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Finding $p$ such that $\sum\limits_{n=1}^\infty n(1+n^2)^p$ converges. Check my work.
Given series
$$
\sum\limits_{n=1}^\infty n(1+n^2)^p
$$
Find the values of $p$, such that the series is convergent.
To find $p$, I use integral test,
assume $f(x)=x(1+x^2)^p$,
\begin{eqnarray}
\int\limits_1^\infty x(1+x^2)^pdx &=& \lim\limits_{b\to\infty} \int\limits_1^b x(1+x^2)^pdx\\
&=& \lim\limits_{b\to\infty} \left[\dfrac{1}{2(p+1)}\left((1+b^2)^{p+1}-2^{p+1}\right)\right]
\end{eqnarray}
*
*If $p=1$, the integral is divergent,
*If $p>-1$, the integral is divergent,
*If $p<-1$, the integral is convergent to $-\dfrac{1}{p+1}2^p$,
So, I conclude the series is convergent while $p<-1$.
This answer is correct or incorrect?
| Alternative:
If $p\ge -1$, then $(1+n^2)^p \ge (1+n^2)^{-1}$
$$\sum_{n=1}^\infty n(1+n^2)^p \ge\sum_{n=1}^\infty \frac{n}{1+n^2} \ge \sum_{n=1}^\infty \frac{n}{2n^2}= \sum_{n=1}^\infty \frac{1}{2n}$$
Hence it diverges.
If $p<-1$, then $-p > 1$, $(1+n^2)^{-p} \ge (n^2)^{-p}$. Also, we have $-2p-1>2-1=1$
$$\sum_{n=1}^\infty \frac{n}{(1+n^2)^{-p}} \le \sum_{n=1}^\infty \frac{n}{n^{-2p}}= \sum_{n=1}^\infty \frac{1}{n^{-2p-1}}$$
Hence, by $p$-series and comparison test, it converges.
| {
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"source": "stackexchange",
"question_score": "4",
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Quotient $\mathbf{F}_3[X]/(X^5+1)$
Factor $X^5+1\in\mathbf{F}_3[X]$ into irreducibles. What does the quotient $\mathbf{F}_3[X]/(X^5+1)$ look like?
Since $-1$ is a zero, we divide $X^5+1$ by $X+1$ using long division, to obtain $X^5+1=(X+1)(X^4-X^3+X^2-X+1)$. Now we claim that $X^4-X^3+X^2-X+1$ is irreducible. It has no zeros in $\mathbf{F}_3$, and we can check that it can not be written as product of quadratic factors.
I am hesitating about the following part:
$(X^5+1)=(X+1,X^4-X^3+X^2-X+1)$.
By long division, we have $X^4-X^3+X^2-X+1=(X+1)(X^3-2X^2+3X-4)+5$, therefore $(X^5+1)=(X+1,5)$. We have $$\frac{\mathbf{F}_3[X]}{(X^5+1)}\cong\frac{\mathbf{F}_3[X]}{(X+1,2)}\cong\frac{\mathbf{Z}[X]/(3)}{(X+1,2)}\cong \frac{\mathbf{Z}[X]/(X+1)}{(3,2)}\cong\mathbf{Z}.$$
But $\mathbf{Z}$ is not a field, and $\frac{\mathbf{F}_3[X]}{(X^5+1)}$ is I think. Could someone help me?
| I think the mistake you made is that the ideal $(X^5 + 1) \neq (X+1, X^4 - X^3 + X^2 - X + 1)$. Similarly $(X^5 + 1) \neq (X + 1,5)$. Long division only gives us "$\subset$" in both cases, but not the other direction.
What you want to do is factor $X^5 + 1 = (X + 1) \cdot (X^4 - X^3 + X^2 - X + 1)$. Note the ideal generated by $X^5 + 1$ is also the product of the ideal generated by the two factors, but not the sum.
If you want to compute the quotient $\mathbb{F}[X] / (X^5 + 1)$, you can now apply the Chinese remainder theorem, that tells us
$$\mathbb{F}[X] / (X^5 + 1) \cong \mathbb{F}[X] / (X+1) \times \mathbb{F}[X]/(X^4 - X^3+X^2-X+1).$$
Also note that this is not a field, because $X^5+1$ is not irreducible. This is because $X+1$ and $X^4 - X^3 + X^2 - X + 1$ are not in the ideal, i.e. they are not $0$ in the quotient, but if multiplied they give $0$, so they are zero divisors of the quotient.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\int_{0}^{\infty} \frac{1}{1 + x^r}\:dx = \frac{1}{r}\Gamma\left( \frac{r - 1}{r}\right)\Gamma\left( \frac{1}{r}\right)$ As part of a recent question I posted, I decided to try and generalise for a power of $2$ to any $r \in \mathbb{R}$. As part of the method I took, I had to solve the following integral:
\begin{equation}
I = \int_{0}^{\infty} \frac{1}{1 + x^r}\:dx
\end{equation}
I believe what I've done is correct, but I'm concerned that I may missed something (in particular whether it holds for all $r \neq 0$). So, here I have two questions (1) Is what I've done correct? and (2) What other methods can be employed that doesn't rely on complex analysis?
Here is the method I took:
First make the substitution $u = x^{\frac{1}{r}}$ to arrive at
\begin{equation}
I = \frac{1}{n} \int_{0}^{\infty} \frac{1}{1 + u} \cdot u^{1 -\frac{1}{r}}\:du
\end{equation}
We now substitute $t = \frac{1}{1 + u}$ to arrive at:
\begin{align}
I &= \frac{1}{r} \int_{1}^{0} t \cdot \left(\frac{1 - t}{t}\right)^{\frac{1}{r} -1}\frac{1}{t^2}\:dt = \frac{1}{r}\int_{0}^{1}t^{-\frac{1}{r}}\left(1 - t\right)^{ \frac{1}{r} - 1}\:dt \\
&= \frac{1}{r}B\left(1 - \frac{1}{n}, 1 + \frac{1}{r} - 1\right) = \frac{1}{r} B\left(\frac{r - 1}{r},\frac{1}{r}\right) \\
&= \frac{1}{r} B\left(\frac{r - 1}{r},\frac{1}{r}\right)
\end{align}
Wheer $B(a,b)$ is the Beta function. Using the relationship between the Beta and Gamma function we arrive at:
\begin{equation}
I = \frac{1}{r} \frac{\Gamma\left( \frac{r - 1}{r}\right)\Gamma\left( \frac{1}{r}\right)}{\Gamma\left(\frac{r - 1}{r} + \frac{1}{r}\right)} = \frac{1}{r}\Gamma\left( \frac{r - 1}{r}\right)\Gamma\left( \frac{1}{r}\right)
\end{equation}
And so, we arrive at:
\begin{equation}
I = \int_{0}^{\infty} \frac{1}{1 + x^r}\:dx = \frac{1}{r}\Gamma\left( \frac{r - 1}{r}\right)\Gamma\left( \frac{1}{r}\right)
\end{equation}
for $r > 1$
As per KemonoChen's comment and others, we can employ Euler's Reflection Formula to position this result for $\frac{1}{r} \not \in \mathbb{Z}$ Here, as $r \in \mathbb{R}, r > 1 \rightarrow \frac{1}{r} \not \in \mathbb{Z}$ and so our formula holds.
\begin{equation}
I = \int_{0}^{\infty} \frac{1}{1 + x^r}\:dx = \frac{1}{r}\Gamma\left( \frac{r - 1}{r}\right)\Gamma\left( \frac{1}{r}\right) = \frac{\pi}{r\sin\left(\frac{\pi}{r} \right)}
\end{equation}
Thank you also to Winther, Jjagmath, and MrTaurho's for their comments and corrections/clarifications.
| Once again I will offer up a method that first converts the integral to a double integral.
For $r > 0$, we begin by enforcing a substitution of $x \mapsto x^{1/r}$. Doing so yields
$$I = \frac{1}{r} \int_0^\infty \frac{x^{1/r - 1}}{1 + x} \, dx.$$
Now noting that
$$\frac{1}{1 + x} = \int_0^\infty e^{-u(1 + x)} \, du,$$
our integral can be rewritten as
$$I = \frac{1}{r} \int_0^\infty x^{1/r - 1} \int_0^\infty e^{-u (1 + x)} \, du \, dx,$$
or
$$I = \frac{1}{r} \int_0^\infty e^{-u} \int_0^\infty x^{1/r - 1} e^{-ux} \, dx \, du,$$
after changing the order of integration.
Next we enforce a substitution of $x \mapsto x/u$. This gives
\begin{align}
I &= \frac{1}{r} \int_0^\infty u^{- 1/r} e^{-u} \, du \int_0^\infty x^{1/r - 1} e^{-x} \, dx\\
&= \frac{1}{r} \Gamma \left (1 - \frac{1}{r} \right ) \Gamma \left (\frac{1}{r} \right )\\
&= \frac{\pi}{r \sin \left (\frac{\pi}{r} \right )},
\end{align}
where in the last line we have made use of Euler's reflexion formula for the gamma function.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the value of $1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\frac{1}{23}+\frac{1}{25}....$
Find the value of this :$$1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\frac{1}{23}+\frac{1}{25}....$$
Try: We can write the above series as
$${S} = \int^{1}_{0}\bigg[1-x^6+x^8-x^{14}+x^{16}-x^{22}+\cdots\bigg]dx$$
$$S = \int^{1}_{0}(1-x^6)\bigg[1+x^{8}+x^{16}+\cdots \cdots \bigg]dx$$
So $$S = \int^{1}_{0}\frac{1-x^6}{1-x^{8}}dx = \int^{1}_{0}\frac{x^4+x^2+1}{(x^2+1)(x^4+1)}dx$$
Now i am struck in that integration.
Did not understand how to solve it
could some help me to solve it. Thanks in advance
| HINT: Notice that
$$\frac{x^4+x^2+1}{(x^2+1)(x^4+1)}=\frac{1}{4}\frac{1}{x^2+x\sqrt{2}+1}+\frac{1}{4}\frac{1}{x^2-x\sqrt{2}+1}+\frac{1}{2}\frac{1}{x^2+1}$$
Now we just have to evaluate the integrals
$$I_1=\frac{1}{4}\int_0^1 \frac{dx}{x^2+x\sqrt{2}+1}$$
$$I_2=\frac{1}{4}\int_0^1 \frac{dx}{x^2-x\sqrt{2}+1}$$
$$I_3=\frac{1}{2}\int_0^1 \frac{dx}{x^2+1}$$
and your sum is given by $I_1+I_2+I_3$. Can you evaluate these?
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify the determinant of a $4 \times 4$ matrix. I have to find the determinant of the following 4x4 matrix:
$\quad A=\begin{bmatrix}3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{bmatrix}$
So I apply the Gaussian elimination to obtain an upper-triangle matrix:
$$det\begin{bmatrix}3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{bmatrix}=\begin{vmatrix}3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{vmatrix}\xrightarrow{3R_3-R_1}\begin{vmatrix}3&0&1&0\\0&2&0&0\\0&0&8&0\\0&0&0&-4\end{vmatrix}$$
Since I know from the solutions that the determinant is -64, I suppose that I need to simplify the third row in the reduced form to $\quad 0 \quad 0 \quad 2 \quad 0 \quad$ and then multiply the elements in the upper-left-to-bottom-right diagonal, which is indeed -64. But this doesn't make much sense since there's also a $-4$ that we can simplify. Can someone explain me the actual rules we need to follow?
| Rather than applying row operations, expand the minors.
$\det A = -4\begin{vmatrix} 3&0&1\\0&2&0\\1&0&3\end{vmatrix} = (-4)(18-2) = -64$
If you apply row operations, you don't want your row operations to change the determinant.
I think of multiplying by an elementary matrix.
$\begin{vmatrix} 1\\&1\\-\frac 13&&1\\&&&1\end{vmatrix}\begin{vmatrix} 3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{vmatrix} = \begin{vmatrix} 3&0&1&0\\0&2&0&0\\0&0&\frac {8}{3}&0\\0&0&0&-4\end{vmatrix}$
keeps the determinant unchanged.
while
$\begin{vmatrix} 1\\&1\\-1&&3\\&&&1\end{vmatrix}\begin{vmatrix} 3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{vmatrix} = \begin{vmatrix} 3&0&1&0\\0&2&0&0\\0&0&8&0\\0&0&0&-4\end{vmatrix}$
will change the determinant by a factor of $3.$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate $\int \frac{x^3}{\sqrt {x^2+1}}dx$ Evaluate $$\int \frac{x^3}{\sqrt {x^2+1}}dx$$
Let $u={x^2}+1$. Then $x=\sqrt {u-1}$ and $dx=\frac{1}{2\sqrt{u-1}}du$.
Therefore, $$\int \frac{(u-1)\sqrt{u-1}}{\sqrt u}\frac{1}{2\sqrt {u-1}}du$$
$$=\frac{1}{2}\int {\sqrt u}-{\frac {1}{\sqrt u}du}$$
$$={\frac{2}{3}}{{(x^2+1})}^{3/2}-{\frac{1}{2}}{\sqrt {x^2+1}}+C$$for some constant $C$
| Here I give you an alternative approach. If we make $x = \sinh(u)$, we get
\begin{align*}
\int\frac{x^{3}}{\sqrt{x^{2}+1}}\mathrm{d}x & = \int\sinh^{3}(u)\mathrm{d}u = \int(\cosh^{2}(u)-1)\sinh(u)\mathrm{d}u\\
& = \frac{\cosh^{3}(u)}{3} - \cosh(u) + K
\end{align*}
Since $x = \sinh(u) = \pm\sqrt{\cosh^{2}(u)-1}$, we conclude that $\cosh(u) = \sqrt{x^{2} + 1}$. Finally, it results that
\begin{align*}
\int\frac{x^{3}}{\sqrt{x^{2}+1}}\mathrm{d}x = \frac{(x^{2}+1)^{3/2}}{3} - \sqrt{x^{2}+1} + K
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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With $z\in \mathbb C$ find the maximum value for $\lvert z\rvert$ such that $\lvert z+\frac{1}{z}\rvert=1$
With $z\in \mathbb C$ find the maximum value for |z| such that
$$\left\lvert z+\frac{1}{z}\right\rvert=1.$$
Source: List of problems for math-contest training.
My attempt: it is easy to see that the given condition is equivalent
$$\lvert z^2+1\rvert=\lvert z\rvert$$
and if $z=a+bi$,
\begin{align*}
\lvert z\rvert&=\lvert a^2-b^2+1+2ab i\rvert=\sqrt{(a^2-b^2+1)^2+4 a^2b^2}\\
&=\sqrt{(a^2-b^2)^2+2(a^2-b^2)+1+4a^2b^2}\\
&=\sqrt{(a^2+b^2)^2+2(a^2-b^2)+1}\\
&=\sqrt{\lvert z\rvert^4+2(a^2-b^2)+1}
\end{align*}
I think it is not leading to something useful... the approach I followed is probably not useful.
Hints and answers are welcomed.
| We can maximize $g(a,b) = a^2+b^2$ subject to $(a^2+b^2)^2 =3b^2-1$. Solve for $a^2$ we have: $a^2= \sqrt{3b^2-1} -b^2\implies g(a,b) = \sqrt{3b^2-1}$. Thus the problem now is the find the maximum value of $b^2$. To this end, let $c = b^2 \ge 0$, we have: $a^4+2a^2c+c^2 - 3c+ 1 = 0\implies c^2-3c+1 \le 0\implies c \le \dfrac{3+\sqrt{5}}{2}\implies 3b^2 -1 = 3c-1\le \dfrac{7+3\sqrt{5}}{2}\implies |z|_{\text{max}} = \sqrt{g(a,b)}_{\text{max}} = \sqrt[4]{\dfrac{7+3\sqrt{5}}{2}}= \dfrac{1+\sqrt{5}}{2}$,and this max occurs when $a = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of a series of a number raised to incrementing powers-- for any number Sum of a series of a number raised to incrementing powers. I have a sub question on this. As I dont have reputation to ask for in comment. The formula being derived for series of summation of powers to 2 i.e $2^{n+1}-1$ doesnt apply for other numbers suppose $5^0+5^1+5^2+5^3+5^4$ is not equal to $5^{4+1}-1$. Hence I request if there is any generalized formula i.e that will give the summation of series of powers of the number available.
If any such formula is available need your help!!
| This is a geometric series
$$S=5^0+5^1+5^2+5^3+5^4+\cdots+5^n$$
$$5S=5^1+5^2+5^3+5^4+5^5+\cdots+5^{n+1}$$
$$4S=5^{n+1}-1$$
$$S=\frac{5^{n+1}-1}{4}$$
In general to find $1+x+x^2+\cdots+x^n$,
$$\sum_{j=0}^{n}x^j=\frac{x^{n+1}-1}{x-1}$$
| {
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sum which have binomial coefficients
Finding value of
$\displaystyle \binom{50}{6}-\binom{5}{1}\binom{40}{6}+\binom{5}{2}\binom{30}{6}-\binom{5}{3}\binom{20}{6}+\binom{5}{4}\binom{10}{6}$
Try: Equation coefficient of $x^6$ on both side
$$\bigg[(1+x)^{10}-1\bigg]^5=(1+x)^{50}-\binom{5}{1}(1+x)^{40}+\binom{5}{2}(1+x)^{30}-\binom{5}{3}(1+x)^{20}+\binom{5}{4}(1+x)^{10}$$
So our required sum is coefficient of $x^6$ in $$x^5\cdot \bigg[1+(1+x)+(1+x)^2+\cdots +(1+x)^9\bigg]^5.$$
So coefficient of $x$ in $$\bigg[1+(1+x)+(1+x)^2+\cdots \cdots +(1+x)^{9}\bigg]^5$$
$$=5(10)^4 (1+2+\cdots +9)=5\cdot (10)^4\cdot 45$$
But I am also trying to solve it using the principle of inclusion -exclusion.
I have seems that we will form $5$ groups and selecting some persons from each group, but I did not understand how I can solve it
Could some help me to solve it , thanks
| Let's consider the following problem: Given 50 balls colored in five colors, 10 per color, how many ways can we choose 6 balls so that each color is represented?
Denote the answer $X$.
We will solve this problem in two different ways. Both should give the same result.
First, let's solve the above using inclusion-exclusion principle.
*
*We count how many ways we can choose 6 balls. That is $50 \choose 6$
*Now we remove all combinations that miss a color. How many are there? There are 5 possibilities for the missing color, and 40 balls colored differently. Thus, subtract ${5 \choose 1}{40 \choose 6}$ combinations.
*But arrangements missing two colors were counted twice in step 2, so we need to add them back once. There are $5 \choose 2$ ways to select two missing colors. This leaves 20 balls, so we have \choose 6$ ways to choose the balls.
*To complete the inclusion and exclusion, we have to subtract arrangements missing 3 colors, and add arrangements missing 4 colors (there are no arrangements missing 5 colors).
In short, from inclusion-exclusion principle, we have
$$X = \binom{50}{6}-\binom{5}{1}\binom{40}{6}+\binom{5}{2}\binom{30}{6}-\binom{5}{3}\binom{20}{6}+\binom{5}{4}\binom{10}{6}$$
There is, however, an easier way to solve this problem. If we select 6 balls so that there is at least one from each color, we'll have exactly one ball per color, except for one color which will have two.
So we have 5 ways to choose the color that has two balls, $10 \choose 2$ ways to choose those 2 balls, and then 10 ways per color to choose the balls from the remaining colors. Thus,
$$X = 5 \times {10 \choose 2} \times 10^4 = 5 \times 45 \times 10 000 = 2 250 000$$.
Thus, the sum we need to compute is 2 250 000.
| {
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How do I solve $\lim \limits_{x \to \frac{π}{3}} \frac{2 \sin x - \sqrt{3}}{\cos \frac{3x}{2}}$ Alright, I know, there are easier ways to solve this, like L'hopitals Rule etc.
But I'm not solving it for the answer, just doing it for the fun so I tried using substitution method.
Put $t= x- \dfrac{π}{3}$
$\lim \limits_{x \to \frac{π}{3}} \dfrac{2 \sin x - \sqrt{3}}{\cos \frac{3x}{2}}$
$= \lim\limits_{t \to 0} \dfrac{2 \sin \left(t+\frac{π}{3} \right) - \sqrt{3}}{\cos \left( \frac{3t}{2} + \frac{π}{2}\right)}$
$= \lim\limits_{t \to 0} \dfrac{\sqrt{3}- \sin t - \sqrt{3} \cos t}{ \sin \frac{3t}{2}}$
Where do I go from here, I'm not able to eliminate the $t$ fully from the Nr and Dr, any help?
Or any other alternative way that uses only the fact that $\lim\limits_{x \to 0} \dfrac{ \sin x}{x} = 1$?
Thanks :)
| To end your computation, you can split your expression in two:
$$\frac{\sqrt{3}- \sin t - \sqrt{3} \cos t}{ \sin \frac{3t}{2}} =-\frac{\sin t}{ \sin \frac{3t}{2}}+\sqrt 3\,\frac{1- \cos t}{\sin \frac{3t}{2}} $$
Now it is standard that $\;\lim_{t\to 0}\dfrac{\sin at}{\sin bt}=\dfrac ab$. On the other hand
$$\frac{1-\cos t}{\sin\frac{3t}{2}}=\frac{2\sin^2t}{\sin\frac{3t}{2}}=2\sin t\,\frac{\sin t}{\sin\frac{3t}{2}}.$$
Can you conclude now?
| {
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Proof that $x^{(n+4)} \bmod 10 = x^n \bmod 10\,$ for $\,n\ge 1$ While solving a programming challenge in which one should efficiently compute the last digit of $a^b$, I noticed that apparently the following holds (for $n > 0$)
$x^{(n+4)} \mod 10 = x^n \mod 10$
How can this be proven?
| If you can prove $x^4 \equiv 1 \pmod {10}$ that's enough as $x^{n+4}\equiv x^n\cdot x^4 \equiv x^n\cdot 1\equiv x^n\pmod {10}$
If you google Euler's theorem that will be true for all numbers relatively prime to $10$. Intuitively (but hand wavy) if $x$ and $10$ are relatively prime then $x^n$ will be relatively prime. There are only $4$ possible last digits that are relatively prime to $10$ so $x^k$ will cycle through the same four digits. (More or less.)
If $x$ is even or divisible by $5$. Well, if it's both then $x \equiv 0 \mod 10$ so $x^k \equiv 0 \pmod {10}$ and $x^{n+4} \equiv x^n \equiv 0\pmod {10}$. Likewise if it's odd but divisible by $5$ then $x^k \equiv 5 \pmod{10}$ and so $x^{n+4} \equiv x^n \equiv 5 \pmod {10}$.
Now if $x$ is even but not divisible by $5$ then. Well, $x^n$ is even and there are only $4$ even digits it could end with and it cycles through them.
More to the point $2*j \pmod 10 \equiv 2*(j \pmod 5)$ and there are only $4$ non-zero classes $\pmod 5$ so $x^k$ just cycles through those.
| {
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Show that $\frac{x}{\sqrt{1-a^2}}=\frac{y}{\sqrt{1-b^2}}=\frac{z}{\sqrt{1-c^2}}$
Given $3$ equations:
*
*$ x-cy-bz=0$
*
*$-cx+y-az=0$
*$-bx-ay+z=0$
Show that $\dfrac{x}{\sqrt{1-a^2}}=\dfrac{y}{\sqrt{1-b^2}}=\dfrac{z}{\sqrt{1-c^2}}$
Now solving the 3 equations I got:
$\dfrac{x}{{1-a^2}}=\dfrac{y}{{ab+c}}=\dfrac{z}{ac+b}$
$\dfrac{x}{ac+b}=\dfrac{y}{a+bc}=\dfrac{z}{1-c^2}$
$\dfrac{x}{a+bc}=\dfrac{y}{1-b^2}=\dfrac{z}{a+bc}$
How to prove the required fact?
Any way to prove it ?
| Hint:
By $(1),(2)$ $$0=x-cy-bz+c(-cx+y-az)=x(1-c^2)-z(b+ca)$$
By $(2),(3)$ $$0=a(-cx+y-az)-bx-ay+z=z(1-a^2)-x(ca+b)$$
$$\dfrac{x(1-c^2)}{z(1-a^2)}=\dfrac{z(b+ca)}{x(ca+b)}$$
What if $b+ca\ne0$
| {
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"url": "https://math.stackexchange.com/questions/3051807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If the equations $ax^3+(a+b)x^2+(b+c)x+c=0$ and $2x^3+x^2+2x-5=0$ have a common root,then $a+b+c$ can be equal to If the equations $ax^3+(a+b)x^2+(b+c)x+c=0$ and $2x^3+x^2+2x-5=0$ have a common root, then $a+b+c$ can be equal to(where $a,b,c\in R,a\ne0$)
$(1)\;5a$
$(2)\;3b$
$(3)\;2c$
$(4)\;0$
As $x=1$ is the root of the equation $2x^3+x^2+2x-5=0$ and its other two roots are complex in nature. So the common root is $x=1$
Put $x=1$ in $ax^3+(a+b)x^2+(b+c)x+c=0$ we get $2a+2b+2c=0$ so $(4)$ option is correct. But $(1)$ and $(3)$ options are also correct. I don't know how they are correct.
| Note that
$$ax^3+(a+b)x^2+(b+c)x+c=(x+1)(ax^2+bx+c)$$
and
$$2x^3+x^2+2x-5=(x-1)(2x^2+3x+5)$$
So we can also have the case when the roots in common are the two complex roots: $a=2k$, $b=3k$ and $c=5k$, that is $a+b+c=5a=2c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3052813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\sin a + \sin b + \sin(a+b) = 4 \sin\frac12(a+b) \cos \frac12a \cos\frac12b$ Helping my son with his trigonometry review. We know that
$$\sin(a+b) = \sin a \cos b + \cos a \sin b$$
We also know that
$$\sin a \cos b = \frac12 \left(\sin(a-b) + \sin(a+b)\right)$$
And we have
$$\sin a + \sin b = 2 \sin\frac12(a+b)\cos\frac12(a-b)$$
From there, we seem to be missing how to get to the right-hand side of the equation.
We first expand
$$\sin a + \sin b = 2 \sin\frac12(a+b) \cos\frac12(a-b)$$
Then we add $\sin(a+b)$, which is $\sin a \cos b + \cos a \sin b$.
We now have:
$$2 \sin\frac12(a+b) \cos\frac12(a-b) + \frac12 \left(\sin(a-b) + \sin (a+b)\right) + \frac12 \left( \sin(b-a) + \sin (a+b)\right)$$
From there, we can't see how to obtain the right-hand side of the equation which is
$$4 \sin\frac12(a+b) \cos\frac12a \cos\frac12b$$
| An alternative strategy uses $1+\cos x=2\cos^2\tfrac{x}{2}$ to obtain $$\sin a+\sin b+\sin (a+b)=2\left(\sin a\cos^2\tfrac{b}{2}+\sin b\cos^2\tfrac{a}{2}\right).$$To finish, use $\sin x=2\sin\tfrac{x}{2}\cos\tfrac{x}{2}$ to obtain$$\sin a+\sin b+\sin (a+b)=4\cos\tfrac{a}{2}\cos\tfrac{b}{2}\left(\sin\tfrac{a}{2}\cos\tfrac{b}{2}+\sin\tfrac{b}{2}\cos\tfrac{a}{2}\right)=4\cos\tfrac{a}{2} \cos\tfrac{b}{2}\sin\tfrac{a+b}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3053978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove $\frac{\tan (A)}{\tan (A)}+\frac{\cot (A)}{\cot (A)}=\frac{1}{1-2\cos(A)^2}$ I am unable to prove this trigonometric identity
$$\frac{\tan (A)}{\tan (A)}+\frac{\cot (A)}{\cot (A)}=\frac{1}{1-2\cos^2(A)}$$
I have tried to transform the left-hand side and stuck with this
$$\frac{2\sin(A)\cos(A)}{\sin(A)\cos(A)}$$
And I have tried to transform the right-hand side by changing the $$2\cos^2(A)$$ to $$\frac{2}{\sec^2(A)}$$, and used the trigonometric identity $$1+\tan^2(A)=\sec^2(A)$$ and got this instead
$$\frac{1+\tan^2(A)}{\tan^2(A)-1}$$ which I can transform to $$\frac{\cot(A)+\tan(A)}{\tan(A)-\cot(A)}$$.
I cannot get both sides equal, help please?
| One way we can prove the identity false is as follows:
$$\begin {align}
\dfrac {\tan A} {\tan A} + \dfrac{\cot A}{\cot A} = \dfrac {1}{1-2\cos^2 2A} \\
2 = \dfrac {1}{1-2\cos^2 2A} \\
2 (1-2\cos^2 2A) = 1 \\
2 - 4\cos^2 2A = 1 \\
- 4\cos^2 2A = \dfrac {1}{2} \\
\cos^2 2A = -\dfrac {1}{8}
\end {align}$$
Since the last line would require us to take the square root of a negative number, $A$ does not exist, and the identity is false.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Evaluate $\lim_{n\to\infty}(1+\frac{1}{a_1})\cdots(1+\frac{1}{a_n})$, where $a_1=1$, $a_n=n(1+a_{n-1})$
Evaluate $\lim_{n\to\infty}(1+\frac{1}{a_1})(1+\frac{1}{a_2})\cdots(1+\frac{1}{a_n})$,
where $a_1 = 1$, $a_n = n(1+a_{n-1})$
\begin{align*}
&\lim_{n\to\infty} \left(1+\frac{1}{a_1}\right)\left(1+\frac{1}{a_2}\right)\cdots\left(1+\frac{1}{a_n}\right) \\
&= \lim_{n\to\infty} \frac{1}{a_1} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdots \frac{1}{n} \cdot (a_n + 1) \\
&= \lim_{n\to\infty} \frac{a_n + 1}{n!}.
\end{align*}
Then I'm stuck. How to proceed?
| Observe that
$$a_n=n+na_{n-1}$$
$$a_n=n+n(n-1)+n(n-1)a_{n-2}$$
and so on, giving us eventually
$$a_n=\sum_{k=1}^n\frac{n!}{(k-1)!}$$
Now we have
$$\lim_{n\to\infty}\frac{1}{n!}(a_n+1)=\lim_{n\to\infty}\frac{a_n}{n!}$$
$$=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{(k-1)!}$$
$$=\sum_{k=0}^\infty\frac{1}{k!}=e$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3058562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Magnitude of Complex Number And Other Equation Sorry for the unclear question, but I have no idea how to solve this problem:
For certain positive integers $a,b,c$ and $d$, the set of complex numbers $z$ that satisfy $|z-5\sqrt{3}-5i|=5$ is equivalent to the set of complex numbers $z$ that satisfy $\left| \frac{1}{z} - \frac{1}{a \sqrt{b}} + \frac{i}{c} \right| = \frac{1}{d}$, where $b$ is not divisible by the square of a prime. Find $a+b+c+d$.
My approach was to graph a circle on the imaginary plane with center $(5\sqrt{3},5)$ with radius 5. The equation of this circle is $(x-5\sqrt{3})^2+(y-5i)^2=25$ (I think?). I'm not sure how to relate the set of $z$ to the equation. Any help is appreciated.
| Note that if you replace $\frac 1z$ with $u$ in your second equation $\left| \frac{1}{z} - \frac{1}{a \sqrt{b}} + \frac{i}{c} \right| = \frac{1}{d}$ then the locus for $u$ in the complex plane is another circle whose radius is $\frac 1d$ and center is $\frac{1}{z} - \frac{1}{a \sqrt{b}}$. Thus the desired circle is the "inversion" of your first circle under the operation $z \to \frac 1z$.
You have the equation of the given circle, so find the equation of the inverted circle. There are several ways to do this. One way is to find three points (complex numbers) on the original circle, find the reciprocals of those three points, then find the circle that goes through those three points. Once you find that circle you find its center and radius then can find the values of $a, b, c,$ and $d$.
Or better, find the point on the first circle that is closest to the origin (point $0+0i$) and the point that is farthest from the origin. Those two points are the endpoints of a diameter of the first circle. The closest point is $\frac 52\sqrt{3} + \frac 52i$ and the farthest is $\frac {15}2\sqrt{3} + \frac {15}2i$, which can be seen easily geometrically (see the diagram, and recognize that the right triangle is a 30°-60°-90° triangle). Now invert those two points. The closest point on the first circle goes to the farthest point in the inverted circle, and vice versa. So the inverted points determine a diameter of the inverted circle. The average of those two inverted points is then the diameter of the inverted circle, and the radius is the modulus of the difference between the center and either of the two inverted points. This is much easier than using three general points on the circles.
Here are the final answers: when we invert the near point $z_1=\frac 52\sqrt{3} + \frac 52i$ we get $u_1=\frac 1{10}\sqrt 3-\frac 1{10}i$ and when we invert the far point $z_2=\frac {15}2\sqrt{3} + \frac {15}2i$ we get $u_2=\frac 1{30}\sqrt 3-\frac 1{30}i$. The center of the inverted circle is the average of those two points, $u_c=\frac 1{15}\sqrt 3-\frac 1{15}i$. If we rationalize the numerator we get
$$u_c=\frac 1{5\sqrt 3}-\frac 1{15}i$$
The radius of the circle is
$$|u_1-u_c|=\sqrt{\left[\frac 1{10}-\frac 1{15}\right]^2\sqrt 3^2+\left[\frac 1{10}-\frac 1{15}\right]^2}=\sqrt{\frac 3{900}+\frac 1{900}}=\frac 1{15}$$
So the final result is
$$a+b+c+d = 5+3+15+15 = 38$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The sum of the series $\frac{15}{16}+\frac{15}{16} \times \frac{21}{24}+\frac{15}{16} \times \frac{21}{24} \times \frac{27}{32}+\dots$ Suppose $S=\frac{15}{16}+\frac{15}{16} \times \frac{21}{24}+\frac{15}{16} \times \frac{21}{24} \times \frac{27}{32}+\dots \dots$
Does it converge? If so find the sum.
What I attempted:- On inspection of the successive terms, it easy to deduce that the $n^{th}$ term of the series is $t_{n}=\left(\frac{3}{4}\right)^n \quad\frac{5.7.9.\dots \dots (2n+3)}{4.6.8. \dots \dots (2n+2)}$
Thus $\frac{t_{n+1}}{t_n}=\frac{3}{4} \times \frac{2n+5}{2n+4}$. As $n \to \infty$ this ratio tends to $\frac{3}{4}<1$. Hence by Ratio test it turns out to be convergent.
A similar type of question has already been asked here. One of the commenters has provided a nice method to evaluate the sum of such series using recurrence relation and finally using the asymptotic form of Catalan Number.
To proceed exactly in the similar way, I wrote $t_n$ as follows:-
$t_n=\frac{1}{3} \left(\frac{3}{8}\right)^n \quad\frac{1.3.5.7.9.\dots \dots (2n+3)}{(n+1)!}=\frac{1}{3} \left(\frac{3}{8}\right)^n \frac{n+2}{2^{n+2}} \binom{2n+4}{n+2}\approx \left(\frac{3}{4}\right)^{n-1}\frac{\sqrt{n+2}}{\sqrt{\pi}} \quad (\mbox{For large $n$})$.
I have used the recurrence relation $S_n=S_{n-1}+T_n$, along with the initial condition $S_1=\frac{15}{16}$, in order to get a solution like this $$S_n=7.5+\frac{T_n^2}{T_n-T_{n-1}}$$.
I am getting trouble in evaluating the limit of the second term as $n \to \infty$.
I haven't cross checked all the steps. Hope I would be pointed in case of any mistake.
| Let
$$f(x)=\sum_{n=1}^\infty t_nx^n=\sum_{n=1}^\infty\frac{5\times7\times\cdots
\times(2n+3)}{4\times6\times\cdots
\times(2n+2)}x^n.$$
Then
$$t_n=\frac23\frac1{(n+1)!}\left(\frac32\right)\left(\frac52\right)\cdots
\left(\frac{2n+3}2\right)=\frac23u_{n+1}$$
where
$$u_n=\frac{(3/2)(5/2)\cdots((2n+1)/2)}{n!}.$$
Then, for $|x|<1$,
$$\sum_{n=0}^\infty u_nx^n=\frac1{(1-x)^{3/2}}$$
by the binomial theorem.
Then
$$f(x)=\frac23\sum_{n=1}^\infty u_{n+1}x^n=\frac23\sum_{n=2}^\infty
u_nx^{n-1}=\frac2{3x}\left(\frac1{(1-x)^{3/2}}-1-\frac{3x}2\right)$$
Now insert $x=3/4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Several ways to prove that $\sum\limits^\infty_{n=1}\left(1-\frac1{\sqrt{n}}\right)^n$ converges I believe there are several ways to prove that $\sum\limits^{\infty}_{n=1}\left(1-\frac{1}{\sqrt{n}}\right)^n$ converges. Can you, please, post yours so that we can learn from you?
HERE IS ONE
Let $n\in\Bbb{N}$ be fixed such that $a_n=\left(1-\frac{1}{\sqrt{n}}\right)^n.$ Then,
\begin{align} a_n&=\left(1-\frac{1}{\sqrt{n}}\right)^n \\&=\exp\ln\left(1-\frac{1}{\sqrt{n}}\right)^n\\&=\exp \left[n\ln\left(1-\frac{1}{\sqrt{n}}\right)\right] \\&=\exp\left[ -n\sum^{\infty}_{k=1}\frac{1}{k}\left(\frac{1}{\sqrt{n}}\right)^k\right]\\&=\exp\left[ -n\left(\frac{1}{\sqrt{n}}+\frac{1}{2n}+\sum^{\infty}_{k=3}\frac{1}{k}\left(\frac{1}{\sqrt{n}}\right)^k\right)\right]\\&=\exp \left[-\sqrt{n}-\frac{1}{2}-\sum^{\infty}_{k=3}\frac{n}{k}\left(\frac{1}{\sqrt{n}}\right)^k\right]\\&\equiv\exp \left(-\sqrt{n}\right)\exp \left(-\frac{1}{2}\right)\end{align}
Choose $b_n=\exp \left(-\sqrt{n}\right)$, so that
\begin{align} \dfrac{a_n}{b_n}\to\exp \left(-\frac{1}{2}\right).\end{align}
Since $b_n \to 0$, there exists $N$ such that for all $n\geq N,$
\begin{align} \exp \left(-\sqrt{n}\right)<\dfrac{1}{n^2}.\end{align}
Hence, \begin{align}\sum^{\infty}_{n=N}b_n= \sum^{\infty}_{n=N}\exp \left(-\sqrt{n}\right)\leq \sum^{\infty}_{n=N}\dfrac{1}{n^2}<\infty,\end{align}
and so, $\sum^{\infty}_{n=1}b_n<\infty\implies \sum^{\infty}_{n=1}a_n<\infty$ by Limit comparison test.
| For $n\ge2$,
$$
\begin{align}
\left(1-\frac1{\sqrt{n}}\right)^n
&=\left(\left(1+\frac1{\sqrt{n}-1}\right)^{\sqrt{n}}\right)^{-\sqrt{n}}\tag{1a}\\
&\le\left(1+\frac{\sqrt{n}}{\sqrt{n}-1}\right)^{-\sqrt{n}}\tag{1b}\\[9pt]
&\le2^{-\sqrt{n}}\tag{1c}
\end{align}
$$
Explanation:
$\text{(1a)}$: $1-\frac1{\sqrt{n}}=\left(1+\frac1{\sqrt{n}-1}\right)^{-1}$
$\text{(1b)}$: Bernoulli's Inequality
$\text{(1c)}$: $1+\frac{\sqrt{n}}{\sqrt{n}-1}\ge2$
Since $(1)$ holds for $n=1$, it is true for $n\ge1$.
$$
\begin{align}
\sum_{n=1}^\infty\left(1-\frac1{\sqrt{n}}\right)^n
&\le\sum_{n=1}^\infty2^{-\sqrt{n}}\tag{2a}\\
&=\sum_{k=1}^\infty\sum_{n=(k-1)^2+1}^{k^2}2^{-\sqrt{n}}\tag{2b}\\
&\le\sum_{k=1}^\infty(2k-1)\,2^{-k+1}\tag{2c}\\[9pt]
&=6\tag{2d}
\end{align}
$$
Explanation:
$\text{(2a)}$: apply $(1)$
$\text{(2b)}$: group the terms
$\text{(2c)}$: $2k-1$ terms of size less than $2^{-k+1}$
$\text{(2d)}$: compute sum using
$\phantom{\text{(2d):}}$ $\sum\limits_{k=1}^\infty x^k=\frac{x}{1-x}\quad$ and $\quad\sum\limits_{k=1}^\infty kx^k=\frac{x}{(1-x)^2}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is $\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$ $15\sqrt{5}$ and not $15\sqrt[4]{25}$? I have an expression I am to simplify:
$$\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$$
I arrived at $15\sqrt[4]{25}$. My textbook tells me that the answer is in fact $15\sqrt{5}$. Here is my thought process:
$\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$ = $\frac{15*\sqrt[4]{5}*\sqrt[4]{25}}{\sqrt[4]{5}}$ =
(cancel out $\sqrt[4]{5}$ present in both numerator and denominator)
leaving:
$$15\sqrt[4]{25}$$
Where did I go wrong and how can I arrive at $15\sqrt{5}$?
| $$\frac{\sqrt[4] {125}}{\sqrt[4] 5} = \frac{\sqrt[4] {5^3}}{\sqrt[4] 5} = \sqrt[4]{5^2} = 5^{\frac{2}{4}} = 5^{\frac{1}{2}} = \sqrt 5$$
Even quicker, $\sqrt[4]{25}$ means $\sqrt{\sqrt{25}}$, which becomes $\sqrt{5}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating an improper integral $\int_{0}^{\infty}\frac{x^2}{(x^4+1)^2}dx$ I tried to solve the integral: $$\int_{0}^{\infty}\frac{x^2}{(x^4+1)^2}dx$$
using $ x = \sqrt{\tan(u)}$ and $dx = \frac{ \sec^2(u)}{2\sqrt{\tan(u)}} du,$
but I ended up with an even worse looking integral $$ \int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\tan(u)}}{\sec^2(u)}du.$$
Wolfram gave an answer of $ \dfrac{\pi}{8\sqrt{2}},$ but how would one get to that answer?
| Let us start with a step of integration by parts:
$$ \int_{0}^{+\infty}\frac{1}{4x}\cdot\frac{4x^3}{(x^4+1)^2}\,dx =\int_{0}^{+\infty}\frac{1}{4x^2}\left(1-\frac{1}{1+x^4}\right)\,dx=\frac{1}{4}\int_{0}^{+\infty}\frac{dx}{x^2+\frac{1}{x^2}}$$
and finish with Glasser's master theorem:
$$ \frac{1}{8}\int_{-\infty}^{+\infty}\frac{dx}{\left(x-\frac{1}{x}\right)^2+2}\stackrel{\text{GMT}}{=}\frac{1}{8}\int_{-\infty}^{+\infty}\frac{dx}{x^2+2} = \frac{\pi}{8\sqrt{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How to show that $\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\frac{\pi^2}{6}$ Wolfram Alpha shows that
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\zeta(2)=\frac{\pi^2}{6}$$
I want to prove this.
Attempt:
I tried to treat this as a telescoping series:
$$\begin{align}
\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}&=\sum_{n=1}^{\infty}H_n\left(\frac{1}{n}-\frac{1}{n+1}\right)\\
&=H_{1}\left(1-\frac{1}{2}\right)+H_{2}\left(\frac{1}{2}-\frac{1}{3}\right)+H_{3}\left(\frac{1}{3}-\frac{1}{4}\right)\\
&=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{6}+\frac{1}{3}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{9}-\frac{1}{12}
\end{align}$$
I think this method is not quite useful, so I tried another one:
$$H_n=\int_{0}^{1}\frac{1-t^n}{1-t}dt$$
Then,
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\sum_{n=1}^{\infty}\frac{1}{n^2+n}\int_{0}^{1}\frac{1-t^n}{1-t}dt$$
At this point, I do not know how to proceed.
| Beside telescoping, all you need is to reverse the order of the double sum: $$\sum_{n\ge 1}\frac{\sum_{k=1}^n k^{-1}}{n(n+1)}=\sum_{k\ge 1}k^{-1}\sum_{n\ge k}\frac{1}{n(n+1)}=\sum_{k\ge 1}k^{-2}=\frac{\pi^2}{6}.$$Edit to explain the reversal: we're summing all terms of the form $\frac{k^{-1}}{n(n+1)}$ with $n,\,k$ integers in the set $\{(n,\,k)|n,\,k\ge 1,\,k\le n\}$. But I could equivalently describe this set as $\{(n,\,k)|k\ge 1,\,n\ge k\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Relation of Roots of Bi-quadratic equation without using Vieta's formula
If $ a, b, c, d $ are roots of the equation
$$ x^4 + px^3 + qx^2 + rx + s=0 $$
Show that
$$ (1+a^2)(1+b^2)(1+c^2)(1+d^2)=(1+s-q)^2 + (p-r)^2 $$
I was able to show this by using Vieta's formulas but the book from which I am doing this has not yet taught, till now it has just taught reminder, factor theorem and fundamental theorem of algebra so I want to ask that is there any other solution using only these or some more interesting way to do
| Also, we can use $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2.$$
Indeed,
$$(1+a^2)(1+b^2)(1+c^2)(1+d^2)=((a+b)^2+(1-ab)^2)((c+d)^2+(1-cd)^2)=$$
$$=((a+b)(c+d)-(1-ab)(1-cd))^2+((a+b)(1-cd)+(c+d)(1-ab))^2=$$
$$=(ab+ac+ad+bc+bd+cd-abcd-1)^2+(a+b+c+d-abc-abd-acd-bcd)^2=$$
$$=(q-s-1)^2+(-p+r)^2=(1+s-q)^2+(p-r)^2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the inverse of the following matrix: $A$ Find the inverseof the following matrix:
$$A= \begin{bmatrix}
0 & 1 & 0 & 0 &0 \\
1 & 0 & 0 & 0 &0 \\
0 & 0 & 0 & 0&1 \\
0 & 0 & 1 & 0 &0 \\0&0&0&1&0
\end{bmatrix}$$
My attempt : I think the inverse will not exist because $\det A=0 $
Is it True ?
Any hints/solution will be appreciated.
| Interchange rows three times ($R_1-R_2, R_3-R_4,R_4-R_5$):
$$\det(A)= \begin{vmatrix}
0 & \color{blue}1 & 0 & 0 &0 \\
\color{red}1 & 0 & 0 & 0 &0 \\
0 & 0 & 0 & 0&\color{green}1 \\
0 & 0 & \color{brown}1 & 0 &0 \\0&0&0&\color{purple}1&0
\end{vmatrix}=(-1)^3\begin{vmatrix}
\color{red}1 & 0 & 0 & 0 &0 \\
0 & \color{blue}1 & 0 & 0 &0 \\
0 & 0 & \color{brown}1 & 0& 0 \\
0 & 0 & 0 & \color{purple}1&0 \\0&0&0&0&\color{green}1
\end{vmatrix}=-1\ne 0.$$
Consider the expanded matrix and perform the same row interchanges:
$$\begin{vmatrix}
0 & \color{blue}1 & 0 & 0 &0 \\
\color{red}1 & 0 & 0 & 0 &0 \\
0 & 0 & 0 & 0&\color{green}1 \\
0 & 0 & \color{brown}1 & 0 &0 \\0&0&0&\color{purple}1&0
\end{vmatrix}\begin{vmatrix}
1 & 0 & 0 & 0 &0 \\
0 & 1 & 0 & 0 &0 \\
0 & 0 & 1 & 0&0 \\
0 & 0 & 0 & 1 &0 \\0&0&0&0&0
\end{vmatrix} \Rightarrow \begin{vmatrix}
\color{red}1 & 0 & 0 & 0 &0 \\
0 & \color{blue}1 & 0 & 0 &0 \\
0 & 0 & \color{brown}1 & 0& 0 \\
0 & 0 & 0 & \color{purple}1&0 \\0&0&0&0&\color{green}1
\end{vmatrix}\begin{vmatrix}
0 & 1 & 0 & 0 &0 \\
1 & 0 & 0 & 0 &0 \\
0 & 0 & 0 & 1&0 \\
0 & 0 & 0 & 0 &1 \\
0 & 0 & 1 & 0 &0
\end{vmatrix} \ \ \ \ $$
Hence, the inverse matrix is:
$$A^{-1}=\begin{vmatrix}
0 & 1 & 0 & 0 &0 \\
1 & 0 & 0 & 0 &0 \\
0 & 0 & 0 & 1&0 \\
0 & 0 & 0 & 0 &1 \\
0 & 0 & 1 & 0 &0
\end{vmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Prove the identity for $\tan3\theta$
Prove the identity for $$\tan3\theta= \frac{3\tan\theta - \tan^3 \theta}{1-3\tan^2 \theta}$$
Using de Moivre's theorem I have found that:
$$\cos3\theta = 4\cos^3\theta - 3\cos \theta$$
$$\sin 3\theta = 3\sin \theta-4\sin^3\theta$$
therefore:
$$\tan 3\theta = \frac{\sin 3\theta}{\cos 3 \theta}=\frac{3\sin \theta-4\sin^3\theta}{4\cos^3\theta - 3\cos \theta}$$
To then try and get the whole expression in terms of $\tan\theta$ I multiplied top and bottom of the fraction by $(4\cos^3\theta)^{-1}$. This gave me the following but I'm not now sure how to finish it off
$$\tan3\theta =\frac{\frac{3\sin \theta}{4\cos^3\theta} - \tan^3\theta}{1-\frac{3\cos \theta}{4\cos^3 \theta}}$$
| You have achieved the $cos 3x$ and $sin 3x$ answers by matching the Real and Imaginary components to show them in terms of only $cos x$ and $sin x$ I presume, leading on from an earlier question.
It should be a lot easier if you simply do exactly what you've done for the answers before matching the components
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3067903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
What is the radius and center of the image of $|z|=1$ under $ f(z) = \frac{3z+2}{4z+3}$?
I would like to compute the image of the circle $|z|=1$ about the fractional linear transformation:
$$ f(z) = \frac{3z+2}{4z+3} $$
In particular, I'd like to compute the new center and radius.
The Möbius transformation can be turned into inversion as well:
*
*$C_1= 4|z|^2+3\overline{z}-3z-2 $
*$C_2 =|z|^2 - 1$
Or we could turn the second circle into a fractional linear transforation $g(z) = - \frac{1}{z}.$
Then I could multiply the two transformations:
$$
\left[ \begin{array}{cc} 3 & 2 \\ 4 & 3 \end{array} \right]
\left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right]
=
\left[ \begin{array}{cc} 2 & 3 \\ 3 & 4 \end{array} \right]
$$
and this could turn back into a circle:
*
*$ C_1C_2 = 3|z|^2 + 4 \overline{z} + 3z + 2 $
I found this technique in a somewhat dated geometry textbook from the 1930's and I'm still figuring out their notation. I definitely like the idea that Möbius transformations and circles can be identified.
| Here's another solution I was able to find. Notice the matrix factorization:
$$
\left[ \begin{array}{cc} 3 & 2 \\ 4 & 3 \end{array} \right]
=
\left[ \begin{array}{cc} \frac{1}{5} & 0\\ 0 & 5 \end{array} \right] \times
\left[ \begin{array}{cc} 1 & 18 \\ 0 & 1 \end{array} \right] \times
\left[ \begin{array}{cr} \frac{3}{5}& -\frac{4}{5}\\ \frac{4}{5}& \frac{3}{5} \end{array} \right] = A \times B \times C
$$
The geometry behind this is that we have a Möbius transformation that factors into three parts:
$$ \text{Möbius} = rotation \times translation \times dilation $$
Now we have that $|z|=1$ is a circle centered at the origin passing through the points $z = \pm 1$ and $ z = i$. In fact, all these transformation will map to circles symmetric about the real axis. Here are the endpoints after the respective transformations:
$$ (-1,1) \stackrel{C}{\to} (7, - \frac{1}{7}) \stackrel{B}{\to} (25,\frac{125}{7})\stackrel{A}{\to} (1, \frac{5}{7}) $$
This corresponds to a circle centered at $z = \frac{6}{7}$ with radius $\frac{1}{7}$.
One possibility for computing this image circle is to notice the circle $|z|=1$ is a geodesic curve in the upper-half plane (with metric $ds^2 = \frac{dx^2 +dy^2}{y^2}$) and passing through the point $(z, \vec{u}) = (i, (1,0)) \in T_1(\mathbb{H}) $.
A Möbius transformation on $\mathbb{H}$ can be "lifted" to a Möbius transformation on $T_1(\mathbb{H})$ like this:
$$
\left[ z \mapsto \frac{az+b}{cz+d} \right] \to \left[
(z, \vec{u}) \mapsto \left( \frac{az+b}{cz+d} , \frac{\vec{u}}{(cz+d)^2} \right) \right] $$
Let's see what happens when I try the previous example here:
$$ \big(i, (1,0)\big) \mapsto \left( \frac{3i+2}{4i+3}, \frac{(1,0)}{(4i+3)^2}\right) = \left( \frac{18+i}{25} , \frac{1}{25}(24,-7) \right) $$
The factor of $\frac{1}{25}$ can be discarded since we only need the unit vector. This map is an isometry in hyperbolic space. The vector $\vec{u}$ would be tangent to a semi-circle with radius in the direction $\vec{u}_\perp$ passing through the point $f(z)=(\frac{18}{25}, \frac{1}{25})$. Therefore the center would be:
$$ (\frac{18}{25}, \frac{1}{25}) + \frac{1}{7 \times 25}(24,-7) = (\frac{1}{7},0) $$
agreeing with the previous answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3068729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 5
} |
Explicit calculation of $\int_0^{\infty} \frac{\sin(2\arctan x)}{(1+x^2)(\exp(2\pi x)-1)}\mathrm{d}x$ Is it possible to confirm the value of this integral using the methods of complex analysis or similar?
$$
\int_0^{\infty} \frac{\sin(2\arctan x)}{(1+x^2)(\exp(2\pi x)-1)}\mathrm{d}x=\frac{\pi^2-9}{12}
$$
Of course, one can reduce it to the definition of the polylogarithm and a $\zeta$-function, but I was looking for an explicit calculation.
| Here is an approach that first converts the integral to a double integral.
Let
$$I = \int_0^\infty \frac{\sin (2 \tan^{-1} x)}{(1 + x^2) (e^{2 \pi x} - 1)} \, dx = 2 \int_0^\infty \frac{x}{(1 + x^2)^2 (e^{2 \pi x} - 1)} \, dx.$$
Observe that
$$\frac{1}{2} \int_0^\infty y e^{-y} \sin (xy) \, dy = \frac{x}{(1 + x^2)^2}.$$
Thus the integral can be rewritten as
$$I = \int_0^\infty y e^{-y} \int_0^\infty \frac{\sin (xy)}{e^{2 \pi x} - 1} \, dx \, dy, \tag1$$
after a change of order has been made.
For the inner $x$-integral
\begin{align}
\int_0^\infty \frac{\sin (xy)}{e^{2 \pi x} - 1} \, dx &= \int_0^\infty \frac{e^{-2 \pi x} \sin (xy)}{1 - e^{-2 \pi x}} \, dx\\
&= \int_0^\infty \sum_{n = 0}^\infty e^{-2 \pi x} \sin (xy) \cdot e^{-2\pi n x} \, dx\\
&= \sum_{n = 1}^\infty \int_0^\infty e^{-2\pi n x} \sin (xy) \, dx,
\end{align}
where a shift in the index of $n \mapsto n - 1$ has been made. Now integrating by parts twice leads to
$$\int_0^\infty \frac{\sin (xy)}{e^{2 \pi x} - 1} \, dx = \sum_{n = 1}^\infty \frac{y}{4 \pi^2 n^2 + y^2} = \frac{1}{2 \pi} \sum_{n = 1}^\infty \frac{\frac{y}{2\pi}}{\left (\frac{y}{2 \pi} \right )^2 + n^2}. \tag2$$
Using the well-known result of
$$\pi \coth (\pi z) = \frac{1}{z} + 2 \sum_{n = 1}^\infty \frac{z}{z^2 + n^2} , \quad z \neq 0,$$
the sum in (2) can be expressed as
$$\int_0^\infty \frac{\sin (xy)}{e^{2 \pi x} - 1} \, dx = \frac{1}{4} \coth \left (\frac{y}{2} \right ) - \frac{1}{2y}.$$
On returning to our integral in (1), we have
\begin{align}
I &= -\frac{1}{2} \int_0^\infty e^{-y} + \frac{1}{4} \int_0^\infty y e^{-y} \coth \left (\frac{y}{2} \right ) \, dy = -\frac{1}{2} + \frac{1}{4} J,
\end{align}
where
$$J = \int_0^\infty y e^{-y} \coth \left (\frac{y}{2} \right ) \, dy.$$
Finding $J$ we have
\begin{align}
J &= \int_0^\infty y e^{-y} \frac{e^{y/2} + e^{-y/2}}{e^{y/2} - e^{-y/2}} \, dy\\
&= \int_0^\infty y e^{-y} \frac{1 + e^{-y}}{1 - e^{-y}} \, dy\\
&= \int_0^\infty y e^{-y} \left (1 + \frac{2 e^{-y}}{1 - e^{-y}} \right ) \, dy\\
&= \int_0^\infty y e^{-y} \, dy + 2 \int_0^\infty \frac{y e^{-2y}}{1 - e^{-y}} \, dy\\
&= 1 + 2 \sum_{n = 0}^\infty \int_0^\infty y e^{-y(n + 2)} \, dy\\
&= 1 + 2 \sum_{n = 0}^\infty \frac{1}{(n + 2)^2} \qquad \text{(by parts)}\\
&= 1 + 2 \sum_{n = 2}^\infty \frac{1}{n^2} \\
&= 1 + 2 \sum_{n = 1}^\infty \frac{1}{n^2} - 2\\
&= -1 + 2 \cdot \frac{\pi^2}{6}\\
&= -1 + \frac{\pi^2}{3}.
\end{align}
So finally we have
$$I = -\frac{1}{2} + \frac{1}{4} \left (-1 + \frac{\pi^2}{3} \right ),$$
or
$$\int_0^\infty \frac{\sin (2 \tan^{-1} x)}{(1 + x^2) (e^{2 \pi x} - 1)} \, dx = \frac{\pi^2 - 9}{12},$$
as expected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3072149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Power tower last digits Can anyone solve the following problem of finding the 6th last digit from the right of the decimal representation of the following number:
$6^{6^{6^{6^{6^{6}}}}}$
Essentially it means reducing this modulo $10^6$ and supposedly Chinese Remainder Theorem should be used, but I have no idea how to solve this. Can anyone help?
| We're interested in $6^x \mod 10^6$; this is determined by $6^i \mod 2^6$ and $6^i \mod 5^6$. Mod $2^6$ it's easy: $6^x$ is divisible by $2^6$ if $x \ge 6$. $6$ is coprime to $5^6$ with multiplicative order $3125 = 5^5$ mod $5^6$. In fact, the multiplicative order of $6$ mod $5^m$ seems to be $5^{m-1}$ (prove!). So
$$\eqalign{6 &\equiv 1 \mod 5\cr 6^6 &\equiv 6^1 = 6 \mod 5^2\cr 6^{6^6} &\equiv 6^6 \equiv 31 \mod 5^3\cr
6^{6^{6^6}} &\equiv 6^{31} \equiv 531 \mod 5^4\cr
6^{6^{6^{6^6}}} &\equiv 6^{531}\equiv 1156 \mod 5^5\cr
6^{6^{6^{6^{6^6}}}} &\equiv 6^{1156} \equiv 4281 \mod 5^6\cr
}$$
and using Chinese Remainder, $$6^{6^{6^{6^{6^6}}}} \equiv 238656 \mod 10^6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3072436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Trying to solve this triple integral: $\iiint (x-1)(y-1) \,dx\,dy\,dz$ Here's the question
$$\iiint (x-1)(y-1) \,dx\,dy\,dz.$$
I am asked to evaluate this integral over the region $$D:=\left \{ (x,y,z) \in\mathbb{R}^3 :x^2+y^2 \leq z \leq 2x+2y+2 \right \}.$$
There are the bounds of integration in set D (the variable $z$ is isolated) and well I tried to find the solution of this integral :
\begin{align*}
&\iint_{Pr_{(y,x)}(D)}\int_{x^2+y^2}^{2x+2y+2}(x-1)(y-1) \,dx\, dy\, dz \\
=&\iint_{Pr_{(y,x)}(D)}\int_{x^2+y^2}^{2x+2y+2}(xy-x-y+1) \,dx\, dy\, dz,
\end{align*}
and integrate only with respect to $z.$ I have that:
\begin{align*}
\int_{x^2+y^2}^{2x+2y+2}(xy-x-y+1) \,dz&=(xy-x-y+1)*(2x+2y+2-(x^2+y^2)) \\
&=3x^2y+3xy^2-2x^3y-2xy^3-3x^2+x^3-2xy-3y^2+y^3+2.
\end{align*}
It looks like this way is too long. The second thing that came to mind when I saw the set $D$ was to apply cylindrical coordinates, but this doesn't make easier the left member of the set $D.$
What can I do or what have I done wrong up until now?
Any support for this question would be appreciated.
| Hint. Note that the intersection of paraboloid $z=x^2+y^2$ and the plane $z=2x+2y+2$ projected onto $xy$-plane is given by the circle $x^2+y^2=2x+2y+2$ that is
$$(x-1)^2+(y-1)^2=2^2.$$
Hence the given triple integral becomes
$$\iint_{(x-1)^2+(y-1)^2\leq 2^2}\int_{x^2+y^2}^{2x+2y+2}(x-1)(y-1) dx dy dz,$$
and after integrating with respect to $z$ we get
$$\iint_{(x-1)^2+(y-1)^2\leq 2^2}(x-1)(y-1)(4-(x-1)^2-(y-1)^2) dx dy.$$
Now use the polar coordinates centered at $(1,1)$. Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3077372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I solve for $a$ and $b$ in $\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2$? Given $\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2$
I need to solve for $a$ and $b$, so here we go,
$\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right)$
$= \lim\limits_{x \to ∞} x \left(2 +(3+x) \left(\ln (1+\frac{a}{x}) - \ln(1+\frac{b}{x}) \right) \right)$
$=\lim\limits_{x \to ∞} x \left(2 +(3+x)\left( \dfrac{a-b}{x} \right) \right)$
$=\lim\limits_{x \to ∞} \left(2x +(3+x)\left( a-b \right) \right)$
$=\lim\limits_{x \to ∞} \left(2x + 3(a-b) + x(a-b) \right) $
Since the limit exists, $2x$ and $x(a-b)$ must cancel out so $a-b = 2$
But if I do this, I'm left with $3(a-b) = 2$ which gives me a contradiction since $a-b$ is supposed to be $2$.
What do I do now? And where exactly have I gone wrong?
Thank you!
| Hint
First of all, let $x=\frac 1y$ to make the expression
$$A=\frac{(3 y+1) \log \left(\frac{1+a y}{1+b y}\right)+2 y}{y^2}$$ Now, using Taylor series, you should have
$$A=\frac{a-b+2}{y}+\frac{1}{2} \left(-a^2+6 a+b^2-6 b\right)+\frac{1}{6} y \left(2
a^3-9 a^2-2 b^3+9 b^2\right)+O\left(y^2\right)$$ So
$$a-b+2=0 \qquad \text{and} \qquad \frac{1}{2} \left(-a^2+6 a+b^2-6 b\right)=2$$ This is easy to solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3079016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
$a+b=c+d$ , $a^3+b^3=c^3+d^3$ , prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$
Let $a, b, c, d$ be four numbers such that $a + b = c + d$ and $a^3 + b^3 = c^3 + d^3$. Prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$.
I've got $a^2+b^2-ab=c^2+d^2-cd$.
I tried squaring or cubing it repeatedly but I didn't get what I wanted.
Now how do I proceed?
| Hint: the case $a+b=c+d=0$ is obvious, so you can assume $a+b=c+d\ne0$; prove that $ab=cd$. Further hint: $x^2-xy+y^2=(x+y)^2-{\color{red}{?}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
Prove that for all positive real numbers $x$ and $y$, $(x+y)(\frac1x+\frac1y)\ge4$. Prove that for all positive real numbers $x$ and $y$,
$(x+y)(\frac1x+\frac1y)\ge4$.
Any help would be appreciated thanks.
| To begin with, notice that
\begin{align*}
(x+y)\left(\frac{1}{x} + \frac{1}{y}\right) = 2 + \frac{x}{y} + \frac{y}{x}
\end{align*}
According to the inequality $AM\geq GM$, we have
\begin{align*}
\frac{x}{y} + \frac{y}{x} \geq 2\sqrt{\frac{x}{y}\cdot\frac{y}{x}} = 2
\end{align*}
from whence we obtain the claimed result.
EDIT
Due to JavaMan's observation, I provide you another approach to solve the second part. Indeed, for any $a\in\textbf{R}_{>0}$, we have
$$(a-1)^{2}\geq 0 \Leftrightarrow a^{2} - 2a + 1 \geq 0 \Leftrightarrow a^{2} + 1 \geq 2a\Leftrightarrow a + \frac{1}{a} \geq 2$$
In the present case, it suffices to take $a = x/y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Determine Fourier series expansion for $f(\theta)=\cos^4\!\theta$ Q: The function $f(\theta) = \cos^4\! \theta$ is a nice smooth function, so will have a Fourier series expansion. That is, it will have an expansion as a sum of functions $\cos j \theta$ and $\sin j \theta$ with real coefficients. Determine what the expansion is.
For reference, the Fourier coefficients are:
\begin{align*}
c_n &= \frac{1}{2\pi} \int_0^{2 \pi} f(t) e^{-int} \, dt = \frac{1}{2\pi} \int_0^{2 \pi} f(t) \left( \cos (nt) - i \sin (nt) \right) \, dt \\
\end{align*}
The coefficients give the Fourier Series expansion:
\begin{align*}
f(t) &= \sum\limits_{n=0}^\infty c_n e^{int} = \sum\limits_{n=0}^\infty c_n \left( \cos (nt) + i \sin (nt) \right) \\
\end{align*}
One route is to plug $f(\theta)$ into the equation for the Fourier series coefficients:
\begin{align*}
c_n &= \frac{1}{2\pi} \int_0^{2 \pi} \cos^4 t \cdot \left[ \cos (nt) - i \sin (nt) \right] \, dt \\
\end{align*}
That integral is looking complex. I suspect there is an easier solution to this problem?
| Using $$\cos^2(\theta )=\frac{1+\cos(2\theta )}{2},$$ one get $$\cos^4(\theta )=\frac{1+2\cos(2\theta )+\cos^2(2\theta )}{4}=\frac{1+2\cos(2\theta )+\frac{1+\cos(4\theta )}{2}}{4}$$
$$=\frac{3}{8}+\frac{1}{2}\cos(2\theta )+\frac{1}{8}\cos(4\theta ).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Cannot find angle for trigonometry problem
A right angle triangle is entrapped within a circle. The triangle entraps within it a circle of its own. The ratio between the big radius and the little radius is $\frac{13}{4}$. What are the angles of the triangle?
Here is the problem as I understand it, given that:
1. $\measuredangle ABC = 90^\circ \rightarrow AC = 2R$
2. The smaller radii are perpendicular to the triangle.
3. The origin of the smaller circle is the intersection between the angle bisectors of triangle ABC.
Let $\measuredangle ACB = 2\alpha \rightarrow \measuredangle ACG = \alpha$
$\measuredangle BAC = 90 - 2\alpha \rightarrow \measuredangle GAC = 45 - \alpha$
From here we can construct the following system:
$$\begin{cases} 2R = \frac{r}{\tan \alpha} + \frac{r}{\tan(45 - \alpha)} \\ \frac{R}{r} = \frac{13}{4} \rightarrow R = \frac{13r}{4} \end{cases}$$
$$\downarrow \\ \frac{26r}{4} = r\left(\frac{1}{\tan \alpha} + \frac{1}{\tan(45 - \alpha)}\right) \quad / \div r \\ 6.5 = \frac{\cos \alpha}{\sin \alpha} + \frac{\cos(45 - \alpha)}{\sin(45 - \alpha)} \quad / {\sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \sin\beta\cos\alpha\\ \cos(\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta} \\ 6.5 = \frac{\cos \alpha}{\sin \alpha} + \frac{\cos \alpha \cos 45 + \sin \alpha \sin 45}{\sin 45 \ cos \alpha - \sin \alpha \cos 45} \\ 6.5 = \frac{\cos \alpha}{\sin \alpha } + \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} \cdot \frac{\cos \alpha + \sin \alpha}{\cos \alpha - \sin \alpha} \\ 6.5 = \frac{\cos^2 \alpha - \sin\alpha \cos \alpha + \cos \alpha \sin \alpha + \sin^2 \alpha}{\sin(\cos\alpha - \sin \alpha)} \quad / \sin^2\alpha + \cos ^2 \alpha = 1\\ 6.5 = \frac{1}{\sin\alpha (\cos\alpha - \sin \alpha)}$$
From this point on, I have no identity that I can think of which can be applied here to help me solve for $\alpha$. Amusingly enough, I got an answer which was very close to the correct one, but that was only because I made a mistake in the signs of the identity, and the second fracture turned into 1.
I am aware that they may be an identity which can solve this, but I am restricted to only use the identities available here. I have either made a mistake somewhere, or just don't know how to use the available identities to solve the problem.
| We will use that$$2r=a+c-b$$ and $$2R=b$$ then $$16b=13a+13c-13b$$ so $$29b=13(a+c)$$ and with $$a^2+c^2=b^2$$ we can eliminate $b$ so we get
$$\frac{a}{c}+\frac{c}{a}=\frac{169}{336}$$ from here you will get the quotient of $$\frac {a}{c}$$ and thus we can calculate the angles of triangle $$\Delta ABC$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3084842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
For the equation $y = 4x^2 + 8x + 5$, what are the integer values of x such that y/13 is an integer? For the equation $y = 4x^2 + 8x + 5$, what are the integer values of x such that y/13 is an integer?
For example, if x = 3, $y = 4(3^2) + 8(3) + 5$ = 65 which is divisible by 13
if x = 8, y = 325 which is divisible by 13
if x = 16, y = 1157 which is divisible by 13
if x = 21, y = 1937 which is divisible by 13
I am guessing that values of x = 13i + 3 or x = 13i + 8 where i is an integer will result in a value of y that is evenly divisible by 13.
How do you prove that x = 13i + 3 or x = 13i + 8 will result in a value of y that is evenly divisible by 13?
Is there a general proof to find values of x that will result in a value of y that is evenly divisible by an odd integer p?
| You're on the right track.
When $x=13i+3$, $y=4(13i+3)^2+8(13i+3)+5= 4(169i^2+78i+9)+8(13i+3)+5$.
Leaving out some multiples of $13$, this is $4\times 9+8\times3+5=36+24+5=65=5 \times 13$.
This shows that, when $x=13i+3$, $y$ is a multiple of $13.$
(I'll leave $x=13i+8$ as an exercise.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Prove $\sqrt{b} - \sqrt{a} < \sqrt{b-a}$
Prove that if $0 < a < b$ then
$$\sqrt{b} - \sqrt{a} < \sqrt{b-a}$$
This is what I have so far:
square both sides to get $a + b -2\sqrt{ab} < b-a$
subtract $b$ from both sides $a-2\sqrt{ab} < -a$
add $a$ to both sides $2a-2\sqrt{ab} < 0$
than add $2 \sqrt{ab}$ to both sides and get $2a < 2\sqrt{ab}$
divide by $2$ and get $a < \sqrt{ab}$
Thus we know $\sqrt{ab}$ is bigger than $a$ so that $\sqrt{a \cdot a} < \sqrt{ab}$ which means $\sqrt{a} < \sqrt{b}$.
Therefore together with the given $a < b$, we have $\sqrt{b} - \sqrt{a} < \sqrt{b - a}$
I'm confused as to where I went wrong and how to fix this.
| It seems the OPs is somewhat awkwardly presented, and so ascertaining its validity is a bit of a challenge. If the proof were more concise, it might be easier to decipher. Here's a suggestion:
We recall that
$0 < a < b; \tag 0$
then, note that
$b = a + (b - a) < a + 2\sqrt a \sqrt{b - a} + (b - a) = (\sqrt a + \sqrt{b - a})^2; \tag 1$
thus,
$\sqrt b < \sqrt a + \sqrt{b - a}, \tag 2$
whence
$\sqrt b - \sqrt a < \sqrt{b - a}. \tag 3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
} |
maximum and minimum of $x^2+y^4$ for real $x,y$ If $y^2(y^2-6)+x^2-8x+24=0$ then maximum and minimum value of $x^2+y^4$ is
what i try
$y^4-6y^2+9+x^2-8x+16=1$
$(x-4)^2+(y^2-3)^2=1\cdots (1)$
How i find maximum and minimum of $x^2+y^4$ from $(1)$ relation
help me to solve it please
| You could compare equation (1) with a standard ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ where $(x, y)$ has parametric co-ordinates $(acos\theta, bsin\theta)$.
So in your case $(x, y^2)$ could be represented by $(4+cos\theta, 3+sin\theta)$.
Then $x^2 + y^4 = 25 + 1 + 8cos\theta+6sin\theta$ which has a maximum of $26+10$ and a minimum of $26-10$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding all of the solutions of the functional equation $P(x+2)+P(x-2) = 8x+6$
Let $P(x+2)+P(x-2) = 8x+6$. Find all of the solutions for $P(x)$.
If we put $P(x) = ax + b $ the answer is obvious but how to determine all of the solutions?
| This difference equation is linear so it's solution can be reprasented as
$$
P(x) = P_h(x) + P_p(x)
$$
with
$$
P_h(x+2)+P_h(x-2) = 0\\
P_p(x+2)+P_p(x-2) = 8x+6
$$
for the homogeneous we propose $P_h(x) = \gamma^x$ and after substituting
$$
\gamma^x\left(\gamma^2+\frac{1}{\gamma^2}\right)=0
$$
and the solutions are
$$
\gamma_h = \left\{-\frac{1+i}{\sqrt{2}},\frac{1+i}{\sqrt{2}},\frac{1-i}{\sqrt{2}},-\frac{1-i}{\sqrt{2}}\right\}
$$
so the homogeneous solution is
$$
P_h(x) = C_1\sin\left(\frac{\pi x}{4}\right)+C_2\cos\left(\frac{\pi x}{4}\right)+C_3\sin\left(\frac{3\pi x}{4}\right)+C_4\cos\left(\frac{3\pi x}{4}\right)
$$
now for the particular making $P_p(x) = 4x+3$ we conclude
$$
P(x) = 4x+3+C_1\sin\left(\frac{\pi x}{4}\right)+C_2\cos\left(\frac{\pi x}{4}\right)+C_3\sin\left(\frac{3\pi x}{4}\right)+C_4\cos\left(\frac{3\pi x}{4}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3099690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find rational numbers $\alpha $ and $\beta$ in $\sqrt[3]{7+5\sqrt{2}}=\alpha+\beta\sqrt{2}$ How should we find two rational numbers $\alpha$ and $\beta$ such that
$$\sqrt[3]{7+5\sqrt{2}}=\alpha+\beta\sqrt{2}$$
The answer I got $\alpha = 1 $ and $\beta = 1$. If I'm wrong, please correct me. Thank you
| You are right.
We can get your answer by the following way.
Let $\sqrt[3]{7+5\sqrt2}=x$.
Thus, $x>2$ and $$x^3=7+5\sqrt2,$$ which gives
$$(x^3-7)^2=50$$ or
$$x^6-14x^3-1=0$$ or
$$x^3-\frac{1}{x^3}-14=0.$$
Now,let $x-\frac{1}{x}=t$.
Thus, $t>0$ and we obtain:
$$t^3+3t-14=0$$ or
$$t^3-2t^2+2t^2-4t+7t-14=0$$ or
$$(t-2)(t^2+2t+7)=0,$$ which gives $$t=2,$$
$$x-\frac{1}{x}=2$$ or
$$x^2-2x-1=0$$ or
$$x=1+\sqrt2,$$ which says $$\alpha=\beta=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3099920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Prove that $1+b+(1+c)+1/c+1+a \ge 3$ if $a, b,$ and $c$ are positive real numbers. Let $a, b, c$ be positive real numbers. prove that
$$
\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\ge\frac{3}{1+abc},
$$
and that equality occurs if and only if $a = b = c = 1$.
What I tried:
$1$st part: I tried a brute force approach where I make a common denominator for all $4$ fractions, but the number get real big real fast.
| Let $$(a;b;c)\rightarrow (\frac{kx}{y};\frac{ky}{z};\frac{kz}{x})\Rightarrow abc=k^3$$
Then we will need to prove $$\frac{yz}{kx\left(ky+z\right)}+\frac{xz}{ky\left(kz+x\right)}+\frac{xy}{kz\left(kx+y\right)}\ge \frac{3}{k^3+1}$$
By Cauchy-Schwarz:
$$LHS=\sum_{cyc}\frac{y^2z^2}{kxyz\left(ky+z\right)}\ge \frac{\left(xy+yz+xz\right)^2}{xyz\left(x+y+z\right)k\left(k+1\right)}$$
$$\ge\frac{3xyz\left(x+y+z\right)}{xyz\left(x+y+z\right)k\left(k+1\right)}=\frac{3}{k\left(k+1\right)}$$
So we will prove $$\frac{3}{k\left(k+1\right)}\ge \frac{3}{k^3+1}\Leftrightarrow \frac{3\left(k-1\right)^2}{k\left(k+1\right)\left(k^2-k+1\right)}\ge 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3100653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integral $\int_0^1 \frac{\arctan x}{x^2-x-1}dx$ After seeing this integral I've decided to give a try to calculate:
$$I=\int_0^1 \frac{\arctan x}{x^2-x-1}dx$$
That is because it's common for many integrals to have a combination of a polynomial in the denominator and a logarithm or an inverse trig function in the numerator.
Mostly I tried standard ways such as integrating by parts, random substitutions, or using:
$$\frac{\arctan x}{x}=\int_0^1 \frac{dy}{1+x^2y^2}$$
But I realised that is not a great idea since it gives some mess after partial fractions, so I decided to prepare the integral a little for a Feynman's trick using probably the only helpful thing with that denominator, it won't change while using $x\mapsto 1-x$.
$$I=\int_0^1 \frac{\arctan x}{x^2-x-1}dx=\int_0^1 \frac{\arctan (1-x)}{x^2-x-1}dx$$
$$\Rightarrow 2I=\frac{\pi}{2 \sqrt 5}\ln\left(\frac{3-\sqrt 5}{3+\sqrt 5}\right)-\int_0^1 \frac{\arctan(x^2-x+1)}{x^2-x-1}dx$$
$$\small J(a)=\int_0^1 \frac{\arctan(a(x^2-x-1)+2)}{x^2-x-1}dx\Rightarrow J'(a)=\int_0^1 \frac{1}{1+(a(x^2-x-1)+2)^2}dx$$
But I am stuck now. I would like to see a method which finds a closed form for this integral, hopefully something decent comes out. I already imagine there will be some special functions.
| Here is one approach. As a warning, the final answer I find is not pretty.
Let
$$I = \int_0^1 \frac{\tan^{-1} x}{x^2 -x - 1} \, dx.$$
Start by using a self-similar substitution of
$$x = \frac{1 - u}{1 + u}, \,\, dx = -\frac{2}{(1 + u)^2} \, du.$$
So , after having reverted the dummy variable $u$ back to $x$, we have
$$I = 2 \int_0^1 \frac{\tan^{-1} \left (\frac{1 - x}{1 + x} \right )}{x^2 - 4x - 1} \, dx.$$
Noting that for $0 < x < 1$
$$\tan^{-1} \left (\frac{1 - x}{1 + x} \right ) = \frac{\pi}{4} - \tan^{-1} x,$$
then
\begin{align}
I &= \frac{\pi}{2} \int_0^1 \frac{dx}{x^2 - 4x - 1} - 2 \int_0^1 \frac{\tan^{-1} x}{x^2 - 4x - 1} \, dx\\
&= -\frac{\pi}{2 \sqrt{5}} \coth^{-1} \left (\frac{3}{\sqrt{5}} \right ) - 2 J,
\end{align}
where
$$J = \int_0^1 \frac{\tan^{-1} x}{x^2 - 4x - 1} \, dx.$$
To find $J$ we begin by noting that $\tan^{-1} x = \operatorname{Im} \ln (1 + ix)$. Thus
$$J = \operatorname{Im} \int_0^1 \frac{\ln (1 + ix)}{x^2 - 4x - 1} \, dx.$$
Making a substitution of $t = 1 + ix$ we have
$$J = - \operatorname{Re} \int_1^{1+i} \frac{\ln t}{(t - \alpha)(t - \beta)} \, dt,$$
where $\alpha = 1 + i(2 - \sqrt{5})$ and $\beta = 1 + i(2 + \sqrt{5})$. After performing a partial fraction decomposition we are left with
$$J = \frac{1}{2 \sqrt{5}} \operatorname{Im} \left [\int_1^{1 + i} \frac{\ln t}{\alpha - t} \, dt - \int_1^{1 + i} \frac{\ln t}{\beta - t} \, dt \right ].$$
Now, as
$$\int \frac{\ln x}{z - x} \, dx = - \ln \left (1 - \frac{x}{z} \right ) \ln x - \operatorname{Li}_2 \left (\frac{x}{z} \right ),$$
(for a proof of this result see the appendix below), one has
$$J = \frac{1}{2 \sqrt{5}} \operatorname{Im} \left [\ln (1 + i) \ln \left (\frac{\alpha}{\beta} \cdot \frac{\beta - 1 - i}{\alpha - i - i} \right ) + \operatorname{Li}_2 \left (\frac{1}{\alpha} \right ) - \operatorname{Li}_2 \left (\frac{1}{\beta} \right ) + \operatorname{Li}_2 \left (\frac{1 + i}{\beta} \right ) - \operatorname{Li}_2 \left (\frac{1 + i}{\alpha} \right ) \right ]$$
or after performing a huge amount of algebra
\begin{align}
J &= \frac{\pi}{8 \sqrt{5}} \ln (\sqrt{5} - 1) + \frac{1}{2 \sqrt{5}} \operatorname{Im} \left [\operatorname{Li}_2 \left (\frac{1}{2} + \frac{1}{\sqrt{5}}+ \frac{i}{2 \sqrt{5}} \right ) - \operatorname{Li}_2 \left (\frac{1}{2} - \frac{1}{\sqrt{5}} - \frac{i}{2 \sqrt{5}} \right ) \right.\\
& \quad+ \left. \operatorname{Li}_2 \left (\frac{1}{2} -\frac{1}{2 \sqrt{5}} - i \left (\frac{3}{2 \sqrt{5}} - \frac{1}{2} \right ) \right ) - \operatorname{Li}_2 \left (\frac{1}{2} +\frac{1}{2 \sqrt{5}} + i \left (\frac{3}{2 \sqrt{5}} + \frac{1}{2} \right ) \right ) \right ]\\
&= \frac{\pi}{8 \sqrt{5}} \ln (\sqrt{5} - 1) + \frac{1}{2 \sqrt{5}} \operatorname{Im} \frak{w},
\end{align}
where $\frak{w}$ is the term containing the four dilogarithms with complex arguments. Thus
$$\int_0^1 \frac{\tan^{-1} x}{x^2 - x - 1} \, dx = -\frac{\pi}{4 \sqrt{5}} \left (\ln 2 + \sinh^{-1} (2) \right ) - \frac{1}{\sqrt{5}} \operatorname{Im} \frak{w}.$$
Note that as $\operatorname{Im} {\frak{w}} = -0.8363170651979\ldots$ we see that $I \approx -0.376513$.
Appendix
Proof of
$$\int \frac{\ln x}{z - x} \, dx = - \ln \left (1 - \frac{x}{z} \right ) \ln x - \operatorname{Li}_2 \left (\frac{x}{z} \right ) + C$$
Setting $t = x/z, dt = dx/z$, we have
\begin{align}
\int \frac{\ln x}{z - x} \, dx &= \int \frac{\ln (zt)}{1 - t} \, dt\\
&= -\ln (1 - t) \ln (zt) + \int \frac{\ln (1 - t)}{t} \, dt \qquad \text{(by parts)}\\
&= -\ln (1 - t) \ln (zt) - \operatorname{Li}_2 (t) + C\\
&= - \ln \left (1 - \frac{x}{z} \right ) \ln x - \operatorname{Li}_2 \left (\frac{x}{z} \right ) + C,
\end{align}
as required to show.
| {
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"url": "https://math.stackexchange.com/questions/3105528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Integration: Integrating then substituting a number has a different answer to substituting a number then integrating; for trigonometric functions This might be a very dumb question but I am really stuck as to why this happens. Now, the answer to the integral
$$B_n = 2 \int_0^1 k\sin(3\pi x)\sin(n\pi x) dx \qquad (k = 0.01,
n=1,2,3,...)$$
is $k$ when $n=3$ and $0$ when $n\neq3 $.
But when I try to solve for a general solution I get
$$
\begin{align}
B_n &= 2k\int_0^1 \frac12[\cos(3\pi x-n\pi x)-\cos(3\pi x+n\pi x)]dx
\\ &= k\int_0^1 \cos(3\pi x-n\pi x)-cos(3\pi x+n\pi x)dx
\\ &= k[\frac{1}{3\pi - n\pi}\sin(3\pi x - n\pi x) - \frac{1}{3\pi+ n\pi}\sin(3\pi x+ n\pi x)]_0^1
\\ &= k(\frac{1}{3\pi - n\pi}\sin(3\pi - n\pi) - \frac{1}{3\pi+ n\pi}\sin(3\pi+ n\pi)
)\end{align}
$$
which gives $0$ for all integer $n$.
Why is that?
| Note that $\dfrac{1}{3\pi - n\pi}$ is not defined at $n=3$. So you need to solve the integral for $n=3$ separately.
When $n=3$, we have
$$k\int_0^1 \cos(3\pi x-n\pi x)-\cos(3\pi x+n\pi x)dx=k\int_0^1 \cos(0)-cos(6\pi x)dx=k\int_0^1 1-\cos(6\pi x)dx.$$
Can you complete this?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculate limit with squares $\lim_{n \to \infty}\left(\sqrt[3]{n^3+2n-1}-\sqrt[3]{n^3+2n-3}\right)^6 \cdot (1+3n+2n^3)^4$
$$\lim_{n \to \infty}\left(\sqrt[3]{n^3+2n-1}-\sqrt[3]{n^3+2n-3}\right)^6 \cdot (1+3n+2n^3)^4 $$
What I did was to multiply it and I got $\frac{1}{2}$ as the final result. Could someone confirm if it's correct?
| Use the binomial expansion or, better, Taylor series to show that
$$\sqrt[3]{n^3+2n-1}=n+\frac{2}{3 n}-\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\sqrt[3]{n^3+2n-3}=n+\frac{2}{3 n}-\frac{1}{ n^2}+O\left(\frac{1}{n^3}\right)$$
$$\sqrt[3]{n^3+2n-1}-\sqrt[3]{n^3+2n-3}=\frac{2}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ The remaining looks to be simple.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3106079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How do we factor out $x^2 - x -2$ in this expression? Suppose we're given that
$$x^4 - 2x^3 +x-2$$
How do we factor out $x^2 - x -2$ in this expression?
$$(x^4 -x^3- 2x^2)-(x^3-x^2-2x)+(x^2-x-2) = x^2(x^2 -x-2)-x(x^2-x-2)+(x^2-x-2) = (x^2-x-2)(x^2-x+1)$$
This satisfies with what we want to get. However, I do not seem to have understood what is done there. Could I get your assistance in order to understand it? Perhaps there's better way of factoring.
Regards
| $$\begin{align}
x^4-2x^3+x-2&= x^4 + (\color{red}{-2x^3+x-2})\\
&= x^4 + \color{red}{P(x)}\\
&= x^4 + \color{green}{-x^3-2x^2+}(\color{green}{x^3+2x^2}+\color{red}{P(x)})\\
&= \color{blue}{(x^4 + -x^3-2x^2)}+(x^3+2x^2+\color{red}{P(x)})\\
&= (x^2-x-2)(x^2)+(x^3+2x^2+\color{red}{P(x)})\\
&= (x^2-x-2)(x^2)+(x^3+2x^2+\color{red}{-2x^3+x-2})\\
&= (x^2-x-2)(x^2)+(-x^3+2x^2+x-2)\\
&= \dots\\
\end{align}$$
Try comparing the line with blue highlighting with your original problem (e.g. $x^4-x^3-2x^2$). Now, keep going, and try to split up the rest of the polynomial, using the same method.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3106729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 5
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Relation involving generalized Laguerre polynomials Playing around with different approaches to solve the radial part of the Schrodinger equation for the hydrogen-like atom, I have obtained the following expression ($l$ and $n$ are non-negative integers)
$$ \tilde L^{(2l+1)}_n(x) = \frac{1}{n!} x^{-(l+1)}e^{x/2} \left(x \frac{d}{dx}-x \frac{d^2}{dx^2} + \frac{l(l+1)}{x} -\frac{x}{4} \right)^n \bigl(x^{l+1} e^{-x/2} \bigr)
$$
which I believe is a
representation of the generalized Laguerre polynomials.
The latter are conventionally defined as via de Rodrigues formula
$$L_n^{(2l+1)}(x) = \frac{1}{n!} x^{-(2l+1)} e^{x} \frac{d^n}{dx^n} \Bigl(x^{2l+1+n} e^{-x} \Bigr)\,. $$
I have the feeling that there is a simple way to show that the two expressions are the same with $L= \tilde L$. Can somebody point me in the right direction?
| Denoting
\begin{equation}
h(x)=x^{l+1}e^{-x/2};\quad \varphi(x)=x^{2l+1}e^{-x}; \quad\Theta=x \frac{d}{dx}-x \frac{d^2}{dx^2} + \frac{l(l+1)}{x} -\frac{x}{4}
\end{equation}
both polynomial definitions can be written as
\begin{align}
\tilde{L}^{(2l+1)}_n(x) &= \frac{1}{n!h(x)}\Theta^n\left[h(x)\right]\\
L^{(2l+1)}_n(x) &= \frac{1}{n!\varphi(x)}\frac{d^n}{x^n}\left[x^n\varphi(x)\right]
\end{align} We introduce the function $$F(x)=x^{-l}e^{x/2}$$
It can easily be checked that $\Theta\left[F(x)\right]=0$. Then, for any function $z(x)$,
\begin{equation}
\Theta\left[F(x)z(x)\right]=xF(x)z'(x)-xF(x)z''(x)-2xF'(x)z'(x)
\end{equation}
But $F'(x)=\frac{1}{2}F(x)-\frac{l}{x}F(x)$, hence
\begin{align}
\Theta\left[F(x)z(x)\right]&=F(x)\left[2lz'(x)-xz''(x)\right]\\
&=F(x)\frac{d}{dx}\left[ 2l+1 -x\frac{d}{dx}\right]z(x)
\end{align}
Denoting
\begin{equation}
M= 2l+1 -x\frac{d}{dx}
\end{equation}
we have shown that
\begin{equation}
\Theta\left[F(x)z(x)\right]=F(x)\frac{d}{dx}M\left[z(x)\right]
\end{equation}
and thus, for $n=1,2,3\ldots$,
\begin{equation}
\Theta^n\left[F(x)z(x)\right]=F(x)\left( \frac{d}{dx}M \right)^n\left[z(x)\right]
\end{equation}
Choosing $z(x)=\varphi(x)=x^{2l+1}e^{-x}$, we have $F(x)\varphi(x)=h(x)$ and then
\begin{equation}
\Theta^n\left[h(x)\right]=F(x)\left( \frac{d}{dx}M \right)^n\left[\varphi(x)\right]
\end{equation}
or, dividing both sides by $n!h(x)$,
\begin{equation}
\frac{1}{n!h(x)}\Theta^n\left[h(x)\right]=\frac{1}{n!\varphi(x)}\left( \frac{d}{dx}M \right)^n\left[\varphi(x)\right]
\end{equation}
It can be checked that $M\left[\varphi(x)\right]=x\varphi(x)$ which gives, for $n=1$,
\begin{equation}
\frac{1}{h(x)}\Theta\left[h(x)\right]=\frac{1}{\varphi(x)}\frac{d}{dx}\left( x\varphi(x) \right)
\end{equation}
and thus $\tilde{L}^{(2l+1)}_1(x)=L^{(2l+1)}_1(x)$. To go further, we use induction. Suppose that for some $n$
\begin{equation}
\left( \frac{d}{dx}M \right)^n\left[\varphi(x)\right]=\frac{d^n}{dx^n}\left( x^n\varphi(x) \right)
\end{equation}
then
\begin{equation}
\left( \frac{d}{dx}M \right)^{n+1}\left[\varphi(x)\right]= \frac{d}{dx}M \frac{d^n}{dx^n}\left( x^n\varphi(x) \right)
\end{equation}
For any function $z(x)$,
\begin{align}
M\frac{d}{dx}z(x)&=\left( 2l+1 -x\frac{d}{dx}\right)\frac{d}{dx}z(x)\\
&=\frac{d}{dx}\left(2l+2-x\frac{d}{dx}\right)z(x)\\
&=\frac{d}{dx}\left( M+1 \right)z(x)
\end{align}
More generally,
\begin{equation}
M\frac{d^n}{dx^n}=\frac{d^n}{dx^n}\left( M+n \right)
\end{equation}
Then
\begin{align}
\left( \frac{d}{dx}M \right)^{n+1}\left[\varphi(x)\right]&=\frac{d^{n+1}}{dx^{n+1}}\left( M+n \right)\left( x^n\varphi(x) \right)\\
&=\frac{d^{n+1}}{dx^{n+1}}\left( 2l+n+1-x\frac{d}{dx} \right)\left(x^{2l+n+1}e^{-x} \right)\\
&=\frac{d^{n+1}}{dx^{n+1}}\left( x^{2l+n+2}e^{-x} \right)\\
&=\frac{d^{n+1}}{dx^{n+1}}\left( x^{n+1}\varphi(x) \right)\\
\end{align}
As the induction hypothesis is true for $n=1$, we have shown that
\begin{equation}
\frac{1}{n!h(x)}\Theta^n\left[h(x)\right]=\frac{1}{n!\varphi(x)} \frac{d^n}{dx^n}\left(x^n\varphi(x)\right)
\end{equation}
and thus $\tilde{L}^{(2l+1)}_n(x)=L^{(2l+1)}_n(x)$, both series of polynomials are identical.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3106871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve this recurrence relation using generating functions: $a_n = 10 a_{n-1}-25 a_{n-2} + 5^n\binom{n+2}2$? How can we solve the following recurrence relation using GF?
$a_n = 10 a_{n-1}-25 a_{n-2} + 5^n {n+2 \choose 2}$ , for each $n>2, a_0 = 1, a_1 = 15$
I think that most of it is pretty straightforward. What really concerns me is this part
$5^n{n+2 \choose 2}$
Problem further analyzed
After trying to create generating functions in the equation we end up in this
$\sum_{n=2}^\infty a_n x^n = 10 \sum_{n=2}^\infty {a_{n-1}} \sum_{n=2}^\infty a_{n-2} x^n + \sum_{n=2}^\infty {n+2 \choose 2}5^nx^n$
Lets take this part:
$ \sum_{n=2}^\infty {n+2 \choose 2}5^nx^n = \sum_{n=2}^\infty \frac{(n+1)(n+2)}{2} 5^nx^n $
then?
| $\sum_{n=0}^\infty {n+2 \choose 2}5^nx^n = \frac12\sum_{n=0}^\infty(n^2+3n+2)(5x)^n=\frac12[\sum_{n=0}^\infty n^2y^n+3\sum_{n=0}^\infty ny^n+2\sum_{n=0}^\infty y^n]$
where $y=5x$. You will need the following series sums:
$(1)\sum_{n=0}^\infty n^2y^n=\dfrac{y(1+y)}{(1-y)^3}\\(2)\sum_{n=0}^\infty ny^n=\dfrac y{(y-1)^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3108550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Minimum point of $x^2+y^2$ given that $x+y=10$ How do you I approach the following question:
Find the smallest possible value of $x^2 + y^2$ given that $x + y = 10$.
I can use my common sense and deduce that the minimum value is $5^2 + 5^2 = 50$. But how do you approach this mathematically?
Thanks in advance.
| Most answers use functions or derivatives... I'll use another approach: Inequalities!
It's easy to prove that the minimum will be achieved for positive values of $x$ and $y$. Thus, in virtue of the QM-AM inequality:
$$\sqrt{\frac{x^2+y^2}{2}}\geq \frac{x+y}{2}=5\iff x^2+y^2\geq 50$$
Proof of the Quadratic Mean - Arithmetic Mean inequality ($x,y\geq0)$: $$(x-y)^2\geq 0\iff x^2+y^2\geq 2xy\iff 2(x^2+y^2)\ge (x+y)^2\iff\color{red}{\sqrt{\frac{x^2+y^2}{2}}\geq \frac{x+y}{2}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Finding the generating function of a sequence for $n$ pennies Let $a_n$ count the number of different ways you can pay a sum of n pennies with 1p and 2p coins. Find the generating function and closed formula for $a_n$. I have attempted this question, finding $a_n$ to be $(1, 1, 2, 2, 3, 3, \ldots)$, and have started calculating the generating function $1 + x + 2 x^2 + 2 x^3 + \ldots$ but am not sure how to proceed from here? Hope someone can help me.
| A direct way of thinking about the generating function is $(1+x+x^2+...)(1+x^2+x^4+x^6+...)={1\over1-x}\times{1\over1-x^2}$ where the first sequence represents the $1$ pennies and the second represents the $2$ pennies.
Now to find the closed form formula for $a_n$ basically you use the partial sum method to split ${1\over1-x}\times{1\over1-x^2}= {a\over 1-x}+{b\over(1-x)^2}+{c\over 1+x}$ and solve for $a,b$ to get $a+b+c=1,-a+c=0,b-2c=0$ and therefore $b={1\over 2},c={1\over 4},a={1\over 4}$ so the sequence is equal to ${1\over 4}(1+x+x^2+...)+{1\over 2}(1+2x+3x^2+4x^3+...)+{1\over 4}(1-x+x^2-x^3+...)$
When you look at the coefficient of $x^n$ you get ${1\over 4}+{1\over2}(n+1)+{1\over 4}(-1)^n$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simultaneous congruences $3x \equiv 2 \pmod{5}$, $3x \equiv 4 \pmod{7}$, $3x \equiv 6 \pmod{11}$ I am stuck in a simultaneous linear congruence problem:
\begin{cases}
3x \equiv 2 \pmod{5} \\[4px]
3x \equiv 4 \pmod{7} \\[4px]
3x \equiv 6 \pmod{11}
\end{cases}
Using the Chinese remainder theorem,
I started with the 'highest' divisor: $11$. Since $(3, 11) =1$ there is a unique solution. $x= 6 \cdot 3 ^{\phi (11) -1} \equiv 6 \cdot 3^4\pmod{11}$ But to be honest, I have no clue how to continue.
Perhaps, cancel out the last equation to: $x \equiv 2 \pmod{11}$?
| First, you could observe that
\begin{cases}
3x \equiv 2 \pmod{5} \Leftrightarrow 3x \equiv 2+10 \equiv 12=3 \cdot 4 \pmod{5} \Leftrightarrow x \equiv \color{red}{4} \pmod{5} \\[4px]
3x \equiv 4 \pmod{7} \Leftrightarrow 3x \equiv 4+14 \equiv 18=3 \cdot 6\pmod{7} \Leftrightarrow x \equiv \color{lime}{6} \pmod{6} \\[4px]
3x \equiv 6 \pmod{11} \Leftrightarrow x \equiv \color{blue}{2} \pmod{11} \\[4px]
\end{cases}
Next, you could find integers $a, b, c$ with
\begin{cases}
a \cdot 7 \cdot 11 \equiv 1 \pmod{5} \Leftrightarrow 2a \equiv 1 \pmod{5} \Leftrightarrow a \equiv 3 \pmod{5}\\[4px]
b \cdot 5 \cdot 11 \equiv 1 \pmod{7} \Leftrightarrow 6b \equiv 1 \pmod{7} \Leftrightarrow b \equiv 6 \pmod{7} \\[4px]
c \cdot 5 \cdot 7 \equiv 1 \pmod{11} \Leftrightarrow 2c \equiv 1 \pmod{11} \Leftrightarrow c \equiv 6 \pmod{11} \\[4px]
\end{cases}
Then, all solutions of your simultaneous congruences are $$x \equiv a \cdot 7 \cdot 11 \cdot \color{red}{4} + b \cdot 5 \cdot 11 \cdot \color{lime}{6} + c \cdot 5 \cdot 7 \cdot \color{blue}{2} \pmod{5 \cdot 7 \cdot 11}$$ so $x \equiv 244 \pmod{385}$.
Btw: Did anybody notice that this way of constructing a solution is exactly the same as finding an interpolation polynomial using Lagrange polynomials?
| {
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"url": "https://math.stackexchange.com/questions/3111894",
"timestamp": "2023-03-29T00:00:00",
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Proving that $\lim_{v\rightarrow \infty}\left[\frac{v^2}{v^3 + 1} + \frac{v^2}{v^3 + 2} +\cdots + \frac{v^2}{v^3+v}\right]= 1 $ I wonder if my solution that $\lim_{v\to\infty}\left[\frac{v^2}{v^3 + 1} + \frac{v^2}{v^3 + 2} + \cdots + \frac{v^2}{v^3+v}\right]= 1 $ is correct.
$$\frac{v^2}{v^3 + 1} + \frac{v^2}{v^3 + 2} + ... + \frac{v^2}{v^3+v} = \frac{\frac{1}{v}}{1 + \frac{1}{v^3}} + \frac{\frac{1}{v}}{1 + \frac{2}{v^3}} + ... + \frac{\frac{1}{v}}{1+\frac{1}{v^2}} \rightarrow 0 $$
So, the limit is not equal to 1.
Where is my mistake ?
| Although each term approaches zero, there are an infinite number of terms as indicated by there being exactly $v$ terms in the addition. The denominator of each term approaches $v^3$ as $v \to \infty$ as $v^3$ is much greater than $v$ or any value less than $v$. This means that the limit can simply be thought of as the following:
$$\lim_{v\to\infty} v\times \frac{v^2}{v^3}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114947",
"timestamp": "2023-03-29T00:00:00",
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Prove that $a^2+b^2+c^2\geqslant\frac{1}{3}$ given that $a\gt0, b\gt0, c\gt0$ and $a+b+c=1$, using existing AM GM inequality Using the AM and GM inequality, given that
$a\gt0, b\gt0, c\gt0$ and $a+b+c=1$ prove that
$$a^2+b^2+c^2\geqslant\frac{1}{3}$$
| $$a^2+{1\over 9} + b^2+{1\over 9} + c^2+{1\over 9}\geq {2\over 3}(a+b+c)$$ by AM-GM.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find at least $5$ integers $n$ such that $\varphi(n)=16$
Let $\varphi(n)$ denote Euler's totient function. Find all integers such that $\varphi(n)=16$.
Answers given were $17,32,34,40,48.$
I am thinking a generalisation of this problem: is there a way to find all positive integers $n$ such that $\varphi(n)=k$ for a specific $k$? Is there a way to do it other than trial and error? (which is what I did BTW)
| If $$n= p_1^{a_1}p_2^{a_2}...p_k^{a_k}$$ where $a_i\geq 1$ and $p_1<p_2<...<p_k$ then we have $$\phi(n) = p_1^{a_1-1}p_2^{a_2-1}...p_k^{a_k-1} (p_1-1)...(p_k-1)$$
$$16= p_1^{a_1-1}p_2^{a_2-1}...p_k^{a_k-1} (p_1-1)...(p_k-1)$$
If $p_1=2$ and $k>1$ $$2^4= 2^{a_1-1}p_2^{a_2-1}...p_k^{a_k-1} (p_2-1)...(p_k-1)$$
then $a_i=1$ for all $i>1$, so we have $$2^4= 2^{a_1-1} (p_2-1)...(p_k-1)$$
Now if $a_1 = 1$ then $k\leq 5$ so $$ 2^4= (p_2-1)...(p_k-1)$$
and thus $p_k\leq 17$.
We see that $k=2$ and $p_2 = 17$ does works, so $\boxed{ n_1=34}$
and it is the only one if $a_1=1$
If $a_1 = 2$ then $k\leq 4$ so $$ 2^3= (p_2-1)...(p_k-1)$$
and thus $p_k\leq 7$. We see that $p_2 =3$ and $p_3=5$ works, so $\boxed{ n_2 = 60}$
If $a_1 = 3$ then $k\leq 3$ so $$ 2^2= (p_2-1)...(p_k-1)$$
and thus $p_k\leq 5$. We see that $p_2 =5$ works, so $\boxed{ n_3 = 40}$
If $a_1 = 4$ then $k=2$ so $$ 2= p_2-1$$
so $p_2 =3$ works, so $\boxed{n_4 = 48}$
If $a_1 =5$ then o $\boxed{n_5 = 32}$ and we have 5 numbers
If $p_1 >2$ then ...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is my integral in fully reduced form?
I have to integrate this:
$$\int_0^1 \frac{x-4}{x^2-5x+6}\,dx$$
Now
$$\int_0^1 \frac{x-4}{(x-3)(x-2)}\, dx$$
and by using partial fractions we get
$$\frac{x-4}{(x-3)(x-2)} = \frac{A}{x-3} + \frac{B}{x-2}$$
$$x-4 = A(x-2) + B(x-3)$$
$$= Ax - 2A + Bx - 3B$$
$$x-4 = (A+B)x - 2A - 3B$$
so
$$A+B=1$$ or $$2X+2B = 2$$
$-B = -2$ and $B = 2$ and $A = -1$
Then
$$\int_0^1 \frac{x-4}{x^2-5x+6} = \int_0^1 \frac{-1}{x-3}dx + \int_0^1 \frac{2}{x-2} dx$$
usub using $u = x-3$ and $du = dx$
so $$ -1 \ln | x-3 | \rbrack_0^1 + 2 \ln |x-2| \rbrack_0^1$$
$$-1 ( \ln 2 - \ln 3) + 2 (\ln 1 - \ln 2)$$
$$-\ln 2 + \ln 3 + 2\ln 1 - 2\ln 2 = -3 * \ln 2 + \ln 3 + 2\ln 1$$
Is there anyway to simplify this further? Wolfram has the answer at $-\ln(8/3)$ and I'm not sure how to simplify to here? How does my work look?
| Use $ln(a) + ln(b) = ln(a\cdot b)$ and $aln(b) = ln(b^a)$
You get $ln(\frac{1}{2} \cdot \frac{3}{1} \cdot \frac{1^2}{1} \cdot \frac{1}{2^2})$
| {
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"timestamp": "2023-03-29T00:00:00",
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A more natural solution to finding the general terms of a recurrence relation in $2$ variables A high school contest math problem in a problem book:
Find the general terms of
$$a_{1}=a,\quad b_{1}=b,\quad a_{ n + 1 }=\frac { 2 a _ { n } b _ { n } } { a _ { n } + b _ { n } },\quad b_{ n + 1 }=\sqrt{ a_{n+1} b _ {n}}$$
The easiest method is to use the following trigonometric identities:
$$2 \tan \frac { \theta } { 2 } = \frac { 2 \tan \theta \sin \theta } { \tan \theta + \sin \theta } , \quad 2 \sin \frac { \theta } { 2 } = \sqrt { 2 \tan \frac { \theta } { 2 } \sin \theta }$$
We make the substitution
$$a _ { 1 } = a = c \cdot \tan \theta , \quad b _ { 1 } = b = c \cdot \sin \theta$$
Observe that
$$a_{2}=\frac{2a_1b_1}{a_1+b_1}=\frac{2c\tan\theta\sin\theta}{\tan\theta+\sin\theta}=2c\tan\frac{\theta}{2}$$
$$a_{3}=\frac{2a_2b_2}{a_2+b_2}=\frac{2(2c)\tan\frac{\theta}{2}\sin\frac{\theta}{2}}{\tan\frac{\theta}{2}+\sin\frac{\theta}{2}}=2^2c\tan\frac{\theta}{2^2}$$
$$...$$
$$a_n= 2^{n-1} c \tan \frac { \theta } { 2 ^ { n - 1 } }$$
It's possible to express the above formula in terms of $a$ and $b$, just notice
$$\theta=\cos^{-1}\left(\frac{b}{a}\right), c=\frac{a}{\tan\theta}$$
Similarly,
$$b_n=2 ^ { n - 1 } c\sin \frac { \theta } { 2 ^ { n - 1 } }$$
Although I consider the above solution as elegant, it's unnatural at first glance, as the $2$ trigonometric identities are rarely used in practice, at least for those who are mediocre in trigonometry. So my question is:
Is there an alternative solution (non-trigonometric) to the above recurrence relation that is more natural in a certain sense?
Edit: The above method has a defect: complex numbers may pop up for some real values of $a$ and $b$ in the substitution step.
| The approach is nice.
Note that
$$\cot\left(\arccos\dfrac ba\right) = \frac b{\sqrt{a^2-b^2}}$$
(see also Wolfram Alpha).
If $\underline{\frac ba >1},$ then the hyperbolic functions can be used instead of trigonometric ones, because
$$2 \tanh \frac { \theta } { 2 } = \frac { 2 \tanh \theta \sinh \theta } { \tanh \theta + \sinh \theta }$$
(see also Wolfram Alpha),
$$ \quad 2 \sinh \frac { \theta } { 2 } = \sqrt { 2 \tanh \frac { \theta } { 2 } \sinh \theta }$$
(see also Wolfram Alpha),
where
$$\theta = \cosh^{-1}\frac ba = \log\left(\frac ba - \sqrt{\frac{b^2}{a^2}-1}\right),$$
$$c= a\coth\theta = -\dfrac{ab}{\sqrt{b^2-a^2}}$$
(see also Wolfram Alpha)
If $\underline{b=a},$ then $a_n=b_n=a.$
Besides, the substitutions
$$u_n = \frac1{a_n},\quad v_n = \frac1{v_n}$$
lead to the recurrence relation
$$u_{n+1} = \frac{u_n+v_n}2,\quad v_{n+1}=\sqrt{u_{n+1}v_n}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the number of pairs of $(a,b)$ of positive real numbers satisfying $a^4+b^4<1$ and $a^2+b^2>1$?
The number of pairs of $(a,b)$ of positive real numbers satisfying
$a^4+b^4<1$ and $a^2+b^2>1$ is -
$(i)$0
$(ii)$1
$(iii)$2
$(iv)$ more than 2
Solution:We have $a,b>0$,
According to the given situation,$0<a^4+b^4<1<a^2+b^2\implies a^4+b^4<a^2+b^2\implies 0<a^2(a^2-1)<b^2(1-b^2)\implies b^2(1-b)(1+b)<0 ;a^2(a-1)(a+1)>0\implies a\in(-\infty,-1)\cup (0,1);b\in(-1,0)\cup(1,\infty)$
But, in particular, if we choose $a=-1/2,b=1/2$,then $a^4+b^4=1/16+1/16=1/8<1$ and $a^2+b^2=1/4+1/4=1/2<1$.Which contradicts the given condition $a^2+b^2>1$
Where is the mistake in my approach? How should I approach this problem(means how to think ?). Can Triangle inequality be used here?
Please provide some hint...
| $x^2+y^2>1$, and $x^4+y^4 <1$.
Consider $y=x$ in the first quadrant.
$2x^2 >1$, and $2x^4 <1;$
$x ^2 >1/2$ , and $x^2 <1/√2$, i .e.
$1/2=0.5 < x^2 < √2/2=0.7071..$
Points on $y=x$ , line through origin, with
$1<x^2+y^2 < √2 $, satisfy
$x^4+y^4 <1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Can you help me solve this algebra problem? Hi I need to solve this problem and I don’t know how. I’d appreciate your help.
If $x - \frac{ayz}{x^2} = y - \frac{azx}{y^2} = z - \frac{axy}{z^2}$ and $x\neq y\neq z$, then $$x - \frac{ayz}{x^2} = y - \frac{azx}{y^2} = z - \frac{axy}{z^2} = x + y + z - a$$
I think I need to
$x^3 - ayz = x^2k$
$y^3 - azx = y^2k,$
$z^3 - axy = z^2k$
then to multiply both sides of each equality by some quantity, add them all together and factor but I don’t know how to find that quantity.
| $$x-\frac{ayz}{x^2}=y-\frac{axz}{y^2}$$ gives
$$x-y+az\left(\frac{x}{y^2}-\frac{y}{x^2}\right)=0$$ or
$$1+\frac{az(x^2+xy+y^2)}{x^2y^2}=0.$$
Similarly, $$\frac{ax(y^2+yz+z^2)}{y^2z^2}=-1$$ and
$$\frac{ay(x^2+xz+z^2)}{x^2z^2}=-1.$$
Thus, $$x^3(y^2+yz+z^2)=y^3(x^2+xz+z^2)$$ or
$$(x-y)(x^2y^2+xyz(x+y)+z^2(x^2+xy+y^2))=0$$ or
$$\sum_{cyc}(x^2y^2+x^2yz)=0$$ or
$$\sum_{cyc}z^2(x+y)^2=0,$$ which gives $$x+y=x+z=y+z=0$$ or
$$x=y=z=0,$$ which is impossible.
Id est, the given is wrong, which says that
$$x - \frac{ayz}{x^2} = y - \frac{azx}{y^2} = z - \frac{axy}{z^2}\Rightarrow x - \frac{ayz}{x^2} = y - \frac{azx}{y^2} = z - \frac{axy}{z^2} = x + y + z - a$$ is true.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find shortest distance from point on ellipse to focus of ellipse.
Find the point $(x,y)$ on the ellipse $b^2x^2 + a^2y^2 = a^2b^2$ such that the distance to the focus $(c,0)$ is a minimum.
My book I got this problem out of gave a suggestion saying to express the distance as a function of $x$ and work the problem and then express the distance as a function of $y$ and work the problem.
So, I tried to work it out:
$$x^2 = \frac{a^2b^2 - a^2y^2}{b^2}$$
$$y^2 = \frac{a^2b^2 - b^2x^2}{a^2}$$
Now, I write the distance function:
$$d = \sqrt{(x-c)^2 + y^2}$$
Then, I express the distance function as a function of x:
$$d = \sqrt{(x-c)^2 + \frac{a^2b^2 - b^2x^2}{a^2}}$$
Then, I took the derivative w.r.t $x$:
$$d' = 2x - 2c - \frac{2b^2}{a^2}$$
Now, I did as the book told me, and expressed the distance function as a function of y:
$$d' = -\frac{2a^2}{b^2} + \frac{4a^2c}{bx} + 2y$$
I tried to set both derivatives to zero and attempted to solve, but I couldn't really figure out what to do (my algebra is poor).
Any help would be appreciated.
Thanks.
Edit: I need a calculus solution (I'm working out of a calculus book).
| Assuming $a>b$, we have $c=\sqrt{a^2-b^2}$.
Using the parametric equations
$$\begin{cases}x=a\cos\theta,\\y=b\sin\theta\end{cases}$$
we minimize
$$(a\cos\theta-c)^2+(b\sin\theta)^2.$$
The derivative of this expression is
$$-a\sin\theta(a\cos\theta-c)+b\cos\theta(b\sin\theta).$$
There is a root for $$\sin\theta=0$$ (on the major axis) and for
$$(a^2-b^2)\cos\theta=ac,$$ or
$$\cos\theta=\frac a{\sqrt{a^2-b^2}}>1 (!).$$
Hence,
$$d=a-\sqrt{a^2-b^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3125576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Stuck on a Geometry Problem
$ABCD$ is a square, $E$ is a midpoint of side $BC$, points $F$ and $G$ are on the diagonal $AC$ so that $|AF|=3\ \text{cm}$, $|GC|=4\ \text{cm}$ and $\angle{FEG}=45 ^{\circ}$. Determine the length of the segment $FG$.
How can I approach this problem, preferably without trigonometry?
| Calling
$$
AF = a\\
FG=x\\
GC = b\\
EC = c\\
E = (X_0, Y_0)\\
O = \left(a+\frac x2, \frac x2\right)
$$
Considering in the plane $X\times Y$ the square diagonal as the $X$ axis with $A$ at the origin, we have
$$
\frac{\sqrt 2}{2}(a+x+b)= 2c\\
x = \sqrt2 r\\
\left(X_0-\left(a+\frac x2\right)\right)^2+\left(Y_0-\frac x2\right)^2= r^2\\
X_0 = a+x+b -\frac{\sqrt 2}{2}c\\
Y_0 = \frac{\sqrt 2}{2}c
$$
Here $r$ represents the radius for the circle which intersects the $X$ axis at $F, G$ such that $\angle FEG = \frac{\pi}{4}$ is the angle subtended by arc $AB$
Solving for $x,r,X_0,Y_0$ we get at
$$
\left\{
\begin{array}{rcl}
x& =& \frac{1}{3} \left(b-a+2 \sqrt{a^2-2 b a+4 b^2}\right) \\
c& =& \frac{a+2 b+\sqrt{a^2-2 b a+4 b^2}}{3 \sqrt{2}} \\
r& =& \frac{b-a+2 \sqrt{a^2-2 b a+4 b^2}}{3 \sqrt{2}} \\
X_0&=&\frac{1}{2} \left(a+2 b+\sqrt{a^2-2 b a+4 b^2}\right) \\
Y_0&=&\frac{1}{6} \left(a+2 b+\sqrt{a^2-2 b a+4 b^2}\right) \\
\end{array}
\right.
$$
giving after substitution $x = 5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 12,
"answer_id": 5
} |
integrate sin(x)cos(x) using trig identity. Book tells me the answer is:
$$ \int \sin(x)\cos(x) dx = \frac{1}{2} \sin^{2}(x) + C $$
however, I get the result:
$$
\sin(A)\cos(B) = \frac{1}{2} \sin(A-B)+\frac{1}{2}\sin(A+B)
$$
$$
\begin{split}
\int \sin(x)\cos(x) dx
&= \int \left(\frac{1}{2}\sin(x-x) + \frac{1}{2}\sin(x+x)\right) dx \\
&= \int \left(\frac{1}{2}\sin(0) + \frac{1}{2}\sin(2x)\right) dx \\
&= \int \left(\frac{1}{2}\sin(2x)\right) dx \\
&= -\frac{1}{2} \frac{1}{2}\cos(2x) +C\\
&= -\frac{1}{4} \cos(2x) +C
\end{split}
$$
How did the book arrive at the answer $\frac{1}{2}\sin^2(x)$?
| Observe that $\int \sin 2x\ \text {dx} = -\frac {\cos 2x} {2} + C$ and use the fact that $\cos 2x =1-2{\sin}^2 x.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126227",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Find $\lim\limits_{t \to \infty} \int_{0}^{t} \frac{\mathrm dx}{(x^2+a^2)(x^2+b^2)(x^2+c^2)}$
Find $$\lim_{t \to \infty} \int_{0}^{t} \frac{1}{(x^2+a^2)(x^2+b^2)(x^2+c^2)}\mathrm dx$$ where $a,b,c$ are strict positive and dinstinct real numbers.
I know it should be something with arctangent but I don't know how to get there. Can somebody give me some tips, please?
| Hints
Method 1: Partial fractions Since $a, b, c$ are positive and distinct, a straightforward application of the method of partial fractions gives that the integrand is $$\sum \frac{1}{(a^2 - b^2)(a^2 - c^2)} \frac{1}{x^2 + a^2},$$ where the sum is over the cyclic permutations of the parameters $(a, b, c)$. By linearity we can write the integral as a linear combination of integrals of the form $$\int_0^{\infty} \frac{dx}{x^2 + k^2} ,$$ and like you say, these can be evaluated with the usual antiderivative formula $$\int \frac{dx}{x^2 + k^2} = \frac{1}{k} \arctan \frac{x}{k} + C .$$
Method 2: Contour integration Since the integrand in even, it coincides with $$\frac{1}{2} \int_{-\infty}^\infty \frac{dx}{(x^2 + a^2) (x^2 + b^2) (x^2 + c^2)}.$$ This suggests integrating over a semicircle $\Gamma_R$ of radius $R$ with diameter on the real axis and centered at zero, say, in the upper half-plane. For large $R$, the value of the integral over the circular arc is $O(R^{-5})$, so in the limit its contribution to the integral is zero and hence $$\frac{1}{2} \int_{-\infty}^\infty \frac{dx}{(x^2 + a^2) (x^2 + b^2) (x^2 + c^2)} = \lim_{R \to \infty} \frac{1}{2} \oint_{\Gamma_R} \frac{dz}{(z^2 + a^2) (z^2 + b^2) (z^2 + c^2)} ,$$ but (for large $R$, and since $a, b, c$ are positive and distinct) the poles inside $\Gamma_R$ are simple poles at $ia, ib, ic$, so we can apply the residue formula, which gives that the integral is $$\frac{1}{2} \cdot 2 \pi i (\operatorname{Res}(f, ia) + \operatorname{Res}(f, ib) + \operatorname{Res}(f, ic)),$$ where $f$ is the integrand.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Integrate $\int 2x\sin{(x^2+x)} \,dx$ Evaluate $\int 2x\sin{(x^2+x)} \,dx$ using integration by parts.
Let $u=2x$ and $dv=\sin(x^2+x)dx$ and I got that $du=x^2dx$. I'm stuck when integrating $\sin{(x^2+x)}$. Then I tried the integral calculator and it says:
$$-\cos(x^2+x)-\frac{\sqrt{\pi}\cos{\frac{1}{4}}S\left(\frac{\sqrt 2\left(x+\frac{1}{2}\right)}{\sqrt\pi}\right)}{\sqrt2}+\frac{\sqrt{\pi}\sin{\frac{1}{4}}C\left(\frac{\sqrt 2\left(x+\frac{1}{2}\right)}{\sqrt\pi}\right)}{\sqrt2}.$$
But I don't understand how the calculator got the answer. Maybe there's an easier method for a calculus 2 student like me.
| Took me a few tries but here goes. First off, we'll integrate $\sin(x^2 + x)$. Then we'll use that to build the original integral in question.
This one requires a trick. You will need to first "complete the square" on $x^2 + x$. Note that (this can be "reverse-engineered" from Wolfram(?)s output by noting the $x + \frac{1}{2}$ business)
$$x^2 + x = \left(x^2 + x + \frac{1}{4}\right) - \frac{1}{4} = \left(x + \frac{1}{2}\right)^2 - \frac{1}{4}$$
Now you can break up the sine by using the angle summation formula:
$$\sin(a + b) = \sin(a) \cos(b) + \cos(a) \sin(b)$$
so that
$$\begin{align}
\sin(x^2 + x) &= \sin\left(\left(x + \frac{1}{2}\right)^2 - \frac{1}{4}\right)\\
&= \sin\left(\left(x + \frac{1}{2}\right)^2\right) \cos\left(-\frac{1}{4}\right) + \cos\left(\left(x + \frac{1}{2}\right)^2\right) \sin\left(-\frac{1}{4}\right)\\
&= \sin\left(\left(x + \frac{1}{2}\right)^2\right) \cos\left(\frac{1}{4}\right) - \cos\left(\left(x + \frac{1}{2}\right)^2\right) \sin\left(\frac{1}{4}\right)\end{align}$$
Yeah that looks nasty. Now it becomes two integrals:
$$\int \sin(x^2 + x)\ dx = \int \sin\left(\left(x + \frac{1}{2}\right)^2\right) \cos\left(\frac{1}{4}\right)\ dx - \int \cos\left(\left(x + \frac{1}{2}\right)^2\right) \sin\left(\frac{1}{4}\right)\ dx$$
pulling the constant stuff out front,
$$\int \sin(x^2 + x)\ dx = \cos\left(\frac{1}{4}\right) \int \sin\left(\left(x + \frac{1}{2}\right)^2\right) \ dx - \sin\left(\frac{1}{4}\right) \int \cos\left(\left(x + \frac{1}{2}\right)^2\right)\ dx$$
The real meat here, though, is now that you will need to use two functions you likely haven't seen before, called the Fresnel sine and cosine integral, which are what "$S$" and "$C$" in the given answer are. They are defined by
$$S(x) := \int_{0}^{x} \sin(\frac{1}{2}\pi x'^2)\ dx'$$
$$C(x) := \int_{0}^{x} \cos(\frac{1}{2}\pi x'^2)\ dx'$$
There is no simpler way to express these integrals - it is like the integral of $\frac{1}{x}$, which is used to define the natural logarithm, or of $\frac{1}{\sqrt{1 - x^2}}$, which defines the inverse trig. No shame in seeing something new: I remember when I first studied calculus a long, long time ago I was fascinated by the earlier versions of Wolfram's integrator and these "exotic" kinds of functions were thus something I came across relatively early.
The factors of $\frac{1}{2}\pi$ are normalization factors included to make the functions have nice limits as their input goes to infinity - sometimes, these functions are defined with these factors omitted. This gives (multiplying factors inside and out to cancel/create $du$)
$$\int \sin(x^2)\ dx = \sqrt{\frac{\pi}{2}} S\left(\sqrt{\frac{2}{\pi}} x\right)$$
$$\int \cos(x^2)\ dx = \sqrt{\frac{\pi}{2}} C\left(\sqrt{\frac{2}{\pi}} x\right)$$
so
$$\int \sin\left(\left(x + \frac{1}{2}\right)^2\right)\ dx = \sqrt{\frac{\pi}{2}} S\left(\sqrt{\frac{2}{\pi}} \left(x + \frac{1}{2}\right)\right)$$
$$\int \cos\left(\left(x + \frac{1}{2}\right)^2\right)\ dx = \sqrt{\frac{\pi}{2}} C\left(\sqrt{\frac{2}{\pi}} \left(x + \frac{1}{2}\right)\right)$$
then going back to the original expression and saving the common factor $\sqrt{\frac{\pi}{2}}$,
$$\int \sin(x^2 + x)\ dx = \sqrt{\frac{\pi}{2}} \left[\cos\left(\frac{1}{4}\right) S\left(\sqrt{\frac{2}{\pi}} \left(x + \frac{1}{2}\right)\right) - \sin\left(\frac{1}{4}\right) C\left(\sqrt{\frac{2}{\pi}} \left(x + \frac{1}{2}\right)\right)\right] + \mathrm{constant}$$
which is, essentially, what you have given in the original answer. To get the integral of
$$\int 2x\ \sin(x^2 + x)\ dx$$
first convert $2x$ by adding and subtracting $1$: $2x = (2x + 1) - 1$ and then
$$
\begin{align}
\int 2x\ \sin(x^2 + x)\ dx &= \int [(2x + 1) - 1] \sin(x^2 + x)\ dx\\
&= \int (2x + 1) \sin(x^2 + x)\ dx - \int \sin(x^2 + x)\ dx
\end{align}$$
The first integral is a simple $u$-substitution of $u = x^2 + x$, with $du = 2x + 1$, integrating to $-\cos(x^2 + x)$. The second is what we just did, and thus the ultimate answer is
$$\int 2x\ \sin(x^2 + x)\ dx =\\
-\cos(x^2 + x) - \sqrt{\frac{\pi}{2}} \left[\cos\left(\frac{1}{4}\right) S\left(\sqrt{\frac{2}{\pi}} \left(x + \frac{1}{2}\right)\right) - \sin\left(\frac{1}{4}\right) C\left(\sqrt{\frac{2}{\pi}} \left(x + \frac{1}{2}\right)\right)\right] + \mathrm{constant}$$
which is, up to some reformatting, exactly what you gave in your answer.
Sadly, not everything has a simple method, not in maths or in anything else. Many things just take a lot of hard slogging to get right. Efficiency is in learning to be able to strike the fine balance between spending time searching for a simpler method and abandoning that search for the direct, but ugly, guaranteed method.
For what it's worth, the $S(x)$ and $C(x)$ functions have beautiful graphs:
(Courtesy: https://commons.wikimedia.org/wiki/File:Fresnel_integrals.gif, author unnamed, cc-by-sa 3.0.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Factoring a matrix as the product of block triangular and diagonal matrices. How can I check that the matrix
$$K = \left[\begin{array}{c|cc} 1 & 0^{\mathrm T}_m & m \mathbf u^{\mathrm T} \\ \hline 0_m & I_m & I_m \\ m \mathbf u & I_m & O_m \end{array}\right] = \begin{bmatrix}
1 & \textbf{w}^T \\
\textbf{w} & B \\
\end{bmatrix}$$
can be factored as the following product of block triangular and diagonal matrices:
\begin{equation} K =
\begin{bmatrix}
1 & \textbf{w}^{T}B^{-1} \\
\textbf{0} & I_{2m}\\
\end{bmatrix}
\begin{bmatrix}
s & \textbf{0}^T\\
\textbf{0} & B\\
\end{bmatrix}
\begin{bmatrix}
1 & \textbf{0}^{T}\\
B^{-1}\textbf{w} & I_{2m}
\end{bmatrix}
\end{equation}
where $$B = \begin{bmatrix} I_m & I_m \\ I_m & O_m \end{bmatrix} = \begin{bmatrix} 1 I_m & 1 I_m \\ 1 I_m & 0 I_m \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \otimes I_m.$$ and
$$\mathbf w = \begin{bmatrix} 0_m \\ m \mathbf u \end{bmatrix} = \begin{bmatrix} 0 \mathbf u \\ m \mathbf u \end{bmatrix} = \begin{bmatrix} 0 \\ m \end{bmatrix} \otimes \mathbf u$$ where $$\mathbf u = \begin{bmatrix}
1 \\
\vdots \\
1
\end{bmatrix} \in \mathbb R^m$$ and I need to specify $s$. I think i have calculated the inverse of the Kronecker product $B$ correctly as $$B^{-1} = \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix} \otimes I_m = \begin{bmatrix} 0I_{m} & 1I_{m} \\ 1 I_{m} & -1I_{m} \end{bmatrix}$$ but I don't know where to go from here. Any help would be great!
| It works. To multiply
$\begin{equation} K =
\begin{bmatrix}
1 & \textbf{w}^{T}B^{-1} \\
\textbf{0} & I_{2m}\\
\end{bmatrix}
\begin{bmatrix}
s & \textbf{0}^T\\
\textbf{0} & B\\
\end{bmatrix}
\begin{bmatrix}
1 & \textbf{0}^{T}\\
B^{-1}\textbf{w} & I_{2m}
\end{bmatrix}
\end{equation},$
start, e.g., by multiplying the last two matrices. Perform block-wise multiplication like regular matrix multiplication. You get:
$\begin{align} K &=
\begin{bmatrix}
1 & \textbf{w}^{T}B^{-1} \\
\textbf{0} & I_{2m}\\
\end{bmatrix}
\begin{bmatrix}
s + \textbf{0}^T B^{-1} \textbf{w} & s \textbf{0}^T + \textbf{0}^T I\\
\textbf{0} 1 + B B^{-1} w & \textbf{0}\textbf{0}^T + B I \\
\end{bmatrix} \\
& = \begin{bmatrix}
1 & \textbf{w}^{T}B^{-1} \\
\textbf{0} & I_{2m}\\
\end{bmatrix}
\begin{bmatrix}
s & \textbf{0}^T \\
w & B \\
\end{bmatrix}
\end{align}.$
Keep going the same way:
$\begin{align}
& K &= \begin{bmatrix}
s + \textbf{w}^{T}B^{-1}\textbf{w} & 1\textbf{0}^T + \textbf{w}^T B^{-1} B \\
\textbf{0} s + I \textbf{w} & \textbf{0}\textbf{0}^T + B\\
\end{bmatrix} =
\begin{bmatrix}
s + \textbf{w}^{T}B^{-1}\textbf{w} & \textbf{w}^T \\
\textbf{w} & B \\
\end{bmatrix}
\end{align}.$
This is equal to your desired $K$ if $s + \textbf{w}^{T}B^{-1}\textbf{w} =1$. Let us evaluate $\textbf{w}^{T}B^{-1}\textbf{w}$:
$\begin{align}
\textbf{w}^{T}B^{-1}\textbf{w}
&= \left( \begin{bmatrix}0\\m\end{bmatrix} \otimes \textbf{u}\right)^T
\left( \begin{bmatrix}0 & 1\\1&-1\end{bmatrix} \otimes I_m\right)
\left( \begin{bmatrix}0\\m\end{bmatrix} \otimes \textbf{u}\right)\\
&= \left( \begin{bmatrix}0, m\end{bmatrix} \cdot \begin{bmatrix}0 & 1\\1&-1\end{bmatrix}
\cdot \begin{bmatrix}0\\m\end{bmatrix} \right) \otimes \left(
\textbf{u}^T I_m \textbf{u}\right) \\
& = [m, -m] \cdot \begin{bmatrix}0\\m\end{bmatrix}\cdot m = -m^2 \cdot m = -m^3.
\end{align},$
where we used the fact that $(A \otimes B) \cdot (C \otimes D) = (AC \otimes BD)$ if dimensions comply.
Therefore, your desired $s$ must satisfy $s -m^3=1$, i.e., $s = 1+m^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the number of ordered triplets satisfying $5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right)$ Find the number of ordered triplets $(x,y,z)$ of real numbers satisfying $$5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right)$$
and
$$xy+yz+zx=1$$
My try:
Letting:
$$\left(x+\frac{1}{x}\right)=\frac{k}{5}\tag{1}$$
$$\left(y+\frac{1}{y}\right)=\frac{k}{12}\tag{2}$$
$$\left(z+\frac{1}{z}\right)=\frac{k}{13}\tag{3}$$
By $AM-GM$
$$\left(z+\frac{1}{z}\right)^2 \ge 4$$
So $$k^2 \ge 676$$
Adding $(1),(2),(3)$ we get:
$$x+y+z+\frac{xy+yz+zx}{xyz}=\frac{281k}{780}$$
$$x+y+z+\frac{1}{xyz}=\frac{281k}{780}$$
Any clue from here?
| For every triplet of real numbers $(x,y,z)$, which satisfies $$xy+yz+zx=1,$$ then there exists an unique triplet of angles $(\alpha, \beta, \gamma)$, such that $0 \le \alpha, \beta, \gamma < \pi $ and $\alpha + \beta + \gamma= \pi $, i.e. they are angles of a triangle, and $$tg(\frac{\alpha}{2}) = x, \text{ } tg(\frac{\beta}{2}) = y, \text{ }tg(\frac{\gamma}{2}) = z$$
Indeed, you can check that in every triangle with angles $\alpha, \beta, \gamma$ you have $tg(\frac{\alpha}{2})tg(\frac{\beta}{2}) + tg(\frac{\beta}{2})tg(\frac{\gamma}{2}) + tg(\frac{\alpha}{2})tg(\frac{\gamma}{2}) =1 $
We also know $$sin(\alpha)=\frac{2*tg(\frac{\alpha}{2})}{tg(\frac{\alpha}{2})^2+1}= \frac{2x}{x^2+1}$$
The other condition, written such at it helps our intuition, now becomes:
$$5\left(\frac{x^2+1}{2x}\right)=12\left(\frac{y^2+1}{2y}\right)=13\left(\frac{z^2+1}{2z}\right)$$ which is just
$$ \frac{sin (\alpha)}{5}= \frac{sin(\beta)}{12} = \frac{sin(\gamma)}{13} =k_0 $$
From Sine law, we know lengths of sides in a triangle are directly proportional to the sines of the opposing angles, therefore if we denote $a,b,c$ the sides, we have
$$ \frac{a}{5}= \frac{b}{12} = \frac{c}{13} =k $$so we obtain
$$ a=5k, b=12k, c= 13k $$
Using cosines law, we get $$ cos(\alpha)=\frac{b^2+c^2-a^2}{2bc} $$ and the analogous formulae for $\beta $ and $\gamma$ give us
$$ cos(\alpha) = \frac{12}{13}, cos(\beta) = \frac{5}{13}, cos(\gamma)= 0 $$
which means that $$x= tg(\frac{\alpha}{2}) = \sqrt{\frac{1-cos(\alpha)}{1+cos(\alpha)}} = \frac{1}{5}$$
$$y= tg(\frac{\beta}{2}) = \sqrt{\frac{1-cos(\beta)}{1+cos(\beta)}} = \frac{2}{3}$$
$$z= tg(\frac{\gamma}{2}) = \sqrt{\frac{1-cos(\gamma)}{1+cos(\gamma)}} = 1$$
We have chosen the positive root from $tg^2(\frac{\alpha}{2}) = \frac{1-cos(\alpha)}{1+cos(\alpha)}$, and the analagous, because $0 \le \alpha, \beta, \gamma < \pi $, therefore $0 <\frac{\alpha}{2}, \frac{\beta}{2}, \frac{\gamma}{2} < \frac{\pi}{2} $, which means that $tg(\frac{\alpha}{2}), tg(\frac{\beta}{2}), tg( \frac{\gamma}{2}) $ are positive .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Having the sequence $(a_{n})_{n\geq1}$, $a_{n}=\int_{0}^{1} x^{n}(1-x)^{n}dx$, find $\lim\limits_{n{\rightarrow}\infty} \frac{a_{n+1}}{a_{n}}$ Having the sequence $(a_{n})_{n\geq1}$, $a_{n}=\int_{0}^{1} x^{n}(1-x)^{n}dx$, find $\lim\limits_{n{\rightarrow}\infty} \frac{a_{n+1}}{a_{n}}$. Is this limit equal to $\lim\limits_{n{\rightarrow}\infty}(a_{n})^{\frac{1}{2}}$? I am unsure if I can apply l'Hospital to this limit, as the integral is a real value. I tried applying integration by parts but I ended nowhere.
Later edit: I am so sorry, it is $(1-x)^{n}$, not $(1-x^{n})$, I did this mistake out of tiredness.
| Starting with integration by parts we have
$$\begin{aligned} \displaystyle a_n \displaystyle & = \int_0^1 x^n(1-x)^n\,{dx} \\& = \frac{1}{n+1}\int_0^1 (x^{n+1})'(1-x)^n\,{dx} \\& = \frac{x^{n+1}}{n+1}(1-x)^n\,\bigg|_0^1 -\frac{1}{n+1}\int_0^1 x^{n+1}[(1-x)^n]'\,{dx} \\& = \frac{n}{n+1}\int_0^1 x^{n+1}(1-x)^{n-1}\,{dx} \end{aligned} $$
But also by symmetry this is the same as $\displaystyle a_n = \frac{n}{n+1}\int_0^1 x^{n-1}(1-x)^{n+1}\,{dx}$, therefore
$$\begin{aligned} \displaystyle 2a_{n} & = \frac{n}{n+1}\int_0^1 x^{n+1}(1-x)^{n-1}\,d{x}+ \frac{n}{n+1}\int_0^1 x^{n-1}(1-x)^{n+1}\,{dx} \\& = \frac{n}{n+1}\int_0^1 (2x^2-2x+1)x^{n-1}(1-x)^{n-1}\,{dx} \\& = -\frac{2n}{n+1} a_n+\frac{n}{n+1}a_{n-1}\end{aligned}$$
Therefore $\displaystyle a_{n} = \frac{na_{n-1}}{4n+2}$, hence $\displaystyle a_{n+1} = \frac{(n+1)a_{n}}{4n+6}$ so $\displaystyle \frac{a_{n+1}}{a_n} = \frac{(n+1)}{4n+6} \to \frac{1}{4}. $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Help finding the derivative of $f(x) = \cos{\left(\sqrt{e^{x^5} \sin{x}}\right)}$ I am trying to find the derivative of $f(x) = \cos(\sqrt{(e^{x^5} \sin(x)})$.
I keep getting the wrong answer, and I'm not sure what I'm doing wrong.
$$\frac{d}{dx} e^{x^5} = e^{x^5} \cdot 5x^4$$
$$\frac{d}{dx} \sin(x) = \cos(x)$$
$$\frac{d}{dx} (e^{x ^5} \cdot \sin(x)) = [(e^{x^5} \cdot 5x^4) \cdot \sin(x)] + [\cos(x) \cdot e^{x^5}]$$
$$\frac{d}{dx} \sqrt{x} = \frac{1}{2\sqrt{x}}$$
$$\frac{d}{dx} \sqrt{(e^{x^5} \sin(x))} = \frac{[(e^{x^5} \cdot 5x^4) \cdot \sin(x)] + [\cos(x) \cdot e^{x^5}]}{2 \sqrt{e^{x^5} \sin(x)}}$$
Therefore, since $\frac{d}{dx} \cos(x) = -\sin(x)$, I have
$$ f'(x) = -\sin(\sqrt{e^{x^5} \sin(x)}) \cdot \frac{[(e^{x^5} \cdot 5x^4) \cdot \sin(x)] + [\cos(x) \cdot e^{x^5}]}{2 \sqrt{e^{x^5} \sin(x)}}$$
However, the website I'm using, "WeBWorK", says this is incorrect.
| Use the Chain Rule:
$\dfrac{\mathrm d}{\mathrm dx}\cos(\sqrt{e^{x^5}\sin x})$
$=\dfrac{\mathrm d}{\mathrm d(\sqrt{e^{x^5}\sin x})}\cos(\sqrt{e^{x^5}\sin x})\cdot \dfrac{\mathrm d}{\mathrm d(e^{x^5}\sin x)}(\sqrt{e^{x^5}\sin x})\cdot \dfrac{\mathrm d}{\mathrm dx}(e^{x^5}\sin x)$
$=\dfrac{\mathrm d}{\mathrm d(\sqrt{e^{x^5}\sin x})}\cos(\sqrt{e^{x^5}\sin x})\cdot \dfrac{\mathrm d}{\mathrm d(e^{x^5}\sin x)}(\sqrt{e^{x^5}\sin x})\cdot \left(\sin x\dfrac{\mathrm d}{\mathrm dx}e^{x^5}+e^{x^5}\dfrac{\mathrm d}{\mathrm dx}\sin x\right)$
$=\dfrac{\mathrm d}{\mathrm d(\sqrt{e^{x^5}\sin x})}\cos(\sqrt{e^{x^5}\sin x})\cdot \dfrac{\mathrm d}{\mathrm d(e^{x^5}\sin x)}(\sqrt{e^{x^5}\sin x})\cdot \left[\sin x\left(\dfrac{\mathrm d}{\mathrm d(x^5)}e^{x^5}\cdot\dfrac{\mathrm d}{\mathrm dx}x^5\right)+e^{x^5}\dfrac{\mathrm d}{\mathrm dx}\sin x\right]$
$=-\sin(\sqrt{e^{x^5}\sin x})\cdot\dfrac{1}{2\sqrt{e^{x^5}\sin x}}\cdot \left[\sin x\cdot e^{x^5}\cdot 5x^4+e^{x^5}\cdot\cos x\right]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3138417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
solve $\int_{-\sqrt{3}}^{\sqrt{3}} 4 \sqrt{3-y^2}dy$ $\int_{-\sqrt{3}}^{\sqrt{3}} 4 \sqrt{3-y^2}dy$
trig sub
$y = \sqrt{3}\sin(u)$
$dy = \sqrt{3}\cos(u)du$
\begin{align}\int_{\sqrt{3}\sin(-\sqrt{3})}^{\sqrt{3}\sin(\sqrt{3})}& 4 \sqrt{3-3\sin^2(u)}\sqrt{3}\cos(u)\,du = 12 \int_{\sqrt{3}\sin(-\sqrt{3})}^{\sqrt{3}\sin(\sqrt{3})} \cos^2(u)\,du\\
&= 6\int_{\sqrt{3}\sin(-\sqrt{3})}^{\sqrt{3}\sin(\sqrt{3})} (1+\cos(2u))du\\
&=\left[6u + 3\sin(2u)\right]_{\sqrt{3}\sin(-\sqrt{3})}^{\sqrt{3}\sin(\sqrt{3})}
\end{align}
If I plug in the limits to you I get some insanely low number while the answer should be $6 \pi$
what do I do ?
| Your endpoints are wrong. If $y = \pm \sqrt{3}$, $\sin(u) = \pm 1$. So take $u$ from $-\pi/2$ to $\pi/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3139558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integral of $x^2 \sin(x^2)$ I was playing around learning the SymPy syntax (python library) e then I saw an example that intrigued me.
$$\int x^2\sin(x^2)dx = \frac{5x\cos(x^2)\Gamma(\frac{5}{2})}{8\Gamma({\frac{9}{4})}} + \frac{5\sqrt{2\pi}C(\frac{x\sqrt{2}}{\sqrt{\pi}})\Gamma{(\frac{5}{4})}}{16\Gamma{(\frac{9}{4})}}$$
Does anyone know which integration method (series expansions, transforms or whatever) is used to calculate it ? And what this $C$ means ?
| With change of variable $X=x^2\quad;\quad dx=\frac{dX}{2\sqrt{X}}$ :
$$I=\int x^2\sin(x^2)dx=\int X\sin(X)\frac{dX}{2\sqrt{X}}=\frac12\int \sqrt{X}\sin(X)dX$$
Integration by part :
$$I=-\frac12\sqrt{X}\cos(X)+\frac14\int\frac{\cos(X)}{\sqrt{X}}dX$$
$$I=-\frac12x\cos(x^2)+\frac14\int\frac{\cos(x^2)}{x}2xdx$$
$$I=-\frac12x\cos(x^2)+\frac12\int\cos(x^2)dx$$
$\int\cos(x^2)dx=\sqrt{\frac{\pi}{2}}\text{C}\left(\sqrt{\frac{2}{\pi}}\:x\right)+$constant
C$(t)$ is a special function called Fresnel Integral : http://mathworld.wolfram.com/FresnelIntegrals.html
In math softwares this function is commonly named FresnelC.
In the present case $t=\sqrt{\frac{2}{\pi}}\:x$ .
$$I=\int x^2\sin(x^2)dx=-\frac12x\cos(x^2)+\frac12\sqrt{\frac{\pi}{2}}\text{C}\left(\sqrt{\frac{2}{\pi}}\:x\right)+\text{constant}$$
This is exactly the same result than
$$\int x^2\sin(x^2)dx = -\frac{5x\cos(x^2)\Gamma(\frac{5}{4})}{8\Gamma({\frac{9}{4})}} + \frac{5\sqrt{2\pi}C(\frac{x\sqrt{2}}{\sqrt{\pi}})\Gamma{(\frac{5}{4})}}{16\Gamma{(\frac{9}{4})}}+\text{constant}$$
because $\frac{\Gamma(\frac54)}{\Gamma(\frac94)}=\frac45$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3140253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to find the roots of $\frac{\sqrt{3}-1}{\sin x}+\frac{\sqrt{3}+1}{\cos x}=4\sqrt{2}$? Find all $x$ in the interval $(0,\pi/2)$ such that $\frac{\sqrt{3}-1}{\sin x}+\frac{\sqrt{3}+1}{\cos x}=4\sqrt{2}$.
The options are (i)$\pi/9,2\pi/7$, (ii)$\pi/36,11\pi/12$ (iii)$\pi/12,11\pi/36$ (iv) All
I have been able to find one value of $x$, $\pi/12$. How do I find the other root(s)?
My attempt:
$\frac{\sqrt{3}-1}{\sin x}+\frac{\sqrt{3}+1}{\cos x}=4\sqrt{2}$
or, $\frac{\sin\pi/3-\sin\pi/6}{\sin x}+\frac{\cos\pi/6+\cos\pi/3}{\cos x}=2\sqrt{2}$
or, $\frac{\sin(\pi/4)cos(\pi/12)}{\sin x}+\frac{\cos(\pi/4)cos(\pi/12)}{\cos x}=\sqrt{2}$
or, $\sin(x+\pi/12)=\sin2x$
or, $x=\pi/12$
| Be careful that your final equation has more potential solutions. The equation
$$ \sin \left(x + \frac{\pi}{12}\right) = \sin 2x$$
implies in fact
$$ x + \frac{\pi}{12} = 2x + 2k \pi$$
or
$$ x + \frac{\pi}{12} = \pi - 2x + 2k \pi.$$
Also recall that you can always check the number of solutions by intersecting
$$ \frac{\sqrt 3 -1}{Y} + \frac{\sqrt 3 +1}{X}=4 \sqrt 2$$
with the unit circle
$$X^2+Y^2 = 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3142283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
A simpler method to prove $\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c} = 1?$ This is the way I think about it:
$
1 = \log_aa = \log_bb = \log_cc \\~\\
\textbf{Using the ‘change of base rule':} \\
\log_{a}{b} = \frac{\log_{b}{b}}{\log_{b}{a}}, \hspace{0.3cm} \log_{b}{c} = \frac{\log_{c}{c}}{\log_{c}{b}}, \hspace{0.3cm} \log_{c}{a} = \frac{\log_{a}{a}}{\log_{a}{c}}\\~\\
\rightarrow \log_{b}{a} = \frac{\log_{b}{b}}{\log_{a}{b}}, \hspace{0.3cm} \log_{c}{b} = \frac{\log_{c}{c}}{\log_{b}{c}}, \hspace{0.3cm} \log_{a}{c} = \frac{\log_{a}{a}}{\log_{c}{a}}\\~\\
\rightarrow \large\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c}\rightarrow\\~\\
\rightarrow\frac{\log_{b}{b}}{\log_{a}{b}} \times \frac{\log_{c}{c}}{\log_{b}{c}} \times \frac{\log_{a}{a}}{\log_{c}{a}} \overset? = 1\\
\rightarrow \frac{1}{\log_{a}{b} \times \log_{b}{c} \times \log_{c}{a}} \overset? = 1\\
\rightarrow \log_{a}{b} \times \log_{b}{c} \times \log_{c}{a} = 1\\
\rightarrow \log_{a}{b} \times \log_{b}{c} \times \log_{c}{a} \overset? = \log_{b}{a} \times \log_{c}{b} \times \log_{a}{c}\\
\rightarrow \frac{\log_{a}{b} \times \log_{b}{c} \times \log_{c}{a}}{\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c}} = 1 \\
\rightarrow \frac{\log_{a}{b}}{\log_{b}{a}} \times \frac{\log_{b}{c}}{\log_{c}{b}} \times \frac{\log_{c}{a}}{\log_{a}{c}} = 1\\~\\
\small \text{------ Using the ‘change of base rule' again ------} \\~\\
\Large \therefore \hspace{0.2cm}\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c} = 1
$
Any other methods?
Thanks in advance.
| $$\log_{b}{a}\log_{c}{b}\log_{a}{c}=e^{\ln\left(\log_{b}{a}\log_{c}{b}\log_{a}{c}\right)}=$$
$$e^{\ln\ln{a}-\ln\ln{b}+\ln\ln{b}-\ln\ln{c}+\ln\ln{c}-\ln\ln{a}}=e^0=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
If complex $z$ satisfies $z+\frac{1}{z}=\cos x$, then evaluate $z^n+\frac{1}{z^n}$ for integer $n$
Let $z$ be a complex number such that
$$
z + \frac{1}{z} = \cos x
$$
Then what is the value of the expression
$$
z^n + \frac{1}{z^n}
$$
where $n$ is an integer?
Please help me. I have tried somehow using the trigonometric way of defining complex numbers but still didn't manage to get anywhere.
| \begin{cases}
z+\frac1z=\cos x\\
z\cdot\frac1z=1
\end{cases}
Considering $t^2-\cos x\cdot t+1=0$, and we get
\begin{equation}
\{z,\frac1z\}=\{\frac{\cos x+\sqrt{\cos^2x-4}}{2},\frac{\cos x-\sqrt{\cos^2x-4}}{2}\}
\end{equation}
So
\begin{align}
z^n+\frac{1}{z^n}
&= (\frac{\cos x+\sqrt{\cos^2x-4}}{2})^n+(\frac{\cos x-\sqrt{\cos^2x-4}}{2})^n\\
&= \frac{\sum_{k=0}^{\lfloor{n/2}\rfloor}\cos^{n-2k}x\cdot(\cos^2x-4)^k}{2^n}.
\end{align}
not so beautiful :(
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3148375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Equilateral triangle with vertices on 3 concentric circles Is my idea correct? 3 concentric circles of radius 1, 2 and 3 are given. An equilateral triangle is formed having its vertices lie on the side of the three concentric circles. What is the length of the equilateral triangle?
My idea is to set a point at the middle of the triangle, then use the distance of it to the vertices given that the three concentric circles are set as $$x^2 + y^2 = 1$$$$x^2 + y^2 = 4$$ and $$x^2 + y^2 = 9$$ I will manipulate the formula afterwards,,,
| Following the notations of the figure by @Dr. Mathva, knowing that $AD=3, BD=2, CD=1,$ and setting $x=AB=AC=BC$, let us express that the Cayley-Menger determinant of the four points $A,B,C,D$ is zero (see here):
$$\begin{vmatrix}0&x^2&x^2&3^2&1\\
x^2&0&x^2&2^2&1\\
x^2&x^2&0&1^2&1\\
3^2&2^2&1^2&0&1\\
1&1 &1 &1& 0\end{vmatrix}=0\tag{1}$$
Expanding and factoring, one obtains :
$$-2x^2(x^2 - 7)^2=0\tag{2}$$
therefore with unique "geometrically acceptable" answer $x=\sqrt{7}$.
Generalization, with radii $a, b, c$.
$$\begin{vmatrix}0&x^2&x^2&c^2&1\\
x^2&0&x^2&b^2&1\\
x^2&x^2&0&a^2&1\\
c^2&b^2&a^2&0&1\\
1&1 &1 &1& 0\end{vmatrix}=0\tag{2}$$
giving, by setting $X=x^2$, the equation :
$$2X(X^2 - sX+ \underbrace{(s^2-3r)}_q)=0 \ \text{where} \ r:=a^2b^2+a^2c^2+b^2c^2, \ s:=a^2+b^2+c^2\tag{3}$$
which, setting apart solution $X=0$, is a quadratic equation. Therefore, conditionned by the following constraint on the sign of its discriminant:
$$\Delta=3(4r-s^2)\ge 0\tag{4}$$
warranting real solutions.
$\Delta$ possesses a remarkable factorization:
$$\Delta=(a+b+c)(a+b-c)(a-b+c)(-a+b+c)\tag{5}$$
As a consequence, when $\Delta > 0$, $\sqrt{\Delta}$ is 4 times the area of the triangle with sides $a,b,c$, due to Heron formula.
The signs of the roots are governed by the sign of $q$ which is in fact always positive, meaning that there are two possible sidelengths for the equilateral triangle if $\Delta >0$,
$$x=\sqrt{\frac12 (s \pm \sqrt{\Delta})}\tag{6}$$
Graphical representation: We can, without loss of generality, work in a plane with coordinates $(a/c,b/c)$ or said otherwise in a plane with coordinates $(a,b)$ with $c=1$. Therefore, equation $\Delta=0$ becomes the equation of a curve... which is plainly the union of four straight lines, 3 of them appearing in the first quadrant. Here is the result of a simulation : in green, resp. red: values of $(a/c, b/c)$ giving a (double) solution, resp. not yielding a solution.
Different remarks:
*
*3 lines are materialized in this simulation : $a+b-1=0, a-b+1=0, and -a+b+1=0$
*the particular case $a=3, b=2, c=1$ considered in the first part corresponds to a limit case between the green and the red zone.
*Here is a case where we have two solutions $a=1.8, \ b=1.5, \ c=1$ with two different values of sidelength $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Convergence of $a_n = \frac{1}{n} \cdot \frac{1\cdot3\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot\ldots\cdot(2n)}$ I am not finding any path to solve it. Online calculators are not helping either... The answer is it converges and converges to 0.
I believe it is the same as the result of
$$\lim_{x\to\infty}\frac1x\times\frac{(2x)!}{4^x\times(x!)^2}$$
I got the idea above by multiplying $a_n$ by ($\frac{2\cdot4\cdot\ldots\cdot2n}{2\cdot4\cdot\ldots\cdot2n}$)
| If
$a_n
= \frac{1}{n} \cdot \frac{1\cdot3\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot\ldots\cdot(2n)}
= \frac{\prod_{k=1}^n(2k-1)}{n\prod_{k=1}^n(2k)}
$
then
$a_{n+1}
= \frac{\prod_{k=1}^{n+1}(2k-1)}{(n+1)\prod_{k=1}^{n+1}(2k)}
$
so
$\begin{array}\\
\frac{a_n}{a_{n+1}}
&=\frac{(n+1)2(n+1)}{n(2n+1)}\\
&=\frac{2(n^2+2n+1)}{2n^2+n}\\
&=\frac{n^2+2n+1}{n^2+n/2}\\
&=\frac{n^2+n/2+(3/2)n+1}{n^2+n/2}\\
&=1+\frac{(3/2)n+1}{n(n+1/2)}\\
&=1+\frac{(3/2)(n+1/2)+1/4}{n(n+1/2)}\\
&=1+\frac{3}{2n}+\frac{1}{4n(n+1/2)}\\
&>1+\frac{3}{2n}\\
\end{array}
$
Since,
if $c_k > 0$,
then
$\prod_{k=1}^n (1+c_k)
\ge 1+\sum_{k=1}^n c_k
$
(easily proved by induction),
$\begin{array}\\
\frac{a_1}{a_{m+1}}
&=\prod_{n=1}^m\frac{a_n}{a_{n+1}}\\
&>\prod_{n=1}^m(1+\frac{3}{2n})\\
&\gt 1+\sum_{n=1}^m\frac{3}{2n}\\
&\gt \frac32\sum_{n=1}^m\frac1{n}\\
&\gt \frac32\ln(m)
\qquad\text{(well-known harmonic sum)}\\
\end{array}
$
so
$a_{m+1}
\lt \frac{2a_1}{3\ln(m)}
= \frac{1}{3\ln(m)}
\to 0$
as
$m \to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How to derive this sequence: $1^3+5^3+3^3=153,16^3+50^3+33^3=165033,166^3+500^3+333^3=166500333,\cdots$? I found it is interesting but I don't know how the R.H.S is coming from the L.H.S,
i.e, how to derive this sequence?
The sequence is as follows:
$$1^3+5^3+3^3=153$$
$$16^3+50^3+33^3=165033$$
$$166^3+500^3+333^3=166500333$$
$$1666^3+5000^3+3333^3=166650003333$$
$$.$$
$$.$$
$$.$$
$$\text{and so on}$$
So,any help please...
| The $n$th such equation is $$\left(\tfrac{5\times 10^{n-1}-2}{3}\right)^3+(5\times 10^{n-1})^3+\left(\tfrac{10^n-1}{3}\right)^3=10^{3n-1}+\tfrac{2}{3}(10^{n-1}-1)10^{2n}+5\times 10^{2n-1}+\tfrac{10^n-1}{3},$$or with $x=10^n$ we can write it as $$\left(\tfrac{x-4}{6}\right)^3+\left(\tfrac{x}{2}\right)^3+\left(\tfrac{x-1}{3}\right)^3=\tfrac{x^3}{10}+\tfrac{2}{3}\left(\tfrac{x}{10}-1\right)x^2+\tfrac{x^2}{2}+\tfrac{x-1}{3},$$which we can verify algebraically (both sides are $\tfrac{(x-1)(x^2+2)}{6}$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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What is wrong with this proof of $3\arcsin x$? We know that
\begin{align}
2\arcsin x&= \arcsin \left(2x\sqrt{1-x^2}\right) \tag{1}\\
\arcsin x + \arcsin y &= \arcsin \left[x\sqrt{1-y^2}+y\sqrt{1-x^2}\right] \tag{2}\\
3\arcsin x &= \arcsin x + 2\arcsin x \tag{3}
\end{align}
Thus $x=x, y=2x\sqrt{1-x^2}$
using ($1$), ($2$) and ($3$):
\begin{align}
3\arcsin x &= \arcsin \left[x\sqrt{1-2x\sqrt{1-x^2}^2}+ 2x(1-x^2)\right]\\
&= \arcsin \left[x\sqrt{1-4x^2(1-x^2)}+ 2x(1-x^2)\right]\\
&= \arcsin \left[x\sqrt{1-2(2x^2)+(2x^2)^2}+ 2x(1-x^2)\right]\\
&= \arcsin \left[x\sqrt{(2x^2-1)^2}+ 2x(1-x^2)\right]\\
&= \arcsin \left[x|2x^2-1|+ 2x(1-x^2)\right]
\end{align}
If $2x^2-1$ is positive, then $|2x^2-1|$ is $2x^2 -1$.
If $2x^2-1$ is negative, then $|2x^2-1|$ is $-2x^2+1$.
Range of $x$ is $-1\leq x \leq 1 \implies 0\leq x^2 \leq 1 \implies 0\leq 2x^2 \leq 2$.
For $x\in\left(\frac{-1}{\sqrt2}, \frac{+1}{\sqrt2}\right)$, then $2x^2-1$ is negative.
For $x\in\left(-1, \frac{-1}{\sqrt2}\right) \cup \left(\frac{+1}{\sqrt2}\ , 1\right)$, then $2x^2-1$ is positive.
Thus for $x\in\left( \frac{-1}{\sqrt2}, \frac{+1}{\sqrt2}\right)$
\begin{align}
3\arcsin x &= \arcsin [x|2x^2-1|+ 2x(1-x^2)]\\
&= \arcsin \left[-2x^3 +x+ 2x(1-x^2)\right]\\
&= \arcsin [3x - 4x^3]
\end{align}
Thus for $x\in\left(-1, \frac{-1}{\sqrt2}\right) \cup \left(\frac{+1}{\sqrt2}, 1\right)$
\begin{align}
3\arcsin x &= \arcsin [x|2x^2-1|+ 2x(1-x^2)]\\
&= \arcsin [2x^3- x+2x-2x^3]\\
&= \arcsin[x]
\end{align}
But clearly, $3\arcsin x = \arcsin[3x-4x^3]$.
So what is wrong whith this proof?
| The problem is a conflict between these two lines:
$2\arcsin x= \arcsin(2x\sqrt{1-x^2}) \tag1$
and
Thus for $x\in\left(-1, \frac{-1}{\sqrt2}\right) \cup \left(\frac{+1}{\sqrt2}, 1\right)$
The problem is that if $\frac1{\sqrt2} < \lvert x \rvert < 1,$
then $\frac\pi4 < \lvert\arcsin x\rvert < \frac\pi2,$ and therefore the left-hand side of equation $(1)$ obeys
$$ \frac\pi2 < \lvert2\arcsin x\rvert < \pi.$$
But the right-hand side of $(1)$ is just an application of the arc sine to a number, which must obey the conditions
$$ \lvert \arcsin (2x\sqrt{1-x^2}) \rvert \leq \frac\pi2.$$
Hence it is impossible for $(1)$ to be true when
$x\in\left(-1, -\,\frac{1}{\sqrt2}\right) \cup \left(\frac{1}{\sqrt2}, 1\right).$
That means that everything you concluded by using $(1)$ -- which is to say, everything after the first few lines -- is inapplicable to the case
$x\in\left(-1, -\,\frac{1}{\sqrt2}\right) \cup \left(\frac{1}{\sqrt2}, 1\right).$
By using $(1),$ you restrict the validity of your argument to only the case
where $-\frac\pi4 \leq \arcsin x \leq \frac\pi4,$
that is, $x \in \left(-\,\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right).$
The line
$ 3\arcsin x= \arcsin\left[x\sqrt{1-\left(2x\sqrt{1-x^2}\right)^2}+ 2x(1-x^2)\right] $
restricts the applicability of your argument even further, because again the right-hand side has magnitude at most $\frac\pi2,$ which means $\arcsin x$ can only have magnitude at most $\frac\pi6.$
So even the line
Thus for $x\in\left(\frac{-1}{\sqrt2},\frac{+1}{\sqrt2}\right)$
is not right; everything you wrote after the first few lines is applicable only for
$-\frac12 \leq x \leq \frac12,$ so you really should have written,
"Thus for $x \in \left[-\frac12,\frac12\right]$".
For other $x$ you can follow Yves Daoust's answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that for $a,p,q \in \Bbb R$ the solutions of: $\frac{1}{x-p} + \frac{1}{x-q} = \frac {1}{a^2}$ are real numbers. Prove that for $a,p,q \in \Bbb R$ the solutions of: $$\frac{1}{x-p} + \frac{1}{x-q} = \frac {1}{a^2}$$
are real numbers.
I tried manipulating the expression, getting rid of the denominators, but i can't factor $x$.
Any hints?
| We need $a\ne0$. The equation can be written as $(x-p)(x-q)=a^2(2x-p-q)$, or $x^2-(2a^2+p+q)x+a^2(p+q)+pq=0$.
The discriminant is
$(2a^2+p+q)^2-4a^2(p+q)-4pq=(p-q)^2+4a^4>0$
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
} |
If $f(a)=f(b)=0$ and $|f''(x)|\le M$ prove $|\int_a^bf(x)\mathrm{d}x| \le \frac{M}{12}(b-a)^3$ If $f(a)=f(b)=0$ and $|f''(x)|\le M$. Prove
$$|\int_a^bf(x)\mathrm{d}x| \le \frac{M}{12}(b-a)^3$$
I have thought about that since $f(a) = f(b) = 0 $ there is $\xi$ such that $f'(\xi) = 0$. Then when $x \le \xi$, $|f'(x)| \le (\xi - x)M$ and when $x \ge \xi$, $|f'(x)| \le (x-\xi)M$. After that $|f(x)| \le \frac{\xi^2 - (\xi -x)^2}{2}$ when $x \le \xi$ and $|f(x)| \le \frac{(b-\xi)^2 - (x - \xi)^2}{2}$ when $x \ge \xi$. Therefore
$$|\int_a^b f(x)\mathrm{d}x| \le \int_a^\xi |f(x)| + \int_\xi^b |f(x)| = \frac{\xi^3}{3} + \frac{(b-\xi)^3}{3}$$
If $x = \frac{a+b}{2}$, we have
$$\frac{x^3}{3}+\frac{(b-x)^3}{3} = \frac{(b-a)^3}{12}$$
But in this situation $\frac{x^3}{3}+\frac{(b-x)^3}{3}$ is the minimal value. So I can't go on.
| Show that $f(a)=f(b)=0$ and $|f''(x)|\le M$ implies
$$
\tag{*}
|f(x)| \le \frac M2 (x-a)(b-x)
$$
for $a \le x \le b$, i.e. $f$ is bounded by the parabola with constant second derivative $-M$ and zeros at $x=a$ and $x=b$.
Then
$$
\left|\int_a^b f(x)\, dx \right| \le \frac M2\int_a^b (x-a)(b-x) \, dx = \frac{M}{12}(b-a)^3
$$
follows.
In order to prove $(*)$, consider for fixed $y \in (a, b)$ the function
$$
h(x) = f(x) (y-a)(b-y) - f(y) (x-a)(b-x) \, .
$$
$h$ satisfies $h(a) = h(y) = h(b) = 0$, so that a repeated application of Rolle's theorem implies
$$
h''(\xi) = f''(\xi)(y-a)(b-y) +2f(y) = 0
$$
for some $\xi \in (a, b)$.
Alternatively note that the two functions
$$
\frac M2 (x-a)(b-x) \pm f(x)
$$
are concave in $[a, b]$ because their second derivative is $\le 0$. A concave functions attains its minimum on an interval at one of the boundary points, and therefore
$$
\frac M2 (x-a)(b-x) \pm f(x) \ge 0 \, .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3152826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
GCD of cubic polynomials I would appreciate some help finding $GCD(a^3-3ab^2, b^3-3ba^2)$; $a,b \in \mathbb{Z}$. So far I've got here: if $GCD(a,b)=d$ then $\exists \alpha, \beta$ so that $GCD(\alpha, \beta)=1$ and $\alpha d=a$, $\beta d=b$.
Therefore we know that $GCD(a^3-3ab^2, b^3-3ba^2)=d^3 GCD(\alpha^3-3\alpha \beta^2, \beta^3-3\beta \alpha^2)$. However I don't know to figure out $GCD(\alpha^3-3\alpha \beta^2, \beta^3-3\beta \alpha^2)$ given that $GCD(\alpha,\beta)=1$.
| As you've already shown, factoring out the cube of the GCD of $a$ and $b$ gives a new equation
$$e = \gcd\left(\alpha^3-3\alpha \beta^2, \beta^3-3\beta \alpha^2 \right) \tag{1}\label{eq1}$$
where
$$\gcd\left(\alpha,\beta \right) = 1 \tag{2}\label{eq2}$$
Update: Here is a simpler solution than what I originally wrote. First, note that no factor of $e$ may divide $\alpha$ or $\beta$. If any do, let's say $\alpha$, then it must divide $\beta^3 - 3\beta\alpha^2$ and, thus, must divide $\beta^3$, which is not possible due to \eqref{eq2}. Thus, from the first term of \eqref{eq1}, since $\alpha^3-3\alpha \beta^2 = \alpha\left(\alpha^2 - 3\beta^2\right)$, this means that $e \mid \alpha^2 - 3\beta^2$. Similarly, for the second term, $\beta^3-3\beta \alpha^2 = \beta\left(\beta^2 - 3\alpha^2\right)$ gives that $e \mid \beta^2 - 3\alpha^2$. Also, $e$ must divide any linear combination of these values, including $e | \alpha^2 - 3\beta^2 + 3\left(\beta^2 - 3\alpha^2\right) = -8\alpha^2$. Thus, $e$ can only be a power of $2$. To finish this off, go to the second last paragraph. Otherwise, for the rest of the original, longer solution, continue reading.
$ $
Next, note that if $f = \gcd(g,h)$, then $f$ divides $g$ and $h$ and, thus, will also divide any linear combination of $g$ and $h$, including their sum & difference. From \eqref{eq1}, first check the sum of the $2$ inside values:
\begin{align}
\alpha^3-3\alpha \beta^2 + \beta^3-3\beta \alpha^2 & = \alpha^3 + \beta^3 -3\left(\alpha \beta\right)\beta - 3\left(\alpha \beta\right)\alpha \\
& = \left(\alpha + \beta\right)\left(\alpha^2 - \alpha\beta + \beta^2\right) - 3\alpha\beta\left(\alpha + \beta\right) \\
& = \left(\alpha + \beta\right)\left(\alpha^2 - 4\alpha\beta + \beta^2\right) \tag{3}\label{eq3}
\end{align}
Suppose there's a factor $m \gt 1$ which divides $e$ and $\alpha + \beta$. Then, $\alpha \equiv -\beta \pmod m$, so $\alpha^3-3\alpha \beta^2 \equiv 2\beta^3 \pmod m$. From \eqref{eq2}, this means that $m = 2$, and that any other factors of $e$ must divide $\alpha^2 - 4\alpha\beta + \beta^2$.
From \eqref{eq1}, next check the difference of the $2$ inside values:
\begin{align}
\alpha^3-3\alpha \beta^2 - \beta^3 + 3\beta \alpha^2 & = \alpha^3 - \beta^3 -3\left(\alpha \beta\right)\beta + 3\left(\alpha \beta\right)\alpha \\
& = \left(\alpha - \beta\right)\left(\alpha^2 + \alpha\beta + \beta^2\right) + 3\alpha\beta\left(\alpha - \beta\right) \\
& = \left(\alpha - \beta\right)\left(\alpha^2 + 4\alpha\beta + \beta^2\right) \tag{4}\label{eq4}
\end{align}
Suppose there's a factor $n \gt 1$ which divides $e$ and $\alpha - \beta$. Then, $\alpha \equiv \beta \pmod n$, so $\alpha^3-3\alpha \beta^2 \equiv -2\beta^3 \pmod m$. From \eqref{eq2}, this means that $n = 2$, and that any other factors of $e$ must divide $\alpha^2 + 4\alpha\beta + \beta^2$.
This shows that any factor, other than $2$, which divides $e$ must divide both $\alpha^2 - 4\alpha\beta + \beta^2$ and $\alpha^2 + 4\alpha\beta + \beta^2$. Thus, it must also divide their difference, which is $8\alpha\beta$. This can only be true for $2$, $4$ or $8$.
At this shows overall, only powers of $2$ may possibly divide $e$. Since $e$ is relatively prime to $\alpha$ & $\beta$, this means they must both be odd. From $\alpha^3 - 3\alpha\beta^2 = \alpha\left(\alpha^2 - 3\beta^2\right)$, note that $\alpha^2 \equiv \beta^2 \equiv 1 \pmod 4$, so $\alpha^2 - 3\beta^2 \equiv -2 \pmod 4$. In other words, there will only be $1$ factor of $2$.
In summary, with your original equation of $d = \gcd\left(a,b\right)$, we get that $\gcd\left(a^3-3ab^2, b^3-3ba^2\right)$ is $2d^3$ if both $\frac{a}{d}$ and $\frac{b}{d}$ are odd, else it's $d^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3156949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solving combinatorial problems with symbolic method and generating functions I am trying to solve the following problems:
a) Let $\mathcal{F}$ be the family of all finite 0-1-sequences that have no 1s directly behind each other. Let the weight of each sequence be its length.
How can $\mathcal{F}$ be constructed with simpler objects? How does the generating function look like?
b) Show with generating functions: The number of partitions of n into different summands equals the number of partitions of n into odd summands.
c) Show with generating functions: The number of compositions of n into summands being 1 or 2 equals the number of compositions of n+2 into summands greater than or equal 2.
My solutions:
a) I have no idea here.
b) Let $\mathcal{P}$ be the partition in different summands. Then $\mathcal{P} = (\{\epsilon\}+\{1\}) \times (\{\epsilon\}+\{2\})\times (\{\epsilon\}+\{3\})\times ...$
$\Rightarrow P(z) = (1+z)\cdot (1+z^2) \cdot (1+z^3) \cdot \dotsc = \frac{1}{(1-z)\cdot(1-z^3)\cdot(1-z^5)\cdot \dotsc}$
Now let $\tilde{\mathcal{P}}$ be the partition in odd summands. Then $\tilde{\mathcal{P}} = \{1\}^{\ast}\times\{3\}^{\ast}\times\{5\}^{\ast}\times\dotsc$
$\Rightarrow \tilde{P(z)} = \frac{1}{1-z}\cdot\frac{1}{1-z^3}\cdot\frac{1}{1-z^5}\cdot \dotsc$.
Therefore $P(z) = \tilde{P}(z)$ and so $[z^n]P(z) = [z^n]\tilde{P}(z)$, which proofs that the numbers of partitions of n are the same.
c) Let $\mathcal{K}$ be the number of compositions of n into 1s and 2s.
Then $\mathcal{K} = \{1,2\}^{\ast}$ and so $K(z) = \frac{1}{1-(z+z^2)}$.
Let $\tilde{\mathcal{K}}$ be the number of compositions of n+2 into 2,3,4,5,6,7,... Then $\tilde{\mathcal{K}} = \{2,3,4,5,6,...\}^{\ast}$ and therefore $\tilde{K}(z) = \frac{1}{1-(z^2+z^3+z^4+z^5+...)}$.
I am not sure if I have determined $\mathcal{K}, \tilde{\mathcal{K}}, K(z)$ and $\tilde{K}(z)$ correctly and if so, I don't know how to show that $[z^n]K(z) = [z^{n+2}]\tilde{K}(z)$.
So I'd very much appreciate your help on a) and c). Thanks in advance!
|
Ad c.)
Your approach is fine. With ${\mathcal{K}} = \{1,2\}^{\ast}$ we obtain
\begin{align*}
K(z) &= \frac{1}{1-\left(z+z^2\right)}\\
&=\frac{1}{1-z-z^2}\tag{1}
\end{align*}
and with $\tilde{\mathcal{K}} = \{2,3,4,5,6,...\}^{\ast}$ we obtain
\begin{align*}
\tilde{K}(z) &= \frac{1}{1-(z^2+z^3+z^4+z^5+\cdots)}\\
&=\frac{1}{1-\frac{z^2}{1-z}}\tag{2}\\
&=\frac{1-z}{1-z-z^2}\\
&=1+\frac{z^2}{1-z-z^2}\tag{3}
\end{align*}
Comment:
*
*In (2) we use the geometric series expansion.
We obtain from (1) and (3) for $n\geq 1$
\begin{align*}
\color{blue}{[z^{n+2}]\tilde{K}(z)} &=[z^{n+2}]\left(1+\frac{z^2}{1-z-z^2}\right)\\
&=[z^{n+2}]\frac{z^2}{1-z-z^2}\tag{4}\\
&=[z^n]\frac{1}{1-z-z^2}\tag{5}\\
&\,\,\color{blue}{=[z^n]K(z)}
\end{align*}
and the claim follows.
Comment:
*
*In (4) we skip the term $1$ which does not contribute to the coefficient of $z^{n+2}$ since $n\geq 1$.
*In (5) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
Note the coefficients of
\begin{align*}
K(z)&=\frac{1}{1-z-z^2}\\
&=\color{blue}{1}+\color{blue}{1}z+\color{blue}{2}z^2+\color{blue}{3}z^3+\color{blue}{5}z^4+\color{blue}{8}z^5+\cdots
\end{align*}
are the Fibonacci numbers.
ad a.)
We start with binary sequences with no consecutive equal characters at all. See example III.24 Smirnov words from Analytic Combinatorics by P. Flajolet and R. Sedgewick for more information.
A generating function for the number of Smirnov words over a binary alphabet is given by
\begin{align*}
\left.\left(1-\frac{u}{1+u}-\frac{w}{1+w}\right)^{-1}\right|_{u=w=z}\tag{6}
\end{align*}
where $u$ represents occurrences of $0$ and $w$ occurrences of $1$.
Since there are no restrictions stated for zeros we replace occurrences of $0$ in a Smirnov word by one or more zeros.
\begin{align*}\
u\longrightarrow u+u^2+u^3+\cdots=\frac{u}{1-u}\tag{7}
\end{align*}
We substitute (7) in (6) evaluate at $z$ and obtain
\begin{align*}
\left(1-\frac{\frac{z}{1-z}}{1+\frac{z}{1-z}}-\frac{z}{1+z}\right)^{-1}
&=\left(1-z-\frac{z}{1+z}\right)^{-1}\\
&=\frac{1+z}{1-z-z^2}\\
&=1+2z+3z^2+5z^3+8z^4+\cdots
\end{align*}
which is again a generating function of (shifted) Fibonacci numbers.
| {
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"url": "https://math.stackexchange.com/questions/3157169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove $f(x, y) = \frac{x^2}{y}$ is a convex function on the set $\{(x, y) \in \mathbb{R}^2 : y > 0\}$
Prove $f(x, y) = \frac{x^2}{y}$ is a convex function on the set $\{(x,y) \in \mathbb{R}^2 : y > 0\}$.
Attempt:
I start with the basic convexity, i.e.,
\begin{align}
f( \alpha_1 x_1 + \alpha_2 x_2) \leq \alpha_1 f(x_1) + \alpha_2f(x_2) \ ,
\end{align}
where $\alpha_1 + \alpha_2 = 1$.
Let $z = (x,y)$, then
\begin{align}
f( \alpha z_1 + (1-\alpha) z_2) \leq \alpha f(z_1) + (1-\alpha)f(z_2) \ ,
\end{align}
How to proceed from here? Thank you so much.
Question: Can this be proved by perspective function? If yes, how to prove that.
| A twice continuously differentiable function of several variables is convex on a convex set if and only if its Hessian matrix of second partial derivatives is positive semidefinite on the interior of the convex set.
$$f(x,y) = \frac{x^2}{y}$$
Partial derivatives:
$$\frac{\partial f(x,y)}{\partial x} = \frac{2x}{y}, \frac{\partial f(x,y)}{\partial y} = -\frac{x^2}{y^2}$$
$$\frac{\partial^2 f(x,y)}{\partial x^2} = \frac{2}{y}, \frac{\partial^2 f(x,y)}{\partial x \partial y} = -\frac{2x}{y^2}$$
$$\frac{\partial^2 f(x,y)}{\partial y \partial x} = -\frac{2x}{y^2}, \frac{\partial^2 f(x,y)}{\partial y^2} = \frac{2x^2}{y^3}$$
Hessian matrix:
$$H =
\begin{bmatrix}
\frac{\partial^2 f(x,y)}{\partial x^2} & \frac{\partial^2 f(x,y)}{\partial x \partial y} \\
\frac{\partial^2 f(x,y)}{\partial y \partial x} & \frac{\partial^2 f(x,y)}{\partial y^2}
\end{bmatrix}
$$
$$H =
\begin{bmatrix}
\frac{2}{y} & -\frac{2x}{y^2} \\
-\frac{2x}{y^2} & \frac{2x^2}{y^3}
\end{bmatrix}
$$
Sylvester's criterion is used to prove that matrix is positive semidefinite:
matrix is positive semidefinite if and only if all leading minors are non-negative.
$$\Delta_{(1)} = \frac{\partial^2 f(x,y)}{\partial x^2} = \frac{2}{y} >= 0$$
$$\Delta_{(2)} = \frac{\partial^2 f(x,y)}{\partial y^2} = \frac{2x^2}{y^3} >= 0$$
$$\Delta_{(1,2)} =
\begin{vmatrix}
\frac{2}{y} & -\frac{2x}{y^2} \\
-\frac{2x}{y^2} & \frac{2x^2}{y^3}
\end{vmatrix} =
\frac{2}{y} * \frac{2x^2}{y^3} - (-\frac{2x}{y^2}) * (-\frac{2x}{y^2}) =
\frac{4x^2}{y^4} - \frac{4x^2}{y^4} = 0$$
So, all leading minors are non-negative on $\lbrace (x,y) \in \mathbb{R}^2 : y>0 \rbrace$.
Hence, Hessian matrix is positive semidefinite.
Hence, function $f(x,y)$ is a convex function on the set $\lbrace (x,y) \in \mathbb{R}^2 : y>0 \rbrace$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluating expression with Integer part and Fraction part of a nested radical
Let
$$A= \sqrt{93+28\sqrt{11}}$$
if $B$ is the integer part of $A$ and $C$ is the fraction part of $C$, what is the value of
$$B+C^2$$
I tried manipulating it by setting
$$ A=B+C$$
but I can't transform it into the expression, do I need approximate the integer part of A in order to solve this or is there another way?
| Let's find intgers $x,y$ such that $$(x+y\sqrt{11})^2= 93+28\sqrt{11}$$
Then $$x^2+11y^2 =93 \;\;\;\wedge \;\;\; 2xy = 28$$
Since $11y^2<99\implies y^2 <9 \implies |y|\leq 2$ and $xy=14$. Playing with numbers we see that $x=7$ and $y=2$ works well, so $$\boxed{\sqrt{93+28\sqrt{11}} = 7+2\sqrt{11}}$$
Since $$13= 7+2\cdot 3<7+2\sqrt{11} <7+2\cdot \sqrt{49\over 4} = 14$$
we see $B =13$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3159345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Need to compute $\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+2)3^{n+1}}$. Is my solution correct? $$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+2)3^{n+1}}$$
Since $$\tan^{-1}x=\int \frac{1}{1+x^{2}} dx=\int (1-x^{2}+x^{4}+...)dx=x-\frac{x
^3}{3}+\frac{x^5}{5}+...$$
so$$\int \tan^{-1}x dx=\int (x-\frac{x^3}{3}+\frac{x^5}{5}+...)dx=\frac {x^2}{2}-\frac{x^4}{3\cdot4}+\frac {x^6}{5\cdot 6}+...=\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+2}}{(2n+1)(2n+2)}$$
Therefore, $$\int_{0}^{1/{\sqrt 3}}\tan^{-1}xdx=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)(2n+2)3^{n+1}}$$
$$\Rightarrow \frac{\tan^{-1}(\frac{1}{\sqrt 3})}{\sqrt 3}-\frac{\ln(1+\frac{1}{3})}{2}=\frac{\pi}{6\sqrt3}-\frac{\ln\frac{4}{3}}{2}$$
Is my method correct?
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over
\pars{2n + 1}\pars{2n + 2}3^{n + 1}}} =
{1 \over 3}\sum_{n = 0}^{\infty}{\pars{-1/3}^{n} \over 2n + 1} - {1 \over 3}\sum_{n = 0}^{\infty}
{\pars{-1/3}^{n} \over 2n + 2}
\\[5mm] = &\
{1 \over 3}\sum_{n = 0}^{\infty}{\pars{-1/3}^{n} \over 2n + 1} + \sum_{n = 1}^{\infty}{\pars{-1/3}^{n} \over 2n}
\\[5mm] = &\
{1 \over 3}\,{\root{3} \over \ic}\sum_{n = 0}^{\infty}{\pars{\ic/\root{3}}^{2n + 1} \over 2n + 1} +
\sum_{n = 1}^{\infty}{\pars{\ic/\root{3}}^{2n} \over 2n}
\\[5mm] = &\
{1 \over 3}\,{\root{3} \over \ic}\sum_{n = 1}^{\infty}{\pars{\ic/\root{3}}^{n} \over n}\,{1 - \pars{-1}^{n} \over 2} +
\sum_{n = 1}^{\infty}{\pars{\ic/\root{3}}^{n} \over n}
\,{1 + \pars{-1}^{n} \over 2}
\\[5mm] = &\
{\root{3} \over 3}\,\Im\sum_{n = 1}^{\infty}{\pars{\ic/\root{3}}^{n} \over n} +
\Re\sum_{n = 1}^{\infty}{\pars{\ic/\root{3}}^{n} \over n}
\end{align}
Since
$\ds{\sum_{n = 1}^{\infty}{\pars{\ic/\root{3}}^{n} \over n} =
-\ln\pars{1 - {\root{3} \over 3}\,\ic} =
-\,{1 \over 2}\,\ln\pars{4 \over 3} + {\pi \over 6}\,\ic}$:
\begin{align}
&\bbox[10px,#ffd]{\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over
\pars{2n + 1}\pars{2n + 2}3^{n + 1}}} =
{\root{3} \over 3}\,{\pi \over 6} -
{1 \over 2}\ln\pars{4 \over 3}
\\[5mm] = &\
\bbx{{\root{3} \over 18}\,\pi - {1 \over 2}\ln\pars{4 \over 3}}
\approx 0.1585
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3162027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
induction proof: number of compositions of a positive number $n$ is $2^{n-1}$ So my problem was to find the formula for the number of composition for a positive number $n$. Then prove its validity. I found the formula which was $2^{n-1}$. Having trouble with the induction.
$$Claim:2^0+2^1+2^2+...+(2^{n-2}+1)=2^{n-1}$$
When $n=5$ $$ 1+2+4+(2^{5-2}+1)=2^{5-1}$$
$$ 1+2+4+(2^{3}+1)=2^{4}$$
$$ 1+2+4+(8+1)=16$$
$$ 16=16$$
Assume that $n=k$
$$2^0+2^1+2^2+...+(2^{k-2}+1)=2^{k-1}$$
Now show for $k+1$
$$2^0+2^1+2^2+...+(2^{k-1}+1)=2^{k}$$
Don't know what to do after. Help is appreciated.
$$2^0+2^1+2^2+...+(2^k)+(2^{k-1}+1)=2^{k}$$
| $n=k$ $$2^0+2^1+2^2+...+(2^{k-2}+1)=2^{k-1}$$ so
$$\underbrace{2^0+2^1+2^2+...+2^{k-2}}=\overbrace{2^{k-1} -1}$$
Now show for k+1
$$\underbrace{2^0+2^1+2^2+...+2^{k-2}}+(2^{k-1}+1)=\overbrace{2^{k-1} -1}+(2^{k-1}+1)$$
$$=2^{k-1}+2^{k-1}=2*2^{k-1}=2^{k}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3163924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Find minimum perimeter of the triangle circumscribing semicircle The following diagram shows triangle circumscribing a semi circle of unit radius. Find minimum perimeter of triangle
My try:
Letting $$AP=AQ=x$$
By power of a point we have:
$$BP^2=OB^2-1$$ where $O$ is center of the circle.
Also $$CQ^2=OC^2-1$$
Let $$OB=y$$
$$OC=z$$
Then $$BP=\sqrt{y^2-1}$$
$$CQ=\sqrt{z^2-1}$$
So the perimeter of triangle is:
$$P=2x+y+z+\sqrt{y^2-1}+\sqrt{z^2-1}$$
Any help from here?
| The problem with that is that there are only really two degrees of freedom; the variables $x,y,z$ are related in some way. We could try to find what that relation is - or we could find another way to express things, one with easier relations to spot.
I'd go with the angles of the triangle. Labeling $O$ as you did, $OP$ is perpendicular to $AB$ and $OQ$ is perpendicular to $AC$. Therefore, from right triangle $OBP$, $OB=\csc B$ and $BP=\cot B$. Similarly, $OC=\csc C$ and $CQ=\cot C$. At vertex $A$, note that triangles $AOP$ and $AOQ$ are congruent - so $AO$ bisects angle $A$, and $AP=AQ=\cot\frac{A}{2}$.
We now seek to minimize the perimeter $p$ subject to the constraints $A+B+C=\pi$ and $A,B,C > 0$. There's one more thing to note: $\csc \theta + \cot \theta = \cot\frac{\theta}{2}$. Applying this identity,
$$p(A,B,C) = \csc B+\cot B +\csc C+\cot C+2\cot\frac A2 = 2\cot \frac A2+ \cot \frac B2+\cot \frac C2$$
Now, we solve this constrained optimization problem. First, the boundary constraints $A,B,C\ge 0$. Since each of the half-angles $\frac A2, \frac B2, \frac C2$ is in $(0,\pi)$, all of those cotangents are positive. As any of these angles goes to zero, its cotangent goes to $\infty$ and so too does the perimeter. No minimum there.
In the interior triangular region, we apply Lagrange multipliers. Differentiating,
$$\nabla p(A,B,C) = -\frac12\left(2\csc^2\frac A2, \csc^2 \frac B2, \csc^2 \frac C2\right)$$
must be a constant multiple of $(1,1,1)$, the gradient of the constraint. Since $\csc^2$ decreases monotonically on $(0,\frac{\pi}{2})$, we must have $B=C < A$. If the angles summed to $2\pi$, we would have a nice solution $\frac A2=\frac{\pi}{2}, \frac B2=\frac C2 = \frac{\pi}{4}$ - but no, we're not that lucky. Instead, we apply $\frac A2 = \frac{\pi}{2}-B$ to eliminate $A$:
$$1-\cos B=2\sin^2 \frac B2 = \sin^2 \frac A2 = \cos^2 B$$
so $\cos^2 B + \cos B - 1 = 0$ and $\sin \frac A2 = \cos B = \dfrac{\sqrt{5}-1}{2}$. Then
\begin{align*}\cos\frac A2 &= \sqrt{1-\left(\frac{\sqrt{5}-1}{2}\right)^2} = \sqrt{\frac{\sqrt{5}-1}{2}}\\
\sin\frac B2 &= \sqrt{\frac{3-\sqrt{5}}{4}} = \frac1{\sqrt{2}}\cdot \frac{\sqrt{5}-1}{2}\\
\cos\frac B2 &= \sqrt{\frac{1+\sqrt{5}}{4}} = \frac1{\sqrt{2}}\cdot\sqrt{\frac{\sqrt{5}+1}{2}}\\
\cot\frac A2 &= \sqrt{\frac{2}{\sqrt{5}-1}} = \sqrt{\frac{\sqrt{5}+1}{2}}\\
\cot\frac B2 &= \sqrt{\frac{\sqrt{5}+1}{3-\sqrt{5}}} = \sqrt{\sqrt{5}+2}\\
p &= 2\cot\frac A2 + 2\cot\frac B2 = 2\sqrt{\frac{\sqrt{5}+1}{2}}+2\sqrt{\sqrt{5}+2}\approx 6.66\end{align*}
A truly fiendish problem.
Numerically, the minimizing angles are about $76^\circ$ at $A$ and $52^\circ$ at $B$ and $C$.
| {
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"url": "https://math.stackexchange.com/questions/3166836",
"timestamp": "2023-03-29T00:00:00",
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How to prove the inequality $a/(b+c)+b/(a+c)+c/(a+b) \ge 3/2$ Suppose $a>0, b>0, c>0$.
Prove that:
$$a+b+c \ge \frac{3}{2}\cdot [(a+b)(a+c)(b+c)]^{\frac{1}{3}}$$
Hence or otherwise prove:
$$\color{blue}{\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge \frac{3}{2}}$$
| Using AM-GM inequality:$$\frac{x_1+\dots+x_n}{n}\geq (x_1\dots x_n)^{1/n}$$
let $n=3$ and $x_1=a+b,\; x_2=b+c,\; x_3=c+a$:
$$\frac{2(a+b+c)}{3}\geq [(a+b)(b+c)(c+a)]^{1/3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that the equation $x^5+x^4=1$ has a unique solution. Show that the equation $x^5 + x^4 = 1 $ has a unique solution.
My Attempt:
Let $f(x)=x^5 + x^4 -1 $
Since $f(0)=-1$ and $f(1)=1$, by the continuity of the function $f(x)$ has at least one solution in $(0,1)$.
| If $-1<x<0$ then $x^{5}+x^{4}=x^{4}(1+x) <1$ because $x^{4}$ and $1+x$ both belong to $(0,1)$. If $ x \leq -1$ then $x^{5}+x^{4}=x^{4}(1+x) \leq 0<1$. Hence there is no root with $x <0$. For $x>0$ the function is strictly increasing so it cannot have a second root.
| {
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"source": "stackexchange",
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Given $\cos 3\theta = 4(\cos\theta)^3 - 3\cos\theta$, solve $4x^3 - 9x - 1 = 0$, correct to 3 decimal places. I am trying to solve the following problem:
Given $\cos 3\theta = 4(\cos\theta)^3 - 3\cos\theta$, solve the equation $4x^3 - 9x - 1 = 0$, correct to 3 decimal places.
Assuming that $x = \cos 3\theta$, you can substitute it into the equation so that you get:
$4(\cos\theta)^3 - 9(\cos\theta) = 1$
The problem with the above is that there is a $9$ in front of the $\cos\theta$. If there was a $3$ instead, I could have replaced the LHS of the equation with $\cos 3\theta$, and then solved for $\theta$. How can I solve this equation? Any insights are appreciated.
| Render $x=a\cos\theta$. Thereby
$4x^3-9x=4a^3\cos^3\theta-9a\cos\theta$
$=a^3(4\cos^3\theta-(9/a^2)\cos\theta)$
If we put $a=\sqrt{3}$ then $9/a^2=3$ and then
$4x^3-9x=3\sqrt{3}\cos(3\theta)=1$
where the "=1" cones from the constant term of the original equation. So $\cos(3\theta)=\sqrt{3}/9$, from which we can get values of $\theta$ to four decimal places. That should be enough accuracy as $x=a\cos\theta=\sqrt{3}\cos\theta$ has a derivative within $\pm 2$ for all $\theta$.
| {
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Solve $ \int_0^a x^4 \sqrt{a^2-x^2}dx$
Solve $$ \int_0^a x^4 \sqrt{a^2-x^2}dx.$$
Substitution: $x = a\sin t \Rightarrow dx = a\cos t dt$
$$ \int_0^a x^4 \sqrt{a^2-x^2}dx = a^6 \int_0^{\pi/2}\sin^4t\cos^2 tdt = \frac{a^6}6 \frac{\Gamma(5/2)\Gamma(3/2)}{\Gamma(3)} = \frac{a^6 \pi}{288}.$$
When evaluating this integral in Wolfram Alpha, I get another result. Also plugging in values and checking if they correspond to the answer I would get with my result, give contradictions. Where is my mistake?
Thanks.
EDIT: I found my mistake. Just had a little brain fart: $\Gamma(3) = 2!$ and not $3!$. Thus we get $$\frac{a^6}{12}\cdot \frac{3\pi}{8} = \frac{a^6 \pi}{32},$$ which is correct. Thank you for your time and responses!
| A direct path to the Beta function can be made by considering the substitution $x = a \, \sqrt{u}$ in the integral:
\begin{align}
I &= \int_{0}^{a} x^{4} \, (a^2 - x^2)^{1/2} \, dx \\
&= \int_{0}^{1} a^4 \, u^2 \, (a^2 - a^2 u)^{1/2} \, \frac{a \, du}{2 \, \sqrt{u}} \\
&= \frac{a^6}{2} \, \int_{0}^{1} u^{3/2} \, (1-u)^{1/2} \, du \\
&= \frac{a^6}{2} \, B\left(\frac{5}{2}, \frac{3}{2}\right) \\
&= \frac{a^6 \, \pi}{32}.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the maximum values of the function $f(x,y,z)=x^2y^2z^2$ subject to the constraint $x^2+y^2+z^2=289$. I have already figured out the majority of the problem. I took the derivative of $x$ and got: $x=0,y=0,z=0$. Initially I put that there were infinitely many solutions, but that is not true. I do not know where I went wrong as I took the derivatives in respect to $x$ and subtracted from the derivative of x to the $g(x)$. However, it gave me zero for all of my solutions.
I successfully found the minimum though.
| With AM-GM$$x^2+y^2+z^2=289\ge3\cdot \sqrt[3]{x^2y^2z^2}$$ Thus $$\bigg(\frac{289}{3}\bigg)^3\geq x^2y^2z^2$$
Equality holds iff $\pm x=\pm y=\pm z$
Detailed Explanation
Theorem 1: Arithmetic Mean-Geometric Mean (or simply AM-GM) inequality for three variables
Let $a,b,c\in \mathbb R^+$, then $$\frac{a+b+c}{3}\ge\sqrt[3]{a\cdot b\cdot c}$$
Proof
Substitute $a=d^3, b=e^3, c=f^3$. Observe now that
\begin{align*}
&d^3+e^3+f^3-3\cdot def=\frac{1}{2}\cdot (d+e+f)\cdot \big((d-e)^2+(e-f)^2+(f-d)^2\big)\ge0\tag{1}\\
\iff & d^3+e^3+f^3\ge 3\cdot def\\
\iff & a+b+c\ge 3\cdot \sqrt[3]{abc}
\end{align*}
You can check the equation in $(1)$ here or prove it yourself - it's up to you.
Sidenote: Note that you'll only achive equality in $(1)$ if and only if $$d+e+f=0\qquad \text{or} \qquad (d-e)^2+(e-f)^2+(f-d)^2=0$$
i.e. if and only if $d=e=f\implies a=b=c$.
Now, back to your problem, observe that the function $f(x, y, z):=x^2\cdot y^2\cdot z^2$ can only take non-negative values as long as $f:\mathbb R\to \mathbb R^+$ - which I assume.
Thus - in virtue of the Theorem 1 $$\frac{289}{3}=\frac{x^2+y^2+z^2}{3}\ge \sqrt[3]{x^2y^2z^2} $$
| {
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How to solve this with mathematical induction? $$\sum_{i=0}^n \frac{i}{2^i} = 2 - \frac{n+2}{2^n} $$
Let's skip the check, since when n = 1, I have $\frac{1}{2} = \frac{1}{2}$
What i will next do ? What for expression i may receive ?
$$\sum_{i=0}^{n+1}\frac{i}{2^i} = 2 - \frac{n+2}{2^n} $$
My trying :
$$ \biggl(\sum_{i=0}^{n+1}\frac{i}{2^i} = 2 - \frac{n+2}{2^n}\biggl) = \frac{n+1}{2^{n+1}} $$
or
$$ \biggl(\sum_{i=0}^{n+1}\frac{i}{2^i} = 2 - \frac{n+2}{2^n}\biggl) = \frac{2^{n+2}-(n+1)+2}{2^{n+1}} $$
Which answer should I use? And how will I decide next?
| You should prove that$$\sum_{i=0}^n\frac i{2^i}=2-\frac{n+2}{2^n}\implies\sum_{i=0}^{n+1}\frac i{2^i}=2-\frac{n+3}{2^{n+1}}.$$In order to do that, you do:\begin{align}\sum_{i=0}^{n+1}\frac i{2^i}&=\left(\sum_{i=0}^n\frac i{2^i}\right)+\frac{n+1}{2^{n+1}}\\&=2-\frac{n+2}{2^n}+\frac{n+1}{2^{n+1}}\\&=2-\frac{2n+4}{2^{n+1}}+\frac{n+1}{2^{n+1}}\\&=2-\frac{n+3}{2^{n+1}}.\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $s_n$ denote the sum of the first $n$ primes. Prove that for each $n$ there exists an integer whose square lies between $s_n$ and $s_{n+1}$. Let $s_n$ denote the sum of the first $n$ primes. Prove that for each $n$ there exists an integer whose square lies between $s_n$ and $s_{n+1}$.
I cannot give a proof to this, although I have try on some small examples.
I also notice that $\pi(x)\sim x/\log x$ and there are approximately $\sqrt x$ perfect square smaller than $x$. I have a feel that because $\sqrt x=\mathcal O(x/\log x)$, there will be more perfect squares, two or three, between $s_n$ and $s_{n+1}$ when $n$ gets large.
Any suggestion?
| After some discussion with my professor, we have solved this problem by using some elementary arguments.
We first prove the following lemma.
Lemma : For $n\geq 4$ we have $$s_n<\bigg(\frac{p_{n+1}-1}{2}\bigg)^2.$$
To show this we first see $s_4=2+3+5+7=17$ and $(p_{5}-1)/2=5$, so obviously $17< 5^2$. Now if there is $m\geq 4$ such that $s_m<[(p_{m+1}-1)/2]^2$, then $$s_{m+1}=s_m+p_{m+1}<\bigg(\frac{p_{m+1}-1}2\bigg)^2+p_{m+1}=\bigg(\frac{p_{m+1}+1}2\bigg)^2\leq \bigg(\frac{p_{m+2}-1}{2}\bigg)^2$$
since for primes $\geq 11$, every pair of consecutive primes has least gap equals to $2$, the lemma's proof is completed now.
Now choose $n\geq 4$, let $a$ be the largest integer satisfy $a^2\leq s_n$, then by definition we have $(a+1)^2> s_n$. Now we have $$a^2\leq s_n< \bigg(\frac{p_{n+1}-1}2\bigg)^2\implies a< \frac{p_{n+1}-1}{2}$$
therefore $2a+1< p_{n+1}$. Combining these we have $$(a+1)^2=a^2+2a+1< s_n+p_{n+1}=s_{n+1}$$
So we have proved that $s_n< (a+1)^2< s_{n+1}$.
We have $s_1=2, s_2=5, s_3=10, s_4=17$, so $$s_1< 2^2< s_2< 3^2< s_3< 4^2< s_4$$
completely proved this statement.
| {
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Use generating functions to solve the recurrence relation Use generating functions to solve $a_n = 3a_{n-1} - 2a_{n-2} + 2^n + (n+1)3^n$.
What I have so far, not sure if I forgot to do something or am missing out on something obvious:
Define $$G(x) = \sum_{n=0}^{\infty} a_nx^n$$
Then, $G(x) = a_0 + a_1x + a_2x^2 + \sum_{n=3}^{\infty} a_nx^n$
= $a_0 + a_1x + a_2x^2 + \sum_{n=3}^{\infty} (3a_{n-1} - 2a_{n-2} + 2^n + (n+1)3^n)x^n$
= $a_0 + a_1x + a_2x^2 + \sum_{n=3}^{\infty} 3a_{n-1}x^n - \sum_{n=3}^{\infty} 2a_{n-2}x^n + \sum_{n=3}^{\infty} 2^nx^n + \sum_{n=3}^{\infty} (n+1)3^nx^n$
= $a_0 + a_1x + a_2x^2 + 3\sum_{n=3}^{\infty} a_{n-1}x^n - 2\sum_{n=3}^{\infty} a_{n-2}x^n + \sum_{n=3}^{\infty} (2x)^n + \sum_{n=3}^{\infty} (n+1)(3x)^n$
=$a_0 + a_1x + a_2x^2 + 3x\sum_{n=3}^{\infty} a_{n-1}x^{n-1} - 2x^2\sum_{n=3}^{\infty} a_{n-2}x^{n-2} + \sum_{n=3}^{\infty} (2x)^n +
\sum_{n=3}^{\infty} (n+1)(3x)^n$
= $a_0 + a_1x + a_2x^2 + 3x\sum_{n=2}^{\infty} a_nx^n - 2x^2\sum_{n=1}^{\infty} a_nx^n + \sum_{n=3}^{\infty} (2x)^n +
\sum_{n=3}^{\infty} (n+1)(3x)^n$
| Hint: if $G(x)$ is the generating function for $a_n$, what are $x G(x)$ and $x^2 G(x)$ the generating functions for?
| {
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Evaluating $\sum_{n=1}^\infty\frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n}$
Question: How can we evaluate $$\sum_{n=1}^\infty\frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n},$$where $H_n=\frac11+\frac12+\cdots+\frac1n$?
Quick Results
This series converges because $$\frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n}=O\left(\frac{\ln^2n}{n^{3/2}}\right).$$
My Attempt
Recall the integral representation of harmonic number $$H_n=\int_0^1\frac{1-x^n}{1-x}d x$$
we have $$
S=\sum_{n=1}^\infty\frac1n\frac{\binom{2n}n}{4^n}\iint_{[0,1]^2}\frac{(1-x^n)(1-y^n)}{(1-x)(1-y)}d xd y\\
=\tiny\iint_{[0,1]^2}\frac{x y \log (4)-2 x y \log \left(\sqrt{1-x}+1\right)-2 x y \log \left(\sqrt{1-y}+1\right)+2 x y \log \left(\frac{1}{2} \left(\sqrt{1-x y}+1\right)\right)}{\left(\sqrt{1-x y}-1\right) \left(\sqrt{1-x y}+1\right)}dxdy\\
$$
This integral is too hard for me and Mathematica to compute. Numerical integration returns $12.6178$, it agrees with the numerical summation of the original series. I tried to integrate with respect to $x$, but failed.
| First, we prove a lemma on the integral representation of $(H_n)^2$.
$$I_n=\int_0^1\left(nx^{n-1}\ln^2(1-x)-\frac{x^n\ln x}{1-x}\right)d x-\zeta(2)=(H_n)^2$$
Let's prove by induction. $\displaystyle I_0=-\int_0^1\frac{\ln x}{1-x}dx=\zeta(2)=\zeta(2)+(H_0)^2$.\
Assume the equation holds for $n-1$,
$$\begin{aligned}
I_n&=\int_0^1\left(2(x^n-1)\frac{\ln(1-x)}{1-x}-\frac{x^n\ln x}{1-x}\right)d x-\zeta(2)\\
&=I_{n-1}+\int_0^1\left(2(x^n-x^{n-1})\frac{\ln(1-x)}{1-x}-\frac{(x^n-x^{n-1})\ln x}{1-x}\right)d x\\
&=(H_{n-1})^2+\int_0^1\left(-2x^{n-1}\ln(1-x)+x^{n-1}\ln x\right)d x\\
&=\left(H_n-\frac1n\right)^2-\frac1{n^2}+2\cdot\frac{H_n}n=(H_n)^2
\end{aligned}$$
Result Therefore, and by integrating $\displaystyle\sum_{n=1}^\infty\frac{\binom{2n}n}{4^n}x^n=\frac{1}{\sqrt{1-x}}-1$ from $0$ with respect to $x$, we have
$$\begin{aligned}
S&=\sum_{n=1}^\infty\frac1n\frac{\binom{2n}n}{4^n}\left(\int_0^1\left(nx^{n-1}\ln^2(1-x)-\frac{x^n\ln x}{1-x}\right)d x-\zeta(2)\right)\\
&=\int_0^1\left(\frac{1}{x\sqrt{1-x}}-\frac1x\right)\ln^2(1-x)d x-\int_0^12\ln\frac{2}{1+\sqrt{1-x}}\frac{\ln x}{1-x}d x-2\ln2\zeta(2)\\
&=I_1-I_2-2\ln2\zeta(2)
\end{aligned}$$
$I_1=12\zeta(3)$ can be easily deduced by substitution $x\mapsto 1-x^2$. $-2\ln2\zeta(2)+\frac32\zeta(3)$, the value of $I_2$, can also be deduced by the same substitution.
By combining these results, $S=\frac{21}2\zeta(3)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Definite Integral of $\int_0^1\frac{dx}{\sqrt {x(1-x)}}$ We have to calculate value of the following integral :
$$\int_0^1\cfrac{dx}{\sqrt {x(1-x)}} \qquad \qquad (2)$$
What i've done for (2) :
\begin{align}
& = \int_0^1\cfrac{dx}{\sqrt {x(1-x)}} \\
& = \int_0^1\cfrac{dx}{\sqrt {x-x^2}} \\
& = \int_0^1\cfrac{dx}{\sqrt {(x^2-x+\frac 14)-\frac 14 }} \\
& = \int_0^1\cfrac{dx}{\sqrt {(x-\frac 12)^2-(\frac 12)^2 }} \\
& = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {(\frac 12\sec \theta)^2-(\frac 12)^2 }} I\ used\ trigonometric\ substitution \ u=a\sec \theta, by \ it's \ form \ u^2-a^2 \\
& = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {(\frac 14\sec^2 \theta)-\frac 14 }} \\
& = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {\frac 14(\sec ^2\theta-1)}} \ using \\tan^2\theta=\sec^2\theta-1 \\
& = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {\frac 12(\sqrt{\tan^2\theta) }}} \\
& = \int_0^1\sec\theta d\theta = \sec\theta \tan \theta |_0^1 \\
\end{align}
But i got problems calculating $\theta$ value, using trigonometric substitution, any help?
| You must convert the limits to be in terms of $\theta$. When $x=0,1$, what is $\theta$?
Also if you want another method consider partial fractions
| {
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Solution to $\sqrt{\sqrt{x + 5} + 5} = x$ There are natural numbers $a$, $b$, and $c$ such that the solution to the equation
\begin{equation*}
\sqrt{\sqrt{x + 5} + 5} = x
\end{equation*}
is $\displaystyle{\frac{a + \sqrt{b}}{c}}$. Evaluate $a + b + c$.
I am not sure where I saw this problem. My guess is that it was from a high school math competition. The solution to the equation is $\frac{1 + \sqrt{21}}{2}$. This suggests use of the quadratic formula.
The solution set to the given equation is a subset of the solution set to
\begin{equation*}
x^{2} - 5 = \sqrt{x + 5} ,
\end{equation*}
\begin{equation*}
x^{4} - 10x^{2} + 25 = x + 5
\end{equation*}
\begin{equation*}
x^{4} - 10x^{2} - x + 20 = 0 .
\end{equation*}
Using the quartic equation (or Wolfram), the solutions to this equation are computed to be
\begin{equation*}
\frac{1 \pm \sqrt{21}}{2} , \qquad \frac{-1 \pm \sqrt{17}}{2} .
\end{equation*}
| *
*You can get the four apparent solutions from $x^{4} - 10x^{2} - x + 20 = (x^2 - x - 5) (x^2 + x - 4)$ and then solving two quadratic equations
*Squaring can produce spurious results:
*
*for example if $x=\frac{-1 - \sqrt{17}}{2}$ then $\sqrt{\sqrt{x + 5} + 5} = -x$ rather than $+x$
*but you lost that distinction when you went to ${\sqrt{x + 5} + 5} = x^2$
*similarly $x=\frac{-1 + \sqrt{17}}{2}$ or $x=\frac{1 - \sqrt{21}}{2}$ then $\sqrt{x+5} = -(x^2-5)$ rather than $+(x^2-5)$ but you lost that distinction with the second squaring
When you use transformations which are not $1-1$ then you should check any apparent answers in the original question to see whether they are spurious or not
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3191768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How does one prove such an equation? The problem occurred to me while I was trying to solve a problem in planimetry using analytic geometry.
for $b$ between $-\frac{1}2$ and $1$ :
$\sqrt{2+\sqrt{3-3b^2}+b} = \sqrt{2-2b}+ \sqrt{2-\sqrt{3-3b^2}+b}$
| Hint:
Square
$$\sqrt{2+\sqrt{3-3b^2}+b}- \sqrt{2-\sqrt{3-3b^2}+b} = \sqrt{2-2b}.$$
This gives
$$2+\sqrt{3-3b^2}+b+2-\sqrt{3-3b^2}+b-2\sqrt{(2+\sqrt{3-3b^2}+b)(2-\sqrt{3-3b^2}+b)}= 2-2b,$$
$$4+2b-2\sqrt{(2+b)^2-(3-3b^2)}= 2-2b,$$
$$2+4b-2\sqrt{4b^2+4b+1}= 0.$$
Remains to discuss the domain and the signs.
| {
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"url": "https://math.stackexchange.com/questions/3193543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Complex number cannot arive at $\frac{9}{2}-\frac{9}{2}i$ with problem $\frac{4+i}{i}+\frac{3-4i}{1-i}$ I am asked to evaluate: $\frac{4+i}{i}+\frac{3-4i}{1-i}$
The provided solution is: $\frac{9}{2}-\frac{9}{2}i$
I arrived at a divide by zero error which must be incorrect. My working:
$\frac{4+i}{i}$, complex conjugate is $-i$ so:
$\frac{-i(4+i)}{-i*i}$
= $\frac{-4i+i^2}{i^2}$
= $\frac{-4i--1}{-1}$
= $-4i+1$
Then the next part:
$\frac{3-4i}{1-i}$ complex conjugate is $1+i$ so:
$\frac{(1+i)(3-4i)}{(1+i)(1-i)}$
= $\frac{3-4i+3i-4i^2}{1-i^2}$
= $\frac{7-i}{0}$ # 1 + -1 = 0
How can I arrive at $\frac{9}{2}-\frac{9}{2}i$?
| Hint: It is $$\frac{4+i}{i}+\frac{3-4i}{1-i}=\frac{(4+i)(1-i)+(3-4i)i}{i(1-i)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3193823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to integrate problem I've tried u substitution and partial fraction doesn't seem to work either.
$\int \frac{1}{x^2+3}$
Can I take $\int \frac{1}{x^2+\sqrt(3)^2}$ in order to use $x=a\tan\theta$
| We are asked to evaluate $ \int \frac{1}{x^2+3}dx$. We need to know $\int \frac{1}{1+x^2}dx= \arctan(x)+C$ to answer this question.
We need to modify the denominator $x^2+3$. Note $x^2+3 = 3(\frac{x^2}{3}+1)$. So we have
$$ \int \frac{1}{x^2+3}dx = \int \frac{1}{3(\frac{x^2}{3}+1)}dx = \frac{1}{3} \int \frac{1}{\frac{x^2}{3}+1}dx $$
Now we want to make use of substitution rule. In doing so, we want $u^2 = \frac{x^2}{3}$, which means we can let $u = \frac{x}{\sqrt 3}$. After a bit of algebra you can conclude
$$ \int \frac{1}{x^2+3}dx = \frac{\sqrt 3}{3} \arctan( \frac{\sqrt 3 \ x}{3}) + C $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3196586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Clarification in proof of perpendicular bisectors meeting at a point Context:
My crude drawing from Paint to illustrate triangle OAB:
My working:
$\begin{align*}
(z-\frac{1}{2}(x+y)) \cdot (y-x) &= z\cdot(y-x) + \frac{1}{2}(-x-y) \cdot (y-x)\\
&= z\cdot y -z\cdot x + \frac{1}{2}(\lVert x\rVert^{2} - \lVert y\rVert^{2})\\
&= (\frac{1}{2}y+b\rho(y))\cdot y - (\frac{1}{2}x+c\rho(x))\cdot x+ \frac{1}{2}(\lVert x\rVert^{2} - \lVert y\rVert^{2})\\
&= \frac{1}{2}y\cdot y - \frac{1}{2}x\cdot x+ \frac{1}{2}(\lVert x\rVert^{2} - \lVert y\rVert^{2}) \\
&= \frac{1}{2}\lVert y\rVert^{2} - \frac{1}{2}\lVert y\rVert^{2} - \frac{1}{2}\lVert x\rVert^{2} + \frac{1}{2}\lVert x\rVert^{2}\\
&= 0
\end{align*}$
However, I can't see how this helps me "show that $z$ lies on the perpendicular bisector of $\vec{AB}$ " as we have only shown $(z-\frac{1}{2}(x+y))$ (whatever this point is) is perpendicular to $(y-x)=\vec{AB}$.
Cheers!
| Let $K$ be the point of intersection of the perpendicular bisectors of the sides $OA$ and $OB.$ Let $\overrightarrow {OK} = z.$ Let $M$ be the midpoint of the side $AB.$ Then observe that if we can show that $\overrightarrow {MK} \perp \overrightarrow {AB}$ we are through. So we need only to show that $$\overrightarrow {MK} \cdot \overrightarrow {AB} = 0.$$ Now what is $\overrightarrow {MK}$?
Observe that $$\overrightarrow {MK} = \overrightarrow {OK} - \overrightarrow {OM}.$$ Now observe that $$\overrightarrow {AM} = \frac {1} {2} (y-x)\ \ \text {and}\ \overrightarrow {OM} = \overrightarrow {OA} + \overrightarrow {AM} = x+ \frac {1} {2} (y-x) = \frac {1} {2} (x+y).$$ Also $\overrightarrow {OK} = z.$ So we have $$\overrightarrow {MK} = z - \frac {1} {2} (x+y).$$ Again $\overrightarrow {AB} = y-x.$ So we need only to show that $$\left (z - \frac {1} {2} (x+y) \right ) \cdot (y-x) = 0$$ as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3196864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Given three non-negatve numbers $a,b,c$. Prove that $1+a^{2}+b^{2}+c^{2}+4abc\geqq a+b+c+ab+bc+ca$ .
Given three non-negatve numbers $a, b, c$. Prove that:
$$1+ a^{2}+ b^{2}+ c^{2}+ 4abc\geqq a+ b+ c+ ab+ bc+ ca$$
Let $t= a+ b+ c$, we have to prove
$$\left(\!\frac{1}{t^{3}}- \frac{1}{t^{2}}+ \frac{1}{t}\!\right)\sum a^{3}+ \left(\!\frac{3}{t^{3}}- \frac{3}{t^{2}}\!\right)\left (\!\sum a^{2}b+ \sum a^{2}c\!\right )+ \left(\!\frac{6}{t^{3}}- \frac{6}{t^{2}}- \frac{3}{t}+ 4\!\right)abc\geqq 0$$
If $0< t< 1$ so
$${\rm LHS}\geqq \left(\frac{3}{t}+ 1\right)\left(\frac{3}{t}- 2\right)^{2}abc\geqq 0$$
If $1< t$ so
$${\rm LHS}= \left(\!\frac{3}{t^{2}}- \frac{3}{t^{3}}\!\right)(\!{\rm Schur.3}\!)+ \frac{1}{t}\left(\!\frac{2}{t}- 1\!\right)^{2}(\!{\rm a.m.}- {\rm g.m.}\!)+ \left(\!\frac{3}{t}+ 1\!\right)+ \left(\!\frac{3}{t}- 2\!\right)^{2}abc\geqq 0$$
(Can you find the way without deviding two cases as above?)
| It can be characterized as a sum of non-negative polynomials. This decompostion writes it as follow
$$\!1\!+\!a^{2}\!+\!b^{2}\!+\!c^{2}\!+\!4 abc\!-\!a\!-\!b\!-\!c\!-\!ab\!-\!bc\!-\!ca\!=\\=\!abc\left\{\!\left(\!1\!+\!\dfrac{3}{a\!+\!b\!+\!c}\!\right)\left(\!2\!-\!\dfrac{3}{a\!+\!b\!+\!c}\!\right)^{2}\!+\!\dfrac{3\left(\!\sum a^{2}\!-\!\sum ab\!\right)}{(\!a+ b+ c\!)^{3}}\!\right\}\!$$
$$+\sum\limits_{cyc}\left\{\!\left(\!\frac{3}{2}+ 3\times \frac{a}{a+ b+ c}\!\right)\left(\!\frac{1}{2}- \frac{a}{a+ b+ c}\!\right)^{2}
\left(\!b- c\!\right)^{2}\right\}$$
$$+\sum\limits_{cyc}\left\{\left(\!\frac{1}{2}+ \frac{3a}{a+ b+ c}\!\right)\left(\!\frac{1}{2}- \frac{1}{a+ b+ c}\!\right)^{2}\!(\!b- c\!)^{2}\right\}\geqq 0$$
q.e.d / You can also see here $\lceil$ https://h-a-i-d-a-n-g-e-l.hatenablog.com/entry/2019/05/12/145452 $\rfloor$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3198229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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How to compute a Jacobian using polar coordinates? Consider the transformation $F$ of $\mathbb R^2\setminus\{(0,0)\}$ onto itself defined as
$$
F(x, y):=\left( \frac{x}{x^2+y^2}, \frac{y}{x^2+y^2}\right).$$
Its Jacobian matrix is
$$\tag{1}
\begin{bmatrix} \frac{y^2-x^2}{(x^2+y^2)^2} & -\frac{2xy}{(x^2+y^2)^2} \\ -\frac{2xy}{(x^2+y^2)^2} & \frac{x^2-y^2}{(x^2+y^2)^2} \end{bmatrix},\quad \text{and its determinant equals}\ \frac{-1}{(x^2+y^2)^2}.$$
The following alternative computation is wrong at (!) and (!!), and I cannot see why.
Let $\phi\colon (0, \infty)\times (-\pi, \pi)\to \mathbb R^2$ be the map $$\phi(r, \theta) =(r\cos \theta, r\sin \theta).$$ Let moreover $$\tag{2}\tilde{F}:=\phi^{-1}\circ F\circ \phi;$$ then, by an easy direct computation, $$\tilde{F}(r, \theta)=\left( \frac1r, \theta\right).$$The Jacobian matrix of $\tilde{F}$ is, thus, $$\tag{!}\begin{bmatrix} \frac{-1}{r^2} & 0 \\ 0 & 1\end{bmatrix} , \quad \text{and its determinant equals }\ \frac{-1}{r^2}.$$On the other hand, by (2) and by the chain rule, the Jacobian determinants of $F$ and $\tilde{F}$ are equal. We conclude that the Jacobian determinant of $F$ is $$\tag{!!} \frac{-1}{r^2}=\frac{-1}{x^2+y^2}.$$
The result (!!) is off by a factor of $r^{-2}$ from the correct one, which is given in (1). Equation (!) must also be wrong. Indeed, computing the Jacobian matrix from (2) using the chain rule, and using that
$$
D\phi = \begin{bmatrix} \cos \theta & \sin \theta \\
-r\sin \theta & r\cos \theta\end{bmatrix}$$
and that
$$\tag{!!!}
D(\phi^{-1})= \begin{bmatrix} \frac{x}{\sqrt{x^2+y^2}} & \frac{y}{\sqrt{x^2+y^2}} \\ -\frac{y}{x^2+y^2} & \frac{x}{x^2+y^2}\end{bmatrix},$$
I obtain the result
$$
\begin{bmatrix} \frac{x}{\sqrt{x^2+y^2}} & \frac{y}{\sqrt{x^2+y^2}} \\ -\frac{y}{x^2+y^2} & \frac{x}{x^2+y^2}\end{bmatrix} \begin{bmatrix} \frac{y^2-x^2}{(x^2+y^2)^2} & -\frac{2xy}{(x^2+y^2)^2} \\ -\frac{2xy}{(x^2+y^2)^2} & \frac{x^2-y^2}{(x^2+y^2)^2} \end{bmatrix}\begin{bmatrix} \cos\theta & -r\sin\theta \\ \sin \theta & r\cos \theta\end{bmatrix} = \begin{bmatrix} -\frac1{r^2} & 0 \\ 0 & \frac{1}{r^2}\end{bmatrix},$$
which is different from the matrix in (!), and which gives the correct determinant of $-1/r^4$, as it should be.
Can you help me spot the mistake?
SOLUTION (added at a later time). As answerers pointed out, there is a mistake in (!!!). The correct matrix to be used is
$$ D(\phi^{-1})|_{F\circ \phi(r, \theta)}
= \begin{bmatrix} \frac{\frac{x}{x^2 + y^2}}{\sqrt{\left(\frac{x}{x^2 + y^2} \right)^2+\left( \frac{y}{x^2 + y^2}\right)^2}} & \frac{\frac{y}{x^2 + y^2}}{\sqrt{\left(\frac{x}{x^2 + y^2} \right)^2+\left( \frac{y}{x^2 + y^2}\right)^2}} \\ -\frac{\frac{y}{x^2 + y^2}}{\left(\frac{x}{x^2 + y^2} \right)^2+\left( \frac{y}{x^2 + y^2}\right)^2} & \frac{\frac{x}{x^2 + y^2}}{\left(\frac{x}{x^2 + y^2} \right)^2+\left( \frac{y}{x^2 + y^2}\right)^2}\end{bmatrix} = \begin{bmatrix}\cos\theta & \sin \theta \\ - r\sin \theta & r\cos\theta \end{bmatrix}.$$
Had I used this matrix, I would have found the correct result for the Jacobian matrix of $\tilde{F}$, which is the equation marked (!). Thus, (!) is actually correct.
My fundamental misunderstanding was the assumption that, because of (2), the Jacobian determinant should be invariant for coordinate changes. This is not true; what follows from (2) is only that
$$
\det D\tilde{F}|_{(r, \theta)}= \det D\phi^{-1}|_{F\circ\phi(r, \theta)}\det D\phi|_{(r, \theta)} \det DF|_{\phi(r, \theta)}.
$$
The first two factors in the right-hand side need not cancel, as I erroneously thought.
| The Jacobians of the two functions aren't equal by the chain rule.
In actual fact, $D(\phi(\frac{1}{r}, \cos\theta)) × D\tilde{F}(r, \theta)= DF \times D(\phi(r, \theta))$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3198750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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