While easy to calculate, the range is sensitive to outliers and may not be representative of the variability of the majority of the data. More details in a separate Article.
Interquartile Range (IQR)
The interquartile range is a more robust measure of spread that is less sensitive to outliers. It is calculated by subtracting the 25th percentile (Q1) from the 75th percentile (Q3) of the dataset. The IQR represents the range of the middle 50% of the data. More details in a separate Article.
Standard Deviation
The standard deviation is a widely used measure of spread that takes into account the distance of each data point from the mean. It is calculated by taking the square root of the variance, which is the average of the squared deviations from the mean. The standard deviation is sensitive to outliers, but less so than the range. More details in a separate Article.
Variance
The variance is similar to the standard deviation in that it measures the spread of data around the mean. However, it is calculated by taking the average of the squared deviations from the mean, rather than the square root of that average. The variance is more sensitive to outliers than the standard deviation. More details in a separate Article.
How Measures of Spread are Used
Measures of spread are used in a variety of ways in data analysis. Here are some examples:
Identifying Outliers
Outliers are data points that are significantly different from the rest of the data. Measures of spread can be used to identify outliers by highlighting values that fall outside a certain range.
Comparing Datasets
Measures of spread can be used to compare the variability of different datasets. For example, if we want to compare the variability of test scores in two different schools, we can use measures of spread to see which school has more consistent scores.
Estimating Confidence Intervals
Measures of spread can be used to calculate confidence intervals, which are ranges of values that are likely to contain the true population parameter. The width of the confidence interval is determined by the level of variability in the data.
Conclusion
Measures of spread are essential tools in data analysis that allow us to understand the variability of data. By quantifying the spread of data, we can make more informed decisions and draw more accurate conclusions. Whether we are identifying outliers, comparing datasets, or estimating confidence intervals,
If you want to learn more about statistical analysis, including central tendency measures, check out our comprehensive statistical course. Our course provides a hands-on learning experience that covers all the essential statistical concepts and tools, empowering you to analyze complex data with confidence. With practical examples and interactive exercises, you’ll gain the skills you need to succeed in your statistical analysis endeavors. Enroll now and take your statistical knowledge to the next level!
If you’re looking to jumpstart your career as a data analyst, consider enrolling in our comprehensive Data Analyst Bootcamp with Internship program. Our program provides you with the skills and experience necessary to succeed in today’s data-driven world. You’ll learn the fundamentals of statistical analysis, as well as how to use tools such as SQL, Python, Excel, and PowerBI to analyze and visualize data. But that’s not all – our program also includes a 3-month internship with us where you can showcase your Capstone Project.
One Response
1. Great line up. We will be linking to this great article on our site. Keep up the good writing.
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https://www.answers.com/Q/How_much_water_is_in_an_18_x_33_foot_above_ground_pool_that_is_7_ft_at_its_deepest_hold
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0
# How much water is in an 18 x 33 foot above ground pool that is 7 ft at its deepest hold?
Updated: 9/27/2023
Wiki User
15y ago
There are 231 cu.in. in a gallon of water 18'= 216", 33'=396", multiplied together you have 85,539 sq.in. Multiply by the height (in inches) and devide by 231 and you get the gallons in the pool. 3'high=36" 85,539x36= 3,079,296 cu. in. 3,079,296 divided by 231 = 13,330.26 gallons.
Wiki User
12y ago
Wiki User
13y ago
Cannot be answered with only these measurements. You need to specify what shape the pool is. Also, for a rectangular pool, the length, width and depth are needed. For a round pool, the diameter and depth are needed.
Wiki User
12y ago
18-ft x 33-ft x 4-ft = 2,376 cubic feet = 17,773.7 gallons
That's the volume of the pool. We have no way to know
how much water there may be in it.
Wiki User
15y ago
Go to http://www.havuz.org/pool-calculators.htm
Wiki User
15y ago
An above ground pool 18x33 - 4ft deep holds about 14k gallons of water. An 18x33 - 4ft above ground pool has about 14k gallons of water.
Wiki User
14y ago
Formula for Volume: Length X width X depth X 7.48 = gal
Wiki User
14y ago
Pool gallonage equation is: L x W x Average Depth X 7.5 = ???? gallons of water
Wiki User
14y ago
26658.72 gallons
Wiki User
13y ago
Earn +20 pts
Q: How much water is in an 18 x 33 foot above ground pool that is 7 ft at its deepest hold?
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Algebra Tutorials!
Tuesday 23rd of January
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Depdendent Variable
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Number of inequalities to solve: 23456789
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# Dividing Polynomials
## Dividing a Polynomial by a Polynomial
Example
Use long division to find (6x3 + 7x2 + 4x - 2) ÷ (2x + 1)
Solution Step 1 Write the problem in long division form. Algebra The terms of each polynomial are in descending order. Step 2 Divide the first term of the dividend by the first term of the divisor. Here’s a long division problem from arithmetic to help you see the similarities between the algebra and the arithmetic. Divide 6x3 by 2x to get 3x2. Write 3x2 in the quotient line above 7x2, the x2-term of the dividend. Step 3 Multiply the divisor by the term you found in Step 2. Multiply (2x + 1) by 3x2 to get 6x3 + 3x2. Step 4 Subtract the expression you found in Step 3 from the dividend. Subtract (6x3 + 3x2) from (6x3 + 7x2). The result is 4x2. Step 5 Bring down the next term from the dividend. Write + 4x to the right of 4x2. Step 6 Repeat Steps 2 through 5 until the degree of the remainder is less than the degree of the divisor. Divide 4x2 by 2x to get 2x. Write 2x in the quotient line.Multiply (2x + 1) by 2x to get 4x2 + 2x. Subtract (4x2 + 2x) from (4x2 + 4x). The result is 2x. Write -2 to the right of 2x. Divide 2x by 2x to get 1. Write +1 in the quotient line. Multiply (2x + 1) by 1 to get 2x + 1. Subtract (2x + 1) from (2x - 2). The result is -3. The degree of the remainder, -3, is less than the degree of the divisor, 2x + 1. So we stop. Step 7 Write the quotient.The quotient is: Quotient is
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# Radii of Two Circles Are 6. 3 Cm and 3.6 Cm. State the Distance Between Their Centres If : They Touch Each Other Externally - Mathematics
Course
#### Question
Radii of two circles are 6. 3 cm and 3.6 cm. State the distance between their centres if:
they touch each other externally
#### Solution
Radius of bigger circle = 6.3 cm
and of smaller circle = 3.6 cm
Two circles are touching each other at P externally. O and O ' are the centers of the circles. Join
OP and O'P
OP = 6.3 cm, O'P = 3.6 cm
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GEOMETR. PRACT. nor B, à maiori ADF, differt, ad BD, differentiam ſtationum: ita ſinus anguli B,
in remotiori ſtatione. ad D A.
### 69.1.
10. triang,
rectil.
2. Distantia verò à puncto A, ad pedem menſoris C, hoc eſt, recta AC,
cognoſcetur per Problema 12. triang. rectil. cap. 3. lib. 1. cum in triangulo obli-
quangulo ABC, duo latera AB, BC, nota ſint, nimirum diſtantia inuenta, & ſta-
tura menſoris, comprehendantque angulum ABC, notum, vtpote conflatum
ex recto CBD, & angulo obſeruationis ABD.
3. Sit deinde punctum A, vt in muro HI, infra oculum B. Inſpecto pun-
cto A, obſeruetur angulus CBA. quem latus pinnacidiorum cum perpendiculi
filo, vel dio ptræ linea fiduciæ cum BC, facit: Deinde accede verſus A. vſque ad
D, & iterum conſidera angulum EDA: exiſtentque rectæ AB, AD BD, BC, DE,
in vno eodemque plano, in eo videlicet, quod per ſta-
turas menſoris BC, DE, & per punctum A, ducitur. Et quoniam angulus FDA, complementi anguli ob- ſeruationis in propinquiore ſtatione æqualis eſt duo-
bus DBA, DAB; ſi DBA, angulus complementi an-
guli remotioris ſtationis dematur ex angulo A D F,
complementi anguli ſtationis propinquioris, reliquus
fiet notus BAD. Si ergo fiat,
### 69.1.
32. primi.
10. Triang.
rectil.
Vt ſinus anguli BAD, dif- \\ ferentiæ inter angulos com \\ plementorum angulorum \\ obſeruationum # ad B D, diffe- \\ rentiam ſta- \\ tionum: # Ita ſinus anguli ADB, con- \\ flati ex recto B D E, & \\ ex angulo obſeruationis \\ A D E, in propinquiore \\ ſtatione # ad AB di- \\ ſtantiam \\ quæſitã.
cognita erit diſtantia A B, quam quærimus, in partibus differentiæ ſtatio-
num B D.
Qvod ſi oculus ponatur in D, & recedatur à puncto D, vſque ad B, repe-
rietur eodem modo diſtantia DA, ſi pro angulo BDA, aſſumes angulum DBA,
complementianguli ABC, obſeruationis in remotiore ſtatione, vt manifeſtum
eſt. Nam eſt, vt ſinus anguli BAD, differentiæ inter angulos complemento- rum angulorum obſeruationum, ad BD, differentiam ſtationum: ita ſinus angu-
li DBA, complementi anguli ABC, in remotiore ſtatione, ad DA.
### 69.1.
10. Triang.
rectil.
4. Vt autem diſtantia CA, à pede ad punctum A, inueniatur, ita progredie-
mur. Quoniam in triangulo rectangulo ABG, (ſi ex puncto A, concipiatur
ducta ad BC, ſtaturam menſoris perpendicularis AG,) baſis AB, nota eſt per in-
uentionem, & angulus BAG, notus, quippe cum ſit complementum anguli
obſeruationis ABG; Si fiat,
2. Triang.
rectil.
Vt ſinus \\ totus # ad baſem A B, proximè \\ inuentam: # Ita ſinus anguli B A G, complemen- \\ ti anguli obſeruationis, # ad B G,
cognoſcetur BG, in partibus baſis AB, hoc eſt, in partibus differentiæ ſtationum
BD, in quibus AB, inuenta fuit. Ablata autem BG, ex menſoris ſtatura BC, no-
ta fiet reliqua CG. Item ſi fiat,
2. Triang.
rectil.
Vt ſinus to- \\ tus # ad baſem A B, nu- \\ per inuentam: # Ita ſinus anguli obſeruatio- \\ nis A B G, # ad A G,
nota etiam fiet A G, in partibus eiuſdem baſis A B, vel differentiæ ſtationum
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Q. 18 C4.4( 7 Votes )
Length of height of rhombus is 4 cm. and length of side is 5 cm. What is the area of field in the shape of rhombus.
Given, Length of height of rhombus is 4 cm. and length of side is 5 cm.
Hence, DB= 4cm and then BO = 2cm
By Pythagoras theorem,
BO2 + AO2 = AB2
22 + AO2 = 52
AO2 = 25 – 4
AO =
AO = 4.6cm
Hence, AC = 2.AO
AC = 2×4.6
AC = 9.2 cm
Area of rhombus = ×4×9.2
= 18.4 sq. cm
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# Convert 20 from octal to binary
What is 20 octal in binary? 20 from octal to binary is 10000. Here we show you how to write 208 in binary and how to convert 20 from base-8 to base-2.
Number:
From:
To:
208 in binary
100002
In numeral system, we know octal is base-8 and binary is base-2. To convert octal 20 to binary, you follow these steps:
To do this, first convert octal into decimal, then the resulting decimal into binary
1. Start from one's place in octal : multiply ones place with 8^0, tens place with 8^1, hundreds place with 8^2 and so on from right to left
2. Add all the products we got from step 1 to get the decimal equivalent of given octal value.
3. Then, divide decimal value we got from step-2 by 2 keeping notice of the quotient and the remainder.
4. Continue dividing the quotient by 2 until you get a quotient of zero.
5. Then just write out the remainders in the reverse order to get binary equivalent of decimal number.
First, convert 208 into decimal, by using above steps:
= 208
= 2 × 810 × 80
= 1610
Now, we have to convert 1610 to binary
16 / 2 = 8 with remainder 0
8 / 2 = 4 with remainder 0
4 / 2 = 2 with remainder 0
2 / 2 = 1 with remainder 0
1 / 2 = 0 with remainder 1
Then just write down the remainders in the reverse order to get the answer, The octal number 20 converted to binary is therefore equal to :
10000
Here are some more examples of octal to binary conversion
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## Trigonometry (11th Edition) Clone
$1.4426$
$\sin s=0.9918$ can be written as $s=\sin^{-1} (0.9918)$. Ensuring that the calculator is in radians, we type $\sin^{-1} (0.9918)$ into the calculator and solve: $s=\sin^{-1} (0.9918)\approx1.4426$
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# 3x^2 + 12x + 39 = 0
Posted on
You need to solve for x the quadratic equation, hence, you should use quadratic formula, such that:
`x_(1,2) = (-b+-sqrt(b^2 - 4ac))/(2a)`
Identifying coeficients a,b,c, yields:
`a = 3, b =12, c = 39`
`x_(1,2) = (-12+-sqrt(12^2 - 4*3*39))/(2*3)`
`x_(1,2) = (-12+-sqrt(144 - 468))/6`
`x_(1,2) = (-12+-sqrt(-324))/6`
Since the problem does not specify the nature of the solutions to equation, hence, you may also consider complex roots, such that:
`x_(1,2) = (-12+-18*i)/6`
Notice that complex number theory states that `sqrt(-1) = i.`
`x_(1,2) = (-2+-3*i)`
Hence, evaluating the complec solutions to the given quadratic equation, yields `x_(1,2) = (-2+-3*i)` .
Posted on
`3x^2+12x+39=0`
`x^2+4x+13=0`
`x^2+4x +4+9=0`
`x^2+4x+4=-9`
`(x+2)^2=-9`
`x+2=+-3i`
`x=-2+-3i`
indeed `Delta= 16- 4 xx 13= 16-52=-36 <0`
So has two complex coniugate roots
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# The Matching Rounds Problem--Proof by induction
Can someone please explain how we skipped from line (3) to (4)
This problem deals with the Hats Problem which state that n men throw their hats into the center of a room. The hats are mixed up and each man randomly selects one. Then the expected number of any number of men who select their own hats is always $1$.
Now we suppose that those choosing their own hats leave the room, while the others (those without a match) put their selected hats in the center of the room again, mix them up, and then re-select. This process continues until each individual has his own hat.
The question is: Find $E[R_n]$ where $R_n \Doteq \$ is the number of rounds that are necessary when $n$ individuals are initially present.
Solution: We use induction to prove this:
on average, there will be one match per round. Hence, one might suggest that $E[R_n] = n$. This turns out to be true, and an induction proof will now be given.
Because it is obvious that $E[R_1] = 1$, assume that $E[R_k] = k \ for \ k = 1, . . . , n − 1$.
To compute $E[R_n]$, we start by conditioning on $X_n$, the number of matches that occur in the first round. This gives
$E[R_n] = \sum_{i=0}^n E[R_n|X_n = i]P[X_n = i]$
Now, given a total of $i$ matches in the initial round, the number of rounds needed will equal $1$ plus the number of rounds that are required when $n − i$ persons are to be matched with their hats. Therefore,
$E[R_n] = \sum_{i=0}^n (1 + E[R_{n−i}])P[X_n = i]$ $= 1 + E[R_n]P[X_n = 0] + \sum_{i=0}^n E[R_n−i]P[X_n = i]$ $= 1 + E[R_n]P[X_n = 0] + \sum_{i=0}^n (n − i)P[X_n = i]$ $(3)$
This is where I get stuck. by the induction hypothesis we can get $= 1 + E[R_n]P[X_n = 0] + n(1 − P[X_n = 0]) − E[X_n]$ $(4)$
in the end we get: $E[R_n]= E[R_n]P[X_n = 0] + n(1 − P[X_n = 0])$
Question 2: by the way, what is $P[X_n = 0]?$
Question 2: Why can't we just say: $E[R_n] = E[R_{n-1}+R_{n^{th}}]$ where $R_{n^{th}}$ is the nth case which equal $1$ by hats problem expectation.
Hence $E[R_n] = E[R_{n-1}]+E[R_{n^{th}}]$ by inductive hypothesis: $E[R_n] = n-1+E[R_{n^{th}}]$
$E[R_n] = n-1+1 = n$ what's wrong with that?
It would be easier to follow if you transcribed the calculation correctly:
\begin{align*} E[R_n]&=\sum_{i=0}^n(1+E[R_{n-i}])P[X_n=i]\\ &=1+E[R_n]P[X_n=0]+\sum_{i=1}^nE[R_{n-i}]P[X_n=i]\\ &=1+E[R_n]P[X_n=0]+\sum_{i=1}^n(n-i)P[X_n=i]\;. \end{align*}
The last step above is where you use the induction hypothesis that $E[R_k]=k$ for $k<n$. Now expand that last summation:
\begin{align*} \sum_{i=1}^n(n-i)P[X_n=i]&=n\sum_{i=1}^nP[X_n=i]-\sum_{i=1}^niP[X_n=i]\\ &=n\left(\sum_{i=0}^nP[X_n=i]-P[X_n=0]\right)-\sum_{i=0}^niP[X_n=i]\\ &=n(1-P[X_n=0])-E[X_n]\;, \end{align*}
since the probabilities of the possible values of $X_n$ sum to $1$, and the last summation is by definition $E[X_n]$. You now have
$$E[R_n]=1+E[R_n]P[X_n=0]+n(1-P[X_n=0])-E[X_n]\;,$$
and you've already shown that $E[X_n]=1$, so
$$E[R_n]=E[R_n]P[X_n=0]+n(1-P[X_n=0])\;.$$
Thus,
$$E[R_n](1-P[X_n=0])=n(1-P[X_n=0])\;,$$
and $E[R_n]=n$ (since clearly $P[X_n=0]\ne 1$).
The probability that no one gets his hat back is a bit messy to compute, but it's approximately $\frac1e$; for more information you should read about derangements.
Your proposed alternative calculation mixes two very different things: $E[R_{n-1}$ is an expected number of rounds, while your $E[R_{n^{th}}]$ is just another name for $E[X_n]$, an expected number of people getting the right hats. You can't expect to add rounds and hats and get rounds.
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# Rule of twelfths
Rule of twelfths
The Rule of Twelfths is a rule of thumb for estimating the height of the tide at any time, given only the time and height of high and low water. This is important when navigating a boat or a ship in shallow water and when launching and retrieving boats on slipways on a tidal shore.
The rule assumes that the rate of flow of a tide increases smoothly to a maximum halfway between high and low tide before smoothly decreasing to zero again and that the interval between low and high tides is approximately six hours. The rule states that in the first hour after low tide the water level will rise by one twelfth of the range, in the second hour two twelfths, and so on according to the sequence - 1:2:3:3:2:1.
Example calculation
If a tide table gave us the information that tomorrow's low water would be at noon and that the water level at this time would be two metres above chart datum and further, that at the following high tide the water level would be 14 metres. We could work out the height of water at 3:00 p.m. as follows:
*The total increase in water level between low and high tide would be: 14 - 2 = 12 metres.
*In the first hour the water level would rise by 1 twelfth of the total (12 m) or: 1 m
*In the second hour the water level would rise by 2 twelfths of the total (12 m) or: 2 m
*In the third hour the water level would rise by 3 twelfths of the total (12 m) or: 3 m
*This gives us the increase in the water level by 3:00 p.m. as 6 metres.
This represents only the increase - the total depth of the water (relative to chart datum) will include the 2 m depth at low tide: 6 m + 2 m = 8 metres.
Obviously the calculation can be simplified by adding twelfths together and reducing the fraction beforehand i.e.
Rise of tide in three hours $=left\left(\left\{1over12\right\}+\left\{2over12\right\}+\left\{3over12\right\} ight\right) imes 12 mathrm\left\{m\right\}=left\left(\left\{6over12\right\} ight\right) imes12 mathrm\left\{m\right\}=left\left(\left\{1over2\right\} ight\right) imes12 mathrm\left\{m\right\}=6 mathrm\left\{m\right\}$
Caveats
The rule is a rough approximation only and should be applied with great caution when used for navigational purposes. Officially produced tide tables should be used in preference whenever possible.
The rule assumes that all tides behave in a regular manner, this is not true of some geographical locations, such as Poole Harbour where there is "double" high water or Weymouth Bay where there is a double low water.
The rule assumes that the period between high and low tides is six hours but this is an underestimate and can vary anyway.
Wikimedia Foundation. 2010.
### Look at other dictionaries:
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# Pathfinder Optimization Algorithm
• Last Updated : 06 Oct, 2021
Nature is full of social behaviours for performing different tasks. Although the ultimate goal of all individuals and collective behaviours is survival, creatures cooperate and interact in groups, herds, schools, colonies, and flocks for several reasons: hunting, defending, navigating, and foraging. In order to mimic these characteristics of animals, swarm-intelligence based optimization algorithms are introduced.
## Inspiration
A pathfinder simply is the individual which leads a swarm. This entity leads the swarm and this entity leads various acts. In addition, this entity takes the swarm to destinations including pastures, water and feeding areas. The group of animals frequently decide the movement between members via social order. Such animals may either have to decide with the leader or with no leader. Leadership however is temporary, with few people knowing the place, hunting area, route, etc.
## Mathematical Model
The population basically follows the pathfinder for various activities. However, in order to do so, we need to specify the position of both the pathfinder and any arbitrary population member. The position of a member says Xi is defined as:
Where Xi is the position vector of ith follower at k iteration, r1 and r2 are random variables uniformly generated in the range of [0,1], α is the coefficient for interaction and β is the coefficient of attraction. The value of α, ß are set in such a way that there is the perfect balance between interaction (i.e. the magnitude of movement of any member together with its neighbour) and attraction (i.e. the random distance for keeping the herd roughly with the leader). The optimum values for α, ß should be around 1. ε is the vector of vibration and it is defined as follows:
where, , u1 is random variable in range [-1,1]
The position of pathfinder is defined as:
Xp is the position vector of the pathfinder, K is the current iteration, r3 if a random variable in the range of [0,1], A is the vector of fluctuation rate. A is defined as follows:
where u2 is a random variable in the range [-1,1], Kmax is the maximum iteration.
ε, A can provide random movement (walk) for all members. Therefore setting different values for them ensures exploration and exploitation. The term in ε, A ensures exploration and exploitation phases of a metaheuristic technique. At first is very small, thereby resulting in rapid change constituting to exploration. Then at later stages, its value becomes eventually small or even zero, constituting exploitation.
## Algorithm
1. Define the parameters such as r1, r2, ε, A
2. Initialize the population and calculate fitness for each member
3. Set the best fitness value as the pathfinder
4. while iteration condition is not met or till max iteration do
5. Generate α, ß in range [0, 1]
6. Using the pathfinder equation update the position of pathfinder
7. If fitness of current pathfinder is better than old pathfinder Then
8. Update fitness of the pathfinder
9. For i=1 to population size
10. Update position of followers using followers equation.
11. Update fitness values of each member
12. Find the best fitness
13. If best fitness is better than old pathfinder Then
14. Update fitness of the pathfinder
15. For i=1 to population size
16. If new fitness of the member is better than old fitness Then
17. Update fitness of member
18. Generate ε, A
19. End
The best position and fitness is returned and can be used to solve optimization problems.
My Personal Notes arrow_drop_up
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# Themes, topics - math word problems
1. Two vans
At 8:30 a. M. , two vans leave at a parking area. They travel in opposite directions: one van travels at 48 km/h and the other at 62 km/h. At what time will the cars be 550 km apart?
2. Working together
One worker performs work in 3 days, the second in 5 days. How much do they work together?
3. Mixing water
The 30-liter container should we fill with water at 60 degrees Celsius. How many liters of water 80 degrees C hot and how many liters of water 20 degrees Celsius warm we have to mix?
4. Walking
Lucie can walk about 3 4/5 miles each hour. How far can she walk in 2 hours 45 minutes?
5. Effective and mean voltage
A voltage divider consisting of resistors R1 = 103000 Ω and R2 = 197000 Ω is connected to the ideal sine wave voltage source, R2 is connected to a voltmeter which measures the mean voltage and has an internal resistance R3 = 200300 Ω, the measured value i
6. Two workers
The first worker would himself completed the work in 8 hours, the other in 6 hours. After two hours of joint work left the first worker to a doctor and other completed work himself. How many hours worked the other worker himself?
7. Express train
International express train drove from Kosice to Teplice. In the first 279 km, the track was repaired, and therefore it was moving at a speed of 10km/h less than it was scheduled to drive. The rest of the 465 km trip has increased the speed by 8 km/h than
8. Two trains
From station A, the freight train traveled at a speed of 40 km/h in 9h. When he drove 15km, the fast train started from station A in the same direction at a speed of 70km/h. When will it the freight train catch up?
9. Orlík hydroelectric plant
The Orlík hydroelectric power plant, built in 1954-1961, consists of four Kaplan turbines. For each of them, the water with a flow rate of Q = 150 m3/s is supplied with a flow rate of h = 70.5 m at full power. a) What is the total installed power of the
10. The factory
The factory z can complete an order in 12 days, factory p in 18 days. For how many days does an order completed when the first two days work only the factory z and then both factories?
11. Beds
At the summer camp, there are 41 chalets. Some rooms are 3-beds, some 4-beds. How many campers from 140 are living in 3-bed?
12. Water reservoir
One drain hole emptied the water from the tank for 20 hours, with the second drain for 30 hours. How long will the water drain from this tank if we open two drains simultaneously?
13. Year 2018
The product of the three positive numbers is 2018. What are the numbers?
14. Harvesters
The first harvester will harvest the grain from the field in 20 hours, the second, the more powerful in 10 hours. How long will take harvest this field by both harvesters, if the second harvester has to be set up first and this takes 2 hours? The first ha
15. Bob traveled
Bob traveled 20 meters to Dave's house. He arrived in 3 minutes. What was his average speed for the trip.
16. A report
Maricris can type a committee report in 5 hours. Jane helped Maricris and together they finished the report in 3 hours. How long would it take Jane to complete the report she had worked alone?
17. A car
A car weighing 1.05 tonnes driving at the maximum allowed speed in the village (50 km/h) hit a solid concrete bulkhead. Calculate height it would have to fall on the concrete surface to make the impact intensity the same as in the first case!
18. Free fall
The free fall body has gone 10m in the last 0.5s. Find the body speed at the moment of impact.
19. Home cleaning
Mr. Smith is cleaning up a big mess at home. In the closet, he finds a solution that is 5% bleach, and another stronger solution that is 20% bleach. For this particular job, he needs 600mL of 15% bleach. How much of each type (to the nearest mL) should he
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## Raise Your Scores Now !
Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
#### SIGN UP FOR A FREE TRIAL
1. If (x + y)2 = p, (x – y)2 = q, then what is (x2 + y2) only in terms of ‘p’ and ‘q’?
pq
p + q
(p + q)/ 2
p – q
(pq)2
Show me the answers!
2. If f(x) = 4x – 6, then what is the value of f(-2) – f(2)?
16
-16
-14
2
14
Show me the answers!
3. Given two points (-1, 3) and (2, k) and the slope of the line passing these through two points is -2/3. What is the value of k?
0
1
2
3
4
Show me the answers!
4. If an object is travelling with a speed of 30m/sec, then how much distance would it cover in a time interval of 1.5minutes?
45m
900m
2700m
450m
20m
Show me the answers!
5. If the width of a rectangle is 12m and the length of its diagonal is 20m, then what is the area of the rectangle in ‘m2’?
240
320
480
144
192
Show me the answers!
6. In triangle ABC, the measure of the exterior angle ABD is 4x – 26, and the measure of the angles A and C are x + 11 and x + 13 respectively as shown in the figure below. What is the value of x?
25°
50°
38°
36°
74°
Show me the answers!
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# Two grandmothers
Two grandmothers went to sell eggs at the market, and they had a total of 100. When they sold all the eggs, they made the same money. The first grandmother says to the second: "If I sold my eggs for your price, I would earn 15 crowns. " The other grandmother replies: "If I sold my eggs for your price, I would get 6 and 2/3 crowns. " How many eggs did each of the bucks have? Try to find the result without solving a complex quadratic equation.
Correct result:
a = 60
b = 40
#### Solution:
Our quadratic equation calculator calculates it.
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
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Re: Compute only certain number of hours between two date/time
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5 - Automation Enthusiast
Hello! Trying to figure out a formula to compute for hours worked between two dates/time given a condition. The formula only needs to compute for hours worked between 10:00PM - 6:00AM. Here are the scenarios:
Scenario 1
Time in: May 4, 2021 4:00AM
Time out: May 4, 2021 1:00PM
Hours worked: 2 hours
Scenario 2
Time in: May 3, 2021 11:00PM
Time out: May 4, 2021 8:00AM
Hours worked: 7 hours
Trying to compute for “night differential” as part of our payroll. Hoping someone can help or lead me towards the right direction. Thank you!
6 Replies 6
18 - Pluto
Welcome to the Airtable community!
This is an interesting question. There are a few different ways to solve it.
Will the two dates always be either the same date or consecutive dates? If so, the formula will be much simpler than if the two dates could be several days apart.
5 - Automation Enthusiast
To answer your question, it can be both. same date or consecutive date! Here is the scenario for reference:
Scenario 1
Time in: May 4, 2021 4:00AM
Time out: May 4, 2021 1:00PM
Hours worked: 2 hours
Scenario 2
Time in: May 3, 2021 11:00PM
Time out: May 4, 2021 8:00AM
Hours worked: 7 hours
18 - Pluto
Can it be non-consecutive days? For example, time in is a Monday, and time out is a Wednesday?
5 - Automation Enthusiast
Hi @kuovonne! There won’t be such scenarios but would like to have airtable flag it if it does :slightly_smiling_face:
18 - Pluto
I’m sorry, but the more that I look at this, the more complicated it gets and I don’t have the time to create this formula for fun right now.
There are many different scenarios that the formula would have to take into account. Even if the time in and out are on the same days, the formula would need to take into account all of these scenarios:
• both start and end times before 6am
• both start and end times between 6am and 10pm
• both start and end times after 10pm
• start time before 6am, and end time before 10pm
• start time before 6am, and end time after 10pm
• start time between 6 and 10, and end time after 10pm
• both begin and end times
Once you add in the possibility of start and end times being on different days, things get even more complex.
This doesn’t even take time zones into account. Airtable stores all date/times in GMT time, and formulas will need to account for this.
The end result will likely be a very long, very complex formula that is very hard to maintain.
I think that this value is better computed in a script.
5 - Automation Enthusiast
Hi @kuovonne thanks for providing inputs on how I can approach this formula. I created a sample set of data based from your inputs. I think I got it but would be more than happy to get your thoughts :slightly_smiling_face:
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# Differentiation – Implicit Function
## Differentiation – Implicit Function
i) In given implicit function, we solve this for dy/dx directly
ii) Moderate level sometime in question, we asked dy/ dx at some particular point x.
So, after getting dy/dx, we put the given particular value of x in dy/ dx and get the desired result.
If the relation between the variable x and y is given by an equation containing both of the form φ (x, y) = 0 and this equation is not immediately solvable for y i, e., it is not passible to express y as a function of x in the form y = φ(x), then y is called an implicit function of x. In such a case to find dy/dx differentiate both sides of the given relation with respect to ‘x’ keeping in the mind that the derivative of f(y) with respect ‘x’ is (dφ/dy). (dy/dx).
Examples:
• $$\frac{d}{dx}(\tan y)={{\sec }^{2}}y.\frac{dy}{dx}$$.
• $$\frac{d}{dx}({{y}^{4}})=4{{y}^{3.}}.\frac{dy}{dx}$$.
Shortcuts for implicit function: For implicit function, put
$$\frac{d}{dx}\{f(x,y)\}=\frac{-\partial f/\partial x}{\partial f/\partial y}$$,
Where $$\frac{\partial f}{\partial x}$$ is a partial differential of given function with respect to ‘x’ and $$\frac{\partial f}{\partial y}$$ means partial differential of given function with respect to ‘y’.
Problem: If sin²y + cos (xy) = k, then dy/dx =?
Solution: Given that
sin²y + cos (xy) = k,
On differentiation both sides with respect to ‘x’ we get
$$\frac{d}{dx}\left( {{\sin }^{2}}x+\cos xy \right)=\frac{d}{dx}(k)$$.
$$\frac{d}{dx}({{\sin }^{2}}x)+\frac{d}{dx}(\cos xy)=0$$.
$$2\sin y\cos y.\frac{dy}{dx}+(-\sin xy)\frac{d}{dx}(xy)=0$$.
$$\sin 2y\frac{dy}{dx}-\sin xy\left( x\frac{dy}{dx}+y.1 \right)=0$$.
$$\sin 2y.\frac{dy}{dx}-x\sin xy.\frac{dy}{dx}=y\sin xy$$.
$$\frac{dy}{dx}=\frac{y\sin (xy)}{\sin 2y-x\sin (xy)}$$.
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### Polynomials & Rational Functions
Many ways problem
## Things you might have tried
Can you find a function that gives the following graph?
There are many ways to approach this problem. We suggest three possibilities below. These are complementary: we could use parts of each of them to reach a conclusion.
• What are the main features of the graph? How can we represent these algebraically?
If we take a step back and look at the graph from a distance, we see that it looks roughly like $y=x$ when $x$ is not near zero.
This gives us some hope that we might be able to draw this graph using a function based around polynomials.
Two main features of the graph are the asymptotes (two vertical and one oblique) and the axis crossings, including at the origin. A third key feature is that the function is odd: the graph has rotational symmetry about the origin.
As we have no scale on the graph, we have a certain amount of freedom to choose values for these, as long as they are reasonably consistent with what we see in the graph.
We often need to decide where to focus our attention and how to rework the problem: this is one of the choice points we face when working on a piece of mathematics. Our three approaches each start with the above observations, but take different routes from there.
#### Approach 1
In this approach, we will focus on the asymptotes.
We will take the asymptotes to be $x=1$, $x=-1$ and $y=x$. (We could have used other numbers, such as $y=2x$ or $x=\pm3$, for example.)
Looking at the $x=1$ asymptote, we see that the graph tends to $+\infty$ on the left of it and to $-\infty$ on its right. We can think through our repertoire of graphs related to polynomials: which graphs have a single asymptote with this sort of behaviour?
The graph of $y=\dfrac{1}{x}$ comes to mind, though we will have to transform it to get it to the right place; something like $y=\dfrac{1}{x-1}$, or perhaps it is $y=-\dfrac{1}{x-1}$, should give us the asymptotes we want. We can do something similar near the asymptote $x=-1$.
These two graphs both tend to $0$ as $x$ gets large, though, and we want our graph to tend to $y=x$ for large $x$. So maybe we also need to think about the graph of $y=x$ itself?
Here are the graphs of these three functions:
We now have three functions, each of which displays a piece of the behaviour we want from the whole graph. Now all that we need to do is to figure some way of putting these three functions together…
How can you convince yourself that your final answer does have the properties of the graph in the problem?
#### Approach 2
In this approach, we begin with our initial observation that once $x$ is large (either positive or negative), the graph looks roughly like $y=x$. So we can try representing the function as $$$\label{eq:xtimes} y=x\cdot\dfrac{\text{polynomial}}{\text{polynomial}},$$$
where the fraction part tends to $1$ as $x$ gets large. For this to happen, we need the two polynomials to have the same degree and also the same leading term.
The denominator has to be zero at the vertical asymptotes, which we will take to be $x=1$ and $x=-1$, so the polynomial $(x-1)(x+1)=x^2-1$ will do.
This means that the numerator would have to be a quadratic of the form $x^2+\text{something}$ if this approach works.
Let us now reflect on another question asked in the suggestion:
• Where does the curve cross the axes? How can we represent this information algebraically?
What will the resulting equation of the graph be? Does this work when you plot it with graph-plotting software?
#### Approach 3
Similarly to the previous approach, we note that the graph looks like $y=x$ when $x$ is large. So we can write the equation of the graph as $$$y=x+\dfrac{\text{polynomial}}{\text{polynomial}},$$$
where this time the fraction part tends to zero as $x$ gets large. For this to happen, the degree of the numerator needs to be less than the degree of the denominator.
To get the asymptotes, we need the denominator to be zero at $x=\pm1$ as before, so we can again take the denominator to be $x^2-1$.
We could go two ways from here.
Firstly, we could add the $x$ and the fraction parts together. This will give us $y=\dfrac{x(x^2-1)+g(x)}{x^2-1}=\dfrac{x^3-x+g(x)}{x^2-1}$ for some polynomial $g(x)$ of degree $0$ or $1$, so $y$ will be a cubic divided by $x^2-1$.
If we can work out the roots of this cubic, we can then write down the whole equation of the graph, and we should be done.
Alternatively, we can think about the fraction part in isolation. Its numerator must be a polynomial of degree $0$ or $1$, and the denominator is $x^2-1$. Also, the whole function is odd, and the initial $x$ is odd, so the fraction must also be odd.
The denominator is $x^2-1$, which is an even function; what does this tell us about the numerator?
Does this give us enough information to find an equation for the graph?
• How do these approaches illuminate different aspects of the graph?
• Which aspects are most easily generalised?
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# What is a sack of cement?
In the United States, a sack refers the amount of cement that occupies a bulk volume of 1.0 ft3. For most portland cement, including API classes of cement, a sack weighs 94 pounds. The sack is the basis for slurry design calculations and is often abbreviated as sk.
Considering this, what is cement unit?
Bag (unit) Cement is commonly sold in bags of 94 pounds weight, because this is about 1 cubic foot of powdered cement.
Also, how many cubic feet is a sack of concrete? Water divided by 7.5 ( gallons of water per cubic foot) equals . 80 cubic foot. One bag of cement, 94 pounds, divided by 196.6 pounds (solid weight of cement ) = . 48 cubic foot.
Besides, how do you calculate concrete sacks?
Calculate Total Bulk Cement and Water Required
1. Let’s learn from the example.
2. Cement Density = Total weight per sack of cement ÷ Total volume of cement slurry.
3. Total sack of cement = 6300 ÷ 11.257 = 560 sacks.
4. Total water = 2501 ÷ 42 = 59.5 bbl.
How much does a sack of sand weigh?
A sandbag should be filled 1/2 to 2/3 full and will weigh 35-40 pounds. 30 sandbags would be about 1,000 pounds and would be the maximum legal limit for a standard pickup truck.
### What is RM in civil engineering?
engineering, building. RM. Room. architecture, internet slang, transportation. architectural, technology.
### How many kg is one bag of cement?
1 bag [portland cement] is equal to 42.63768278 kilogram.
### What is the unit weight of cement?
Unit Weight Of Building Materials
Material Unit Weight
Cement(ordinary) 1440 Kg/ m3
Cement (rapid hardening) 1250 Kg/ m3
Cement Mortar 2000 Kg/ m3
Cement Concrete (Plain) 2400 Kg/ m3
### What is the volume of 50kg cement bag?
The volume of cement bag is usually considered as 35 litres or 0.035 cum or 1.226 cft approximately. To know volume of 50 kg cement (1 bag). 1 m^3 = 35.3147 cubic feet. 1 m^3 = 10^3 liters.
### What is the unit of measurement for concrete?
Because concrete is measured in cubic yards, you will have to convert your cubic-foot measurement to cubic yards. Below is a cubic yard (remember 3 feet = 1 yard). A cubic yard equals 27 cubic feet.
### What is specific gravity of cement?
The specific gravity is normally defined as the ratio between the weight of a given volume of material and weight of an equal volume of water. The portland cement have a specific gravity of value around 3.15. To determine the specific gravity of cement, kerosene which does not recent with cement is used.
### What is the unit of plastering?
Methods of Measurements and Units of Civil Construction Works
Particulars of item Units of measurement
Plastering, points and finishing
1. Plastering – cement or lime mortar (thickness and proportion specified) SQM
2. Pointing SQM
3. White washing, color washing, cement washing, (number of coats specified) SQM
### How much does a bag of cement cost?
Originally Answered: How much does a bag of cement cost? A bag of cement, portland cement, 94 pound in the USA, less in Canada, used with sand, rock and water to make concrete runs right at \$10 around here. Prices vary of course.
### What is cement yield?
The volume occupied by one sack of dry cement after mixing with water and additives to form a slurry of a desired density. Yield is commonly expressed in US units as cubic feet per sack (cu.
### How do you calculate slurry volume?
Slurry volume (in gallon per sack, gal/sk)
Volume = (94 lb / Specific Gravity of cement x 8.33 lb/gal) + (weight of additive, Ib / Specific Gravity of additive x 8.33 lb/gal) + water volume, gal.
4.5 cubic feet
### How do you calculate the balance of a concrete plug?
Balanced Cement Plug Calculation
1. Determine volume of cement that you need.
2. Determine height of cement and spacer when pipe in hole.
3. Determine displace volume to balance the hydrostatic both sides.
4. When you pull cement stinger, you will have the balance set properly like this.
### How much is concrete per post?
Remember, the depth of the post hole should be one-half of the above-ground post height. (Example: For a 6 feet above ground post, use a post with an overall height of 9 feet and place 3 feet in the ground). The calculator will indicate the number of 50 lb. bags of QUIKRETE® Fast-Setting Concrete you need.
### How many pounds is a sack?
1 sack = 364.00029691882 lbs. 1 x 364.00029691882 lbs = 364.00029691882 Pounds. In relation to the base unit of [mass weight] => (kilograms), 1 Sacks (sack) is equal to 165.10762268 kilograms, while 1 Pounds (lbs) = 0.453592 kilograms.
### How many 80lb bags of concrete makes a yard?
How many 80 lb bags of concrete do I need for 1 yard? The finished volume of an 80# bag of Sacrete or Quikrete (pre-mixed cement, sand and gravel) is 0.6 cubic foot (stated on the bag). There are 27 cubic feet in a cubic yard. Dividing 27 cubic feet by the volume of the bag will give you the number of bags you need.
### How do you find out cubic feet?
To find the cubic feet of a square or rectangular shape, measure the length, width and height of the object in feet or inches. Once you have these three numbers, multiply them together to get the volume. If you measured in inches, divide your answer by 1728 to convert the measurement to feet.
### What is the meaning of cubic feet?
The cubic foot (symbol ft3) is an imperial and US customary (non-metric) unit of volume, used in the United States and the United Kingdom. It is defined as the volume of a cube with sides of one foot (0.3048 m) in length. Its volume is 28.3168 L (about ?135 of a cubic metre or meter).
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Math
How many real roots does f(x) = x^3+5x+1 have?
x^3 = odd polynomial = at least 1 real root
Can I follow Descartes' rule for this? No sign change, so there aren't any positive zeros.
when I sub in -x for x
(-x)^3+ 5(-x) + 1
= -x^3-5x+1
Then there is one sign change, so 1 negative root.
So my conclusion is that there is one real root. Is this right?
1. 👍 0
2. 👎 0
3. 👁 205
1. I agree. The Descartes rule says that there cannot be more than one real root in this case, and that it will be negative. There is another rule that says there must be at least one real root. Therefore the number of real roots is one.
1. 👍 0
2. 👎 0
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# If 0 < r < 1 < s < 2, which of the following must be less
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If 0 < r < 1 < s < 2, which of the following must be less [#permalink]
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25 Oct 2010, 11:56
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If 0 < r < 1 < s < 2, which of the following must be less than 1?
I. r/s
II. rs
III. s - r
A. I only
B. II only
C. III only
D. I and II
E. I and III
OPEN DISCUSSION OF THIS QUESTION IS HERE: if-0-r-1-s-2-which-of-the-following-must-be-less-than-128112.html
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Re: GMAT Prep - If 0 < r < 1 < s < 2, which of the following..?? [#permalink]
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25 Oct 2010, 12:07
6
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n2739178 wrote:
GMAT Prep question... What's the easiest way to solve this using MGMAT style case-charts?
Attachment:
q2.jpg
Not sure what you mean by MGMAT style case charts. But this question is quite straight forward
For r,s positive : r/s is less than 1, when r<s. which we know
rs : No need for this to be less than 1 (r could be almost 1 and s could be 1.5 or 1.7)
s-r : No need for this to be less than 1 (r could be almost 0 and s could be 1.5 or 1.7)
Think of extreme cases in such questions, you should always be able to find the exception like that
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Re: GMAT Prep - If 0 < r < 1 < s < 2, which of the following..?? [#permalink]
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25 Oct 2010, 14:28
aaa yes of course - extreme cases - that's what I remember MGMAT going on about for these sorts of questions - awesome, thanks!
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Re: GMAT Prep - If 0 < r < 1 < s < 2, which of the following..?? [#permalink]
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10 May 2012, 06:03
Is the following true:
If r and s are positive numbers and s > r, then r/s will always be < 1.
Math Expert
Joined: 02 Sep 2009
Posts: 47981
Re: GMAT Prep - If 0 < r < 1 < s < 2, which of the following..?? [#permalink]
### Show Tags
10 May 2012, 06:14
13
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honsmos wrote:
Is the following true:
If r and s are positive numbers and s > r, then r/s will always be < 1.
If given that r>0, s>0 and s>r then divide s>r by s (we can safely do that since s>0): 1>r/s.
Complete solution:
If 0 < r < 1 < s < 2, which of the following must be less than 1?
I. r/s
II. rs
III. s - r
A. I only
B. II only
C. III only
D. I and II
E. I and III
Notice that we are asked "which of the following MUST be lees than 1, not COULD be less than 1. For such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.
Given: $$0 < r < s$$ --> divide by $$s$$ (we can safely do this since we know that $$s>0$$) --> $$\frac{r}{s}<1$$, so I must be true;
II. rs: if $$r=\frac{9}{10}<1$$ and $$1<(s=\frac{10}{9})<2$$ then $$rs=1$$, so this statement is not alway true;
III. s-r: if $$r=0.5<1$$ and 1$$<(s=1.5)<2$$ then $$s-r=1$$, so this statement is not alway true.
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Re: If 0 < r < 1 < s < 2, which of the following must be less [#permalink]
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# Greatest integer function grapher software
Greatest integer function, step function in trigonometry i am a mother who is trying to home school my high school senior through trigonometry. The greatest integer function is a function such that the output is the greatest integer that is less than or equal to the input. Many functions can be graphed by accessing commands commonly used in the tinspire calculator application. The greatestinteger function is denoted by y x for all real values of x, the greatestinteger function returns the largest integer less than or equal to x. Floor x can be entered in standardform and inputform as. If necessary, press 1 to activate the first category. Greatest integer function study material for iit jee. So every point on the real line has a right derivative with the greatest integer. Let me show you how, i broke the question into parts and solved in an easy way. Greatest integer function, step function in trigonometry i am a mother who is trying to home school my high school. Theres a way of looking at things, where you can say that a function has a left derivative and a right derivative at any point. The first variation shows the function that replaces the x coordinate with it subtracted by three. Kuta works option to control how long choices are hidden.
Interactive, free online graphing calculator from geogebra. About graphing greatest integer function graphing greatest integer function is the stuff which is needed to the children who study high school math. Multiple functions at once you may enter up to 5 functions simultaneously. The greatest integer function has its own notation and tells us to round whatever decimal number it is given down to the. How to graph a rational function when there is a common factor in the numerator. In mathematics and computer science, the floor function is the function that takes as input a real number x \displaystyle x x and gives as output the greatest integer less than or equal to x \displaystyle. It will only take the value of 1 at isolated points. The greatest integer function is denoted by f x x and is defined as the greatest integer less or equal to x. The greatest integer function has its own notation and tells us to round whatever decimal.
Follow 41 views last 30 days vipul sharma on 8 jun 2016. Rewrite each absolute value expression as a piecewise function. Greatest integer function domain and range with graph. For all real numbers, x, the greatest integer function returns the largest integer less than or equal to x. It is also called the floor function and step function symbol of greatest integer function is. Floorx gives the greatest integer less than or equal to x. A step function of x which is the least integer greater than or equal to x. Graphing such functions is an easy task, if you know the basics of each type of graphs. Graph a step function in tinspire graphs application dummies.
The video then shows the variations of this function. Solutionthe greatest integer function is denoted by y x. Start graphing four function and scientific check out the newest additions to the desmos calculator family. The greatest integer function one of the most commonly used step functions is the greatest integer function. Nov 11, 2016 sketch a graph of the greatest integer function ycosx 2192971. Mar 16, 2020 you can use floor for the greatest integer function. The greatestinteger function is denoted by y x for all real values of x, the greatestinteger. The greatest integer function is a quite standard name for what is also known as the floor function. Graphs with holes precalculus polynomial and rational functions.
Graphing the greatest integer or floor function duration. You can find both of these functions under the misc tab. Which equation matches the graph of the greatest integer. In this activity, you will create a function similar to the greatest integer function graph by having a group of students stand in a line in front of a motion detector. In the above graph, the left endpoint in every step is blockeddark dot to show that the. If the number is not an integer, use the next smaller integer. How to plot the graph of greatest integer function. The graph of the greatest integer function resembles an ascending staircase. Sep 24, 2008 thanks to all of you who support me on patreon. I believe the easiest way to solve this problem would be to graph the function.
In essence, it rounds down a real number to the nearest lowest integer. The greatest integer function is a piecewise defined. Graphing the greatest integer or floor function youtube. Position the cursor next to the first available function line. What is the greatest integer function, and how do you integrate it. Chart graph greatest integer function greatest integer function greatest integer function example. The greatest integer function, y intx, is one such graph, which you create by following these steps.
Intermediate value theorem calculus 1 ab precalculus duration. Alright, weve got this greatest integer function concept down, but this may leave you curious about what. Sep 30, 2012 what does the graph of the derivative of the greatest integer function look like. The video first describes the basic greatest integer function. This means the greatest integer less than or equal to the number gave. But it is easy to get a little confused when we apply the greatest integer function to negative numbers. A timesaving video on the greatest integer function, or the step function. In essence, it rounds down to the the nearest integer. Walkaround activities domino like matching sorts interactive bulletin board ideas practice. Kuta works option to hide answers and results from students until after due date new. It could only be solved of you have the upper and lower limits. For exact numeric quantities, floor internally uses numerical approximations to establish.
Greatest integer function desmos graphing calculator. The greatest integer brackets should be a tip off that some sort of step function graph is appropriate. Nov 16, 2012 it returns the greatest integer less than or equal to x. What does the graph of the derivative of the greatest integer function look like. The greatest integer function has its own notation and tells us to round whatever decimal number it is given down to the nearest integer, or the greatest integer that is less than the number. Floorx, a gives the greatest multiple of a less than or equal to x. The integer function is sometimes called the step function because of the stair step effect obtained when graphing it. Sketch a graph of the greatest integer function ycosx 2192971 1. Graph a step function in tinspire graphs application. Someone suggested its because of some editing software that comes with microsoft. Trial software how to plot the graph of greatest integer function. The greatest integer function is a function that takes an input, increases it by two.
How to draw the graph of sin x where greatest integer function. How to graph the greatest integer or floor function math. Be able to graph each of the above kinds of functions with translations warm up. Graph functions, plot data, evaluate equations, explore transformations, and much more for free. Below is the process to find the graph of the greatest integer function. For each value of x, f x assumes the value of the greatest integer, less than or equal to x.
The greatest integer function concept precalculus video by. The graph of a greatest integer function is shown in. You may also want to use dot mode instead of connected mode when graphing the integer function. Examples, solutions, videos, worksheets, and activities to help precalculus students learn about the greatest integer function. Four function and scientific check out the newest additions to the desmos calculator family.
The greatest integer function, also called step function, is a piecewise function whose graph looks like the steps of a staircase. The greatest integer function, y intx, is one such graph, which you create by following. To understand the behavior of this function, in terms of a graph, lets construct a table of values. Or, you can use ceil for the least integer function. For all real numbers, x, the greatest integer function returns the largest. You can use floor for the greatest integer function. Greatest integer function definition, examples, and graph. Free functions calculator explore function domain, range, intercepts, extreme points and asymptotes stepbystep this website uses cookies to ensure you get the best experience. The greatest integer function is a function that takes an input, adds an integer to it, and this is the output. Similarly, the ceiling function maps x \displaystyle x to the least integer greater than or equal to x \displaystyle x, denoted ceil. Oct 09, 2002 no, the derivative of 2 would be undefined. If you graph the expression inside the brackets you can visually see the general path that will be involved. The graph of a greatest integer function is shown in figure given below.
You just need to split it on integer points or in other words wherever it changes its value. The greatest integer function concept precalculus video. Click on the plot to clear the graph, or drag again to clear and begin a new graph. The greatest functions are defined piecewise its domain is a group of real numbers that are divided into intervals like 4, 3, 3, 2, 2, 1, 1, 0 and so on. How to graph a rational function when there is a common factor in the numerator and denominator. Greatest integer function x indicates an integral part of the real number which is nearest and smaller integer to. Apr 18, 2017 it could only be solved of you have the upper and lower limits. Greatest integer function gives out greatest integer less than or equal to x. I need to generate something like what i generated in sage. Sketch a graph of the greatest integer function ycosx. Be sure to use a decimal setting when graphing the greatest integer function, or you will. Graphing greatest integer function is the stuff which is needed to the children who study high school math. The greatest integer function is a function in which the goal is to find the. We are studying special functions and i am stumped with how i can explain greatest integer function to him.
Of course building the diagram then will take more time, possibly several seconds. Floor x returns an integer when is any numeric quantity, whether or not it is an explicit number. Greatest integer function greatest integer function. Greatest integer function x indicates an integral part of the real number x. The greatest integer function is a piecewise defined function.
If the two derivatives are the same, then the function is differentiable at that point. Wouldnt it be just a horizontal line with open circles at integers. Multiple functions at once you may enter up to 5 functions simultaneously, each terminated by a semicolon. Area of greatest integer function mathematics stack exchange. Be sure to use a decimal setting when graphing the greatest integer function, or you will get weird results. The greatest integer function examples, solutions, videos. Sorry brother,but this is showing a continuous graph whereas the greatest integer function is a discontinuous function,sorry but this plot is wrong. Greatest integer function or step funtion definition. Greatest integer function or step funtion definition, graph. The graph shows that it is increasing not strictly manytoone function. Greatest integer function and graph math warehouse.
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# Find (–7) – 8 – (–25) Find (–7) – 8 – (–25) Find (–7) – 8 – (–25)
Harshit Singh
3 years ago
Dear Student
(-7)–8–(-25) = -7 -8 + 25
= -15 + 25
= 10
Thanks
| 87 | 158 |
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+0
# please help
0
77
1
Kevin and randy mouse have a jar containing 97 coins, all of which are either quarters or nickels. The total value of the coins in the jar is \$14.25. How many of each type of coin do they have?
Mar 1, 2020
### 1+0 Answers
#1
+1
Let the number of nickels =N
The number of quarters = 97 - N
0.05N + 0.25[97 -N] = 14.25, solve for N
N = 50 Nickels
97 - 50 = 47 Quarters
Mar 1, 2020
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Anda di halaman 1dari 45
# Angle Modulation Frequency Modulation
## Consider again the general carrier ( ) ( )
c c c c
+ t V = t v cos
( )
c c
+ t represents the angle of the carrier.
There are two ways of varying the angle of the carrier.
a) By varying the frequency, e
c
Frequency Modulation.
b) By varying the phase, |
c
Phase Modulation
Frequency Modulation
In FM, the message signal m(t) controls the frequency f
c
of the carrier. Consider the
carrier
( ) ( ) t V = t v
c c c
cos
then for FM we may write:
FM signal
( ) ( ) ( ) t + f V = t v
c c s
deviation frequency 2 cos
,where the frequency deviation
will depend on m(t).
Given that the carrier frequency will change we may write for an instantaneous
carrier signal
( ) ( ) ( )
i c i c i c
V = t f V = t V cos 2 cos cos
where |
i
is the instantaneous angle =
t f = t
i i
2
and f
i
is the instantaneous
frequency.
Frequency Modulation
Since
t f =
i i
2 then
dt
d
= f f =
dt
d
i
i i
i
2
1
or 2
i.e. frequency is proportional to the rate of change of angle.
If f
c
is the unmodulated carrier and f
m
is the modulating frequency, then we may
deduce that
( )
dt
d
= t f + f = f
i
m c c i
2
1
cos
Af
c
is the peak deviation of the carrier.
Hence, we have
( ) t f + f =
dt
d
m c c
i
cos
2
1
,i.e.
( ) t f + f =
dt
d
m c c
i
cos 2 2
Frequency Modulation
After integration i.e.
( ) ( )
}
dt t f +
m c c
cos 2
( )
m
m c
c i
t f
+ t =
sin 2
( ) t
f
f
+ t =
m
m
c
c i
sin
Hence for the FM signal, ( ) ( )
i c s
V = t v cos
( ) ( )
|
|
.
|
\
|
t
f
f
+ t V = t v
m
m
c
c c s
sin cos
Frequency Modulation
The ratio
m
c
f
f
is called the Modulation Index denoted by | i.e.
frequency modulating
deviation frequency Peak
=
Note FM, as implicit in the above equation for v
s
(t), is a non-linear process i.e.
the principle of superposition does not apply. The FM signal for a message m(t) as a
band of signals is very complex. Hence, m(t) is usually considered as a 'single tone
modulating signal' of the form
( ) ( ) t V = t m
m m
cos
Frequency Modulation
The equation ( ) ( )
|
|
.
|
\
|
t
f
f
+ t V = t v
m
m
c
c c s
sin cos
may be expressed as Bessel
series (Bessel functions)
( ) ( ) ( )
= n
m c n c s
t n + J V = t v cos
where J
n
(|) are Bessel functions of the first kind. Expanding the equation for a few
terms we have:
+ + + +
+ + + =
+
t J V t J V
t J V t J V t J V t v
m c m c
m c m c
c
f f
m c c
f f
m c c
f f
m c c
f f
m c c
f
c c s
2 Amp
2
2 Amp
2
Amp
1
Amp
1
Amp
0
) 2 ( cos ) ( ) 2 ( cos ) (
) ( cos ) ( ) ( cos ) ( ) ( cos ) ( ) (
e e | e e |
e e | e e | e |
FM Signal Spectrum.
The amplitudes drawn are completely arbitrary, since we have not found any value for
J
n
(|) this sketch is only to illustrate the spectrum.
Generation of FM signals Frequency
Modulation.
An FM demodulator is:
a voltage-to-frequency converter V/F
a voltage controlled oscillator VCO
In these devices (V/F or VCO), the output frequency is dependent on the input voltage
amplitude.
V/F Characteristics.
Apply V
IN
, e.g. 0 Volts, +1 Volts, +2 Volts, -1 Volts, -2 Volts, ... and measure the
frequency output for each V
IN
. The ideal V/F characteristic is a straight line as
shown below.
f
c
, the frequency output when the input is zero is called the undeviated or nominal
carrier frequency.
V
f
is called the Frequency Conversion Factor,
denoted by o per Volt.
V/F Characteristics.
Consider now, an analogue message input,
( ) ( ) t V = t m
m m
cos
As the input m(t) varies from
m m
V V + 0
the output frequency will vary from a
maximum, through f
c
, to a minimum
frequency.
V/F Characteristics.
For a straight line, y = c + mx, where c = value of y when x = 0, m = gradient, hence
we may say
IN OUT
V + f = f
c
and when V
IN
= m(t)
( ) t m + f = f
c OUT
,i.e. the deviation depends on m(t).
Considering that maximum and minimum input amplitudes are +V
m
and -V
m
respectively, then
m c
m c
V f = f
V + f = f
min
max
on the diagram on the previous slide.
The peak-to-peak deviation is f
max
f
min
, but more importantly for FM the peak
deviation Af
c
is
Peak Deviation,
m c
V = f Hence, Modulation Index,
m
m
m
c
f
V
=
f
f
=
Summary of the important points of FM
In FM, the message signal m(t) is assumed to be a single tone frequency,
( ) ( ) t V = t m
m m
cos
The FM signal v
s
(t) from which the spectrum may be obtained as
( ) ( ) ( )
= n
m c n c s
t n + J V = t v cos
where J
n
(|) are Bessel coefficients and Modulation Index,
m
m
m
c
f
V
=
f
f
=
o Hz per Volt is the V/F modulator, gradient or Frequency Conversion Factor,
o per Volt
o is a measure of the change in output frequency for a change in input amplitude.
Peak Deviation (of the carrier frequency from f
c
)
m c
V = f
FM Signal Waveforms.
The diagrams below illustrate FM signal waveforms for various inputs
At this stage, an input digital data
sequence, d(t), is introduced
the output in this case will be FSK,
(Frequency Shift Keying).
FM Signal Waveforms.
Assuming
s 0' for
s 1' for ) (
V
V t d
=
+ =
)
`
= =
+ = =
s 0' for
s 1' for
0
1
V f f f
V f f f
c OUT
c OUT
o
o the output switches
between f
1
and f
0
.
FM Signal Waveforms.
The output frequency varies gradually from f
c
to (f
c
+ oV
m
), through f
c
to
(f
c
- oV
m
) etc.
FM Spectrum Bessel Coefficients.
In the series for v
s
(t), n = 0 is the carrier component, i.e. ) cos( ) (
0
t J V
c c
e | , hence the
n = 0 curve shows how the component at the carrier frequency, f
c
, varies in amplitude,
with modulation index |.
|
J
n
(|)
| = 2.4 | = 5
FM Spectrum Bessel Coefficients.
Hence for a given value of modulation index |, the values of J
n
(|) may be read off the
graph and hence the component amplitudes (V
c
J
n
(|)) may be determined.
A further way to interpret these curves is to imagine them in 3 dimensions
Examples from the graph
| = 0: When | = 0 the carrier is unmodulated and J
0
(0) = 1, all other J
n
(0) = 0, i.e.
| = 2.4: From the graph (approximately)
J
0
(2.4) = 0, J
1
(2.4) = 0.5, J
2
(2.4) = 0.45 and J
3
(2.4) = 0.2
Effect of increase
in modulation
index , , due to
increasing
amplitude
Effect of increase
in modulation
index , , due to
decrease in f
m
Significant Sidebands Spectrum.
As may be seen from the table of Bessel functions, for values of n above a certain
value, the values of J
n
(|) become progressively smaller. In FM the sidebands are
considered to be significant if J
n
(|) > 0.01 (1%).
Although the bandwidth of an FM signal is infinite, components with amplitudes
V
c
J
n
(|), for which J
n
(|) < 0.01 are deemed to be insignificant and may be ignored.
Example: A message signal with a frequency f
m
Hz modulates a carrier f
c
to produce
FM with a modulation index | = 1. Sketch the spectrum.
n J
n
(1) Amplitude Frequency
0 0.7652
0.7652V
c
f
c
1 0.4400 0.44V
c
f
c
+f
m
f
c
- f
m
2 0.1149 0.1149V
c
f
c
+2f
m
f
c
- 2f
m
3 0.0196 0.0196V
c
f
c
+3f
m
f
c
-3 f
m
4 0.0025 Insignificant
5 0.0002 Insignificant
Significant Sidebands Spectrum.
As shown, the bandwidth of the spectrum containing significant
components is 6f
m
, for | = 1.
Significant Sidebands Spectrum.
The table below shows the number of significant sidebands for various modulation
indices (|) and the associated spectral bandwidth.
| No of sidebands > 1% of
unmodulated carrier
Bandwidth
0.1 2 2f
m
0.3 4 4f
m
0.5 4 4f
m
1.0 6 6f
m
2.0 8 8f
m
5.0 16 16f
m
10.0 28 28f
m
e.g. for | = 5,
16 sidebands
(8 pairs).
Carsons Rule for FM Bandwidth.
An approximation for the bandwidth of an FM signal is given by
BW = 2(Maximum frequency deviation + highest modulated
frequency)
) ( 2 Bandwidth
m c
f f + A =
Carsons Rule
Narrowband and Wideband FM
From the graph/table of Bessel functions it may be seen that for small |, (| s 0.3)
there is only the carrier and 2 significant sidebands, i.e. BW = 2fm.
FM with | s 0.3 is referred to as narrowband FM (NBFM) (Note, the bandwidth is
the same as DSBAM).
For | > 0.3 there are more than 2 significant sidebands. As | increases the number of
sidebands increases. This is referred to as wideband FM (WBFM).
Narrowband FM NBFM
Wideband FM WBFM
VHF/FM
m c
V f o = A
VHF/FM (Very High Frequency band = 30MHz 300MHz) radio transmissions, in the
band 88MHz to 108MHz have the following parameters:
Max frequency input (e.g. music) 15kHz
f
m
Deviation 75kHz
Modulation Index | 5
m
c
f
f A
= |
For | = 5 there are 16 sidebands and the FM signal bandwidth is 16fm = 16 x 15kHz
= 240kHz. Applying Carsons Rule BW = 2(75+15) = 180kHz.
The FM spectrum contains a carrier component and an infinite number of sidebands
at frequencies f
c
nf
m
(n = 0, 1, 2, )
FM signal,
=
+ =
n
m c n c s
t n J V t v ) cos( ) ( ) ( e e |
In FM we refer to sideband pairs not upper and lower sidebands. Carrier or other
components may not be suppressed in FM.
The relative amplitudes of components in FM depend on the values J
n
(|), where
m
m
f
V o
| =
thus the component at the carrier frequency depends on m(t), as do all the
other components and none may be suppressed.
Components are significant if J
n
(|) > 0.01. For |<<1 (| ~ 0.3 or less) only J
0
(|) and
J
1
(|) are significant, i.e. only a carrier and 2 sidebands. Bandwidth is 2f
m
, similar to
DSBAM in terms of bandwidth - called NBFM.
Large modulation index
m
c
f
f A
= | means that a large bandwidth is required called
WBFM.
The FM process is non-linear. The principle of superposition does not apply. When
m(t) is a band of signals, e.g. speech or music the analysis is very difficult
(impossible?). Calculations usually assume a single tone frequency equal to the
maximum input frequency. E.g. m(t) band 20Hz 15kHz, fm = 15kHz is used.
Power in FM Signals.
From the equation for FM
=
+ =
n
m c n c s
t n J V t v ) cos( ) ( ) ( e e |
we see that the peak value of the components is V
c
J
n
(|) for the n
th
component.
Single normalised average power =
2
2
) (
2
RMS
pk
V
V
=
|
|
.
|
\
|
then the n
th
component is
( )
2
) (
2
) (
2
2
| |
n c n c
J V J V
= |
.
|
\
|
Hence, the total power in the infinite spectrum is
Total power
=
=
n
n c
T
J V
P
2
)) ( (
2
|
2
)) ( (
2
2
2
2
c
n
n
c
V
J
V
= =
=
|
Power in FM Signals.
Now consider if we generate an FM signal, it will contain an infinite number of
sidebands. However, if we wish to transfer this signal, e.g. over a radio or cable, this
implies that we require an infinite bandwidth channel. Even if there was an infinite
channel bandwidth it would not all be allocated to one user. Only a limited bandwidth
is available for any particular signal. Thus we have to make the signal spectrum fit into
the available channel bandwidth. We can think of the signal spectrum as a train and
the channel bandwidth as a tunnel obviously we make the train slightly less wider
than the tunnel if we can.
Power in FM Signals.
However, many signals (e.g. FM, square waves, digital signals) contain an infinite
number of components. If we transfer such a signal via a limited channel bandwidth,
we will lose some of the components and the output signal will be distorted. If we put
an infinitely wide train through a tunnel, the train would come out distorted, the
question is how much distortion can be tolerated?
Generally speaking, spectral components decrease in amplitude as we move away
from the spectrum centre.
Power in FM Signals.
In general distortion may be defined as
spectrum in total Power
spectrum d Bandlimite in Power - spectrum in total Power
= D
T
BL T
P
P P
D
=
With reference to FM the minimum channel bandwidth required would be just wide
enough to pass the spectrum of significant components. For a bandlimited FM
spectrum, let a = the number of sideband pairs, e.g. for | = 5, a = 8 pairs
(16 components). Hence, power in the bandlimited spectrum P
BL
is
=
=
a
a n
n c
BL
J V
P
2
)) ( (
2
|
= carrier power + sideband powers.
Power in FM Signals.
Since
2
2
c
T
V
P =
=
=
=
=
a
a n
n
c
a
a n
n
c c
J
V
J
V V
D
2
2
2
2 2
)) ( ( 1
2
)) ( (
2 2
|
|
Distortion
Also, it is easily seen that the ratio
=
= = =
a
a n
n
T
BL
J
P
P
D
2
)) ( (
spectrum in total Power
spectrum d Bandlimite in Power
|
= 1 Distortion
i.e. proportion p
f
power in bandlimited spectrum to total power =
=
a
a n
n
J
2
)) ( ( |
Example
Consider NBFM, with | = 0.2. Let V
c
= 10 volts. The total power in the infinite
2
2
c
V
spectrum
= 50 Watts, i.e.
=
a
a n
n
J
2
)) ( ( |
= 50 Watts.
From the table the significant components are
n J
n
(0.2) Amp = V
c
J
n
(0.2)
Power =
2
) (
2
Amp
0 0.9900 9.90 49.005
1 0.0995 0.995 0.4950125
P
BL
= 49.5 Watts
i.e. the carrier + 2 sidebands contain 99 . 0
50
5 . 49
= or 99% of the total power
Example
Distortion =
01 . 0
50
5 . 49 50
=
T
BL T
P
P P
or 1%.
Actually, we dont need to know V
c
, i.e. alternatively
Distortion =
1
1
2
)) 2 . 0 ( ( 1
n
n
J
(a = 1)
D = 01 . 0 ) 0995 . 0 ( ) 99 . 0 ( 1
2 2
=
Ratio
99 . 0 1 )) ( (
1
1
2
= = =
=
D J
P
P
n
n
T
BL
|
FM Demodulation General Principles.
An FM demodulator or frequency discriminator is essentially a frequency-to-voltage
converter (F/V). An F/V converter may be realised in several ways, including for
example, tuned circuits and envelope detectors, phase locked loops etc.
Demodulators are also called FM discriminators.
Before considering some specific types, the general concepts for FM demodulation
will be presented. An F/V converter produces an output voltage, V
OUT
which is
proportional to the frequency input, f
IN
.
FM Demodulation General Principles.
If the input is FM, the output is m(t), the analogue message signal. If the input is FSK,
the output is d(t), the digital data sequence.
In this case f
IN
is the independent variable and V
OUT
is the dependent variable (x and
y axes respectively). The ideal characteristic is shown below.
We define V
o
as the output when f
IN
= f
c
, the nominal input frequency.
FM Demodulation General Principles.
f
V
A
A
is called the voltage conversion factor
i.e. Gradient = Voltage Conversion Factor, K volts per Hz.
Considering y = mx + c etc. then we may say V
OUT
= V
0
+ Kf
IN
from the frequency
modulator, and since V
0
= V
OUT
when f
IN
= f
c
then we may write
IN OUT
V K V V o + =
0
where V
0
represents a DC offset in V
OUT
. This DC offset may be removed by level
shifting or AC coupling, or the F/V may be designed with the characteristic shown next
FM Demodulation General Principles.
The important point is that V
OUT
= KoV
IN
. If V
IN
= m(t) then the output contains the
message signal m(t), and the FM signal has been demodulated.
FM Demodulation General Principles.
Often, but not always, a system designed so that
o
1
= K
, so that Ko = 1 and
V
OUT
= m(t). A complete system is illustrated.
FM Demodulation General Principles.
Methods
Tuned Circuit One method (used in the early days of FM) is to use the slope of a
tuned circuit in conjunction with an envelope detector.
Methods
The tuned circuit is tuned so the f
c
, the nominal input frequency, is on the slope, not at
the centre of the tuned circuits. As the FM signal deviates about f
c
on the tuned circuit
slope, the amplitude of the output varies in proportion to the deviation from f
c
. Thus
the FM signal is effectively converted to AM. This is then envelope detected by the
diode etc to recover the message signal.
Note: In the early days, most radio links were AM (DSBAM). When FM came along,
with its advantages, the links could not be changed to FM quickly. Hence, NBFM was
used (with a spectral bandwidth = 2fm, i.e. the same as DSBAM). The carrier
frequency fc was chosen and the IF filters were tuned so that fc fell on the slope of the
filter response. Most FM links now are wideband with much better demodulators.
A better method is to use 2 similar circuits, known as a Foster-Seeley Discriminator
Foster-Seeley Discriminator
This gives the composite characteristics shown. Diode D
2
effectively inverts the f
2
tuned circuit response. This gives the characteristic S type detector.
Phase Locked Loops PLL
A PLL is a closed loop system which may be used for FM demodulation. A full
analytical description is outside the scope of these notes. A brief description is
presented. A block diagram for a PLL is shown below.
Note the similarity with a synchronous demodulator. The loop comprises a multiplier,
a low pass filter and VCO (V/F converter as used in a frequency modulator).
Phase Locked Loops PLL
The input f
IN
is applied to the multiplier and multiplied with the VCO frequency output
f
O
, to produce E = (f
IN
+ f
O
) and A = (f
IN
f
O
).
The low pass filter passes only (f
IN
f
O
) to give VOUT which is proportional to (f
IN
f
O
).
If f
IN
~ f
O
but not equal, V
OUT
= V
IN
, of
IN
f
O
is a low frequency (beat frequency) signal to
the VCO.
This signal, V
IN
, causes the VCO output frequency f
O
to vary and move towards f
IN
.
When f
IN
= f
O
, V
IN
(f
IN
f
O
) is approximately constant (DC) and f
O
is held constant, i.e
locked to f
IN
.
As f
IN
changes, due to deviation in FM, f
O
tracks or follows f
IN
. V
OUT
= V
IN
changes to drive
f
O
to track f
IN
.
V
OUT
is therefore proportional to the deviation and contains the message signal m(t).
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CC-MAIN-2019-26
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latest
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en
| 0.845349 |
https://www.teacherspayteachers.com/Product/Math-Doodle-Sum-of-Angle-Measures-EASY-to-Use-Notes-PPT-included-3850962
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Math Doodle - Sum of Angle Measures - EASY to Use Notes - PPT included!
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Product Description
Sum of Angle Measures Math Doodle Notes
Guided Practice Sheet
PowerPoint – to show students the KEY
– two versions of both sheets included: INB and large 8.5 x 11 size
TEKS and CC Aligned – plan for the FULL year included!
This download is part of a larger bundle of 68 Doodle Notes found here:
Doodle Notes - 6th Grade Math ALL the DOODLE NOTES ~ So Fun and Engaging!
These 6th grade math doodle notes are a great way to help the students learn math concepts! Students are engaged as they take creative notes and decorate to make them their own. Doodling helps content retention and students love these! Students will have the most complete creative math interactive notebook ever! As an added bonus, so many times parents want to help but don’t understand how to do math anymore and these descriptive step by step sheets help students and parents understand math. This interactive notebook works as an amazing resource to refer to for the whole year!
LIST of 68 Doodle Notes plus Guided Practice Sheets
1. The Number System
2. Integers and Their Opposites
3. Comparing and Ordering Integers
4. Absolute Value
5. Rational Numbers
6. Opposites and Absolute Value of Rational Numbers
7. Comparing and Ordering Rational Numbers
8. Multiplying Fractions
9. Multiplying Mixed Numbers
10. Dividing Fractions
11. Dividing Mixed Numbers
12. Multiplying Decimals with Models
13. Multiplying Decimals with the Standard Algorithm
14. Dividing Decimals with Models
15. Dividing Decimals with the Standard Algorithm
16. Multiplying and Dividing Fractions and Decimals
17. Checking with the Digital Root
18. Adding Integers with the Same Sign
19. Adding Integers with Different Signs
20. Subtracting Integers
22. Multiplying Integers
23. Dividing Integers
24. Integer Operations
25. Ratios
26. Rates
27. Comparing Additive and Multiplicative Relationships
28. Ratios and Rates in Tables and Graphs
29. Converting Measurements
30. Understanding Percent
31. Benchmark Fractions to Decimals
32. Percent with Fractions and Decimals
33. Percent Problems
34. Exponents
35. Prime Factorization
36. Order of Operations
37. Modeling Equivalent Expressions
38. Evaluating Equivalent Expressions
39. Generating Equivalent Expressions
40. Writing Equations
42. Multiplication and Division Equations
43. Writing Inequalities
45. Multiplication & Division of Inequalities with Positive Numbers
46. Multiplication & Division of Inequalities with Rational Numbers
47. Graphing and the Coordinate Plane
48. Independent and Dependent Variables
49. Writing Equations from Tables
50. Algebraic Relationships in Tables and Graphs
51. Forming Triangles
52. Sum of Angle Measures
53. Relationships of Sides and Angles
55. Area of Triangles
56. Solving Area Equations
57. Solving Volume Equations
58. Measures of Center (Mean, Median, and Range)
59. Box Plots
60. Dot Plots
61. Stem and Leaf Plots & Histograms
62. Categorizing Data
63. Checking Accounts
64. Balancing a Check Register
65. Credit and Debit Cards
66. Credit Reports
67. Paying for College
68. Wages, Salaries and Careers
Total Pages
20 pages
Included
Teaching Duration
45 minutes
Report this Resource
\$3.00
| 813 | 3,407 |
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| 4.3125 | 4 |
CC-MAIN-2018-43
|
latest
|
en
| 0.793723 |
http://www.mathscore.com/math/practice/Counting%20Objects%202/
| 1,369,114,393,000,000,000 |
text/html
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crawl-data/CC-MAIN-2013-20/segments/1368699730479/warc/CC-MAIN-20130516102210-00069-ip-10-60-113-184.ec2.internal.warc.gz
| 593,077,785 | 4,634 |
## Counting Objects 2 - Sample Math Practice Problems
The math problems below can be generated by MathScore.com, a math practice program for schools and individual families. References to complexity and mode refer to the overall difficulty of the problems as they appear in the main program. In the main program, all problems are automatically graded and the difficulty adapts dynamically based on performance. Answers to these sample questions appear at the bottom of the page. This page does not grade your responses.
See some of our other supported math practice problems.
### Complexity=10, Mode=rect
Count the number of objects.
1. How many squares are there?
2. How many squares are there?
### Complexity=10, Mode=rand
Count the number of objects.
1. How many squares are there?
2. How many squares are there?
### Complexity=15, Mode=rect
Count the number of objects.
1. How many squares are there?
2. How many objects are there?
### Complexity=20, Mode=rect
Count the number of objects.
1. How many objects are there?
2. How many objects are there?
### Complexity=20, Mode=line
Count the number of objects.
1. How many objects are there?
2. How many squares are there?
### Complexity=10, Mode=rect
Count the number of objects.
1 How many squares are there?
Solution
1 2 3 4 5 6
2 How many squares are there?
Solution
1 2 3 4 5 6 7 8 9 10
### Complexity=10, Mode=rand
Count the number of objects.
1 How many squares are there?
Solution
1 2 3 4 5 6 7 8 9
2 How many squares are there?
Solution
1 2 3 4 5 6 7
### Complexity=15, Mode=rect
Count the number of objects.
1 How many squares are there?
Solution
1 2 3 4 5 6 7 8 9 10 11 12 13 14
2 How many objects are there?
Solution
1 2 3 4 5 6 7 8 9 10 11 12 13
### Complexity=20, Mode=rect
Count the number of objects.
1 How many objects are there?
Solution
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
2 How many objects are there?
Solution
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
### Complexity=20, Mode=line
Count the number of objects.
| 607 | 2,036 |
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| 3.65625 | 4 |
CC-MAIN-2013-20
|
longest
|
en
| 0.927071 |
https://learn.saylor.org/mod/page/view.php?id=88
| 1,679,371,528,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-14/segments/1679296943625.81/warc/CC-MAIN-20230321033306-20230321063306-00778.warc.gz
| 391,836,336 | 19,481 |
## Unit 1 Activities
View
Completing this unit should take approximately 5 hours and 30 minutes.
• Subunit 1.1: 50 minutes
• Subunit 1.2: 35 minutes
• Subunit 1.3: 1 hour and 30 minutes
• Subunit 1.4: 2 hours and 35 minutes
##### Learning Outcomes
Upon successful completion of this unit, you will be able to:
• Distinguish between quantitative and categorical variables.
• Describe the difference between a population and a sample and be able to distinguish between a parameter and a statistic.
• Given a type of measurement, identify the correct level of measurement: nominal, ordinal, interval, or ratio.
• Calculate the mean, median, and mode for a set of data, and compare and contrast these measures of center.
• Identify the symbols and know the formulas for sample and population means.
• Calculate the midrange, weighted mean, percentiles, and quartiles for a data set.
• Calculate the range, the interquartile range, the standard deviation, and the variance for a population and a sample, and know the symbols, formulas, and uses of these measures of spread.
##### Standards Addressed (Common Core and AP):
1.1 Statistical Terminology
The intent of this first section of the first unit is to introduce you to the terminology of statistics. Please make note of the following: quantitative variable, categorical variable, sample, population, statistic, and parameter. These terms will be used frequently throughout the course.
Explanation: Saylor Academy's Flexbook: Jill Schmidlkofer's Advanced Probability and Statistics"Section 1.1: Statistical Terminology”; partially adapted from David Lane's Online Statistics Education: An Interactive Multimedia Course of Study and CK-12: Advanced Probability and Statistics (PDF)
Instructions: Click on the link above and read Section 1.1. Statistics has its own language, and getting a foundation in the course requires that you become fluent with the basic vocabulary. Take notes about the new statistics terms that you learn as you read about Galapagos tortoise reintroduction.
Reading this section and taking notes should take approximately 30 minutes.
Standards Addressed (Common Core and AP):
Checkpoint: Saylor Academy's Flexbook: Jill Schmidlkofer's Advanced Probability and Statistics: "Section 1.1: Statistical Terminology”; partially adapted from David Lane's Online Statistics Education: A Multimedia Course of Study and CK-12: Advanced Probability and Statistics (PDF)
Instructions: Answer the review questions at the end of Section 1.1. You will use the vocabulary words learned in the section to answer them. A short answer key is provided at the end of the problem set.
Completing the review questions should take approximately 20 minutes.
1.2 An Overview of Data
The way we collect, display, and analyze data depends on the type of data in which we are interested. Therefore, we learn in this subunit more vocabulary about the four main types of data: nominal, ordinal, interval, and ratio.
Explanation: Saylor Academy's Flexbook: Jill Schmidlkofer's Advanced Probability and Statistics: "Section 1.2: An Overview of Data”; partially adapted from David Lane's Online Statistics Education: A Multimedia Course of Study and CK-12: Advanced Probability and Statistics (PDF)
Instructions: Read Section 1.2. In statistics, we analyze and interpret different types of data. For example, exam grades of A, B, C, D, and F are different from exam scores of 92, 85, 76, 61, and 45. Each type is valid, but we analyze the types differently. Therefore, a good foundation of the types of data available and their classifications is important.
Reading this section and taking notes should take approximately 25 minutes.
Standards Addressed (Common Core and AP):
Checkpoint: Saylor Academy's Flexbook: Jill Schmidlkofer's Advanced Probability and Statistics: "Section 1.2: An Overview of Data”; partially adapted from David Lane's Online Statistics Education: A Multimedia Course of Study and CK-12: Advanced Probability and Statistics (PDF)
Instructions: Answer the review questions at the end of Section 1.2. You will identify the level of data that is described in several scenarios. A short answer key is provided at the end of the problem set.
Completing the review questions should take approximately 10 minutes.
1.3 Measures of Center
A set of raw data is just a bunch of numbers. We learn to organize and summarize the raw data in this subunit by calculating the familiar mean, median, mode, and weighted mean. But we also learn to group the raw data into quartiles and percentiles. Pay close attention to the symbols and formulas in this subunit. You will see them again!
Explanation: Saylor Academy's Flexbook: Jill Schmidlkofer's Advanced Probability and Statistics"Section 1.3: Measures of Center”; partially adapted from David Lane's Online Statistics Education: A Multimedia Course of Study and CK-12: Advanced Probability and Statistics (PDF)
Instructions: Read Section 1.3. Take notes and work the examples provided as you read. Detailed instructions for using the TI-83/84 calculator are provided.
Reading this Section, taking notes, and working through the examples should take approximately 40 minutes.
Standards Addressed (Common Core and AP):
Instructions: Click on the link above and take notes as you watch the video. This is an easy topic. You should mentally work ahead of the instructor, anticipating his next step.
Watching this lecture and taking notes should take approximately 10 minutes.
Standards Addressed (Common Core and AP):
Activity: Shodor: "Plop It! An Interactive Study of the Mean and the Median”
Instructions: Begin this exercise by entering six data values into the box on the right (Example: 3, 3, 4, 4, 5, 6). A histogram of the data will appear. Look below the histogram and note the values of the mean and the median. Now, click on the "drag block” radio button, and click on the "6” box, moving it up to the value 14. What has happened to the median? What has happened to the mean? When you moved the box to the value 14, you created an outlier, and the median didn't change, but the mean did. This property of the median is called resistance. The median is resistant to the effects of an outlier, but the mean is not resistant, because its value is affected by an outlier.
Now, spend some time playing with the data and the histogram. See what happens if you pile all the boxes on top of each other. Enter a new set of data values, half of which are very small and half of which are very large. Then move the boxes around, seeing what happens to the mean and the median after each move.
Completing this activity should take approximately 10 minutes.
Standards Addressed (Common Core and AP):
Checkpoint: Saylor Academy's Flexbook: Jill Schmidlkofer's Advanced Probability and Statistics: "Section 1.3: Measures of Center”; partially adapted from David Lane's Online Statistics Education: A Multimedia Course of Study and CK-12: Advanced Probability and Statistics (PDF)
Instructions: Answer the review questions at the end of Section 1.3. Use your calculator to perform some of the statistical calculations. A short answer key is provided at the end of the problem set. For a detailed solution, click here.
Completing the review questions should take approximately 30 minutes.
Standards Addressed (Common Core and AP):
Assume that two cities have an average year-round temperature of 76 degrees. You want to go sightseeing in February, and you think that either city would be a good choice. Hold on! What if City A has a temperature of 76 degrees every day, whereas City B has summer temperatures of 100+ degrees and winter days as cold as 10 degrees below zero? Both can boast the same average temperature, but we also have to consider how far from the average the really cold or really hot days vary. That is what we call the spread of the data, and that is the subject of this subunit. Pay close attention to the standard deviation and the interquartile range, both of which will be used in later units in this course.
Explanation: Saylor Academy's Flexbook: Jill Schmidlkofer's Advanced Probability and Statistics: "Section 1.4: Measures of Spread”; partially adapted from David Lane's Online Statistics Education: A Multimedia Course of Study and CK-12: Advanced Probability and Statistics (PDF)
Instructions: Read Section 1.4. Take notes, and work the examples as you read. You will be given additional instructions for calculating the standard deviation and interquartile range with the TI-83/84 calculator.
Reading this section, taking notes, and practicing new calculator skills should take approximately 90 minutes.
Standards Addressed (Common Core and AP):
Instructions: Click on the link above and take notes as you watch the video. This is an introduction to the variance of a population. Pay attention to the concept, and note the relationship between the variance and the standard deviation.
Watching this lecture and taking notes should take approximately 10 minutes.
Standards Addressed (Common Core and AP):
Instructions: Click on the link above and take notes as you watch the video. This is an introduction to the variance of a sample. Pay attention to the concept, and note the relationship between the variance and the standard deviation. The standard deviation will be used throughout the second half of the course, so it will be to your benefit to master its definition, uses, symbol, and calculation.
Watching this lecture and taking notes should take approximately 15 minutes.
Standards Addressed (Common Core and AP):
| 2,049 | 9,599 |
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CC-MAIN-2023-14
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latest
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| 0.855346 |
https://studylib.net/doc/5779104/rotational-motion-and-equilibrium
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# Rotational Motion and Equilibrium
```-Angular and Linear Quantities
-Rotational Kinetic Energy
-Moment of Inertia
AP Physics C
Mrs. Coyle
Tangential (Linear) Speed and
Angular Speed
s=qr
v
ds
dt
r
dq
r
dt
v r
Tangential Acceleration and
Angular Acceleration
at
dv
dt
r
d
r
dt
a r
Angular and Linear Quantities
• Displacements
• Speeds
• Accelerations
s qr
v r
a r
• Every point on the rotating object has the
same angular motion but not the same linear
motion. The a and v are functions of r.
Remember:
Centripetal Acceleration
aC
v
2
r
r
v is the tangential speed
2
Resultant Linear Acceleration
• The net acceleration is
the sum of the
tangential and
centripetal
accelerations.
a
a a
2
r
2
t
Rotational Kinetic Energy
• A particle in a rotating object has rotational
kinetic energy:
Ki = ½ mivi2 , vi = i r (tangential velocity)
For the
Object:
KR
K
i
KR
i
1
2m r
i i
2
2
i
1
1
2
2
2
m i ri I
2 i
2
Rotational Kinetic Energy and
Moment of Inertia
• The total rotational kinetic energy of the rigid
object is the sum of the energies of all its
particles
1
2
2
K R K i m i ri
i
i 2
KR
1
1
2
2
2
m i ri I
2 i
2
• I is called the moment of inertia
Moment of Inertia, I
• Moment of Inertia, I, is a measure of
the resistance of an object to
changes in its rotational motion.
• Moment of Inertia is analogous to
mass in translational motion.
Example #20
Rigid rods of negligible mass lying along the y
axis connect three particles. If the system
rotates about the x-axis with an angular speed
the x-axis and the total rotational kinetic energy
evaluated from ½ I ω2 and b) the tangential
speed of each particle and the total kinetic
energy evaluated from ½ mi vi 2
Ans: a)92kg m2 , 184J , b) 6m/s,4m/s,8m/s,184J
```
– Cards
– Cards
– Cards
– Cards
– Cards
| 740 | 1,851 |
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| 3.90625 | 4 |
CC-MAIN-2019-09
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latest
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https://builtin.com/machine-learning/polynomial-regression
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| 132,229,849 | 16,428 |
# Polynomial Regression: An Introduction
Polynomial regression is an important method in machine learning. It is an extension of linear regression that models non-linear relationships between outcome and predictor variables.
Written by Rory Spanton
Published on Apr. 03, 2023
Image: Shutterstock / Built In
Linear regression is a fundamental method in statistics and machine learning. It allows a data scientist to model the relationship between an outcome variable and predictor variables. From this, the model can make predictions about test data. Yet, as the name suggests, linear regression assumes that outcome and predictor variables have a linear relationship, which isn’t the case in all data sets.
Luckily, polynomial regression allows for the accurate modeling of non-linear relationships. In this article, I’ll explain what polynomial regression models are and how to fit and evaluate them.
## What Is Polynomial Regression?
Polynomial regression is an extension of a standard linear regression model. Polynomial regression models the non-linear relationship between a predictor and an outcome variable using the Nth-degree polynomial of the predictor
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## Polynomial Regression Equation
To understand the structure of a polynomial regression model, let’s consider an example where one is appropriate.
In the simulated data above, the predictor variable on the x-axis is not linearly related to the outcome variable on the y-axis. Instead, the relationship between these variables is better described by a curve. If we fit a linear regression model to these data, the model equation would be as follows:
where a is the intercept (the value at which the regression line cuts through the y-axis), b is the coefficient, and ϵ is an error term.
As seen in the plot above, this straight-line equation doesn’t do a good job of capturing the non-linear relationship in the data. To address this, we can fit a polynomial regression model. This model is an extension of the previous one, but X is now added again as a second-degree polynomial. The equation for this model is
It’s clear from a quick visual inspection that the polynomial model gives a closer fit to the curved data. This will lead to more accurate predictions of new values in test data.
If necessary, more polynomial components can be added to the regression equation to fit more complex non-linear relationships. These would make the regression equation take this form:
So, how can you fit a polynomial regression model, and how can you tell when it includes too many components?
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## How to Fit a Polynomial Regression Model
The standard method for fitting both linear and polynomial regression in R is the method of least squares. This involves minimizing the sum of the squared residuals in the model by adjusting the values of the intercept and coefficients. The `lm` function in R minimizes the sum of squares for us, so all we need to do is specify the model. We can start by fitting a simple linear regression model to our example data.
``````model1 <- lm(Y ~ X, data = example_data)
summary(model1)``````
The summary above shows us the adjusted R² value for the model, which is a measure of how well the model predicts our outcome. An R2 equal to zero means the model accounts for none of the variance in the outcome, whereas one would mean it accounts for all the variance. In this case, R2 equals 0.34, meaning that our regression model accounts for 34 percent of the variance in the outcome variable. This is OK, but given the shape of the data, it makes sense to try adding a polynomial term to the model.
It’s easy to specify a polynomial regression model in R. It’s the same as linear regression, but we use the `poly` function to state that we want to add a polynomial term to our predictor and the power in the term itself. In this example, we fit a model with a quadratic component — a second-degree polynomial.
``````model2 <- lm(Y ~ poly(X, 2), data = example_data)
summary(model2)``````
It’s clear from the model summary that the polynomial term has improved the fit of the regression. R² now equals 0.81, a large increase from the previous model. But, just like in multiple regression, adding more terms to a model will always improve the fit. It is therefore essential to test whether this improvement in model fit is substantial enough to be considered meaningful.
## How to Evaluate a Polynomial Regression Model
To justify adding polynomial components to a regression model, it’s important to test whether each one significantly improves the model fit. To do this, we can compare the fit of both models using an analysis of variance (ANOVA).
``anova(model1, model2)``
The results of this ANOVA are significant. The p-value (shown under “Pr(>F)” in the output) is very small and well below 0.05, the typical threshold for statistical significance. This means that adding the polynomial term helped the second regression model give a substantially better fit to the data than the first.
We can keep expanding the model and testing whether successive terms improve the fit. When more advanced terms no longer significantly improve the model fit, we have our final model specification. Testing whether a cubic polynomial term (a third-degree polynomial) to the model demonstrates this outcome.
``````model3 <- lm(Y ~ poly(X, 3), data = example_data)
summary(model3)
anova(model2, model3)``````
Here, the ANOVA is no longer significant, meaning that the cubic component didn’t substantially improve the model fit. This means we can leave out the cubic component and choose model2 as our final model.
## How to Fit and Evaluate Polynomial Regression Models With Bayesian Methods
To fit polynomial regression models using Bayesian methods, you’ll need the BayesFactor R package. This includes the `lmBF` function; the Bayesian equivalent of the `lm` function.
To specify a polynomial regression equation in `lmBF`, we can’t use the poly function like in the `lm` example. This is because an error occurs if we try to use poly inside `lmBF`. To get around this, we can create a new column in our data that contains a polynomial term and then insert that as a coefficient in the model as shown below.
``````library(BayesFactor)
example_data\$X_poly2 <- poly(example_data\$X, 2)[,"2"]
BF_poly2 <- lmBF(Y ~ X + X_poly2, data = as.data.frame(example_data))
BF_poly2``````
This outputs a Bayes factor for the regression, which is a measure of the evidence for our regression model versus a model with no coefficients. This gives us an idea of whether or not all of the predictors do a good job of explaining variance in our outcome. Bayes factors above three are often interpreted as being sufficient evidence in a model’s favor. Ours in this case is much greater, meaning the model is 2.05 × 1031 times more likely than one with no predictors.
This Bayes factor doesn’t tell us how useful each individual predictor is at improving the model fit, however. For this, we’ll need to compare models. To test whether the quadratic polynomial component improves our model fit, we can fit a simpler linear model with `lmBF`. Then, we divide the Bayes factor of our polynomial model by the Bayes factor of the simpler model. The resulting Bayes factor can be interpreted as the ratio of evidence for the complex model versus the simpler one.
``````BF_simple <- lmBF(Y ~ X, data = as.data.frame(example_data))
BF_poly2 / BF_simple``````
This value is once again very large, indicating sufficient evidence that the polynomial component is reliably improving the model fit. Based on this, we’d choose the polynomial regression model over the simple linear regression model.
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## Strengths and Limitations of Polynomial Regression
While polynomial regression is useful, it should be used with caution. Although it’s possible to add lots of polynomial components to a model to improve its fit, this increases the risk of overfitting. Overfitting is when a model fits the training data set very closely but is not generalizable enough to make accurate predictions about test data.
In an extreme case, a model with many polynomial terms could fit a training data set nearly perfectly, drawing a wavy line through all the data points. A model like this would be unable to generalize to new data, however, and would give all sorts of inaccurate predictions because it picked up so much of the random variation in the training data. This inaccuracy could lead you to make misinformed conclusions from your model, so you must avoid it.
To avoid overfitting, it’s important to test that each polynomial component in a regression model makes a meaningful difference to the model fit. By doing this, your model will include only the essential components needed to predict the outcome of interest. Because it avoids unnecessary complexity, it will therefore return more accurate predictions about test data.
Polynomial regression is also more sensitive to outliers than linear regression. In linear regression, outliers don’t usually have substantial effects on the model coefficients unless the outlying values themselves are very large. In contrast, one or two outlying values might change the whole specification of a polynomial regression model. Again, this can lead polynomial regression models to make inaccurate predictions.
With these limitations in mind, polynomial regression is a useful method for modelling non-linear relationships between predictor and outcome variables. As demonstrated, polynomial regression models give more accurate predictions about test data than linear regression in these cases. When used carefully, it is a powerful and versatile tool that belongs in any data scientist’s skill set.
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## 556936
556,936 (five hundred fifty-six thousand nine hundred thirty-six) is an even six-digits composite number following 556935 and preceding 556937. In scientific notation, it is written as 5.56936 × 105. The sum of its digits is 34. It has a total of 5 prime factors and 16 positive divisors. There are 271,824 positive integers (up to 556936) that are relatively prime to 556936.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 34
• Digital Root 7
## Name
Short name 556 thousand 936 five hundred fifty-six thousand nine hundred thirty-six
## Notation
Scientific notation 5.56936 × 105 556.936 × 103
## Prime Factorization of 556936
Prime Factorization 23 × 43 × 1619
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 5 Total number of prime factors rad(n) 139234 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 556,936 is 23 × 43 × 1619. Since it has a total of 5 prime factors, 556,936 is a composite number.
## Divisors of 556936
16 divisors
Even divisors 12 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 1.0692e+06 Sum of all the positive divisors of n s(n) 512264 Sum of the proper positive divisors of n A(n) 66825 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 746.281 Returns the nth root of the product of n divisors H(n) 8.33425 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 556,936 can be divided by 16 positive divisors (out of which 12 are even, and 4 are odd). The sum of these divisors (counting 556,936) is 1,069,200, the average is 66,825.
## Other Arithmetic Functions (n = 556936)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 271824 Total number of positive integers not greater than n that are coprime to n λ(n) 67956 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 45735 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 271,824 positive integers (less than 556,936) that are coprime with 556,936. And there are approximately 45,735 prime numbers less than or equal to 556,936.
## Divisibility of 556936
m n mod m 2 3 4 5 6 7 8 9 0 1 0 1 4 2 0 7
The number 556,936 is divisible by 2, 4 and 8.
## Classification of 556936
• Arithmetic
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
## Base conversion (556936)
Base System Value
2 Binary 10000111111110001000
3 Ternary 1001021222021
4 Quaternary 2013332020
5 Quinary 120310221
6 Senary 15534224
8 Octal 2077610
10 Decimal 556936
12 Duodecimal 22a374
20 Vigesimal 39c6g
36 Base36 bxqg
## Basic calculations (n = 556936)
### Multiplication
n×i
n×2 1113872 1670808 2227744 2784680
### Division
ni
n⁄2 278468 185645 139234 111387
### Exponentiation
ni
n2 310177708096 172749132036153856 96210210599687383945216 53582929850547492864912818176
### Nth Root
i√n
2√n 746.281 82.2751 27.3182 14.0981
## 556936 as geometric shapes
### Circle
Diameter 1.11387e+06 3.49933e+06 9.74452e+11
### Sphere
Volume 7.2361e+17 3.89781e+12 3.49933e+06
### Square
Length = n
Perimeter 2.22774e+06 3.10178e+11 787626
### Cube
Length = n
Surface area 1.86107e+12 1.72749e+17 964641
### Equilateral Triangle
Length = n
Perimeter 1.67081e+06 1.34311e+11 482321
### Triangular Pyramid
Length = n
Surface area 5.37244e+11 2.03587e+16 454736
## Cryptographic Hash Functions
md5 4f00587c163b2957e0ba6ed4e72fbce1 1461bfd7a9e398994eca106ee1e637d760aad602 334e2620f76af4ca1a1d5cf561c06f9631575be3006fb03edd2d3e80ec8fef91 d69c194fa6bc32ebe1153ef58d83fe135f668d45e77698b2f6b5e4274769fce80f99343fc529e19624552d069864ff92884bd585cd534bfc8839b4655f6c9b88 47a6f60032c7c301b1ed669f41f4787996320e50
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# One sample Wilcoxon signed-rank test - overview
This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table
One sample Wilcoxon signed-rank test
Kruskal-Wallis test
Independent variableIndependent/grouping variable
NoneOne categorical with $I$ independent groups ($I \geqslant 2$)
Dependent variableDependent variable
One of ordinal levelOne of ordinal level
Null hypothesisNull hypothesis
H0: $m = m_0$
Here $m$ is the population median, and $m_0$ is the population median according to the null hypothesis.
If the dependent variable is measured on a continuous scale and the shape of the distribution of the dependent variable is the same in all $I$ populations:
• H0: the population medians for the $I$ groups are equal
Else:
Formulation 1:
• H0: the population scores in any of the $I$ groups are not systematically higher or lower than the population scores in any of the other groups
Formulation 2:
• H0: P(an observation from population $g$ exceeds an observation from population $h$) = P(an observation from population $h$ exceeds an observation from population $g$), for each pair of groups.
Several different formulations of the null hypothesis can be found in the literature, and we do not agree with all of them. Make sure you (also) learn the one that is given in your text book or by your teacher.
Alternative hypothesisAlternative hypothesis
H1 two sided: $m \neq m_0$
H1 right sided: $m > m_0$
H1 left sided: $m < m_0$
If the dependent variable is measured on a continuous scale and the shape of the distribution of the dependent variable is the same in all $I$ populations:
• H1: not all of the population medians for the $I$ groups are equal
Else:
Formulation 1:
• H1: the poplation scores in some groups are systematically higher or lower than the population scores in other groups
Formulation 2:
• H1: for at least one pair of groups:
P(an observation from population $g$ exceeds an observation from population $h$) $\neq$ P(an observation from population $h$ exceeds an observation from population $g$)
AssumptionsAssumptions
• The population distribution of the scores is symmetric
• Sample is a simple random sample from the population. That is, observations are independent of one another
• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2, $\ldots$, group $I$ sample is an independent SRS from population $I$. That is, within and between groups, observations are independent of one another
Test statisticTest statistic
Two different types of test statistics can be used, but both will result in the same test outcome. We will denote the first option the $W_1$ statistic (also known as the $T$ statistic), and the second option the $W_2$ statistic. In order to compute each of the test statistics, follow the steps below:
1. For each subject, compute the sign of the difference score $\mbox{sign}_d = \mbox{sgn}(\mbox{score} - m_0)$. The sign is 1 if the difference is larger than zero, -1 if the diffence is smaller than zero, and 0 if the difference is equal to zero.
2. For each subject, compute the absolute value of the difference score $|\mbox{score} - m_0|$.
3. Exclude subjects with a difference score of zero. This leaves us with a remaining number of difference scores equal to $N_r$.
4. Assign ranks $R_d$ to the $N_r$ remaining absolute difference scores. The smallest absolute difference score corresponds to a rank score of 1, and the largest absolute difference score corresponds to a rank score of $N_r$. If there are ties, assign them the average of the ranks they occupy.
Then compute the test statistic:
• $W_1 = \sum\, R_d^{+}$
or
$W_1 = \sum\, R_d^{-}$
That is, sum all ranks corresponding to a positive difference or sum all ranks corresponding to a negative difference. Theoratically, both definitions will result in the same test outcome. However:
• Tables with critical values for $W_1$ are usually based on the smaller of $\sum\, R_d^{+}$ and $\sum\, R_d^{-}$. So if you are using such a table, pick the smaller one.
• If you are using the normal approximation to find the $p$ value, it makes things most straightforward if you use $W_1 = \sum\, R_d^{+}$ (if you use $W_1 = \sum\, R_d^{-}$, the right and left sided alternative hypotheses 'flip').
• $W_2 = \sum\, \mbox{sign}_d \times R_d$
That is, for each remaining difference score, multiply the rank of the absolute difference score by the sign of the difference score, and then sum all of the products.
$H = \dfrac{12}{N (N + 1)} \sum \dfrac{R^2_i}{n_i} - 3(N + 1)$
Here $N$ is the total sample size, $R_i$ is the sum of ranks in group $i$, and $n_i$ is the sample size of group $i$. Remember that multiplication precedes addition, so first compute $\frac{12}{N (N + 1)} \times \sum \frac{R^2_i}{n_i}$ and then subtract $3(N + 1)$.
Note: if ties are present in the data, the formula for $H$ is more complicated.
Sampling distribution of $W_1$ and of $W_2$ if H0 were trueSampling distribution of $H$ if H0 were true
Sampling distribution of $W_1$:
If $N_r$ is large, $W_1$ is approximately normally distributed with mean $\mu_{W_1}$ and standard deviation $\sigma_{W_1}$ if the null hypothesis were true. Here $$\mu_{W_1} = \frac{N_r(N_r + 1)}{4}$$ $$\sigma_{W_1} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{24}}$$ Hence, if $N_r$ is large, the standardized test statistic $$z = \frac{W_1 - \mu_{W_1}}{\sigma_{W_1}}$$ follows approximately the standard normal distribution if the null hypothesis were true.
Sampling distribution of $W_2$:
If $N_r$ is large, $W_2$ is approximately normally distributed with mean $0$ and standard deviation $\sigma_{W_2}$ if the null hypothesis were true. Here $$\sigma_{W_2} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{6}}$$ Hence, if $N_r$ is large, the standardized test statistic $$z = \frac{W_2}{\sigma_{W_2}}$$ follows approximately the standard normal distribution if the null hypothesis were true.
If $N_r$ is small, the exact distribution of $W_1$ or $W_2$ should be used.
Note: if ties are present in the data, the formula for the standard deviations $\sigma_{W_1}$ and $\sigma_{W_2}$ is more complicated.
For large samples, approximately the chi-squared distribution with $I - 1$ degrees of freedom.
For small samples, the exact distribution of $H$ should be used.
Significant?Significant?
For large samples, the table for standard normal probabilities can be used:
Two sided:
Right sided:
Left sided:
For large samples, the table with critical $X^2$ values can be used. If we denote $X^2 = H$:
• Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
• Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
Example contextExample context
Is the median mental health score of office workers different from $m_0 = 50$?Do people from different religions tend to score differently on social economic status?
SPSSSPSS
Specify the measurement level of your variable on the Variable View tab, in the column named Measure. Then go to:
Analyze > Nonparametric Tests > One Sample...
• On the Objective tab, choose Customize Analysis
• On the Fields tab, specify the variable for which you want to compute the Wilcoxon signed-rank test
• On the Settings tab, choose Customize tests and check the box for 'Compare median to hypothesized (Wilcoxon signed-rank test)'. Fill in your $m_0$ in the box next to Hypothesized median
• Click Run
• Double click on the output table to see the full results
Analyze > Nonparametric Tests > Legacy Dialogs > K Independent Samples...
• Put your dependent variable in the box below Test Variable List and your independent (grouping) variable in the box below Grouping Variable
• Click on the Define Range... button. If you can't click on it, first click on the grouping variable so its background turns yellow
• Fill in the smallest value you have used to indicate your groups in the box next to Minimum, and the largest value you have used to indicate your groups in the box next to Maximum
• Continue and click OK
JamoviJamovi
T-Tests > One Sample T-Test
• Put your variable in the box below Dependent Variables
• Under Tests, select Wilcoxon rank
• Under Hypothesis, fill in the value for $m_0$ in the box next to Test Value, and select your alternative hypothesis
ANOVA > One Way ANOVA - Kruskal-Wallis
• Put your dependent variable in the box below Dependent Variables and your independent (grouping) variable in the box below Grouping Variable
Practice questionsPractice questions
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# Associative algebra
Associative algebra is a fundamental concept in mathematics, characterising structures where multiplication is associative, meaning the equation (ab)c = a(bc) holds for all elements a, b, and c. This principle is pivotal in understanding complex algebraic systems, including various types of rings and fields. By internalising the associative property, students can adeptly navigate algebraic operations, laying a solid foundation for advanced mathematical studies.
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## What Is Associative Algebra?
Associative algebra stands as an integral branch of mathematics that delves into the study of algebraic structures known as algebras. These structures play a pivotal role in facilitating the understanding of various mathematical concepts, particularly in areas such as abstract algebra. By exploring associative algebra, you embark on a journey through the intricacies of algebraic operations and their properties, which are foundational to grasping more complex mathematical theories.Let's dive deeper into what associative algebra entails, starting with its definition and exploring its key components.
Associative Algebra: A branch of algebra focusing on algebraic structures where the associative property holds for a specific binary operation. This means, for any three elements a, b, and c within the algebra, the equation $$a \cdot b) \cdot c = a \cdot (b \cdot c)\ always holds true. In simple terms, associative algebra examines the behaviour of elements under a certain operation where grouping them differently does not affect the outcome of their combination. This characteristic is paramount in the construction and analysis of various algebraic structures, including vector spaces and groups, often making associative algebra a cornerstone topic in advanced mathematics education.An understanding of associative algebra thereby not only enhances your algebraic skills but also broadens your perspective on how mathematical operations interact and transform within different contexts. ### Key Components of Associative Algebra To genuinely appreciate the scope and utility of associative algebra, it's crucial to identify and understand its key components. These include: • Algebras: The primary focus of associative algebra, algebras are structures that combine elements of both ring theory and vector spaces, offering a richer framework for exploring algebraic operations. • Binary Operations: Operations involving two inputs, such as addition or multiplication, are central to associative algebra. The associative property ensures the consistency of these operations regardless of how the elements are grouped. • Associative Property: This property stipulates that the way elements are grouped in an operation (e.g., whether \(a \cdot b)\ first, then \(\cdot c)\, or \(b \cdot c)\ first, then \(a \cdot$$ does not alter the result. It is fundamental to the structure of associative algebra.
• Elements: The individual components or entries within an algebra that are subjected to algebraic operations. Understanding how these elements interact under the associative property is key to mastering associative algebra.
An effective way to conceptualise the associative property is by visualising the grouping of elements with parentheses, which signifies the order of operation does not change the outcome.
## Exploring Examples of Associative Algebra
Through the lens of associative algebra, it becomes clear how mathematical principles are not just abstract concepts but practical tools that pervade our daily lives. Associative algebra, distinguished by its operations where grouping does not impact the outcome, offers a multitude of real-life applications and examples that underscore its relevance and utility. Let's delve into some instances where associative algebra plays a pivotal role.
### Associative Algebra Examples in Real Life
Associative algebra finds application in several areas outside the classroom, illustrating the profound impact of mathematics on our everyday experiences. From the technology we rely on to the financial systems that govern economies, associative algebra's principles underpin a wide range of practical scenarios.One illustrative example can be found in computer science, particularly in the design and optimisation of algorithms. The associative property of certain operations allows for more efficient data processing and manipulation, enabling computers to execute complex tasks more quickly.
Example: In data encryption, algorithms utilise the associative property to securely transform original data into encrypted information. For instance, let's consider the encryption process involving two steps, encrypting with key A ($$E_A ext{}$$) and then with key B ($$E_B ext{}$$). The associative property ensures that $E_A(E_B(data)) = E_B(E_A(data))$, allowing for the flexible application of encryption steps without affecting the security or outcome of the process.
### Applying the Algebra Associative Property
The algebra associative property stands as a cornerstone concept in mathematics, offering insights and strategies to simplify complex problems. By understanding and applying this property, you can navigate mathematical challenges more effectively and with greater confidence.One area where the associative property is particularly beneficial is in the simplification of algebraic expressions. This property allows for the re-arrangement and grouping of terms in a manner that renders the expression more manageable, thus facilitating its solution.
Example: Consider the algebraic expression $3 \times (4 \times 5)$. Applying the associative property allows us to assert that $3 \times (4 \times 5) = (3 \times 4) \times 5$. This re-grouping does not change the result, which remains 60, but it can simplify computation by breaking down the problem into smaller, more manageable parts.
Remember, the associative property applies to addition and multiplication, but not to subtraction or division. This distinction is crucial in correctly applying the property to solve problems.
Thinking beyond the straightforward applications, the associative property in algebra paves the way for exploring more complex mathematical concepts like group theory and ring theory. These areas, integral to abstract algebra, further elaborate on how operations can be structured and manipulated within different mathematical systems.Moreover, the understanding and application of the associative property foster analytical thinking and problem-solving skills. It encourages a flexible approach to mathematical operations, allowing for innovative solutions to emerge in both theoretical and applied mathematics.
## The Associative Property of Multiplication in Algebra
The associative property of multiplication in algebra is a fundamental concept that simplifies the way we approach and solve mathematical problems. It states that when multiplying three or more numbers, the way in which the numbers are grouped does not change the result. This property ensures consistency and provides a reliable foundation for more complex algebraic operations.Understanding and applying this property effectively can greatly enhance your mathematical prowess. Let's delve into the details of this property and explore its practical applications.
The Associative Property of Multiplication states that for any three real numbers, a, b, and c, the equation $$a \times b) \times c = a \times (b \times c)\ holds true. This property guarantees that the product remains unchanged regardless of how the numbers are grouped. Example: Consider the multiplication operation among the numbers 2, 3, and 4. Applying the associative property of multiplication, we can group these numbers in two different ways without altering the result: $(2 \times 3) \times 4 = 2 \times (3 \times 4)$. Both expressions will yield the same product, 24, demonstrating the property's validity. A useful tip when dealing with multiple numbers in multiplication is to group numbers that are easier to multiply together. This can significantly simplify the calculation, thanks to the associative property. ### Practical Applications of Multiplication's Associative Property The associative property of multiplication extends beyond the classroom and is utilised in various real-world scenarios and disciplines. Its utility can be observed in fields ranging from computer science and engineering to everyday tasks such as cooking or financial planning. Understanding how this property operates in practical scenarios can illuminate its importance and versatility. For instance, in computer programming, efficient computation is paramount. The associative property allows algorithms to re-group operations for optimised processing, potentially saving significant computational resources. Similarly, in construction or engineering, this property enables professionals to calculate large quantities or measurements in parts, facilitating easier handling and more accurate results. Diving deeper, the associative property of multiplication is not just a rule; it's a principle that underpins much of algebra. It allows for the development of algebraic expressions and equations in a way that is logical and consistent. Furthermore, this property plays a critical role in more advanced topics, such as polynomial multiplication and matrix operations, where the re-grouping of terms according to the associative property can greatly simplify complex calculations. Acknowledging the wide-ranging applications and the foundational role it plays in mathematics highlights the associative property's significance. ## Beyond Basics: Non Associative and Free Associative Algebra Moving beyond the conventional frameworks of associative algebra, the mathematical landscape broadens to include non associative and free associative algebra. These advanced topics challenge the standard rules we've become accustomed to, presenting new perspectives on algebraic structures and their operations.Exploring these areas not only deepens understanding of algebra but also opens doors to diverse mathematical theories and applications. Let's delve into the distinctions and intricacies of non associative and free associative algebra. ### Distinguishing Non Associative Algebra Non Associative Algebra: A type of algebra where the associative property does not necessarily hold for all operations. This means that for some elements a, b, and c, it's possible that \((a \cdot b) \cdot c \neq a \cdot (b \cdot c)$$.
Non associative algebra introduces structures where the conventional rule of operation grouping having no effect on the outcome does not apply. This deviation offers a unique framework for exploring mathematical phenomena that aren't adequately described by associative algebras.Key examples of non associative algebra include structures like Lie algebras and Jordan algebras, which are pivotal in various fields, including theoretical physics and differential geometry.
Example: Consider the cross product in vector algebra, a common operation in physics and engineering. The cross product is inherently non associative, meaning for vectors a, b, and c, the equation $$(a \times b) \times c \neq a \times (b \times c)$$ generally holds true. This property necessitates a careful approach when performing sequential operations with vector cross products.
### Introduction to Free Associative Algebra
Free Associative Algebra: An algebraic structure where the associative property is preserved without any relations imposed on the generators of the algebra, beyond those necessary for association. In simpler terms, elements can be freely combined under the operation, respecting associativity.
Free associative algebra plays a crucial role in the study and development of algebraic theories. By allowing elements to associate freely, it facilitates the construction of polynomial algebras and other complex algebraic structures without the restriction of specific identities or relations among the elements.Its applications span various mathematical disciplines, acting as a foundational tool in group theory, ring theory, and combinatorial algebra. Understanding free associative algebras enables mathematicians to explore and construct abstract algebraic systems with greater flexibility.
An effective approach to grasp the concept of free associative algebra is to compare it with the familiar setting of polynomials in variables, where the variables can be combined in any associative manner without constraints on their relationships.
Exploring further into free associative algebra reveals its significance in the context of universal algebraic constructions, such as free groups and free rings. These constructions serve as the 'building blocks' for more complex algebraic structures, providing insight into the underpinnings of algebraic theory.Moreover, the study of free associative algebra intersects with areas of computer science, particularly in the development of algorithms for symbolic computation. This demonstrates the broad applicability and interdisciplinary nature of free associative algebra, reinforcing its importance in both theoretical and practical mathematics.
## Associative algebra - Key takeaways
• Associative Algebra Definition: A branch of algebra dealing with algebraic structures where the associative property is maintained for a binary operation, meaning that (a · b) · c = a · (b · c) for any elements a, b, and c.
• Associative Property: A fundamental property where the grouping of elements does not affect the outcome of their combination, vital in studying algebraic structures like vector spaces and groups.
• Associative Algebra Examples: Includes its application in computer science for optimising algorithms and data encryption, where operations' outcomes are independent of element grouping.
• Associative Property of Multiplication: A specific case of the associative property in algebra stating that the product of three or more numbers is independent of how the numbers are grouped (a × b) × c = a × (b × c).
• Non Associative and Free Associative Algebra: Extensions of associative algebra where the associative property may not hold (non associative algebra), and structures where elements can be freely combined while respecting associativity (free associative algebra).
#### Flashcards in Associative algebra 24
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##### Frequently Asked Questions about Associative algebra
What is the definition of an associative algebra?
An associative algebra is a vector space equipped with a bilinear product that is associative, meaning that for any three elements $$a$$, $$b$$, and $$c$$ in the algebra, the product $$a(bc)$$ equals the product $$(ab)c$$.
How can one apply the concept of associative algebra in real-world problems?
Associative algebra can be applied in real-world problems through cryptography for secure communication, in coding theory for error detection and correction, and in computer graphics for efficient rendering and manipulation of digital images and animations.
What are some examples of associative algebra in mathematics?
Some examples of associative algebra in mathematics include matrix algebra, polynomial algebra, group algebra, and function algebras. Each of these structures follows the associative property, meaning that the equation (ab)c = a(bc) holds for all elements a, b, and c within the algebra.
What are the main properties of an associative algebra that differentiate it from other algebraic structures?
The main properties of an associative algebra that differentiate it from other algebraic structures include the presence of a bilinear product that is associative, i.e., (ab)c = a(bc) for all elements, and the structure being both an algebra over a field and a ring, allowing for scalar multiplication alongside the ring operations.
How does the associative property impact the solving of equations in an associative algebra?
The associative property in an associative algebra simplifies the solving of equations by allowing the regrouping of terms without affecting the outcome, enabling strategies like factorisation and the combination of like terms to efficiently find solutions and manipulate expressions.
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How does the Associative Property of Multiplication simplify calculations?
What defines a Non Associative Algebra?
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