Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Evaluate the Integral $\int^2_{\sqrt{2}}\frac{1}{t^3\sqrt{t^2-1}}dt$ $\int^2_{\sqrt{2}}\frac{1}{t^3\sqrt{t^2-1}}dt$
I believe I've done everything right; however, my answer does not resemble the answer in the book. I think it has something to do with my Algebra. Please tell me what I am doing wrong.
| $$\int\frac{1}{t^3\sqrt{t^2-1}}\space\space\text{d}t=$$
Substitute $t=\sec(u)$ and $\text{d}t=\tan(u)\sec(u)\space\space\text{d}u$. Then $\sqrt{t^2-1}=\sqrt{\sec^2(u)-1}=\tan(u)$ and $u=\sec^{-1}(t)$:
$$\int\cos^2(u)\space\space\text{d}u=$$
$$\int\left(\frac{1}{2}\cos(2u)+\frac{1}{2}\right)\space\space\text{d}u=$$
$$\frac{1}{2}\int\cos(2u)\space\space\text{d}u+\frac{1}{2}\int 1\space\space\text{d}u=$$
Substitute $s=2u$ and $\text{d}s=2\space\space\text{d}u$:
$$\frac{1}{4}\int\cos(s)\space\space\text{d}s+\frac{1}{2}\int 1\space\space\text{d}u=$$
$$\frac{1}{4}\sin(s)+\frac{u}{2}+\text{C}=$$
$$\frac{\sin\left(s\right)}{4}+\frac{u}{2}+\text{C}=$$
$$\frac{\sin\left(2u\right)}{4}+\frac{u}{2}+\text{C}=$$
$$\frac{\sin\left(2\sec^{-1}(t)\right)}{4}+\frac{\sec^{-1}(t)}{2}+\text{C}$$
Filling in your boundaries gives us:
$$\left(\frac{\sin\left(2\sec^{-1}(2)\right)}{4}+\frac{\sec^{-1}(2)}{2}\right)-\left(\frac{\sin\left(2\sec^{-1}(\sqrt{2})\right)}{4}+\frac{\sec^{-1}(\sqrt{2})}{2}\right)=$$
$$\left(\frac{\sin\left(2\cdot\frac{\pi}{3}\right)}{4}+\frac{\frac{\pi}{3}}{2}\right)-\left(\frac{\sin\left(2\cdot\frac{\pi}{4}\right)}{4}+\frac{\frac{\pi}{4}}{2}\right)=$$
$$\left(\frac{\sin\left(\frac{2\pi}{3}\right)}{4}+\frac{\pi}{6}\right)-\left(\frac{\sin\left(\frac{\pi}{2}\right)}{4}+\frac{\pi}{8}\right)=$$
$$\left(\frac{\frac{\sqrt{3}}{2}}{4}+\frac{\pi}{6}\right)-\left(\frac{1}{4}+\frac{\pi}{8}\right)=$$
$$\left(\frac{\sqrt{3}}{8}+\frac{\pi}{6}\right)-\left(\frac{1}{4}+\frac{\pi}{8}\right)=$$
$$\frac{\pi}{24}+\frac{\sqrt{3}}{8}-\frac{1}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1506645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Evaluate the Integral $\int \sqrt{1-4x^2}\ dx$ $\int \sqrt{1-4x^2}\ dx$
I am confused as I get to the end. Why would I use a half angle formula? And why is it necessary to use inverses?
| In general case whenever you have something like $1-a^2x^2$. Remember $\sin^2x+\cos^2x=1$ and so we should multiply it by $a$. Thus take $\sin y=\frac{1}{a}x$. Now we have
$\sqrt{1-a^2x^2}=\sqrt{1-\sin^2y}=\cos y$.
Hence one can see that
$\int\sqrt{1-a^2x^2}dx=\int\sqrt{1-\sin^2y}\times a\cos y \,dy=\int a\cos^2 y dy$
Now we have $\int \cos^2y$. The standard technique to deal with this is using the gold formula $\cos^2y=\frac{1}{2}(1+\cos 2y)$. In fact the gold formula is revealing the connection between $\cos^2$ and $\cos$.
The most controversial part is the integral of $\int\cos 2ydy$.
It is $\frac{1}{2}\sin 2y$ or in other words it is equal to $\sin y\cos y$. Now recall that $\sin y=\frac{1}{2}x$. Thus one can see that $\cos y=\sqrt{1-\frac{1}{4}x^2}$ and we are done, as we only need to have $\sin y$ and $\cos y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1507501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
A quotient of trigonometric expressions in complex analysis How is $$\frac{\cos\frac{20\pi}{3}+i\sin\frac{20\pi}{3}}{\cos{\frac{15\pi}{4}}{+i\sin\frac{15 \pi}{4}}}=\cos\frac{35\pi}{12}+i\sin\frac{35 \pi }{12}\ \ \ ?$$
I think that I am having trouble understanding a fundamental concept in complex analysis, but cannot pinpoint which.
| $$\frac{\cos\left(\frac{20\pi}{3}\right)+\sin\left(\frac{20\pi}{3}\right)i}{\cos\left(\frac{15\pi}{4}\right)+\sin\left(\frac{15\pi}{4}\right)i}=\frac{e^{\frac{20\pi}{3}i}}{e^{\frac{15\pi}{4}i}}=e^{\left(\frac{20\pi}{3}-\frac{15\pi}{4}\right)i}=e^{\frac{35\pi}{12}i}=\cos\left(\frac{35\pi}{12}\right)+\sin\left(\frac{35\pi}{12}\right)i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1507839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving degree three inequality
$$\frac{2}{(x^2+1)}\geq x.$$
I thought to do this:$$2-x^3-x\geq 0$$
But we haven't learnt how to solve equations of third grade. Could you help me maybe by factorizing sth?
| \begin{align}
\frac{2}{x^2+1} &\ge x \iff \\
2 &\ge x(x^2+1) = x^3 + x \iff \\
0 &\ge x^3 + x - 2
\end{align}
Trying some easy roots one gets $x=1$:
\begin{align}
0 &\ge (x-1)(x^2+x+2) \quad (*)
\end{align}
Then we have
$$
x^2 +x+ 2 = (x+1/2)^2 + 7/4 \ge 7/4
$$
which means no real roots for this part, no change of sign. So we can divide both sides by the above and simplify $(*)$ to:
\begin{align}
0 &\ge x - 1 \iff \\
1 &\ge x
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1508894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve $ \left|\frac{x}{x+2}\right|\leq 2 $ I am experiencing a little confusion in answering a problem on Absolute Value inequalities which I just started learning. This is the problem: Solve:
$$
\left|\frac{x}{x+2}\right|\leq 2
$$
The answer is given to be $x\leq-4$ or $x \geq-1$
This is my attempt to solve the problem:
By dividing, $\left|\frac{x}{x+2}\right|\leq 2$ is equivalent to $\left |1-\frac{2}{x+2}\right|\leq2$ which is also equivalent to $\left |\frac{2}{x+2}-1\right|\leq2$
So, $-2\leq\frac{2}{x+2}-1\leq2$ which is equivalent to $-\frac{1}{2}\leq\frac{1}{x+2}\leq\frac{3}{2}$
Case 1: $x+2>0$. Solving $-\frac{1}{2}\leq\frac{1}{x+2}\leq\frac{3}{2}$, I get $x\geq-4$ and $x\geq-\frac{4}{3}$ which is essentially $x\geq-\frac{4}{3}$.
Case 2: $x+2<0$. Solving $-\frac{1}{2}\times(x+2)\geq{1}\geq\frac{3}{2}\times(x+2)$, I get $x\leq-4$ and $x\leq-\frac{4}{3}$ which is essentially $x\leq-4$.
So, the solutions are: $x\leq-4$ or $x\geq-\frac{4}{3}$.
I couldn't get $x \geq-1$ as a solution. Did I do anything wrong? The book I am using is Schaum's Outlines-Calculus. Another question I would like to ask is that am I using 'and' and 'or' correctly in the above attempt to solve the problem? I have had this problem many times.
| So you have
$$
\lvert x \rvert \leq 2\lvert x+2\rvert.
$$
If $x \geq 0$, then $x + 2 > 0$. And then the equation reads $x \leq 2x + 4$ implying that $x \geq -4 $. This in all gives you the solutions $x \geq 0$.
If $x\in [-2, 0)$, then the equation becomes $-x \leq 2x + 4$. So $-4\leq 3x$. So $\frac{-4}{3} \leq x$. In all you get $x\in [-4/3, 0)$.
Now the last case is where $x <-2$. Then $-x \leq -2x - 4$. So $4 \leq -x$. So $-4\geq x$. This gives the set of solutions $(-\infty, -4]$.
Putting it all together gives you the closed interval:
$$
(-\infty, -4] \cup [-4/3,\infty)
$$
It looks like your solution is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1515986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Solve $z^3=(sr)^3$ where $r,s,z$ are integers? Let $x,y,z$ be 3 non-zero integers defined as followed:
$$(x+y)(x^2-xy+y^2)=z^3$$
Let assume that $(x+y)$ and $(x^2-xy+y^2)$ are coprime
and set $x+y=r^3$ and $x^2-xy+y^3=s^3$
Can one write that $z=rs$ where $r,s$ are 2 integers?
I am not seeing why not but I want to be sure.
| Yes, there exist such integers $r$ and $s$. It is simplest to use the Fundamental Theorem of Arithmetic (Unique Factorization Theorem). The result is easy to prove for negative $z$ if we know the result holds for positive $z$. Also, the result is clear for $z=1$. So we may assume that $z\gt 1$.
Let $z=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$, where the $p_i$ are distinct primes. Then $z^3=p_1^{3a_1}\cdots p_k^{3a_k}$.
Because by assumption $x+y$ and $x^2-xy+y^2$ are relatively prime, the primes in the factorization of $z^3$ must split into two sets, the ones that "belong to" $x+y$ and the ones that belong to $x^2-xy+y^2$. Because any prime has exponent divisible by $3$, each of $x+y$ and $x^2-xy+y^2$ is a perfect cube.
The rest is easy. From $z^3=r^3s^3$, it immediately follows that $z=rs$.
Remark: Note that $x$ and $y$ relatively prime does not imply $x+y$ and $x^2-xy+y^2$ are relatively prime. They could be both divisible by $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1516925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
A Basic Probability Question - I am getting the wrong answer Problem:
In a deck of $52$ cards there are $4$ kings. A card is drawn at random
from the deck and its face value noted; then the card is returned. This
procedure is followed $4$ times. Compute the probability that there are
exactly $2$ kings in the $4$ selected cards if it is known that there is at
least one king in those selected.
Answer:
Let $p$ be the probability we seek. Let $p_2$ be the probability that we
draw exactly $2$ kings. Let $p_1$ be the probability that we draw at least
$1$ king.
\begin{eqnarray*}
p &=& \frac{p_2}{p_1} \\
p_2 &=& {13 \choose 2}{(\frac{4}{52})^2}{(\frac{48}{52})}^2 \\
{13 \choose 2} &=& \frac{13(12)}{2(1)} = 13(6) \\
p_2 &=& 13(6){(\frac{1}{13})^2}{(\frac{12}{13})}^2 = \frac{6(12)^2}{13^3} \\
\end{eqnarray*}
Let $p_0$ be the probability that we draw no kings.
\begin{eqnarray*}
p_0 &=& (\frac{48}{52})^4 = (\frac{12}{13})^4 \\
p_1 &=& 1 - p_0 = 1 - (\frac{12}{13})^4 = \frac{13^4 - 12^4}{13^4} \\
p &=& \frac{ \frac{6(12)^2}{13^3} }{ \frac{13^4 - 12^4}{13^4} } =
\frac{6(13)(12)^2}{13^4 - 12^4} \\
\end{eqnarray*}
However, the book gets:
\begin{eqnarray*}
p = \frac{6(12)^2}{13^4 - 12^4} \\
\end{eqnarray*}
I am hoping that somebody can tell me where I went wrong.
~
| $p_2$ the probability that exactly two of four draws will be kings (one of thirteen cards in a suit) is:$$p_2= \binom{\color{red}{4}}{2}\left(\frac{1}{13}\right)^2\left(\frac{12}{13}\right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1519680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Limit of a sum of infinite series How do I find the following?
$$\lim_{n\to\infty} \frac{\left(\sum_{r=1}^n\sqrt{r}\right)\left(\sum_{r=1}^n\frac1{\sqrt{r}}\right)}{\sum_{r=1}^n r}$$
The lower sum is easy to find. However, I don't think there is an expression for the sums of the individual numerator terms... Nor can I think of a way to get the combined sum.
I just need a hint.
| HINTS:
*
*Summations:
$$\sum_{r=1}^{n}\sqrt{r}=\text{H}_n^{-\frac{1}{2}}$$
$$\sum_{r=1}^{n}\frac{1}{\sqrt{r}}=\text{H}_n^{\frac{1}{2}}$$
$$\sum_{r=1}^{n}r=\frac{1}{2}n(1+n)$$
With $\text{H}_n^r$ is the generalized harmonic number.
*
*Fraction:
$$\frac{\text{H}_n^{-\frac{1}{2}}\cdot\text{H}_n^{\frac{1}{2}}}{\frac{1}{2}n(1+n)}=\frac{2\text{H}_n^{-\frac{1}{2}}\cdot\text{H}_n^{\frac{1}{2}}}{n(1+n)}$$
*Limit:
$$\lim_{n\to\infty}\frac{2\text{H}_n^{-\frac{1}{2}}\cdot\text{H}_n^{\frac{1}{2}}}{n(1+n)}=2\lim_{n\to\infty}\frac{\text{H}_n^{-\frac{1}{2}}\cdot\text{H}_n^{\frac{1}{2}}}{n(1+n)}=$$
$$2\lim_{n\to\infty}\frac{\frac{\text{d}}{\text{d}n}\left(\text{H}_n^{-\frac{1}{2}}\cdot\text{H}_n^{\frac{1}{2}}\right)}{\frac{\text{d}}{\text{d}n}n(1+n)}=2\lim_{n\to\infty}\frac{\frac{1}{2}\left(\text{H}_n^{-\frac{1}{2}}\zeta\left(\frac{3}{2},n+1\right)-\text{H}_n^{\frac{1}{2}}\zeta\left(\frac{1}{2},n+1\right)\right)}{2n+1}=$$
$$\lim_{n\to\infty}\frac{\text{H}_n^{-\frac{1}{2}}\zeta\left(\frac{3}{2},n+1\right)-\text{H}_n^{\frac{1}{2}}\zeta\left(\frac{1}{2},n+1\right)}{2n+1}=\lim_{n\to\infty}\frac{\frac{\text{d}}{\text{d}n}\left(\text{H}_n^{-\frac{1}{2}}\zeta\left(\frac{3}{2},n+1\right)-\text{H}_n^{\frac{1}{2}}\zeta\left(\frac{1}{2},n+1\right)\right)}{\frac{\text{d}}{\text{d}n}\left(2n+1\right)}=$$
$$\lim_{n\to\infty}\frac{\frac{1}{2}\left(3\zeta\left(\frac{5}{2},1+n\right)\left(\zeta\left(-\frac{1}{2},1+n\right)-\zeta\left(-\frac{1}{2}\right)\right)+\zeta\left(\frac{3}{2},1+n\right)\left(-3\zeta\left(\frac{1}{2},1+n\right)+\zeta\left(\frac{1}{2}\right)\right)\right)}{2}=$$
$$\frac{1}{4}\lim_{n\to\infty}\left(3\zeta\left(\frac{5}{2},1+n\right)\left(\zeta\left(-\frac{1}{2},1+n\right)-\zeta\left(-\frac{1}{2}\right)\right)+\zeta\left(\frac{3}{2},1+n\right)\left(-3\zeta\left(\frac{1}{2},1+n\right)+\zeta\left(\frac{1}{2}\right)\right)\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1520866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Evaluate the Integral: $\int^\pi_0\cos^6\theta\ d\theta$ $\int^\pi_0\cos^6\theta\ d\theta$
So I split the trig value into:
$\int^\pi_o\cos^5\theta\ cos\theta\ d\theta$
Then I utilized the Pythagorean theorem for $cos^5\theta$
$\int^\pi_o(1-sin^5\theta)\ cos\theta$
I utilized u-substitution:
$u=sin\ \theta$
$du=cos\ \theta$
Thus:
$\int^{x=\pi}_{x=0}\ (1-u^5)\ d\theta$
I intergated
$(\frac{1}{6}u^6)+(\frac{1}{6}u^6)$
$-(\frac{\pi^6}{6})+(0)$
$-(\frac{\pi^6}{6})$
Is my answer right?
| An easy way is to use Euler formula:
$$\cos \theta = \frac{1}{2}(e^{i\theta} + e^{-i\theta})$$
Then use the binomial expansion:
$$\cos^6\theta = \frac{1}{2^6}\sum_{k = 0}^6 \binom{6}{k}e^{i(6 - k)\theta}e^{-ik\theta} = \frac{1}{2^6}\sum_{k = 0}^6 \binom{6}{k}e^{i(6 - 2k)\theta}.$$
If $k = 3$, $\int_0^\pi e^{i(6 - 2k)\theta} d\theta = \pi.$ Otherwise
$$\int_0^\pi e^{i(6 - 2k)\theta} d\theta = \frac{1}{6 - 2k}(e^{i(6 - 2k)\pi} - 1) = 0$$
in view of $6 - 2k$ is always an even number. Therefore,
$$\int_0^\pi \cos^6\theta d\theta = \frac{1}{2^6} \binom{6}{3}\pi = \frac{5}{16}\pi.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Estimating the number of integers relatively prime to $6$ between $1$ and some integer $x$? I am trying to understand the standard way to estimate the number of integers relatively prime to $6$ where we don't know which congruence class $x$ belongs to.
For a given $x$, if we know the congruence class, it is straight forward to determine the number of integers between $1$ and $x$ that are relatively prime to $6$.
If $x \equiv 0 \pmod 3$, the answer is $\frac{x}{3}$: i.e., between $1$ and $9$, there are $3$ that are relatively prime: $\{ 1, 5, 7\}$
If $x \equiv 1 \pmod 6$, the answer is $\frac{x+2}{3}$: ie., between $1$ and $7$, there are $3$.
If $x \equiv 2 \pmod 3$, the answer is $\frac{x+1}{3}$
if $x \equiv 4 \pmod 6$, the answer is $\frac{x-1}{3}$
What be the standard answer for $x$ where we don't know the congruence class?
Would we say between $\frac{x-1}{3}$ and $\frac{x+2}{3}$?
What would be the right way for any $x$ to estimate the number of integers relatively prime to $6$?
| The number of integers relatively prime to $6$ for each $x$ is given by the expression
$$\left\lfloor \frac{x-1}{6}\right\rfloor +\left\lfloor \frac{x+1}{6}\right\rfloor +1$$
which is clearly $\sim x/3.$
Check:
rp6[x_] := Floor[(x - 1)/6] + Floor[(x + 1)/6] + 1
{Length@Select[Range@#, GCD[#, 6] == 1 &] & /@ Range@40,
rp6@# & /@ Range@40} // ListLinePlot
Generalisation:
The number of integers relatively prime to $n$ for each $x$ is $\sim x \cdot \phi(n)/n,$ where $\phi$ is the Euler totient function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1523105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$
I would like to prove
$$ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$$
*
*I'm interested in more ways of proving it
My thoughts:
\begin{align}
\sqrt{x+2}-\sqrt{x+1}\neq \sqrt{x+1}-\sqrt{x}\\
\frac{x+2-x-1}{\sqrt{x+2}+\sqrt{x+1}}&\neq \frac{x+1-x}{\sqrt{x +1}+\sqrt{x}}\\
\frac{1}{\sqrt{x+2}+\sqrt{x+1}}&\neq \frac{1}{\sqrt{x +1}+\sqrt{x}}\\
\sqrt{x +1}+\sqrt{x} &\neq \sqrt{x+2}+\sqrt{x+1}\\
\sqrt{x} &\neq \sqrt{x+2}\\
\end{align}
*
*Is my proof correct?
*I'm interested in more ways of proving it.
| Assume the statement is true
\begin{align}
\sqrt{x + 2} - \sqrt{x + 1} &= \sqrt{x + 1} - \sqrt{x}\\
\sqrt{x + 2} + \sqrt{x} &= 2\sqrt{x + 1}\\
2x + 2 + (2\sqrt{x^2 + 2x}) &= 4x + 4\\
\sqrt{x^2 + 2x} &= x + 1\\
x^2 + 2x &= x^2 + 2x + 1\\
0 &= 1\\
\end{align}
Which is a contradiction. Now I am not sure if I'm allowed to manipulate the equation like this so the proof might be invalid. The reason I though I could was I had seen simillar logic in proving the irrationality of $\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1523179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
What is the method for finding $\int\frac {x^3 + 5x^2 +2x -4}{(x^4-1)}\mathrm{d}x$? $$
\int\frac {x^3 + 5x^2 +2x -4}{(x^4-1)}dx
$$
A bit confused with how to integrate this question. I though it was partial fractions but was unsure about the what to do after that.
| The integral can be written as
$$I=\frac{1}{4}\int\frac{4x^3}{x^4-1}+5\int\frac{x^2}{(x^2-1)(x^2+1)}+\int\frac{2x}{(x^2-1)(x^2+1)}-4\int\frac{dx}{(x^2-1)(x^2+1)}=I_1+I_2+I_3+I_4$$
$$I_1=\frac{1}{4}ln|x^4-1|+C$$
$$I_2=\frac{5}{2}\int\frac{1}{x^2-1}+\int\frac{1}{x^2+1}$$
$$I_3=\int\frac{dt}{(t-1)(t+1)}$$ where $t=x^2$
$$I_4=-2\int\frac{1}{x^2-1}-2\int\frac{1}{x^2+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Convergence of the series $\sum\limits_{n = 4}^\infty {\frac{{n + 1}}{{(n + 5)(n + 4)(n - 3)}}}$? To analyze the convergence of the
$$\sum\limits_{n = 4}^\infty {\frac{{n + 1}}{{(n + 5)(n + 4)(n - 3)}}}$$
series I used the criterion of integral $$\displaystyle\int_4^\infty {\frac{{x + 1}}{{(x + 5)(x + 4)(x - 3)}}dx},$$ but calculate this improper integral is a very laborious task.
Is there a shorter way? What criteria of convergence would be most effective or simple?
| In the same spirit as Leg's answer, starting with the partial fraction decomposition $$\dfrac{n+1}{(n+5)(n+4)(n-3)} = -\dfrac12\cdot\dfrac1{n+5} + \dfrac37\cdot\dfrac1{n+4} + \dfrac1{14} \cdot\dfrac1{n-3}$$ and using harmonic numbers definition $$\sum_{n=p}^m \dfrac1{n+5}=H_{m+5}-H_{p+4}$$ $$\sum_{n=p}^m \dfrac1{n+4}=H_{m+4}-H_{p+3}$$ $$\sum_{n=p}^m \dfrac1{n-3}=H_{m-3}-H_{p-4}$$ which make $$A_p=\sum_{n=p}^m \dfrac{n+1}{(n+5)(n+4)(n-3)} =\frac{1}{14} \left(H_{m-3}+6 H_{m+4}-7 H_{m+5}-H_{p-4}+H_{p+3}+\frac{7}{p+4}\right)$$ When $m\to \infty$, $H_{m-3}+6 H_{m+4}-7 H_{m+5}\to 0$ and then, at the limit, $$A_p=\frac{1}{14}
\left(\frac{1}{p-3}+\frac{1}{p-2}+\frac{1}{p-1}+\frac{1}{p}+\frac{1}{p+1}+\frac{1}{p+2}+\frac{1}
{p+3}+\frac{7}{p+4}\right)$$ which gives $$A_4=\frac{971}{3920}$$ $$A_5=\frac{6289}{35280}$$ $$A_6=\frac{5113}{35280}$$ as so forth.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1526806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
question on right angle triangle Let ABC and DBC be two equilateral triangle on the same base BC,a point P is taken on the circle with centre D,radius BD. Show that PA,PB,PC are the sides of a right triangle.
|
First let's assume $AB=1$ and call $PB=a;PC=b;PA=c$, we have to prove that $a^2+b^2=c^2$. Applying the law of cosines to $PBC$ with respect to angle $\angle BPC=\pi/6$ we get
$$a^2+b^2-\sqrt3ab=1$$
Now, call $\angle PCB=y$. Applying the law of cosines to $PBC$ with respect to angle $y$ we get
$$ 1+b^2-2b\cos y=a^2$$
hence $\cos y=\frac{1+b^2-a^2}{2b}$. From the law of sines on $PBC$ we also get
$\sin y=a/2.$
With these information we can comupte the cosine of angle $\angle PCA=y+\pi/3$:
$$cos(y+\pi/3)=\cos y/2-\sin y \sqrt 3/2.$$
Finally we apply the cosine law on $PAC$ with angle $\angle PCA$:
\begin{align}
c^2&=1+b^2-2b\cos (y+\pi/3) \\
&=1+b^2-2b(\frac{\cos y}{2}-\frac{\sin y \sqrt 3}{2})\\
&=1+b^2-b\big(\frac{1+b^2-a^2}{2b}-\frac{a\sqrt3}{2}\big)\\
&=\frac{1+b^2-a^2}{2}+\frac{ab\sqrt3}{2}
\end{align}
and plugging in the first equation we find:
$$c^2=\frac{1+b^2+a^2}{2}+\frac{a^2+b^2-1}{2}=a^2+b^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to use boundary condition for Euler-Bernoulli equation How to use enter boundary condition into a Euler-Bernoulli fourth order equation
| OK, first integrate step by step to obtain
$$\begin{array}{*{20}{l}}
{\frac{{{d^4}y}}{{d{x^4}}} = F} \\
{\frac{{{d^3}y}}{{d{x^3}}} = Fx + A} \\
{\frac{{{d^2}y}}{{d{x^2}}} = \frac{F}{2}{x^2} + Ax + B} \\
{\frac{{dy}}{{dx}} = \frac{F}{6}{x^3} + \frac{A}{2}{x^2} + Bx + C} \\
{y = \frac{F}{{24}}{x^4} + \frac{A}{6}{x^3} + \frac{B}{2}{x^2} + Cx + D}
\end{array}$$
Next, apply your boundary conditions to obtain a linear algebraic system for the unknown constants $A$, $B$, $C$, $D$
$$\begin{array}{*{20}{l}}
{y(0) = 0}& \to &{D = 0} \\
{y(l) = 0}& \to &{\frac{F}{{24}}{l^4} + \frac{A}{6}{l^3} + \frac{B}{2}{l^2} + Cl + D = 0} \\
{\frac{{dy}}{{dx}}(0) = 0}& \to &{C = 0} \\
{\frac{{dy}}{{dx}}(l) = 0}& \to &{\frac{F}{6}{l^3} + \frac{A}{2}{l^2} + Bl + C = 0}
\end{array}$$
and hence
$$\left\{ \begin{gathered}
\frac{A}{6}{l^3} + \frac{B}{2}{l^2} = - \frac{F}{{24}}{l^4} \hfill \\
\frac{A}{2}{l^2} + Bl = - \frac{F}{6}{l^3} \hfill \\
\end{gathered} \right.\,\,\,\, \to \,\,\,\,\left\{ \begin{gathered}
\frac{l}{6}A + \frac{1}{2}B = - \frac{{F{l^2}}}{{24}} \hfill \\
\frac{l}{2}A + B = - \frac{{F{l^2}}}{6} \hfill \\
\end{gathered} \right.$$
Solve for $A$ and $B$. I will leave obtaining the constants to you. :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How many numbers have unit digit $1$?
Let $f(n)$ be the number of positive integers that have exactly $n$ digits and whose
digits have a sum of $5$. Determine, with proof, how many of the $2014$ integers
$f(1), f(2), . . . , f(2014)$ have a units digit of $1.$
HINTS ONLY
EDIT:
for $f(2)$ we have: $x + y = 5$ with $x \ne 0$, which gives: $\binom{5}{1}$
$f(5)$ gives: $x + y + z + t +w = 5$ so: $\binom{8}{4}$
$f(4)$ gives, $\binom{7}{3}$
For $f(n)$ we have: $f(n) = \binom{n + 3}{n-1}$
$f(n) = \frac{(n+3)(n+2)(n+1)(n)(n-1)!}{24(n-1)!} = \frac{(n+3)(n+2)(n+1)(n)}{24}$
PROOF CHECK:
Consider $N = \overline{a_1 a_2 ... a_n}$ with $a_i$ as digits such that $a_1 \ge 1$. We have a string of $n$ characters, $a_1, a_2, a_3, ..., a_n$ with $a_1 \ne 0$, thus this string with $a_n$ gives the $n$th digit numbers. the number of numbers corresponds to the solutions of: $a_1 + a_2 + ... + a_n = 5$ for nonnegative integers with $a_1 > 0$. By the stars-and-bars principle there exist: $\binom{n + 4 - 1}{n-1} = \binom{n+3}{n-1} = \frac{(n+3)(n+2)(n+1)(n)}{24}$ numbers, thus giving an explicit formula for $f(n)$.
Is this enough for a proof? Or is induction needed for a stronger argument?
So we have: $f(n) \equiv 1\pmod{10}$.
Since $f(n)$ is a combinatoric coefficient, $f(n) \in \mathbb{Z+}$, which implies that: $\frac{(n+3)(n+2)(n+1)(n)}{24} \equiv (n+3)(n+2)(n+1)(n) \pmod{10}$. Since the numerator was divisible by the denominator for all $n$.
Now help is needed:
How to solve this modular arithemetic expression?
| If you want to solve $$\frac{n(n+1)(n+2)(n+3)}{24}\equiv 1\pmod{10},$$
You can multiple both sides by $3$, since it is relatively prime to $10$, but when you multiply by $8$, you have to apply that to the modulus, to. So, you are trying to solve:
$$n(n+1)(n+2)(n+3)\equiv 24\pmod{80}$$
Solve this in pairs, using Chinese remainder theorem:
$$n(n+1)(n+2)(n+3)\equiv 24\pmod{5}$$
$$n(n+1)(n+2)(n+3)\equiv 24\pmod{16}$$
As it turns out, the second only depends on $n\pmod 8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding $\int _0^a\sqrt{1+\frac{1}{4x}}dx$ to calculate arclength So I'm trying to find the arclength of $x^{0.5}$ and its tougher than I thought. Tried substitutions like $\dfrac{\cot^2x}{4}$ and some other trig subs but they got me nowhere. Any tips?
$$\int _0^a\sqrt{1+\frac{1}{4x}}dx$$
Edit:
This is what I got so far: $\int_{0}^{a}\sqrt{1+\left(\left(\sqrt{x}\right)'\right)^{2}}dx=\int_{0}^{a}\sqrt{1+\frac{1}{4x}}dx=\left[\begin{array}{cc}
t^{2}=1+\frac{1}{4x} & \sqrt{1+\frac{1}{4N}},\sqrt{1+\frac{1}{4a}}\\
2tdt=-\frac{1}{8x^{2}}dx & x=\frac{1}{4\left(t^{2}-1\right)}
\end{array}\right]=\\\lim_{N\to0}-\int_{\sqrt{1+\frac{1}{4N}}}^{\sqrt{1+\frac{1}{4a}}}\frac{t^{2}dt}{\left(t^{2}-1\right)^{2}}=\lim_{N\to0}\int_{\sqrt{1+\frac{1}{4N}}}^{\sqrt{1+\frac{1}{4a}}}\left(\frac{1}{4\left(t+1\right)}-\frac{1}{4\left(t+1\right)^{2}}-\frac{1}{4\left(t-1\right)}-\frac{1}{4\left(t-1\right)^{2}}\right)dt=\\=\lim_{N\to0}\frac{1}{4}\left[\ln\left(t+1\right)+\frac{1}{t+1}-\ln\left(t-1\right)+\frac{1}{t-1}\right]_{\sqrt{1+\frac{1}{4N}}}^{\sqrt{1+\frac{1}{4a}}}=\\=\left[\ln\left(\frac{\sqrt{1+\frac{1}{4a}}+1}{\sqrt{1+\frac{1}{4a}}-1}\right)-\frac{2\sqrt{1+\frac{1}{4a}}}{1+\frac{1}{4a}-1}\right]-\lim_{N\to0}\left[\ln\left(\frac{\sqrt{1+\frac{1}{4N}}+1}{\sqrt{1+\frac{1}{4N}}-1}\right)-\frac{2\sqrt{1+\frac{1}{4N}}}{1+\frac{1}{4N}-1}\right]=\\=\left[\ln\left(\frac{\sqrt{1+\frac{1}{4a}}+1}{\sqrt{1+\frac{1}{4a}}-1}\right)-8a\sqrt{1+\frac{1}{4a}}\right]-\lim_{N\to0}\left[\ln\left(\frac{\sqrt{1+\frac{1}{4N}}+1}{\sqrt{1+\frac{1}{4N}}-1}\right)-8N\sqrt{1+\frac{1}{4N}}\right]=\\=\ln\left(\frac{\sqrt{1+\frac{1}{4a}}+1}{\sqrt{1+\frac{1}{4a}}-1}\right)-8a\sqrt{1+\frac{1}{4a}}-0+0=\ln\left(4a\left(\sqrt{\frac{1}{a}+4}+2\right)+1\right)-8a\sqrt{1+\frac{1}{4a}}$
But it doesn't seem right... any Ideas what went wrong?
| Hint:
Use differential binomial. Note that $$\sqrt{1+\frac{1}{4x}}=0.5x^{-0.5}(1+4x)^{0.5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Express $\frac{9x}{(2x+1)^2(1-x)}$ as a sum of partial fractions with constant numerators. Answer doesn't match with solution provided in book. Express $\frac{9x}{(2x+1)^2(1-x)}$ as a sum of partial fractions with constant numerators. Answer doesn't match with solution provided in book.
My method:
$\frac{9x}{(2x+1)^2(1-x)}\equiv\frac{A}{2x+1}+\frac{B}{(2x+1)^2}+\frac{C}{1-x}$
Comparing L.H.S. numerators with R.H.S. numerators:
$9x=A(2x+1)+B(1-x)+C(2x+1)^2$
$x=-\frac12, B=-3$
$x=0, A+B+C=0 \Rightarrow A=3+C$
$x=1, 3A+9C=9 \Rightarrow 3(3+C)+9C=9\Rightarrow C=0$
This is the point where my answer differs from the one given in the book. And I don't understand why.
Answer in book: $\frac{1}{1-x}+\frac{2}{2x+1}-\frac{3}{(2x+1)^2}$
| Spotted mistake. $A(2x+1)$ should be $A(2x+1)(1-x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1535277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
calculate $\lim\limits_{x \to 1}(1 - x)\tan \frac{\pi x}{2}$ I need to calculate $$\lim_{x \to 1}\left((1 - x)\tan \frac{\pi x}{2}\right)$$.
I used MacLaurin for $\tan$ and got $\frac{\pi x} {2} + o(x)$. Then the full expression comes to $$\lim_{x \to 1}\left(\frac {\pi x} {2} - \frac {\pi x^2} {2} + o(x)\right) = 0$$But WolframAlpha says it should be $\frac 2 \pi$. What am I doing wrong?
| We use that $\tan\theta =\cot\left (\frac{\pi}{2}-\theta\right)$.
Letting $y=\frac{\pi(1-x)}{2}=\frac{\pi}{2}-\frac{\pi}{2}x$, we get that we are seeking:
$$\lim_{y\to 0} \frac{2}{\pi}y\cot y = \frac{2}{\pi}\lim_{y\to 0}\left(\frac{y}{\sin y}\cdot \cos y\right)$$
From there, it should be easy.
Alternatively, note that:
$$\frac{\cot\frac{\pi x}{2}}{1-x} = -\frac{\cot\frac{\pi x}{2}-\cot\frac{\pi}{2}}{x-1}\tag{1}$$
So as $x\to 1$, the limit of $(1)$ as $x\to 1$ is the definition of $-f'(1)$, where $f(x)=\cot\frac{\pi x}{2}$.
But $$(1-x)\cot \frac{\pi x}{2} = \frac{1-x}{\cot\frac{\pi x}{2}}$$ and thus your limit is $-\frac{1}{f'(1)}$. So if you know the derivative of cotangent, you are done.
So if $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Show that there is a common line of intersection of the three given planes. Let $x-y\sin\alpha-z\sin\beta=0,x\sin\alpha-y+z\sin\gamma=0$ and $x\sin\beta+y\sin\gamma-z=0$ be the equations of the planes such that $\alpha+\beta+\gamma=\frac{\pi}{2}$,(where $\alpha,\beta,\gamma\neq0$).Then show that there is a common line of intersection of the three given planes.
We know that three given planes,$a_1x+b_1y+c_1z+d_1=0,a_2x+b_2y+c_2z+d_2=0,a_3x+b_3y+c_3z+d_3=0$ meet in a common line if $\begin{vmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3 \\
\end{vmatrix}=0$
if $\begin{vmatrix}
1 & -\sin\alpha & -\sin\beta \\
\sin\alpha & -1 & \sin\gamma \\
\sin\beta & \sin\gamma & -1 \\
\end{vmatrix}=0$
I simplified this determinant to get $1-\sin^2\alpha-\sin^2\beta-\sin^2\gamma-2\sin\alpha\sin\beta\sin\gamma=0$
We have to prove the trigonometric expression true in order to prove the question.But i could not prove the trigonometric expression true.
Please help me.
| Rewrite the required result as
$$ \cos^2\gamma - \sin^2 \alpha - \sin^2 \beta \overset{?}= 2\sin \alpha \sin \beta \sin \gamma $$
We'll try to show that the LHS equals the RHS. Note that $\cos \gamma = \sin (\alpha + \beta)$ and vice versa, as $\gamma = \pi/2 - (\alpha + \beta)$
\begin{equation}
\sin^2 (\alpha + \beta) - \sin^2 \alpha - \sin^2 \beta \\
\overset{(a)}= ~\sin^2\alpha \cos^2\beta - \sin^2\alpha + \cos^2\alpha\sin^2\beta - \sin^2\beta + 2\sin \alpha \sin \beta \cos \alpha \cos \beta \\
\overset{(b)}= ~-2\sin^2\alpha \sin^2 \beta + 2\sin \alpha \sin \beta \cos \alpha \cos \beta \\
= ~2\sin \alpha \sin \beta (\cos \alpha \cos \beta - \sin \alpha \sin \beta )\\
\overset{(c)}= 2 \sin \alpha \sin \beta \cos (\alpha + \beta)\\
\end{equation}
where
$(a)$ follows from squaring $\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$ and some regrouping.
$(b)$ is $\sin^2\alpha \cos^2\beta - \sin^2\alpha = \sin^2 \alpha (\cos^2 \beta -1) = -\sin^2 \alpha \sin^2 \beta$ applied a couple of times.
$(c)$ is the standard formula for $\cos (\alpha + \beta)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
A general approach for this type of questions I've come across multiple questions like these. $\left(\iota=\sqrt{-1}\right)$
If $f\left(x\right)=x^4-4x^3+4x^2+8x+44$, find the value of $f\left(3+2\iota\right)$.
If $f\left(z\right)=z^4+9z^3+35z^2-z+4$, find $f\left(-5+2\sqrt{-4}\right)$.
Find the value of $2x^4+5x^3+7x^2-x+41$, when $x=-2-\sqrt{3}\iota$.
...and so on. The answers to these are real numbers. And the methods given to solve these include random manipulations of $x=\text{<the given complex number>}$ until we reach the given $f\left(x\right)$ or a term which $f\left(x\right)$ can be rewritten in terms of, so that we get the remainder. But all these approaches are completely random, and I see no logical approach. Can someone explain me exactly how would you solve a question of this format?
EDIT : Here's what I mean by random manipulations. Here's the solution to the third one.
$x+2=-\sqrt{3}\iota\Rightarrow x^2+4x+7=0$
Therefore, $2x^4+5x^3+7x^2-x+41$ $=\left(x^2+4x+7\right)\left(2x^2-3x+5\right)+6$$=0\times\left(2x^2-3x+5\right)+6=6$
Now how are we supposed to observe that the given quartic could be factorized? We have not been taught any method to factorise degree four polynomials, unless one root is known, in which case I can reduce it into a product of a linear and a cubic. Further if a root of the cubic is visible.
| Ill give a guide line we have $x=-2-\sqrt{3}i$ now bring two on lhs and square both sides. So we get $x^2+4x+4=-3$ so $x^2+4x+7=0$ ..(1).now just divide the given expression by this qyadratic so we have . $\frac{2x^4+5x^3+7x^2-x+41}{x^2+4x+7}$ . I hope now you know this type of division . If not look up on net. Sowe get the answer as 6 on division. Now we have $2x^4+5x^3+7x^2-x+41=(x^2+4x+7)(2x^3-3x+5)+6$ but according to eqn(1) we have $x^2+4x+7=0$ . Thus the value of $2x^4+5x^3+7x^-x+41=6$ . Hope now you can get ideas for other problems. Look up on the net for division of polynomials and you will know how did i divide the quartic with a quadratic. For first take value as $x=3+2i$ and for second we have $z=-{5}+2.2i$ so we have $z=-{5}+4i$. Suppose if we have $x=a+bi$ the best trick is to bring the iota independent term together with x here 'a' and squaring both the sides . This removes the iota thus dqn becomes $x^2-a^2=-b^2$ note $i^2=-1$ . Then we divide the given polynomial with this quadratic and get the answer. As a referrence take the example of my answer. Thats what a general solution can be like.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1539379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Converting a sum of trig functions into a product Given,
$$\cos{\frac{x}{2}} +\sin{(3x)} + \sqrt{3}\left(\sin\frac{x}{2} + \cos{(3x)}\right)$$
How can we write this as a product?
Some things I have tried:
*
*Grouping like arguments with each other. Wolfram Alpha gives $$\cos{\frac{x}{2}} + \sqrt{3}\sin{\frac{x}{2}} = 2\sin{\left(\frac{x}{2} + \frac{\pi}{6}\right)}$$but I don't know how to derive that myself or do a similar thing with the $3x$.
*Write $3x$ as $6\frac{x}{2}$ and then using the triple and double angle formulas, but that is much too tedious and there has to be a more efficient way.
*Rewriting $\sqrt{3}$ as $2\sin{\frac{\pi}{3}}$ and then expanding and trying to use the product-to-sum formulas, and then finally grouping like terms and then using the sum-to-product formulas, but that didn't work either.
I feel like I'm overthinking this, so any help or insights would be useful.
| $$\cos{\frac{x}{2}} +\sin(3x) + \sqrt{3}\left(\sin\frac{x}{2} + \cos(3x)\right)$$
$$=\cos{\frac{x}{2}} + \sqrt{3}\sin\frac{x}{2} +\sin(3x) + \sqrt{3}\cos(3x)$$
$$=2\left(\frac{1}{2}\cos\frac{x}{2} + \frac{\sqrt{3}}{2}\sin\frac{x}{2} +\frac{1}{2}\sin(3x) + \frac{\sqrt{3}}{2}\cos(3x)\right)$$
Note that $\frac{1}{2}=\sin\frac{\pi}{6}$ and $\frac{\sqrt{3}}{2}=\cos\frac{\pi}{6}$ so:
$$=2\left(\sin\frac{\pi}{6}\cos\frac{x}{2} + \cos\frac{\pi}{6}\sin\frac{x}{2} +\sin\frac{\pi}{6}\sin(3x) + \cos\frac{\pi}{6}\cos(3x)\right)$$
Then using Addition Theorem:
$$=2\left(\sin\left(\frac{x}{2}+\frac{\pi}{6}\right)+\cos\left(3x-\frac{\pi}{6}\right)\right)$$
$$=2\left(\sin\left(\frac{x}{2}+\frac{\pi}{6}\right)+\sin\left(3x+\frac{\pi}{3}\right)\right)$$
Then using Sums to Products:
$$=4\left(\sin\left(\frac{\frac{x}{2}+\frac{\pi}{6}+3x+\frac{\pi}{3}}{2}\right)\cos\left(\frac{\frac{x}{2}+\frac{\pi}{6}-3x-\frac{\pi}{3}}{2}\right)\right)$$
$$=4\sin\left(\frac{7x+\pi}{4}\right)\cos\left(\frac{-15x-\pi}{12}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
how can i find the matrix $P$ that diagonalizes the matrix $A$? I want to find matrix $P$ that diagonalizes the matrix $A$:
$$
\begin{bmatrix}
4 & 0 & 1 \\
2 & 3 & 2 \\
1 & 0 & 4 \\
\end{bmatrix}
$$
| You need to calculate the eigenvectors which are (1,2,1), (-1,0-1) and (0,1,0). So we have the passage matrix
\begin{equation}P=\left(\begin{array}{ccc}
1 & -1& 0\\
2 & 0 & 1\\
1 & 1 & 0
\end{array}\right)
\end{equation}
\begin{equation}P^{-1}=\left(\begin{array}{ccc}
1/2 & 0& 1/2\\
-1/2 & 0 & 1/2\\
-1 & 1 & -1
\end{array}\right)
\end{equation}
The inverse of P is
\begin{equation}D=P^{-1}AP=\left(\begin{array}{ccc}
5 & 0& 0\\
0 & 3 & 0\\
0 & 0 & 3
\end{array}\right)
\end{equation}
So we have the diagonalization
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Evaluating the limit $\mathop {\lim }\limits_{x \to 0} \frac{{x(1 - 0.5\cos x) - 0.5\sin x}}{{{x^3}}}$ For evaluating the limit $\lim\limits_{x \to 0} \frac{x(1 - 0.5\cos x) - 0.5\sin x}{x^3}$, I proceeded as follows:
$$\lim_{x \to 0} \left(\frac{x(1 - 0.5\cos x)}{x^3} - \frac{0.5}{x^2}\left(\frac{\sin x}{x}\right)\right)$$
Using the fact that $\lim\limits_{x \to 0} \frac{\sin x}{x}=1$ and $\lim\limits_{x \to a} \{ f(x) - g(x)\} = \lim\limits_{x \to a} f(x) - \lim\limits_{x \to a} g(x)$, I got
$$\lim\limits_{x \to 0} \frac{1 - 0.5\cos x}{x^2} - \lim\limits_{x \to 0} \frac{0.5}{x^2}$$
$$ = \lim_{x \to 0} \left(\frac{1 - 0.5\cos x - 0.5}{x^2} \right)$$
Now, after applying L'Hopital's Rule I got the final answer as $0.25$. However, on evaluating the original limit using Mathematica, I got the answer as $\frac{1}{3}$.
Can someone please tell me where am I going wrong.
Somehow, I believe that you cannot use the fact that $\lim\limits_{x \to a} \{ f(x) - g(x)\} = \lim \limits_{x \to a} f(x) - \lim\limits_{x \to a} g(x)$ in case of indeterminate forms.
Thanks in advance!
| If you want to compute this limit by making use of known basic limits then
use the following
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}}=-\frac{1}{6},\ \ \ \ and\ \ \ \
\ \ \lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}}=\frac{1}{2}
\end{equation*}
as follows
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{x(1-\frac{1}{2}\cos x)-\frac{1}{2}\sin x}{x^{3}}
&=&\lim_{x\rightarrow 0}\frac{1}{2}\frac{2x(1-\frac{1}{2}\cos x)-\sin x}{%
x^{3}} \\
&=&\lim_{x\rightarrow 0}\frac{1}{2}\frac{x+x-x\cos x-\sin x}{x^{3}} \\
&=&\lim_{x\rightarrow 0}\frac{1}{2}\left( \frac{x-\sin x}{x^{3}}+\frac{%
1-\cos x}{x^{2}}\right) \\
&=&\frac{1}{2}\left( \lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}}%
+\lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}}\right) \\
&=&\frac{1}{2}\left( -\frac{1}{6}+\frac{1}{2}\right) =\frac{1}{6}.
\end{eqnarray*}
Warning: When we write
\begin{equation*}
\lim_{x\rightarrow 0}\left( \frac{x-\sin x}{x^{3}}+\frac{1-\cos x}{x^{2}}%
\right) =\lim_{x\rightarrow 0}\left( \frac{x-\sin x}{x^{3}}\right)
+\lim_{x\rightarrow 0}\left( \frac{1-\cos x}{x^{2}}\right)
\end{equation*}
it is because we $\textbf{previously}$ know that the limits of the RHS exist
as real numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
What is $\frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1}$ given $x^2 + x - 1 = 0$?
Given that $x^2 + x - 1 = 0$, what is $$V \equiv \frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1} = \; ?$$
I have reduced $V$ to $\dfrac{x^8 + 1}{(x^4 + 1) (x^4 - x^2 + 1)}$, if you would like to know.
| $$\frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1} =\dfrac{(x^2+1)(x^8+1)}{(x^4+1)(x^6+1)}$$
$$=\dfrac{\left(x+\dfrac1x\right)\left(x^4+\dfrac1{x^4}\right)}{\left(x^2+\dfrac1{x^2}\right)\left(x^3+\dfrac1{x^3}\right)}$$
$$=\dfrac{\left(x^2+\dfrac1{x^2}\right)^2-2}{\left(x^2+\dfrac1{x^2}\right)\left(x^2+\dfrac1{x^2}+1\right)}$$
Now $x^2+x-1=0\implies x-\dfrac1x=-1$ as $x\ne0$
Now $x^2+\dfrac1{x^2}=\left(x-\dfrac1x\right)^2+2=1+2$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Solving $a_n=3a_{n-1}-2a_{n-2}+3$ for $a_0=a_1=1$ I'm trying to solve the recurrence $a_n=3a_{n-1}-2a_{n-2}+3$ for $a_0=a_1=1$. First I solved for the homogeneous equation $a_n=3a_{n-1}-2a_{n-2}$ and got $\alpha 1^n+\beta 2^n=a_n^h$. Solving this gives $a_n^h =1$. The particular solution, as I understand, will be $a_n^*=B$ since $f(n)=3\times 1^n$. But then I get $B=a^*=3a^*_{n-1}-2a^*_{n-2}+3=B+3$. This has to be a mistake, but I don't see what I did wrong.
| Let $f(x) = \sum_{n=0}^\infty a_nx^n$. Multiplying the recurrence by $x^n$ and summing over $n\geqslant 2$ we find that the LHS is
$$\sum_{n=2}^\infty a_nx^n = f(x)-1-x $$
and the RHS is
$$3\sum_{n=2}^\infty a_{n-1}x^n - 2\sum_{n=2}^\infty a_{n-2}x^n + 3\sum_{n=2}^\infty x^n = 3x(f(x)-1) - 2x^2f(x)+\frac{3x^2}{1-x}. $$
Equating the above and $f(x)$ we find that
$$f(x) = \frac{1-2x}{1-3x+2x^2} + \frac{3x^2}{(1-x)(1-3x+2x^2)}. $$
Partial fraction decomposition yields
$$f(x) = \frac1{1-x} + \frac3{1-2x} - \frac3{(1-x)^2}, $$
which has series representation
$$f(x) = \sum_{n=0}^\infty(1+3\cdot(2^n) - 3(n+1))x^n. $$
It follows that
$$a_n = 1+3\cdot(2^n) - 3(n+1),\ n\geqslant0. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
The sum of the first $n$ squares $(1 + 4 + 9 + \cdots + n^2)$ is $\frac{n(n+1)(2n+1)}{6}$ Prove that the sum of the first $n$ squares $(1 + 4 + 9 + \cdots + n^2)$ is
$\frac{n(n+1)(2n+1)}{6}$.
Can someone pls help and provide a solution for this and if possible explain the question
| *
*check if the statement holds true for $n=1$:
$$1^2=1=\frac{1(1+1)(2\cdot 1+1)}{6}=\frac{6}{6}=1$$
*Inductive step: show that if statement holds for $k$, then it also holds for $k+1$. This is done as follows:
$$1^2+2^2+3^2+\ldots+k^2=\frac{k(k+1)(2k+1)}{6}$$
$$1^2+2^2+3^2+\ldots+k^2+(k+1)^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$$
Note that the last equation may be rewritten as:
$$\frac{k(k+1)(2k+1)}{6}+(k+1)^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$$
It remains that both sides are indeed the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find GCD$(A_0,A_1,...,A_{2015})$ where $A_n=2^{3n}+3^{6n+2}+5^{6n+2}$ ; $n=0,...,2015$ Find GCD$(A_0,A_1,...,A_{2015})$ where $A_n=2^{3n}+3^{6n+2}+5^{6n+2}$ ; $n=0,...,2015$
Is there some intuitive method or formula for finding GCD of $n$ integers?
| Fisrtly, $A_0 = 35$, so $d:=\gcd(A_0\ldots A_{2015})\mid 35$, that means that $d \in \{1,5,7,35\}$.
Now let's try to factorize the $\gcd$. Starting with $5$, notice that $2^{4n} \equiv 3^{4n} \equiv 1\pmod{5}$ for every $n\geqslant 0$ (Fermat's little theorem). So:
\begin{align}A_n = 2^{3n}+3^{6n+2}+5^{6n+2} \equiv 2^{3n}+ 3^{2n+2} \pmod{5}\end{align}
By the previous observation, one can see that $A_{1} \equiv 3+1 \equiv 4\pmod{5}$, so $d\in\{1,7\}$. Again, having Fermat's little thm in mind:
\begin{align}A_n = 2^{3n}+3^{6n+2}+5^{6n+2} &\equiv 8^n+ 9\cdot 3^{6n} + 25\cdot 5^{6n} \pmod{7} \\ &\equiv 1^n + 2\cdot 1^n + 4\cdot 1^{n} \pmod{7} \\ &\equiv 7 \pmod{7} \\ &\equiv 0\pmod{7}\end{align}
So $7\mid A_n,~\forall n\geqslant 0$, thus $d$ must be $7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How can I find $\cos(\theta)$ with $\sin(\theta)$?
If $\sin^2x$ + $\sin^22x$ + $\sin^23x$ = 1, what does $\cos^2x$ + $\cos^22x$ + $\cos^23x$ equal?
My attempted (and incorrect) solution:
*
*$\sin^2x$ + $\sin^22x$ + $\sin^23x$ = $\sin^26x$ = 1
*$\sin^2x = 1/6$
*$\sin x = 1/\sqrt{6}$
*$\sin x =$ opposite/hypotenuse
*Therefore, opp = 1, hyp = $\sqrt6$, adj = $\sqrt5$ (pythagoras theorum)
*$\cos x$ = adjacent/hypotenuse = $\sqrt5/\sqrt6$
*$\cos^2x$ = 5/6
*$\cos^2x + \cos^22x + \cos^23x = \cos^26x = 5/6 * 6 = 5$
I think I did something incorrectly right off the bat at step-2, and am hoping somebody will point me in the right direction.
| You know that
\begin{equation}
\textrm{sin}^2x + \textrm{sin}^2 2x + \textrm{sin}^2 3x = 1 \qquad (\textrm{Eq.} 1)
\end{equation}
Notice that
\begin{equation}
\begin{aligned}
\textrm{sin}^2x + \textrm{cos}^2 x = 1 \Rightarrow \textrm{sin}^2x = 1 - \textrm{cos}^2 x
\end{aligned}
\end{equation}
Substituting the identity in Equation (1):
\begin{equation}
\begin{aligned}
\underbrace{\textrm{sin}^2x}_{1-\textrm{cos}^2x} & + \underbrace{\textrm{sin}^2 2x}_{1-\textrm{cos}^2 2x } + \underbrace{\textrm{sin}^2 3x}_{1-\textrm{cos}^2 3x} = 1 \\ \\ & \Rightarrow 1-\textrm{cos}^2x + 1-\textrm{cos}^2 2x + 1-\textrm{cos}^2 3x = 1 \\ & \Rightarrow 3 -\left( \textrm{cos}^2x + \textrm{cos}^2 2x + \textrm{cos}^2 3x \right) = 1 \\ & \Rightarrow 2 = \textrm{cos}^2x + \textrm{cos}^2 2x + \textrm{cos}^2 3x
\end{aligned}
\end{equation}
Therefore:
\begin{equation}
\textrm{cos}^2x + \textrm{cos}^2 2x + \textrm{cos}^2 3x = 2
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1545182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to eliminate $\theta$? While doing a sum I was stuck in a particular step:
$$r_1= \frac{4a \cos \theta }{\sin^2 \theta}$$ and $$r_2=\frac{4a \sin \theta }{\cos^2 \theta}$$
How to eliminate $\theta$ ?
| $$r_1\cdot r_2 = \frac{(4a)^2}{\sin \theta \cos \theta}$$
$$\left(\frac{r_2}{r_1}\right)^{\frac{1}{3}}=\frac{\sin \theta}{\cos \theta}$$
$$\frac{\left(\frac{r_2}{r_1}\right)^{\frac{1}{3}}}{r_1\cdot r_2 }=\frac{\sin^2 \theta}{(4a)^2}=r_1^{-\frac{4}{3}}r_1^{-\frac{2}{3}}$$
$$\frac{1}{\left(\frac{r_2}{r_1}\right)^{\frac{1}{3}}r_1\cdot r_2 }=\frac{\cos^2 \theta}{(4a)^2}=r_1^{-\frac{2}{3}}r_1^{-\frac{4}{3}}$$
$$\therefore r_1^{-\frac{4}{3}}r_2^{-\frac{2}{3}}+r_1^{-\frac{2}{3}}r_2^{-\frac{4}{3}}=\frac{1}{(4a)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1546217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Prove that $ \sqrt{\frac{a}{b+3} } +\sqrt{\frac{b}{c+3} } +\sqrt{\frac{c}{a+3} } \leq \frac{3}{2} $ For all $a, b, c>0$ and $a+b+c=3$ Prove that $$ \sqrt{\frac{a}{b+3} } +\sqrt{\frac{b}{c+3} } +\sqrt{\frac{c}{a+3} } \leq \frac{3}{2} $$
I tried cauchy-schwarz inequality for the L. H. S like and I get
$ [\left( \sqrt{a} \right) ^{2}+\left( \sqrt{b} \right) ^{2}+\left( \sqrt{c} \right) ^{2}]\left[ \left( \frac{1}{\sqrt{b+3} } \right) ^{2}+\left( \frac{1}{\sqrt{c+3} } \right) ^{2}+\left( \frac{1}{\sqrt{a+3} } \right) ^{2}\right] \geq \left( \frac{\sqrt{a} }{\sqrt{b+3} } +\frac{\sqrt{b} }{\sqrt{c+3} } +\frac{\sqrt{c} }{\sqrt{a+3} } \right) ^{2}$
Then I get by AM-GM the maximum value of $abc=1$ and the inequality have the value 3... How I can prove inequality?.
| First, we use Cauchy-Schwartz inequality:
$$
A^2=\left(\sqrt{\frac{a}{b+3} } +\sqrt{\frac{b}{c+3} } +\sqrt{\frac{c}{a+3} }\right)^2 =\\
\left(\sqrt{\frac{a}{(a+3)(b+3)}(a+3) } +\sqrt{\frac{b}{(b+3)(c+3)} (b+3)} +\sqrt{\frac{c}{(c+3)(a+3)}(c+3) }\right)^2 \leq \\
\left({\frac{a}{(a+3)(b+3)} } +{\frac{b}{(b+3)(c+3)} } +{\frac{c}{(c+3)(a+3)} }\right)\times (a+3+b+3+c+3) .
$$
Since $a+3+b+3+c+3=12$, it is enough to prove:
$$
\left({\frac{a}{(a+3)(b+3)} } +{\frac{b}{(b+3)(c+3)} } +{\frac{c}{(c+3)(a+3)} }\right)\leq \frac 3{16}. (\star)
$$
From this step on, I went the ugly way, since the manipulation was not so frightening. With a simple and nice proof of this step, the whole proof would become much easier.
Anyway, after the simple multiplications, we get to the following inequality:
$$
18-7(ab+ac+bc)+3abc\geq 0.
$$
This can be further simplified, using the fact such as $a^2+b^2+c^2=9-2(ab+ac+bc)$ and using the famous following identity:
$$
a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).
$$
We can further simplify the inequality to the following:
$$
a^3+b^3+c^3\geq a^2+b^2+c^2.
$$
This follows from the following steps:
*
*Cauchy-Schwartz: $(a+b+c)(a^3+b^3+c^3)\geq (a^2+b^2+c^2)^2 \implies 3(a^3+b^3+c^3)\geq (a^2+b^2+c^2)^2$
*Cauchy-Schwartz: $(1+1+1)(a^2+b^2+c^2)\geq (a+b+c)^2 \implies (a^2+b^2+c^2)\geq 3.$
*Combine two previous steps:
$$
3(a^3+b^3+c^3)\geq (a^2+b^2+c^2)^2\geq 3(a^2+b^2+c^2).
$$
And therefore, this proves $(\star)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1548990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Min and max of a two variables function
I consider the function
$$f:\mathbb{R}^2 \to \mathbb{R}, \
(x,y) \mapsto \max\left(1-\sqrt{x^2+y^2},2-\sqrt{(x-6)^2+y^2},0\right).$$
I have to find $\max$ and $\min$ (local and global).
I calculate where $1-\sqrt{x^2+y^2}=2-\sqrt{(x-6)^2+y^2}$ is an ellipse but I don't understand how find $\max$ and $\min$.
| Note that $1 - \sqrt{x^2 + y^2} > 0$ inside a circle of radius $1$ about the origin, and is $< 0$ outside that circle. Meanwhile $2-\sqrt{(x-6)^2+y^2} > 0$ inside a circle of radius $2$ about the point $(6, 0)$ and is $< 0$ outside. Since the centers of these circles are a distance of $6$ apart while the sum of the radii is only $3$. So these circles are disjoint.
Ergo, outside both circles and on the circles themselves, $0$ is the maximum. Inside the circle about $(0,0), 1 - \sqrt{x^2 + y^2}$ is the maximum. Inside the circle about $(6, 0), 2-\sqrt{(x-6)^2+y^2}$ is the maximum.
The minimum problem is more difficult. Since for any point one of the two functions is negative, $0$ is never the minimum. Now connect any point where $1 - \sqrt{x^2 + y^2} <2-\sqrt{(x-6)^2+y^2}$ to any point where $1 - \sqrt{x^2 + y^2} >2-\sqrt{(x-6)^2+y^2}$ with a line segment, and by the continuity of the two functions, somewhere along that line segment the two functions are equal. Thus if we identify where the two functions are equal, that curve will divide the plane into regions where either one or the other is less.
$$1 - \sqrt{x^2 + y^2} = 2-\sqrt{(x-6)^2+y^2}\\
\sqrt{(x-6)^2+y^2} = 1 + \sqrt{x^2 + y^2}\\
x^2-12x + 36+y^2 = 1 + x^2 + y^2 + 2\sqrt{x^2 + y^2}\\
35 - 12x = 2\sqrt{x^2 + y^2}\\
1225 - 840x + 144x^2 = 4x^2 + 4y^2\\
140(x^2 - 6x + 9) - 4y^2 = (9)(140) - 1225\\
140(x - 3)^2 - 4y^2 = 35$$
So the two functions are equal on a hyperbola. But note that the calculation isn't reversable. So not every point on the hyperbola may be a place the two functions are equal. In particular, when $y = 0$, we get that $|x - 3| = 1/2$
Now $$1 - \sqrt{2.5^2 + 0^2} = -1.5\\2-\sqrt{(2.5-6)^2+0^2} = -1.5$$ but $$1 - \sqrt{3.5^2 + 0^2} = -2.5\\2-\sqrt{(3.5-6)^2+0^2} = -0.5$$
So the two functions are equal only on the arc of the hyperbola through $(2.5, 0)$. Namely when $$x = 3 - \sqrt\frac{35 - 4y^2}{140}$$
Ergo, when $x < \sqrt\frac{35 - 4y^2}{140}, 2-\sqrt{(x-6)^2+y^2}$ is minimum, when $x > \sqrt\frac{35 - 4y^2}{140}, 1 - \sqrt{x^2 + y^2}$ is minimum, and when $x = 3 - \sqrt\frac{35 - 4y^2}{140}$, the two are equal, so either one could be taken as minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that the curve has two tangents I'm a little stuck on a math problem that reads as follows:
Show that the curve $x = 5\cos(t), y = 3\sin(t)\, \cos(t)$ has two tangents at $(0, 0)$ and find their equations
What I've Tried
*
*$ \frac{dx}{dt} = -5\sin(t) $
*$ \frac{dy}{dt} = 3\cos^2(t) - 3\sin^2(t) $ because of the product rule. We can simplify this to $ 3(\cos^2(t) - \sin^2(t)) \rightarrow 3\cos(2t) $
*In order to get the slope $ m $, we can write $$ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} $$
*Solving for $ \frac{dy}{dx} $ as follows:
*
*$ \frac{3\cos(2t)}{-5\sin(t)} $ can be rewritten as
*$ \frac{-3}{5}(\cos(2t)\csc(t)) = m $
*Plugging $ (0,0) $ back into the equations of $ x $ and $ y $ we have as follows:
*
*$ 5\cos(t) = 0 \rightarrow t = \frac{\pi}{2} $
*
*Note: I'm unsure what happens to the $ 5 $
*$ \frac{dx}{dt} $ evaluated at $ t = \frac{\pi}{2} $ gives us $ -5\sin(\frac{\pi}{2}) = -5 $
*$ \frac{dy}{dt} $ evaluated at $ t = \frac{\pi}{2} $ gives us $ 3\cos(\frac{2\pi}{2}) \rightarrow 3\cos(\pi) = -3 $
*$ \frac{dy}{dx} = \frac{3}{5} $
*Continuing on, if we add $ \pi $ to the value of $ t $ we get $ t = \frac{3\pi}{2} $. Plug the new value of $ t $ into the equations of $ x $ and $ y $
*
*$ \frac{dx}{dt} $ evaluated at $ t = \frac{3\pi}{2} $ gives us $ -5sin(\frac{3\pi}{2}) = 5 $
*$ \frac{dy}{dt} $ evaluated at $ t = \frac{3\pi}{2} $ gives us $ 3cos(\frac{6\pi}{2}) \rightarrow 3cos(3\pi) = -3 $
*$ \frac{dy}{dx} = -\frac{3}{5} $
*We now have our two slopes of the tangent lines:
*
*$ y = -\frac{5}{3}x $
*$ y = \frac{5}{3}x $
The issue is that webassign is claiming that the slopes are wrong as can be seen here:
Here is the solution in graph form that is correct:
p.s. My apologies if this is a repost. I've seen this response Show that the curve x = 5 cos t, y = 4 sin t cos t has two tangents at (0, 0) and find their equations. and followed it already with no avail.
| The user kccu already found your mistake. Here is a shorter way to do the question.
The curve is $(5\cos t, 3\sin t\,\cos t)$. With $0\leq t<2\pi$, it goes through the origin two times: when $t=\pi/2$ and when $t=3\pi/2$. Then derivative is, as you say,
$$
(-5\sin t, 3\cos 2t).
$$
At the two values of $t=\pi/2$ and $t=3\pi/2$ we get
$$
(-5,-3),\ \ \text{ and }\ \ (5,-3).
$$
The lines are then $(x,y)=t(-5,-3)$ and $(x,y)=t(5,-3)$. That is
$$
y=-\frac35\,x \ \ \ \text{ and } y=\frac35\,x.
$$
So the slopes are $3/5$ and $-3/5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Integrate $\int \frac{\arctan\sqrt{\frac{x}{2}}dx}{\sqrt{x+2}}$ $$\int \frac{\arctan\sqrt{\frac{x}{2}} \, dx}{\sqrt{x+2}}$$
I've tried substituting $x=2\tan^2y$, and I've got:
$$\frac{1}{\sqrt2}\int\frac{y\sin y}{\cos^4 y} \, dy$$
But I'm not entirely sure this is a good thing as I've been unable to proceed any further from there.
| Notice, $$\int \frac{\tan^{-1}\sqrt{\frac{x}{2}}}{\sqrt{x+2}}\ dx=\int \frac{\tan^{-1}\sqrt{\frac{x}{2}}}{\sqrt 2\sqrt{\frac{x}{2}+1}}\ dx$$
now, let $\frac{x}{2}=\tan^2\theta\implies dx=2\tan\theta\sec^2\theta \ d\theta$, $0\le \theta\le \pi/2$
$$=\frac{1}{\sqrt2}\int \frac{\tan^{-1}\left(\tan\theta\right)}{\sqrt{\tan^2\theta+1}}(2\tan\theta\sec^2\theta \ d\theta)$$
$$=\frac{2}{\sqrt2}\int \frac{\theta\sec^2\theta\tan\theta}{\sec\theta}\ d\theta$$
$$=\sqrt 2\int \theta(\sec\theta\tan\theta)\ d\theta$$
$$=\sqrt 2\left(\theta\sec\theta-\int \sec\theta \ d\theta\right)$$
$$=\sqrt 2\left(\theta\sec\theta-\ln\left|\sec\theta+\tan\theta\right|\right)+C$$
$$=2\sqrt{x+2} \tan^{-1} \sqrt{\frac x 2} - 4\sqrt{x+2} + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1550405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Prove that functional equation doesn't have range $\Bbb R.$ Prove that any solution $f: \mathbb{R} \to \mathbb{R}$ of the functional equation
$$ f(x + 1)f(x) + f(x + 1) + 1 = 0 $$
cannot have range $\mathbb{R}$.
I transformed it into
$$ f(x) = \frac {-1} {f(x + 1)} - 1 = \frac {-1 - f(x + 1)} {f(x + 1)} $$
I tried to evaluate $f(x + 1)$ and $f(x + 2)$ and put them into the transformed equation:
1) after $f(x + 1)$
$$ f(x) = \frac {1} {-1 - f(x + 2)}$$
2) after $f(x + 2)$
$$ f(x) = \frac {1} {\frac {-1}{f(x + 3)}} = -f(x + 3) $$
What am I supposed to do next?
| You should always be careful when dividing, because you'll run into trouble if you try and divide by 0.
We have that for all $x$, $$f(x+1)(f(x) + 1) = -1$$
Notice that $f$ can never be zero, then, because that would mean $0$ on the left-hand side was $-1$ on the right-hand side. This justifies your division step, and proves that $0$ is never in the range of $f$; that is all you need.
As an appendix, we prove that there are, in fact, solutions to the functional equation.
If we fix $x=0$, obtain what is effectively the recurrence $$x_{n+1} = \frac{-1}{x_n + 1}$$
where $x_n = f(n)$. This certainly does have solutions (for example, if $x_0 = 1$, then it cycles $1, -\frac{1}{2}, -2, 1, -\frac{1}{2}, -2, \dots$), so the recurrence does have a solution over the reals.
One such solution is therefore $$f(x) = \begin{cases}
1 & \lfloor x \rfloor \equiv 0 \pmod{3} \\
-\frac{1}{2} & \lfloor x \rfloor \equiv 1 \pmod{3} \\
-2 & \lfloor x \rfloor \equiv 2 \pmod{3}
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1551600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Suppose $n$ is prime and $x \in Z$ satisfies $x^2 \equiv 1 \mod n.$ Prove that $x \equiv 1 \mod n$ or $x \equiv -1 \mod n$. Suppose $n$ is prime and $x \in Z$ satisfies $x^2 \equiv 1 \mod n.$ Prove that $x \equiv 1 \mod n$ or $x \equiv -1 \mod n$.
So far I have done the following proof, but I am unsure how to complete it:
We will begin by explaining why $x^2 \equiv 1 \mod n$ iff $n|(x^2 - 1)$. Then use this to prove that $x \equiv 1 \mod n$ or $x \equiv -1 \mod n$.
If $x^2 \equiv 1 \mbox{ mod } n$, then $x^2 = nk +1 $, for some $k \in Z$. Therefore $nk = x^2 -1$, and $n|(x^2 -1)$.
On the other hand, if $n|(x^2 -1)$, then $nk = x^2 -1$ for some $k \in Z$, and then we can say $x^2 = nk +1$, and thus $x^2 \equiv 1 \mod n$.
From here, however I am not sure how to demonstrate that this shows $x \equiv 1 \mod n$ or $x \equiv -1 \mod n$.
| An essential property of primes is the following:
If $p \mid ab$ then $p \mid a$ or $p \mid b$.
In your case, $n \mid (x+1)(x-1)$, so $n \mid (x+1)$ or $n \mid (x-1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1551929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int \frac{x+1}{(x^2-x+8)^3}\, dx$ Could you give me a hint on how to find $$\int \frac{x+1}{(x^2-x+8)^3}\, dx$$
It doesn't seem like partial fractions are the way to go with here and using the integration by parts method seems to be tedious.
I have also tried substituting $(x^2-x+8)$ but it gets even more complicated then.
Is there a way to solve this using only basic formulas?
| $$\int \frac{x+1}{\left(x^2-x+8\right)^3}\, dx=\frac{1}{2}\left(\int \frac{d\left(x^2-x+8\right)}{\left(x^2-x+8\right)^3}+\int\frac{3}{\left(x^2-x+8\right)^3}\, dx\right)$$
$$=\frac{1}{2}\left(\frac{1}{-2\left(x^2-x+8\right)^2}+96\int \frac{d(2x-1)}{\left((2x-1)^2+31\right)^3}\right)$$
Let $2x-1=\sqrt{31}\tan t$. Then $d(2x-1)=\frac{\sqrt{31}}{\cos^2 t}\, dt$.
$$\int \frac{d(2x-1)}{\left((2x-1)^2+31\right)^3}=\frac{1}{961\sqrt{31}}\int \cos^4 t\, dt$$
$$=\frac{1}{961\sqrt{31}}\int \left(\frac{1+\cos (2t)}{2}\right)^2 \, dt$$
$$=\frac{1}{3844\sqrt{31}}\left(\int dt+\int \cos(2t)\, d(2t)+\int \frac{1+\cos(4t)}{2}\, dt\right)$$
$$=\frac{1}{3844\sqrt{31}}\left(t+\sin(2t)+\frac{t}{2}+\frac{\sin(4t)}{8}\right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1553924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find all values of $n$ for which the Diophantine equation $n=a^2-b^2$ has a solution Let $n$ be an integer. Find all values of $n$ for which the Diophantine equation $n=a^2-b^2$ has a solution for integers $a$ and $b$. For those values of $n$ found in the previous part find all solutions of $n=a^2-b^2$ for integers $a$ and $b$.
The first part is pretty easy. It is just if $n$ is the product of at least $2$ factors of $2$ or $n$ is the product of solely odd factors.
For the second part, I used this argument. If $n$ is the product of at least two factors of $2$, then let $a+b$ be any factor of $n$ with at least one power of $2$ but less than or equal to one minus the maximum power of $2$ dividing $n$. The value of $a-b$ is just the other part of $n$. On the other hand, if $n$ is the product of solely odd numbers, let $a+b$ be an odd factor of $n$. Then the value of $a-b$ follows and this yields all integer solutions $a, b$.
Is it necessary to formalize my argument into algebra and number theory or is how I put it fine?
| The equation $$n=a^2-b^2=(a-b)(a+b)$$ is solveable over the integers if and only if there are numbers $u,v$ with $uv=n$, such that the system $$a-b=u$$ $$a+b=v$$
is solveable over the integers.
The system is solveable if and only if $u$ and $v$ have the same parity. We can find suitable $u$ and $v$, if and only if $n\ne 2\ (\ mod\ 4\ )$. If $n$ is odd, we can simply choose $u=1$ , $v=n$. If $n$ is divisble by $4$, we can choose two even numbers $u,v$ with $uv=n$. In the case $n\equiv 2\ (\ mod\ 4\ )$, one of the factors will be odd and the other even.
So, $n=a^2-b^2$ is solveable over the integers, if and only if $n\ne 2\ (\ mod\ 4\ )$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Volume bounded by elliptic paraboloids Find the volume bounded by the elliptic paraboloids given by $z=x^2 + 9 y^2$ and $z= 18- x^2 - 9 y^2$.
First I found the intersection region, then I got $x^2+ 9 y^2 =1$. I think this will be area of integration now what will be the integrand. Please help me.
| You probably got something looking like this
where the required volume is the dark volume inside the two parabolas. Of course there are many ways to find that volume; I'll stick to the slice method: essentially you fix the third variable $z$ and calculate the area of desired region intersecting the plane $z = z_0$, then you sum up all this areas. Note that if you intersect the region with a plane parallel to the $XY$ plane you get an ellipse, with major semi-axis $\sqrt{\frac{18-z}{9}}$ and minor semi-axis $\sqrt{18-z}$ if $z \ge 9$ (that is when the parabolas intersect each other), and major semi-axis $\sqrt{z}$ and minor semi-axis $\sqrt{\frac{z}{9}}$ if $z \ge 9$ when $z \ge 9$. So the area of the intersection between the region and the plane at height $z$ is
$$
\begin{cases}
\sqrt{\frac{18-z}{9}} \cdot \sqrt{18-z} \cdot \pi = \frac{18 - z}{3} \pi \iff z \ge 9\\
\sqrt{\frac{z}{9}} \cdot \sqrt{z} \cdot \pi= \frac{z}{3} pi \iff z \le 9
\end{cases}
$$
So the volume desired is
$$\int \limits_0 ^{9} \frac{z}{3} \pi dz + \int \limits_9 ^{18} \frac{18 - z}{3} \pi dz = \pi \left(\frac13 \int \limits_0 ^9 z dz + \int \limits_9 ^{18} 6 dz - \frac13 \int \limits_9 ^{18} zdz \right) = \pi \left(\frac13 \frac{z^2}{2} \big|_0 ^9 + 6 \big|_9 ^{18} - \frac13 \frac{z^2}{2} \big|_9 ^{18} \right) = 27\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Zero divided by zero must be equal to zero What is wrong with the following argument (if you don't involve ring theory)?
Proposition 1: $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}$ is not equal to $0$
$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac{0}{0}) = 2x$ $\Rightarrow$ $\frac{2\cdot 0}{0} = 2x$ $\Rightarrow$ $\frac{0}{0} = 2x$ $\Rightarrow$ $x = 2x$ $\Rightarrow$ $ x = 0$ $\Rightarrow$[because $x$ is not equal to $0$]$\Rightarrow$ contradiction
Therefore, it is not the case that $\frac{0}{0}$ is not equal to $0$
Therefore, $\frac{0}{0} = 0$.
Q.E.D.
Update (2015-12-01) after your answers:
Proposition 2: $\frac{0}{0}$ is not a real number
Proof [Update (2015-12-07): Part 1 of this argument is not valid, as pointed out in the comments below]:
Suppose that $\frac{0}{0}= x$, where $x$ is a real number.
Then, either $x = 0$ or $x$ is not equal to $0$.
1) Suppose $x = 0$, that is $\frac{0}{0} = 0$
Then, $1 = 0 + 1 = \frac{0}{0} + \frac{1}{1} = \frac{0 \cdot 1}{0 \cdot 1} + \frac{1 \cdot 0}{1 \cdot 0} = \frac{0 \cdot 1 + 1 \cdot 0}{0 \cdot 1} = \frac{0 + 0}{0} = \frac{0}{0} = 0 $
Contradiction
Therefore, it is not the case that $x = 0$.
2) Suppose that $x$ is not equal to $0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow$ contradiction
Therefore, it is not the case that $x$ is a real number that is not equal to $0$.
Therefore, $\frac{0}{0}$ is not a real number.
Q.E.D.
Update (2015-12-02)
If you accept the (almost) usual definition, that for all real numbers $a$, $b$ and $c$, we have $\frac{a}{b}=c$ iff $ a=cb $, then I think the following should be enough to exclude $\frac{0}{0}$ from the real numbers.
Proposition 3: $\frac{0}{0}$ is not a real number
Proof: Suppose that $\frac{0}{0} = x$, where $x$ is a real number.
$\frac{0}{0}=x \Leftrightarrow x \cdot 0 = 0 = (x + 1) \cdot 0 \Leftrightarrow \frac{0}{0}=x+1$
$ \therefore x = x + 1 \Leftrightarrow 0 = 1 \Leftrightarrow \bot$
Q.E.D.
Update (2015-12-07):
How about the following improvement of Proposition 1 (it should be combined with a new definition of division and fraction, accounting for the $\frac{0}{0}$-case)?
Proposition 4: Suppose $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, and that the rule $a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$ holds for all real numbers $a$, $b$ and $c$.
Then, $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}=x$, where $x \ne 0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
Suggested definition of division of real numbers:
If $b \ne 0$, then
$\frac{a}{b}=c$ iff $a=bc$
If $a=0$ and $b=0$, then
$\frac{a}{b}=0$
If $a \ne 0$ and $b=0$, then $\frac{a}{b}$ is undefined.
A somewhat more minimalistic version:
Proposition 5. If $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, then $\frac{0}{0}=0$.
Proof: Suppose $\frac{0}{0} \in \mathbb{R}$ and that $\frac{0}{0}=a \ne 0$.
$a = \frac{0}{0} = \frac{2 \cdot 0}{0} = 2a \Rightarrow a = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
| Well, then $n \cdot 0=0\; \forall \mathbb{R} \implies \frac{0}{0}=n=\frac{1}{n}$ so, $\frac{0}{0}$ is pretty much any real number. This is definitely not the case. This is why it is undefined.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 16,
"answer_id": 14
} |
How to show that $f(x) = \frac{\sqrt{\cos x}}{1-x^2}$ is convex and has a minimum value I want to show that $f(x) = \frac{\sqrt{\cos x}}{1-x^2}$ is convex on the interval $]-1,1[$. How do I have to proceed?
I did take the derivative of the function which is
$f'(x) = \frac{2x\sqrt{\cos x }}{(1-x^2)^2}-\frac{\sin x}{2\sqrt{\cos x}(1-x^2)}$
For a function to be convex I need to show that it is increasing on the interval?
I'm a little stuck to proof that rigorously. Thanks for your help!
| Let $f$ be given by
$$f(x)=\frac{\sqrt{\cos x}}{1-x^2}$$
for $|x|<1$. Then, the first derivative of $f$ is
$$f'(x)=\frac{\sqrt{\cos x}}{2(1-x^2)}\left(\frac{4x}{1-x^2}-\tan x\right) \tag 1$$
From $(1)$ we can write
$$\frac{f'(x)}{f(x)}=\frac12\left(\frac{4x}{1-x^2}-\tan x\right) \tag 2$$
whereupon differentiating both sides of $(2)$ and solving for $f''(x)$ reveals that
$$f''(x)=\frac{\left(f'(x)\right)^2}{f(x)}+f(x)\left(\frac{4(1+x^2)}{(1-x^2)^2}-\sec^2 x\right)$$
It is evident that $f(x)>0$ and $\left(f'(x)\right)^2\ge 0$ for $|x|<1$. We will now show that the term $\left(\frac{4(1+x^2)}{(1-x^2)^2}-\sec^2 x\right)$ is positive thereby establishing that $f''(x)> 0$.
Recall that the sine function satisfies the inequality
$$|\sin x|\le |x| \tag 2$$
From $(2)$ it is easy to show that for $|x|\le 1$
$$\cos x\ge \sqrt{1-x^2} \tag 3$$
Using $(3)$ reveals
$$\begin{align}
\left(\frac{4(1+x^2)}{(1-x^2)^2}-\sec^2 x\right) &\ge \left(\frac{4(1+x^2)}{(1-x^2)^2}-\frac{1}{1-x^2}\right)\\\\
&=\frac{3+5x^2}{(1-x^2)^2}\\\\
&>0
\end{align}$$
Since $f''(x)>0$, $f$ is convex on $-1<x<1$.
Since $f$ is even and convex, it attains it minimum at $x=0$ where we have $f(0)=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Help me evaluate this infinite sum I have the following problem:
For any positive integer n, let $\langle n \rangle$ denote the integer nearest to $\sqrt n$.
(a) Given a positive integer $k$, describe all positive integers $n$ such that $\langle n \rangle = k$.
(b) Show that $$\sum_{n=1}^\infty{\frac{2^{\langle n \rangle}+2^{-\langle n \rangle}}{2^n}}=3$$
My progress: The first one is rather easy. As $$\left( k-\frac{1}{2} \right) < \sqrt n < \left( k+\frac{1}{2} \right) \implies \left( k-\frac{1}{2} \right)^2 < n < \left( k+\frac{1}{2} \right)^2 \implies \left( k^2-k+1 \right) \leq n \leq \left( k^2+k \right)$$
Actually, there would be $2k$ such integers.
But, I have no idea how to approach the second problem. Please give me some hints.
| Direct computation seems to work. I recall this problem being an old Putnam problem.
We have $\langle n \rangle = k$, iff $n \in [k^2 - k + 1, k^2 + k]$.
Now, let's compute this sum.
$$
\sum_{n = 1}^\infty \frac{2^{\langle n \rangle} + 2^{-\langle n \rangle}}{2^n}
= \sum_{k=1}^\infty \sum_{i=k^2 - k + 1}^{k^2 + k} \frac{2^k + 2^{-k}}{2^i}
= \sum_{k=1}^\infty \frac{2^{2k}+1}{2^k}\sum_{i=k^2 - k + 1}^{k^2 + k} \frac1{2^i}
= \sum_{k=1}^\infty \frac{2^{2k}+1}{2^{k^2+1}}\sum_{i=0}^{2k-1} \frac{1}{2^i}
= \sum_{k=1}^\infty \frac{2^{2k}+1}{2^{k^2 + 1}}\frac{1 - \frac{1}{4^k}}{1-\frac{1}{2}}
= \sum_{k=1}^\infty \frac{2^{2k}+1}{2^{k^2 + 1}}\frac{2^{2k}-1}{2^{2k-1}}
= \sum_{k=1}^\infty \frac{2^{4k}-1}{2^{k^2 + 2k}}
= \sum_{k=1}^\infty \left(2^{1-(k-1)^2} - 2^{1-(k+1)^2}\right)
= 2^1 + 2^0 = 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
A question on summation of series...
If $$\frac{\left(1^4+\frac14\right)\left(3^4+\frac14\right)\ldots\left((2n-1)^4+\frac14\right)}{\left(2^4+\frac14\right)\left(4^4+\frac14\right)\ldots\left((2n)^4+\frac14\right)}=\frac1{k_1n^2+k_2n+k_3}$$ then $k_1-k_2+k_3$ is equal to . . . ?
What I basically tried was simply putting the value of $n$ from $1$ to $3$. By doing so I got $3$ simultaneous equations and hence I calculated the desired answer.
What I want to know is that is there any other elegant way to solve the above problem because my process is very time consuming ..looking forward to here from my friends thanks in advance......
| Use the Sophie-Germain identity: $$a^4+4b^4=(a^2+2b^2)^2-(2ab)^2=(a^2-2ab+2b^2)(a^2+2ab+2b^2)$$Using this, we can factor an expression of the form $m^4+\frac{1}{4}$ into $$m^4+4\bigg(\frac{1}{2}\bigg)^4=(m^2-m+\frac{1}{2})(m^2+m+\frac{1}{2})$$Therefore, the LHS of your equation is equal to $$\frac{\displaystyle\prod_{i=1}^{n}\bigg((2i-1)^2-(2i-1)+\frac{1}{2}\bigg)\bigg((2i-1)^2+(2i-1)+\frac{1}{2}\bigg)}{\displaystyle\prod_{i=1}^{n}\bigg((2i)^2-(2i)+\frac{1}{2}\bigg)\bigg((2i)^2+(2i)+\frac{1}{2}\bigg)}$$Note that $$(2i-1)^2+(2i-1)+\dfrac{1}{2}=(2i)^2-4i+1+2i-1+\dfrac{1}{2}=(2i)^2-(2i)+\dfrac{1}{2}$$and $$(2i-2)^2+(2i-2)+\frac{1}{2}=(2i)^2-8i+4+2i-2+\frac{1}{2}=(2i-1)^2-(2i-1)+\frac{1}{2}$$So after cancelling, our expression above is just $$\frac{(2-1)^2-(2-1)+\frac{1}{2}}{(2n)^2+(2n)+\frac{1}{2}}=\frac{1}{8n^2+4n+1}$$and the answer is $8-4+1=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Easy way of memorizing or quickly deriving summation formulas My math professor recently told us that she wants us to be familiar with summation notation. She says we have to have it mastered because we are starting integration next week. She gave us a bunch of formulas to memorize. I know I can simply memorize the list, but I am wondering if there is a quick intuitive way of deriving them on the fly. It has to be a really quick derivation because all of her test are timed.
Otherwise is there an easy way, you guys remember these formulas.
$\begin{align}
\displaystyle
&\sum_{k=1}^n k=\frac{n(n+1)}{2} \\
&\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6} \\
&\sum_{k=1}^n k^3=\frac{n^2(n+1)^2}{4} \\
&\sum_{k=1}^n k(k+1)=\frac{n(n+1)(n+2)}{3} \\
&\sum_{k=1}^n \frac{1}{k(k+1)}=\frac{n}{n+1} \\
&\sum_{k=1}^n k(k+1)(k+2)=\frac{n(n+1)(n+2)(n+3)}{4} \\
&\sum_{k=1}^n \frac{1}{k(k+1)(k+2)}=\frac{n(n+3)}{4(n+1)(n+2)} \\
&\sum_{k=1}^n (2k-1)=n^2
\end{align}$
Note: Sorry if there is an easy and obvious answer to this question. Most of the students in my class already know these formulas from high school, but I only went up to Algebra 2 and Trig when I was in high school. PS: This is for a calculus class in college.
| Perhaps reorganizing them this way :
Class 1 : clear pattern (without end in fact!)
\begin{align}
\sum_{k=1}^n k&=\frac{n(n+1)}2\\
\sum_{k=1}^n k(k+1)&=\frac{n(n+1)(n+2)}3\\
\sum_{k=1}^n k(k+1)(k+2)&=\frac{n(n+1)(n+2)(n+3)}4\\
&\cdots\\
\end{align}
$$\boxed{\displaystyle\sum_{k=1}^n k^{(p)}=\frac{n^{(p+1)}}{p+1}}$$
(the rising factorial $\;k^{(p)}:=k(k+1)(k+2)\cdots(k+p-1)\;$ and the analogy between sum and integration should help to remember all these and... make nice mathematics!)
Class 2 : using simple relations as $\,\sum k^2=\sum k(k+1)-\sum k\;$ and previous results...
\begin{align}
\sum_{k=1}^n k&=\frac{n(n+1)}2\\
\sum_{k=1}^n k^2&=\frac{n(n+\frac 12)(n+\frac 22)}3\\
\\
\sum_{k=1}^n k^3&=\left(\sum k\right)^2\\
\end{align}
Class 3 : reciprocal rising factorial :
\begin{align}
\sum_{k=1}^n \frac 1{k(k+1)}&=\frac 1{1}-\frac 1{n+1}\qquad \text{}\\
\sum_{k=1}^n \frac 2{k(k+1)(k+2)}&=\frac 1{1\cdot 2}-\frac 1{(n+1)(n+2)}\\
&\cdots\\
\end{align}
$$\boxed{\displaystyle\sum_{k=1}^n \frac p{k^{(p+1)}}=\frac 1{1^{(p)}}-\frac 1{(n+1)^{(p)}}}\qquad 1^{(p)}=p!$$
Now remember that $\quad \displaystyle\int_{k=1}^{n+1} \frac p{k^{p+1}}\,dk=\frac 1{1^p}-\frac 1{(n+1)^p}$
(our upper-bound had to be changed from $n\to n+1$ while for class $1$ the lower-bound $1\to 0$)
Concerning the last one (sum of odds) remember this picture :
$\qquad\displaystyle\sum_{k=1}^n (2k-1) =n^2\quad $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1561570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 0
} |
Family of Circles A system of Circles pass through $(2,3)$ and have their centers on the line $x+2y-7=0$.
Show that the chords in which the circles of the given system intersects the circle $S_1:x^2+y^2-8x+6y-9=0$ are concurrent and also find the point of concurrency.
ATTEMPT:
The circles of the given system must intersect the line $x+2y-7=0.$
Therefore the equation of family of circles passing through $(2,3)$ and $(7-2k,k)$ , for any arbitrary $k$ can be written as $S$:
$(x-2)(x-7+2k) +(y-3)(y-k) +c $$\begin{vmatrix}
1 & x & y \\
1 & 7-2k & k \\
1 & 2 & 3 \\
\end{vmatrix}
$$ =0$ ,where $c$ is any arbitrary constant.
Now using the condition that the center lies on the line $x+2y-7=0$ we can get a relation in $k$ and $c.$
And the equation of common chords can be given by $S-S_1=0.$
But how to prove that the chords are concurrent?
| The center of the system of circle lies on $x+2y-7=0$, if the center is $(h,k)$ then $h+2k-7=0$
And as the circles passes through $(2,3)$, therefore the radius is $\sqrt{(h-2)^2+(k-3)^2}$
Therefore, the equation of the circle is
$(x-h)^2+(y-k)^2=(h-2)^2+(k-3)^2$ i.e., $x^2+y^2-2xh-2yh+4h+6k+13=0$ we call this $S$
And the fixed circle is $x^2+y^2-8x+6y-9=0$----$S-1$
The common chord of $S$ and $S_1$ is $S-S_1=0$
i.e., $-8x+6y-9+2xh+2yk-4h-6k-13=0$
or, $-8x+6y-9+2xh+(7-h)y-4h-3(7-h)-13=0$
or, $-8x+6y+7y-9-21-13+h(2x-y-4-3)$
or, $-8x+13y-43+h(2x-y-7)=0$
This is a system of lines which always passes through through the intersection of $-8x+13y-43=0$ and $(2x-y-7)=0$
Therefore, the common chords are concurrent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1561845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $a^2+b^2\geq \frac{c^2}{2}$ and friends if $a+b\geq c\geq0$ Sorry for my inequality spam, but I got to prepare for my exams today :( Here's another:
Problem:
Prove$$a^2+b^2\geq \frac{c^2}{2}$$$$a^4+b^4\geq \frac{c^4}{8}$$$$a^8+b^8\geq \frac{c^8}{128}$$ if $a+b\geq c\geq0$
Attempt:
Working backwards:
$$a^2+b^2\geq \frac{c^2}{2}$$
$$\implies 2a^2+2b^2\geq c^2$$
I am stuck on the first one, let alone the others. I know by AM-GM, $\frac{a^2+b^2}{2}\geq ab$ but how can I use it here?
| $(a^2+b^2)+(2ab)\ge c^2\implies $ at least one of the two terms in the LHS must be $\ge c^2/2$. Evidently this must be the larger one, viz $(a^2+b^2)$.
Now the second and third inequalities follow from the first (replace $a,b,c$ with $a^2, b^2, c^2/2$ etc).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1561962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Question regarding $f(n)=\cot^2\left(\frac\pi n\right)+\cot^2\left(\frac{2\pi}n\right)+\cdots+\cot^2\left(\frac{(n-1)\pi}n\right)$ $$f(n)=\cot^2\left(\frac\pi n\right)+\cot^2\left(\frac{2\pi}n\right)+\cdots+\cot^2\left(\frac{(n-1)\pi}n\right)$$ then how to find limit of $\dfrac{3f(n)}{(n+1)(n+2)}$ as $n\to\infty$?
I don't know any series like that. Riemann sum is not working. what to do?
| Recall the expansion: $$\begin{align}\tan nx &= \dfrac{\binom{n}{1}\tan x - \binom{n}{3}\tan^3 x + \cdots }{1-\binom{n}{2}\tan^2 x + \binom{n}{4}\tan^4 x + \cdots } \\&= \frac{\binom{n}{1}\cot^{n-1} x - \binom{n}{3}\cot^{n-3} x + \cdots }{\cot^{n} x - \binom{n}{2}\cot^{n-2} x + \binom{n}{4}\cot^{n-4} x + \cdots }\end{align}$$
Now, $\displaystyle \left\{\frac{k\pi}{n}\right\}_{k=1}^{n-1}$ are the roots of the equation: $\displaystyle \tan nx = 0$
Thus, $\displaystyle \binom{n}{1}\cot^{n-1} x - \binom{n}{3}\cot^{n-3} x + \cdots = 0$ whenever, $x = \dfrac{k\pi}{n}$ for $1 \le k \le n-1$.
Thus, using Vieta's formula we might write:
$$\begin{align*}\sum\limits_{k=1}^{n-1} \cot^2 \frac{k\pi}{n} &= \left(\sum\limits_{k=1}^{n-1} \cot \frac{k\pi}{n}\right)^2 - 2\sum\limits_{1 \le k_1 < k_2 \le n-1} \cot \frac{k_1\pi}{n}\cot \frac{k_2\pi}{n}\\&= 0 + 2\frac{\binom{n}{3}}{\binom{n}{1}}\\&= \frac{(n-1)(n-2)}{3}\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
Area bounded between the curve $y=x^2 - 4x$ and $y= 2x/(x-3)$ I've determined the intersects to be $x = 0, 2, 5$ and that $\frac{2x}{x-3}$, denoted as $f(x)$, is above $x(x-4)$, denoted as $g(x)$, so to find the area, I'll need to find the integral from $0$ to $2$ of $f(x) - g(x)$. But I've been stuck for a while playing around with this question.
| First, simplify the difference between the two functions:
$$
\begin{eqnarray}
f(x) - g(x) &=& \frac{2x}{x-3} - x(x - 4) \\
&=& \frac{-x^3 + 7x^2 - 10x}{x-3} \\
\end{eqnarray}
$$
Then, integrate by substituting $u = x-3$:
$$
\begin{eqnarray}
\int_0^2 \! \frac{-x^3 + 7x^2 - 10x}{x-3} \, \textrm{d}x &=& \int_{-3}^{-1} \! \frac{-(u+3)^3 + 7(u+3)^2 - 10(u+3)}{u} \, \textrm{d}u \\
&=& \int_{-3}^{-1} \! \frac{-u^3 - 2u^2 + 5u + 6}{u} \, \textrm{d}u \\
&=& -\frac{u^3}{3} - u^2 + 5u + 6\ln(u) \bigg|_{-3}^{-1} \\
&=& \frac{28}{3} - 6\ln(3) \\
\end{eqnarray}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
On the complete solution to $x^2+y^2=z^k$ for odd $k$? While trying to answer this question, I was looking at a computer output of solutions to $x^2+y^2 = z^k$ for odd $k$ and noticed certain patterns. For example, for $k=5$ we have $x,y,z$,
$$10, 55, 5\\25, 50, 5\\38, 41, 5\\117, 598, 13\\122, 597, 13\\338, 507, 13\\799, 884, 17$$
Question: Is it true that all integer solutions to $x^2+y^2=z^k$ for odd $k>1$ are given by just two formulas, namely,
*
*Primitives $\gcd(x,y)=1$:
$$A^2+B^2 = (a^2+b^2)^k\tag1$$
where $A,B$ is the expansion of $(a+bi)^k = A+Bi$. Example, $$(a+bi)^3 =(a^3 - 3 a b^2)+ (3 a^2 b - b^3)i$$ hence, $$(a^3 - 3 a b^2)^2+ (3 a^2 b - b^3)^2 = (a^2+b^2)^3$$ and so on for other $k$.
*Non-primitives $\gcd(x,y)\neq1$:
$$a^2(a^2+b^2)^{k-1}+b^2(a^2+b^2)^{k-1} = (a^2+b^2)^k\tag2$$
where, for both Forms $1$ and $2$, we use some rational $a,b$?
Example. The first three solutions for $k=5$ above use:
$$a,b = 2/5,\;11/5\quad \text{Form 2}$$
$$a,b = 1,\;2\quad \text{Form 2}$$
$$a,b = -2,\;1\quad \text{Form 1}$$
| Must write more correctly. $Z$ representation as a sum of squares gives the standard approach.
$$X^2+Y^2=Z^{k}$$
Uses the standard formula of the Brahmaputra. View the sum of the squares piece.
$$X^2+Y^2=(ab+cd)^2+(cb-ad)^2=(a^2+c^2)(b^2+d^2)$$
So the sum of squares is converted until
$$(b_1^2+d_1^2)...(b_{k}^2+d_{k}^2)=Z^{2k}$$
$$Z^2=b_1^2+d_1^2=...=b_{k}^2+d_{k}^2$$
Odd.
$$(a^2+b^2)(b_1^2+d_1^2).....(b_{k}^2+d_{k}^2)=Z*Z^{2k}$$
$$Z^2=b_1^2+d_1^2=...=b_{k}^2+d_{k}^2$$
$$Z=a^2+b^2$$
$$b_1=b_2=...=b_{k}=a^2-b^2$$
$$d_1=d_2=...=d_{k}=2ab$$
Will as an example this equation.
$$X^2+Y^2=Z^5$$
$$(xb+yd)^2+(yb-xd)^2=(x^2+y^2)(b^2+d^2)=(a^2+c^2)(b^2+d^2)(b^2+d^2)=Z*Z^2*Z^2$$
$$Z=a^2+c^2$$
$$x=ab+cd$$
$$y=cb-ad$$
$$d=a^2-c^2$$
$$b=2ac$$
And then the decision on the record.
$$X=a^5-10c^2a^3+5ac^4$$
$$Y=c^5-10a^2c^3+5ca^4$$
So these numbers are multiples of each its option.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Series' convergence - making my ideas formal
Find the collection of all $x \in \mathbb{R}$ for which the series $\displaystyle \sum_{n=1}^\infty (3^n + n)\cdot x^n$ converges.
My first step was the use the ratio test:
$$ \lim_{n \to \infty} \dfrac{(3^{n+1}+n+1) \cdot |x|^{n+1}}{(3n+n) \cdot |x|^n} = \lim_{n \to \infty} \dfrac{3^{n+1}|x|+n|x|+|x| }{3^n+n} = \lim_{n \to \infty} ( \dfrac{3^{n+1}|x|}{3^n+n} + \dfrac{n|x|}{3^n+n} + \dfrac{|x|}{3^n+n}) $$
$$ = \lim_{n \to \infty} (\dfrac{3|x|}{1+\dfrac{n}{3^n}} + \dfrac{|x|}{\dfrac{3^n}{n} + 1} + \dfrac{|x|}{3^n + n}) = 3|x| + 0 + 0 = 3|x| $$
So we want $3|x| < 1$, i.e. $|x| < \dfrac{1}{3}$. But we have to also check the points $x=\dfrac{1}{3}, -\dfrac{1}{3}$, where the limit equals $1$.
For $x=\dfrac{1}{3}$ we have $\displaystyle \sum_{n=1}^\infty (3^n + n)\cdot (\dfrac{1}{3})^n = \displaystyle \sum_{n=1}^\infty (1 + n \cdot \dfrac{1}{3}^n) $ which obviously diverges.
For $x=-\dfrac{1}{3}$ we have $\displaystyle \sum_{n=1}^\infty (3^n + n)\cdot (-\dfrac{1}{3})^n = \displaystyle \sum_{n=1}^\infty ((-1)^{n} + n \cdot (-\dfrac{1}{3})^n)$. Now I know this diverges because $(-1)^n$ and $(-\dfrac{1}{3})^n$ have alternating coefficients. But is there a theorem that I can use here?
So we conclude that $|x| < \dfrac{1}{3}$, but is the above work enough to show it conclusively?
| Once you've reduced to
$$\sum_{n = 1}^{\infty} \big((-1)^n + n \left(-\frac 1 3\right)^n\big)$$
you can simply note that the $n$th term of this series does not tend to zero; the right-hand portion tends to $0$ (and hence, is eventually less than $1/2$ in absolute value), while the left hand portion is always $1$ in absolute value. This is bounded away from zero for all $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1564354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Express the following complex numbers in standard form $$\left(\frac{\sqrt 3}{2}+\frac{i}{2}\right)^{25}$$
I know that you have to put it in the form $\cos\theta+i\sin \theta$ but I'm not sure how to go about it.
| Applying De-moivre's theorem, $(\cos\theta+i\sin \theta)^n=\cos n\theta+i\sin n\theta$
hence, $$\left(\frac{\sqrt 3}{2}+\frac{i}{2}\right)^{25}$$
$$=\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)^{25}$$
$$=\cos\frac{25\pi}{6}+i\sin\frac{25\pi}{6}$$
$$=\cos\left(4\pi+\frac{\pi}{6}\right)+i\sin\left(4\pi+\frac{\pi}{6}\right)$$
$$=\cos\left(\frac{\pi}{6}\right)+i\sin\left(\frac{\pi}{6}\right)$$
$$=\color{red}{\frac{\sqrt 3}{2}+\frac{i}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1568690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Help me with the result of this determinant.. $$
D =
\begin{vmatrix}
1 & 1 & 1 & \dots & 1 & 1 \\
2 & 1 & 1 & \dots & 1 & 0 \\
3 & 1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\
n-1 & 1 & 0 & \dots & 0 & 0 \\
n & 0 & 0 & \dots & 0 & 0 \\
\end{vmatrix}
=n*1*(-1)^\frac{n(n-1)}{2}
$$
I don't quite understand the solution of this determinant. I do understand that if we use Laplace expansion along the last row we get
$$
D = n*
\begin{vmatrix}
1 & 1 & 1 & \dots & 1 & 1 \\
1 & 1 & 1 & \dots & 1 & 0 \\
1 & 1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\
1 & 1 & 0 & \dots & 0 & 0 \\
1 & 0 & 0 & \dots & 0 & 0 \\
\end{vmatrix}
$$
But how does the remaining determinant euqal: $1*(-1)^\frac{n(n-1)}{2}$?
Edit:
$$
\begin{vmatrix}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0 \\
\end{vmatrix}
=(-1)^{4}
\begin{vmatrix}
0 & 1 \\
1 & 0 \\
\end{vmatrix}
=(-1)^{4+3+2}
$$
I thought it should go: $(-1)^{3+2+1}$ or is the power actually the sum of row and column coordinates?
| First do an elementary transformation that does not change the determinant, namely add to the first column the sum of all the others. You are left with
$$
\begin{vmatrix}
n & 1 & 1 & \dots & 1 & 1 \\
n & 1 & 1 & \dots & 1 & 0 \\
n & 1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\
n & 1 & 0 & \dots & 0 & 0 \\
n & 0 & 0 & \dots & 0 & 0 \\
\end{vmatrix}
=
n \cdot \begin{vmatrix}
1 & 1 & 1 & \dots & 1 & 1 \\
1 & 1 & 1 & \dots & 1 & 0 \\
1 & 1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\
1 & 1 & 0 & \dots & 0 & 0 \\
1 & 0 & 0 & \dots & 0 & 0 \\
\end{vmatrix}
=
n \cdot (-1)^{s}.
$$
Here $s$ is the sign of the permutation $\sigma$ on $\{1, 2, \dots, n\}$ that swaps $i$ with $n-i$, since exchanging the $i$-th row with the $(n-i)$-th row of
$$
\begin{bmatrix}
1 & 1 & 1 & \dots & 1 & 1 \\
1 & 1 & 1 & \dots & 1 & 0 \\
1 & 1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\
1 & 1 & 0 & \dots & 0 & 0 \\
1 & 0 & 0 & \dots & 0 & 0 \\
\end{bmatrix},
$$
for $1 \le i \le n/2$, we get a matrix which has visibly determinant $1$.
The permutation $\sigma$ has $s = n/2$ two-cycles if $n$ is even, and $s = (n-1)/2$ two-cycles if $n$ is odd. So indeed
$$
(-1)^{s} = (-1)^{n (n-1)/2}
=
\begin{cases}
(-1)^{n/2} & \text{if $n$ is even, as $(-1)^{n-1} = -1$,}\\
(-1)^{(n-1)/2} & \text{if $n$ is odd, as $(-1)^{n} = -1$.}\\
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
prove inequality $2(x+z)^3<(3x+z)(x+3z)$ Let $0<x<1,0<z<1$. Then
$$
2(x+z)^{3}<(3x+z)(x+3z)
$$
This checks out numerically, but I don't know why.
|
Expanding, our claim is that
$$ 2x^3 + 6xz^2 + 6x^2z + 2z^3 < 3x^2 + 3z^2 + 10xz$$
$0 < x < 1, 0 < z < 1$.
Note that we can write the RHS as $$2x^2 + 2z^2 + x^2 + z^2 + 10xz $$
We see that
$$ 2x^3 < 2x^2 \\ 2z^3 < 2z^2 \\ 6xz^2 < 5xz + z^2 \\ 6x^2z < 5xz + x^2$$
By adding these together our claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Determinant of Tridiagonal matrix I'm a bit confused with this determinant.
We have the determinant
$$\Delta_n=\left\vert\begin{matrix}
5&3&0&\cdots&\cdots&0\\
2&5&3&\ddots& &\vdots\\
0&2&5&\ddots&\ddots&\vdots\\
\vdots&\ddots&\ddots&\ddots&\ddots&0\\
\vdots& &\ddots&\ddots&\ddots&3\\
0&\cdots&\cdots&0&2&5\end{matrix}
\right\vert$$
I compute $\Delta_2=19$, $\Delta_3=65$.
Then I would like to find a relation for $n\geq 4$ which links $\Delta_n, \Delta_{n-1}$ and $\Delta_{n-2}$ and thus find an expression of $\Delta_n$. How could we do that for $n\geq 4$?
Thank you
| Let prove the theorem. Suppose the determinant of tri-diagonal matrix as $\Delta_{n}$, and operate the following calculation.
$$
\begin{align}
\Delta_{n}=& \det
\begin{bmatrix}
a_{1} & b_{1} & 0 & \cdots & 0 & 0 & 0 \\
c_{1} & a_{2} & b_{2} & \ddots & \vdots & \vdots & \vdots \\
0 & c_{2} & a_{3} & \ddots & a_{n-2} & b_{n-2} & 0 \\
\vdots & \vdots & \vdots & \ddots & c_{n-2} & a_{n-1} & b_{n-1} \\
0 & 0 & 0 & \cdots & 0 & c_{n-1} & a_{n}
\end{bmatrix}
\\ \\ =& \det
\begin{bmatrix}
a_{1} & b_{1} & 0 & \cdots & 0 & 0 & 0 \\
c_{1} & a_{2} & b_{2} & \ddots & \vdots & \vdots & \vdots \\
0 & c_{2} & a_{3} & \ddots & a_{n-2} & b_{n-2}-c_{n-1}\frac{b_{n-1}}{a_{n-1}} & 0 \\
\vdots & \vdots & \vdots & \ddots & c_{n-2} & a_{n-1}-c_{n-1}\frac{b_{n-1}}{a_{n-1}} & 0 \\
0 & 0 & 0 & \cdots & 0 & c_{n-1} & a_{n}
\end{bmatrix}
\\ \\ =& a_{n} \Delta_{n-1} - b_{n-1}c_{n-1} \Delta_{n-2}
\end{align}
$$
Based on this formula, it can be described as below using matrix and vector product. Then, we try to estimate $\Delta_{n}$.
$$
\begin{align}
\begin{bmatrix}
\Delta_{n} \\
\Delta_{n-1}
\end{bmatrix}
=&
\begin{bmatrix}
a_{n} & -b_{n-1}c_{n-1} \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
\Delta_{n-1} \\
\Delta_{n-2}
\end{bmatrix}
\\ =&
\prod_{k=4}^n
\begin{bmatrix}
a_{n+4-k} & -b_{n-k+3}c_{n-k+3} \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
\Delta_{3} \\
\Delta_{2}
\end{bmatrix}
\end{align}
$$
This problem's case, these elements are identity each diagonal factors like $a_{i}=5$ $b_{i}=3$, $c_{i}=2$. Therefore this equation can be simplified as follows.
$$
\begin{bmatrix}
\Delta_{n} \\
\Delta_{n-1}
\end{bmatrix}
=
\begin{bmatrix}
5 & -6 \\
1 & 0
\end{bmatrix}
^{n-3}
\begin{bmatrix}
65 \\
19
\end{bmatrix}
$$
After that, we get the eigenvalues, eigenvectors and diagonalization of the matrix.
$$
\begin{align}
\begin{bmatrix}
\Delta_{n} \\
\Delta_{n-1}
\end{bmatrix}
=&
\begin{bmatrix}
3 & 2 \\
1 & 1
\end{bmatrix}
\begin{bmatrix}
3^{n-1} & 0 \\
0 & 2^{n-1}
\end{bmatrix}
\begin{bmatrix}
3 & 2 \\
1 & 1
\end{bmatrix}
^{-1}
\begin{bmatrix}
65 \\
19
\end{bmatrix}
\\ =&
\begin{bmatrix}
3^n-2^n & 3 \cdot 2^{n-2} -2 \cdot 3^{n-2} \\
3^{n-3} - 2^{n-3} & 3 \cdot 2^{n-3} - 2 \cdot 3^{n-3}
\end{bmatrix}
\begin{bmatrix}
65 \\
19
\end{bmatrix}
\end{align}
$$
Eventually, $\Delta_{n}$ is
$$
\begin{align}
\Delta_{n}=&
65(3^{n-2}-2^{n-2})+19(3 \cdot 2^{n-2} -2 \cdot 3^{n-2}) \\ =&
3^{n+1} - 2^{n+1}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1571038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Express the real root of $x^3-3x+7$ using radicals I want to express the real root of $x^3-3x+7=0$ using radicals.
My attempt is contained in the answer below.
| To express the real root of $x^3-3x+7$ using radicals, we can apply Cardano's method:
\begin{align*}
x^3-3x+7=0&,\text{ let } x=u+v,\quad 3uv=3\implies uv=1\\
\implies& (u+v)^3-3(u+v)+7=0\\
\implies& (u^2+2uv+v^2)(u+v)-3(u+v)+7=0\\
\implies& u^3+2u^2v+uv^2+u^2v+2uv^2+v^3-3(u+v)+7=0\\
\implies& u^3 + v^3+uv(3u+3v)-(3u+3v)+7=0\\
\implies& u^3 + v^3+(uv-1)(3u+3v)+7=0\\
\end{align*}
Since $uv=1$ we simplify to:
$$\implies u^3+v^3+7=0$$
Notice also that $uv=1\implies v=\frac1u$
$$\implies u^3+u^{-3}+7=0$$
let $w=u^3$
\begin{align*}
\implies& w+w^{-1}+7=0\\
\implies& w^2+7w+1=0\\
\implies& w=\frac{-7\pm \sqrt{49-4}}{2}\\
\implies& w=\frac{-7\pm 3\sqrt{5}}{2}\\
\end{align*}
Lets take $w=\frac{-7+ 3\sqrt{5}}{2}$, then $u=\sqrt[3]{\frac{-7+ 3\sqrt{5}}{2}}$
$$x=\sqrt[3]{\frac{-7+ 3\sqrt{5}}{2}}+ \frac{1}{\sqrt[3]{\frac{-7+ 3\sqrt{5}}{2}}}$$
$$x=\sqrt[3]{\frac{-7+ 3\sqrt{5}}{2}}+ \sqrt[3]{\frac{2}{-7+ 3\sqrt{5}}}\approx -2.426\in \Bbb R$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1571813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Estimate of n factorial: $n^{\frac{n}{2}} \le n! \le \left(\frac{n+1}{2}\right)^{n}$ on our lesson at our university, our professsor told that factorial has these estimates
$n^{\frac{n}{2}} \le n! \le \left(\dfrac{n+1}{2}\right)^{n}$
and during proof he did this
$(n!)^{2}=\underbrace{n\cdot(n-1)\dotsm 2\cdot 1}_{n!} \cdot \underbrace{n\cdot(n-1) \dotsm 2\cdot 1}_{n!}$
and then:
$(1 \cdot n) \cdot (2 \cdot (n-1)) \dotsm ((n-1) \cdot 2) \cdot (n \cdot 1)$
and it is equal to this
$(n+1)(n+1) \dotsm (n+1)$
why it is equal, I didn't catch it. Do you have any idea? :)
| Not equal, but by a standard inequality: $\sqrt{ab} \le \frac{a+b}{2}$, so $ab \le \frac{(a+b)^2}{4}$.
So all products $i\cdot ((n+1)-i)$ are estimated above by $\frac{(i + ((n+1)-i))^2}{4} = \frac{(n+1)^2}{4}$. So no equality, but upper bounded by. Because we have $(n!)^2$ we get rid of the square roots again so $n! \le (\frac{n+1}{2})^n$ (where the $n$-th power comes form the fact that we have $n$ terms in the $(n!)^2$ expression.
The first equality is just rearranging terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1572094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Arc length of the squircle The squircle is given by the equation $x^4+y^4=r^4$. Apparently, its circumference or arc length $c$ is given by
$$c=-\frac{\sqrt[4]{3} r G_{5,5}^{5,5}\left(1\left|
\begin{array}{c}
\frac{1}{3},\frac{2}{3},\frac{5}{6},1,\frac{4}{3} \\
\frac{1}{12},\frac{5}{12},\frac{7}{12},\frac{3}{4},\frac{13}{12} \\
\end{array}
\right.\right)}{16 \sqrt{2} \pi ^{7/2} \Gamma \left(\frac{5}{4}\right)}$$
Where $G$ is the Meijer $G$ function. Where can I find the derivation of this result? Searching for any combination of squircle and arc length or circumference has led to nowhere.
| You can do this using this empirical formula to find perimeter of general super-ellipse
$$L=a+b\times\left(\frac{2.5}{n+0.5}\right)^\frac{1}{n}\times \left( b+a\times(n-1)\times\frac{\frac{0.566}{n^2}}{b+a\times\left(\frac{4.5}{0.5+n^2}\right)}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Using Demoivre's Theorem prove that $ {\cos5 \theta} = 16{\cos^5 \theta} - 20{\cos^3 \theta} + 5{\cos \theta} $ . $ {\cos5 \theta} = 16{\cos^5 \theta} - 20{\cos^3 \theta} + 5{\cos \theta} $ .
Demoivre's Theorem
$$ \{\cos \theta + i \sin \theta \}^n = \cos n\theta + i\sin n\theta $$
Where n is an integer .
I tried
$$ \{\cos \theta + i \sin \theta \}^5 - i\sin 5\theta = \cos 5\theta $$
But then I cant eliminate sines .
| Notice, $$\cos5\theta=(\cos\theta+i\sin\theta)^5-i\sin 5\theta$$
$$=(\cos\theta+i\sin\theta)^2(\cos\theta+i\sin\theta)^3-i\sin 5\theta$$
$$=(\cos^2\theta-\sin^2\theta+2i\sin\theta\cos\theta)(\cos^3\theta-3\sin^2\theta\cos\theta-i\sin^3\theta+3i\sin\theta\cos^2\theta)-i\sin 5\theta$$
comparing the real parts on both the sides, one should get
$$\cos5\theta=(\cos^2\theta-\sin^2\theta)(\cos^3\theta-3\sin^2\theta\cos\theta)$$$$+2\sin\theta\cos\theta(\sin^3\theta-3\sin\theta\cos^2\theta)$$
$$\cos5\theta=(2\cos^2\theta-1)(4\cos^3\theta-3\cos\theta)+2\cos\theta(1-\cos^2\theta)(1-4\cos^2\theta)$$
$$\cos5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that if $a>1$ then $\log a - \int_a^{a+1} \log x dx$ differs from $\frac{-1}{2a}$ by less than $\frac{1}{6a^2}$
Show that if $a>1$ then $\log a - \int_a^{a+1} \log x dx$ differs from
$\frac{-1}{2a}$ by less than $\frac{1}{6a^2}$.
For some $\theta$ between $a-1$ and $a$ and odd $n\in\mathbb{N}$ we have equality:
$$\log a - \int_a^{a+1} \log x dx=\log a - \int_{a-1}^atdt+\int_{a-1}^a \frac{t^2}{2}dt-\int_{a-1}^a \frac{t^3}{3}dt+ \ldots -\int_{a-1}^a \frac{\theta^{n+1}}{n+1}dt$$
I don't know how to move further. Any hint please?
| Direct solution
$\int_a^{a+1} \ln(x)\ dx = [ x \ln(x) - x ]_a^{a+1}$
$ = (a+1) \ln(a+1) - a \ln(a) - 1$
$ = \ln(a) + (a+1) \ln(1+\frac{1}{a}) - 1$
$ = \ln(a) + (a+1) ( \frac{1}{a} - \frac{1}{2a^2} + \frac{1}{3a^3} - \frac{1}{4a^4} + \cdots ) - 1$
$ = \ln(a) + \frac{1}{2a} - \frac{1}{6a^2} + \frac{1}{12a^3} - \cdots$ [which is clearly an alternating series since $a \ge 1$].
Thus you get the answer immediately, and much more, since the general term in the asymptotic expansion is trivial to find.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the number of ordered pairs that have a product of $10!$ and a least common multiple of $9!$ Problem
How many ordered pairs $(a,b)$ of positive integers have a product of $10!$ and a least common multiple of $9!$?
I was told my answer of $16$ to this problem was wrong. I don't see how since we just have to distribute the max powers of $2,3,5,$ and $7$ and there are $16$ ways to do that. We don't need to divide by $2$ since $(3,4)$ is different from $(4,3)$, for example.
| You can rephrase it as: how many ordered pairs $(a,b)$ of positive integers satisfy $ab=10!$ and $\gcd(a,b)=10$, because $ab=\gcd(a,b)\text{lcm}(a,b)$.
I.e. let $a=10a_1, b=10b_1$, and you're asking how many ordered pairs $(a_1,b_1)$ of positive integers satisfy $a_1b_1=\frac{10!}{100}=2^6\cdot 3^4\cdot 7$ and $\gcd(a_1,b_1)=1$, so all the cases are $$a_1\in\{1,2^6,3^4,7,2^6\cdot 3^4,3^4\cdot 7,2^6\cdot 7,2^6\cdot 3^4\cdot 7\},$$
so the answer is $8$ ordered pairs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show by induction that $(n^2) + 1 < 2^n$ for intergers $n > 4$ So I know it's true for $n = 5$ and assumed true for some $n = k$ where $k$ is an interger greater than or equal to $5$.
for $n = k + 1$ I get into a bit of a kerfuffle.
I get down to $(k+1)^2 + 1 < 2^k + 2^k$ or equivalently:
$(k + 1)^2 + 1 < 2^k * 2$.
A bit stuck at how to proceed at this point
| First, show that this is true for $n=5$:
$5^2+1<2^5$
Second, assume that this is true for $n$:
$n^2+1<2^n$
Third, prove that this is true for $n+1$:
$(n+1)^2+1=$
$n^2+\color\green{2}\cdot{n}+1+1<$
$n^2+\color\green{n}\cdot{n}+1+1=$
$n^2+n^2+1+1=$
$2\cdot(\color\red{n^2+1})<$
$2\cdot(\color\red{2^n})=$
$2^{n+1}$
Please note that the assumption is used only in the part marked red.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show that $\sqrt{x^2 +1 }$ is uniformly continuous. $x \in \mathbb{R}$ Let $x,y \in \mathbb{R} $ such that;
If $|x| \leq \delta$ then $|y| = |x| +(|y|-|x|) \leq |x| +||y|-|x|| \leq |x| + |y-x| < 2 \delta$
$$|\sqrt{x^2 +1 } - \sqrt{y^2 +1 }| = \sqrt{y^2 +1 } - \sqrt{x^2 +1 } \leq \sqrt{y^2 +1 } - 1 \leq y^2 +1 -1 < 2\delta := \epsilon $$
Now, this is the part I am having trouble with
if $|x| > \delta$
$$|\sqrt{x^2 +1 } - \sqrt{y^2 +1 }| \leq |\frac{x^2 - y^2}{\sqrt{x^2 +1 } + \sqrt{y^2 +1 }}| \color{red}{\leq \frac{(x+y)^2}{2\sqrt{\delta^2 + 1}}}...$$
I don't think the red term makes much sense. As the $2xy$ term could very well be negative.
Could someone suggest a next step please.
| You came close. We have
$$\left|\sqrt{1+x^2}-\sqrt{1+y^2}\right|=\frac{|x^2-y^2|}{\sqrt{1+x^2}+\sqrt{1+y^2}} =|x-y|\frac{|x+y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}. $$
Now
$$\frac{|x+y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}\le \frac{|x|+|y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}\lt 1. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Division in finite fields Let's take $GF(2^3)$ as and the irreducible polynomial $p(x) = x^3+x+1$ as an example. This is the multiplication table of the finite field
I can easily do some multiplication such as $$(x^2+x)\cdot(x+1) = x^3 + x^2 + x +1 = x+1+x^2+x+2 = x^2$$
I am wondering how to divide some random fields such as $x^2 / (x+1)$. The result is $x^2+x$ (compare above). But how do I actually calculate this. Polynomial long division does not help be:
*
*Why don't I get $x+1$ as result?
*How can I calculate $x / (x^2+x+1)$? The result should be $x+1$
| Since any element of this field can be written as $ax^2+bx+c$ we write:
$\frac{x}{x^2+x+1}=ax^2+bx+c$
$x=(ax^2+bx+c)*(x^2+x+1)$
$x=ax^4+(a+b)x^3+(a+b+c)x^2+(b+c)x+c$
Use the fact that $x^3=x+1$ to further simplify.
$x=a(x^2+x)+(a+b)(x+1)+(a+b+c)x^2+(b+c)x+c$
$x=(b+c)x^2+(c)x+(a+b+c)$
Now solve for $a$, $b$, and $c$ by comparing each power of $x$.
$b+c=0$
$c=1$
$a+b+c=0$
Thus, $a=0$, $b=1$, and $c=1$.
So $\frac{x}{x^2+x+1}=x+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find $\lim\limits_{x\to \infty}\frac{\ln(1+4^x)}{\ln(1+3^x)}$ Find $\lim\limits_{x\to \infty}\frac{\ln(1+4^x)}{\ln(1+3^x)}$
Using Taylor series:
$$\ln(1+4^x)=\frac{2\cdot 4^x-4^{2x}}{2}+O(4^{2x}),\ln(1+3^x)=\frac{2\cdot 3^x-3^{2x}}{2}+O(3^{2x})\Rightarrow$$
$$\lim\limits_{x\to \infty}\frac{\ln(1+4^x)}{\ln(1+3^x)}=\lim\limits_{x\to \infty}\frac{2\cdot 4^x-4^{2x}}{2\cdot 3^x-3^{2x}}=\infty$$
The limit should be $0$. Could someone point out what is wrong?
| Note that the series $\log(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\,x^n$ is valid only for $-1<x\le 1$. We can write for $|x|>1$,
$$\begin{align}
\log(1+x)&=\log x+\log \left(1+\frac1x\right) \\\\
&=\log x +\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\,x^{-n} \tag 1
\end{align}$$
Using $(1)$, we can write for $a>1$
$$\begin{align}
\log(1+a^x)&=\log(a^x)+\log(1+a^{-x})\\\\
&=x\log a+O(a^{-x}) \tag 2
\end{align}$$
From $(2)$ we can write
$$\begin{align}
\frac{\log(1+4^x)}{\log(1+3^x)}&=\frac{x\log 4+O(4^{-x})}{x\log 3+O(3^{-x})}\\\\
&=\frac{\log 3+O\left(\frac{4^{-x}}{x}\right)}{\log 4+O\left(\frac{3^{-x}}{x}\right)}
\end{align}$$
The limit as $x\to \infty$ is evident now as the numerator tends to $\log 4$ and the denominator tends to $\log 3$. Therefore, we obtain
$$\lim_{x\to \infty}\frac{\log(1+4^x)}{\log(1+3^x)}=\frac{\log 4}{\log 3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1583582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
If $\arcsin x+\arcsin y+\arcsin z=\pi$,then prove that $(x,y,z>0)x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$ If $\arcsin x+\arcsin y+\arcsin z=\pi$,then prove that $(x,y,z>0)$
$x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$
$\arcsin x+\arcsin y+\arcsin z=\pi$,
$\arcsin x+\arcsin y=\pi-\arcsin z$
$\arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2})=\pi-\arcsin z$
$x\sqrt{1-y^2}+y\sqrt{1-x^2}=z$
Similarly,$y\sqrt{1-z^2}+z\sqrt{1-y^2}=x$
Similarly,$x\sqrt{1-z^2}+z\sqrt{1-x^2}=y$
Adding the three equations,we get
$x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=\frac{x+y+z}{2}$
I am stuck here,please help me.Thanks.
| All we need to facilitate analysis is the double angle formula
$$\sin(2a)=2\sin(a)\,\cos(a) \tag 1$$
the Prosthaphareis Identity
$$\cos (a-b)-\cos(a+b)=2\sin (a)\,\sin (b) \tag 2$$
along with the Reverse Prosthaphareis Identity
$$\sin (a)+\sin (b)=2\sin\left(\frac{a+b}{2}\right)\,\cos\left(\frac{a-b}{2}\right) \tag 3$$
Then, letting $x=\sin \alpha$, $y=\sin \beta$, and $z=\sin \gamma$ gives
$$\alpha +\beta +\gamma = \pi \tag 4$$
We wish now to transform the function $F(\alpha,\beta,\gamma)$ as given by
$$\begin{align}
F(\alpha,\beta,\gamma)&=\frac12 \sin (2\alpha)+\frac12 \sin (2\beta)+\frac12 \sin (2\gamma) \tag 5\\\\
&=x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}
\end{align}$$
First, use $(1)$ with $a=2\gamma$, $(3)$ with $a=2\alpha$ and $b=2\beta$, and $(4)$ to write $(5)$ as
$$\begin{align}
F(\alpha,\beta,\gamma)&=\sin(\alpha+\beta)\,\cos(\alpha-\beta)+\sin(\gamma)\,\cos (\gamma) \\\\
&=\sin(\pi-\gamma)\,\cos(\alpha-\beta)+\sin(\gamma)\,\cos (\pi-\alpha-\beta)
\end{align}$$
Next, substituting $(4)$ into $(6)$ and recalling that $\sin (\pi -a)=\sin(a)$ and $\cos (\pi-a)=-\cos(a)$ reveals
$$F(\alpha,\beta,\gamma)=\sin(\gamma)\left(\cos(\alpha-\beta)-\cos (\alpha+\beta)\right) \tag 7$$
Finally, using $(2)$ in $(7)$ yields
$$\begin{align}
F(\alpha,\beta,\gamma)&=\sin(\gamma)\left(2\sin(\alpha)\,\sin(\beta)\right)\\\
&=2\sin(\alpha)\,\sin(\beta)\,\sin)\gamma)\\\\
&=2xyz
\end{align}$$
Therefore, we obtain the identity
$$x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$$
as was to be shown!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How to compute the area of that portion of the conical surface $x^2+y^2=z^2$ which lies between the two planes $z=0$ and $x+2z=3$? How to compute the area of that portion of the conical surface $x^2+y^2=z^2$ which lies between the two planes $z=0$ and $x+2z=3$ ? I can't even figure out what the integrand will be ( should it be $\sqrt{z^2-x^2}$ ?
) and not even the limits . Please help . thanks in advance
| The slope of the plane $x+2z=3$ is smaller than the slope of the cone, so their intersection curve is an ellipse, drawn with brown in the figure. The plane $z=0$ passes through the apex. Therefore, the region we are interested in is a ''hat''.
The projection of the surface region onto to the $x-y$ co-ordinate plane is another ellipse (the red one).
Between the two planes we have $0\le z\le \frac{3-x}2$, so the red ellipse is detemined by
$$
x^2+y^2\le\left(\frac{3-x}2\right)^2 \\
\frac{(x+1)^2}4 + \frac{y^2}3 \le 1.
$$
So, the area of the red ellipse is $2\sqrt3\pi$.
The slope of the cone is $1$, so the are of the two areas is $\sqrt2$. Therefore, the area of the conic region is $\sqrt2\cdot 2\sqrt3\pi=2\sqrt6\pi$.
(The question did not involve the area of the brown ellipse; just in order to verify the previous answer, from the slope of the plane $z=\frac{3-x}2$ it is $\sqrt{1^2+(\frac12)^2}\cdot 2\sqrt3\pi=\sqrt{15}\pi$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1587440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve for Rationals $p,q,r$ Satisfying $\frac{2p}{1+2p-p^2}+\frac{2q}{1+2q-q^2}+\frac{2r}{1+2r-r^2}=1$. Find all rational solutions $(p,q,r)$ to the Diophantine equation
$$\frac{2p}{1+2p-p^2}+\frac{2q}{1+2q-q^2}+\frac{2r}{1+2r-r^2}=1\,.$$
At least, determine an infinite family of $(p,q,r)\in\mathbb{Q}^3$ satisfying this equation. If you can also find such an infinite family of $(p,q,r)$ with the additional condition that $0<p,q,r<1+\sqrt{2}$, then it would be of great interest.
This question is related to Three pythagorean triples.
| If you combine the system into a single equation, you find that $r$ satisfies the quadratic equation
\begin{equation*}
r^2-\frac{4(p^2q+p(q^2-4q-1)-q)}{p^2(q^2-1)-4pq-q^2+1}r-1=0
\end{equation*}
For $r \in \mathbb{Q}$, the discriminant must be a rational square, so there must exist $d \in \mathbb{Q}$ such that
\begin{equation*}
d^2=(p^2+1)^2q^4-32p^2q^3+(2p^4-32p^3+68p^2+32p+2)q^2+32p^2q+(p^2+1)^2
\end{equation*}
The quartic has a rational solution when $q=0$, and so is birationally equivalent to an elliptic curve.
Using Mordell's method with a symbolic algebra package, we find this elliptic curve to be
\begin{equation*}
V^2=U^3+2(p^4-4p^3+10p^2+4p+1)U^2+(p^2-2p-1)^4U
\end{equation*}
with the reverse transformation
\begin{equation*}
q=\frac{(p^2+1)V+8p^2U}{(p^2+1)^2U+(p^2-2p-1)^4}
\end{equation*}
Numerical experiments suggest that the torsion subgroup of the curve is isomorphic to $\mathbb{Z4}$, and we find $(0,0)$ is a point of order $2$, and
$(-(p^2-2p-1)^2, \pm 4p(p^2-2p-1)^2)$ are of order $4$.
These experiments also suggest that the rank of the curves is always at least $1$, and it is not hard to find that $(-(p^2+1)^2, \pm 8p^2(p^2+1)\,)$
are of infinite order.
Taking the negative $V$ value gives
\begin{equation*}
q=\frac{2p(p^2+1)^2}{(p+1)(p-1)(p^4-2p^3+2p^2+2p+1)}
\end{equation*}
and, if we slot this into the quadratic for $r$ we get
\begin{equation*}
r=\frac{(p-1)(p^6-2p^5+7p^4-p^2+2p+1)}{(p+1)(p^6-2p^5-p^4+7p^2+2p+1)}
\end{equation*}
Because there is a point of infinite order, we can generate an infinite number of parametric solutions, though they become increasingly complicated.
Hope this helps, and a Merry Christmas to all.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Prove $1^2-2^2+3^2-4^2+......+(-1)^{k-1}k^2 = (-1)^{k-1}\cdot \frac{k(k+1)}{2}$ I'm trying to solve this problem from Skiena book, "Algorithm design manual".
I don't know the answer but it seems like the entity on the R.H.S is the summation for series $1+2+3+..$. However the sequence on left hand side is squared series and seems to me of the form:
$-3-7-11-15\ldots $
I feel like its of the closed form:
$\sum(-4i+1)$
So how do I prove that the equality is right?
| i) $k=2n$:
\begin{align}
\sum_{m=1}^{2n}(-1)^{m-1}m^2&=\sum_{m=1}^{2n}m^2-2\sum_{m=1}^{n}(2m)^2\\
&=\sum_{m=1}^{2n}m^2-8\sum_{m=1}^{n}m^2\\
&=\frac{2n(2n+1)(4n+1)}{6}-8\frac{n(n+1)(2n+1)}{6}\\
&=\frac{2n(2n+1)(4n+1-4n-4)}{6}\\
&=\frac{2n(2n+1)(-1)}{2}=-\frac{k(k+1)}{2}=(-1)^{k-1}\frac{k(k+1)}{2}
\end{align}
ii)$k=2n+1$:
\begin{align}
\sum_{m=1}^{2n+1}(-1)^{m-1}m^2&=(2n+1)^2+\sum_{m=1}^{2n}(-1)^{m-1}m^2\\
&=(2n+1)^2+\frac{2n(2n+1)(-1)}{2}\\
&=\frac{(2n+1)(4n+2-2n)}{2}=\frac{k(k+1)}{2}=(-1)^{k-1}\frac{k(k+1)}{2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
How to solve an irrational equation? I want to solve this equation
$$2 (x-2) \sqrt{5-x^2}+(x+1)\sqrt{5+x^2} = 7 x-5.$$
I tried
The given equation equavalent to
$$2 (x-2) (\sqrt{5-x^2}-2)+(x+1)(\sqrt{5+x^2}- 3)=0$$
or
$$(x-2)(x+1)\left [\dfrac{x+2}{\sqrt{5+x^2} + 3} - \dfrac{2(x-1)}{\sqrt{5-x^2} + 2}\right ] = 0.$$
I see that, the equation
$$\dfrac{x+2}{\sqrt{5+x^2} + 3} - \dfrac{2(x-1)}{\sqrt{5-x^2} + 2} = 0$$
has unique solution $x = 2$, but I can not solve. How can I solve this equation or solve the given equation with another way?
| The usual way to do something like this is to square to get rid of one radical, rearrange, and square again to get rid of the remaining radical, as @vhspdfg says in a comment. I won’t detail the intermediate steps, but the octic polynomial I got was
$$
2000 - 1600x - 3480x^2 + 2960x^3 + 825x^4 - 1340x^5 + 510x^6 - 140x^7 + 25x^8\\
=5(x+1)^2(x-2)^3(5x^3 - 8x^2 + 65x - 50)\,.
$$
The cubic seems to have no rational roots, and if this is correct, it’s irreducible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Range of function $ f(x) = x\sqrt{x}+\frac{1}{x\sqrt{x}}-4\left(x+\frac{1}{x}\right),$ Where $x>0$
Find the range of the function $\displaystyle f(x) = x\sqrt{x}+\frac{1}{x\sqrt{x}}-4\left(x+\frac{1}{x}\right),$ where $x>0$
$\bf{My\; Try::}$ Let $\sqrt{x}=t\;,$ Then $\displaystyle f(t) = t^3+\frac{1}{t^3}-4\left(t^2+\frac{1}{t^2}\right)\;,$
Now After Simplification, We get $\displaystyle f(t) = \left(t+\frac{1}{t}\right)^3-3\left(t+\frac{1}{t}\right)-4\left[\left(t+\frac{1}{t}\right)^2-2\right]$
Now Put $\displaystyle t+\frac{1}{t} = u\;,$ Then $\displaystyle \sqrt{x}+\frac{1}{\sqrt{x}} = u\;,$ So we get $u\geq 2$ (Using $\bf{A.M\geq G.M}$)
And our function convert into $\displaystyle f(u) = u^3-4u^2-3u+8\;,$ Where $u\geq 2$
Now Using Second Derivative Test, $f'(u) = 3u^2-8u-3$ and $f''(u) = 6u-8$
So for Max. and Min., We put $\displaystyle f'(u)=0\Rightarrow u=3$ and $f''(3)=10>0$
So $u=3$ is a point of Minimum.
So $f(2)=8-4(4)-3(2)+8 = -6$ and $f(3) = -10$
and Graph is Like this
So Range is $$\displaystyle \left[-10,\infty \right)$$
My question is can we solve it any other way, Like using Inequality
If yes, Then plz explain here
Thanks
| HINT: $$f'(x)=3/2\,\sqrt {x}-3/2\,{x}^{-5/2}-4+4\,{x}^{-2}$$
solving for $x$ we obtain
$$1,1/4\, \left( \sqrt {5}-3 \right) ^{2},1/4\, \left( \sqrt {5}+3 \right) ^{2}$$
plugging $1$ in $f(x)$ we obtain $-6$ plugging the other two terms in $f$ we get $$-10$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Calculation of $\max$ and $\min$ value of $f(x) = \frac{x(x^2-1)}{x^4-x^2+1}.$
Calculation of $\max$ and $\min$ value of $$f(x) = \frac{x(x^2-1)}{x^4-x^2+1}$$
My try: We can write $$f(x) = \frac{\left(x-\frac{1}{x}\right)}{\left(x^2+\frac{1}{x^2}\right)-1} = \frac{\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+1}$$
Now put $\left(x-\frac{1}{x}\right)=t,x\ne0$. Then we get $$f(t) = \frac{t}{t^2+1} = \frac{1}{2}\left(\frac{2t}{1+t^2}\right)$$
Now put $t=\tan \theta$. Then $$f(\theta) = \frac{1}{2}\frac{2\tan \theta}{1+\tan^2 \theta} = \frac{1}{2}\sin 2\theta$$
So we get $$-\frac{1}{2}\leq f(\theta)\leq \frac{1}{2}\Rightarrow f(\theta)\in \left[-\frac{1}{2}\;,\frac{1}{2}\right]$$
My question is: Is my solution right? If not, then how can we solve it?
| Your solution is almost correct. You should note that $x=0$ is neither a point of maximum or minimum, because $f(0)=0$, but $f(2)>0$ and $f(-2)<0$. So considering
$$
f(x) = \frac{\left(x-\frac{1}{x}\right)}{\left(x^2+\frac{1}{x^2}\right)-1} = \frac{\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+1}
$$
for $x\ne0$ is good: the restriction to $x\ne0$ has the same maximum and minimum values. Then the maximum and minimum values of $f$ coincide with the maximum and minimum values of
$$
g(t)=\frac{t}{t^2+1}
$$
because $t=x-1/x$ takes on every real number (twice). Since the substitution $\theta=2\arctan t$ is bijective from $\mathbb{R}$ onto $(-\pi,\pi)$, we can conclude that the function
$$
h(\theta)=\frac{1}{2}\sin\theta,
$$
on the interval $(-\pi,\pi)$, has the same maximum and minimum values as $g$ and so also as $f$. The maximum and minimum values are attained for $\theta=\pi/2$ and $\theta=-\pi/2$, respectively.
This corresponds to $t=\tan(\pi/4)=1$ and $t=\tan(-\pi/4)=-1$, respectively. From
$$
x-\frac{1}{x}=1
$$
we get
$$
x=\frac{1-\sqrt{5}}{2}\qquad\text{or}\qquad x=\frac{1+\sqrt{5}}{2}
$$
for the points of maximum for $f$. From
$$
x-\frac{1}{x}=-1
$$
we get
$$
x=\frac{-1-\sqrt{5}}{2}\qquad\text{or}\qquad x=\frac{-1+\sqrt{5}}{2}
$$
for the points of minimum for $f$.
This can be confirmed with a more tedious computation. We have
$$
f'(x)=\frac{(3x^2-1)(x^4-x^2+1)-(x^3-x)(4x^3-2x)}{(x^4-x^2+1)^2}
$$
and the numerator is
$$
N(x)=3x^6-x^4-3x^4+x^2+3x^2-1-4x^6+4x^4+2x^4-2x^2
$$
or
\begin{align}
N(x)&=-x^6+2x^4+2x^2-1\\[6px]
&=-(x^2+1)(x^4-x^2+1)+2x^2(x^2+1)\\[6px]
&=-(x^2+1)(x^4-3x^2+1)
\end{align}
Thus the derivative vanishes for
$$
x^2=\frac{3\pm\sqrt{5}}{2}=\frac{6\pm2\sqrt{5}}{4}=
\frac{(1\pm\sqrt{5})^2}{4}
$$
giving back the same points as before.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
$(x^2+1)(y^2+1)(z^2+1) + 8 \geq 2(x+1)(y+1)(z+1)$ The other day I came across this problem:
Let $x$, $y$, $z$ be real
numbers. Prove that
$$(x^2+1)(y^2+1)(z^2+1) + 8 \geq 2(x+1)(y+1)(z+1)$$
The first thought was power mean inequality, more exactly : $AM \leq SM$ ( we noted $AM$ and $SM$ as arithmetic and square mean), but I haven't found anything helpful.
(To be more specific, my attempts looked like this :
$\frac{x+1}{2} \leq \sqrt{\frac{x^2+1}{2}}$)
I also take into consideration Cauchy-Buniakowsky-Scwartz or Bergström inequality, but none seems to help.
Some hints would be apreciated.
Thanks!
| Its just a matter of factorization:
$$
(x^2+1)(y^2+1)(z^2+1)+8-2(x+1)(y+1)(z+1)=(x^2y^2z^2-2xyz+1)+\left(\sum_{cyc}x^2y^2-2xy+1\right)+\left(\sum_{cyc}x^2-2x+1\right)=(xyz-1)^2+\left(\sum_{cyc}(xy-1)^2\right)+\left(\sum_{cyc}(x-1)^2\right)≥0
$$
With equality only if $x=y=z=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solving Fourier series of $ f(x)=\begin{cases} x+1 ;-1Please take a look at below Fourier series :
$ f(x)=\begin{cases} x+1 &-1<x<0\\ 1-x & 0<x<1 \end{cases} $
I tried to solve it as follows :
$
a_n=\displaystyle \dfrac{1}{1}\int_{-1}^1f(x)\cos (n\pi x) dx=2\int_{0}^{1} (1-x)\cos( n\pi x) dx
$
solving $a_n$ comes to :
$
a_n=2\left[\dfrac{(-1)^{n}}{n^2\pi^2}+\dfrac{1}{n^2\pi^2}\right]
$
also solving $b_n=0$ then we have Fourier series as below :
$
\displaystyle f(x)=\sum_{n=1}^{\infty}\left(2\left[\dfrac{(-1)^{n}}{n^2\pi^2}+\dfrac{1}{n^2\pi^2}\right]\cos\frac{n\pi}{1}x\right)
$
I have doubts about my recognition about function if it is odd or even.
(Is my solution right?)
| As asked in a comment, I compute the Fourier coefficients :
\begin{align}
a_0(f) & = \int_{-1}^{1}f(x)\mathrm{d}x \\
& = \int_{-1}^{0}\left(x+1\right)\mathrm{d}x + \int_{0}^{1}\left(1-x\right)\mathrm{d}x \\
& = \left[\frac{x^2}{2}+x\right]_{-1}^{0} + \left[x-\frac{x^2}{2}\right]_{0}^{1} \\
& = \left[-\frac{1}{2}+1\right] + \left[1-\frac{1}{2}\right] \\
& = \frac{1}{2} + \frac{1}{2} \\
& = 1.
\end{align}
For all integer $n\geq 1$,
\begin{align}
a_n(f) & = \int_{-1}^{1}f(x)\cos\left(n\pi x\right)\mathrm{d}x \\
& = \int_{-1}^{0}\left(x+1\right)\cos\left(n\pi x\right)\mathrm{d}x + \int_{0}^{1}\left(1-x\right)\cos\left(n\pi x\right)\mathrm{d}x \\
& = \left[\left(x+1\right)\frac{\sin\left(n\pi x\right)}{n\pi}\right]_{-1}^{0} - \frac{1}{n\pi}\int_{-1}^{0}\sin\left(n\pi x\right)\mathrm{d}x \\
& + \left[\left(1-x\right)\frac{\sin\left(n\pi x\right)}{n\pi}\right]_{0}^{1} + \frac{1}{n\pi}\int_{0}^{1}\sin\left(n\pi x\right)\mathrm{d}x \\
& = - \frac{1}{n\pi}\int_{-1}^{0}\sin\left(n\pi x\right)\mathrm{d}x + \frac{1}{n\pi}\int_{0}^{1}\sin\left(n\pi x\right)\mathrm{d}x \\
& = + \frac{1}{n\pi}\left[\frac{\cos\left(n\pi x\right)}{n\pi}\right]_{-1}^{0} - \frac{1}{n\pi}\left[\frac{\cos\left(n\pi x\right)}{n\pi}\right]_{0}^{1} \\
& = \frac{1}{n^2\pi^2}\left[1-(-1)^n\right] - \frac{1}{n^2\pi^2}\left[(-1)^n-1\right] \\
& = \frac{2}{n^2\pi^2}\left(1-(-1)^n\right).
\end{align}
Since $f$ is odd (i.e. $b_n(f)=0$ for all integer $n\geq 1$), the Fourier's series is then formally given by
\begin{align}
S_n(f)(x) & = \frac{a_0(f)}{2} + \sum_{n=1}^{+\infty}a_n(f)\cos\left(n\pi x\right) \\
& = \frac{1}{2} + \sum_{n=1}^{+\infty}\frac{2}{n^2\pi^2}\left(1-(-1)^n\right)\cos\left(n\pi x\right) \\
& = \frac{1}{2} + 4\sum_{p=0}^{+\infty}\frac{\cos\left((2p+1)\pi x\right)}{(2p+1)^2\pi^2}
\end{align}
where the last inequality is due to the fact that $1-(-1)^n=0$ if $n$ is even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What will be the remainder when $2^{31}$ is divided by $5$? The question is given in the title:
Find the remainder when $2^{31}$ is divided by $5$.
My friend explained me this way:
$2^2$ gives $-1$ remainder.
So, any power of $2^2$ will give $-1$ remainder.
So, $2^{30}$ gives $-1$ remainder.
So, $2^{30}\times 2$ or $2^{31}$ gives $3$ remainder.
Now, I cannot understand how he said the last line. So, please explain this line.
Also, how can I do this using modular congruency?
| We have that $5$ is a prime, so, $2^4 \equiv 1 \pmod 5$ (see Fermat's little theorem).
Then, $(2^4)^{7} = 2^{28} \equiv 1 \pmod 5$. Multiplying this by $2^3$, we get $2^{31} \equiv 8 \equiv 3 \pmod 5$.
Your friend is saying that:
$$2^{30} \equiv -1 \pmod 5 \implies 2^{31} \equiv -1 \times 2 \equiv -2 \equiv -2 + 5 \equiv 3 \pmod 5.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
Prove that $\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n} \geq (n-1) \left (\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}+\cdots+\frac{1}{\sqrt{x_n}} \right )$
Let $x_1,x_2,\ldots,x_n > 0$ such that $\dfrac{1}{1+x_1}+\cdots+\dfrac{1}{1+x_n}=1$. Prove the following inequality.
$$\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n} \geq (n-1) \left (\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}+\cdots+\dfrac{1}{\sqrt{x_n}} \right ).$$
This is Exercice 1.48 in the book "Inequalities-A mathematical Olympiad approach".
Attempt
I tried using HM-GM and I got $\left ( \dfrac{1}{x_1x_2\cdots x_n}\right)^{\frac{1}{2n}} \geq \dfrac{n}{\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}+\cdots+\dfrac{1}{\sqrt{x_n}}} \implies \dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_1}}+\cdots+\dfrac{1}{\sqrt{x_n}} \geq n(x_1x_2 \cdots x_n)^{\frac{1}{2n}}$. But I get stuck here and don't know if this even helps.
| Note that $\displaystyle \sum_{i = 1}^n \dfrac{1}{1+a_i} = 1 \implies \sum_{i = 1}^n \dfrac{a_i}{1+a_i} = n-1$. Then, $$\displaystyle \sum_{i = 1}^n \sqrt{a_i}-(n-1)\sum_{i = 1}^n \dfrac{1}{\sqrt{a_i}} = \sum_{i = 1}^n \dfrac{1}{1+a_i}\sum \sqrt{a_i} - \sum_{i = 1}^n \dfrac{a_i}{1+a_i} \sum_{i = 1}^n \dfrac{1}{\sqrt{a_i}} = \displaystyle \sum_{i,j} \dfrac{a_i-a_j}{(1+a_j)\sqrt{a_i}} = \sum_{i>j} \dfrac{(\sqrt{a_i}\sqrt{a_j}-1)(\sqrt{a_i}-\sqrt{a_j})^2(\sqrt{a_i}+\sqrt{a_j})}{(1+a_i)(1+a_j)\sqrt{a_i}\sqrt{a_j}}$$
But $1 \geq \dfrac{1}{1+a_i}+\dfrac{1}{1+a_j} = \dfrac{2+a_i+a_j}{1+a_i+a_j+a_ia_j}$ thus $a_ia_j \geq 1$. Hence the terms of the last sum are positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 2,
"answer_id": 0
} |
Prove for every odd integer $a$ that $(a^2 + 3)(a^2 + 7) = 32b$ for some integer $b$. I've gotten this far:
$a$ is odd, so $a = 2k + 1$ for some integer $k$.
Then $(a^2 + 3).(a^2 + 7) = [(2k + 1)^2 + 3] [(2k + 1)^2 + 7]$
$= (4k^2 + 4k + 4) (4k^2 + 4k + 8) $
$=16k^4 + 16k^3 + 32k^2 + 16k^3 + 16k^2 + 32k + 16k^2 + 16k + 32$
$=16k^4 + 32k^3 + 64k^2 + 48k + 32$
But this isn't a multiple of 32, at most I could say $(a^2 + 3)(a^2 + 7) = 16b$ for some integer $b$
| Use modular arithmetic (this is not the most simple way but it also works very well for similar, more complicated problems). Notice that for any odd $a$ we have $a^2 \equiv 1 \ \text{(mod 4)}$ and $a^2 \equiv 1 \ \text{(mod 8)}$. So $a^2 + 3 \equiv 0\ \text{(mod 4)}$, which means that $a^2 + 3$ is divisable by $4$, and $a^2 + 7 \equiv 0\ \text{(mod 8)}$ (idem dito). So the product is divisible by 32.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Laurent expansion on an annulus problem I have $f(z)= \frac{1}{\sin(z)}$ and am required to show that on the disc {$0<|z|<\pi$} the Laurent expansion is equal to:
$$c_{-1}z^{-1}+\sum_{n=0}^\infty{c_nz^n}$$
My plan is use the expansion of $$\sin(z)= z-\frac{z^3}{3!}+\frac{z^5}{5!}...$$ and rewrite $\frac{1}{\sin(z)}$ as $$\frac{1}{z(1-(\frac{z^2}{3!}-\frac{z^4}{5!}+...))} = \frac{1}{z} \times \frac{1}{(1-(\frac{z^2}{3!}-\frac{z^4}{5!}+...))}.$$
Usually I would ignore the condition about the disc that {$0<|z|<\pi$}
and use the fact for small enough $z$, $$\left|\frac{z^2}{3!}-\frac{z^4}{5!}+...\right|$$ is smaller than $1$ we can use $\frac{1}{1-z}=1+z+z^2...$
so $$\frac{1}{(1-(\frac{z^2}{3!}-\frac{z^4}{5!}+...))}=1+(\frac{z^2}{3!}-\frac{z^4}{5!}+...)+(\frac{z^2}{3!}-\frac{z^4}{5!}+...)^2...$$ so
\begin{align}
\frac{1}{z}\times\frac{1}{(1-(\frac{z^2}{3!}-\frac{z^4}{5!}+...))}&= \frac{1}{z} \times [1+(\frac{z^2}{3!}-\frac{z^4}{5!}+...)+(\frac{z^2}{3!}-\frac{z^4}{5!}+...)^2...] \\ &=\frac{1}{z}+0+\frac{z}{3!}...
\end{align}
and the rest of the terms when you expand out. But do I need to show that $|\frac{z^2}{3!}-\frac{z^4}{5!}+...|$ is smaller than $1$ using the fact that $z$ lies in the disc first?
| $$
z\frac{1}{\sin(z)}
$$
has no singularities in $0 < |z| < \pi$ and has a limit at $z=0$ of $1$, which means that this function has a removable singularity at $z=0$. So you have a power series expansion of the following in $|z| < \pi$:
$$
z\frac{1}{\sin(z)} = 1 + a_1z + a_2 z^2+ a_3z^3+\cdots, \\ \frac{1}{\sin(z)}=\frac{1}{z}+a_1 + a_2 z + a_3 z^2 + \cdots.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A formula which gives the maximum of a series of numbers This formula gives the maximum of 3 numbers:
$$\frac{a}{2} + \frac{b}{4} + \frac{c}{4} + \frac{|b-c|}{4} + \frac{1}{2}\left|a -\frac{b}{2} - \frac{c}{2} - \frac{|b-c|}{2}\right| = \max(a,b,c)$$
I've found this over the internet, I have no idea how can one develop such a formula, and I wonder how.
What would it look like for 4 numbers ? and 5 etc.
Is it possible to have formula which gives the maximum of a series of $n$ numbers?
| Let's start with $2$ numbers. The best way to see this is to imagine the numbers on the number line.
We find the midpoint of the $2$ numbers: $\frac{a+b}{2}$.
Next, we can find half the distance of the $2$ numbers: $\frac{|a-b|}{2}$
Adding them up, we have $\max(a, b)=\frac{a+b}{2}+\frac{|a-b|}{2}$.
The formula you have for $3$ numbers is found by expanding $\max(a, \max(b, c))$.
You can continue with any number of variables, if we want to find the maximum of $a_1, a_2, a_3, \dots, a_n$, we can evaluate it slowly:
$$\max(a_1, \max(a_2, \max(a_3, \dots\max(a_{n-1}, a_n)\dots)))$$
Because this post may take up too much space if I added too many expressions, I would write the expression for $n=5$:
$\frac{a_{1}}{2}+\frac{1}{2}\left(\frac{a_{2}}{2}+\frac{1}{2}\left(\frac{a_{3}}{2}+\frac{1}{2}\left(\frac{a_{4}}{2}+\frac{1}{2}\left(a_{5}\right)+\frac{1}{2}\left|a_{4}-a_{5}\right|\right)+\frac{1}{2}\left|a_{3}-\frac{a_{4}}{2}-\frac{1}{2}\left(a_{5}\right)-\frac{1}{2}\left|a_{4}+a_{5}\right|\right|\right)+\frac{1}{2}\left|a_{2}-\frac{a_{3}}{2}-\frac{1}{2}\left(\frac{a_{4}}{2}+\frac{1}{2}\left(a_{5}\right)+\frac{1}{2}\left|a_{4}-a_{5}\right|\right)-\frac{1}{2}\left|a_{3}+\frac{a_{4}}{2}-\frac{1}{2}\left(a_{5}\right)-\frac{1}{2}\left|a_{4}+a_{5}\right|\right|\right|\right)+\frac{1}{2}\left|a_{1}-\frac{a_{2}}{2}-\frac{1}{2}\left(\frac{a_{3}}{2}+\frac{1}{2}\left(\frac{a_{4}}{2}+\frac{1}{2}\left(a_{5}\right)+\frac{1}{2}\left|a_{4}-a_{5}\right|\right)+\frac{1}{2}\left|a_{3}-\frac{a_{4}}{2}-\frac{1}{2}\left(a_{5}\right)-\frac{1}{2}\left|a_{4}+a_{5}\right|\right|\right)-\frac{1}{2}\left|a_{2}+\frac{a_{3}}{2}-\frac{1}{2}\left(\frac{a_{4}}{2}+\frac{1}{2}\left(a_{5}\right)+\frac{1}{2}\left|a_{4}-a_{5}\right|\right)-\frac{1}{2}\left|a_{3}+\frac{a_{4}}{2}-\frac{1}{2}\left(a_{5}\right)-\frac{1}{2}\left|a_{4}+a_{5}\right|\right|\right|\right|$
Here I include the C++ program used to generate the expression in $\LaTeX$.
#include <cstdio>
#include <algorithm>
using namespace std;
/**
Prints out the maximum of the variables
a_{printed+1}...a_n
sign is used to switch the + and - signs.
**/
void printMax(int n, int printed, int sign) {
if (printed+1 == n) {
printf("a_{%d}", n);
return;
}
char p = '+', m = '-';
if (sign < 0) p = '-', m = '+';
printf("\\frac{a_{%d}}{2}%c\\frac{1}{2}\\left(", printed+1, p);
if (printed<n) printMax(n, printed+1, 1);
printf("\\right)%c\\frac{1}{2}\\left|a_{%d}%c", p, printed+1, m);
if (printed<n) printMax(n, printed+1, -1);
printf("\\right|");
}
int main(void) {
//freopen("maxClosedFormOut.txt", "w", stdout);
int N;
scanf("%d", &N);
printf("$");
printMax(5, 0, 1);
printf("$");
return 0;
}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
General solution for matrix inverse I don't know if anyone know about this,
but solving gpcm(generalized partial credit model) requires the inverse of the matrix of the form below.
in Mathetmatica langauge,
{{b1, -1, 0},{0, b2, -1},{1,1,1}} ^-1
{{b1, -1, 0, 0},{0, b2, -1, 0},{0,0,b3,-1},{1,1,1,1}} ^-1
cf) http://www.wolframalpha.com/input/?i=+%7B%7Bb1%2C+-1%2C+0%7D%2C%7B0%2C+b2%2C+-1%7D%2C%7B1%2C1%2C1%7D%7D+%5E-1
Actually What we need is the last column...
Anyhow, does anyone know how to inverse the matrix of the form above...
in other words, how to inverse the matrix below?
$\left(\begin{matrix} B1 & -1 & 0 & 0 & ... & 0 & 0 \\
0 & B2 & -1 & 0 & ... & 0 & 0\\
0 & 0 & B3 & -1 & ... & 0 & 0\\
0 & 0 & 0 & B4 & ... & 0 & 0\\
& & &\vdots & \ddots \\
0 & 0 & 0 & 0 & ... & B_n & -1\\
1 & 1 & 1 & 1 & ... & 1 & 1 \end{matrix}\right)$
| Denote your matrix by $B$ and the inverse matrix by $B^{-1} = A$. Let us write
$$ B \begin{pmatrix} x_1 \\ \vdots \\ x_n \\ x_{n+1} \end{pmatrix} = \begin{pmatrix} b_1 x_1 - x_2 \\ \vdots \\ b_n x_n - x_{n-1} \\ \sum_{i=1}^{n+1} x_i \end{pmatrix} = \begin{pmatrix} y_1 \\ \vdots \\ y_n \\ y_{n+1} \end{pmatrix}. $$
We also have by definition
$$ A \begin{pmatrix} y_1 \\ \vdots \\ y_n \\ y_{n+1} \end{pmatrix} = \begin{pmatrix} a_{1,1}y_1 + \cdots + a_{1,n+1}y_{n+1} \\ \vdots \\ a_{n+1,1}y_1 + \cdots + a_{n+1,n+1}y_{n+1} \end{pmatrix} = \begin{pmatrix} x_1 \\ \vdots \\ x_n \\ x_{n+1} \end{pmatrix}. $$
Since we want to find only the last column of $A$, we need to express $x_i$ in terms of $y_i$ and find the coefficient of $y_{n+1}$ - this will be $a_{i,n+1}$. We have the equations
$$ b_i x_i - x_{i+1} = y_i \implies x_{i+1} = b_i x_i + y_i, \,\,\, \forall 1 \leq i \leq n. \tag{1} $$
Applying them recursively, we find that
$$ x_{i+1} = b_i x_i + y_i = b_i (y_{i-1} + b_{i-1} x_{i-1}) + y_i = b_i b_{i-1} x_{i-1} + b_i y_{i-1} + y_i = \ldots \\
= \left( \prod_{j=1}^i b_j \right) x_1 - \sum_{j=1}^i \left( \prod_{k=j+1}^i b_k \right) y_j.$$
Plugging the $x_i$ into the equation
$$ x_1 + \ldots + x_{n+1} = y_{n+1} $$
we get
$$ (1 + b_1 + b_1 b_2 + \ldots + b_1 \ldots b_n)x_1 = y_{n+1} + \star $$
where $\star$ involves only $y_n, \ldots, y_1$. Thus, we have
$$ a_{1,n+1} = \frac{1}{1 + b_1 + b_1 b_2 + \ldots + b_1 \ldots b_n}. $$
Returning back to equation $(1)$, we see that the expression $x_{i+1}$ depend on $y_{n+1}$ only through $x_1$ and so
$$ a_{i,n+1} = \frac{\prod_{j=1}^{i-1} b_j}{1 + b_1 + b_1 b_2 + \ldots + b_1 \ldots b_n} $$
for all $1 \leq i \leq n + 1$. This can be used to find all the other entries of $A = B^{-1}$ as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that if $a,b,$ and $c$ are positive real numbers, then $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \geq ab + bc + ca$.
Prove that if $a,b,$ and $c$ are positive real numbers, then $\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a} \geq ab + bc + ca$.
I tried AM-GM and it doesn't look like AM-GM or Cauchy-Schwarz work here. The $ab+bc+ca$ reminds of a cyclic expression, so that may help by factoring the inequality and getting a true statement.
| First I'll prove a Lemma: $a^2+b^2+c^2\ge ab+bc+ca$ for all $a,b,c\in\mathbb R$.
Proof: it follows from the Rearrangement Inequality, because $(a,b,c)$ and $(a,b,c)$ are similarly sorted.
Or notice that it's equivalent to $\frac{1}{2}\left((a-b)^2+(b-c)^2+(c-a)^2\right)\ge 0$, which is true; or add the following inequalities: $a^2+b^2\ge 2ab$, $b^2+c^2\ge 2bc$, $c^2+a^2\ge 2ca$. $\ \square$
Your inequality is cyclic. Wlog there are two cases:
*
*$a\ge b\ge c$. Then $(a,b,c)$ and $(1/a,1/b,1/c)$ are oppositely sorted, so
$$\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge \frac{a^3}{a}+\frac{b^3}{b}+\frac{c^3}{c}=a^2+b^2+c^2$$
Now use the Lemma.
*$a\ge c\ge b$. Then $(a,c,b)$ and $(1/a,1/c,1/b)$ are oppositely sorted, so $$\frac{a^3}{b}+\frac{c^3}{a}+\frac{b^3}{c}\ge \frac{a^3}{a}+\frac{c^3}{c}+\frac{b^3}{b}=a^2+c^2+b^2$$
Now use the Lemma.
In your inequality, equality holds if and only if $a=b=c$.
Another proof: By Hölder's inequality:
$$\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge \frac{(a+b+c)^3}{3(a+b+c)}=\frac{(a+b+c)^2}{3}$$
Also $(a+b+c)^2\ge 3(ab+bc+ca)$, because this is equivalent to $a^2+b^2+c^2\ge ab+bc+ca$ (see the Lemma).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 0
} |
Infinite primes proof There is a proof for infinite prime numbers that i don't understand.
right in the middle of the proof:
"since every such $m$ can be written in a unique way as a product of the form:
$\prod_{p\leqslant x}p^{k_p}$. we see that the last sum is equal to: $\prod_{\binom{p\leqslant x}{p\in \mathbb{P}}}(\sum_{k\leqslant 0}\frac{1}{p^k})$.
I don't see that. can anyone can explain this step to me?
| Suppose the prime numbers not exceeding $x$ are $2$, $3$, and $5$. Then
\begin{align}
& \prod_{\begin{smallmatrix} p\in\mathbb P \\ p\le x \end{smallmatrix}} \sum_{k\ge 0} \frac 1 {p^k} \\[10pt] = {} & \left( 1 + \frac 1 2 + \frac 1 4 + \frac 1 8 + \cdots+ \frac 1 {2^k} + \cdots \right) \left( 1 + \frac 1 3 + \frac 1 9 + \frac 1 {27} + \cdots+ \frac 1 {3^k} + \cdots \right) \times {} \\
{} & {} \times \left( 1 + \frac 1 5 + \frac 1 {25} + \frac 1 {125} + \cdots+ \frac 1 {5^k} + \cdots \right) \tag a \\[12pt]
= {} & 1 + \frac 1 2 + \frac 1 3 + \frac 1 4 + \frac 1 5 + \frac 1 6 + \frac 1 8 + \frac 1 9 + \frac 1 {10} + \frac 1 {12} + \frac 1 {15} + \frac 1 {16} \\[6pt]
{} & {} + \frac 1 {18} + \frac 1 {20} + \frac 1 {24} + \frac 1 {25} + \frac 1 {30} + \frac 1 {32} + \frac 1 {36} + \cdots
\end{align}
In this last sum we exclude $1/7$, $1/11$, $1/13$, $1/14$, etc. and include only numbers $1/m$ where $m$ has no prime factors other than $2$, $3$, and $5$. This last series must converge to a finite number because all three of the series in the line labeled (a) converge to finite numbers.
By contrast, the harmonic series $\displaystyle\sum_{m=1}^\infty \frac 1 m$ diverges to $+\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
If $2$ is subtracted from each root,the results are reciprocals of the original roots.Find the value of $b^2+c^2+bc.$ The equation $x^2+bx+c=0$ has distinct roots .If $2$ is subtracted from each root,the results are reciprocals of the original roots.Find the value of $b^2+c^2+bc.$
Let $\alpha$ and $\beta$ are the roots of the equation $x^2+bx+c=0$.
According to the question,
$\alpha-2=\frac{1}{\alpha}$ and $\beta-2=\frac{1}{\beta}$
$\alpha^2-2\alpha-1=0$ and $\beta^2-2\beta-1=0$
Adding the two equations,
$\alpha^2+\beta^2-2\alpha-2\beta-2=0$
$(\alpha+\beta)^2-2\alpha\beta-2\alpha-2\beta-2=0$
$(-b)^2-2c-2(-b)-2=0$
$b^2+2b-2c-2=0$
But i am not able to find $b^2+c^2+bc=0.$What should i do now?I am stuck here.
| Hint...$\alpha^2-2\alpha-1=0\Rightarrow b=-2, c=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
To find constant term of polynomial defined recursively
Let $\{f_n(x)\}$ be a sequence of polynomials defined inductively as
\begin{align*}
f_1(x) &= (x-2)^2, \\
f_{n+1}(x) &= (f_n(x)-2)^2, \quad n \geq 1.
\end{align*}
Let $a_n$ and $b_n$ respectively denote the constant term and the coefficient of $x$ in $f_n(x)$. Then
*
*$a_n = 4$, $b_n = -4^n$,
*$a_n = 4$, $b_n = -4n^2$,
*$a_n = 4^{(n-1)!}$, $b_n = -4^n$,
*$a_n = 4^{(n-1)!}, b_n = -4n^2$.
I see that general or $n$-th term of this will be $((x-2)^2 - 2)^2-2)^2...$ But how do I simplify this to get to the constant term
| To expand on the comment by Vinod, for any polynomial, we get the constant term by plugging 0 in for $x$. In fact, that's pretty much the definition of the "constant term". Doing this we see that we will get
$$\begin{align*}
(\cdots (0-2)^2 - 2)^2 - \cdots - 2)^2 &= (\cdots ((-2)^2-2)^2 - 2)^2 - \cdots - )^2 \\
&= (\cdots (4-2)^2 - 2)^2 - \cdots - 2)^2 \\
&= (\cdots (2)^2 - 2)^2 - \cdots - 2)^2 \\
&= (\cdots (4 - 2)^2 - \cdots - 2)^2 \\
&= \cdots \\
&= 4
\end{align*}$$
(We continually remove a set of parentheses in each step, and eventually end up with $4$ as our constant term.)
Now onto the coefficient of $x$, we can perform a similar analysis. We first note that the coefficient of $x$ for $n=1$ is $-4$:
$$f_1(x) = (x-2)^2 = x^2 - 4x + 4$$
Now we prove by induction that the coefficient is $-4^n$.
Assume that the coefficient of $x$ is $-4^n$ for $f_n$. Then we know $f_{n+1}(x)$ is defined as
$$ f_{n+1}(x) = (f_n(x) - 2)^2$$
We know three very important things about this equation. First, by the inductive hypothesis, we know that the coefficient of $x$ for $f_n$ is $-4^n$, we know by the previous step that the constant term of $f_n$ is $4$, and we know that we don't care about the other terms. Observe:
$$\begin{align*}
(f_n(x) - 2)^2 &= (c_n x^n + \cdots + c_2 x^2 - 4^n x + 4 - 2)^2 \\
&= (c_n x^n + \cdots + c_2 x^2 - 4^n x + 2)^2 \\
&= (c_n x^n + \cdots + c_2 x^2 - 4^n x + 2) \cdot (c_n x^n + \cdots + c_2 x^2 - 4^n x + 2) \\
&= \cdots - 4^n x \cdot 2 - 4^n x \cdot 2 + 2 \cdot 2 \\
&= \cdots - 4 \cdot 4^n x + 4 \\
&= \cdots - 4^{n+1} x + 4
\end{align*}$$
as desired. Note that the $\cdots$ in the last few equations are all higher order terms, i.e. when multiplying $(c_n x^n + \cdots + c_2 x^2 - 4^n x + 2)$ by itself, the $\cdots$ represent all of the terms whose exponents are 2 or greater.
Finally, it is worth noting that once you know the constant term is 4, you can manually calculate the $f_1$, $f_2$, and $f_3$ by hand and see that the coefficients for $x$ are $-4^1 = -4$, $-4^2 = -16$, and $-4^3 = -64$ respectively, and then conclude (since it is a multiple choice problem), that the answer is $a_n = 4$ and $b_n = -4^n$, but where's the fun in that?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the integral $\int \frac{1}{x^3\sqrt{x^2+x-1}}\mathrm dx$ What substitution to use in this integral? I tried to factorize $x^2+x-1$ and use a substitution $u=\sqrt{\frac{a(x-\alpha)}{x-\beta}}$.
| $$I=\int \frac{1}{x^3\sqrt{x^2+x-1}}dx$$
Euler Substitution:
Let $x=\frac{u^2+1}{2u+1}$, $$I=2\int \frac{(2u+1)^2}{(u^2+1)^3}du$$
Partial fraction get $I=2\int \frac{4u-3}{(u^2+1)^3}du+8\int\frac{1}{(u^2+1)^2}du$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the integral $\int x^{\frac{-4}{3}}(-x^{\frac{2}{3}}+1)^{\frac{1}{2}}\mathrm dx$ $$x^{\frac{-4}{3}}(-x^{\frac{2}{3}}+1)^{\frac{1}{2}}=\frac{\sqrt{(-\sqrt[3]{x^2}+1)}}{\sqrt[3]{x^4}}$$
Is it necessary to simplify the function further? What substitution is useful?
$u=\sqrt[n]{\frac{ax+b}{cx+d}}$ doesn't work.
| Notice, $$\int x^{-4/3}\left(-x^{2/3}+1\right)^{1/2}\ dx$$$$=\int \frac{1}{x}\left(-x^{2/3}+1\right)^{1/2}(x^{-1/3}\ dx)$$
Let $-x^{2/3}+1=\sin^2\theta\implies -\frac{2}{3}x^{-1/3}\ dx=2\sin\theta\cos\theta\ d\theta$ or $x^{-1/3}\ dx=-3\sin\theta\cos\theta\ d\theta $
$$=\int\frac{1}{(1-\sin^2\theta)^{3/2}}(\sin\theta)(-3\sin\theta\cos\theta\ d\theta )$$
$$=-3\int\frac{\sin^2\theta\cos\theta}{\cos^3\theta}\ d\theta$$
$$=-3\int\tan^2\theta\ d\theta$$
$$=-3\int(\sec^2\theta-1)\ d\theta$$
$$=-3(\tan\theta-\theta)+C$$
$$=3\theta-3\tan\theta+C$$
substituting back the value of $\theta$,
$$=\color{red}{3\sin^{-1}(\sqrt{1-x^{2/3}}) \ - \ 3 \frac{\sqrt{x^{-2/3}-1}}{x^{1/3}} \ + \ C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\sum_{i = 1}^n \frac{x_i}{i^2} \geq \sum_{i = 1}^n \frac{1}{i}.$
Let $x_1,x_2,\ldots,x_n$ be distinct positive integers. Prove that $$\displaystyle \sum_{i = 1}^n \dfrac{x_i}{i^2} \geq \sum_{i = 1}^n \dfrac{1}{i}.$$
Attempt
I tried using Cauchy-Schwarz and I got that $$(x_1^2+x_2^2+\cdots+x_n^2) \left (\dfrac{1}{1^2}+\dfrac{1}{2^2}+\cdots+ \dfrac{1}{n^2} \right ) \geq \left ( \dfrac{x_1}{1}+\dfrac{x_2}{2}+\cdots+\dfrac{x_n}{n} \right)^2 = \displaystyle \left (\sum_{i = 1}^n \dfrac{x_i}{i} \right)^2,$$ but this doesn't seem to help.
| Let $0 < y_1 < y_2 < \ldots < y_n$ be an rearrangement of $x_1, x_2, \ldots x_n$ sorted in ascending order.
Since $y_k$ are distinct positive integers, one can show $y_k \ge k$ for $k = 1,\ldots, n$ by induction.
Since the finite sequence $\frac{1}{1^2}, \frac{1}{2^2}, \ldots, \frac{1}{n^2}$ is ascending, by rearrangement inequality, we have
$$\sum_{k=1}^n \frac{x_k}{k^2} \ge \sum_{k=1}^n \frac{y_k}{k^2} \ge \sum_{k=1}^n \frac{k}{k^2} = \sum_{k=1}^n \frac{1}{k}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Why is $(1+\frac{3}{n})^{-1}=(1-\frac{3}{n}+\frac{9}{n^2}+o(\frac{1}{n^2}))$ and how to get around the Taylor expansion?
Let be $(u_n)$ a real sequence such that $u_0>0$ and that $\forall n \in \mathbb{R}$:
$$\frac{u_{n+1}}{u_n}=\frac{n+1}{n+3}$$
Let be $(v_n)$ a real sequence such that $\forall n \in \mathbb{R}$:
$$v_n=n^2u_n$$
Let's determine the nature of $\sum\ln(\frac{v_{n+1}}{v_n})$
I did:
\begin{align*}
\frac{v_{n+1}}{v_n} &=\left(\frac{n+1}{n}\right)^2\frac{u_{n+1}}{u_n}\\
&=\left(1+\frac{1}{n}\right)^2\frac{n+1}{n+3}\\
&=\left(1+\frac{1}{n}\right)^3\left(1+\frac{3}{n}\right)^{-1}\\
\end{align*}
and there I was stuck. A friend of mine gave me this tip:
$$=\left(1+\frac{3}{n}+\frac{3}{n^2}+\frac{1}{n^3}\right)\left(1-\frac{3}{n}+\frac{9}{n^2}+o\left(\frac{1}{n^2}\right)\right)$$
But I don't understand this notation $o\left(\frac{1}{n^2}\right)$ I know that it means that its negligible. I checked on wikipedia that it describes the limiting behavior of a function when the argument tends towards a particular value or infinity
But still, I'm stuck:
*
*what does that mean, how do we end up to it?
I think it is related to the Taylor expansion of $\left(1+\frac{3}{n}\right)^{-1}$
because $(1+x)^\alpha=1+\alpha x+\frac{\alpha(\alpha -1)}{(2)!}x^2 + x^n\epsilon(x)$
then $\left(1+\frac{3}{n}\right)^{-1}=...$
but still I'm not very smart at usual Taylor developments so...
*is there a way to get around?
| As jordan's comment notes, the sum is telescoping: $$\begin{align*} \sum_{n=1}^N \log \frac{v_{n+1}}{v_n} &= \sum_{n=1}^N \log v_{n+1} - \log v_n \\ &= \log v_{N+1} + \sum_{n=2}^{N} \log v_n - \sum_{n=2}^N \log v_n - \log v_1 \\ &= \log v_{N+1} - \log v_1 \\ &= \log \frac{v_{N+1}}{v_1}. \end{align*}$$ Next, it is easy to see that $$\frac{u_{N+1}}{u_1} = \prod_{n=1}^N \frac{u_{n+1}}{u_n} = \prod_{n=1}^N \frac{n+1}{n+3} = \frac{(2)(3)}{(N+2)(N+3)}.$$ Therefore, the finite sum is $$\log \frac{(N+1)^2u_{N+1}}{u_1} = \log \frac{6(N+1)^2}{(N+2)(N+3)},$$ and the limit as $N \to \infty$ is simply $$\lim_{N \to \infty} \log \frac{6(N^2 + 2N + 1)}{N^2 + 5N + 6} = \log 6 + \lim_{N \to \infty} \log \frac{1 + 2N^{-1} + 1N^{-2}}{1 + 5N^{-1} + 6N^{-2}} = \log 6. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A trigonometric identities with the ratio of four terms like $1+(\frac{\tan x}{\sin y})^2$ Prove:
$$\frac{1+\left(\frac{\tan x}{\sin y}\right)^2}{1+\left(\frac{\tan x}{\sin z}\right)^2}=\frac{1+\left(\frac{\sin x}{\tan y}\right)^2}{1+\left(\frac{\sin x}{\tan z}\right)^2}$$
I started by opening the brackets and squaring but did not get the required answer.
| Notice, the given equality can be easily proved by simplifying $LHS$ $$LHS=\frac{1+\left(\frac{\tan x}{\sin y}\right)^2}{1+\left(\frac{\tan x}{\sin z}\right)^2}$$
$$=\frac{1+\left(\frac{\sin x}{\sin y\cos x}\right)^2}{1+\left(\frac{\sin x}{\sin z\cos x}\right)^2}$$
$$=\frac{\sin^2z(\sin^2 y\cos^2 x+\sin^2 x)}{\sin^2y(\sin^2 z\cos^2 x+\sin^2 x)}$$
$$=\frac{\sin^2z((1-\cos^2 y)(1-\sin^2 x)+\sin^2 x)}{\sin^2y((1-\cos^2 z)(1-\sin^2 x)+\sin^2 x)}$$
$$=\frac{\sin^2z(1-\cos^2 y-\sin^2 x+\sin^2 x\cos^2y+\sin^2 x)}{\sin^2y(1-\cos^2 z-\sin^2 x+\sin^2 x\cos^2 z+\sin^2 x)}$$
$$=\frac{\sin^2z((1-\cos^2 y)+\sin^2 x\cos^2y)}{\sin^2y((1-\cos^2 z)+\sin^2 x\cos^2 z)}$$
$$=\frac{\sin^2z(\sin^2 y+\sin^2 x\cos^2y)}{\sin^2y(\sin^2 z+\sin^2 x\cos^2 z)}$$
$$=\frac{\sin^2z\sin^2 y\left(1+\frac{\sin^2 x\cos^2y}{\sin^2 y}\right)}{\sin^2y\sin^2 z\left(1+\frac{\sin^2 x\cos^2 z}{\sin^2 z}\right)}$$
$$=\frac{1+\frac{\sin^2 x}{\tan^2 y}}{1+\frac{\sin^2 x}{\tan^2 z}}$$
$$=\frac{1+\left(\frac{\sin x}{\tan y}\right)^2}{1+\left(\frac{\sin x}{\tan z}\right)^2}=RHS$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Line integral of a vector field $$\int_{\gamma}ydx+zdy+xdz$$
given that $\gamma$ is the intersection of $x+y=2$ and $x^2+y^2+z^2=2(x+y)$ and its projection in the $xz$ plane is taken clockwise.
In my solution, I solved the non-linear system of equations and I found that $x^2+y^2+z^2 = 4$. Given the projection in the $xz$ plane is taken clockwise, the parametrization is: $\gamma(t) = (2\sin t, 2 - 2\sin t,2\cos t)$, $0<t<2\pi$. But when I evaluate this integral, I keep getting wrong results:
$\int_{0}^{2\pi}[(2-2\cos t)2\cos t+2\cos t(-2\cos t)+2\sin t(-2\sin t)]dt$, I've only susbstituted $x,y,z$ and $dx,dy,dz$; which leads me to $-8\pi$, but the answer on my textbook is $-2\pi \sqrt{2}$.
Is there any conceptual mistake in my solution?
| Let us just rewrite some equations first
$$\begin{align}
x^2+y^2+z^2&=2(x+y)\\
(x^2-2x+1)+(y^2-2y+1)+z^2-2&=0 \\
(x-1)^2+(y-1)^2+z^2&=2 \\
\end{align}$$
So this is a sphere of radius $R=\sqrt{2}$ centered at $(1,1,0)$. It's parametric equations are
$$\begin{align}
x&=1+\sqrt{2}\sin u \cos v \\
y&=1+\sqrt{2}\sin u \sin v \\
z&=\sqrt{2} \cos u
\end{align}, \qquad 0 \le u \le \pi, \quad 0 \le v \lt 2\pi$$
Now, put these parametric equations into $x+y=2$ to get
$$\begin{align}
x+y=2+\sqrt{2} &\sin u (\cos v + \sin v)=2 \\
&\sin u (\cos v + \sin v) =0 \\
\end{align}$$
and hence according to the range of $u$ and $v$ we have
$$\begin{array}{}
u=0 & \text{Not acceptable as it just corresponds to a special point $(1,1,\sqrt{2})$} \\
v=\frac{3\pi}{4}, \frac{7\pi}{4} & \text{You can choose any of them as they just effect the orientation of the intersection curve}
\end{array}$$
and finally by choosing $v=\frac{7\pi}{4}$ which gives a clock orientation we get the parametric equations as
$$\boxed{
\begin{align}
x&=1+\sin u \\
y&=1-\sin u \\
z&=\sqrt{2} \cos u
\end{align}, \qquad 0 \le u \lt 2\pi
}$$
so your parametric equations were wrong. Here is a picture which helps you to visualize better
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Let $0 \le a \le b \le c$ and $a+b+c=1$. Show that $a^2+3b^2+5c^2 \ge 1$
Let $0 \le a \le b \le c$ and $a+b+c=1$. Show that $a^2+3b^2+5c^2
\ge 1$.
My solution: since $a+b+c=1$ we have to show that $a^2+3b^2+5c^2\ge1=a+b+c$
Since $a,b,c \ge 0 $ the inequality is true given that every term on the left hand side of the inequality is greater or equal to the corresponding term on the right.
However I am not sure if I am reasoning correctly, as the hint from my book seems to depict the problem in a harder way than I am ,as it suggests to square the expression $a+b+c=1$ and so on...
So my question is wheter I am overlooking some detail in the problem which makes my solution inadequate.
| Based on the inequality
\begin{align*}
\frac{x^2}{a} + \frac{y^2}{b} +\frac{z^2}{c} \ge \frac{(x+y+z)^2}{a+b+c}.
\end{align*}
we have shown in This Question, for any positive $x$, $y$, $z$, $a$, $b$, and $c$, we have that
\begin{align*}
a^2 + 3b^2 + 5c^2 &\ge 3a^2 + 3 b^2+3c^2\\
&\ge \frac{(a+b+c)^2}{\frac{1}{3} + \frac{1}{3}+\frac{1}{3}} =1.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding the common integer solutions to $a + b = c \cdot d$ and $a \cdot b = c + d$ I find nice that $$ 1+5=2 \cdot 3 \qquad 1 \cdot 5=2 + 3 .$$
Do you know if there are other integer solutions to
$$ a+b=c \cdot d \quad \text{ and } \quad a \cdot b=c+d$$
besides the trivial solutions $a=b=c=d=0$ and $a=b=c=d=2$?
| Here's a solution when $a$, $b$, $c$, and $d$ are positive; I'm not sure how to tackle the more general problem though (although something tells me it's probably similar).
Note that substituting $b=cd-a$ into the second equation yields $$a(cd-a)=c+d\implies a^2-cad+c+d=0.$$ By the Quadratic Formula on $a$, we have $$a=\dfrac{cd\pm\sqrt{(cd)^2-4(c+d)}}2.$$ This can only be an integer when $(cd)^2-4(c+d)$ is a perfect square.
Note that since $(cd)^2-4(c+d)\equiv (cd)^2\pmod 2$, we have $$(cd)^2-4(c+d)\leq(cd-2)^2.$$ This rearranges to $$-4(c+d)\leq -4cd+4\implies 2\geq (c-1)(d-1).$$ This is great news, because it means we actually don't have many cases to check!
CASE 1: $c=1$ (or $d=1$, for that matter).
Plugging back in, we see that for some integer $K$ we have $$d^2-4(d+1)=(d-2)^2-8=K^2.$$ The only case that works here is $d=5$ (since $d=6$ fails and past that point perfect squares differ by at least $9$). By symmetry, $c=5$, $d=1$ also works.
CASE 2: $c=2$ and $d\leq 3$.
Plugging back in to the original expression, we need $$(2d)^2-4(2+d)=(2d-1)^2-9$$ to be a perfect square. Trial and error works since $d$ is small, but here $d=2$ and $d=3$ both work.
CASE 3: $c=3$ and $d=2$.
Oh look, we've already shown this works! (We don't need to check $d=1$ because that was already covered in Case 1.)
Hence the only possible pairs for $(c,d)$ are $(2,3)$, $(2,2)$, $(5,1)$, arrangements. By a similar argument, we see that the same exact possible pairs are the only possible ones for $(a,b)$ as well.
From here, it's not hard to see that $(a,b,c,d)=\boxed{(2,2,2,2),(2,3,5,1)}$ and permutations are the only possible solutions for $a,b,c,d>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 0
} |
Finding the sum of the infinite series whose general term is not easy to visualize: $\frac16+\frac5{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\cdots$ I am to find out the sum of infinite series:-
$$\frac{1}{6}+\frac{5}{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+...............$$
I can not figure out the general term of this series. It is looking like a power series as follows:-
$$\frac{1}{6}+\frac{5}{6^2\cdot2!}+\frac{5\cdot8}{6^3\cdot3!}+\frac{5\cdot8\cdot11}{6^4\cdot4!}+.....$$
So how to solve it and is there any easy way to find out the general term of such type of series?
| Notice $$
\begin{align}
& \frac{1}{6} + \frac{5}{6\cdot 12} + \frac{5\cdot 8}{6\cdot 12\cdot 18} + \cdots\\
= & \frac12 \left[ \frac{3-1}{6} + \frac{(3-1)(6-1)}{6^2 \cdot 2!} + \frac{(3-1)(6-1)(9-1)}{6^33!} + \cdots\right]\\
= & \frac12 \left[ \frac{\frac23}{2} + \frac{\frac23(\frac23+1)}{2^2\cdot 2!}
+ \frac{\frac23(\frac23+1)(\frac23+2)}{2^3\cdot 3!} + \cdots \right]
\end{align}
$$
The sum we want has the form
$\displaystyle\;\frac12 \sum_{k=1}^\infty \frac{\prod_{\ell=0}^{k-1}(\frac23 + \ell)}{2^k k!}$.
Compare this with the expansion
$$\frac{1}{(1-z)^\alpha} = \sum_{k=0}^\infty \frac{(\alpha)_k}{k!} z^k
\quad\text{ where }\quad
(\alpha)_k = \prod\limits_{\ell=0}^{k-1}( \alpha + \ell )$$
We find the desired sum equals to
$$\frac12 \left[\frac{1}{\left(1-\frac12\right)^{2/3}} - 1\right] = \frac12\left( 2^{2/3} - 1 \right)
\approx
0.293700525984...
$$
Notes
After I finish this answer, I have a déjà vu feeling that I have seen this before
(more than once).
Following are some similar questions I can locate, I'm sure there are more...
*
*Sum of a Hyper-geometric series. (NBHM 2011)
*Find a closed form for this infinite sum: $ 1+\frac 1 2 +\frac{1 \times2}{2 \times 5}+\frac{1 \times2\times 3}{2 \times5\times 8}+ \dots$
*Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
If $a^2 + b^2 = 1$, show there is $t$ such that $a = \frac{1 - t^2}{1 + t^2}$ and $b = \frac{2t}{1 + t^2}$ My question is how we can prove the following:
If $a^2+b^2=1$, then there is $t$ such that $$a=\frac{1-t^2}{1+t^2} \quad \text{and} \quad b=\frac{2t}{1+t^2}$$
| The easiest way is by brute checking.
\begin{align*}
a^2&=\frac{(t^2-1)^2}{(t^2+1)^2}\\
b^2&=\frac{4t^2}{(t^2+1)^2}\\
a^2+b^2&=\frac{t^4-2t^2+1+4t^2}{(t^2+1)^2}\\
&=\frac{t^4+2t^2+1}{(t^2+1)^2}\\
&=\frac{(t^2+1)^2}{(t^2+1)^2}\\
&=1
\end{align*}
So we know that if we can find a $t$ everything works out fine. Now observe that if $t=\frac{b}{a+1}$ the two expressions for $a$ and $b$ in terms of $t$ hold, so for any point in the circle other than $(-1,0)$ this works.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Prove that $\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b} \geq 1$
For three positive real numbers $a,b,$ and $c$, prove that $$\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b} \geq 1.$$
Attempt
Rewritting we obtain $\dfrac{2 a^3+2 a^2 b-3 a^2 c-3 a b^2-3 a b c+2 a c^2+2 b^3+2 b^2 c-3 b c^2+2 c^3}{(a+2b)(2a+c)(b+2c)} \geq 0$. Then to I proveed to use rearrangement, AM-GM, etc. on the numerator?
| Use Cauchy–Schwarz's inequality as follows :
$$(a(b+2c)+b(c+2a)+c(a+2b)) \left (\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b} \right ) \geq (a+b+c)^2$$ and then :
$$\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b} \geq \frac{(a+b+c)^2}{3(ab+bc+ca)}$$
Now it's easy to see that $$\frac{(a+b+c)^2}{3(ab+bc+ca)} \geq 1$$ this being equivalent with $$a^2+b^2+c^2 \geq ab+bc+ca$$ which is in turn equivalent with $$(a-b)^2+(b-c)^2+(c-a)^2 \geq 0$$
(after a multiplication with $2$ and rearranging )
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1602418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
$AB$ is any chord of the circle $x^2+y^2-6x-8y-11=0,$which subtend $90^\circ$ at $(1,2)$.If locus of mid-point of $AB$ is circle $x^2+y^2-2ax-2by-c=0$ $AB$ is any chord of the circle $x^2+y^2-6x-8y-11=0,$which subtend $90^\circ$ at $(1,2)$.If locus of mid-point of $AB$ is circle $x^2+y^2-2ax-2by-c=0$.Find $a,b,c$.
The point $(1,2)$ is inside the circle $x^2+y^2-6x-8y-11=0$.I let the points $A(x_1,y_1)$ and $B(x_2,y_2)$ are the end points of the chord $AB$.As $AB$ subtend $90^\circ$ at $(1,2)$
So $\frac{y_1-2}{x_1-1}\times \frac{y_2-2}{x_2-1}=-1$
But i do not know how to find the locus of mid point of chord $AB$ $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$.
| Let $C(3,4),D(1,2)$. Also, let $E(X,Y)$ be the midpoint of $AB$.
$\qquad\qquad\qquad$
Since $\triangle{CAE}$ is a right triangle with
$$|AC|=6,\quad |CE|=\sqrt{(X-3)^2+(Y-4)^2}$$
we have
$$|AE|^2=|AC|^2-|CE|^2=36-(X-3)^2-(Y-4)^2\tag1$$
Also, since we can see that $D$ is on the circle whose diameter is the line segment $AB$, we have $$|DE|=|AE|\tag 2$$
From $(1)(2)$, we have
$$36-(X-3)^2-(Y-4)^2=(X-1)^2+(Y-2)^2,$$
i.e.
$$X^2+Y^2-2\cdot 2X-2\cdot 3Y-3=0.$$
Hence, $a=2,b=3,c=3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to compute $\lim _{x\to 0}\frac{x\bigl(\sqrt{3e^x+e^{3x^2}}-2\bigr)}{4-(\cos x+1)^2}$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it. Is it possible to compute this limit with the McLaurin expansion? Can you explain the method and the steps used? Thanks. (I prefer to avoid to use L'Hospital's rule.)
$$\lim _{x\to 0}\frac{x\bigl(\sqrt{3e^x+e^{3x^2}}-2\bigr)}{4-(\cos x+1)^2}$$
| Hint do you know about Maclaurin series?
You can write
$$\lim _{x\to 0}\left(\frac{x\left(\sqrt{3e^x+e^{3x^2}}-2\right)}{4-\left(\cos x+1\right)^2}\right) = \lim _{x\to 0}\left(\frac{x\left(\sqrt{3 \cdot (1 + x + \frac{x^2}{2} + O(x^2)) +(1 + 3x^2 + O(3x^2))}-2\right)}{4-\left((1 - \frac{x^2}{2} + O(x^2))+1\right)^2}\right) = \lim _{x\to 0} \frac{\frac34 x^2 + \frac{63}{64} x^3 + O(x^3)}{2x^2-\frac{5}{12}x^4 + O(x^6)} = \lim _{x\to 0} \frac{\frac34 x^2 + O(x^2)}{2x^2 + O(x^2)} = \frac{\frac34}{2}\\ = \frac38$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
$g(Z^3,Z^2) = 0$ implies $g(X,Y) = (X^2 - Y^3)h(X,Y)$ Let $F$ be a field and $g(X,Y) \in F[X,Y]$. Suppose that $g(Z^3,Z^2) = 0$ where $g(Z^3, Z^2) \in F[Z]$. I want to show that $g(X,Y) = (X^2 - Y^3) h(X,Y)$ for some $h(X,Y) \in F[X,Y]$.
My first idea is that $g(X,Y)$ can be expressed as $g(X,Y) = X^2 p(X,Y) +Y^3 q(X,Y) + a_1XY^2 + a_2XY + a_3Y^2 + a_4X + a_5Y + a_6$. Using the condition, we obtain $$0 = Z^6 (p(Z^3, Z^2) + q(Z^3, Z^2) + a_1Z) + a_2Z^5 + a_3 Z^4 + a_4 Z^3 + a_5 Z^2 + a_6.$$
It follows that all $a_k$ are $0$. So $p(Z^3, Z^2) + q(Z^3, Z^2) = 0$. Moreover, $1, Z^2, Z^3, Z^4, Z^5$ in these two polyomials can only be obtained be substituting for $1, Y, X, Y^2, XY$ respectively. Thus the coefficients of these monomials in $p(X,Y)$ and $q(X,Y)$ must match. Let $\tilde{p}(X,Y)$ and $\tilde{q}(X,Y)$ be the sum of the monomials which have either $X^2$ or $Y^3$ as a factor. It remains to show that $p(X,Y) = q(X,Y) \mod (X^2 - Y^3)$.
However, from here onwards I'm stuck. I would like to solve it with basic algebra, no geometry.
| It is better to start with (by division algorithm)
$$g(X,Y)=(X^2-Y^2)h_1(X,Y)+X\cdot h_2(Y)+h_3(Y)$$
Plugging $X=Z^3, Y=Z^2$ gives:
$$g(Z^3,Z^2)=(Z^6-Z^6)h_1(Z^3,Z^2)+Z^3\cdot h_2(Z^2)+h_3(Z^2)$$
which implies
$$Z^3\cdot h_2(Z^2)+h_3(Z^2)=0\quad\quad \quad \text{for all }Z $$
The first component contains odd order terms only, the second component contains even order terms only. So they must both be zeros.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do you find the value of $m$ and $n$ if $x+y+z=\frac{m}{\sqrt n}$ given certain conditions on x,y,z? Problem:
Let $x,y$ and $z$ be real numbers satisfying:
$$x=\sqrt{y^2 - \frac{1}{16}} + \sqrt{z^2 - \frac{1}{16}}$$
$$y=\sqrt{z^2 - \frac{1}{25}} + \sqrt{x^2 - \frac{1}{25}}$$
$$z=\sqrt{x^2 - \frac{1}{36}} + \sqrt{y^2 - \frac{1}{36}}$$
If $x+y+z=\frac{m}{\sqrt n}$, where $m,n \in N$ and $n$ is not divisible by the square of any prime number, then find $m$ and $n$
I don't know how to even begin with this question, so any help will be appreciated.
| Working on The hint of joey we can make this triangle and take x,y,z as the sides of triangle as follows without the loss of generality....
Now just to make a detailed figure ....
Just to sum these up ...from the figure we have ...
$$DC=x=\sqrt{y^2 - \frac{1}{16}} + \sqrt{z^2 - \frac{1}{16}}$$
$$BD=y=\sqrt{z^2 - \frac{1}{25}} + \sqrt{x^2 - \frac{1}{25}}$$
$$BC=z=\sqrt{x^2 - \frac{1}{36}} + \sqrt{y^2 - \frac{1}{36}}$$
Now note an interesting thing...the triangles $ABC,ACD,ABD$ all have equal sides that is they are congruent so there areas must also be equal ... So
Area of ABC $\frac{1}{2}×z×\frac{1}{6}$ equal to($=$)
area of ACD $\frac{1}{2}×x×\frac{1}{4}$ equal to($=$)
area of ABD $\frac{1}{2}×y×\frac{1}{5}$ equal to ($=$)
That is we can say ... $$\frac{1}{2}×y×\frac{1}{5} = \frac{1}{2}×z×\frac{1}{6}=\frac{1}{2}×x×\frac{1}{4}$$ solving which gives $x+y+z=\frac{15x}{4}$ ...now we just need to find $x$ ...any ideas from your side...
Edit: Thanks to the hint of Joey that is also easily done by using herons formula and then equating it to the (area of ACD $=1/2×x×1/4$ )
After calculating the area by Herons formula as shown in the image you will get Area as $\frac{15x^2\sqrt{7}}{64}$ which is equal to $1/2×x×1/4$ solving this you will get $x=8/15\sqrt{7}$ put this in the value of $x+y+z$ to get answer $2/\sqrt{7}$ so answer is $m=2,n=7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Limit of: $\lim_{n \to \infty}(\sqrt[3]{n^3+\sqrt{n}}-\sqrt[3]{n^3-1})\cdot \sqrt{(3n^3+1)}$ I want to find the limit of: $$\lim_{n \to \infty}(\sqrt[3]{n^3+\sqrt{n}}-\sqrt[3]{n^3-1})\cdot \sqrt{(3n^3+1)}$$
I tried expanding it by $$ \frac{(n^3+n^{1/2})^{1/3}+(n^3-1)^{1/3}}{(n^3+n^{1/2})^{1/3}+(n^3-1)^{1/3}} $$
but it didn't help much. Wolfram says the answer is $\frac{3^{1/2}}{3}$.
Any help would be greatly appreciated.
| Using the identity $a^3-b^3=(a-b)\left(a^2+ab+b^2\right)$, we get
$$
\begin{align}
&\left(\sqrt[3]{n^3+\sqrt{n}}-\sqrt[3]{n^3-1}\right)\sqrt{3n^3+1}\\
&=\frac{\left(n^3+n^{1/2}\right)-\left(n^3-1\right)}{\left(n^3+n^{1/2}\right)^{2/3}+\left(n^3+n^{1/2}\right)^{1/3}\left(n^3-1\right)^{1/3}+\left(n^3-1\right)^{2/3}}\cdot\sqrt{3n^3+1}\\
&=\frac{n^{1/2}\left(1+n^{-1/2}\right)}{n^2\left(\left(1+n^{-5/2}\right)^{2/3}+\left(1+n^{-5/2}\right)^{1/3}\left(1-n^{-3}\right)^{1/3}+\left(1-n^{-3}\right)^{2/3}\right)}\cdot\sqrt3n^{3/2}\sqrt{1+\frac1{3n^3}}\\
&=\frac{\overbrace{\left(1+n^{-1/2}\right)}^{\to1}}{\underbrace{\left(1+n^{-5/2}\right)^{2/3}}_{\to1}+\underbrace{\left(1+n^{-5/2}\right)^{1/3}\left(1-n^{-3}\right)^{1/3}}_{\to1}+\underbrace{\left(1-n^{-3}\right)^{2/3}}_{\to1}}\cdot\sqrt3\,\,\overbrace{\sqrt{1+\frac1{3n^3}}}^{\to1}\\
&\to\frac1{\sqrt3}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1607031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
How can I evaluate $\sum_{i=0}^\infty \frac{1}{k^i} \binom{2i}{i}$ Evaluate $$\sum_{i=0}^\infty \left(\frac{\binom{2i}{i}}{k^i}\right),$$
where $k$ is a whole number.
I can't figure out how to approach this question, as no binomial series has such coefficients.
| Note, that
$$\binom{2i}{i}=(-4)^i\binom{-\frac{1}{2}}{i}$$
So, we can write OPs series as binomial series
\begin{align*}
\sum_{i=0}^{\infty}\binom{2i}{i}\frac{1}{k^i}
&=\sum_{i=0}^{\infty}\binom{-\frac{1}{2}}{i}\left(-\frac{4}{k}\right)^i\\
&=\frac{1}{\sqrt{1-\frac{4}{k}}}\\
&=\sqrt{\frac{k}{k-4}}
\end{align*}
convergent for $\left|-\frac{4}{k}\right|<1$, i.e. $k>4$.
[2016-01-14] Addendum
We can extend the binomial coefficient for arbitrary $\alpha\in\mathbb{C}$ and $n\in\mathbb{N}$
\begin{align*}
\binom{\alpha}{n}:=\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n(n-1)\cdots3\cdot2\cdot1}
\end{align*}
Putting $\alpha=-\frac{1}{2}$ we obtain
\begin{align*}
\binom{-\frac{1}{2}}{n}&=\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\cdots\left(-\frac{1}{2}-n+1\right)}{n!}\\
&=\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\cdots\left(-\frac{2n-1}{2}\right)}{n!}\\
&=\left(-\frac{1}{2}\right)^n\frac{1}{n!}(2n-1)!!\tag{1}\\
&=\left(-\frac{1}{2}\right)^n\frac{1}{n!}\cdot\frac{(2n)!}{(2n)!!}\\
&=\left(-\frac{1}{2}\right)^n\frac{1}{n!}\cdot\frac{(2n)!}{n!2^n}\tag{2}\\
&=\left(-\frac{1}{4}\right)^n\frac{(2n)!}{(n!)^2}\\
&=\left(-\frac{1}{4}\right)^n\binom{2n}{n}
\end{align*}
Comment:
*
*In (1) we use double factorials and the relation $(2n)!=(2n)!!(2n-1)!!$
*In (2) we use $(2n)!!=(2n)(2n-2)\cdots4\cdot2=n!2^n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1610662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits