Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Evaluate $\lim\limits_{x\to \infty}x^2 \left( a^{1 \over x} - a^{1 \over x+1} \right)$ I am trying to evaluate
$$\lim\limits_{x\to \infty}x^2 \left( a^{1 \over x} - a^{1 \over x+1} \right)$$
for $a>0$.
The idea:
$$a^{1 \over x} = e^{\ln{a^{1 \over x}}}=e^{{1 \over x}\ln{a}} = 1 + {\ln{a} \over x} + {1 \over 2}{(\ln{a})^2 \over x^2}+{1 \over 6}{(\ln{a})^3 \over x^3}+\dots$$
$$a^{1 \over x}-a^{1 \over x+1} = e^{\ln{a^{1 \over x}}}-e^{\ln{a^{1 \over x+1}}}=e^{{1 \over x}\ln{a}}-e^{{1 \over x+1}\ln{a}} = \ln{a} \left({1\over x} - {1 \over x+1} \right)+ {1 \over 2}(\ln{a})^2\left({ {1\over x^2} - {1 \over (x+1)^2}}\right)+{1 \over 6}(\ln{a})^3\left({ {1\over x^3} - {1 \over (x+1)^3}}\right)+\dots$$
$$x^2 \left(a^{1 \over x}-a^{1 \over x+1}\right) =\ln{a} \left(x - {x^2 \over x+1} \right)+ {1 \over 2}(\ln{a})^2\left({ 1 - {x^2 \over (x+1)^2}}\right)+{1 \over 6}(\ln{a})^3\left({ {1\over x} - {x^2 \over (x+1)^3}}\right)+\dots$$
This is where I get stuck. I suppose the whole sum should converge to $\ln{a}$ but can't quite find the way to get there. I would appreciate any comments.
| I got it with a different approach. Take the logarithm of the limit,
$\ln(x^2) + \ln(a^{1/x} - a^{1/(1+x)}) = \ln(x^2) + \frac{\ln(a)}{x} + \ln(1 - a^{1/(1+x)-1/x})$
$\ln(x^2) + \frac{\ln(a)}{x} + \ln(1-e^{\frac{-\ln(a)}{x(1+x)}}) $
Since $\frac{-\ln(a)}{x(1+x)} \rightarrow 0$ as $x \rightarrow \infty$ we take a Taylor expansion of $e^{\frac{-\ln(a)}{x(1+x)}}$.
$\ln(x^2) + \frac{\ln(a)}{x} + \ln(1-(1-\frac{\ln(a)}{x(1+x)} + \mathcal{O}(\frac{1}{x^4})) $
$\ln(x^2) + \frac{\ln(a)}{x} + \ln(\frac{\ln(a)}{x(1+x)}) + \mathcal{O}(\frac{1}{x^2}) $
$\ln(\frac{x^2}{x(x+1)}) + \frac{\ln(a)}{x} + \ln(\ln(a)) + \mathcal{O}(\frac{1}{x^2}) \rightarrow \ln(\ln(a))$ as $x \rightarrow \infty$
Thus the limit itself is $\ln(a)$.
| {
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An amusing factorisation of a trivariate polynomial I noticed the following amusing pattern when playing around in Wolfram Alpha:
$a(b-c) + b(c-a) + c(a-b) = 0$
$a^2(b-c) + b^2(c-a) + c^2(a-b) = -(a-b)(b-c)(c-a)$
$a^3(b-c) + b^3(c-a) + c^3(a-b) = -(a-b)(b-c)(c-a)(a+b+c)$
$a^4(b-c) + b^4(c-a) + c^4(a-b) = -(a-b)(b-c)(c-a)(a^2+b^2+c^2 + ab+ bc +ca)$
.
.
.
$a^n(b-c) + b^n(c-a) + c^n(a-b) = -(a-b)(b-c)(c-a)\left(\displaystyle \sum_{i+j+k=n-2} a^ic^jb^k\right)$
I can prove that $(a-b)(b-c)(c-a)$ is a factor. How do we get the last factor? Is it always irreducible?
| Image of Solution
For irreducibility plz see https://mathoverflow.net/questions/98043/is-complete-homogeneous-symmetric-polynomials-an-irreducibile-element-in-polyno
| {
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Finding extreme values of $f(x,y,z)=x^2+2y^2+3z^2$ on unit sphere $x^2+y^2+z^2=1$ I defined $G(x)=x^2+y^2+z^2-1$, such that the gradient of G $\nabla G=(2x, 2y, 2z)$, so that $\nabla f=(2x, 4y, 6z)=\lambda (2x,2y,2z)$.
The conclusion I drew was that the only possible value was either $\lambda=0$, or y and z were both 0 but x can be anything. This seems incorrect to me, but I'm not sure how to proceed.
| Without calculus: $f(x,y,z)=x^2+2y^2+3z^2 = 2(x^2+y^2+z^2) + z^2-x^2= z^2-x^2+2\,$.
Since $0 \le x^2,z^2 \le 1$ it follows that $-1\le z^2-x^2 \le 1$. The extrema are actually attained at $f(\pm 1, 0, 0) = -1+2 = 1$ and $f(0,0,\pm 1)=1+2=3\,$.
| {
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At what point does $ y = 3ln(x)$ have maximum curvature? The formula for the curvature of a $y = f(x)$ function is given by:
$k(x) = \frac{|y'(x)|}{[1+(y''(x))^2]^\frac{3}{2}}$
Given that $y = 3ln(x),$ we know:
$y'(x) = 3/x$
$y''(x) = -3/x^2$
Plugging these in:
$k(x) = \frac{3/x}{[1+\frac{9}{x^4}]^\frac{3}{2}}$
$k(x) = \frac{3}{x[1+\frac{9}{x^4}]^\frac{3}{2}}$
$k(x) = \frac{3}{x[\frac{x^4+9}{x^4}]^\frac{3}{2}}$
$k(x) = \frac{3}{x[\frac{(x^4+9)^\frac{3}{2}}{x^6}]}$
$k(x) = \frac{3}{\frac{x}{x^6}{(x^4+9)^\frac{3}{2}}}$
$k(x) = \frac{3}{\frac{1}{x^5}{(x^4+9)^\frac{3}{2}}}$
$k(x) = \frac{3x^5}{{(x^4+9)^\frac{3}{2}}}$
From here, I need the critical points, which I get by taking the derivative of k(x):
$k'(x) = \frac{15x^4(x^4+9)^\frac{3}{2} - (3x^5)(3/2)(x^4+9)^\frac{1}{2}(4x^3)}{(x^4+9)^3}$
$k'(x) = \frac{15x^4(x^4+9)^\frac{3}{2} - 18x^8(x^4+9)^\frac{1}{2}}{(x^4+9)^3}$
$k'(x) = \frac{3x^4(x^4+9)^\frac{1}{2}[5(x^4+9)^3-6x^4]}{(x^4+9)^3}$
Beyond this, I have no idea what to do. I can solve it graphically, but how can I do it algebraically?
| The correct form of the (unsigned) curvature of a function $y(x)$ is $$\kappa(x) = \frac{|y''(x)|}{(1+(y'(x))^2)^{3/2}}.$$ For $y(x) = 3 \log x$, we then easily find $$y'(x) = \frac{3}{x}, \quad y''(x) = -\frac{3}{x^2},$$ hence $$\kappa(x) = \frac{3x}{(9+x^2)^{3/2}}, \quad x > 0.$$ It follows that $$\kappa'(x) = \frac{3(2x^2-9)}{(9+x^2)^{3/2}}$$ which has a critical point $x = 3/\sqrt{2}$. It remains an exercise to show that this value corresponds to a local maximum of $\kappa$ and therefore the $x$-value of the point on $y = 3 \log x$ that has maximum curvature.
As an additional exercise, what is the location of the center of the osculating circle that corresponds to this maximal curvature?
As a second additional exercise, what is the locus of the centers of the osculating circles for all points on $y = 3 \log x$? Choose a suitable parametrization.
| {
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Find the sum to n term of the series. $1+3x+5x^2+7x^3................, X\ne1$ Here, $a=1, d=2, b=1, r=x$
\begin{align}
S_n&= \frac{ab}{1-r}+\frac{bdr(1-r{^n}^{-1})}{(1-r)^2}-\frac{[a+(n-1)d]br^n}{1-r}\\
S_n&=\frac 1 {1-x}+\frac{ 2x(1-x{^n}^{-1})} {(1-x)^2}-\frac{[1+(n-1)(2)]x^n} {1- x}\\
&= \frac 1 {1-x}+\frac{2x}{(1-x)^2}-\frac {2x.x{^n}^{-1}}{(1-x)^2}-\frac{[1+2n-2]x^n}{(1-x)}\\
&= \frac 1 {1-x}+\frac{2x}{(1-x)^2}-\frac {2x^n}{(1-x)^2}-\frac{[2n-1]x^n}{(1-x)}\\
\end{align}
Is it correct. I have not got the answer, please show me how to move to this answer without skipping any line
$\frac {1-3x} {(1-x)^2}+\frac {2x^n}{(1-x)^2}-\frac{(2n-1)x^n}{(1-x)}$
| Here's how I did it:
let $S = 1 + 3x + 5x^2 + 7x^3 + ....$
By expansion, $S = \displaystyle\sum_{n=0}^{\infty}(2n+1)x^n .... (1)$
but similarly, $S = \displaystyle\sum_{n=1}^{\infty}(2n-1)x^{n-1} ....(2)$
Examining $(2)$, $(2n-1)x^{n-1} = \displaystyle\frac{1}{x}(2n-1)x^n$
$ = \displaystyle\frac{1}{x}(2n+1-2)x^n = \displaystyle\frac{1}{x}[(2n+1)x^n - 2x^n]$
So $(2)$ becomes
$S = \displaystyle\frac{1}{x}\sum_{n=1}^{\infty}[(2n+1)x^n - 2x^n]$
But using $(1)$, we get
$S = \displaystyle\frac{1}{x}(S-1)- \frac{2}{x}\sum_{n=1}^{\infty} x^n ....(3)$
(making term limits of $(1)$ as 1 and infinity and subtracting the first term.)
Now $\displaystyle\sum_{n=1}^{r} x^n = x + x^2 + x^3 + .... = x(1 + x + x^2 + ...)$ is the geometric series $\displaystyle\frac{x(1-x^r)}{1-x}$.
Hence $(3)$ becomes
$S = \displaystyle\frac{1}{x}(S-1)- \displaystyle\frac{2(1-x^r)}{1-x}$
which makes $S = \displaystyle\frac{x+1-2x^{r+1}}{(1-x)^2}$ for $x>1$
or $S = \displaystyle\frac{x+1}{(1-x)^2}$ if $0<x<1$.
| {
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Arc Length of an Ellipse using integration I was thinking about what the arc length of an ellipse is, but throughout my calculations I got stuck. Here is how I approached the problem:$$$$We have an ellipse in the form:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\Rightarrow y=\pm \frac{b}{a}\sqrt{a^2-x^2}$$By applying the formula of the arc length of a function, we get:$$L=4\int_0^a\sqrt{1+\frac{b^2x^2}{a^2(a^2-x^2)}}dx=4\int_0^a\sqrt{\frac{a^4+(b^2-a^2)x^2}{a^2(a^2-x^2)}}dx$$Now I made a little subsitution recalling trigonometry: $$x=a\sin(u)\\dx=a\cos(u)du$$So the Integral now can be expressed as:$$L=4\int_0^{\frac{\pi}{2}}a\cos(u)\sqrt{\frac{a^4+(b^2-a^2)a^2\sin^2(u)}{a^2(a^2-a^2\sin^2(u))}}du=\\4\int_0^{\frac{\pi}{2}}a\cos(u)\sqrt{\frac{a^4+(b^2-a^2)a^2\sin^2(u)}{a^2(a^2\cos^2(u)+a^2\sin^2(u)-a^2\sin^2(u))}}du=\\4\int_0^{\frac{\pi}{2}}\sqrt{a^2+(b^2-a^2)\sin^2(u)}du$$So we have:$$L=4a\int_0^{\frac{\pi}{2}}\sqrt{1+\frac{(b^2-a^2)}{a^2}\sin^2(u)}du$$
Letting $m=\frac{(b^2-a^2)}{a^2}$ we finally get:$$L=4a\int_0^{\frac{\pi}{2}}\sqrt{1+m\sin^2(u)}du$$
However, at this point I do not know any way on how to integrate this function because the $m$ is 'in the way'. Does anyone have any hints?
| This is an elliptical integral.
You can use the elliptical integral tables to look up a value for the integral- much like the standard normal distribution.
| {
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If $2^x=3^y=6^{-z}$ then find:$ \frac{1}{x}+\frac{1}{y}-\frac{1}{z}$ I am a grade 8 student and i am aware that $ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$ is proved by setting all values $= k$.
See the question number 24 in my assignment.
1:
| Actually you have because of misprint mistake in your question.
$2^x = 3^y = 6^{-z} = k $
$ 2 = k^{\frac{1}{x}}$
$ 3 = k^{\frac{1}{y}}$
$ 6 = k^{\frac{1}{-z}}$
As we know,
2.3 = 6
$k^{\frac{1}{x}} \cdot k^{\frac{1}{y}} = k^{\frac{1}{-z}}$
$k^{\frac{1}{x} + \frac{1}{y}} = k^{\frac{1}{-z}}$
On comparing powers we have,
$\frac{1}{x} + \frac{1}{y} = \frac{1}{-z}$
$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$
To prove $\frac{1}{x} + \frac{1}{y} - \frac{1}{z} = 0$
We should have expression $2^x = 3^y = 6^z$
| {
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Proof of the radical expression of $\cos\dfrac {2\pi}{17}$
Question: How would you prove the equation$$\small\cos\dfrac {2\pi}{17}=\dfrac {-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}-\sqrt{34-2\sqrt{17}}-2\sqrt{34+2\sqrt{17}}}}{16}\tag1$$
I'm not too sure how to prove it and I'm not sure where to begin. I started with $\exp(2\pi i)=\cos(2\pi)+i\sin(2\pi)=1\implies \cos(2k\pi)+i\sin(2k\pi)=\exp(2k\pi)$. But I'm not sure what to do next.
| The more natural item is $$ 2 \cos \left( \frac{2 \pi}{17} \right), $$ which is one root of
$$ x^8 + x^7 - 7 x^6 - 6 x^5 + 15 x^4 + 10 x^3 - 10 x^2 - 4 x + 1, $$
from page 18 of Reuschle (1875) using a method due to Gauss, having to do with cyclotomy, and predating Galois. If $\omega \neq 1$ but $\omega^{17} = 1,$ it is not hard to show that $\omega + \frac{1}{\omega}$ is a root of the given polynomial, using
$$ \small 1 + \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 + \omega^7 + \omega^8 + \omega^9 + \omega^{10} + \omega^{11} + \omega^{12} + \omega^{13} + \omega^{14} + \omega^{15} + \omega^{16} =0 $$
| {
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$3\nmid a,b \in \mathbb{Z} \implies a^2+b^2$ is not perfect square
Question: We want to show that if $3\nmid a,b \in \mathbb{Z} \implies a^2+b^2$ is not perfect square of an integer.
Answer: We have:
*
*$3\nmid a \iff a\neq3k,\ \forall k \in \mathbb{Z}$
*$3\nmid b \iff b\neq3l,\ \forall l \in \mathbb{Z}$
From this and the devision algorithm we have:
*
*$a=3q_1+r_1,$ with $1 \leq r_1 \leq2$ for some $\ q_1,r_1\in\mathbb{Z}$
*$b=3q_2+r_2,$ with $1 \leq r_2 \leq2$ for some $\ q_2,r_2\in\mathbb{Z}$
So, there are $4$ possible cases. I will do only the first, because I don't know if I work correctly: $$a=3q_1+1,\ b=3q_2+1$$
Lets assume that $a^2+b^2$ is perfect square of a random number $n$.
Then, $a^2+b^2=(3q_1+1)^2+(3q_2+1)^2=[3(q_1+q_2)]^2+6(q_1+q_2)+2=n^2$, and if $q_1+q_2=x$ we can say that:
$$(3x)^2+6x+2=n^2 \iff (3x+1)^2+1=n^2$$
My question is can we say now that we have contradiction? And why?
Thank you.
| You can continue. Let $3x + 1$ = $m$ so you have $m^2 + 1 = n^2$ or two consecutive integers are both squares. That seems wrong somehow. But we must prove it.
which we can
Notice there are no integers between $n$ and $n+1$. So there are no squares between $n^2$ and $n^2 + 2n + 1$. So if $n^2 < n^2 + 1 = m^2 \le n^2 + 2n + 1$ that is only possible if $n^2 + 1 = n^2 + 2n +1$ which is only possible if $n=0$. but we assumed $3\not \mid n$.
But it might be easier to do it directly.
$a = 3k + r_1; b = 3j + r_2;$ and $m = 3l + q$ and
$a^2 + b^2 = m^2 \implies 9(k^2 + j)^2 + 6(kr_1 + jr_2) + r_1^2 + r_2^2 = 9l^2 + 6l + q^2$ or
$r_1^2 + r_2^3 \equiv q^2 \mod 3$
or $\{1,2\}^2 + \{1,2\}^2 \equiv \{0,1,2\}^2 \mod 3$
so $\{1,4\} + \{1,4\} \equiv \{0,1,4\} \mod 3$
so $\{2,5,8\} \equiv \{0,1 , 4\} \mod 3$
so $2 \equiv \{0,1\} \mod 3$ which is impossible.
And that would be a lot less of a headache if you you used $-1, 0 , + 1$ rather than $0,1 ,2$.
The $a = 3k \pm 1$ and $b = 3j \pm 1$ and $m = 3l \pm 1$ or $ 3l$.
So $a^2 + b^2 = 9(k^2 + j^2) \pm 6(j+k) + 1 + 1 \equiv 2 \mod 3$ while
$m^2 = 9l^2 \pm 6l +\{0, 1\}\equiv 0, 1 \mod 3$. A contradiction.
| {
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A cubic nonlinear Euler sum Any idea how to solve the following Euler sum
$$\sum_{n=1}^\infty \left( \frac{H_n}{n+1}\right)^3 =
-\frac{33}{16}\zeta(6)+2\zeta(3)^2$$
I think It can be solved it using contour integration but I am interested in solutions using real methods.
| \begin{align}
S&=\sum_{n=1}^\infty\left(\frac{H_n}{n+1}\right)^3=\sum_{n=1}^\infty\left(\frac{H_{n-1}}{n}\right)^3\\
&=\sum_{n=1}^\infty\frac{H_n^3}{n^3}-3\sum_{n=1}^\infty\frac{H_n^2}{n^4}+3\sum_{n=1}^\infty
\frac{H_n}{n^5}-\sum_{n=1}^\infty\frac{1}{n^6}
\end{align}
Substituting the following results:
$$\sum_{n=1}^\infty\frac{H_n^3}{n^3}=\frac{93}{16}\zeta(6)-\frac52\zeta^2(3)$$
$$\sum_{n=1}^\infty \frac{H_n^2}{n^4}=\frac{97}{24}\zeta(6)-2\zeta^2(3)$$
$$\sum_{n=1}^\infty\frac{H_n}{n^5}=\frac74\zeta(6)-\frac12\zeta^2(3)$$
We get $$\boxed{S=2\zeta^2(3)-\frac{33}{16}\zeta(6)}$$
Note that the first and second sum are proved here and here respectively. As for the third sum, can be obtained using the Euler identity.
| {
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Triangle lengths inequality
$x$, $y$ and $z$ are lengths of the sides of a triangle. Prove the following and determine when both LHS and RHS are not an inequality but EQUAL.
$$\sqrt{x+y-z} + \sqrt{y+z-x} + \sqrt{z+x-y} \leqslant \sqrt{x}+\sqrt{y}+\sqrt{z}$$
| using the Ravi-substitution $$x=b+c,y=a+c,z=a+b$$ we get
$$\sqrt{2a}+\sqrt{2b}+\sqrt{2c}\le \sqrt {a+b}+\sqrt{a+c}+\sqrt{b+c}$$
but we have
$$\sqrt{2a}+\sqrt{2b}+\sqrt{2c}=\frac{\sqrt{2a}+\sqrt{2b}}{2}+\frac{\sqrt{2a}+\sqrt{2c}}{2}+\frac{\sqrt{2b}+\sqrt{2c}}{2}\le\sqrt{\frac{2a+2b}{2}}+\sqrt{\frac{2a+2c}{2}}+\sqrt{\frac{2c+2b}{2}}=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}$$
| {
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Verify $(y-x)y' = y-x+8$ has an explicit solution $y = x+ 4 \sqrt{x+2}$ Ive solved this problem and both sides of the function don't appear to be equal to me. However the solution in the back of the book says it is a solution. Am I not simplifying enough? Thanks for the help!
(y-x)y' = y-x+8 ; y = x+ 4 sqrt(x+2)
y'=( 2/sqrt(x+2)) +1
Then I plug in the values of y and y' into the function on the left.
(x+4*sqrt(x+2) -x)*(2/sqrt(x+2)) +1 = x+4*sqrt(x+2) -x +8
| Step 1:
$$y = x+ 4 \sqrt{x+2} \implies y' = \dfrac{2}{\sqrt{x+2}}+1$$
Step 2:
$$(y-x)y' = (x+ 4 \sqrt{x+2} - x) \left(\frac{2}{\sqrt{x+2}}+1\right) = 4 \left(\sqrt{x+2}+2\right)$$
Step 3:
$$ y-x+8 = x+ 4 \sqrt{x+2} - x + 8 = 4 \left(\sqrt{x+2}+2\right)$$
| {
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Finding the number of such 3 digits numbers Let xyz be a three digit number such that x,y,z form sides of an equilateral triangle. How many such numbers are there?
The answer is 9 using p n c
9*1*1 = 9
Now, if x,y,z form sides of an isoceles triangle. How many such numbers are there?
My try :
the numbers we can use are 1,2,3,4,5,6,7,8,9 and not 0.
9*1*9 + 9*9*1 + 1*9*9 - .......
I am not getting how to subtract the repeating numbers. Can someone guide?
| Let the equal side length be n and k be the length of the other side
For $n = 1, k = 1 \qquad (1)$
For $n = 2, k = 1,2,3 \qquad (3)$
For $n = 3, k = 1,2,3,4,5 \qquad (5)$
For $n = 4, k = 1,2,3,4,5,6,7 \qquad (7)$
For $n>4, k = 1,2,3,4,5,6,7,8,9 \quad (5\times 9=45)$
Total $ = 1+3+5+7+45 = 61$
Considering combination, total number of such numbers become $61 × (3!)/(2!) = 61×3 = 183$
Now eliminating the repeated numbers, we get $183 - (2×9) = 165$. Considering the Equilateral triangles are also isoceles triangles.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $1+x\ln(x+\sqrt{x^2+1})\geq \sqrt{1+x^2}\;\forall x \geq 0$ without derivative
Prove that $1+x\ln(x+\sqrt{x^2+1})\geq \sqrt{1+x^2}\;\forall x \geq 0$
$\bf{My\; Try::}$ Means we have to prove $x\ln(x+\sqrt{x^2+1})\geq \sqrt{1+x^2}-1$
Put $\sqrt{x^2+1}+x = t\geq 1\;,$ Then $\displaystyle (\sqrt{x^2+1}-x) = \frac{1}{t}.$
So we get $\displaystyle x = \frac{t^2-1}{2t}$ and $\displaystyle \sqrt{x^2+1} = \frac{t^2+1}{2t}$
So we have to prove $\displaystyle \left(\frac{t^2-1}{2t}\right)\ln (t)\geq \frac{t^2+1-2t}{2t}\Rightarrow (t^2-1)\ln t \geq (t-1)^2$
So we have to prove $\displaystyle \ln(t)\geq \frac{t-1}{t+1}\forall t \geq 1$
Help required for proving above statement, Thanks
| Applying the "well-known" inequality $\ln y \le y - 1$
to $y = \frac 1t $ gives for $t \ge 1$
$$
\ln t = - \log \frac 1t \ge -(\frac 1t-1) = \frac{t-1}{t} \ge \frac{t-1}{t+1} \, .
$$
| {
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Limit $\lim_{x\to 0} \frac{\tan ^3 x - \sin ^3 x}{x^5}$ without l'Hôpital's rule. I need to solve $$\lim_{x\to 0} \dfrac{\tan ^3 x - \sin ^3 x}{x^5}$$
I did like this:
$\lim \limits_{x\to 0} \dfrac{\tan ^3 x - \sin ^3 x}{x^5} = \lim \limits_{x\to 0} \dfrac{\tan ^3 x}{x^5} - \dfrac{\sin ^3 x}{x^5}$
$=\dfrac 1{x^2} - \dfrac 1{x^2} =0$
But it's wrong. Where I have gone wrong and how to do it?
| $$\begin{align}\lim_{x\to 0} \dfrac{\tan ^3 x - \sin ^3 x}{x^5}&=\lim_{x\to 0} \dfrac{\frac{\sin ^3 x}{\cos^3x} - \sin ^3 x}{x^5}\\
&=\lim_{x\to 0} \dfrac{\sin ^3 x(1-\cos^3x)}{x^5\cos ^3 x}\\
&=\lim_{x\to 0} \dfrac{\sin ^3 x\Big[(1-\cos x)(1+\cos x+\cos^2x)\Big]}{x^5\cos ^3 x}\\
&=\lim_{x\to 0} \left(\dfrac{\sin ^3 x\Big[(1-\cos x)(1+\cos x+\cos^2x)\Big]}{x^5\cos ^3 x}\cdot\frac{1+\cos x}{1+\cos x}\right), \quad\text{note that }(1-\cos x)(1+\cos x)=\sin^2x\\
&=\lim_{x\to 0} \dfrac{\sin ^5 x\quad(1+\cos x+\cos^2x)}{x^5\cos ^3 x\quad(1+\cos x)}\\
&=\lim_{x\to 0}\left[\left(\frac{\sin x}{x}\right)^5\cdot \frac{1}{\cos^3x}\cdot\frac{1+\cos x+\cos^2x}{1+\cos x}\right]\\
&=1^5\cdot\frac{1}{1^3}\cdot\frac{1+1+1^2}{1+1}=\frac{3}{2}.
\end{align}$$
| {
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If $a^b = b^a$ and $a=2b$ then find the value of $a^2+b^2$ If $a^b = b^a$ and $a=2b$ then find the value of $a^2+b^2$
My Attempt,
$$a^b = b^a$$
$$a^b =b^{2b}$$
$$a^b =b^b.b^b$$.
Now, what should I do further?
| We know that $a=2b $, so we have $$a^b = b^b \cdot b^b \Rightarrow (2b)^b = b^b \cdot b^b \Rightarrow 2^b = b^b $$. Assuming $b\in \mathbb Z $, we get, $b=2$ and thus $a=4$, so thus $\boxed {a^2+b^2=20} $. Hope it helps.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Divisibility by $8$ and $9$. How many ordered pairs $(a,b)$ exist such that the four-digit number, $a04b$, is divisible by both $8$ and $9$?
How should I approach this? (without modular arithmetic?)
| We can write:
$$a04b=1000\cdot a+100\cdot 0+10\cdot 4+b$$
Once $1000$ and $40$ are divisible by $8$ then $b$ must be divisible by $8$ and then $b=8$.
For division by $9$ we can write:
$$a048= 999\cdot a + a+40+8$$
Once $999$ is divisible by $9$ then $a+48$ has to be divisible by $9$. Once $0\le a\le 9$ the only value that fits is $a=6$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of nonnegative integral solutions of $3x+y+z \leq 25$
Find the number of nonnegative integral solutions of $$3x+y+z \leq 25$$
I can get the answer to $3x+y+z=25$ but I can't get the answer with inequality. Please help.
| The generating function approach is that this is the coefficient of $x^{25}$ in:
$$\frac{1}{(1-x)^3(1-x^3)}$$
Which can be rewritten as:
$$\frac{(1+x+x^2)^3}{(1-x^3)^4}=(x^6 + 3 x^5 + 6 x^4 + 7 x^3 + 6 x^2 + 3 x + 1)\sum_{j=0}^{\infty}\binom{j+3}{3}x^{3j}$$
So the coefficient of $x^{25}$ is:
$$3\binom{11}{3}+6\binom{10}{3}$$
More generally, the number of solutions to $3x+y+z\leq 3n-2$ is:
$$3\binom{n+2}{3}+6\binom{n+1}{3}=\frac{3n^2(n+1)}{2}$$
The number of solutions to $3x+y+z\leq 3n-1$ is:
$$6\binom{n+2}{3}+3\binom{n+1}{3}=\frac{3n(n+1)^2}{2}$$
The number of solutions to $3x+y+z\leq 3n$ is:
$$\binom{n+3}{3} + 7\binom{n+2}{3}+\binom{n+1}{3}=\frac{(n+1)(3n^2+6n+2)}{2}$$
| {
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Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega\neq 1$, find ... Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
$$\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} + \frac{\omega^4}{1 - \omega^3}$$
I've been having trouble with this unit, need help on solving this problem.
| You may notice that $\{\omega,\omega^2,\omega^3,\omega^4\}$ are the roots of $\frac{x^5-1}{x-1}$. If we set $Z=\{\omega,\omega^2,\omega^3,\omega^4\}$
we have
$$ \sum_{z\in Z}\frac{z}{1-z^2}=\sum_{z\in Z}\frac{z^3}{1-z}=\sum_{z\in Z}\frac{1}{1-z}-\sum_{z\in Z}(1+z+z^2)=-2+\sum_{z\in Z}\frac{1}{1-z}.$$
If $z\in Z$, $1-z$ is a root of $\frac{1-(1-x)^5}{x}=x^4-5x^3+10x^2-10x+5$.
By Vieta's theorem it follows that
$$ \sum_{z\in Z}\frac{1}{1-z} = \frac{10}{5} = 2$$
hence:
$$ \sum_{z\in Z}\frac{z}{1-z^2} = \color{red}{0}.$$
Key steps:
*
*$z\mapsto z^3$ is a bijection on $Z$
*for any $k\in[1,4]$ we have $\sum_{z\in Z}z^k = -1$.
| {
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Show that $\frac {1} {r-1} = \frac {1} {r+1} + \frac {2} {r^2+1} + \frac {4} {r^4+1} +\cdots$ Problem:
Show that
$$\frac {1} {r-1} = \frac {1} {r+1} + \frac {2} {r^2+1} + \frac {4} {r^4+1} +\cdots $$
for all $r > 1$, with a hint given that $\displaystyle\frac {1} {r-1} - \frac {1} {r+1} = \frac {2} {r^2-1}$.
Thoughts:
I am having difficulty seeing a connection from the hint to the problem. Any insights appreciated.
| Note that if $r\neq1$, $\frac{n}{r^n-1}=\frac{n}{r^n+1}+\frac{2n}{r^{2n}-1}$ for any positive integer $n$.
Now, we can see taht
$\frac{1}{r-1}=\frac{1}{r+1}+\frac{2}{r^2-1}=\frac{1}{r+1}+\frac{2}{r^2+1}+\frac{4}{r^4-1}=\cdots$
i.e. $\frac{1}{r-1}=\frac{2^{n+1}}{r^{2^{n+1}}-1}+\sum_{k=0}^n\frac{2^k}{r^{2^k}+1}$.
Finally, for $r>1$, $\frac{1}{r-1}=\lim_{n\rightarrow\infty}\frac{2^{n+1}}{r^{2^{n+1}}-1}+\sum_{k=0}^n\frac{2^k}{r^{2^k}+1}=\sum_{k=0}^{\infty}\frac{2^k}{r^{2^k}+1}$.
Good luck.
| {
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What is the value of $\frac {1}{2\cdot3} + \frac {1}{4\cdot 5} +\frac {1}{6\cdot 7} + \cdots $? What is value of $\dfrac {1}{2\cdot 3} + \dfrac {1}{4\cdot 5} +\dfrac {1}{6\cdot7} + \cdots =?$
1) $\log \left( \dfrac 2e \right)$
2) $\log \left( \dfrac e2 \right)$
3) $\log \left( 2e \right)$
4) $e-1$
I know it converges but I am not getting how to solve
| We use $$\frac{1}{n(n+1)}=\frac{n+1-n}{n(n+1)}=\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$
Note that your series, $$\dfrac {1}{2 \times 3} + \dfrac {1}{4 \times 5} +\dfrac {1}{6 \times 7} + \cdots=\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\dots$$
Since $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots=\ln 2$$
This result is known, and is discussed as the Alternating Harmonic Series on Wikipedia. So since $$\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\dots=1-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots \right)$$ the answer is $1-\ln 2$. So the answer to your question is $$1-\ln 2=\ln e-\ln 2=\ln \frac{e}{2}$$So the answer is $2)$
| {
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Prove the inequality $\frac{a^8+b^8+c^8}{a^3b^3c^3}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ for $a,b,c>0$ As in the title. Prove the inequality $$\frac{a^8+b^8+c^8}{a^3b^3c^3}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$ for $a,b,c>0$.
Thsi inequality can be proved in a pretty straightforward manner utilizing the Muirhead's inequality, yet I ought to prove it using the rearrangement inequality. I can't however figure out any suitable sequences and permutations of them.
| Multiply everywhere by $a^3b^3c^3$ to get that we want to prove
$$
a^8 + b^8 + c^8 \geq a^2b^3c^3 + a^3b^2c^3 + a^3b^3c^2
$$
Let's start with the right-hand side, and assume that $a\leq b\leq c$. In that case, $a^3b^3\leq a^3c^3\leq b^3c^3$ and $a^2\leq b^2\leq c^2$, so by the rearrangement inequality, we have
$$
a^2b^3c^3 + a^3b^2c^3 + a^3b^3c^2 \leq a^2a^3b^3 + b^2a^3c^3 + c^2c^3b^3\\
= a^5b^3 + a^3b^2c^3 + c^5b^3
$$
And again, by the rearrangement inequality (twice, in brackets), we have
$$\bigg[a^5b^3 + a^3b^2c^3\bigg] + c^5b^3\\
\leq \bigg[a^5a^3 + b^3b^2c^3\bigg] + c^5b^3 \\
= a^8 + \bigg[b^5c^3+c^5b^3\bigg]\\
\leq a^8 + \bigg[b^5b^3 + c^5c^3\bigg] \\
=a^8 + b^8+c^8$$ This finishes our proof.
| {
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Is there a quicker way to solve this integral: $\int \frac{3-\cos(x)}{(1+2\cos(x))\sin^2(x)}dx$? The integral is:
$$ \int \frac{3-\cos(x)}{(1+2\cos(x))\sin^2(x)}dx$$
This is the way I approached it:
$$ \tan\left(\frac{x}{2}\right)=u\\dx=\frac{2}{\sec^2\left(\frac{x}{2}\right)}du$$
By using trigonometric identities we get:
$$ \sin(x)=\frac{2u}{1+u^2};\ \cos(x)=\frac{1-u^2}{1+u^2};\ \sec^2\left(\frac{x}{2}\right)=1+u^2
$$
Therefore the integral now becomes:
$$ 2\int \frac{3-\frac{1-u^2}{1+u^2}}{\left(1+2\left(\frac{1-u^2}{1+u^2}\right)\right)\left(\frac{2u}{1+u^2}\right)^2(1+u^2)}du=$$ $$\int\frac{(1+2u^2)(1+u^2)}{u^2(3-u^2)}du$$ By dividing the two polynomials we get: $$\int\left(-2-\frac{9u^2+1}{u^2(3-u^2)}\right)du$$ Using partial fractions we get to the simplified form: $$\int\left(-2-\frac{1}{3u^2}+\frac{28}{3(3-u^2)}\right)du$$ $$-2u-\frac{1}{3u}+\frac{28\sqrt3}{9}\int\frac{1}{1-\left(\frac{u}{\sqrt3}\right)^2}du \\ -2u-\frac{1}{3u}+\frac{28\sqrt3\tanh^{-1}{\left(\frac{u}{\sqrt3}\right)}}{9}+C$$
By substituting back in for $u$, we get the solution: $$ \bbox[5px,border:2px solid black]{\frac{28\sqrt3\tanh^{-1}{\left(\frac{\tan\left(\frac{x}{2}\right)}{\sqrt3}\right)}}{9}-2\tan\left(\frac{x}{2}\right)-\frac{1}{3}\cot\left(\frac{x}{2}\right)+C}$$
My question is, as you can understand from the title, is there any easier and faster way to solve this integral? If so, how? Thank you.
| HINT
One can look for coefficients of identity
$$f(y)=\frac{3-y}{(1+2y)(1-y^2)}=\frac A{1+2y}+\frac B{1-y}+\frac C{1+y}:$$
$$A=\lim_{y\to -\dfrac12}(1+2y)f(y) =\frac{14}3,$$
$$B=\lim_{y\to 1}(1-y)f(y) = \frac13,$$
$$C=\lim_{y\to-1}(1+y)f(y) = -2$$
and then find the integrals through the universal trigonometric substitution and known integrals
$$\int\dfrac{\mathrm dx}{1-\cos(x)}=\dfrac12\int\dfrac{\mathrm dx}{\sin^2\left(\dfrac x2\right)} = -\cot\left(\dfrac x2\right)+constant,$$
$$\int\dfrac{\mathrm dx}{1+\cos(x)}=\dfrac12\int\dfrac{\mathrm dx}{\cos^2\left(\dfrac x2\right)} = \tan\left(\dfrac x2\right)+constant.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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finding value of $\sum^{10}_{i=0}\frac{1}{a_{i}}$ Let $a_{0},a_{1},a_{2},a_{3},\cdots \cdots a_{n}$ be a sequence of numbers satisying $(3-a_{n+1})(6+a_{n}) = 18$ and $a_{0} = 3.$ then find $\displaystyle \sum^{10}_{i=0}\frac{1}{a_{i}}$
from $18-6a_{n+1}+3a_{n}-a_{n}a_{n+1} = 18.$ So $6a_{n+1}-3a_{n}+a_{n}a_{n+1} =0$
want be able to go further, could some help me, thanks
| From the condition we have $$-6a_{n+1}+3a_n-a_na_{n+1}=0$$ or
$$-\frac{6}{a_n}+\frac{3}{a_{n+1}}-1=0$$ or
$$-6\left(\frac{1}{a_0}+\frac{1}{a_1}+...+\frac{1}{a_{10}}\right)+3\left(\frac{1}{a_1}+...+\frac{1}{a_{11}}\right)-11=0$$
and for the ending it remains to find a value of $a_{11}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate $\int_0^\infty {\frac{x}{{\left( {x + 1} \right)\sqrt {4{x^4} + 8{x^3} + 12{x^2} + 8x + 1} }}dx}$ Prove
$$I=\int_0^\infty {\frac{x}{{\left( {x + 1} \right)\sqrt {4{x^4} + 8{x^3} + 12{x^2} + 8x + 1} }}dx} = \frac{{\ln 3}}{2} - \frac{{\ln 2}}{3}.$$
First note that
$$4{x^4} + 8{x^3} + 12{x^2} + 8x + 1 = 4{\left( {{x^2} + x + 1} \right)^2} - 3,$$
we let
$${x^2} + x + 1 = \frac{{\sqrt 3 }}{{2\cos \theta }} \Rightarrow x = \sqrt { - \frac{3}{4} + \frac{{\sqrt 3 }}{{2\cos \theta }}} - \frac{1}{2},$$
then
$$I=\frac{1}{2}\int_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\frac{{\left( {\sqrt {2\sqrt 3 \sec \theta - 3} - 1} \right)\sec \theta }}{{\left( {\sqrt {2\sqrt 3 \sec \theta - 3} + 1} \right)\sqrt {2\sqrt 3 \sec \theta - 3} }}d\theta } .$$
we have
\begin{align*}
&\frac{{\left( {\sqrt {2\sqrt 3 \sec \theta - 3} - 1} \right)\sec \theta }}{{\left( {\sqrt {2\sqrt 3 \sec \theta - 3} + 1} \right)\sqrt {2\sqrt 3 \sec \theta - 3} }} = \frac{{{{\left( {\sqrt {2\sqrt 3 \sec \theta - 3} - 1} \right)}^2}\sec \theta }}{{\left( {2\sqrt 3 \sec \theta - 4} \right)\sqrt {2\sqrt 3 \sec \theta - 3} }}\\
=& \frac{{\left( {2\sqrt 3 \sec \theta - 2 - 2\sqrt {2\sqrt 3 \sec \theta - 3} } \right)\sec \theta }}{{\left( {2\sqrt 3 \sec \theta - 4} \right)\sqrt {2\sqrt 3 \sec \theta - 3} }} = \frac{{\left( {\sqrt 3 \sec \theta - 1 - \sqrt {2\sqrt 3 \sec \theta - 3} } \right)\sec \theta }}{{\left( {\sqrt 3 \sec \theta - 2} \right)\sqrt {2\sqrt 3 \sec \theta - 3} }}\\
= &\frac{{\left( {\sqrt 3 \sec \theta - 1} \right)\sec \theta }}{{\left( {\sqrt 3 \sec \theta - 2} \right)\sqrt {2\sqrt 3 \sec \theta - 3} }} - \frac{{\sec \theta }}{{\sqrt 3 \sec \theta - 2}}.
\end{align*}
and
$$\int {\frac{{\sec \theta }}{{\sqrt 3 \sec \theta - 2}}d\theta } = \ln \frac{{\left( {2 + \sqrt 3 } \right)\tan \frac{\theta }{2} - 1}}{{\left( {2 + \sqrt 3 } \right)\tan \frac{\theta }{2} + 1}}+ C.$$
while
\begin{align*}&\int {\frac{{\left( {\sqrt 3 \sec \theta - 1} \right)\sec \theta }}{{\left( {\sqrt 3 \sec \theta - 2} \right)\sqrt {2\sqrt 3 \sec \theta - 3} }}d\theta } = \int {\frac{{\sqrt 3 - \cos \theta }}{{\left( {\sqrt 3 - 2\cos \theta } \right)\sqrt {2\sqrt 3 \cos \theta - 3{{\left( {\cos \theta } \right)}^2}} }}d\theta } \\
= &\frac{1}{2}\int {\frac{1}{{\sqrt {2\sqrt 3 \cos \theta - 3{{\left( {\cos \theta } \right)}^2}} }}d\theta } + \frac{{\sqrt 3 }}{2}\int {\frac{1}{{\left( {\sqrt 3 - 2\cos \theta } \right)\sqrt {2\sqrt 3 \cos \theta - 3{{\left( {\cos \theta } \right)}^2}} }}d\theta } .
\end{align*}
But how can we continue? It is related to elliptic integral.
| This is a pseudo-elliptic integral, it has an elementary anti-derivative:
$$\int \frac{x}{(x+1)\sqrt{4x^4+8x^3+12x^2+8x+1}} dx = \frac{\ln\left[P(x)+Q(x)\sqrt{4x^4+8x^3+12x^2+8x+1}\right]}{6} - \ln(x+1) + C$$
where $$P(x) = 112x^6+360x^5+624x^4+772x^3+612x^2+258x+43$$
and
$$Q(x) = 52x^4+92x^3+30x^2-22x-11$$
To obtain this answer, just follow the systematic method of symbolic integration over simple algebraic extension. Alternatively, you can throw it to a CAS with Risch algorithm implemented (not Mathematica), a convenient software is the online Axiom sandbox.
| {
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Prove that: $\frac {4(\cos^4 \frac {\pi}{8} + \cos^4 \frac {3\pi}{8})}{\cos^4 \frac {7\pi}{8} - \cos^4 \frac {11\pi}{8}}=…$ Prove that: $$\frac {4(\cos^4 \frac {\pi}{8} + \cos^4 \frac {3\pi}{8})}{\cos^4 \frac {7\pi}{8} - \cos^4 \frac {11\pi}{8}}=2\sqrt {2} \sin \frac {\pi}{3}.\tan \frac {\pi}{3}$$
My Attempt:
$$L.H.S= \frac {4(\cos^4 \frac {\pi}{8} + \cos^4 \frac {3\pi}{8})}{\cos^4 \frac {7\pi}{8} - \cos^4 \frac {11\pi}{8}}$$
$$=\frac {4((\cos^2 \frac {\pi}{8})^2 + (\cos^2 \frac {3\pi}{8})^2)}{\cos^4 \frac {\pi}{8} - \cos^4 \frac {3\pi}{8}}$$.
What should I do next?.
| HINT:
$$\dfrac\pi8+\dfrac{3\pi}8=\dfrac\pi2$$
So, $\cos\dfrac{3\pi}8=\sin\dfrac\pi8$
Now $\cos^4x+\sin^4x=(\cos^2x+\sin^2x)^2-2\cos^2x+\sin^2x=1-\dfrac{(2\sin x\cos x)^2}2=1-\dfrac{\sin^22x}2$
Similarly, $\dfrac{11\pi}8-\dfrac{7\pi}8=\dfrac\pi2$
$\cos\dfrac{11\pi}8=\cos\left(\dfrac\pi2+\dfrac{7\pi}8\right)=-\sin\dfrac{7\pi}8$
Now $\cos^4y-(-\sin y)^4=(\cos^2y-\sin^2y)(\cos^2y+\sin^2y)=\cos2y$
| {
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} |
If $ab+ac+bc=3$ so $\sum\limits_{cyc}\sqrt{a^2+b^2+7bc}\geq9$. Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc=3$. Prove that:
$$\sqrt{a^2+b^2+7bc}+\sqrt{b^2+c^2+7ca}+\sqrt{c^2+a^2+7ab}\geq9$$
The equality occurs also for $a=2$, $b=\frac{3}{2}$ and $c=0$.
I tried the following Holder.
$$\left(\sum\limits_{cyc}\sqrt{a^2+b^2+7bc}\right)^2\sum_{cyc}(a^2+b^2+7bc)^2(ka+mb+c)^3\geq$$
$$\geq\left(\sum_{cyc}(a^2+b^2+7bc)(ka+mb+c)\right)^3.$$
Thus, it remains to prove that
$$\left(\sum_{cyc}(a^2+b^2+7bc)(ka+mb+c)\right)^3\geq9(ab+ac+bc)^2\sum_{cyc}(a^2+b^2+7bc)^2(ka+mb+c)^3,$$
but I did not find a non-negative values of $k$ and $m$, for which the last inequality would be true.
| Proof
By Holder, we have
\begin{align}
&\Big(\sum_{\mathrm{cyc}} \sqrt{a^2+b^2+7bc}\Big)^2\sum_{\mathrm{cyc}} (a^2+b^2+7bc)^2
(6ab + 3bc + 10ca)^3\\
\ge \ &
\Big(\sum_{\mathrm{cyc}} (a^2+b^2+7bc)(6ab+3bc+10ca)\Big)^3.
\end{align}
Thus, it suffices to prove that
\begin{align}
&\Big(\sum_{\mathrm{cyc}} (a^2+b^2+7bc)(6ab+3bc+10ca)\Big)^3\\
\ge \ & 27(ab+bc+ca)\sum_{\mathrm{cyc}} (a^2+b^2+7bc)^2
(6ab + 3bc + 10ca)^3.
\end{align}
Let $f(a,b,c) = \mathrm{LHS}-\mathrm{RHS}$.
The Buffalo Way works. WLOG, assume that $c = \min(a,b,c)$. There are two possible cases:
1) $c\le b \le a$: Let $b = c + s,\ a = c+s+t; \ s,t\ge 0$. Then
$$f(c+s+t, c+s, c) = g(c,s,t) + s^3(352s^4+1068s^3t+1269s^2t^2+378st^3+81t^4)(s-3t)^2(s+t)^3$$
where $g(c,s,t)$ is a polynomial with non-negative coefficients. True.
2) $c\le a\le b$: Let $a=c+s, \ b = c+s+t; \ s,t\ge 0$. Then
$f(c+s, c+s+t, c)$ is a polynomial in $c, s, t$ with non-negative coefficients. True.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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} |
Find the sum of the series $\sum_{k=1}^{\infty} \frac {1}{(k)(k+2)(k+4)}$. Problem:
Find the sum of the series $\sum_{k=1}^{\infty} \frac {1}{(k)(k+2)(k+4)}$.
Thoughts
I first tried to write out the series to detect some kind of a pattern, I suspect I need a way of rewriting the expression $\frac {1}{(k)(k+2)(k+4)}$ , but not sure how to proceed. To help visualize it looks like:
$\frac {1} {(1)(3)(5)} + \frac {1} {(2)(4)(6)} + \frac {1} {(3)(5)(7)} + ...$
| Telescoping series plus some PFD:
$$\frac1{k(k+2)(k+4)}=\frac{1/8}{k(k+2)}-\frac{1/8}{(k+2)(k+4)}$$
Thus, we have
$$\begin{align}S&=\frac{1/8}{1\cdot3}-\color{#4488dd}{\frac{1/8}{3\cdot5}}\\&+\frac{1/8}{2\cdot4}-\color{#44aa88}{\frac{1/8}{4\cdot6}}\\&+\color{#4488dd}{\frac{1/8}{3\cdot5}}-\color{#dd8822}{\frac{1/8}{5\cdot7}}\\&+\color{#44aa88}{\frac{1/8}{4\cdot6}}-\color{#dd5555}{\frac{1/8}{6\cdot8}}\\&+\dots\end{align}$$
Notice each term on the right cancels with a term on the left two lines below it, giving
$$S=\frac{1/8}{1\cdot3}+\frac{1/8}{2\cdot4}=\frac{11}{96}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2134163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Simplify $ z^4+z^2+1$
$$ z^4+z^2+1$$
I'm trying to get this form: $(z^2+z+1)(z^2-z+1)$ but don't know how.
I tried: $z^2(z^2+1)+1$
| Using just the $\;a^n \pm b^n$ identities:
$$
\require{cancel}
z^4+z^2+1=\frac{z^6-1}{z^2-1}=\frac{(z^3-1)(z^3+1)}{z^2-1}=\frac{\cancel{(z-1)}(z^2+z+1)\,\cancel{(z+1)}(z^2-z+1)}{\cancel{z^2-1}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to find how many real roots of an equation?
Roots of $ax^2 + bx + c = 0$ are real and positive. $a$, $b$ and $c$ are real.
Then $ax^2 + b|x| + c = 0$ has how many real roots?
My try:
I studied one method where we see how many signs are changing in equation. Then we are able to find real roots. But I forget the method name. So find difficult to solve.
Other methods are also appreciated. Thank you.
| It's a weird question but it boils down to: If $k > 0$ then $|k| = k$ and $|-k| = k$ and $(-k)^2 = k^2$ so if $k > 0$ is a solution to $ax^2 + bx + c$ then both $k$ and $-k$ are solutions to $ax^2 + b|x| + c$.
[Another way of putting it is $ax^2 + b|x| + c = a(-x)^2 + b|-x| + c$ so $k$ is a solution if and only if $-k$ is a solution. If $x > 0$ then $a^2 + bx + c = ax^2 + b|x| + c = a(-x)^2 + b|-x| + c$ then $k$ is a positive solution to $a^2 + bx + c$ if and only if $k, -k$ are a positive and negative solution to $ax^2 + b|x| + c$..]
So if $a^2 + bx + c$ has two positive real solutions; $k,j$ then $a^2 + b|x| + c$ will have the same two positive real solutions, $k,j$ as well as their negative counter parts $-k,-j$ as solutions.
... Note: $ax^2 + b|x| + c$ is NOT a polynomial. Hence the rule that there are at most two solutions does not apply. (Actually, $ax^2 + b|x| + c$ is a system of two polynomials: $ax^2 + bx + c$ or $ax^2 - bx + c$ and as each may have up to two solutions, the system will have up to four solutions [to one or the other]. Furthermore, if $k,j$ are the solutions to $ax^2 + bx + c$ then $-k,-j$ are the solutions to $ax^2 -bx +c$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2138252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Evaluation of a Sum Find the sum:
$$\sum_\limits{i=0}^\infty\sum_\limits{j=0}^\infty \frac{1}{a^i\cdot a^j}$$ where $i \not= j$ and $a>1$
| $$\sum_\limits{i=0}^\infty\sum_\limits{j=0 \\ i\not=j}^\infty \frac{1}{a^i\cdot a^j}$$
$$\sum_\limits{i=0}^\infty\sum_\limits{j=0}^\infty \frac{1}{a^i\cdot a^j}-\sum_\limits{j=0}^\infty \frac{1}{a^{2j}}$$
$$=\left(\sum_\limits{i=0}^\infty \frac{1}{a^i}\right)\left(\sum_\limits{j=0}^\infty \frac{1}{a^j}\right)-\frac{1}{1-\frac{1}{a^2}}$$
$$=\frac{1}{1-\frac{1}{a}} \cdot \frac{1}{1-\frac{1}{a}}-\frac{a^2}{a^2-1}$$
$$=\left(\frac{a}{a-1}\right)^2-\frac{a^2}{a^2-1}$$
$$=\frac{a^2}{a-1}\left(\frac{1}{a-1}-\frac{1}{a+1}\right)$$
$$=\frac{2a^2}{(a-1)^2(a+1)}$$
Hope this helps you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
$\lim_{x\to 1} \frac{\tan(\pi x)}{x^2-\sqrt{x}}$ find the limit:
$$\lim_{x\to 1} \frac{\tan(\pi x)}{x^2-\sqrt{x}}$$
my try :
$$\lim_{x\to 1} \frac{\tan(\pi x)}{x^2-\sqrt{x}}\\=\frac{\tan(\pi x)}{x^2(1-\frac{1}{x\sqrt{x}})}\\=\lim_{x\to 1} \frac{\tan(\pi x)}{x^2}.\frac{1}{{1-\frac{1}{x\sqrt{x}}}}$$
now ?
| Set $\sqrt x=y\implies x=y^2$
$$\dfrac{\tan\pi x}{x^2-\sqrt x}=\dfrac{\tan\pi y^2}{y^4-y}=\pi\cdot\dfrac{y^2-1}{y(y-1)(y^2+y+1)}\cdot\dfrac{\tan\pi(y^2-1)}{\pi(y^2-1)}$$
Now $\dfrac{y^2-1}{y(y-1)(y^2+y+1)}=\dfrac{y+1}{y(y^2+y+1)}$ for $y\ne1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What is a common value and how to find it? I have googled this term and did not find anything useful. This term is used multiple times in this problem explanation from the recent AMC. For example, I do not understand how the common value of $x+2$ and $3$ is $3$ and of $x+2$ and $y-4$ is $y-4$.
| If two of the three terms are equal then the common value is just the value of the two terms that are equal.
1) If $3$ and $x + 2$ are the two terms that are equal, the common value is $k = 3 = x+1$. (Which means $x = 1$ and $k =3 = x + 1 = 3$.)
2) If $3$ and $y - 4$ are the two terms that are equal, the common value is $k = 3 = y-4$. (Which means $y = 7$ and $k = 3 = y-4 = 3$.
3) If $x+2$ and $y-4$ are the two terms that are equal, the common value is $k= x+ 2=y-4$. (Which means the common value is 4 less than whatever $y$ is and 2 more than whatever $x$).
We are also told that the third value is at most this common value.
So
1) if $3 = x+2$ then $y-4 \le 3 = x+2$.
2) if $3 = y-4$ then $x+2 \le 3 = y-4$.
3) if $x+2 = y-4$ then $3 \le x+2 = y-4$.
In this case "common value" just means .... common value. It is not a mathematical term. It means exactly what it says.
Let Sam, Fred, and Jane be friends. Two of them live at the same house. The third lives down the street from the common address. "common address" just means the address of the two people who live together.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2140119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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advanced expected value？
*David repeatedly flips a fair coin. Find the expected value of the total number of heads he will flip before flipping two consecutive tails.
I know that expected value is the sum of value multiplied by probability and all the probability sums up to $1$, however I find this somehow hard to approach so can anyone help me..
| $E(X)=(1(\frac{1}{2}+\frac{1}{2^2})+2(\frac{1}{2^2}+2C_1(\frac{1}{2^3})+\frac{1}{2^4})+3(\frac{1}{2^3}+3C_1(\frac{1}{2^4})+3C_2(\frac{1}{2^5})+3C_3(\frac{1}{2^6}))+.......)\frac{1}{2^2}
$
$= (\sum_{n=1}^\infty n(\frac{1}{2^n})(nC_0(\frac{1}{2^0})+nC_1(\frac{1}{2^1})+nC_2(\frac{1}{2^2})+nC_3(\frac{1}{2^3})+.....+nC_n(\frac{1}{2^n})))\frac{1}{2^2}$
$=(\sum_{n=1}^\infty n(\frac{1}{2^n})(\frac{3}{2^n}))\frac{1}{2^2}$
$=(\sum_{n=1}^\infty n(\frac{3}{4^n}))\frac{1}{2^2}$
$=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2142927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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Integrating around a dog bone contour Prove that
$$\int^{1}_{0} \sqrt{x}\sqrt{1-x}\,dx = \frac{\pi}{8}$$
$$\textit{proof}$$
Consider the function
$$f(z) = \sqrt{z-z^2} = e^{\frac{1}{2}\log(z-z^2)}$$
Consider the branch cut on the x-axis
$$x(1-x)\geq 0\,\, \implies \, 0\leq x \leq 1 $$
Consider $ w= z-z^2 $ then
$$\log(w) = \log|w|+i\theta,\,\, \theta\in[0,2\pi)$$
Consider the contour
Consider the integral
$$\int_{c_0}f(z)\,dz+\int_{c_1}f(z)\,dz+\int^{1-\epsilon}_{\epsilon} e^{\frac{1}{2}\log|x-x^2|}\,dx-\int^{1-\epsilon}_{\epsilon} e^{\frac{1}{2}\log|x-x^2| +\pi i}\,dx = 2\pi i \mathrm{Res}(f,\infty)$$
Consider the Laurent expansion of
$$\sqrt{z-z^2} =i \sqrt{z^2} \sqrt{1-\frac{1}{z}}= iz\sum_{k=0}^\infty{\frac{1}{2} \choose k} \left(-\frac{1}{z} \right)^k$$
Hence we deuce that
$$ \mathrm{Res}(f,\infty) = -\frac{i}{8}$$
That implies
$$\int_{c_0}f(z)\,dz+\int_{c_1}f(z)\,dz+2\int^{1-\epsilon}_{\epsilon} \sqrt{x}\sqrt{1-x}\,dx = \frac{\pi}{4}$$
Considering integrals the contours around $c_0$ and $c_1$ go to zero . Finally we get
$$\int^{1}_{0} \sqrt{x}\sqrt{1-x}\,dx = \frac{\pi}{8}$$
Question
I have my concerns about the expansion at infinity
$$\sqrt{z-z^2} =i \sqrt{z^2} \sqrt{1-\frac{1}{z}}= iz\sum_{k=0}^\infty{\frac{1}{2} \choose k} \left(-\frac{1}{z} \right)^k$$
First I am assuming that $\sqrt{z^2} = z$ which seems to be wrong on the chosen branch cut. Also it is wrong to assume that $\sqrt{zw} = \sqrt{z}\sqrt{w}$.
| We may use the method from the following MSE
link for the
residue at infinity. The introduction and the continuity argument can
be copied verbatim and will not be repeated here. We use
$$f(z) = \exp(1/2\times \mathrm{LogA}(z))
\exp(1/2\times \mathrm{LogB}(1-z))$$
with the two logarithms defined as in the linked-to post. Note that
this is not the choice Mathematica or Maple makes, but consistency is
sufficient here and we can re-use a vetted computation.
The difference is that we have $f(z)\sim z$ at infinity so we need to
determine the coefficients on $z,$ the constant coefficient and the
one on $1/z.$ There was less work at the linked-to computation because
the function was $\sim 1/z$ at infinity.
We use $f(z)/z^2$ for the first one and put $z = R\exp(i\theta).$ Now
the modulus of $\mathrm{LogA}(z)$ is $\log R.$ We get for the modulus
of $\mathrm{LogB}(1-z)$
$$\log\sqrt{(1-R\cos(\theta))^2 + R^2\sin(\theta)^2}
\\ = \log\sqrt{1-2R\cos(\theta) + R^2}.$$
We are manipulating logarithms of positive real numbers using the real
logarithm and may continue with
$$\log R + \log\sqrt{1-2\cos(\theta)/R+1/R^2}.$$
Using the leading term and the method from the link we immediately
obtain $f(z) \sim -iz.$ For the constant coefficient we use
$(f(z)+iz)/z.$ We have
$$\exp\left(\frac{1}{2}\log\sqrt{1-2\cos(\theta)/R+1/R^2}\right)
= \sqrt[4]{1-2\cos(\theta)/R+1/R^2}
\\ = 1 - \frac{1}{4} (2\cos(\theta)/R-1/R^2)
- \frac{3}{32} (2\cos(\theta)/R-1/R^2)^2 -\cdots
\\ = 1 - \frac{1}{2}\cos(\theta)\frac{1}{R}
+ \left(1/4 - \frac{3}{8}\cos(\theta)^2\right)\frac{1}{R^2}
- \cdots$$
We also have
$$\arctan\left(\frac{-R\sin(\theta)}{1-R\cos(\theta)}\right)
= \arctan\left(\frac{-\sin(\theta)}{1/R-\cos(\theta)}\right)
\\ = \theta + \sin(\theta)\frac{1}{R} +
\sin(\theta)\cos(\theta)\frac{1}{R^2} + \cdots$$
so that (this actually holds everywhere as long as $\theta$ matches
the range of $\arg\mathrm{LogB}$)
$$\exp\left(\frac{1}{2}i\arg\mathrm{LogB(1-z)}\right)
\\ = \exp\left(\frac{1}{2}i\theta\right)
\left(1 + \frac{1}{2}i\sin(\theta)\frac{1}{R} +
\frac{1}{2}i\sin(\theta)\cos(\theta)\frac{1}{R^2} -
\frac{1}{8} \sin(\theta)^2\frac{1}{R^2} + \cdots\right).$$
Adding in $iz$ cancels the contribution from the first terms of these
two expansions. We get from the second term and collecting the
contributions from the lower and upper half plane
$$\int_0^{2\pi} \frac{1}{R\exp(i\theta)}
\exp(\log R) \exp(i\theta+\pi i/2)
\\ \times \left(\frac{1}{2}i\sin(\theta)\frac{1}{R}
-\frac{1}{2}\cos(\theta)\frac{1}{R}\right)
Ri\exp(i\theta)
\; d\theta
\\ = \int_0^{2\pi}
\exp(\log R) \exp(i\theta+\pi i/2) i
\left(-\frac{1}{2}\exp(-i\theta)\right) \frac{1}{R}
\; d\theta
\\ = \frac{1}{2} \int_0^{2\pi}
\exp(i\theta)\exp(-i\theta)
\; d\theta = \pi.$$
for a residue of $-(\pi)/(2\pi i) = i/2.$ This establishes $f(z) \sim
-iz + i/2.$ We integrate $f(z) + iz - i/2$ to get the coefficient on
$1/z$, obtaining
$$\int_0^{2\pi}
\exp(\log R) \exp(i\theta+\pi i/2)
\\ \times \frac{1}{R^2} \left(\frac{1}{4}-\frac{3}{8}\cos(\theta)^2
+ \frac{1}{2} i\sin(\theta)\cos(\theta)
- \frac{1}{8} \sin(\theta)^2
-\frac{1}{4}i\sin(\theta)\cos(\theta)\right)
\\ \times Ri\exp(i\theta)
\; d\theta$$
The inner term is
$$\frac{1}{8} - \frac{1}{4}\cos(\theta)^2 + \frac{1}{8}i\sin(2\theta)
= - \frac{1}{8}\cos(2\theta) + \frac{1}{8}i\sin(2\theta)
= - \frac{1}{8}\exp(-2i\theta).$$
which leaves for the integral
$$\frac{1}{8} \int_0^{2\pi} \exp(2i\theta)\exp(-2i\theta) \; d\theta
= \frac{\pi}{4}.$$
We get the residue $-(\pi/4)/(2\pi i) = i/8.$
We have established that at infinity,
$$\bbox[5px,border:2px solid #00A000]{ f(z) \sim
-iz + \frac{1}{2} i + \frac{1}{8} i \frac{1}{z}}$$
and hence $\mathrm{Res}_{z=\infty} f(z) = \frac{1}{8} i.$ Taking into
account that the contour used with this branch produces twice the
value of the integral we obtain
$$\frac{1}{2} \times -2\pi i \times \frac{1}{8} i$$
which is
$$\bbox[5px,border:2px solid #00A000]{ \frac{\pi}{8}.}$$
| {
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"url": "https://math.stackexchange.com/questions/2143137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Compute $\lim_\limits{x\to0^+}\frac{\pi/2- \arctan(1/x^2)-\sin(x^3)-1+\cos(x)}{x\tan(x)+e^{x^3}-1}$ I have to compute
$$
\lim_{x\to0^+}\frac{\pi/2- \arctan(1/x^2)-\sin(x^3)-1+\cos(x)}{x\tan(x)+e^{x^3}-1}
$$
I separated the numerator so I got that
$$\dfrac{-1+\cos(x)}{x\tan(x)+e^{x^3}-1} \longrightarrow -\dfrac{1}2;$$
I know that the limit is $\dfrac{1}2$ and I know, by checking on Wolfram Alpha, that
$$\dfrac{-\sin(x^3)}{x\tan(x)+e^{x^3}-1}\longrightarrow 0$$ so
$$\dfrac{π/2- \arctan(1/x^2)}{x\tan(x)+e^{x^3}-1} \longrightarrow 1.$$
I tried using L'Hopital but it gets even more complicated. How can I solve it?
| differentiate numerator
$d/dx(\pi/2 - \arctan(\frac{1}{x^2}) - \sin(\frac{1}{x^3}) - 1 + \cos(x)) $
$= -[(-2) / x^3] / [1 - (\frac{1}{x^2})^2] - 3x^2\cos(x^2) - \sin(x)$
$= 2x / [x^4 + 1] - 3x^2\cos(x^2) - \sin(x)$
you can differentiate that again, but i think that the only non zero term at x = 0 will be $2x / [x^4 +1] - \cos(x)$ - giving the value 1 for the numerator after two differentiations
for the denominator, the first differentiation is
$\tan(x) + x\sec^2(x) + 3x^2 \exp(x^3)$
a second differentiation of that, and setting x =0 is going to give the non-zero terms
$\sec^2(x) + \sec^2(x) + x(\sec(x) \tan(x)) + [zero terms]$
= $\sec^2(x) + \sec^2(x) = 2$
so I make it $\frac{1}{2}$
also I got it as 1/2 as an estimate on a computer, via numerical method
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Prove the equality by introducing a suitable change of variable (double integral) Problem from Apostol's Calculus 2:
Prove the following equality by introducing a suitable change of variable:
$$ \iint_S f(ax + by + c)dxdy = 2\int_{-1}^{1}\sqrt{1-u^2}f(u\sqrt{a^2+b^2} +c)du $$
With $$S= \{(x,y) \space | \space x^2+y^2 \le 1\}$$
And $ a^2+b^2 \neq 0$.
| Assuming $a^2+b^2 \neq 0$. Let,
$$ax+by=u \sqrt{a^2+b^2}$$
$$bx-ay=v \sqrt{a^2+b^2}$$
Note then the boundary for our region becomes $$u^2+v^2=\frac{(ax+by)^2+(bx-ay)^2}{a^2+b^2} $$
$$=\frac{a^2(x^2+y^2)+b^2(x^2+y^2)}{a^2+b^2}=1$$
The absolute value of our Jacobian for our transformation is,
$$1$$
And we then have,
$$\iint_S f(ax + by + c)dxdy$$
$$=\iint_{S_2} f(u\sqrt{a^2+b^2}+c) dv du$$
Where $S_2=\{(u,v) : u^2+v^2 \leq 1 \}$.
$$=\int_{-1}^{1} \int_{-\sqrt{1-u^2}}^{\sqrt{1-u^2}} f(u\sqrt{a^2+b^2}+c) dv du$$
$$=2 \int_{-1}^{1} \sqrt{1-u^2}f(u\sqrt{a^2+b^2}+c) du$$
| {
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"url": "https://math.stackexchange.com/questions/2144889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How does one show that $\sum_{n=0}^{\infty}\left(-{1\over x}\right)^n\cdot{\Gamma\left({2n+1\over 2}\right)\over n!}=\sqrt{{x\pi\over x+1}}?$ Consider $(1)$
$$\sum_{n=0}^{\infty}\left(-{1\over x}\right)^n\cdot{\Gamma\left({2n+1\over 2}\right)\over n!}=S\tag1$$
$x>0$
How does one show that $S=\sqrt{{x\pi\over x+1}}$
An attempt:
$$\Gamma\left({1\over 2}+n\right)={(2n-1)!!\sqrt{\pi}\over 2^n}$$
$(1)$ becomes
$$\sum_{n=0}^{\infty}\left(-{1\over x}\right)^n\cdot{(2n-1)!!\over 2^n}\cdot{1\over n!}=\sqrt{x\over x+1}\tag2$$
further to
$$\sum_{n=0}^{\infty}\left(-{1\over x}\right)^n\cdot{1\over 2^n(2n-2)!!}=\sqrt{x\over x+1}\tag3$$
| We note that
$$\frac{\Gamma \left ( n+\frac12 \right )}{\Gamma \left ( n+1 \right )} = \frac1{\sqrt{\pi}} \int_0^1 dy \, y^{-1/2} (1-y)^{n-1/2} $$
Then
$$\begin{align}\sum_{n=0}^{\infty} \frac{\Gamma \left ( n+\frac12 \right )}{\Gamma \left ( n+1 \right )} w^n &= \frac1{\sqrt{\pi}} \int_0^1 dy \, y^{-1/2} (1-y)^{-1/2} \sum_{n=0}^{\infty} [(1-y) w]^n \\ &= \frac1{\sqrt{\pi}} \int_0^1 dy \, \frac{y^{-1/2} (1-y)^{-1/2}}{1-w (1-y)} \\ &= \frac{2}{\sqrt{\pi}} \int_0^1 dy \frac{(1-y^2)^{-1/2}}{1-w (1-y^2)} \\ &= \frac1{2 \sqrt{\pi}} \int_0^{2 \pi} \frac{d\theta}{1-w \sin^2{\theta}} \end{align}$$
where we assume that $|w| \lt 1$. That last integral may be performed any number of ways, e.g., extension to the complex plane and the residue theorem. The result is $2 \pi/\sqrt{1-w}$. Thus,
$$S = \sqrt{\frac{\pi}{1+\frac1{x}}} = \sqrt{\frac{\pi x}{x+1}} $$
ADDENDUM
Let's illustrate how to do that last integral.
$$\int_0^{2 \pi} \frac{d\theta}{1-w \sin^2{\theta}} = 2 \int_0^{\pi} \frac{d\theta}{1-w \sin^2{\theta}} = 2 \int_0^{\pi} \frac{d\theta}{1-w/2 + (w/2) \cos{2 \theta}} = \int_0^{2 \pi}\frac{d\theta}{1-w/2+(w/2) \cos{\theta}}$$
The above manipulations serve to show how to reduce a potentially messy problem into one a lot less messy. In this case, we have reduced the number of poles in the complex plane from four to two.
Now express as a complex integral using $z=e^{i \theta}$; the integral is equal to
$$-i 2 \oint_{|z|=1} \frac{dz}{(w/2) z^2 + (2-w) z+(w/2)} $$
The poles of the integrand are at $z_{\pm}=[-(2-w) \pm 2 \sqrt{1-w}]/w$, and the only pole inside the unit circle is $z_+$. Thus, by the residue theorem, the integral is equal to $i 2 \pi$ times the residue at that pole, the residue being equal to $1/(2 \sqrt{1-w})$. Thus the integral is equal to $2 \pi/\sqrt{1-w}$ as asserted.
| {
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Prove that $a^{\frac{2n}{m}} - 1 \geq n\big(a^{\frac{n+1}m} - a^{\frac{n-1}{m}}\big)$.
Given natural numbers $m,n,$ and a real number $a>1$, prove the inequality :
$$\displaystyle a^{\frac{2n}{m}} - 1 \geq n\big(a^{\frac{n+1}m} - a^{\frac{n-1}{m}}\big)$$
SOURCE : Inequalities (PDF) (Page Number 2 ; Question Number 153.2)
I have been trying this problem from 2 weeks but still no success. I tried every method I could think of like AM-GM, C-S, Holder and more, but could not find a proof.
Also, is it necessary for $n,m$ to be natural numbers ?
Any help will be gratefully acknowledged.
Thanks in advance ! :)
| Let $x = a^{\frac 1m} > 1$. Using
$$
x^{2n} - 1 = (x-1)(1+ x+x^2 + \ldots + x^{2n-1}) \\
x^{n+1} - x^{n-1} = (x-1) (x^{n-1}+x^n)
$$
we get
$$
x^{2n} - 1 - n(x^{n+1} - x^{n-1}) =
(x-1)\left( 1+ x+x^2 + \ldots + x^{2n-1} - n(x^{n-1}+x^n) \right) \\
= (x-1)\sum_{k=1}^n \left( x^{k-1} + x^{2n-k} - x^{n-1}-x^n\right) \\
= (x-1)\sum_{k=1}^n (x^{n-k}-1)(x^n - x^{k-1}) \\ \ge 0
$$
(with strict inequality for $n \ge 2$), which is the desired
inequality
$$
x^{2n} - 1 \ge n(x^{n+1} - x^{n-1}) \, .
$$
This proof works for positive real $m$ and integer $n \ge 1$.
For $x < 1$ the same inequality with $\ge$ replaced by $\le$
holds.
| {
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} |
Prove that :$\tan 70 - \tan 20 = 2\tan 40 + 4\tan 10$ Prove that :$\tan 70 - \tan 20 = 2\tan 40 + 4\tan 10$
My Attempt,
$$70-20=40+10$$
$$\tan (70-20)=\tan (40+10)$$
$$\dfrac {\tan 70 - \tan 20}{1+\tan 70. \tan 20 }=\dfrac
{\tan 40 + \tan 10 }{1-\tan 40. \tan 10 }$$
How should I move on? Please help
Thanks
| Hint:
Repeatedly use the identity $$\tan x - \cot x = \tan x - \frac{1}{\tan x} = \frac{\tan^2 x-1}{\tan x} = \frac{-2(1-\tan^2 x)}{2\tan x} = \frac{-2}{\frac{2\tan x}{1-\tan^2 x}} = \frac{-2}{\tan 2x}$$ $$ = -2\cot 2x$$ to get, $$\tan 20 - \cot 20 = -2\cot 40 \tag{1}$$ $$2(\tan 40 - \cot 40) = -4\cot 80 \tag{2}$$ Adding $(1)$ and $(2)$ gives us, $$\tan 20 + 2\tan 40 - \cot 20 = -4\cot 80 = -4\tan 10$$ $$\Rightarrow \cot 20 = \tan 70 = \tan 20 + 2\tan 40 + 4\tan 10$$ Hope it helps.
| {
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Mathematical Induction: $3^n +1$ is even for all values of positive integers I am not sure how to go about this "proof by induction" problem. below is my attempt.
Base Case: $n = 0$,substituting the value of $n$ to the equation $3^n+1$
$$= 3^0 + 1$$
$$= 1 + 1 = 2 $$
Thus the equation holds true for initial value of $n$ i.e $0$
Induction Hypothesis: Suppose the equation holds true for all the values of $n$: $1,2,3....k$ therefore, $3^k + 1$ results even.
Induction Step: $n = k$ holds true,
to prove: $3^{k+1}+ 1$ for $n = k+1$
LHS:
$$3^{k+1}+ 1 = (3^k \cdot 3) + 1 = (3^k \cdot (2 + 1)) + 1 = 2 \cdot 3^k + 3^k + 1$$
The above solution results even, because since multiplying any integer with $2$ gives even integer, and from the Induction Hypothesis $3^k+1$ is even.
Hence $3^{k+1} + 1$ is even, thus $3^{n+1}$ is even for all values of $n\ge0$
| Proof I
$$3^n + 1 \implies (2x + 1)+1 \implies 2x+2$$
Case: $n = 0$
$$ (2\cdot0 + 2) = 2 $$
Case: $n = n$
$$(2n + 2) = 2(n+1) $$
Case: $n = n+1$
$$(2(n+1) + 2) = 2n+4 = 2(n+2)$$
$$\Box$$
Proof II
$$(3^n + 1)\text{ mod }2 \implies (3^n + 2 -1) \text{ mod }2 \implies (3^n -1)\text{ mod }2 $$
Case: $n = 0$
$$3^0 \equiv 1 (\text{ mod }2)\implies 3^0 -1 \equiv 0 (\text{ mod }2)$$
Case: $n = n$
$$3^n \equiv 1 (\text{ mod }2)\implies 3^n -1 \equiv 0 (\text{ mod }2)$$
Case: $n = n+1$
$$3^{n+1} \equiv 1 (\text{ mod }2)\implies 3^{n+1} -1 \equiv 0 (\text{ mod }2)$$
$$\Box$$
Case of the Congruence Power Rule
| {
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Find all solutions using the Chinese Remainder Theorem Find all solutions using the Chinese Remainder Theorem.
$$
\begin{cases}x \equiv 3 \pmod{4}\\
x \equiv 5 \pmod{21}\\
x \equiv 7 \pmod{25}
\end{cases}$$
I can see that $4$,$21$, and $25$ are all pairwise relatively prime. I then proceeded to use$ Z=(C_1)\times(X_1)\times(B_1) +(C_2)\times(X_2)\times(B_2) +(C_3)\times(X_3)\times(B_3)$.
Let $C_i $ be the remainder and$ b_i$ be the modulus.
$3=C_1,5=C_2,7=C_3$.
$4=b_1,21=b_2,25=b_3$.
Let $B=b_1\times b_2\times b_3$
Let $B_i=B/b_i$
Then, $525=B_1,100=B_2,84=B_3$
Now I try and solve $(B_i)(X_i)≡ 1 \mod b_i$
I can see that $(B_1)(X_1)≡ 1 \mod b_1 $yields $525(X_1)≡ 1 \mod 4 $with$ X_1 $being $1$.
For $(B_2)(X_2)≡ 1 \mod b_2$, I get $100(X_2)≡ 1 \mod 21$. $X_2$ turns out to be $-1$.
For $(B_3)(X_3)≡ 1 \mod b_3 = 84(X_3)≡ 1 \mod 25$, I needed to use the extended euclidean algorithm to find the gcd and $X_3$, which turned out to be $-11$.
Work:
$$m \;n \; q \; r$$
$$\\84\;25 \;3 \;9$$
$$\\25 \;9 \;2 \;7$$
$$\\9 \;7 \;1\; 2$$
$$\\7 \; 2 \;3 \;1$$
$1=7-(3\times2)
\\=7-(3(9-7))=(4\times7)-(9\times3)
\\=(4(25-(9\times2)))-(9\times3)=(4\times25)-(9\times-11)
\\=(4\times25)-(11(84-(25\times3)))=(37\times25)-(11\times84)$
Thus, $X_3=-11$
It turns out that I computed the wrong $(X_2)$, it is instead $4$.
With this, I can compute the correct $Z$ such that it will satisfy the three congruence equations.
$Z=(3\times525\times1)+(100\times5\times4)+(84\times7\times-11)=-2893$
Using a calculator, I found that
$$Z \equiv 3 \pmod{4}$$
$$Z \equiv 5 \pmod{21}$$
$$Z \equiv 7 \pmod{25}$$
| I should at least mention that there is another way to solve
$\begin{cases}
x \equiv 3 \pmod{4}\\
x \equiv 5 \pmod{21}\\
x \equiv 7 \pmod{25}
\end{cases}$
We need to find integers $\alpha,\beta, \gamma \in \mathbb Z_{2100}$ (where $2100 = 4 \cdot 21 \cdot 25$) such that
$\begin{cases}
\alpha \equiv 1 \pmod{4}\\
\alpha \equiv 0 \pmod{21}\\
\alpha \equiv 0 \pmod{25}
\end{cases}, \qquad
\begin{cases}
\beta \equiv 0 \pmod{4}\\
\beta \equiv 1 \pmod{21}\\
\beta \equiv 0 \pmod{25}
\end{cases},\; \text{and} \qquad
\begin{cases}
\gamma \equiv 0 \pmod{4}\\
\gamma \equiv 0 \pmod{21}\\
\gamma \equiv 1 \pmod{25}
\end{cases}$
It will follow that $x \equiv 3\alpha + 5 \beta + 7 \gamma \pmod{2100}$
where $2100 = 4 \cdot 21 \cdot 25$.
To solve $\begin{cases}
\alpha \equiv 1 \pmod{4}\\
\alpha \equiv 0 \pmod{21}\\
\alpha \equiv 0 \pmod{25}
\end{cases}$
we first note that $\alpha \equiv 0 \pmod{21}$ and
$\alpha \equiv 0 \pmod{25}$ implies $\alpha \equiv 0 \pmod{525}$.
So $\alpha = 525t$ for some integer $t$.
Thus
\begin{align}
\alpha \equiv 1 \pmod 4
&\implies 525t \equiv 1 \pmod 4 \\
&\implies t \equiv 1 \pmod 4 \\
&\implies \alpha \equiv 525 \pmod{2100}
\end{align}.
Similarly
\begin{align}
\beta\equiv 1 \pmod{21}
&\implies 100t \equiv 1 \pmod{21} \\
&\implies -5t \equiv 1 \pmod{21} \\
&\implies t \equiv 4 \pmod{21} \\
&\implies \beta \equiv -20 \pmod{2100}
\end{align}
and
\begin{align}
\gamma \equiv 1 \pmod{25}
&\implies 84t\equiv 1 \pmod{25} \\
&\implies 9t \equiv 1 \pmod{25} \\
&\implies t \equiv 14 \pmod{25} & (9 \cdot 14 = 126 \equiv 1 \pmod{25})\\
&\implies \gamma \equiv 1176 \pmod{2100}
\end{align}.
So
\begin{align}
x
&\equiv 3\alpha + 5 \beta + 7 \gamma \pmod{2100} \\
&\equiv 3(525) + 5(-20) + 7(1176) \pmod{2100} \\
&\equiv 1307 \pmod{2100} \\
\end{align}
| {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Solve $\lim_{x \rightarrow 1}\frac{\sqrt x -1}{2-\sqrt{x+3}}$ without using L'Hôpital's rule. $\lim_{x \rightarrow 1}\frac{\sqrt x -1}{2-\sqrt{x+3}}$
(I have nothing against using L'Hôpital's rule, but my book expects me to solve this without using it which is something I don't quite dominate yet, and so I will try for practice)
I tried the following:
$$\lim_{x \rightarrow 1}\frac{\sqrt x -1}{2-\sqrt{x+3}} \cdot \frac{2+\sqrt{x+3}}{2+\sqrt{x+3}} = \lim_{x \rightarrow 1}\frac{(\sqrt x -1)(2+\sqrt{x+3})}{4-(x+3)} = \lim_{x \rightarrow 1}\frac{2\sqrt x +\sqrt x \sqrt{x+3}-2-\sqrt{x+3}}{1-x} = ???$$
What do I do next?
| In your last step, rather than expand out the numerator you should leave it alone, and instead note that
$$ \frac{(\sqrt{x}-1)(2+\sqrt{x+3})}{1-x}=\frac{(\sqrt{x}-1)(2+\sqrt{x+3})}{(1-\sqrt{x})(1+\sqrt{x})}=-\frac{(2+\sqrt{x+3})}{1+\sqrt{x}} $$
| {
"language": "en",
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"source": "stackexchange",
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"answer_id": 0
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A "simple" trigonometric computation... Show that $$\left(1+4\sin^2\frac{\pi}{18}\right) \left(1+4\sin^2\frac{\pi}{6}\right) \left(1+4\sin^2\frac{5\pi}{18}\right)\left(1+4\sin^2\frac{7\pi}{18}\right)=34.$$
Any ideas about how to tackle this?
| Got it!
Consider $T(x)=256X^4-576X^3+432X^2-120X+9$.
Then $\cos^2\frac{\pi}{18}$, $\cos^2\frac{\pi}{6}$, $\cos^2\frac{5\pi}{18}$, $\cos^2\frac{7\pi}{18}$ are its roots.
$34$ is then found using Viete's formulas.
$T$ is derived from the 9-th Chebyshev polynomial.
| {
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Find Minima and Maxima of $ y = \frac{x^2-3x+2}{x^2+2x+1}$ $$ y = \frac{x^2-3x+2}{x^2+2x+1}$$
I guess I made some mistakes cause after taking the first derivative and simlifying I have
$$y = \frac{2x^3-4x^2+5}{(x+1)^2}$$
but then numerator has complex roots. which should not be, IMO
| The derivative is not correct, unfortunately.
Given that $x^2+2x+1 = (x+1)^2$, observe that if we let $g(x) = (x+1)^2$ and $f(x) = x^2-3x+2$ we have $$\begin{align}{dy\over dx} &= {g(x)f'(x) - f(x)g'(x) \over g^2(x)}\\ &= {(x+1)^2(2x-3) - (x^2-3x+2)2(x+1) \over (x+1)^4}\\ &= {(x+1)(2x-3)-2(x^2-3x+2) \over(x+1)^3}\\&= {2x^2-3x+2x-3 -2x^2+6x-4\over(x+1)^3}\\ &= {5x - 7\over(x+1)^3}. \end{align}$$
Now we set the numerator and the denominator equal to $0$ to find the points of the function where we have global maxima and global minima.
| {
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"source": "stackexchange",
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Equation of three lines forming a equilateral triangle.
Prove that if general equation $$\cos \left(3\alpha\right)\left(x^3 -3xy^2\right) + \sin3\alpha\left(y^3 - 3x^2y\right) + 3a\left(x^2 + y^2\right) - 4a^3 = 0$$ represents three lines then they form a equilateral triangle of area $3a^2\sqrt 3$.
This equation looks like an good candidate to solve in polar coordinates than in cartesian coordinates.
Converting it into polar coordinates by parts
\begin{align}
\left(x^3 -3xy^2\right) &= r^3 \left(\cos^3 \theta - 3\cos \theta \sin^2 \theta\right) =r^3 \left(4\cos^3 \theta - 3\cos \theta\right) = r^3\cos 3\theta \tag1\label1 \\
\left(y^3 - 3x^2y\right) &= r^3\left(4\sin^3 \theta - 3\sin\theta\right) = r^3\left(\cos\left(3\pi/2 - 3\theta\right)\right) = -r^3\sin 3\theta\tag2\label2
\end{align}
Using $\eqref1$, $\eqref2$
\begin{align}
r^3\left(\cos 3\alpha\cos 3\theta - \sin3\alpha\sin 3\theta\right) + 3ar^2 - 4a^3 &= 0 \\
r^3\cos \left(3\alpha+ 3\theta\right) + 3ar^2 - 4a^3 &= 0
\end{align}
On dividing by $a^3$ and substituting $z = r/a$
$$z^3\cos \left(3\alpha+ 3\theta\right) + 3z^2 - 4 = 0.$$
Now this happens very often to me, I am not able to solve this cubic for $z$.I did try to find the factor by putting $\,z = \sin\left(3\alpha + 3\theta\right),\,\tan\left(3\alpha + 3\theta\right),\,\ldots$
What should I do now ? how to solve this cubic ?
| The solutions are $r = a, \theta = -\alpha$, $r = a, \theta = \frac{2\pi}{3} - \alpha$ and $r = a, \theta = \frac{4\pi}{3} - \alpha$
| {
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"answer_id": 2
} |
Finding extremal points on $f(x,y)$ This is the equation:
$$f(x,y) = xye^\left(-\frac{1}{2}(x^2 + y^2)\right)$$
This is what I've done:
$$\nabla f(x,y) = \begin{bmatrix}
(1-x^2)ye^\left(-\frac{1}{2}(x^2 + y^2)\right) \\
(1-y^2)xe^\left(-\frac{1}{2}(x^2 + y^2)\right)
\end{bmatrix}$$
Here's the thing I'm worried about, to find when $\nabla f(x,y) = 0$, i set each equation $= 0$.
$$\begin{align}
(1-x^2)ye^\left(-\frac{1}{2}(x^2 + y^2)\right) &= 0 \\
y e^\left(-\frac{1}{2}(x^2 + y^2)\right) &= x^2ye^\left(-\frac{1}{2}(x^2 + y^2)\right) \\
1 &= x^2 \\
x &= \pm 1\end{align}
$$
Is this legal, or do i lose some solutions when I divide away everything?
| For another approach, to avoid slogging through the partial derivates, we can exploit the symmetry of $f$. If we switch to polar coordinates, we have
$f(r,t) = xye^\left(-\frac{1}{2}(x^2 + y^2)\right)=\frac{r^{2}}{2}\sin 2t\cdot e^\left (-\frac{1}{2}(r^2)\right).$
Inspection shows that $f$ has extrema at $t=\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}\ $ and $\frac{7\pi}{4},\ $ and in each case $f(r,t)=\pm \frac{r^{2}}{2}e^\left (-\frac{1}{2}(r^2)\right).$ Now it's an easy Calc I exercise to show that $f$ has a local maximum at $r=\sqrt 2,t=\frac{\pi}{4},\ $ or $x=1,y=1,\ $ and using the symmetry of $f$ again, we can read off the other local max, and the two local mins.
| {
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Is there any way to Integrate this function? After 2 change of variables, and a $x = \log(u)$ transformation, I have this integral...
$$\int_0^{\infty } \frac{\alpha e^x}{\lambda-e^x \lambda+e^{\alpha x}} \, dx$$
What are some recommendations on how to integrate this?
NOTE: The original integral looked like:
$$\int_{1 }^{\infty } \frac{\alpha \left(\frac{1 }{x}\right)^{\alpha }}{1-\lambda (x-1 ) \left(\frac{1 }{x}\right)^{\alpha }} \, dx$$
and for the integral to converge $\alpha > 1$.
Finally, when $\alpha = 2$, I did get this expression...
$$\frac{8 \tan ^{-1}\left(\frac{\sqrt{\lambda }}{\sqrt{4-\lambda }}\right)}{\sqrt{(4-\lambda ) \lambda }}$$
Thanks,
| For integer $\alpha>1$, I got the following:
Take the Mellin transform of the integrand with respect to $\lambda$,
$$
\int_0^\infty \frac{\alpha e^x}{\lambda-e^x \lambda+e^{\alpha x}} \lambda^{s-1} \; d\lambda = \frac{\alpha e^{x-\alpha x}}{(-e^{-\alpha x}(e^x-1))^s}\pi\csc(\pi s) = I_{\alpha}(s,x)
$$
now integrate over $x$
$$
\int_0^\infty I_{\alpha}(s,x) \; dx = (-1)^{-s}a\pi\csc(\pi s)\frac{\Gamma(1-s)\Gamma(\alpha-1+s-\alpha s)}{\Gamma(\alpha-\alpha s)}
$$
now take the inverse Mellin transform with respect to $s$ to get something that depends on $\alpha$ and $\lambda$,
$$
G_\alpha(\lambda) = \frac{1}{2 \pi i}\int_{c-i \infty}^{c + i \infty} \lambda^{-s}(-1)^{-s}a\pi\csc(\pi s)\frac{\Gamma(1-s)\Gamma(\alpha-1+s-\alpha s)}{\Gamma(\alpha-\alpha s)}\;ds
$$
we can use the Mathematica command
InverseMellinTransform[((-1)^-s a [Pi] Csc[[Pi] s] Gamma[
1 - s] Gamma[-1 + a + s - a s])/Gamma[a - a s] /. a -> 2, s, l]
Where the a -> X part can be changed for various values of $\alpha$. This reproduces your result for $\alpha=2$, but that obscures the hypergeometric nature of the result. It also gives results for higher values of $\alpha$ as hypergeometric functions weighted by products of gamma functions.
The results seem to have a pattern
$$
G_4(\lambda)=\frac{3^2}{2^3}\sqrt{\frac{3}{2}}\frac{\Gamma(\frac{4}{3})\Gamma(\frac{5}{3})}{\Gamma(\frac{5}{4})\Gamma(\frac{7}{4})}\;_4F_3\left(\frac{3}{3},\frac{3}{3},\frac{4}{3},\frac{5}{3}\bigg|\frac{5}{4},\frac{6}{4},\frac{7}{4}\bigg|\frac{3^3}{4^4}\lambda\right)
$$
$$
G_5(\lambda)=\pi\frac{4^3}{5^3}\sqrt{\frac{2}{5}}\frac{\Gamma(\frac{5}{4})\Gamma(\frac{7}{4})}{\Gamma(\frac{6}{5})\Gamma(\frac{7}{5})\Gamma(\frac{8}{5})\Gamma(\frac{9}{5})}\;_5F_4\left(\frac{4}{4},\frac{4}{4},\frac{5}{4},\frac{6}{4},\frac{7}{4}\bigg|\frac{6}{5},\frac{7}{5},\frac{8}{5},\frac{9}{5}\bigg|\frac{4^4}{5^5}\lambda\right)
$$
In general it appears that
$$
G_n=C_n\; _nF_{n-1}\left(\frac{n-1}{n-1},\frac{n-1}{n-1},\frac{n}{n-1},\cdots,\frac{2n-3}{n-1}\bigg|\frac{n+1}{n},\frac{n+2}{n},\cdots,\frac{2n-1}{n}\bigg|\frac{(n-1)^{n-1}\lambda}{n^n}\right)
$$
where
$$
C_n=\sqrt{\frac{2\pi(n-1)}{n}}\frac{(n-1)^{n-2}}{n^{n-2}}\frac{\prod_{k=0}^{n-2} \Gamma(\frac{n+k}{n-1})}{\prod_{k=1}^{n-1} \Gamma(\frac{n+k}{n})}
$$
where we see your result in $C_2=2$ and
$$
_2F_1\left(1,1\bigg|\frac{3}{2}\bigg|\frac{\lambda}{4}\right) = \frac{4 \sin^{-1}(\sqrt{\lambda}/2)}{\sqrt{(4-\lambda)\lambda}}
$$
seems to check out for a few numerical examples.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2160799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Double integral over area bounded by ellipse Evaluate the double integral of $\int \int(x+y)^{2} dxdy$ over the area bounded by the ellipse $(\frac{x}{a})^2+(\frac{y}{b})^2=1$
Please check answer = $\frac{ab\pi}{4}(a^{2}+b^{2})$
| \begin{align}
I= \iint_A (x+y)^{2} dxdy = \iint_A (x^2 +y^2) dxdy + 2\iint_A xy\ dxdy
\end{align}
The latter integral vanishes due to odd symmetry in $x$ and $y$. We didn't need odd symmetry in both for the integral to vanish; just one would have sufficed.
\begin{align}
\iint_A xy\ dxdy = \int_{-b}^b\int_{-a\sqrt{1-(y/b)^2)}}^{a\sqrt{1-(y/b)^2)}} xy\ dxdy =0
\end{align}
The former integral is best evaluated in polar, but we proceed in Cartesian. It turns into a trigonometric integral shortly. First, note the even symmetry in both $x$ and $y$, leading to simplified bounds and a factor of $2^2$ in front.
\begin{align}
I= \iint_A (x^2 +y^2) dxdy &= \int_{-b}^b\int_{-a\sqrt{1-(y/b)^2)}}^{a\sqrt{1-(y/b)^2)}} (x^2 +y^2) dxdy
\\
&= 4\int_0^b\int_0^{a\sqrt{1-(y/b)^2)}} (x^2 +y^2) dxdy
\\
&= 4\int_0^b \left(
\frac13a^3\left(1-\frac{y^2}{b^2}\right)^{3/2} +ay^2\sqrt{1-\frac{y^2}{b^2}}
\right)dy
\end{align}
After reducing to a single integral, we need to make a trig substitution. The trigonometric integrals are then evaluated through repeated use of the half angle formula.
\begin{align}
I &= 4\int_0^{\pi/2} \left(
\frac{a^3b}{3}\cos^4t +ab^3\sin^2 t\cos^2 t
\right) dt
\qquad\text{(letting } \sin t=y/b)
\\
&= 4\int_0^{\pi/2} \left(
\frac{a^3b}{3}\cos^4t +\frac{ab^3}{4}\sin^2(2t)
\right) dt
\qquad\text{(double angle formula)}
\\
&= 4\left(
\frac{a^3b}{3}\frac{3\pi}{16} +\frac{ab^3}{4}\frac{\pi}{4}\right)
\qquad\text{(repeated use of half angle formulae)}
\\
&= \frac{ab\pi}{4}\left(a^2 +b^2\right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Binomial Identity with strange coefficients In my work, I have come across the following equation
$$ \sum_{k = 0}^m \sum_{l = 0}^{n-m} {m \choose k} {n - m \choose l} (-1)^k r_{k + l} $$
where $r_n = 1$ if $n$ is congruent to 0 or 1 modulo 4, and $r_n = -1$ if $n$ is congruent to 2 or 3 modulo 4. I've seen some identities involving the products ${m \choose k} {n - m \choose l}$, but I can't seem to find an algebraic trick to reduce this form to a standard identity. Any tips?
| This is
$$\sum_{k=0}^m {m\choose k} (-1)^k
\sum_{l=0}^{n-m} {n-m\choose l} r_{k+l}.$$
We have that
$$r_{k+l} =
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{k+l+1}} (1+w - w^2-w^3 + \cdots)
\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{k+l+1}}
\frac{1+w - w^2-w^3}{1-w^4}
\; dw.$$
We get for the sum
$$\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w}
\frac{1+w - w^2-w^3}{1-w^4}
\sum_{k=0}^m {m\choose k} \frac{(-1)^k}{w^k}
\sum_{l=0}^{n-m} {n-m\choose l} \frac{1}{w^l}
\; dw
\\ =
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w}
\frac{1+w - w^2-w^3}{1-w^4}
\sum_{k=0}^m {m\choose k} \frac{(-1)^k}{w^k}
\left(1+\frac{1}{w}\right)^{n-m}
\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{(1+w)^{n-m}}{w^{n-m+1}}
\frac{1+w - w^2-w^3}{1-w^4}
\sum_{k=0}^m {m\choose k} \frac{(-1)^k}{w^k}
\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{(1+w)^{n-m}}{w^{n-m+1}}
\frac{1+w - w^2-w^3}{1-w^4}
\left(1-\frac{1}{w}\right)^m
\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{(1+w)^{n-m} (w-1)^m}{w^{n+1}}
\frac{1+w - w^2-w^3}{1-w^4}
\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{(1+w)^{n-m} (w-1)^m}{w^{n+1}}
\frac{1+w}{1+w^2}
\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{(1+w)^{n-m+1} (w-1)^m}{w^{n+1}}
\frac{1}{1+w^2}
\; dw.$$
Now checking on the residue at infinity by a circular contour of
radius $R$ we obtain a term of order $2\pi R \times
R^{n+1}/R^{n+1}/R^2$ which is $2\pi/R$ and vanishes so that residue is
zero. This leaves for the sum the residues at $w=\pm i$ and we get
(remember to flip the sign at the end)
$$\exp(-(n+1)\pi i/2)
\sqrt{2}^{n+1} \frac{1}{2i} \exp(+(n-m+1)\pi i/4)\exp(+3m\pi i/4) \\
- \exp(+(n+1)\pi i/2)
\sqrt{2}^{n+1} \frac{1}{2i} \exp(-(n-m+1)\pi i/4)\exp(-3m\pi i/4)
\\ = \sqrt{2}^{n+1} \sin(-(2n+2)\pi/4 +(n-m+1)\pi/4 + 3m\pi/4).$$
This yields $$- \sqrt{2}^{n+1} \sin((2m-n-1)\pi/4)$$ or
$$\bbox[5px,border:2px solid #00A000]{
\sqrt{2}^{n+1} \sin((n+1-2m)\pi/4).}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2163095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the constant term in the expansion of $\left(x^2 + x + \frac{1}{x} + \frac{1}{x^2}\right)^{15}$ Find the constant term in the expansion of $$\left(x^2 + x + \frac{1}{x} + \frac{1}{x^2}\right)^{15}$$
I know the answer is
$$ \sum _{k = 0}^{5} \binom{15}{5+k}\binom{15}{3k} =68974906 $$
| $$
\begin{align}
\left[x^0\right]\left(x^2+x+\frac1x+\frac1{x^2}\right)^{15}
&=\left[x^0\right]\left(\frac1{x^{30}}(x+1)^{15}\left(x^3+1\right)^{15}\right)\\
&=\left[x^{30}\right]\left((x+1)^{15}\left(x^3+1\right)^{15}\right)\\
&=\left[x^{30}\right]\sum_{k,j}\binom{15}{j}\binom{15}{k}x^{3k+j}\\
&=\sum_{3k+j=30}\binom{15}{j}\binom{15}{k}\\
&=\sum_{k=5}^{10}\binom{15}{30-3k}\binom{15}{k}\\
&=\sum_{k=0}^5\binom{15}{15-3k}\binom{15}{k+5}\\
&=\sum_{k=0}^5\binom{15}{3k}\binom{15}{k+5}\\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2163251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Checking Answers... Is it possible to help me to verify my answer to the questions below?
*
*If you shuffle a standard deck of $52$ cards, what is the probability that the top $2$ cards are both "5"?
ANS: $\frac 4{52} \times \frac 3{51} = \frac1{221}$
*You roll $2$ dice. You get $\$100$ if at least one of them is $6$, or sum of them is $6$. How much do you think this game is worth?
ANS: $(\frac{11}{36} + \frac5{36}) \times 100 = \frac 49\times 100 = \frac{400}9\approx \$44.44$
*A train is approaching a bridge. You are at the $\frac14$ position of the bridge. Running both ways can rescue you (both ways can get you safe, and just before hit by the train). What is the speed ratio between you and the train?
ANS:
Suppose the man has speed of $1$ m/s, train is $x$ m/s,
$\frac{x+4}3=\frac x1$
$x=2$
| *
*Your first answer is correct.
*At least one 6. Means 6 on one die or 6 on both dice.
Probability = $\frac {1}{6} \times \frac {5}{6} + \frac {5}{6} \times \frac {1}{6} + \frac {1}{6} \times \frac {1}{6}$
$ = \frac {5}{36} + \frac {5}{36} + \frac {1}{36}$
$ = \frac {5+5+1}{36} = \frac {11}{36}$
Sum of numbers is 6. Cases are (1,5) , (2,4), (3,3), (4,2), (5,1).
Probability = $\frac {5}{36}$
Total probability = $\frac {11}{36} \times \frac {5}{36}$
So your answer 2 is also correct.
*Question is not enough clear. We have many doubts like how apart train when man start running.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2164666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to show that for $n \in \mathbb{N}$, $n^2 + 4n - 8$ is square if and only if $n=2$ Let $n$ be a positive integer which is bigger than $1$.
How to show that $n^2 + 4n - 8$ is not square if $n \neq 2$ ?
| Assume there is such $n$ that $n>1$ and $n \neq 2$, yet $n^2+4n-8$ is a square, which implies $n>2$.
Then $$(n+2)^2=n^2+4n+4>n^2+4n-8>n^2$$
Since $n>2$. So $n^2+4n-8$ is a square number stuck between $(n+2)^2$ and $n^2$, which means it can only be $(n+1)^2$. So we have$$n^2+4n-8 =(n+1)^2=n^2+2n+1 \implies n=\frac{9}{2}$$
This is a contradiction, since $n \in \mathbb{N}$. So we can only have $n=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2164771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the limit without L'hopital rule $\lim_{ x \to 1}\frac{1-\cot(\frac{π}{4}x)}{\sin πx}=$? Find the limit without L'hopital rule
$$\lim_{ x \to 1}\frac{1-\cot(\frac{π}{4}x)}{\sin πx}=?$$
My Try:
$$1-\cot( \frac{π}{4}x)=1-\frac{1}{\tan( \frac{π}{4}x)}=\frac{\tan( \frac{π}{4}x)-1}{\tan( \frac{π}{4}x)}\\\sin (πx)=\sin (\pi-\pi x)=-\sin \pi(x-1)\\\lim_{ x \to 1}\frac{1-\cot(\frac{π}{4}x)}{\sin πx}=\lim_{ x \to 1}\frac{\frac{\tan( \frac{π}{4}x)-1}{\tan( \frac{π}{4}x)}}{-\sin \pi(x-1)}\\u=x-1⇒x=u+1\\\lim_{ u \to 0}\frac{\frac{\tan( \frac{π}{4}(u+1))-1)}{\tan( \frac{π}{4}(u+1))}}{-\sin \pi u}$$
now ?
| An alternate method -
Convert $1 - \cot{(\dfrac \pi4x)}$ to $\dfrac {\sin {(\dfrac \pi4x)}-\cos({\dfrac \pi4x})}{\sin (\dfrac \pi4x)}$.
Multiply divide by ${\sin {(\dfrac \pi4x)}+\cos({\dfrac \pi4x})}$ and simplify using the formulae
$\sin (2x) = 2\cos(x)\sin(x)$ and $\cos(2x) = \cos^2(x)-\sin^2(x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2165038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
} |
Vieta's Formula - Hard Problem
Let $f(x)=(x^2+6x+9)^{50}-4x+3$, and let $r_1,r_2,\ldots,r_{100}$ be the roots of >$f(x)$.
Compute $(r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}$.
How would I compute this? How to factor it?
| We begin by observing that $x^2+6x+9 = (x+3)^2$
Denote $f(x) = (x^2+6x+9)^{50} - 4x +3 = (x+3)^{100}-4x+3$
$\{r_i\}_{1\leq i \leq 100}$ are roots of this polynomial so $f(r_i)= (r_i+3)^{100}-4r_i+3 = 0$ $\forall 1 \leq i \leq 100$
By Vieta's formula, the sum of roots is negative of coefficient of $x^{99}$ term in $f(x)$ which by Binomial Expansion turns out to be $-\binom{100}{1}3^1 = -300$
Now, we just need to make some manipulations, as follows:
$$(r_i+3)^{100} - 4r_i + 3 = 0 \implies (r_i+3)^{100} = 4r_i - 3$$
Therefore,
$$\sum_{i=1}^{100} (r_i+3)^{100} = \sum_{i=1}^{100} 4r_i - 3 = 4 \left(\sum_{i=1}^{100} r_i\right) -300 = -1500$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2165329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Proving an inequality by mathematical induction I'm trying to solve a problem with inequalities using mathematical induction but I am stuck halfway through the process.
The problem: Use mathematical induction to establish the inequality -
$(1 + \frac{1}{2})^n \ge 1 + \frac{n}{2}$ for n $\in \mathbb{N}$
Steps
1) $n = 1$, $(1 + \frac{1}{2})^1 \ge 1 + \frac{1}{2}$ is TRUE
2) $n = k$, assume that $(1 + \frac{1}{2})^k \ge 1 + \frac{k}{2}$ for n $\in \mathbb{N}$
3) Show the statement is true for $k + 1$
$(1 + \frac{1}{2})^{k+1}$ = $(1 + \frac{1}{2})^k * (1 + \frac{1}{2})$
$\ge$ $(1 + \frac{k}{2}) * (1 + \frac{1}{2})$ - using the assumption in step $2$
My question is, how do I continue this problem? Or did I go wrong somewhere? I just can't figure out what the next step is.
| Continue expanding the product.
$$(1 + \frac{1}{2})^{k+1} =(1 + \frac{1}{2})^k \cdot (1 + \frac{1}{2}) \ge (1+\frac{k}2)(1 + \frac{1}{2}) = 1+\frac{k}2 + \frac12+\frac{k}{4}>1+\frac{k+1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2165958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Finding maximum of $x+y$ Let x and y be real numbers satisfying $9x^{2} + 16y^{2} = 1$. Then $x + y$ is maximum when
a. $y = \frac{9x}{16}$
b. $y = −\frac{9x}{16}$
c. $y = \frac{4x}{3}$
d. $y = −\frac{4x}{3}$
| By C-S $$\frac{25}{144}=\frac{1}{9}+\frac{1}{16}=\left(\frac{1}{9}+\frac{1}{16}\right)(9x^2+16y^2)\geq(x+y)^2,$$
which gives $x+y\leq\frac{5}{12}$.
The equality occurs, when $\left(\frac{1}{3},\frac{1}{4}\right)||(3x,4y)$, id est, for a.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 4
} |
How do you factor this equation that has square roots involved? I'm trying to solve this equation for
$$\sqrt {16+l^2} + \sqrt {4+l^2} = {nv \over f}$$
I know what ${nv \over f}$ is, but no matter what I do I keep getting a weird answer. I've resulted to using Wolfram Alpha, but I'd really like to know how to do this myself.
| Square-rearrange-square again as garserdt216 outlined totally works. I'm not going to solve your problem but here see it applied to an equivalent generalized problem
$$\sqrt{x + a} + \sqrt{x + b} = c$$
Square
$$(x + a) + 2\sqrt{(x + a)(x + b)}+ (x + b) = c^2$$
Isolate square root
$$\sqrt{(x + a)(x + b)} = \frac{c^2 - a - b}{2} - x$$
Square again
$$(x + a)(x + b) = \left (\frac{c^2 - a - b}{2} - x \right )^2$$
Simplify
$$x^2 + (a + b)x + ab = x^2 + (a + b - c^2)x+ \left (\frac{c^2 - a - b}{2}\right)^2 $$
Simplify some more
$$x = \frac{1}{c^2}\left ( \left (\frac{c^2 - a - b}{2}\right)^2 - ab \right ) = \frac{(c^2 - a - b)^2 - 4ab}{4c^2}$$
(Check for consistency with original equation.)
EDIT: Thanks to @Claude Leibovici for pointing out some errors which I've now hopefully corrected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2167813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Prove by mathematical induction Prove the following statement by using mathematical induction: $3+8+ ··· + (n^2 − 1) = \frac{1}{6}n(n − 1)(2n + 5)$, for all integers $n ≥ 2$.
After making a common factor $\frac{1}{6}k$ and expanding I'm left with $2k^2+9k-5 = 2k^2+9k+7$. Where am I going wrong?
|
After making a common factor $\frac{1}{6}k$ and expanding I'm left with $2k^2+9k-5 = 2k^2+9k+7$.
It looks like you lost a term somewhere along the way. Compare to:
$$
\require{cancel}
\begin{align}
3+8+ ··· + (n^2 − 1)+\left((n+1)^2-1\right) &= \frac{1}{6}n(n − 1)(2n + 5) \;+\; n^2+2n + \bcancel{1}-\bcancel{1} \\
&= \frac{1}{6}n(2n^2+3n-5) + n(n+2) \\
&= \frac{1}{6}n(2n^2+3n-5+6n+12) \\
&= \frac{1}{6}n(2n^2+9n+7) \\
&= \frac{1}{6}n(n+1)(2n+7) \\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2168857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Sum of three variables given two equations Given $$x^2+y^2+z^2=121$$
$$x\sqrt{11} + 4y + z\sqrt{22}=77$$
Find $$ \frac{\sqrt{11} + 4 + \sqrt{22}}{x+y+z} $$
I tried to plug in something for z at first, since x and y should have unique values for every value of z, but that didn't seem to work.
The answer is 7/11, which is clearly the second equation divided by the first but I don't understand how or why that would lead the final expression.
| HINT:
$$49\cdot121-77^2=(x\sqrt{11}+4y+z\sqrt{22})^2-(x^2+y^2+z^2)(11+16+22)$$
$$\iff0=(4x-\sqrt{11}y)^2+(\sqrt{22}x-\sqrt{11}z)^2+(\sqrt{22}y-4z)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Prove that BF + CE = BC In $\triangle ABC$ , $\angle A = 60$
$~BE$ is the bisector of $\angle B$ and $~A-E-C.$
$~CF$ is the bisector of $\angle C$ and $~A-F-B$
Prove that $~BF + CE = BC$
| let $AB = c, BC = a, CA = b$. by bisector, we have $\frac{CE}{AE} = \frac{a}{c}, \frac{BF}{AF} = \frac{a}{b}$.
by equal ratios, we have $\frac{CE}{AE+CE} = \frac{a}{a+c}, \frac{BF}{AF+BF} = \frac{a}{a+b}$, as $\frac{CE}{b} = \frac{a}{a+c}, \frac{BF}{c} = \frac{a}{a+b}$, so $CE = \frac{ab}{a+c}, BF = \frac{ac}{a+b}$, then $CE+BF = \frac{a(ab+ac+b^2+c^2)}{(a+b)(a+c)}$.
but $\angle A = 60^\circ$, so $b^2+c^2-2bc\cos60^\circ = a^2$, then $b^2+c^2 = a^2+bc$, so $CE+BF = \frac{a(a^2+ab+ac+bc)}{(a+b)(a+c)} = \frac{a(a+b)(a+c)}{(a+b)(a+c)} = a = BC$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Find $\lim\limits_{x\rightarrow 0} \frac{(\cos{x})^2}{\sin{(x^2)}}-\frac{1}{x^2}$
Find
$$\lim_{x\rightarrow 0} \frac{(\cos{x})^2}{\sin{(x^2)}}-\frac{1}{x^2}$$
I suppose I have to use L'Hopital's rule here but how?
Edit: I cannot use Taylor yet.
| Observe
\begin{align}
\frac{(\cos x)^2}{\sin (x^2)} = \frac{(1-\frac{1}{2!}x^2+\frac{1}{4!}x^4+\dots)^2}{x^2-\frac{1}{3!}x^6+\frac{1}{5!}x^{10}-\ldots}=\frac{1-x^2+\ldots}{x^2(1-\frac{1}{3!}x^4+\frac{1}{5!}x^8-\ldots)}
\end{align}
which means
\begin{align}
\lim_{x\rightarrow 0}\left(\frac{(\cos x)^2}{\sin (x^2)}-\frac{1}{x^2}\right) = -1.
\end{align}
Edit: Rewrite
\begin{align}
\frac{\cos^2x}{\sin(x^2)}-\frac{1}{x^2} = \frac{1-\sin^2 x}{\sin(x^2)}-\frac{1}{x^2}= \left(\frac{1}{\sin(x^2)}-\frac{1}{x^2}\right)-\frac{\sin^2x}{\sin(x^2)}
\end{align}
then observe
\begin{align}
\lim_{x\rightarrow 0}\frac{\sin^2 x}{\sin(x^2)}=\lim_{x\rightarrow 0}\frac{\sin x\cos x}{x\cos(x^2)} = \lim_{x\rightarrow 0}\frac{\sin x}{x}\cdot \frac{\cos x}{\cos(x^2)} = 1.
\end{align}
I will leave it as an exercise to show that
\begin{align}
\lim_{x\rightarrow 0}\left( \frac{1}{\sin(x^2)}-\frac{1}{x^2}\right)=0.
\end{align}
| {
"language": "en",
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How do I evaluate $\int \frac{x^4}{\sqrt{4-x^2}}dx$ using trig substitution? Here is what I have so far:
Let $x = 2\sin\theta$, and $dx=2\cos\theta \mathrm d\theta$
then the new integral is $$\int \frac{2^4\sin^4\theta}{\sqrt{4-4sin^2\theta}}2\cos\theta \mathrm d\theta$$
Then using the identity $1-\sin^2x = \cos^2x$ we can substitute $\cos^2\theta$ and simplify,
$$8\int \sin^4\theta d\theta$$
I then use the formula of $\int \sin^nudu= -\frac{1}{n}\sin^{n-1}u\cos + \frac{n-1}{n}\int \sin^{n-2}udu$, and I end up getting $$4(\theta-\sin^3\theta \cos\theta) +C$$
Is this correct?
How do I go about re-substituting $x$ into the equation?
Thanks in advance!
| Also if $x=2\sin y$
WLOG $-\dfrac\pi2\le y\le\dfrac\pi2\implies\cos y\ge0$
$(2\sin^2y)^2=(1-\cos2y)^2=1-2\cos2y+\cos^22y$
and $2\cos^22y=1+\cos4y$
Now $\sin y=\dfrac x2,\cos y=+\dfrac{\sqrt{4-x^2}}2$ and $y=\arcsin\dfrac x2$
| {
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} |
$P(x) = x^4 + ax^3 +bx^2 +cx+d$ Let $P(x) = x^4 + ax^3 +bx^2 +cx+d$ where $a, b, c, d$ are integers.
$P(x)$ is divided by $x-2012, x-2013, x-2014, x-2015, x-2016$ and has the remainders $24, -6, 4, -6, 24$ respectively.
What is the remainder when $P(x)$ is divided by $x-2017$ ?
Is my answer correct ?
$P(2012) = 24, \;P(2013) = -6, \;P(2014) = 4, \;P(2015) = -6, \;P(2016) = 24$
By Lagrange interpolation,
$P(2017) = \sum{24\cdot\frac{(2017-2013)(2017-2014)(2017-2015)(2017-2016)}{(2012-2013)(2012-2014)(2012-2015)(2012-2016)}} = 24+30+40+60+120=274$
| By symmetry at $x=2014$ we can write $\ P(x) = 4 + a(x\!-\!2014)^2 + b(x\!-\!2014)^4$.
Evaluation implies that $\ a=-15,\ b=5\ $ so $\,P(2017)=4-15(9)+5(81)=274$
| {
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Right triangle : Find $\angle{BPQ}$. $\triangle ABC, \angle{ABC} = 90^{\circ}$.
Let $P$ be the point on $BC$ such that $2 \cdot \angle{BAP} = \angle{CAP} = 14 ^{\circ}$
and $Q$ be the point on $AB$ such that $\angle{BCQ} = 23 ^{\circ}$.
Find $\angle{BPQ}$.
My work :
$\angle{APB} = 83 ^{\circ}$
$\angle{BQC} = 67 ^{\circ}$
$\angle{ACQ} = 46 ^{\circ}$
Let $QC \cap AP= U$, $\angle{QUP} = 120 ^{\circ}$
| Again, no need for trigonometric formulas and clumsy calculations.
Let $D$ be the intersection point of $AP$ and $CQ$. By direct angle chasing, one can calculate that
$$\angle \, ACD = \angle \, ACQ = 46^{\circ} = 2 \cdot 23^{\circ} = 2 \cdot \angle \, BCQ = 2 \cdot \angle \, PCD$$
Moreover, observe that
$$\angle \, CAD = \angle \, CAP = 14^{\circ} = 2 \cdot 7^{\circ} = 2 \cdot \angle \, BAP = 2 \cdot \angle \, QAD$$
Let $I$ be the intersection point of the interior angle bisectors through vertices $A$ and $C$ of triangle $ACD$. Therefore, $DI$ is the interior angle bisector of angle $\angle \, ADC$. Again, some direct angle chasing leads to the conclusion that $\angle \, ADC = 120^{\circ}$ and $\angle \, QDA = \angle \, PDC = 60^{\circ}$.
Now, observe that
$$\angle \, QAD = 7^{\circ} = \frac{1}{2} \cdot 14^{\circ} = \frac{1}{2} \cdot \angle CAD = \angle \, IAD$$
$$\angle \, PCD = 23^{\circ} = \frac{1}{2} \cdot 46^{\circ} = \frac{1}{2} \cdot \angle \, ACD = \angle \, ICD$$
Furthermore, $$\angle \, IDA = \frac{1}{2} \cdot \angle \, ADC = \frac{1}{2} \cdot 120^{\circ} = 60^{\circ} = \angle \, QDA$$
$$\angle \, IDC = \frac{1}{2} \cdot \angle \, ADC = \frac{1}{2} \cdot 120^{\circ} = 60^{\circ} = \angle \, PDC$$ Consequently the triangles $ADQ$ and $ADI$ are congruent, as well as triangles $CDP$ and $CDI$ are also congruent. Therefore, $DQ = DI$ and $DP = DI$. Hence $DQ = DI = DP$ and so triangle $DPQ$ is isosceles with $\angle \, PDQ = 120^{\circ}$. Thus
$$\angle \, QPA = \angle \, QPD = 30^{\circ}$$ By angle chasing, $\angle \, BPA = 83^{\circ}$ and for that reason
$$\angle \, BPQ = \angle \, BPA - \angle \, QPA = 83^{\circ} - 30^{\circ} = 53^{\circ}$$
| {
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"source": "stackexchange",
"question_score": "3",
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differentiation under integral sign How do you find:
$$\int_0^\pi\frac{dx}{(5+3\cos x)^3}$$
and
$$\int_0^\pi\frac{\sin^2x }{(5+3\cos x)^3}dx$$
by differentiating under the integral sign?
I know you can get the result by using $\tan\left(\frac{x}{2}\right)$ substitution but I need to solve it via Feynman's Trick.
I am given the general result that:
$$\int_0^\pi\frac{dx}{a+b\cos x}= \frac{\pi}{\sqrt{a^2-b^2}}$$
| $f(a) = \int_0^\pi\frac{dx}{(a+b\cos x)} = \frac {\pi}{\sqrt {a^2 - b^2}}\\
\frac {df}{da} = \int_0^\pi-(a+b\cos x)^{-2} dx = -a\pi(a^2 - b^2)^{-\frac 32}\\
\frac {d^2f}{da^2} = \int_0^\pi 2(a+b\cos x)^{-3}dx = \pi(-(a^2 - b^2)^{-\frac 32}+3a^2(a^2 - b^2)^{-\frac 52})\\
\int_0^\pi (a+b\cos x)^{-3}dx = \frac \pi2 (-(a^2 - b^2)^{-\frac 32}+3a^2(a^2 - b^2)^{-\frac 52})$
$\\$
$g(b) = \int_0^\pi (a+b\cos x)^{-1} dx = \pi(a^2 - b^2)^{-\frac 12}\\
g'(b) = \int_0^\pi -(\cos x)(a+b\cos x)^{-2} dx = b\pi(a^2 - b^2)^{-\frac 32}\\
\frac 12 g''(b) = \int_0^\pi (\cos^2 x)(a+b\cos x)^{-3} dx = \frac \pi2 ((a^2 - b^2)^{-\frac 32}+3b^2(a^2 - b^2)^{-\frac 52})\\
\int_0^\pi (\sin^2 x)(a+b\cos x)^{-3} dx = \frac 12(f''(a) - g''(b))$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to calculate limit $S_n=\sum\limits_{k=1}^n\sin\Big( \dfrac{k\sqrt{k}}{n^2\sqrt{n}}+\dfrac{1}{n^2}\Big)$ How do I calculate the following limit:
$$\lim\limits_{n \to +\infty} S_n=\lim\limits_{n \to +\infty}\sum\limits_{k=1}^n\sin\Big( \dfrac{k\sqrt{k}}{n^2\sqrt{n}}+\dfrac{1}{n^2}\Big) = \text{?}$$
I think that you need to use Riemann sum, but I don't understand how to get rid of the sine.
Please provide a hint (and not the full solution).
| Using Taylor's theorem with Lagrange form of the remainder, for any $x\in \mathbb R$, $$\sin(x) = x -\frac{x^3}6 \cos(\xi_x)$$
Hence the inequality $$|\sin(x)-x|\leq \frac{|x|^3}6$$
Note next that $\displaystyle \sum_{k=1}^n \left(\frac{k\sqrt k}{n^2\sqrt n}+\frac 1{n^2}\right) = \frac 1n +\underbrace{\frac 1n \sum_{k=1}^n \frac kn \sqrt{\frac kn}}_{\text{Riemann sum}}$
and
$$\begin{align}\left| \sum_{k=1}^n \sin\left(\frac{k\sqrt k}{n^2\sqrt n}+\frac 1{n^2}\right) - \sum_{k=1}^n \left(\frac{k\sqrt k}{n^2\sqrt n}+\frac 1{n^2}\right) \right|
&\leq \sum_{k=1}^n \left|\sin\left(\frac{k\sqrt k}{n^2\sqrt n}+\frac 1{n^2}\right) - \left(\frac{k\sqrt k}{n^2\sqrt n}+\frac 1{n^2}\right)\right| \\
&\leq \sum_{k=1}^n \frac 16 \left(\frac 1n + \frac 1{n^2} \right)^3 \\
&\leq \frac 16 \sum_{k=1}^n \left( \frac 2{n}\right)^3\\
&\leq \frac 43 \frac 1{n^2} \to 0
\end{align}$$
Therefore, both sum have the same limit, that is $\displaystyle \int_0^1 t\sqrt t dt$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Determine the primes p for which $6\in Q_p$(where $Q_p$ denotes quadratic residue) Background This is from Jones Elementary Number theory problem 7.12. I only want the case where $6\in Q_p$ to be worked although the original problem asks for -3,5,6,7,10,169.
Using Legendre's notation and the Quadratic reciprocity law I get:
\begin{align}
\textrm{If }P \equiv & 1 \mod 4&\\
\left(\frac{2}{p}\right)\left(\frac{3}{p}\right) &=\\
\left(\frac{p}{2}\right)\left(\frac{p}{3}\right) &\implies p \equiv 1 \mod 2 \\
&\quad \quad \;\; p \equiv 1 \mod 3 \\
\hline
Thus&\implies p\equiv 1 \mod 6\\
\end{align}
$$
\left.
\begin{array}{l}
p\equiv 1 \mod 4\\
p \equiv 1 \mod 6
\end{array}
\right\}\bbox[5px,border:1px solid red]{p \equiv 1,13 \mod 24}
$$
\begin{align}
\textrm{If }P \equiv & 3 \mod 4&\\
\left(\frac{2}{p}\right)\left(\frac{3}{p}\right) &=\\
\left(\frac{p}{2}\right)\left(-\frac{p}{3}\right) &\implies p \equiv 1 \mod 2 \\
&\quad \quad \;\; p \equiv 2 \mod 3 \\
\hline
Thus&\implies p\equiv 5 \mod 6\\
\end{align}
$$
\left.
\begin{array}{l}
p\equiv 3 \mod 4\\
p \equiv 5 \mod 6
\end{array}
\right\}\bbox[5px,border:1px solid red]{p \equiv 11,23 \mod 24}
$$
The text solution is $p \equiv \pm1 \text{ or } \pm 5 \mod 24$. What did I do wrong?
| The Quadratic Reciprocity law only applies to odd primes $p,q$, and $2$ is not an odd prime. In fact we have
$$\left(\frac{2}{p}\right)=(-1)^{(p^2-1)/8}$$
which does not just depend on the value of $p$ mod $4$. So you will have to split the problem up into more cases.
| {
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Prove the seq $\{\frac{x^4-1}{x^4+x-6}\}$ converges and prove using $N-\epsilon$ proof Prove $$\{\dfrac{x^4-1}{x^4+x-6}\}$$ converges and prove using $N-\epsilon$ proof.
I see that this sequence approach $1$.
So,
I want to show:
$$\forall \epsilon >0, \exists N>0, s.t, n>N \to |\dfrac{x^4-1}{x^4+x-6}-1| < \epsilon$$
I started this proof and was trying to find an ideal $N$.
$$|\dfrac{x^4-1}{x^4+x-6}-1| < \epsilon$$
$$|\dfrac{x^4-1}{x^4+x-6}-\frac{(x^4+x-6)}{x^4+x-6}| < \epsilon$$
$$|\dfrac{-1-x+6}{x^4+x-6}| = |\dfrac{5-x}{x^4+x-6}| = |\dfrac{x-5}{x^4+x-6}| < \epsilon$$
Now I wanted to do some bounding so I can make the function larger:
$$|\dfrac{x-5}{x^4+x-6}| \leq |\dfrac{x-5}{x^4}| \leq |\dfrac{2x}{x^4}| \leq |\dfrac{2}{x^3}|$$
but this is not the case when I graph it. Where is my bounding wrong?
As you can see, the red graph is actually bigger then the blue graph. Why?
| $|\dfrac{x-5}{x^4+x-6}| < \epsilon$
Now start bounding $N$
Suppose $N> 6$
$|x-5| < N\\
|x^4+x-6| > N^4$
$|\dfrac{x-5}{x^4+x-6}| < \frac 1{N^3}$
$\frac 1{N^3} = \epsilon\\
N = \max (6,\frac {1}{\sqrt[3]{\epsilon}})$
| {
"language": "en",
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How to compute characteristic polynomial?
Given a square matrix $\mathbf{X}$, the characteristic polynomial of $\mathbf{X}$ is defined as
$$P_X(t) = \det (t \mathbf{M} - \mathbf{X}).$$(The matrix $\mathbf{M}$ is the identity matrix that has the same dimensions as $\mathbf{X}$. Note that when we expand this determinant, we obtain a polynomial in $t$.)
(a) Compute the characteristic polynomial of the matrix
$$\mathbf{X} = \begin{pmatrix} x & y \\ w & z \end{pmatrix}.$$Express your answer in the form $p_2 t^2 + p_1 t + p_0$.
(b) For the polynomial in part (a), find
$$p_2 \mathbf{X}^2 + p_1 \mathbf{X} + p_0 \mathbf{M}.$$
I am a bit lost where to start. Am I supposed to express $p_2$, $p_1$, and $p_0$ as values of $x$, $y$, $w$, and $z$? Does $t$ mean any constant? Even making these assumptions, I am not sure where to start / how to do. Any help would be appreciated!
| A:
We are given the the Polynomial:
$$P_X(t) = \det (t \mathbf{I} - \mathbf{X}).$$
Substituting $\mathbf{X}$ and the identity matrix $\mathbf{I}$ yields:
$$P_X(t) = \det (t \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} x & y \\ w & z \end{pmatrix})$$ $$\Rightarrow P_X(t) = \det (\begin{pmatrix} t & 0 \\ 0 & t \end{pmatrix} - \begin{pmatrix} x & y \\ w & z \end{pmatrix})$$ $$\Rightarrow P_X(t) = \det (\begin{pmatrix} t-x & -y \\ -w & t-z \end{pmatrix})$$
Now we can solve the determinant:
$$\Rightarrow P_X(t) = \det (\begin{pmatrix} t-x & -y \\ -w & t-z \end{pmatrix})$$$$\Rightarrow P_X(t) = (t-x)(t-z) - yw$$$$\Rightarrow P_X(t) = t^2 -xt -zt +xz - yw$$$$\boxed{ P_X(t) = t^2 -(x+z)t + xz - yw \quad \text{where} \quad p_2 = 1 \quad p_1 = -(x+z) \quad p_0 = xz - yw}$$
B:
Here we need to use $p_2$, $p_1$, and $p_0$ from problem A.
We know $p_2 = 1$, $p_1 = -(x+z)$, and $p_0 = xz - yw$. Subsituting this into what we need to find, we yield:
$$p_2 \mathbf{X}^2 + p_1 \mathbf{X} + p_0 \mathbf{I}.$$ $$\Rightarrow 1 \mathbf{X}^2 -(x+z)\mathbf{X} + (xz - yw) \mathbf{I}$$$$\Rightarrow \begin{pmatrix} x & y \\ w & z \end{pmatrix}^2 -(x+z)\begin{pmatrix} x & y \\ w & z \end{pmatrix} + (xz - wy) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$$$\Rightarrow \begin{pmatrix} x^2 + yw & yx+yz \\ wx+wz & z^2 +yw \end{pmatrix} + \begin{pmatrix} -xz-x^2 & -xy-yz \\ -xw-wz & -xz-z^2 \end{pmatrix} + \begin{pmatrix} xz - yw & 0 \\ 0 & xz - yw \end{pmatrix} $$ $$\Rightarrow \begin{pmatrix} x^2 + (-x - z) x + z x & x y + (-x - z) y + z y \\ x w + (-x - z) w + z w & z^2 + x z + (-x - z) z \end{pmatrix} $$ $$ \Rightarrow \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = \boxed{0}$$
| {
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"source": "stackexchange",
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Computing the Fourier series of $f(\theta)=(\pi-\theta)^2/4$ for $0\leq\theta\leq{2\pi}$ According to p. 36 of Stein and Shakarchi's Fourier Analysis, the Fourier series is $\frac{\pi^2}{12} + \sum_{n=1}^{\infty} \frac {\cos{n\theta}}{n^2}$.
But to calculate the Fourier coefficients, I computed $\int_{-\pi}^{\pi}\frac{(\pi-\theta)^2}{4}e^{-in\theta}d\theta = \frac {1} {4} \Bigg[\frac{4\pi^2e^{in\pi}}{in}-\frac{2}{in}\int_{-\pi}^{\pi}e^{-in\theta}(\pi-\theta)d\theta\Bigg]=\frac {1} {4}\Bigg[\frac{4\pi^2e^{in\pi}}{in}-\frac{4\pi{e^{in\pi}}+2e^{-in\pi}-2e^{in\pi}}{n^2}\Bigg]=\frac{\pi^2ne^{in\pi}+\sin{n\pi}}{in^2}$
This means the Fourier series is $\sum_{n=-\infty}^{\infty}\frac{\pi^2ne^{in\pi}+\sin{n\pi}}{in^2}e^{in\theta}=\sum_{n=-\infty}^{\infty}\frac{\pi^2e^{in(\pi+\theta)}}{in}+\frac{e^{in(\theta-\pi)}}{2n^2}$, but I can't seem to get this to equal that simple thing from the book.
Can anyone help?
| The Fourier coefficients can be calculated as follows.
If $n=0$ then
\begin{align}
\hat{f}(0) &= \frac{1}{2\pi}\int_0^{2\pi}\frac{(\pi-\theta)^2}{4}d\theta\\
&=\frac{1}{8\pi}\int_{-\pi}^{\pi}u^2du\\
&=\frac{1}{8\pi}\cdot\frac{2\pi^3}{3}\\
&=\frac{\pi^2}{12}
\end{align}
If $n\neq0$ then
\begin{align}
\hat{f}(n) &= \frac{1}{2\pi}\int_0^{2\pi}\frac{(\pi-\theta)^2}{4}e^{-in\theta}d\theta\\
&=\frac{e^{-in\pi}}{8\pi}\int_{-\pi}^{\pi}u^2e^{-inu}du\\
&=\frac{e^{-in\pi}}{8\pi}\left[\frac{u^2e^{-inu}}{-in}- \frac{2ue^{-inu}}{(-in)^2}+\frac{2e^{-inu}}{(-in)^3}\right]_{u=-\pi}^{u=\pi}\\
&=\frac{1}{8\pi}\left[\frac{\pi^2}{in}(\underbrace{1-e^{-2\pi in}}_{0})+\frac{2\pi}{n^2}(\underbrace{1+e^{-2\pi in}}_{2})-\frac{2}{in^3}(\underbrace{1-e^{-2\pi in}}_{0})\right]\\
&=\frac{1}{2n^2}
\end{align}
Hence the Fourier series of $f$ is
$$
\frac{\pi^2}{12} + \sum_{\substack{n\in\mathbb{Z}\\n\neq0}}\frac{1}{2n^2}(\cos n\theta+i\sin n\theta) = \frac{\pi^2}{12} + \sum_{n=1}^{\infty}\frac {\cos{n\theta}}{n^2}
$$
Alternatively, the Fourier series of $f$ can be given in the following form:
$$
\frac{1}{2\pi}\int_0^{2\pi}f(\theta)\,d\theta+\sum_{n=1}^{\infty}a_n\cos(n\theta)+\sum_{n=1}^{\infty}b_n\sin(n\theta)
$$
where
$$
a_n:=\frac{1}{\pi}\int_0^{2\pi}f(\theta)\cos(n\theta)\,d\theta\quad(n\geq1)\\
b_n:=\frac{1}{\pi}\int_0^{2\pi}f(\theta)\sin(n\theta)\,d\theta\quad(n\geq1)
$$
In this case we note that $f$ is even hence $b_n=0$ for all $n$. Similar computations as those above lead directly to
$$
\frac{\pi^2}{12} + \sum_{n=1}^{\infty} \frac {\cos{n\theta}}{n^2}
$$
| {
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What is the solution to $x^2=y^2$? What is the solution to $x^2=y^2$? I know it seems simple but I seem to have got myself all confused. Is this correct?
$$x^2=y^2$$
$$\pm\sqrt{x^2}=\pm\sqrt{y^2}$$
$$\pm x=\pm y$$
$$\Rightarrow x_{1}=y,\quad x_{2}=-y, \quad -x_{3}=y, \quad -x_{4}=-y$$
$$\iff x_{1}=x_{4}=y,\quad x_{2}=x_{3}=-y $$
| $x^2 = y^2 \iff x^2 - y^2 = 0 \iff (x-y)(x+y) = 0$
A product is equal to $0$ if and only if one of its factors is equal to $0$
Thus:
$x^2 = y^2 \iff x - y = 0$ or $x + y =0 \iff x = -y$ or $x = y$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\int_0^1\frac{x\ln (1+x)}{1+x^2}dx=\frac{\pi^2}{96}+\frac{\ln^2 2}{8}$ We know that $$\int_0^1\frac{\ln (1+x)}{1+x^2}dx=\frac{\pi}8\ln 2,$$ but how about $$\int_0^1\frac{x\ln (1+x)}{1+x^2}dx?$$Prove that $$\int_0^1\frac{x\ln (1+x)}{1+x^2}dx=\frac{\pi^2}{96}+\frac{\ln^2 2}{8}.$$
| We will be using the following identities :
$$\int_0^1 x^{2n-1}\ln(1+x)dx=\frac{H_{2n}-H_n}{2n}\tag1$$
$$\sum_{n=1}^\infty x^n\frac{H_n}{n}=\frac12\ln^2(1-x)+\operatorname{Li}_2(x)\tag{2}$$
\begin{align}
I&=\int_0^1\frac{x\ln(1+x)}{1+x^2}\ dx=\sum_{n=1}^\infty(-1)^{n-1}\int_0^1x^{2n-1}\ln(1+x)\ dx\\
&\overset{(1)}{=}\sum_{n=1}^\infty(-1)^{n-1}\left(\frac{H_{2n}-H_n}{2n}\right)\\
&=\frac12\sum_{n=1}^\infty(-1)^n\frac{H_n}{n}-\sum_{n=1}^\infty(-1)^n\frac{H_{2n}}{2n}\\
&=\frac12\sum_{n=1}^\infty(-1)^n\frac{H_n}{n}-\Re\sum_{n=1}^\infty(i)^n\frac{H_n}{n}\\
&\overset{(2)}{=}\frac12\left(\frac12\ln^22+\operatorname{Li}_2(-1)\right)-\Re\left(\frac12\ln^2(1-i)+\operatorname{Li}_2(i)\right)\\
&=\frac12\left(\frac12\ln^22-\frac12\zeta(2)\right)-\left(\frac18\ln^22-\frac5{16}\zeta(2)\right)\\
&=\boxed{\frac18\ln^22+\frac1{16}\zeta(2)}
\end{align}
The proof of $(1)$ can be found here and the proof of $(2)$ follows from integrating the generating function $\sum_{n=1}^\infty H_n\ x^n=-\frac{\ln(1-x)}{1-x}$ .
| {
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First-Order Linear Equations I have the equation,
$$\frac{dy}{dx} - \frac{2y}{x} = x^2$$
and I have tried solving it using $e^{G(x)}$ but I get the answer $y(x) = x^4$ which isn't correct. I got the integration of $g(x)$ to be $-2ln|x| + C$ which seems to be right but implementing it in the equation didn´t work for me.
I got
$$\frac{dy}{dx} \frac{1}{x^2} - \frac{2}{x} \frac{1}{x^2} y = \frac{1}{x^2} x^2$$
and then
$$\frac{dy}{dx}\bigg(\frac{1}{x^2} y(x)\bigg) = x^2$$
An ideas?
| Here is your mistake:
$$\frac{dy}{dx}\cdot \frac{1}{x^2} - \frac{2}{x} \cdot \frac{1}{x^2} \cdot y = \frac{1}{x^2} \cdot x^2 $$
This leads to:
$$\frac{d}{dx}\left(\frac{1}{x^2} \cdot y(x)\right) = \color{red}{1}$$
Can you continue from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2183938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Area of an acute triangle, given two sides and an altitude
$12$, $13$, $15$ are the lengths (perhaps not in order) of two sides of an acute-angled triangle and of the height over the third side of triangle. Find the area of the triangle (no calculator allowed).
| Misread the question.
Draw the triangle and the height. Denote by $x,y$ the two segments determined by the height on the triangle.
First, since the perpendicular to a line is smaller than any of the secants, the altitude must be 12.
You have
$$12^2+x^2=13^2 \Rightarrow x^2=13^2-12^2= (13-12)(13+12)=25 \\
12^2+y^2=15^2 \Rightarrow y^2=15^2-12^2= (15-12)(15+12)=3*=81 $$
Therefore the last side is $\sqrt{25}+\sqrt{81}=14$
Then the area is
$$\frac{12 * 14}{2}= 84$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_count": 5,
"answer_id": 3
} |
number of $\lbrace 1,2,...,53 \rbrace$ subsets with member summation divisible by $3$ In $\lbrace 1,2,3,...,53 \rbrace$ how many subsections do we have with this condition:
the summation of subset members must be divisible by $3$.
for example $\lbrace 1,2 \rbrace$ & $\lbrace 1,2,3,4,5 \rbrace$ count.
| The generating function for these is
$$\prod_{q=1}^{53} (1+z^q)$$
and with $\zeta = \exp(2\pi i/3)$ we obtain for the answer
$$\frac{1}{3} \sum_{p=0}^2 \prod_{q=1}^{53} (1+\zeta^{pq})
= \frac{1}{3} 2^{53}
+ \frac{1}{3} \prod_{q=1}^{53} (1+\zeta^{q})
+ \frac{1}{3} \prod_{q=1}^{53} (1+\zeta^{2q}).$$
Now since $(1+\zeta)(1+\zeta^2)(1+\zeta^3)= (2+\zeta+\zeta^2) \times 2
= 2$ this becomes
$$\frac{1}{3} 2^{53}
+ \frac{1}{3} 2^{17} (1+\zeta)(1+\zeta^2)
+ \frac{1}{3} 2^{17} (1+\zeta^2)(1+\zeta)
= \frac{1}{3} 2^{53} + \frac{2}{3} 2^{17}
\\ = 3002399751667712.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2184905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Prove $1/2^1 + 2/2^2 +....+ n/2^n < 2$ using induction This is another problem with induction that I'm sure requires some "thinking outside the box" which is something I cannot apply with my way of solving these problems.
I need to now how I can complete the induction step and I need a solution using induction.
| Suppose
$\sum_\limits{k=1}^{\infty} k r^k$ converges (and it does if r<1)
$S - rS = \sum_\limits{k=1}^{\infty} (n+1 - n) r^k = \sum_\limits{k=1}^{\infty} r^k\\
S = \frac {r}{(1-r)^2}$
$\sum_\limits{k=1}^{\infty} k (\frac 12)^k = \frac {\frac 12}{(1-\frac 12)^2} = 2$
$\sum_\limits{k=1}^{n} k (\frac 12)^k < \sum_\limits{k=1}^{\infty} k (\frac 12)^k$
But that is not a proof by induction, is it?
Rather than prove that
$a_n = \frac 12 + \frac 24 \cdots + \frac {n}{2^n} < 2$ it will be easier if we try to prove something more restrictive.
Proposition:
$a_n = \frac 12 + \frac 24 \cdots + \frac {n}{2^n} = 2 - \frac {n+2}{2^n}$
Proof by induction:
base case:
n = 1
$\frac 12 = 2 - \frac {3}{2}$
Inductive hypothesis:
Suppose $a_n = \frac 12 + \frac 24 \cdots + \frac {n}{2^n} = 2 - \frac {n+2}{2^n}$
we will need to show:
$a_{n+1} <2-\frac {n+3}{2^{n+1}}$
$a_{n+1} = a_n + \frac {n+1}{2^{n+1}}\\
a_{n+1} = 2 - \frac {n+2}{2^n} + \frac {n+1}{2^{n+1}}$
Based on the inductive hypothesis
$a_{n+1} = $$2 - \frac {2(n+2)-(n+1)}{2^{n+1}}\\
2 - \frac {n+3}{2^{n+1}}\\
$
QED
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2185136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How compute $\int_0^{\infty} \frac{(t+n)^2t^2}{[1+(t+n)^2]^3} dt ?$ Put $$I_n= \int_0^{\infty} \frac{(t+n)^2t^2}{[1+(t+n)^2]^3} dt \ (n\in \mathbb N).$$ We note that for each fixed $n\in \mathbb N, $ $I_n$ is convergent.
[ For $t\geq 1, \frac{(t+n)^2t^2}{[1+(t+n)^2]^3} \leq \frac{1}{1+t^2}$, therefor $\int_1^{\infty} \frac{(t+n)^2t^2}{[1+(t+n)^2]^3} dt $ is convergent by comparison test, and near near origin no singularity. So, $I_n$ is convergent for each fixed $n$. ]
Question: Can we compute $I_n$? Is $\{I_n\}_{n\in \mathbb N}$ a bounded sequence in $\mathbb R$?
| $$ I_n = \int_0^\infty \frac{(t+n)^2t^2}{[1+(t+n)^2]^3} dt $$
$$ I_n = \int_0^\infty \frac{n^2+2nt+1}{(n^2+2nt+t^2+1)^3} + \frac{-n^2-2nt-2}{(n^2+2nt+t^2+1)^2} + \frac{1}{n^2+2nt+t^2+1} \, dt$$
using partial fraction expansion and from here you can integrate using $u$-substitutions to find
$$ I_n = \left[\frac{1}{8} \left( (n^2+3)\arctan(n+t) + \frac{n^3+n^2t+3n-5t}{n^2+2nt+t^2+1} - \frac{2(n^3 + n^2t + n -t) }{(n^2+2nt+t^2+1)^2} \right)\right]_{t=0}^{t=\infty}$$
take the limit as $t\to \infty$ and,
$$ I_n = \frac{\pi}{16}(n^2+3) - \frac{1}{8}(n^2+3)\arctan(n) -
\frac{n}{8} $$
and $\underset{n\to\infty}{\lim} I_n = 0$ and let $f : \mathbb{R} \to \mathbb{R}, \,n \mapsto \frac{\pi}{16}(n^2+3) - \frac{1}{8}(n^2+3)\arctan(n) -
\frac{n}{8}$,
$$ \frac{\partial f}{\partial n} = \frac{-(n^2+3)}{8(n^2+1)} + \frac{\pi n}{8} - \frac{n\arctan(n)}{4} - \frac{1}{8} < 0 \quad\forall n$$
So since $I_n = f \, \forall n\in \mathbb{N}$, $\frac{\partial f}{\partial n} < 0$, and $\underset{n\to\infty}{\lim} I_n = 0$ then $I_n$ has a lower bound of $y=0$.
In addition $ \underset{n\to -\infty}{\lim} I_n \to \infty $, so $I_n$ has no upper bound.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Evaluate $\int \frac{dx}{(b^2 + a^2x^2)^{3/2}}$ where $a,b \in \mathbb R^+$
Evaluate $$\int \frac{dx}{(b^2 + a^2x^2)^{3/2}}$$ where $a,b \in \mathbb R^+$
I let $(ax) = b\tan(\theta)$, then $dx = \frac{b}{a}\sec^2(\theta)$
$$= \int \frac{b\sec^2(\theta)}{ab^3(\sec^2(\theta))^{3/2}}$$
$$ = \frac{1}{ab^2} \int \frac{1}{\sec(\theta)}d\theta $$
$$= \frac{1}{ab^2}\sin(\theta) + C$$
$$= \frac{1}{ab^2} \frac{ax}{\sqrt{(ax)^2+b^2}} + C$$
Is this right? Other website did it differently not sure if its right
|
Yes, it is true!
You can always verify your integrals by differentiating them.
The integral can also be solved without using trigonometric functions.
Instead, substitute $ax=by$.
Then $$\int \frac {dx}{(b^2+a^2 x^2)^\frac 3 2} = \frac 1 {ab^2 } \int \frac {dy} {(1+y^2)^\frac 3 2}.$$
This integral can be solved as follows:
Observe that
$$
\frac 1 {(1+y^2)^\frac 3 2}=\frac {1+y^2}{(1+y^2)^\frac 3 2}-\frac {y^2} {(1+y^2)^\frac 3 2}
= \frac {1}{(1+y^2)^\frac 1 2}-\frac {y^2} {(1+y^2)^\frac 3 2}.
$$
Since $$\frac d {dy} \frac 1 {\sqrt{1+y^2}}= -\frac y {(1+y^2)^\frac 3 2}
$$
we can use integration by parts to get
$$
\int \frac {y^2} {(1+y^2)^\frac 3 2} \,dy
=-\int y \,\frac d {dy}\frac 1 {\sqrt{1+y^2}}\, dy
=-\frac y {\sqrt{1+y^2}} + \int \frac 1 {\sqrt{1+y^2}} \, dx.
$$
This implies
$$
\int \frac 1 {(1+y^2)^\frac 3 2} = \frac y {\sqrt{1+y^2}}
$$
Subsituting back and multiplying with the factor $\frac 1 {ab^2}$, we get the desired result
$$\int \frac {dx}{(b^2+a^2 x^2)^\frac 3 2} = \frac 1 {b^2 } \frac x {\sqrt{b^2+a^2x^2)}}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Basic Probability Question about dice with 112233 sides instead of 123456 We have two special $6$-sided dices. each of them has sides $112233$ instead of $123456$.
We roll them. what is the probability that sum of them is $5$? What is the probability that sum of them is odd?
My problem is that I don't know if $n(S) = 6 \times 6 = 36$ or $n(S) = 3 \times 3 = 9$. I want to know should I consider the same sides just one side or not? Is it right to say it is like we have a $3$-sided dice with $123$ on it or not?
| First consider it intuitively:
We can draw up a table showing all possible sums:
$$\begin{array}{|c|cccccc|}\hline&1&1&2&2&3&3\\
\hline
1&2&2&3&3&4&4\\
1&2&2&3&3&4&4\\
2&3&3&4&4&5&5\\
2&3&3&4&4&5&5\\
3&4&4&5&5&6&6\\
3&4&4&5&5&6&6\\
\hline
\end{array}$$
Therefore we can see that $P(S=5) = \frac{8}{36}=\frac 29$
And $P(S \text{ is odd}) = \frac{16}{36}=\frac 49$
Now consider the maths behind this:
We can say that $P(1)=P(2)=P(3)=\frac 26 =\frac 13$
For $S=5$, we must have one dice being $2$ and the other $3$, so we can say that
\begin{align}P(S=5)&=P((D_1=2\text{ and }D_2=3)\text{ or }(D_1=3\text{ and }D_2 =2))\\
&=(P(D_1=2)\times P(D_2=3)) + (P(D_1=3)\times P(D_2=2))\\
&=\left(\frac 13 \times \frac 13\right) + \left(\frac 13 \times \frac 13\right)\\
&=\frac 19 + \frac 19\\
&= \frac 29
\end{align}
Can you work out how to do it for $P(S\text{ is odd})$?
| {
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"url": "https://math.stackexchange.com/questions/2187719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate the following limits? I was reading a proof on the evaluation of $\int_0^\infty e^{-x^2}\ dx$ without advanced techniques and stumbled upon two limits that I can't seem to crack:
$$\lim_{m\to\infty}\left(\sqrt{m}\cdot\prod_{n=1}^m\frac{2n}{2n+1}\right)=\frac{\sqrt{\pi}}2$$
$$\lim_{m\to\infty}\left(\sqrt{m}\cdot\prod_{n=2}^m\frac{2n-3}{2n-2}\right)=\frac1{\sqrt{\pi}}$$
The proof does not go into detail on how these limits were obtained, and since I wanted to understand it completely, I thought this would be the best place to ask. I have not been exposed to infinite products (only summations) and therefore I do not know which rules to apply (I feel as if they are quite similar?). In both cases, I see that an indeterminate form $0\cdot\infty$ presents its self, therefore I am guessing Hospital would be a nice approach? Any help is appreciated! Also, my calculus book does not tackle infinite products, any suggestions on books that might give me a general outlook on the subject?
| Wallis's formula:
$$\frac{\pi}{2}=\prod_{n=1}^\infty \left[\frac{(2n)^2}{(2n+1)(2n-1)}\right].$$
Proof:
Weierstrass factorization of $\sin$ (You can find Euler's semi standard proof of this here) :
$$\sin(x)=x\prod_{n=1}^\infty\left(1-\frac{x^2}{n^2\pi^2}\right).$$
Plug in $x=\pi/2$ and play with the resulting fractions to get the desired result.
For your first product:
\begin{align*}
\prod_{n=1}^m\frac{2n}{2n+1}&=\frac{2\cdot 1}{2\cdot 1+1}\frac{2\cdot 2}{2\cdot 2+1}\frac{2\cdot 3}{2\cdot 3+1}\cdots \frac{2\cdot m}{2\cdot m+1}\\
&=2\cdot 1\frac{2\cdot 2}{2\cdot 1+1}\frac{2\cdot 3}{2\cdot 2+1}\cdots \frac{2\cdot m}{2\cdot (m-1)+1}\frac{1}{2m+1}\\
&=\frac{2}{2m+1}\prod_{n=2}^m\frac{2n}{2n-1}
\end{align*}
Thus:
\begin{align*}
\frac{\pi}{2}&=\lim_{m\rightarrow\infty}\prod_{n=1}^m \left[\frac{(2n)^2}{(2n+1)(2n-1)}\right]\\
&=\lim_{m\rightarrow\infty}\left(\prod_{n=1}^m\frac{2n}{2n+1}\right)\frac{1}{2}\left(\prod_{n=2}^m\frac{2n}{2n-1}\right)\\
&=\lim_{m\rightarrow\infty}\frac{1}{2}\frac{2m+1}{2}\left(\prod_{n=1}^m\frac{2n}{2n+1}\right)^2.
\end{align*}
Now just take the square-root of both sides and notice that $\sqrt{m}/\sqrt{\frac{2m+1}{2}}\rightarrow 1$
For the second question, try a similar trick by shifting the index $n\rightarrow n+2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Is there an elementary proof that $y^2=8x^4+1$ has no integral solution for $x\ge2$?
How can I prove that $y^2=8x^4+1$ has no integral solution with $x\ge 2$ with elementary methods ?
With elementary I mean using only modular arithmetic, the unique factorization theorem and the theory of quadratic residues modulo a prime $p$.
I tried various approaches , but I did not manage to prove the claim :
First of all, if $x\ge 2$, there is a prime factor $p$ dividing $x$.
$y^2-1=(y-1)(y+1)=8x^4$
So, if $y-1$ or $y+1$ is divisible by an odd prime, we have $p^4|y-1$ respective $p^4|y+1$. Does this lead to anywhere ?
$y^2-9=8x^4-8$ , which implies $(y-3)(y+3)=8(x-1)(x+1)(x^2+1)$. Can I make use of the fact that every odd prime $p$ dividing $x^2+1$ has the form $4k+1$ ?
Since the equation is closely related to the triangular numbers (The question whether a triangular number can be a fourth power leads to the given equation), I also studied the convergents of $\sqrt{2}$ (which satisfy the pell-equation $x^2-2y^2=1$).
The sequence $B_n$ of the positive integers, whose squares are triangular satisfy the recurrence relation $B_1=1$ , $B_2=6$ , $B_n=6B_{n-1}-B_{n-2}$. So, showing that no $B_n$ except $1$ is a perfect square would also finish the proof.
Any ideas ?
| The positive integer $y$ must be odd, and letting $z=(y-1)/2$ we get $z(z+1)=2x^4.$
The two consecutive integers $z$ and $z+1$ have no common prime factors, so
one is a fourth power and the other twice a fourth power.
Case 1: If $z=\ell^4$ and $z+1=2m^4$, then $\ell^4+1=2m^4$.
From my answer here,
we see that $\ell=1$ and so $y=3$ and $x=1$.
(This part uses the fact that $z^2=x^4-y^4$ has no solutions in non-zero integers.
This is Exercise 1.6 in Edwards's book on Fermat's Last Theorem.
The proof uses the representation of Pythagorean triples and infinite
descent.)
Case 2: If $z=2m^4$ and $z+1=\ell^4$, then $\ell^4-1=2m^4$.
Since $(\ell^2-1)(\ell^2+1)=2m^4$ and $\gcd(\ell^2-1,\ell^2+1)\leq 2$, one of the factors $\ell^2-1$ or $\ell^2+1$ is a fourth power, in particular a square.
The only two consecutive squares are $0$ and $1$, so we must have $\ell=1$.
This implies $z=0$ and so $y=1$ and $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2189941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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If $a\sin^2\theta+b\cos^2\theta=m, a\cos^2\phi+b\sin^2\phi=n, a\tan\theta=b\tan\phi$, prove that $\frac{1}{n}+\frac{1}{m}=\frac{1}{a}+\frac{1}{b}$
If $\, a\sin^2\theta+b\cos^2\theta=m$, $\, a\cos^2\phi+b\sin^2\phi=n$, $\, a\tan\theta=b\tan\phi$. Prove that
$$\frac{1}{n}+\frac{1}{m}=\frac{1}{a}+\frac{1}{b}$$
I've tried setting :
$$
\frac{a}{b}=\frac{\tan\theta}{\tan\phi}=k
\\ a=bk
\\
\tan\theta=k\tan\phi
$$
and substituting these in the equations. Can anyone tell me what I'm doing wrong ? A shorter path to the answer would also be appreciated.
| $$a\sin^2\theta+b\cos^2\theta=m\to a\tan^2 \theta+b=m\sec^2\theta\\
a\tan^2\theta +b=m(1+\tan^2\theta)\to \tan^2\theta=\frac{m-b}{a-m}$$
so, using the same idea
$$\tan^2\phi=\frac{n-a}{b-n}$$
we also have that
$$\frac{\tan^2 \theta}{\tan^2 \phi}=\frac{b^2}{a^2}$$
Can you finish?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Remainder of $3^n$ modulo $15$ Let $n \in \mathbb{N}_{>0}$. Then we have $$\begin{align*} &3 \equiv 3 \bmod 15\\&3^2 \equiv 9 \bmod 15\\&3^3 \equiv 12 \bmod 15\\&3^4 \equiv 6 \bmod 15\\&3^5 \equiv 3 \bmod 15\end{align*}$$ Hence all remainders modulo $15$ of $3^n$ should be $3,9,12$ and $6$. Why is this the case? Also, How could I calculate $$3^{1291}\equiv ? \bmod 15$$
| The only choices are $3, 6, 9, 12$ because $15$ is divisible by $3$, so what you're really doing is calculating $3^{n-1} \bmod 5$ and multiplying by $3$. That can't be $0$ (powers of $3$ can't be multiples of $5$ by unique factorization), so the only choices are $1, 2, 3, 4$ multiplied by $3$.
As for your second problem, notice that the pattern repeats: $3, 9, 12, 6, 3, 9, 12, 6, 3, 9, 12, 6, \ldots$ So take $u = 1291$ modulo ____ (fill in the blank) and then $3^u \equiv 3^{1291} \bmod 15$.
Oh, in case the reasoning isn't clear, we use the fact that
$$
ab \bmod 3 = (a \bmod 3)(b \bmod 3) \bmod 3
$$
In other words, we can take the modulo before we multiply, making things a lot less computationally intensive.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can we get the closed-form solution of $x_{n+1} = \frac{1+5 x_n}{1+x_n}$? When reading a textbook I encounter a sequence, the closed-form solution of which, claims the author, can be derived. The sequence reads:
\begin{gather*}
x_{n+1} = \frac{1+5 x_n}{1+x_n}
\end{gather*}
and $x_1=1$. The author gives without proof that
$$
x_n = \frac{4+2\sqrt{5}-(\sqrt{5}-1)c^{2-t}}{2+(\sqrt{5}+3)c^{2-t}}
$$
where $c = \frac{1}{2}(7+3\sqrt{5})$.
It seems that we can formulates the problem in a more general framework, say a sequence
$$
y_{n+1} = a+\frac{b}{y_n+c}
$$
I can't see how this kind of sequence can be solved explicitly. Can anyone give a hint on the problem?
| Hint. A standard route is as follows. One may 'linearize' the given recurrence in a 'matrix frame' by writing
$$
\begin{align}
x_{n+1}&=\underbrace{\left(\begin{matrix} 1 & 5 \\ 1 & 1 \end{matrix}\right)}_{\large A}\cdot x_{n}
\end{align}
$$ giving directly
$$
x_n=A^n \cdot x_1
$$ then, by diagonalizing the matrix $A$,
$$
\begin{align}
A=P \cdot J \cdot P^{-1} =\left(\begin{matrix} -\sqrt{5} & \sqrt{5} \\ 1 & 1 \end{matrix}\right)\left(\begin{matrix} 1-\sqrt{5} & 0 \\ 0 & 1+\sqrt{5} \end{matrix}\right)\left(\begin{matrix} -\frac{\sqrt{5}}{10} & \frac12 \\ \frac{\sqrt{5}}{10} & \frac12 \end{matrix}\right)
\end{align}
$$ one easily gets
$$
x_n=P \cdot J^n \cdot P^{-1} \cdot x_1
$$ that is
$$
x_n=\left(\begin{matrix} -\sqrt{5} & \sqrt{5} \\ 1 & 1 \end{matrix}\right)\left(\begin{matrix} (1-\sqrt{5})^n & 0 \\ 0 & (1+\sqrt{5})^n \end{matrix}\right)\left(\begin{matrix} -\frac{\sqrt{5}}{10} & \frac12 \\ \frac{\sqrt{5}}{10} & \frac12 \end{matrix}\right) \cdot x_1
$$ obtaining a closed form of $x_n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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When, where and **how often** do you find polynomials of higher degrees than two in mathematical, pure/applied, research? A formula for solving a polynomial of degree three, see this link; $ax^3+bx^2+cx+d=0$, is
$$\begin{align}
x\quad&=\quad
\sqrt[3]{
\left(
\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}
\right)
+
\sqrt{
\left(
\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}
\right) ^2
+
\left(
\frac{c}{3a} - \frac{b^2}{9a^2}
\right) ^3
}
}\\
&+\quad
\sqrt[3]{
\left(
\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}
\right)
-
\sqrt{
\left(
\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}
\right) ^2
+
\left(
\frac{c}{3a} - \frac{b^2}{9a^2}
\right) ^3
}
}
\;-\;\frac{b}{3a}
\end{align}$$
Unlike quadratic, cubic, and quartic polynomials, the general quintic
cannot be solved algebraically in terms of a finite number of
additions, subtractions, multiplications, divisions, and root
extractions, as rigorously demonstrated by Abel (Abel's impossibility
theorem) and Galois. However, certain classes of quintic equations can
be solved [...] Source: http://mathworld.wolfram.com/QuinticEquation.html
At levels of $5^{\text{th}}$ degree polynomials, things are starting to look really serious in my eyes. My question is:
If it is possible to not answer subjectively: When, where and how often do you find polynomials of higher degrees than two in mathematical, pure/applied, research?
| The characteristic polynomial of an $n \times n$ matrix has degree $n$. We often care about matrices larger than $2 \times 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2197171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 13,
"answer_id": 5
} |
How to solve system of linear equations with negative scalars over finite field? I need to solve the homogeneous system of linear equations over $Z_7$ via matrix:
\begin{cases}
\frac{2}{3}x-\frac{1}{2}y+4z -t = 0 \\
4x - z +3t = 0
\end{cases}
So this is the matrix:
\begin{bmatrix}
\frac{2}{3} & -\frac{1}{2} & 4 & -1 \\
4 & 0 & -1 & 3
\end{bmatrix}
I can't arrive to the correct solution which I checked in Wolfram Alpha http://www.wolframalpha.com/input/?i=((2%2F3)x+-+0.5y%2B4z+-t)+mod+7%3D0,+(4x+-+z+%2B3t)+mod+7%3D0. Can you please point the incorrect transitions I'm making? (I have to use Gaussian elimination for this)
First of all, my assumption is that if we have negative numbers I need to translate them immediately into $Z_7$ so the matrix becomes after switching first and second row:
1)
\begin{bmatrix}
4 & 0 & 6 & 3 \\
\frac{2}{3} & 6.5 & 4 & 6
\end{bmatrix}
2) $R_1 \to 4^{-1}R_1$
\begin{bmatrix}
1 & 0 & 5 & 3 \\
\frac{2}{3} & 6.5 & 4 & 6
\end{bmatrix}
3) $R_2 \to R_2 - \frac{2}{3}R_1$
\begin{bmatrix}
1 & 0 & 5 & 3 \\
0 & 6.5 & \frac{2}{3} & 4
\end{bmatrix}
4) $R_2 \to \frac{10}{65}R_2$ because $6.5^{-1} = \frac{10}{65}$ over $Z_7$
\begin{bmatrix}
1 & 0 & 5 & 3 \\
0 & 1 & \frac{4}{39} & \frac{8}{13}
\end{bmatrix}
Now the matrix looks canonical but I can clearly see that it's not the solution from Worlfram.
| you have 2 equations and 4 unknowns there will be a 2 dimensional solution
$z = z\\
t = t\\
4x = z-3t\\
x = \frac 14 z - \frac 34 t\\
\frac 12 y = \frac 23 x + 4z-t\\
y = 2(\frac 23)(\frac 14 z - \frac 34t) + 8z - 2t\\
y = \frac 13 z - t + 8z - 2t\\
y = \frac {25}{3} z - 3t$
or:
$\begin{bmatrix}x\\y\\z\\t\end{bmatrix} = \begin{bmatrix}\frac 14\\\frac {25}3\\1\\0\end{bmatrix}r + \begin{bmatrix}-\frac 34\\-3\\0\\1\end{bmatrix}s$
Now lets get that mod 7
$\begin{bmatrix}x\\y\\z\\t\end{bmatrix} = \begin{bmatrix}3\\ 100\\12\\0\end{bmatrix}r + \begin{bmatrix}-3\\-12\\0\\4\end{bmatrix}s$
$\begin{bmatrix}x\\y\\z\\t\end{bmatrix} = \begin{bmatrix}3\\ 2\\5\\0\end{bmatrix}r + \begin{bmatrix}4\\2\\0\\4\end{bmatrix}s$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2200055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to prove that $a_{n+1} = 1 + \frac{a_n}{a_n +1}$ is monotone increasing with $a_1=1$ This question has been driving me crazy, and I can't find the answer anywhere.
I tried proving it using induction.
As for the base case:
$a_1=1<1.5=a_2$
Next, suppose $a_k \leq a_{k+1}$
Then,
$$a_k \leq a_{k+1}$$
$$a_k +1 \leq a_{k+1}+1$$
$$ \frac{1}{a_k+1}\geq \frac{1}{ a_{k+1}+1} $$
$$ \frac{a_k}{a_k+1}\geq \frac{a_k}{ a_{k+1}+1} $$
$$ \frac{a_k}{a_k+1} +1\geq \frac{a_k}{ a_{k+1}+1}+1 $$
My goal was to get
$$ \frac{a_k}{a_k+1} +1\leq \frac{a_{k+1}}{ a_{k+1}+1}+1 $$
$$a_{k+1} \leq a_{k+2}$$
But obviously, I was not able to find a way.
I also tried a different inductive approach where I noted
$$a_{k+2}=\frac{5 a_n +3}{3 a_n +2}$$
You can check that this is true by seeing you get the correct $a_3=1.6$ using the given $a_1=1$.
Assuming $a_k \leq a_{k+1}$ I had that
$$ a_k \leq 1 + \frac{a_n}{a_n +1}$$
$$ \frac{a_n}{a_n +1} +1 \leq \frac{{a_n}^2 + 4a_n +2}{{a_n}^2+2a_n+1}$$
From the assumption $a_k \leq 1 + \frac{a_n}{a_n +1}$, we have that ${a_n}^2 \leq a_n +1$, so
$$ \frac{a_n}{a_n +1} +1 \leq \frac{{a_n}^2 + 4a_n +2}{{a_n}^2+2a_n+1} \leq \frac{(a_n+1) +4a_n +2}{{a_n}^2+2a_n+1} =\frac{5a_n+3}{{a_n}^2+2a_n+1}$$
If only I could change the denominator the same way while keeping the direction of the inequality, I could substitute $a_n+1$ for ${a_n}^2$ and I would get the desired $\frac{5 a_n +3}{3 a_n +2}$, proving that $$a_{k+1} \leq a_{k+2}$$ since $$a_{k+2}=\frac{5 a_n +3}{3 a_n +2}$$.
How do I prove the sequence is monotonic increasing? Both of these routes did not get me to the answer.
| You could also show that $1 + \frac{a_i}{a_i+1} > a_i$, by rearranging the inequality to get $2a_i+1>a_i^2+1$ then $2>a_i$. By showing $a_{n+1} = 2 - \frac{1}{1+a_n}$ (as in earlier answer) and that $a_i>0$, you show that $2>a_i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find the complex roots of $4x^2+16$ using QR method $$p(x)=4x^2+16$$
$$m(x)=\frac{p(x)}{4}=x^2+4$$
Now, I can determine the Companion matrix of $x^2+4$:
$$A=A_1=\begin{pmatrix} 0 \ -4 \\ 1 \ \ \ \ \ \ 0 \end{pmatrix}$$
QR decomposition (using Octave):
[Q,R]=qr(A)
$$Q_1\begin{pmatrix} 0 \ -1 \\ -1 \ \ \ \ 0 \end{pmatrix}$$
$$R_1=\begin{pmatrix} -1 \ \ \ 0 \\ 0 \ \ \ \ 4 \end{pmatrix}$$
$$A_2=R_1 Q_1=\begin{pmatrix} 0 \ 1 \\ -4 \ 0 \end{pmatrix}$$
$$R_2=\begin{pmatrix} -4 \ \ \ 0 \\ 0 \ \ \ \ 1 \end{pmatrix}$$
$$Q_2=\begin{pmatrix} 0 \ \ \ 1 \\ 1 \ \ \ \ 0 \end{pmatrix}$$
$$A_3=R_2 Q_2=\begin{pmatrix} 0 \ -4 \\ 1 \ 0 \end{pmatrix}$$
$$R_3=\begin{pmatrix} -1 \ \ \ 0 \\ 0 \ \ \ \ 4 \end{pmatrix}$$
$$Q_3=\begin{pmatrix} 0 \ \ \ -1 \\ -1 \ \ \ \ 0 \end{pmatrix}$$
How can I find the complex roots $2i$,$-2i$ (in general, complex roots of polynomials) using this method?
Thanks!
| With the real QR algorithm, you only get a upper triangular matrix as limit if all eigenvalues are real. With complex eigenvalues, you get $2×2$ matrix blocks on the diagonal for any complex conjugate pair.
In the complex case, you get a unitarian matrix for $Q$ by employing complex shifts. A first strategy is to compute the shift as eigenvalue of the bottom-most $2×2$ diagonal block. Which again makes the matrix formulation for degree $2$ one big circular computation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$(a) \sum_{n=2}^{\infty} \frac{3}{n^2+2n}$ Determine whether it converges and find sum Determine whether the following are convergent or divergent. If it is
convergent find its sum. Make sure to fully justify all of your work.
$$(a) \sum_{n=2}^\infty \frac{3}{n^2+3n}$$
Solution (My attempt):
Let $f(n) = a_n$
On $[2,\infty), f(x) = \frac 3 {x^2+3x} > 0$
$$f'(x) = 3\left(\frac{-(2x+3)}{(x^2+3x)^2}\right) < 0, \forall x \in [2,\infty)$$
Therefore the hypothesis of the integral test is met.
Consider
$$\lim_{A\to\infty} \int_2^A \frac 3 {(x)(x+3)} \, dx$$
$$\lim_{A\to\infty} \int_{2}^{A} \left( \frac{1}{x} - \frac{1}{x+3} \right)dx \text{ By pfd}$$
$$\lim_{A\to\infty} \left( \ln \right(\frac A {A+3}\left) - \ln \right(\frac 2 5 \left)\right) $$
$$= \ln 1 - \ln \left(\frac{2}{5}\right)$$
Therefore by Integral test the original series converges.
I was never thought how to find the sum of this type of series. The only way I know how to find the sum of a series is if its geometric. Does anyone know how too find the sum?
| Consider the partial sum $$S_p=\sum_{n=2}^p\frac{3}{n^2+3n}=\sum_{n=2}^p\frac{1}{n}-\sum_{n=2}^p\frac{1}{n+3}$$ $$S_p=\left(\frac 12+\frac 13+\frac 14+\frac 15+\cdots+\frac 1p\right)-\left(\frac 15+\frac 16+\frac 17+\frac 18+\cdots+\frac 1{p+3}\right)$$ What is left is $$S_p=\frac{13}{12}-\frac{1}{p+1}-\frac{1}{p+2}-\frac{1}{p+3}=\frac{13}{12}-\frac{3 p^2+12 p+11}{p^3+6 p^2+11 p+6}$$ Now, for large $p$, perform the long division (or Taylor series) to get $$S_p=\frac{13}{12}-\frac{3}{p}+\frac{6}{p^2}-\frac{14}{p^3}+O\left(\frac{1}{p^4}\right)$$ For $p=10$, $S_{10}=\frac{119}{143}\approx 0.8322$ while the expansion gives $\frac{311}{375}\approx 0.8293$.
You also could use the harmonic numbers and arrive to $$S_p=\frac{13}{12}+H_p-H_{p+3}$$ and using the asymptotics $$H_n=\gamma +\log \left({n}\right)+\frac{1}{2 n}-\frac{1}{12
n^2}+O\left(\frac{1}{p^4}\right)$$ and arrive to the same result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2204760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
If $p(x)=1+x+x^2+x^3+x^4+x^5$ find the remainder when $p(x^{12})$ is divided by $p(x)$
Question:
If $p(x)=1+x+x^2+x^3+x^4+x^5$ find the remainder when $p(x^{12})$ is divided by $p(x)$
Options:
(A) $0$
(B) $6$
(C) $1+x$
(D) $1+x+x^2+x^3+x^4$
My attempt:
Since $p(x)=1+x+x^2+x^3+x^4+x^5$ is a geometric series assuming $x<1$ with sum
$$S_1=\frac{1-x^6}{1-x}$$
and $p\left(x^{12}\right)$ also is a geometric series with sum
$$S_2=\frac{1-\left(x^{12}\right)^6}{1-x}$$
then
$$\frac{S_2}{S_1}= \frac{1-\left(x^{12}\right)^6}{1-x^6}$$
Now I can't seem to find a way to proceed further.
EDIT
I just found a mistake I made.
$$S_2=\frac{1-\left(x^{12}\right)^6}{1-x^{12}}$$
And not
$$S_2=\frac{1-\left(x^{12}\right)^6}{1-x}$$
So my division was wrong and so was my solution thereafter.
Sorry !
| Hint:
$$
\begin{align}
p(x^{12})-6 & = \big(x^{12}-1\big)+\big((x^{12})^2-1\big)+\big((x^{12})^3-1\big)+\big((x^{12})^4-1\big)+\big((x^{12})^5-1\big) \\
& = (x^{12}-1)( \cdots ) = (x^6-1)(x^6+1)(\cdots) = p(x)\cdot(x-1)(x^6+1)(\cdots)
\end{align}
$$
[ EDIT ] The above can also be derived directly from the more general:
$$a(x) - b(x) \mid p\big(a(x)\big) - p\big(b(x)\big) \implies x^{12} - 1 \mid p(x^{12})-p(1) \implies p(x) \mid p(x^{12}) - 6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2206982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve for $x,y$ and $z$ when: $x+y=\sqrt{4z-1}, \ y+z=\sqrt{4x-1}\ \mathrm{and} \ z+x=\sqrt{4y-1}$ How to Solve for $x,y$ and $z$ when: $x+y=\sqrt{4z-1}, \ y+z=\sqrt{4x-1}\ \mathrm{and} \ z+x=\sqrt{4y-1}$.
I've added the equations and get: $2(x+y+z)=\sqrt{4x-1}+\sqrt{4y-1}+\sqrt{4z-1}$ and after this how I proceed further?
Even if I subtract the first two equations then I'm getting:
\begin{align*}
x-z&=\sqrt{4z-1}-\sqrt{4x-1}\\
\Rightarrow\left(x+\sqrt{4x-1}\right)^2&=\left(z+\sqrt{4z-1}\right)^2\\
\Rightarrow\left(x^2-z^2\right)+4(x-z)&=2z\sqrt{4z-1}-2x\sqrt{4x-1}
\end{align*}
after this also I can't understand how to help with this.
| If there is a solution, the symmetry of the equations suggests we investigate the possibility that $x=y=z$
So assume that $x=y=z=a$.
$$2a=\sqrt{4a-1}$$
Thus $4a^2-4a+1=0$, or $(2a-1)^2=0$. Thus $x=y=z=\frac{1}{2}$.
This is a solution. But, as per the comment of dxiv below, this does not rule out solutions where $x\ne y$, $y\ne x$, $z\ne x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2208107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
How to solve an equation of the form $f(x)=f(a)$ for a fixed real a. I got stuck on this question: find all solutions $x$ for $a\in R$:
$$\frac{(x^2-x+1)^3}{x^2(x-1)^2}=\frac{(a^2-a+1)^3}{a^2(a-1)^2}$$
I see that if we simplify we get:
$$\frac{(x^2-x+1)^3}{x^2(x-1)^2}=\frac{[(x-{\frac 12})^2+{\frac 34}]^3}{[(x-{\frac 12})^2-{\frac 14}]^2}$$
From the expression $(x-{\frac 12})^2$, I see that if $x=x_1$ is a solution, then $x=1-x_1$ is also a solution. But in the solution to this exercise, it was stated that $x=\frac{1}{x_1}$ must also be a solution, and I don't see how.
[EDIT]
Ok, thx for the help guys. What do you think of this solution (doesn't involve any above precalculus math, and needs no long calculations)?
From the above we know that if $x_1=a$ is a solution, then $x_2=1-a$ is also a solution.
Also, from here:
$$\require{cancel}\frac{(x^2-x+1)^3}{x^2(x-1)^2}=\frac{\cancel{x^3}(x+{\frac 1x}-1)^3}{\cancel{x^3}(x+{\frac 1x}-2)}$$
in the expression $x+{\frac 1x}$ we see that if $x=x_1$ is a solution, then $x=\frac{1}{x_1}$ is also a solution, so $x_3=\frac{1}{a}$.
With these two rules we can now keep generating roots until we have 6 total.
If $x=x_2$ is a solution, then $x=\frac{1}{x_2}$ is also a solution, so $x_4=\frac{1}{1-a}$.
If $x=x_3$ is a solution, then $x=1-x_3$ is also a solution, so $x_5=\frac{a-1}{a}$.
Finally, if $x=x_5$ is a solution, then $x=\frac{1}{x_5}$ is also a solution, so $x_6=\frac{a}{a-1}$
The 6 obtained values are distinct, so they cover all the roots.
[EDIT2]
I guess this is answered. No sure whose particular answer to actually select as the right one since they're all correct, so I'll just leave it like this.
| Hint: write the LHS in terms of $z = x+\cfrac{1}{x}$ as follows:
$$
\require{cancel}
\frac{(x^2-x+1)^3}{x^2(x-1)^2} = \frac{(x^2-x+1)^3}{x^2(x^2-2x+1)}= \frac{\bcancel{x^3}\left(x+\cfrac{1}{x}-1\right)^3}{\bcancel{x^3}\left(x+\cfrac{1}{x}-2\right)}=\frac{(z-1)^3}{z-2}
$$
Let $b=a+\cfrac{1}{a}$ and do the same for the RHS, then the equation becomes:
$$
\frac{(z-1)^3}{z-2} = \frac{(b-1)^3}{b-2}
$$
The above is a cubic in $z\,$ with the obvious root $z_1=b\,$, which leaves a quadratic to solve.
$$
\big(\,z-b\,\big) \,\big( \,(b-2)z^2 +(b-2)(b-3)z - (2b^2-6b+5)\,\big) \,=\, 0
$$
After not too pretty calculations, the other roots turn out to be:
$$
z_{2,3} = \frac{b^2 \pm (\sqrt{b^2 - 4} - 5) b \mp \sqrt{b^2 - 4} + 6}{4 - 2 b}
$$
Reverting to the $x$ and $a$ variables, the root $z_1=b$ gives the roots $x=a$ and $x=\cfrac{1}{a}\,$, then the roots $z_{2,3}$ give the other $4$ roots in $x$ after solving the respective equations. The calculations are again not pretty (and not included here), though noting $\sqrt{b^2-4}=a-\cfrac{1}{a}\,$ eliminates the radicals upfront.
[ EDIT ] The following is a shortcut for the final calculations, using OP's observation that:
I see that if $x=x_1$ is a solution, then $x=1-x_1$ is also a solution.
This means that the second root in $z$ is $z_2=1-a + \cfrac{1}{1-a}\,$ corresponding to the roots $x=1-a$, $x=\cfrac{1}{1-a}\,$. The remaining root in $z$ as determined by Vieta's relations is:
$$\require{cancel}z_3=-(b-3)-\left(1-a + \cfrac{1}{1-a}\right)= -\bcancel{a}-\frac{1}{a}+3-1+\bcancel{a}-\frac{1}{1-a}=2 - \frac{1}{a}-\frac{1}{1-a}$$
Solving the equation $x+\cfrac{1}{x}=2 - \cfrac{1}{a}-\cfrac{1}{1-a}$ gives the last two roots $x=\cfrac{a}{a-1}\,$, $x=\cfrac{a-1}{a}$.
[ EDIT #2 ] Even shorter: once established that if $x$ is a root then $1/x$ and $1-x$ are also roots, and given the obvious $x=a$ root, it follows that $\,1/a\,$, $\,1-a\,$, $\,1/(1-a)\,$, $\,1-1/a=(a-1)/a\,$, $\,1/(1-1/a)=a/(a-1)\,$ are also roots. For $a \ne 0,1$ and $a \not \in \{-1, \frac{1}{2},2\}$ the $6$ roots are distinct, and since the equation is of degree $\,6\,$, it follows that these are the only roots. For $a \in \{-1, \frac{1}{2},2\}$ the same can be shown to hold true by a continuity argument.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
} |
Recurrence relation using generating functions to solve? Consider the recurrence relation: $ a_n = −a_{n−1} + 6a_{n−2}, n ≥ 2, a_0 = 0, a_1 = 4.$
(a) Solve the recurrence relation by finding the characteristic
equation.
$r^2+r+6=0 \to (r+ \frac {1}{2})^2 +\frac{23}{4}=0 \to r= -\frac {1}{2}\pm i\frac{ \sqrt {23}}{2}$
$a_n^{(h)} = (-\frac {1}{2}+ (i\frac{ \sqrt {23}}{2})^n + ( -\frac {1}{2}\ -(i\frac{ \sqrt {23}}{2}))^n$
$a_n = c_1(-\frac {1}{2}+ (i\frac{ \sqrt {23}}{2})^n + c_2 ( -\frac {1}{2}\ -(i\frac{ \sqrt {23}}{2}))^n $
$a_0=0 = c_1 + c_2 $
$a_1 =4= c_1(-\frac {1}{2}+ (i\frac{ \sqrt {23}}{2}) + c_2 ( -\frac {1}{2}\ -(i\frac{ \sqrt {23}}{2}))$
(b) Solve the recurrence relation using the method of generating functions.
$ a_2 = −a_{1} + 6a_{0} $
$ a_3 = −a_{2} + 6a_{1} $
$ a_4 = −a_{3} + 6a_{2} $
i know im supposed to multiple $a_1 $ by x etc but im not really sure how to do it... how do i know what $g(x)$='s?
| It is formally very easy, but dizzying sometimes:
Define
$$f(x)=\sum_0^\infty a_nx^n$$
Write the recurrence relation,
$$a_{n+2}=-a_{n+1}+6a_n$$
Multiply by $x^{n+2}$
$$a_{n+2}x^{n+2}=-a_{n+1}x^{n+2}+6a_nx^{n+2}$$
Sum
$$\sum_0^\infty a_{n+2}x^{n+2}=-\sum_0^\infty a_{n+1}x^{n+2}+\sum_0^\infty 6a_nx^{n+2}$$
Do things
$$\sum_2^\infty a_{n}x^{n}=-x\sum_1^\infty a_{n}x^{n}+6x^2\sum_0^\infty a_nx^n$$
$$\sum_0^\infty a_{n}x^{n}-a_1x-a_0=-x\sum_0^\infty a_{n}x^n+a_0x++6x^2\sum_0^\infty a_nx^n$$
Recover $f(x)$
$$f(x)-a_1x-a_0=-xf(x)+a_0x+6x^2f(x)$$
Isolate:
$$(6x^2-x-1)f(x)=-a_1x-a_0x-a_0$$
$$f(x)=-\frac{a_1x+a_0x+a_0}{6x^2-x-1}=-\frac{a_1x+a_0x+a_0}{6x^2-x-1}$$
We can get now $a_0=0$, $a_1=4$
$$f(x)=\frac{-4x}{(2x-1)(3x+1)}=\frac{-4}{5(2x-1)}+\frac{-4}{5(3x+1)}$$
$$f(x)=\frac{4}{5(1-2x)}-\frac{4}{5(1-(-3x))}$$
These can be expanded as the geometric series:
$$f(x)=\sum_0^\infty\frac 45(2x)^n-\sum_0^\infty\frac 45(-3x)^n$$
At last,
$$f(x)=\sum_0^\infty\frac{4}{5}(2^n-(-3)^n)x^n$$
| {
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"url": "https://math.stackexchange.com/questions/2210712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\sum\limits_{cyc}\frac{a-b}{\sqrt{b+c}}\geq0$
Let $a$, $b$, $c$ and $d$ be non-negative numbers such that $\prod\limits_{cyc}(a+b)\neq0$. Prove that:
$$\frac{a-b}{\sqrt{b+c}}+\frac{b-c}{\sqrt{c+d}}+\frac{c-d}{\sqrt{d+a}}+\frac{d-a}{\sqrt{a+b}}\geq0$$
The equality occurs also for $a=c$ and $b=d$.
My trying.
We need to prove that
$$\sum_{cyc}\frac{a+c-b-c}{\sqrt{b+c}}\geq0$$ or
$$(a+c)\left(\frac{1}{\sqrt{b+c}}+\frac{1}{\sqrt{d+a}}\right)+(b+d)\left(\frac{1}{\sqrt{c+d}}+\frac{1}{\sqrt{a+b}}\right)\geq\sum_{cyc}\sqrt{a+b}$$
and what is the rest?
Thank you!
| Since the sign of $\frac{a-b}{\sqrt{b+c}}+\frac{b-c}{\sqrt{c+d}}+\frac{c-d}{\sqrt{d+a}}+\frac{d-a}{\sqrt{a+b}}$ is preserved when $(a,b,c,d)$ is replaced with $(ax,bx,cx,dx)$ for any positive $x$, we can assume that $a+b+c+d=1$ WLOG.
Now let $x=a+b,y=a+c,z=a+d$ and from OP's last inequality, we get $$y\left(\frac{1}{\sqrt {z}}+\frac{1}{\sqrt {1-z}}\right)+(1-y)\left(\frac{1}{\sqrt {x}}+\frac{1}{\sqrt {1-x}}\right) \ge \sqrt x+\sqrt {1-x}+\sqrt z+\sqrt {1-z}$$
To prove this, let's consider this inequality.$$\frac{1}{\sqrt {x}}+\frac{1}{\sqrt {1-x}} \ge \sqrt x+\sqrt {1-x}+\sqrt z+\sqrt {1-z}$$
And $\sqrt z+\sqrt {1-z}$ is maximized at $\sqrt 2$ when $z=1/2$.
$$\frac{1}{\sqrt {x}}+\frac{1}{\sqrt {1-x}} \ge \sqrt x+\sqrt {1-x}+\sqrt 2$$
We can verify this by using calculus, graphing tools, etc.
Therefore, $\frac{1}{\sqrt {x}}+\frac{1}{\sqrt {1-x}} \ge \sqrt x+\sqrt {1-x}+\sqrt z+\sqrt {1-z}$ and similarly $$\frac{1}{\sqrt {z}}+\frac{1}{\sqrt {1-z}} \ge \sqrt x+\sqrt {1-x}+\sqrt z+\sqrt {1-z}$$
Now we get $$\begin{align}&y\left(\frac{1}{\sqrt {z}}+\frac{1}{\sqrt {1-z}}\right)+(1-y)\left(\frac{1}{\sqrt {x}}+\frac{1}{\sqrt {1-x}}\right) \\\ge& y(\sqrt x+\sqrt {1-x}+\sqrt z+\sqrt {1-z})+(1-y)(\sqrt x+\sqrt {1-x}+\sqrt z+\sqrt {1-z})\\=&\sqrt x+\sqrt {1-x}+\sqrt z+\sqrt {1-z}\end{align}$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2212711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Value of n for which f(n) = $\,n^2 + 9n + 30\,$ is a perfect square. I attempted this by setting $f(n) = \,m^2.\,$
So $\,n^2 + 9n + 30 = m ^2\,$.
Then $\,9(n + 10/3) = (m + n)(m - n)\,$.
So $m = 10/3$ and $n = -17/3$ which is incorrect.
| If
$n^2+an+b = m^2$
then
$4n^2+4an+4b = 4m^2$.
Since
$(2n+a)^2
=4n^2+4an+a^2
$,
this means that
$(2n+a)^2-a^2+4b
= 4m^2
$
or
$a^2-4b
=(2n+a)^2-4m^2
=(2n+a-2m)(2n+a+2m)
$.
Looking at all
the possible factorizations
of $a^2-4b$
(positive and negative),
we can get the possible values of
$n$ and $m$.
If $a^2-4b = uv$,
then
$2n+a-2m = u$
and
$2n+a+2m = v$
so
$4m = v-u$,
$4n = v+u-2a$
or
$m = \dfrac{v-u}{4}$
and
$n = \dfrac{v+u-2a}{4}$.
Since we can consider
only non-negative values
of $m$,
we can restrict the factorization
to have
$v \ge u$.
Note that this requires
both $v-u$ and
$v+u-2a$
to be multiples of $4$,
which further restricts the solutions.
Also note that
since
$a^2-4b = uv
=(-v)(-u)
$,
if
$m = \dfrac{v-u}{4}$
and
$n = \dfrac{v+u-2a}{4}$
is a solution
then another solution is
$m' = \dfrac{-u-v}{4}=-m$
and
$n'
= \dfrac{-u-v-2a}{4}
= -\dfrac{u+v+2a}{4}
= -\dfrac{u+v-2a}{4}-\dfrac{4a}{4}
= -n-a
$.
In this case,
$a=9$ and $b=30$
so
$a^2-4b = 81-120 = -39$.
Since $39 = 3\cdot 13$,
the possible factorizations
(with $v \ge u$) are
$(u, v)=
(-1, 39),
(-3, 13),
(-13, 3),
(-39, 1)$.
For these
$(v-u, v+u-2a)=(v-u, v+u-18)=
(40, 20),
(16, -8),
(16, -28),
(40, -56)$.
This are all divisible by $4$,
so we get
$(m, n)=
(10, 5),
(4, -2),
(4, -7),
(10, -14)$.
As expected,
if $n$ is a solution,
then so is
$-n-9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find $\lim\limits_{x \to 1} (x-1)^2 \sin (\frac{1}{\sqrt[3]{x-1}})$ $$\lim\limits_{x \to 1} (x-1)^2 \sin (\frac{1}{\sqrt[3]{x-1}})$$
First I don't know how to solve this problem since there is $\sin (\frac{1}{\sqrt[3]{x-1}})$ in there. But I manage to do this:
$$\lim\limits_{x \to 1} (x-1)^2 \sin (\frac{1}{\sqrt[3]{x-1}}) =
(\lim\limits_{x \to 1} (x-1)^2) (\lim\limits_{x \to 1}\sin (\frac{1}{\sqrt[3]{x-1}}))
$$
$$= ((1-1)^2) (\lim\limits_{x \to 1}\sin (\frac{1}{\sqrt[3]{x-1}}))$$
$$= (0) (\lim\limits_{x \to 1}\sin (\frac{1}{\sqrt[3]{x-1}}))$$
$$= 0$$
But I'm pretty sure that I'm wrong here.
| For all $x\neq 1$, we have
$$\begin{align}
0&\leq\bigg|(x-1)^2\sin\frac{1}{\sqrt[3]{x-1}}\bigg|\\
&=|(x-1)^2|\cdot \bigg|\sin\frac{1}{\sqrt[3]{x-1}}\bigg|\\
&\leq |(x-1)^2|\cdot 1\\
&=(x-1)^2
\end{align}$$
Because $$\lim_{x\to 1} 0=0=\lim_{x\to 1}(x-1)^2,$$ it follows from the Squeeze Theorem that $$\lim_{x\to 1}\bigg|(x-1)^2\sin\frac{1}{\sqrt[3]{x-1}}\bigg|=0$$ and hence, we get
$$\lim_{x\to 1}(x-1)^2\sin\frac{1}{\sqrt[3]{x-1}}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
When $Z$ is divided by $8$ the remainder is $5$. When $Z$ is divided by $8$ the remainder is $5$. What is the remainder when $4Z$ is divided by $8$?
My Attempt:
We know,
$$\textrm{dividend} =\textrm{divisor} \times \textrm{quotient} + \textrm{remainder}$$
$$Z=8\times k+5$$
$$Z-5=8K$$
$$\dfrac {Z-5}{8}=k$$
Now, what should I do further?
| $$\begin{align}Z&=8k+5\to \\4Z&=4(8k)+4\cdot 5\\&=8(4k)+20\\
&=8(4k)+16+4\\&=8(4k+2)+4\end{align}$$
So the remainder is $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2220965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Working with functions
Suppose $f(3-x)=2x^2-5x+4$ and $f(x)=ax^2+bx+c$. What is $a+b+c$?
I don't know how to approach this. I thought of maybe doing $a(3-x)^2+b(3-x)+c=2x^2-5x+4$ and solving for $a+b+c$ but it got messy.
| On continue from your answer.
$a(9+x^2-6x)+b(3-x)+c=2x^2-5x+4$
$\implies 9a+9ax^2-6ax+3b-3bx+c=2x^2-5x+4$
$\implies 9ax^2-(6a+3b)x+(9a+3b+c)=2x^2-5x+4$
Comparing both sides,
$$9a=2$$
$$6a+3b=5$$
$$9a+3b+c=4$$
Find a from first equation, then put a in second equation to find b.
Then put a, b in third equation to find c.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding sum $\frac{1}{2} x^2 + \frac{1}{2} \frac{1}{3} x^3 + \frac{1}{4} \frac{1}{3} x^4 + ... $ I need to find following: for $0 < x < 1$
$$\frac{1}{2} x^2 + \frac{1}{2} \frac{1}{3} x^3 + \frac{1}{4} \frac{1}{3} x^4 + ... $$
My attempt:
I can see that the sum is composed of two infinite sums, one is $\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \sum_{n=1}^{\infty} \left( \frac{1}{n} -
\frac{1}{n+1} \right)$ (Telescoping ) and another is, $\sum_{n=2}^{\infty} x^n$ (it's a G.P.). How can I use these for solving the sum in question?
Any hints will be appreciated...
| Let us set
$$
S(x)=\frac{1}{2} x^2 + \frac{1}{2} \frac{1}{3} x^3 + \frac{1}{4} \frac{1}{3} x^4 + \ldots
$$
then
$$
\frac{dS(x)}{dx}=x + \frac{1}{2}x^2 + \frac{1}{3} x^3 + \ldots.
$$
This is a well-known Taylor series being
$$
x + \frac{1}{2}x^2 + \frac{1}{3} x^3 + \ldots = -\ln(1-x).
$$
We now note that $S(0)=0$, then we have to solve the simple differential equation
$$
\frac{dS(x)}{dx}=-\ln(1-x)
$$
with the given initial condition. This yields
$$
S(x)=(1-x)\ln(1-x)+x
$$
that is the result you were looking for.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2222202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.