Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solving $\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$ Where do I start to solve a equation for x like the one below?
$$\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$$
After squaring it, it's too complicated; but there's nothing to factor or to expand?
Ideas?
| Divide $$(7x-4)-(7x-5)=(4x-1)-(4x-2)\ \ \ \ \ (1)$$ by the original equation, which is $$\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}\ \ \ \ \ (2)$$ You should get:$$\sqrt{7x-4}+\sqrt{7x-5}=\sqrt{4x-1}+\sqrt{4x-2}\ \ \ \ \ (3)$$Now add (2) and (3), you get:
\begin{align}
2\sqrt{7x-4}&=2\sqrt{4x-1}
\\ \\
7x-4&=4x-1
\\ \\
x&=1
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/400395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Find the limit without use of L'Hôpital or Taylor series: $\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)$ Find the limit without the use of L'Hôpital's rule or Taylor series
$$\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2x}\right)$$
| Let $I = \lim_{x\to 0}\left(\frac{1}{x^2} - \frac{1}{\sin^2 x}\right)$, we have:
$$\begin{align}
I = & \lim_{x\to 0}\left(\frac{1}{(2x)^2} - \frac{1}{\sin^2(2x)}\right)\\
= & \lim_{x\to 0} \left(\frac{1}{4 x^2} - \frac{1}{4 \sin^2 x\cos^2 x}\right)\\
\implies 4I = & \lim_{x\to 0}\left(\frac{1}{x^2} - \frac{1}{\sin^2 x\cos^2 x}\right)\\
\implies 3I = & \lim_{x\to 0}\left\{\left(\frac{1}{x^2} - \frac{1}{\sin^2 x\cos^2 x}\right)
-\left(\frac{1}{x^2} - \frac{1}{\sin^2 x}\right)\right\}\\
=&-\lim_{x\to 0} \frac{1-\cos^2 x}{\sin^2 x\cos^2 x} = -\lim_{x\to 0}\frac{1}{\cos^2 x} = -1\\
\implies I = & -\frac{1}{3}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/400541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Telescoping sum of powers
$$
\begin{array}{rclll}
n^3-(n-1)^3 &= &3n^2 &-3n &+1\\
(n-1)^3-(n-2)^3 &= &3(n-1)^2 &-3(n-1) &+1\\
(n-2)^3-(n-3)^3 &= &3(n-2)^2 &-3(n-2) &+1\\
\vdots &=& &\vdots & \\
3^3-2^3 &= &3(3^2) &-3(3) &+1\\
2^3-1^3 &= &3(2^2) &-3(2) &+1\\
1^3-0^3 &= &3(1^2) &-3(1) &+1\\
\underline{\hphantom{(n-2)^3-(n-3)^3}} & &\underline{\hphantom{3(n-2)^2}} &\underline{\hphantom{-3(n-2)}} &\underline{\hphantom{+1}}\\
n^3-0^3 &= & 3f_2(n) &-3f_1(n) &+n
\end{array}
$$
Can somebody explain me how these results are disposed intuitively? I didn't understand why $$(n-1)^3 -(n-2)^3$$ became equals to $$3(n-1)^2 - 3(n-1) + 1$$
How do this transformation was done?
Thanks!
| Notice that
$$(n-2)^3=((n-1)-1)^3=(n-1)^3-3(n-1)^2+3(n-1)-1$$
Anyway, related to the image you posted (which I didn't notice at first sight), the result was really obtained by substitution from the first row ($n\to n-1$), as other people also observed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/402445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
A tricky logarithms problem? $ \log_{4n} 40 \sqrt{3} \ = \ \log_{3n} 45$. Find $n^3$.
Any hints? Thanks!
| My proposal was that
$$\frac{(4n)^p}{(3n)^p} \ = \ (\frac{4}{3})^p \ = \ \frac{40 \sqrt{3}}{45} \ = \ \frac{8}{3 \sqrt{3}} \ = \ (\frac{4}{3})^{3/2} \ . $$
So both logarithms equal 3/2 and the rest follows as Zev Chonoles shows.
EDIT: Just to verify that the other piece checks,
$$(40 \sqrt{3})^2 \ = \ 4800 \ = \ (4n)^3 \ \Rightarrow \ n^3 \ = \ \frac{4800}{64} \ = 75 \ . $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/404389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Need help simplifying complicated rational expression. Studying for my final and I can't figure this out.
Simplify:
$$\large\frac{\frac{3}{x^3y} + \frac{5}{xy^4}}{\frac{5}{x^3y} -\frac{4}{xy}}$$
| The shortest route is to note that the least common multiple of all four denominators of the ‘small’ fractions is $x^3y^4$ and to multiply the expression by $1=\frac{x^3y^4}{x^3y^4}$:
$$\large{\frac{\frac{3}{x^3y} + \frac{5}{xy^4}}{\frac{5}{x^3y} -\frac{4}{xy}}\cdot\frac{x^3y^4}{x^3y^4}}\;.$$
I’ll let you check that this does the trick. It’s analogous to seeing that the least common multiple of $2,3,4$, and $6$ is $12$, so that multiplying
$$\large\frac{\frac12+\frac23}{\frac14+\frac56}$$
by $\frac{12}{12}$ will clean it up nicely.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/405073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How to calculate $\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$? I need to calculate the sum $\displaystyle S=\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$,
where $\displaystyle H_n=\sum\limits_{m=1}^n\frac1m$.
Using a CAS I found that $S=\lim\limits_{k\to\infty}s_k$ where $s_k$ satisfies the recurrence relation
\begin{align}
& s_{1}=-1,\hspace{5mm} s_{2}=\frac18,\hspace{5mm} s_{3}=-\frac{215}{216},\hspace{5mm} s_{4}=\frac{155}{1728},\hspace{5mm} \text{for all} \quad k>4, \\ s_{k} &=\frac1{k^3(2k-3)}\left(\left(-4k^4+18k^3-25k^2+12k-2\right)s_{k-1}+\left(12k^3-39k^2+38k-10\right)s_{k-2} \right.\\
& \hspace{5mm} \left. +\left(4k^4-18k^3+25k^2-10k\right)s_{k-3}\\+\left(2k^4-15k^3+39k^2-40k+12\right)s_{k-4}\right),
\end{align}
but it could not express $S$ or $s_k$ in a closed form.
Can you suggest any ideas how to calculate $S$?
| we have
$$ \frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^{\infty}\left(H_n^2-H_n^{(2)}\right)x^n$$
replace $x$ with $-x$, divide both sides by $x$ then integrate w.r.t $x$ from $0$ to $1$ , we get:
\begin{align*}
S_1&=\sum_{n=1}^{\infty}(-1)^n\left(H_n^2-H_n^{(2)}\right)\int_0^1x^{n-1}\ dx=\sum_{n=1}^{\infty}\left(H_n^2-H_n^{(2)}\right)\frac{(-1)^n}n=\underbrace{\int_0^1\frac{\ln^2(1+x)}{x(1+x)}\ dx}_{x=\frac{1-y}{y}}\\
&=\int_{1/2}^1 \frac{\ln^2x}{1-x}\ dx=\sum_{n=1}^{\infty}\int_{1/2}^1x^{n-1}\ln^2x\ dx=\sum_{n=1}^{\infty}\left(\frac{2}{n^3}-\frac{2}{2^n n^3}-\frac{2\ln2}{2^n n^2}-\frac{\ln^22}{2^n n}\right)\\
&=2\zeta(3)-2\operatorname{Li_3}\left(\frac12\right)-2\ln2\operatorname{Li_2}\left(\frac12\right)-\ln^32
\end{align*}
Now using the identity:
$$\displaystyle \int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac1n\left(H_n^2+H_n^{(2)}\right)$$
multiply both sides by $(-1)^n$ then sum both sides w.r.t $n$ from $1$ to $\infty$, we get
\begin{align*}
S_2&=\sum_{n=1}^{\infty}\left(H_n^2+H_n^{(2)}\right)\frac{(-1)^n}{n}=\int_0^1\frac{\ln^2(1-x)}{x}\sum_{n=1}^{\infty}(-x)^n\ dx=\underbrace{-\int_0^1\frac{\ln^2(1-x)}{1+x}\ dx}_{x=1-y}\\
&=-\int_0^1\frac{\ln^2(x)}{2-x}=-\sum_{n=1}^{\infty}\frac1{2^n}\int_0^1 x^{n-1}\ln^2x\ dx=-2\sum_{n=1}^{\infty}\frac1{2^n n^3}=-2\operatorname{Li_3}\left(\frac12\right)
\end{align*}
we are now ready to calcualate our sum:
\begin{align*}
\frac{S_1+S_2}{2}=\sum\frac{(-1)^n H_n^2}{n}&=\zeta(3)-2\operatorname{Li_3}\left(\frac12\right)-\ln2\operatorname{Li_2}\left(\frac12\right)-\frac12\ln^32\\
&=\frac12\ln2\zeta(2)-\frac34\zeta(3)-\frac13\ln^32
\end{align*}
and as a bonus :
\begin{align*}
\frac{S_2-S_1}{2}=\sum\frac{(-1)^n H_n^{(2)}}{n}&=\ln2\operatorname{Li_2}\left(\frac12\right)-\zeta(3)+\frac12\ln^32\\
&=\frac12\ln2\zeta(2)-\zeta(3)
\end{align*}
where the results of $\operatorname{Li_3}\left(\frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$ and $ \operatorname{Li_2}\left(\frac12\right)=\frac12\zeta(2)-\frac12\ln^22$ were used in the calculations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/405356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 5,
"answer_id": 2
} |
My son's Sum of Some is beautiful! But what is the proof or explanation? My youngest son is in $6$th grade. He likes to play with numbers. Today, he showed me his latest finding. I call it his "Sum of Some" because he adds up some selected numbers from a series of numbers, and the sum equals a later number in that same series. I have translated his finding into the following equation:
$$(100\times2^n)+(10\times2^{n+1})+2^{n+3}=2^{n+7}.$$
Why is this so? What is the proof or explanation? Is it true for any $n$?
His own presentation of his finding:
Every one of these numbers is two times the number before it.
$1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$.
I pick any one of them, times $100$. Then I add the next one, times $10$. Then I skip the next one. Then I add the one after that.
If I then skip three ones and read the fourth, that one equals my sum!
| The result simply holds because
$$\begin{align}
(100*2^n)+(10*2^{n+1})+2^{n+3} &=(2^6+2^5+2^2)*2^n+(2^3+2)*2^{n+1}+2^{n+3} \\
& = 2^{n+6} +2^{n+5}+2^{n+2} +2^{n+4} + 2^{n+2}+2^{n+3} \\
& =2^{n+6}+2^{n+5}+2^{n+4} + 2^{n+3}+(2^{n+2}+2^{n+2})\\
& =2^{n+6}+2^{n+5}+2^{n+4} +(2^{n+3}+2^{n+3})\\
& =2^{n+6}+2^{n+5}+(2^{n+4}+2^{n+4})\\
& =2^{n+6}+(2^{n+5}+2^{n+5})\\
& =(2^{n+6}+2^{n+6})\\
&=\mathbf{2^{n+7}}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/406099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "443",
"answer_count": 10,
"answer_id": 0
} |
Simple number theory problem I found this question in a textbook on number theory:
For which integer c will $\;\displaystyle{\frac{c^6 - 3}{c^2 + 2}}\;$ also be an integer?
I wonder if there is a solution which is not based on trial and error.
| If $(c^6 - 3)/(c^2 + 2)$ is an integer, then so is $$\frac{c^6 - 3}{c^2 + 2} - (c^4 - 2c^2 + 4) = \frac{c^6 - 3}{c^2 + 2} - \frac{c^6 + 8}{c^2 + 2} = \frac{-11}{c^2 + 2},$$ that is, $c^2 + 2$ divides $11$. The only way this can happen is if $c^2 = 9$, so $c = \pm 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/410478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
question about partial fractions why $\frac{6x^2+19x+15}{(x-1)(x-2)^2}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-2)^2}$ can anyone tell me why
$$\frac{6x^2+19x+15}{(x-1)(x-2)^2}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{(x-2)^2}$$
I don't undestand the $\frac{C}{(x-2)^2}$ and also what is wrong according to basic math rules if I write the equation in the following way
$$\frac{6x^2+19x+15}{(x-1)(x-2)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-2}$$
Thanks
| Two things wrong with what you propose.
*
*Those last two terms could be combined into one, $${B\over x-2}+{C\over x-2}={D\over x-2}$$
*It just doesn't work ---- what would you get for $B$ and $C$ if you tried to do $${1\over(x-2)^2}={B\over x-2}+{C\over x-2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/410962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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For what values of $a$ $\int^2_0 \min(x,a)dx=1$ I want to check for what values of $a$ the integral of this function is equal to $1$
$$\int^2_0 \min(x,a)dx=1$$
What I did is to check 2 cases and I am not sure is enough :
case 1:
$$a<x \rightarrow \int^2_0 a = ax|^{2}_{0}=2a=1 \rightarrow a=\frac{1}{2}$$
now I dont know if its enough so say that, I need to check if $a<2$? or what I did?
case 2:
$$a>x \rightarrow \int^2_0 x = \frac{x^2}{2}|^{2}_{0}=2$$
how I need to continue from here?
Thanks!
| Note that $0 \le x \le 2$. Thus, you have considered two cases: when $a<0$ and when $a>2$. It remains to check when $0\le a\le2$. Hence, we split the given integral into the following two integrals:
$$ \begin{align*}
1 &= \int_0^a \min(x,a)dx + \int_a^2 \min(x,a)dx \\
1 &= \int_0^a xdx + \int_a^2 adx \\
1 &= \left[\dfrac{x^2}{2} \right]_0^a + \left[ax \right]_a^2 \\
1 &= \left[\dfrac{a^2}{2} -0\right] + \left[2a-a^2 \right] \\
1 &= \dfrac{-1}{2}a^2+2a \\
2 &= -a^2+4a \\
a^2-4a+4 &= 2 \\
(a-2)^2 &= 2 \\
a-2 &= \pm\sqrt{2} \\
a &= 2 \pm \sqrt{2} \qquad (\approx 0.58579 \text{ or } 3.4142)\\
\end{align*}$$
Hence, since we know that $0 \le a \le 2$, we reject the larger solution and conclude that:
$$
a=2-\sqrt{2}=0.58579...
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/411867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Proving an equality on $\arctan x$ To demonstrate the following equality
\begin{equation}
\arctan x=\pi/2- \arctan(1/x)
\end{equation}
I have proceeded in this way. I know that
\begin{equation}
\arctan x= \int \frac{1}{1+x^2} dx + C
\end{equation}
But:
\begin{equation}
\int \frac{1}{1+x^2} dx=\int \frac{1}{x^2(1+1/x^2)} dx=-\int \frac{1}{1+t^2} dt
\end{equation}
with $t=1/x$. Then:
\begin{equation}
\arctan x= -\arctan\left(\frac{1}{x}\right)+C
\end{equation}
And if we apply the limit
\begin{equation}
\lim_{x\rightarrow0}\ \arctan x=\lim_{x\rightarrow0}\ -\arctan\left(\frac{1}{x}\right)+C
\end{equation}
we obtain: $C=\pi/2$.
What do you think of my proof? Write down your ideas, I'm curious and I want to compare myself with you.
Thank you very much.
| Let $f(x) = \arctan x$. Then $f'(x) = \frac{1}{1+x^2}$. Let $\phi(x) = f(\frac{1}{x})$, then $\phi'(x) = f'(\frac{1}{x}) (- \frac{1}{x^2}) = - \frac{1}{1+x^2}$.
Hence $f'(x)+\phi'(x) = 0$, and so $x \mapsto f(x)+\phi(x)$ is constant on $(-\frac{\pi}{2}, \frac{\pi}{2})$. Since $f(1) = \phi(1) = \frac{\pi}{4}$, we have the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/412014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Vector analysis. Del and dot products I am trying to prove that
$$\nabla(\mathbf{A} \cdot \mathbf{B}) = (\mathbf{A} \cdot \nabla)\mathbf{B} + (\mathbf{B} \cdot \nabla)\mathbf{A} + \mathbf{A} \times (\nabla \times \mathbf{B}) + \mathbf{B} \times (\nabla \times \mathbf{A})$$
I've gotten as far as $\nabla(\mathbf{A} \cdot \mathbf{B}) = \nabla A\cdot B+\nabla B\cdot A$, using subscript summation. I don't know how to proceed.
This is part of proving that
$$\frac{Dv}{Dt}=\frac{\partial v}{\partial t}+\nabla(\frac{v^2}{2})-v\times(\nabla\times v)$$
| Following your effect of using subscript notation: $\newcommand{\b}{\boldsymbol}$
$$
\nabla (\b{A}\cdot \b{B}) = \nabla_{\b{A}}(\b{A}\cdot \b{B}) + \nabla_{\b{B}}(\b{A}\cdot \b{B}),\tag{1}
$$
where $\nabla_{\b{A}}(\b{A}\cdot \b{B})$ is that $\b{A}$ is differentiated while $\b{B}$ is held constant. We can prove:
$$
\nabla_{\b{A}}(\b{A}\cdot \b{B}) = (\b{B}\cdot \nabla) \b{A} + \b{B}\times(\nabla \times\b{A}).\tag{2}
$$
This is a very geometrical identity in that, for $\b{B}$ fixed, the derviate of $\b{A}$ can be decomposed into the normal derivative and the tangential derivative along $\b{A}$:
\begin{align}
&\nabla_{\b{A}}(\b{A}\cdot \b{B}) =
\begin{pmatrix}\partial_x A_1 &\partial_x A_2 & \partial_x A_3
\\ \partial_y A_1& \partial_y A_2 & \partial_y A_3
\\ \partial_z A_1 & \partial_z A_2 & \partial_z A_3
\end{pmatrix}\begin{pmatrix}B_1\\B_2 \\B_3\end{pmatrix}
\\
=&
\begin{pmatrix}\partial_x A_1 &\partial_y A_1 & \partial_z A_1
\\ \partial_x A_2& \partial_y A_2 & \partial_z A_2
\\ \partial_x A_3 & \partial_y A_3 & \partial_z A_3
\end{pmatrix}\begin{pmatrix}B_1\\B_2 \\B_3\end{pmatrix}
\\
&+
\begin{pmatrix}0 & \partial_x A_2-\partial_y A_1 & \partial_x A_3 -\partial_z A_1
\\ \partial_y A_1 - \partial_x A_2& 0 & \partial_y A_3 - \partial_z A_2
\\ \partial_z A_1 - \partial_x A_3 & \partial_z A_2-\partial_y A_3 & 0
\end{pmatrix}\begin{pmatrix}B_1\\B_2 \\B_3\end{pmatrix}
\\
=&(\b{B}\cdot \nabla) \b{A} + \b{B}\times(\nabla \times\b{A}),
\end{align}
extraction of the anti-symmetric part, for the cross product(wedge) can be written as the following anti-symmetric matrix form:
$$
\begin{pmatrix}
0 & -\partial_z & \partial_y
\\
\partial_z & 0 & -\partial_x
\\
-\partial_y & \partial_x & 0
\end{pmatrix}
\begin{pmatrix}
A_1\\A_2\\A_3
\end{pmatrix} = \nabla \times \b{A}.
$$
The rest is plugging (2) into (1), and do the same for the other term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/412260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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help understanding step in derivation of correlation coefficient I'm looking to understand the starred step in the derivation below (also, if someone could help with the LaTex alignment, I'd appreciate it).
The regression line is $y= b_0 + b_1 x$, where $b_0$ and $b_1$ can be found by:
1) taking the difference between each observed value $y_i$ and the expected point regression line, $b_0 + b_1 x_i$
$$\text{ difference } = y_i - b_0 -b_1 x_i$$
2) summing the square of the differences from 1) to get the sum of squares
$$SS = \sum \limits_{i=1}^n (y_i - b_0 -b_1 x_i)^2$$
3) taking the partial derivative with respect to $b_0$ and $b_1$, then solving for each
$$
\begin{align}
\text{ solving for } b_0 \\
SS &= \sum(y_i - b_0 -b_1 x_i)^2\\
SS &= \sum (Y_i ^2 - 2Y_i b_0 - 2 Y_i b_1+ 2b_0 b_1X_i + b_1^2X_i^2+b_0^2) &\text{expand the square}\\
\frac{ \partial }{\partial_{b_0} }SS &= \sum (-2Y_i + 2b_1 X_i + 2b_0) &\text{partial derivative wrt} b_0\\
0 &= \sum 2(-Y_i + b_1 X_i + b_0) &\text{factor out 2 from the sum}\\
0 &= \sum (-Y_i + b_1 X_i + b_0) &\text{divide both sides by 2}\\
0 &= \sum -Y_i + \sum b_1 X_i + \sum b_0 &\text{split summation into parts}\\
\sum Y_i &= \sum b_1 X_i + \sum b_0 \\
\sum Y_i &= b_1 \sum X_i + n b_0 \\
\frac{1}{n}(\sum Y_i - b_1 \sum X_i ) &= b_0 \\
\bar Y - b_1 \bar X &= b_0 \text { rewrite sums as averages since } \frac{1}{n} \sum Y_i = \bar Y\\
\end{align}
$$
$$
\begin{align}
\\
\text{solving for } b_1\\
\frac{ \partial }{\partial_{b_1} }SS &= \sum (-2Y_iX_i + 2b_0 X_i + 2 b_1 X_i^2) &\text{ partial derivative wrt } b_1\\
0 &= 2\sum (-Y_iX_i + b_0 X_i + b_1 X_i^2) \\
0 &= \sum (-Y_iX_i + b_0 X_i + b_1 X_i^2) \\
0 &= -\sum Y_iX_i + b_0 \sum X_i + b_1 \sum X_i^2 &\text{ split summation}\\
0 &= -\sum Y_iX_i + (\bar Y - b_1 \bar X) \sum X_i + b_1 \sum X_i^2 &\text{ substitue } b_0\\
0 &= -\sum Y_iX_i + (\bar Y \sum X_i - b_1 \bar X \sum X_i) + b_1 \sum X_i^2 &\text{ distribute sum}\\
b_1 \bar X \sum X_i - b_1 \sum X_i^2 &= -\sum Y_iX_i + \bar Y \sum X_i &\text{ collect } b_1 \text{ terms}\\
b_1 (\bar X \sum X_i - \sum X_i^2) &= -\sum Y_iX_i + \bar Y \sum X_i \\
b_1 &= { \bar Y \sum X_i -\sum Y_iX_i \over (\bar X \sum X_i - \sum X_i^2) }\\
b_1 &= { \frac{1}{n} \sum Y_i \sum X_i -\sum Y_iX_i \over (\frac{1}{n} \sum X_i \sum X_i - \sum X_i^2) } \biggr ( \frac{-n}{-n} \biggr )\\
b_1 &= { n \sum Y_iX_i - \sum Y_i \sum X_i \over n \sum X_i^2 -(\sum X_i)^2 } \\
\end{align}
$$
$$
\begin{align}
b_0 &= \frac{1}{n} \sum y_i - b_1 \frac{1}{n} \sum x_i\\\\\\
b_1 &= {n \sum x_i y_i - \sum x_i \sum y_i \over n \sum x_i^2-(\sum x_i)^2}
\end{align}
$$
(derivation shown in http://polisci.msu.edu/jacoby/icpsr/regress3/lectures/week2/5.LeastSquares.pdf)
From this point you can use $b_1$ to get the correlation coefficient as follows:
$$
\begin{align}
b_1 &= {\frac{1}{n} \sum x_i y_i - (\frac{1}{n}\sum x_i) (\frac{1}{n} \sum y_i ) \over (\frac{1}{n} \sum x_i^2) -(\frac{1}{n}\sum x_i)^2} & \text{ divide top and bottom by } n^2 \\\\
b_1 &= {\frac{1}{n} \sum x_i y_i - (\bar x) (\bar y ) \over (\frac{1}{n} \sum x_i^2) -(\bar x)^2} & \text{ rewrite product of sums as averages } \\\\
b_1 &= {\frac{1}{n} \sum (x_i - \bar x)(y_i - \bar y ) \over \frac{1}{n} \sum (x_i - \bar x)^2} & \color{red} *\text{application of inscrutably arcane magic} \\\\
b_1 &= { \sum (x_i - \bar x)(y_i - \bar y ) \over \sqrt{\sum (x_i - \bar x)^2} \sqrt{\sum (x_i - \bar x)^2} } & \text{cancel } \frac{1}{n}\text{, factor denominator }\\\\
b_1 &= { \sum (x_i - \bar x)(y_i - \bar y ) \over \sqrt{\sum (x_i - \bar x)^2} \sqrt{\sum ( x_i - \bar x)^2} } \biggr({\sqrt{\sum(y_i - \bar y)^2} \over \sqrt{\sum(y_i - \bar y)^2}}\biggr) & \text{multiply by 1 } \\\\
b_1 &= { \sum (x_i - \bar x)(y_i - \bar y ) \over \sqrt{\sum (x_i - \bar x)^2} \sqrt{\sum(y_i - \bar y)^2}} \biggr({\sqrt{\sum(y_i - \bar y)^2} \over \sqrt{\sum ( x_i - \bar x)^2} }\biggr) & \text{re-arrange } \\\\
b_1 &= R \frac{S_x}{S_y}
\end {align}
$$
| For the numerator, observe that:
$$
\begin{align}
\frac{1}{n} \left( \sum_{i=1}^n x_iy_i \right) - \bar x \bar y
&= \frac{1}{n} \left( \sum_{i=1}^n x_iy_i \right) - \dfrac{n}{n}\bar x \bar y & \text{common denominator}\\
&= \frac{1}{n} \left( \sum_{i=1}^n x_iy_i - n\bar x \bar y \right) & \text{factor out }1/n\\
&= \frac{1}{n} \left( \sum_{i=1}^n x_iy_i - n\bar x \bar y - n\bar x \bar y + n\bar x \bar y \right) & \text{add $0$ in a fancy way }\\
&= \frac{1}{n} \left( \sum_{i=1}^n x_iy_i - \bar x(n\bar y) - \bar y(n\bar x) + n(\bar x \bar y) \right) & \text{rearrange }\\
&= \frac{1}{n} \left( \sum_{i=1}^n x_iy_i - \bar x\sum_{i=1}^ny_i - \bar y \sum_{i=1}^n x_i + \sum_{i=1}^n\bar x \bar y \right) & \text{change back to sigmas}\\
&= \frac{1}{n} \sum_{i=1}^n (x_iy_i - \bar xy_i - \bar y x_i + \bar x \bar y) & \text{combine sigmas}\\
&= \frac{1}{n} \sum_{i=1}^n (x_i- \bar x)(y_i - \bar y) & \text{factor}\\
\end {align}
$$
As for the denominator:
$$
\begin{align}
\left(\frac{1}{n} \sum_{i=1}^n x_i^2 \right) - (\bar x)^2
&= \left(\frac{1}{n} \sum_{i=1}^n x_i^2 \right) - \left(\dfrac{n}{n}(\bar x)^2 \right) & \text{common denominator}\\
&= \dfrac{1}{n}\left(\sum_{i=1}^n x_i^2 - n(\bar x)^2 \right) & \text{factor out $1/n$}\\
&= \dfrac{1}{n} \left( \sum_{i=1}^n x_i^2 - 2n(\bar x)^2 + n(\bar x)^2 \right)& \text{add $0$ in a fancy way}\\
&= \dfrac{1}{n} \left( \sum_{i=1}^n x_i^2 - 2\bar x(n\bar x) + n(\bar x)^2 \right)& \text{rearrange}\\
&= \dfrac{1}{n} \left( \sum_{i=1}^n x_i^2 - 2\bar x \sum_{i=1}^nx_i + \sum_{i=1}^n(\bar x)^2 \right)& \text{change back to sigmas}\\
&= \dfrac{1}{n} \sum_{i=1}^n \left( x_i^2 - 2\bar xx_i + (\bar x)^2 \right)& \text{combine sigmas}\\
&= \dfrac{1}{n} \sum_{i=1}^n ( x_i - \bar x)^2 & \text{factor}\\
\end {align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/413396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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} |
Prove that if $a b c=1$, $a^2+b^2+c^2\ge 3$. (for $a,b,c\in\mathbb{R}$) I can do this problem using calculus minimization techniques, using Lagrange multipliers to find the equations $a^2=b^2=c^2$, so with $abc=1$, $a$, $b$, and $c$, are either $-1$ or $1$ (making sure you don't end up with $abc$ negative). So if the minimum distances are at those points, then $a^2+b^2+c^2=3$ is the closest distance, so all other solutions are further away.
I just pieced together (while writing this) that there's another solution: The arithmetic-geometric inequality gives $3(abc)^{1/3}\le a+b+c$, and by the inequality of this question, we have $9(abc)^{2/3}\le (a+b+c)^2\le 3(a^2+b^2+c^2)$, so since $abc=1$, we have $3\le a^2+b^2+c^2$.
But, is it possible to prove this using other inequality tricks? It's just, the inequality on the link I gave, I had no idea about, so it's uncomfortable to use it.
| By AM/GM, we have
$$\frac{a^2+b^2+c^2}{3}\ge (a^2b^2c^2)^{1/3}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/414419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Given that $x = 4\sin \left( {2y + 6} \right)$ find dy/dx in terms of x My attempt:
$\eqalign{
& x = 4\sin \left( {2y + 6} \right) \cr
& {{dx} \over {dy}} = \left( 2 \right)\left( 4 \right)\cos \left( {2y + 6} \right) \cr
& {{dx} \over {dy}} = 8\cos \left( {2y + 6} \right) \cr
& {{dy} \over {dx}} = {1 \over {8\cos \left( {2y + 6} \right)}} \cr} $
$\eqalign{
& x = 4\sin \left( {2y + 6} \right) \cr
& {{dx} \over {dy}} = \left( 2 \right)\left( 4 \right)\cos \left( {2y + 6} \right) \cr
& {{dx} \over {dy}} = 8\cos \left( {2y + 6} \right) \cr
& {{dy} \over {dx}} = {1 \over {8\cos \left( {2y + 6} \right)}} \cr
& {\cos ^2}\left( {2y + 6} \right) + {\sin ^2}\left( {2y + 6} \right) = 1 \cr
& {\cos ^2}\left( {2y + 6} \right) + {\left( {{x \over 4}} \right)^2} = 1 \cr
& {\cos ^2}\left( {2y + 6} \right) = 1 - {{{x^2}} \over {16}} \cr
& \cos \left( {2y + 6} \right) = \sqrt {{{16 - {x^2}} \over {16}}} \cr
& \cos \left( {2y + 6} \right) = {{\sqrt {16 - {x^2}} } \over 4} \cr
& {{dy} \over {dx}} = {1 \over {2\sqrt {16 - {x^2}} }} = 1 \cr} $
Okay I've got it right, but the official answer confuses me, it says:
${{dy} \over {dx}} = {1 \over {8cos\left( {\arcsin \left( {{x \over 4}} \right)} \right)}} = \left( {\left( \pm \right){1 \over {2\sqrt {\left( {16 - {x^2}} \right)} }}} \right)$
This is the part i'm struggling to get my head around, although I arrive at the same answer.
Okay
$\arcsin {x \over 4} = 2y + 6$
but how does the answer then go :
$\left( {\left( \pm \right){1 \over {2\sqrt {\left( {16 - {x^2}} \right)} }}} \right)$
is there a shortcut or trick I overlooked?
I think I need some sleep, thanks...
| You are on the right track.
Notice that
$$\cos^2(2y+6) = 1-\frac{x^2}{16}$$
implies
$$\cos(2y+6) = \pm\sqrt{1-\frac{x^2}{16}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/415985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Showing that $\int_0^{\pi/3}\frac{1}{1-\sin x}\,\mathrm dx=1+\sqrt{3}$
Show that $$\int_0^{\pi/3}\frac{1}{1-\sin x}\,\mathrm dx=1+\sqrt{3}$$
Using the substitution $t=\tan\frac{1}{2}x$
$\frac{\mathrm dt}{\mathrm dx}=\frac{1}{2}\sec^2\frac{1}{2}x$
$\mathrm dx=2\cos^2\frac{1}{2}x\,\mathrm dt$
$=(2-2\sin^2\frac{1}{2}x)\,\mathrm dt$
How do you get this in the form of $t$ instead of $x$ using $\sin A=\dfrac{2t}{1+t^2}$ ?
$$=\int_0^{1/\sqrt3}\frac{2-2\sin^2\frac{1}{2}x}{1-\frac{2t}{1+t^2}}\mathrm dt\,??$$
| I'd just use the identity $\dfrac 1{1-\sin x}=\dfrac{1+\sin x}{1-\sin^2x}=\dfrac{1+\sin x}{\cos^2x}=\sec^2x+\sec x\tan x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Indefinite Integration : $\int \frac{dx}{(x+\sqrt{(x^2-1)})^2}$ Problem :
Solve : $\int \frac{dx}{(x+\sqrt{(x^2-1)})^2}$......(i)
I tried :
Let $x =\sec\theta$ therefore , (i) will become after some simplification
$$\int \frac{\sin\theta}{(1+\sin\theta)^2}d\theta$$
but i think its wrong method of approaching please suggest further....thanks..
| HINT: $$\frac1{x+\sqrt{x^2-1}}=x-\sqrt{x^2-1}$$
So, $$\frac1{(x+\sqrt{x^2-1})^2}=(x-\sqrt{x^2-1})^2=x^2+x^2-1-2x\sqrt{x^2-1}=2x^2-1-2x\sqrt{x^2-1}$$
Put $x^2-1=y$ for the last part
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/423294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\log _5 7 < \sqrt 2.$
Prove that $\log _5 7 < \sqrt 2.$
Trial : Here $\log _5 7 < \sqrt 2 \implies 5^\sqrt 2 <7.$ But I don't know how to prove this. Please help.
| Want to prove that
*
*$\log_5{7} = \frac{\lg{7}}{\lg{5}} < \sqrt{2}$
Equivalently we can show that
*
*$\lg{7} < \lg{5}\times\sqrt{2}$
*$7 < 5^{\sqrt{2}}$
where $\lg$ is the base 2 logarithm. Notice that
*
*$5\times5^{\frac{2}{5}}= 5^{1.4} <5^{\sqrt{2}}$
So can we show that $\frac{7}{5} < 5^{\frac{2}{5}}$? Sure, since $7<8=32768^{\frac{1}{5}}<78125^{\frac{1}{5}}$. Hence
*
*$7 < 5^{1.4} <5^{\sqrt{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/423348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
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Find the area of the portion of the surface of the sphere $x^2+y^2+z^2=4x$ that is cut off by a nappe of the cone $y^2+z^2=x^2$. I'm having trouble with a Calculus question where I am supposed to use the following formula to calculate the area of a surface:
Suppose that $f$ and its first-order partial derivatives are continuous in the closed region $R$ in the $xy$ plane. So, if $S$ is the measure of the area of the surface $z=f(x,y)$ above $R$,
$$ S = \int_R\int \sqrt{f_x^2(x,y)+f_y^2(x,y)+1} dxdy$$
(Here, $f_x(x,y)$ and $f_y(x,y)$ stand for the first-order partial derivatives with respect to $x$ and $y$, respectively.)
The question asks me the following:
Find the area of the portion of the surface of the sphere $x^2+y^2+z^2=4x$ that is cut off by a nappe of the cone $y^2+z^2=x^2$.
I made the following plot of the situation described in the question. To make things simpler, I plotted only the intersection of the sphere and the cone with the $xy$ plane:
The image above shows the part of the sphere (indicated in cyan (light blue) color) that is cut off by a nappe of the cone. Notice that the intersection of the sphere with the $xy$ plane is the circle $y^2=4x-x^2$, and the intersection of the cone with the $xy$ plane are the lines $y=x$ and $y=-x$.
To apply the given formula, I have to find the region $R$ over which the double integral will be computed. From the image, the region of integration $R$ is the region between the line $x=2$ and the portion of the circumference marked in red. The desired portion of the surface of the sphere is above this region $R$.
So, the integration limits are the following:
$$\int_2^4\int_0^\sqrt{4x-x^2} \cdots dydx$$
In other words, the outer integral is computed with $x$ going from $2$ to $4$, and the inner integral is computed with $y$ going from $0$ to $\sqrt{4x-x^2}$ (notice that I have switched $dx$ and $dy$ from the original formula for convenience).
Now, if I compute the first-order partial derivatives indicated in the formula, I obtain the following results: $f_x(x,y) = \dfrac{2-x}{\sqrt{4x-x^2-y^2}}$ and $f_y(x,y) = -\dfrac{y}{\sqrt{4x-x^2-y^2}}$.
The double integral is, then:
$$\int_2^4\int_0^\sqrt{4x-x^2} \sqrt{\frac{(2-x)^2+y^2}{4x-x^2-y^2} + 1} dydx$$
which, after some simplification, becomes:
$$\int_2^4\int_0^\sqrt{4x-x^2} \sqrt{\frac{4}{4x-x^2-y^2}} dydx$$
I get stuck at this point, because I can't seem to find a way of solving this integral. I have tried to solve the inner integral
$$\int_0^\sqrt{4x-x^2} \sqrt{\frac{4}{4x-x^2-y^2}} dy$$
but without any success. Is the set up of the double integral correct? Or is there a way to solve the above integral and get a real result?
| I will solve this problem again using the suggestion by Ted Shifrin in the comments to this question above. Doing the integral in polar coordinates, the solution becomes much simpler.
I will take the polar plane to be the $yz$ plane.
In $yz$ coordinates, the desired surface is the function $x=f(y,z)=\sqrt{4-y^2-z^2}+2$ (which corresponds to the right hemisphere of the sphere represented in the picture in the question).
As required by the formula, the surface $x=f(y,z)=\sqrt{4-y^2-z^2}+2$ is above a region $R$ which lies on the $yz$ plane (polar plane). In this case, the region of integration $R$ is the projection of the sphere on the polar plane; in other words, $R$ is a circle of radius 2 centered at $(x,y,z)=(2,0,0)$.
Thus, the limits of integration in polar coordinates will be:
$$\int_0^2\int_0^{2\pi} \cdots r d\theta dr$$
The partial derivatives of $f(x,y,z)$ are $f_y = -\dfrac{y}{\sqrt{4-y^2-z^2}}$ and $f_z = -\dfrac{z}{\sqrt{4-y^2-z^2}}$.
The square root inside the double integral is:
$$\sqrt{f_y^2(y,z)+f_z^2(y,z)+1} = \sqrt{ \dfrac{y^2}{4-y^2-z^2} + \dfrac{z^2}{4-y^2-z^2} + 1 }$$
$$= \sqrt{ \dfrac{4}{4-y^2-z^2} }$$
Converting the above expression to polar coordinates (substituting $y^2+z^2=r^2$):
$$\sqrt{ \dfrac{4}{4-r^2} }$$
Thus, the double integral in polar coordinates is:
$$\int_0^2\int_0^{2\pi} \sqrt{ \dfrac{4}{4-r^2} } r d\theta dr$$
Solving the inner integral, the above expression reduces to $\int_0^2 4 \pi r \sqrt{ \dfrac{1}{4-r^2} } dr$, which can be solved by substitution (for example, by setting $u=4-r^2$ and $du=-2rdr$) and gives $8\pi$ as its result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/423767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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problem with partial fraction decomposition I want to do partial fraction decomposition on the following rational function:
$$\frac{1}{x^2(1+x^2)^3}$$
So I proceed as follows:
$$\begin{align}
\frac{1}{x^2(1+x^2)^3} &= \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{1 + x^2} + \frac{Ex + F}{(1 + x^2)^2} + \frac{Gx + H}{(1 + x^2)^3} \\
1 = &Ax(1+x^2)^3 + B(1+x^2)^3 + (Cx + D)x^2(1+x^2)^2 \\
& + (Ex + F)x^2(1+x^2) + (Gx + H)(x^2) \\
1 = &(A+C)x^7 + (B+D)x^6 + (3A+2C+E)x^5 + (3B+2D)x^4 \\
& + (3A+C+E+G)x^3 + (3B+D+H)x^2 +(A)x + (B)\\
\end{align}$$
But from this, $B=1$, $B=-D$, and $3B+2D=0$, which is an inconsistent system. What am I doing wrong?
| Not answering your question, but I see $x^2$ everywhere, so it is better to find a partial fractions expression for $\dfrac{1}{u(1+u)^3}$.
So we are looking for $A$, $B$, $C$, $D$ such that
$$\frac{1}{u(1+u)^3}=\frac{A}{u}+\frac{B}{1+u}+\frac{C}{(1+u)^2}+\frac{D}{(1+u)^3}.$$
When we bring the right-hand side to a common denominator $u(1+u)^3$, the numerator becomes
$$A(1+u)^3+Bu(1+u)^2+Cu(1+u)+Du.$$
This is identically equal to $1$. Put $u=0$. We get $A=1$. Put $u=-1$. We get $D=-1$. The coefficient of $u^3$ is $A+B$ so $B=-1$. The coefficient of $u^2$ is $3A+2B+C$, so $C=-1$.
Now replace $u$ by $x^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\sin^6 x+\cos^6 x$, what am I doing wrong here? I have $\sin 2x=\frac 23$ , and I'm supposed to express $\sin^6 x+\cos^6 x$ as $\frac ab$ where $a, b$ are co-prime positive integers. This is what I did:
First, notice that $(\sin x +\cos x)^2=\sin^2 x+\cos^2 x+\sin 2x=1+ \frac 23=\frac53$ .
Now, from what was given we have $\sin x=\frac{1}{3\cos x}$ and $\cos x=\frac{1}{3\sin x}$ .
Next, $(\sin^2 x+\cos^2 x)^3=1=\sin^6 x+\cos^6 x+3\sin^2 x \cos x+3\cos^2 x \sin x$ .
Now we substitute what we found above from the given:
$\sin^6 x+\cos^6+\sin x +\cos x=1$
$\sin^6 x+\cos^6=1-(\sin x +\cos x)$
$\sin^6 x+\cos^6=1-\sqrt {\frac 53}$
Not only is this not positive, but this is not even a rational number. What did I do wrong? Thanks.
| $\sin^6x + \cos^6x = (\sin^2x)^3 + (\cos^2x)^3 =(\sin^2x + \cos^2x)(\sin^4x + \cos^4x -\sin^2x\cos^2x)$
$\sin^4x+\cos^4x -\sin^2x\cos^2x = (\sin^2x + \cos^2x)^2 - 2\sin^2x\cos^2x -\sin^2x\cos^2x$
or $1-3\sin^2x\cos^2x = 1-3\left(\dfrac13\right)^2 = \dfrac23$.
| {
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"url": "https://math.stackexchange.com/questions/425664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Fractional overlap of 1/2 and 1/3 Given a subset of the natural number sequence (positive integers starting from 1) we could say that $\frac12$ of the numbers in the set are divisible by 2.
e.g if the set were ${[1,2,3,4,5,6,7]}$ we could say that $3\frac12$ of the numbers in it are divisible by 2.
If we now wanted to work out how many numbers are divisible by 3, we could work it out as $\frac73 = 2\frac13$ and we know this is correct because if we look at the set we can see that the numbers 3 and 6 are the 2 numbers that are divisble by 3.
If we wanted to work out how many numbers are divisible by 2 OR 3. At first glace I thought I could add up the 2 fractions and then subtract the overlap.
This would then equate to $\frac12+\frac13-(\frac12*\frac13) = \frac23$
This makes sense $\frac23$ of all natural numbers are divisible by $2$ or $3$. So if we wanted to see how many numbers were divisible by $2$ OR $3$ in the set ${1,2,3,4,5,6,7,8}$ we could say $8 * \frac23 = 5\frac13$ and this makse sense because the $5$ numbers divisible by $2$ or $3$ are ${2,3,4,6,8}$
This is where I get confused, when I test this against the number 10 for example I get $10 * \frac23 = 6\frac23$ BUT there are $7$ numbers under $10$ that are divisible by both $2$ or $3$, so I was expecting the whole number component to be $7$
Please help me understand. Is it possible to create such a fraction that would tell me the number of elements in the set that are divisible by 2 or 3?
Thanks in advance
| Of the numbers $1$ to $n$, $\lfloor n/m \rfloor$ (i.e. the greatest integer $\le n/m$) are divisible by $m$. Now $x$ is divisible by both $m_1$ and $m_2$ if and only if $x$ is divisible by $\text{lcm}(m_1, m_2)$ (the least common multiple of $m_1$ and $m_2$: this is $m_1 m_2$ if $m_1$ and $m_2$ have no common factor $> 1$).
So of the numbers $1$ to $n$, $\lfloor n/m_1 \rfloor + \lfloor n/m_2 \rfloor - \lfloor n/\text{lcm}(m_1, m_2) \rfloor$ are divisible by $m_1$ or $m_2$.
For example, of the numbers $1$ to $7$, there are
$$\lfloor 7/2 \rfloor + \lfloor 7/3 \rfloor - \lfloor 7/6 \rfloor = 3 + 2 - 1 = 4$$ divisible by $2$ or $3$.
Since $\dfrac{n}{m} - 1 < \left\lfloor \dfrac{n}{m} \right\rfloor \le \dfrac{n}{m}$,
$$ \eqalign{\dfrac{n}{m_1} + \dfrac{n}{m_2} - \dfrac{n}{\text{lcm}(m_1,m_2)} - 2 &<
\left\lfloor \dfrac{n}{m_1} \right\rfloor + \left\lfloor \dfrac{n}{m_2} \right\rfloor - \left\lfloor \dfrac{n}{\text{lcm}(m_1, m_2)}\right\rfloor \cr &< \dfrac{n}{m_1} + \dfrac{n}{m_2} - \dfrac{n}{\text{lcm}(m_1,m_2)} + 1\cr}$$
| {
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Calculate the value of x, y and z coordinates System of the equations:
Equation 1:
$$-360 = -6x + x^2 - 40y + y^2 - 20z + z^2$$
Equation 2:
$$-600 = -40x + x^2 - 30y + y^2 - 100z + z^2$$
Equation 3:
$$59.85 = -10x + x^2 - 4y + y^2 - 10z + z^2$$
| It looks like you are trying
to find the intersection of three spheres
(with equation $r^2 = (x-a)^2+(y-b)^2+(z-c)^2$
or $x^2-2ax+y^2-2by+z^2-2cz = r^2-a^2-b^2-c^2$).
Anyway, the equations are
$\begin{align}
a &=-360 = -6x + x^2 - 40y + y^2 - 20z + z^2\\
b&= -600 = -40x + x^2 - 30y + y^2 - 100z + z^2\\
c&= 59.85 = -10x + x^2 - 4y + y^2 - 10z + z^2\\
\end{align}
$
Subtracting the first two,
$a-b = 34x -10y +80z$.
Subtracting the last two,
$b-c = 30x -26y -90z$.
These will let you get
any two of $x, y$, and $z$
in terms of the other.
For example,
you might get
$y = px+q$
and $z = rx+s$
for some $p, q, r, $and $s$.
Substitute these expressions in any of
your equations
and you will get a quadratic for $x$.
For example,
if you substitute these in the first equation,
you get
$\begin{align}
a
&= -6x + x^2 - 40y + y^2 - 20z + z^2\\
&=-6x + x^2 - 40(px+q) + (px+q)^2 - 20(rx+s) + (rx+s)^2\\
&=-6x + x^2 - 40px-40q + p^2x^2+2pxq+q^2 - 20rx-20s + r^2x^2+2rxs+s^2\\
&=-x(6+40p+20r+2pq+2rs) + x^2(1+p^2+r^2) +q^2 -40q-20s +s^2\\
\end{align}
$
Solve this quadratic in $x$
and you will get
0, 1, or 2 real roots depending how
the three spheres intersect.
Get $y$ and $z$ for each $x$ and you have your solutions.
To see how the spheres can intersect,
look at two spheres.
They can be disjoint,
be tangent,
or intersect in a circle.
Adding a third sphere
allows for no common points,
one common point if the sphere
passes through the tangent point,
or two points if the
sphere passes through the circle of intersection.
There is also the possibility of
two of the spheres coinciding,
in which case
the intersection could also be a circle.
In this case, the three original equations
would not be linearly independent.
| {
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Why is $X^4+1$ reducible over $\mathbb F_p$ with $p \geq 3,$ prime I have proven that in $\mathbb F_{p^2}^*$ exists an element $\alpha$ with $\alpha^8 = 1$.
Let $f(X) := X^4+1 \in \mathbb F_p[X]$. How can I prove that $f$ is reducible over $\mathbb F_p$?
Has $f$ a zero in $\mathbb F_p$ ?
| We can use three different factorizations:
If $2$ is a square mod $p$. (which occurs for $p=\pm1$ mod $8$)
$x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-2x^2$ Assume $q^2=2$ mod $p$. Then $x^4+1=(x^2+1+qx)(x^2+1-qx)$.
If $-2$ is a square mod $8$. (which occurs for $p=1$ or $3$ mod $8$)
$x^4+1=x^4-2x^2+1+2x^2=(x^2-1)^2-(-2)x^2=(x^2-1+rx)(x^2-1-rx)$, where $r^2=-2$ mod $8$.
Finally, if $-1$ is a square mod $8$. (which happens for $p=1$ mod $4$)
$x^4+1=x^4-(-1)=(x^2+r)(x^2-r)$, where $r^2=-1$ mod $p$.
Checking if all the cases are covered.
All (odd) primes are congruent to either $1,3,5,7$ mod $8$. The cases $1$ and $7$ are covered by the first factorization. The case $3$ by the second. The case $5$ by the third, since $p=5$ mod $8$ implies $p=1$ mod $4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing that $1^k+2^k + \dots + n^k$ is divisible by $n(n+1)\over 2$
For any odd positive integer $k\geq1$, the sum $1^k+2^k + \dots + n^k$ is divisible by $n(n+1)\over 2$.
I used induction principle for the solution but cannot prove it.
I took $P(k) = 1^k+2^k+\dots+n^k$.
For $P(1)$ it is true.
For $P(n)$ let it be true.
But for $P(n+1)$ I cannot solve it.
| As $k$ is odd,
$r^k+(n+1-r)^k$ is divisible by $r+(n+1-r)=n+1$
and $s^k+(n-s)^k$ is divisible by $s+(n-s)=n$
Putting $r=1,2,\cdots,n-1,n$ and adding them we get
$$(n+1)|\sum_{1\le r\le n}\{r^k+(n+1-r)^k\} \implies (n+1)|2\sum_{1\le r\le n}r^k$$
Putting $s=1,2,\cdots,n-1,n$ and adding them we get
$$n|\sum_{1\le s\le n}\{s^k+(n-s)^k\} \implies (n)|2\sum_{1\le s\le n}s^k \implies n|2\sum_{1\le r\le n}r^k$$
Now, either $n$ or $n+1$ is odd
If $n$ is odd, $$n|2\sum_{1\le r\le n}r^k\implies n|\sum_{1\le r\le n}r^k$$
$$\implies \text{lcm}(n+1,n) | 2\sum_{1\le r\le n}r^k$$
$$\implies n(n+1) | 2\sum_{1\le r\le n}r^k\implies \frac{n(n+1)}2 | \sum_{1\le r\le n}r^k$$
Similarly when $n+1$ is odd
| {
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"timestamp": "2023-03-29T00:00:00",
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Find number of solutions of $|2x^2-5x+3|+x-1=0$
Problem: Find number of solutions of $|2x^2-5x+3|+x-1=0$
Solution:
Case 1: When $2x^2-5x+3 \geq 0$
Then we get, $2x^2-5x+3+x-1=0$
x=1,1
Case 2: When $2x^2-5x+3 < 0$
Then we get, $-2x^2+5x-3+x-1=0$
x=1,2
In both cases, common value of x is 1
Hence solution is x=1
Am I doing right ??
Somebody told me solution is x=2
| Nope, if you have $|f(x)| + g(x) = 0$ $(1)$ you solve it as follows:
*
*assume that $f(x)\geq 0$, find solutions of $f+g = 0$ - say $x_1$. Check whether $f(x_1)\geq 0$. If it holds, then $x_1$ solves $(1)$ as well. If it does not hold, it's not a solution of the original equation.
*assume $f(x)<0$, find solutions of $g-f = 0$ and check whether for them $f<0$.
In your case: for $f+g = 0$ you get a solution $x = 1$, which satisfies $f(1)\geq 0$, so that this is indeed one of the solutions of the original equation. For the case $f<0$ you get $x = 1$ or $x = 2$. We don't have to check $x = 1$ since we know it's already a solution, so we only check whether $f(2)<0$. Since the latter does not hold: $2\cdot 2^2 -5\cdot 2+3 = 1>0$, we obtain that $x=2$ is not a solution, and hence $1$ is the only solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove a Trigonometric Series Question:
$$\cot^{2}\frac{\pi }{2m+1}+\cot^{2}\frac{2\pi }{2m+1}+\cdots+\cot^{2}\frac{m\pi }{2m+1}=\frac{m(2m-1)}{3}$$
$m$ is a positive integer.
Attempt:
I started by showing that
$$\sin(2m+1)\theta =\binom{2m+1}{1}\cos^{2m}\theta \sin\theta -\binom{2m+1}{3}\cos^{2m-2}\theta \sin^{3}\theta +\cdots+(-1)^{m}\sin^{2m+1}\theta$$
By expanding $\left(\text{cis }\theta\right)^{2m+1}$ and then equating the imaginary part.
| Divide this equality by $\sin^{2m+1}\theta$ to get
$$
\frac{\sin(2m+1)\theta}{\sin^{2m+1}\theta}=\sum\limits_{k=0}^m (-1)^k{2m+1 \choose 2k+1}\cot^{2k} \theta\tag{1}
$$
Denote
$$
P_m(x)=\sum\limits_{k=0}^m (-1)^k{2m+1 \choose 2k+1} x^m
$$
This is polynomial of degree $m$. From $(1)$ we see that $P_m$ have $m$ roots $x_k=\cot^2\frac{\pi k}{2m+1}$. By Vietas formula their sum equals to
$$
-\frac{{2m+1 \choose 3}}{{2m+1 \choose 1}}=\frac{m(2m-1)}{3}
$$
So we get
$$
\sum\limits_{k=1}^m\cot^2\frac{\pi k}{2m+1}=\frac{(2m-1)m}{3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/429877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factor $x^4 - 11x^2y^2 + y^4$ This is an exercise from Schaum's Outline of Precalculus. It doesn't give a worked solution, just the answer.
The question is:
Factor $x^4 - 11x^2y^2 + y^4$
The answer is:
$(x^2 - 3xy -y^2)(x^2 + 3xy - y^2)$
My question is:
How did the textbook get this?
I tried the following methods (examples of my working below):
*
*U-Substitution.
*Guess and Check.
*Reversing the question (multiplying the answer out).
Here is my working for each case so far.
(1) U-Substitution.
I tried a simpler case via u-substitution.
Let $u = x^2$ and $v = y^2$. Then $x^4 - 11x^2y^2 + y^4 = u^2 -11uv + v^2$. Given the middle term is not even, then it doesn't factor into something resembling $(u + v)^2$ or $(u - v)^2$. Also, there doesn't appear to be any factors such that $ab = 1$ (the coefficient of the third term) and $a + b = -11$ (the coefficient of the second term) where $u^2 -11uv + v^2$ resembles $(u \pm a)(v \pm b)$ and the final answer.
(2) Guess and Check.
To obtain the first term in $x^4 - 11x^2y^2 + y^4$ it must be $x^2x^2 = x^4$. To obtain the third term would be $y^2y^2 = y^4$. So I'm left with something resembling:
$(x^2 \pm y^2)(x^2 \pm y^2)$.
But I'm at a loss as to how to get the final solution's middle term from guessing and checking.
(3) Reversing the question
Here I took the answer, multiplied it out, to see if the reverse of the factoring process would illuminate how the answer was generated.
The original answer:
$(x^2 - 3xy -y^2)(x^2 + 3xy - y^2) = [(x^2 - 3xy) - y^2][(x^2 + 3xy) - y^2]$
$= (x^2 - 3xy)(x^2 + 3xy) + (x^2 - 3xy)(-y^2) + (x^2 + 3xy)(-y^2) + (-y^2)(-y^2)$
$= [(x^2)^2 - (3xy)^2] + [(-y^2)x^2 + 3y^3x] + [(-y^2)x^2 - 3y^3x] + [y^4]$
$= x^4 - 9x^2y^2 - y^2x^2 + 3y^3x - y^2x^2 - 3y^3x + y^4$
The $3y^3x$ terms cancel out, and we are left with:
$x^4 - 9x^2y^2 - 2x^2y^2 + y^4 = x^4 - 11x^2y^2 + y^4$, which is the original question.
The thing I don't understand about this reverse process is where the $3y^3x$ terms came from. Obviously $3y^3x - 3y^3x = 0$ by additive inverse, and $a + 0 = a$, but I'm wondering how you would know to add $3y^3x - 3y^3x$ to the original expression, and then make the further leap to factoring out like terms (by splitting the $11x^2y^2$ to $-9x^2y^2$, $-y^2x^2$, and $-y^2x^2$).
| \begin{align*}
x^4 - 11x^2y^2 + y^4
&= x^4 - 2x^2y^2 + y^4-9x^2y^2 \\
&= (x^2-y^2)^2-(3xy)^2 \\
&= (x^2-y^2-3xy)(x^2-y^2+3xy).
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit on the expression containing sides of a triangle To find the bounds of the expression $\frac{(a+b+c)^2}{ab+bc+ca}$, when a ,b, c are the sides of the triangle.
I could disintegrate the given expression as $$\dfrac{a^2+b^2+c^2}{ab+bc+ca} + 2$$ and in case of equilateral triangle, the limit is 3.
Now how to proceed further?
| Let $x=p-a$, $y=p-b$, and $z=p-c$, where $p=(a+b+c)/2$. The main expression is equal to $$F:=\frac{4(x+y+z)^2}{x^2+y^2+z^2+3(xy+yz+zx)}=\frac{4\sum x^2+8\sum yz}{\sum x^2 +3\sum yz}\\=4-\frac{4}{\sum x^2/\sum yz +3}.$$ From Cauchy-Schwarz we know that $\sum x^2 \geq \sum yz$. Thus, $F\geq 3$. Furthermore, it is obvious that $F<4$. The lower bound is attainable at $x=y=z$ (i.e., $a=b=c$) while you can only tend to the upper bound by tending two of $x$,$y$, or $z$ two zero while keeping the third fixed.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$ 1 + \sqrt{1 - \sqrt{x^4-x^2}} = x$ can be written in the form $\frac{a}{b}$, find $a+b$. The solution to the equation
$$1 + \sqrt{1 - \sqrt{x^4-x^2}} = x$$
can be written in the form $\frac{a}{b}$ where $a$ and $b$ are coprime positive integers. Find $a+b$.
| I assume the question is not a live problem anymore. What you can do to solve this is just solving for $x$. So assume $x$ is a solution, then $\sqrt{1-\sqrt{x^4-x^2}}=x-1$, so $1-\sqrt{x^4-x^2}=(x-1)^2$. Continuing like this you will get to $-4x^3+5x^2=0$. This has two solutions, but one of the solutions is not a solution to the original equation, an excellent example that squaring will in general not yield to an equivalent equation.
| {
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"timestamp": "2023-03-29T00:00:00",
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Which of the following sets is a complete set of representatives modulo 7? Which of the following sets is a complete set of representatives modulo 7?
1) (1, 8, 27, 64, 125, 216, 343)
1 mod 7 = 1
8 mod 7 = 1
27 mod 7 = 6
64 mod 7 = 1
125 mod 7 = 6
216 mod = 6
343 mod = 0
2) (1, -3, 9, -27, 81, -243, 0)
1 mod 7 = 1
-3 mod 7 = 4
9 mod 7 = 2
-27 mod 7 = 1
81 mod 7 = 4
-243 mod 7 = 2
0 mod 7 = 0
3) (0, 1, -2, 4, -8, 16, -32)
0 mod 7 = 0
1 mod 7 = 1
-2 mod 7 = 5
4 mod 7 = 4
-8 mod 7 = 6
16 mod 7 = 2
-32 mod 7 = 3
i think its only 1. but wanted to make sure.
| Hint: you want a complete set whose members are equivalent, modulo 7, to one and only one element of the set: $\{0, 1, 2, 3, 4, 5, 6\}$, not necessarily in that order. Each and every one of these numbers must be represented by one and only one of the numbers in such a set. That is, you want representatives of the equivalence classes, modulo 7 which will consist of seven distinct integers, each one of which can be expressed in one and only one of the following forms $$(7k_1, 7k_2 + 1, 7k_3 +2, 7k_4 + 3, 7k_5 + 4, 7k_6 +5, 7k_7 + 6), \quad k_i \in \mathbb Z)$$
$(1)$ is not such a set.
Added
Note that the elements of $(3)$ represent each equivalence class, modulo 7, and hence forms a complete set of representatives: $$ 0\equiv 0 \pmod 7,\quad \;1 \equiv 1 \pmod 7, \quad -2\equiv 5\pmod 7,\quad \\ 4\equiv 4 \pmod 7,\quad -8 \equiv 6 \pmod 7,\quad 16\equiv 2 \pmod 7,\quad -32 \equiv 3 \pmod 7$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Unintentional Negative Sign in Limit Evaluation I've been working on evaluating the following limit:
$$\lim_{x\to 0} \left(\csc(x^2)\cos(x)-\csc(x^2)\cos(3x) \right)$$
According to my calculator, the limit should end up being 4.
Though I've tried using the following process to find the limit, I continue to get -4:
$$\begin{align}
\lim_{x\to 0} \left(\frac{\cos(x)-\cos(3x)}{\sin(x^2)}\right)&=\lim_{x\to 0} \left(\frac{\cos(x) - \cos(2x+x)}{\sin(x^2)}\right)\\
&=\lim_{x\to 0} \left(\frac{\cos(x) - \cos(2x)\cos(x) - \sin(2x)\sin(x)}{\sin(x^2)}\right)\\
&=\lim_{x\to 0} \left(\frac{\cos(x) - \left(1-2\sin^2(x)\right)\cos(x) - 2\sin^2(x)\cos(x)}{\sin(x^2)}\right)\\
&=\lim_{x\to 0} \left(\frac{\cos(x) - \cos(x)-2\sin^2(x)\cos(x)-2\sin^2(x)\cos(x)}{\sin(x^2)}\right)\\
&=\lim_{x\to 0} \left(\frac{\cos(x) - \cos(x)-4\sin^2(x)\cos(x)}{\sin(x^2)}\right)\\
&=\lim_{x\to 0} \left(\frac{-4\sin^2(x)\cos(x)}{\sin(x^2)}\right)\\
&=\lim_{x\to 0} \left(\frac{\frac{d}{dx}\left(-4\sin^2(x)\cos(x)\right)}{\frac{d}{dx}\left(\sin(x^2)\right)}\right)\\
&=\lim_{x\to 0} \left(\frac{-4\cos^2(x)\cos(x)-\sin^2(x)\sin(x)}{\cos(x^2)}\right)\\
&=\frac{-4\cdot1 - 0}{1} \\
&=-4 \end{align}$$
Is there a specific place in my steps where I am going wrong with respect to negative sign notation or distribution?
(Also, apologies for the extra steps... and possible incorrect ordering of labels, functions, and symbols.)
| $$ \bigg(\frac{\cos(x) - \cos(2x+x)}{\sin(x^2)}\bigg)=\bigg(\frac{\cos(x) - \cos(2x)\cos(x) + \sin(2x)\sin(x)}{\sin(x^2)}\bigg)$$
| {
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For real numbers $x$ and $y$, show that $\frac{x^2 + y^2}{4} < e^{x+y-2} $
Show that for $x$, $y$ real numbers, $0<x$ , $0<y$
$$\left(\frac{x^2 + y^2}{4}\right) < e^{x+y-2}. $$
Someone can help me with this please...
| From the well known inequality: $1+z\leq e^z$, we replace $z$ by $\frac{z}{2}-1$ to get $\frac{z}{2}\leq e^{\frac{z}{2}-1}$. Square both sides to get $\frac{z^2}{4}\leq e^{z-2}$. Now let $z=x+y$, to get: $\frac{(x+y)^2}{4}\leq e^{x+y-2}$. Hence:
$$\frac{x^2}{4}+\frac{y^2}{4}<\frac{(x+y)^2}{4}\leq e^{x+y-2}$$
Note:
$\frac{x^2}{4}+\frac{y^2}{4}< \frac{(x+y)^2}{4}$ because $x,y> 0$
| {
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"timestamp": "2023-03-29T00:00:00",
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induction proof: $\sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$ I encountered the following induction proof on a practice exam for calculus:
$$\sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$$
I have to prove this statement with induction.
Can anyone please help me with this proof?
| If $P(n): \sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6},$
we see $P(1): 1^2=1$ and $\frac{1(1+1)(2\cdot1+1)}{6}=1$ so, $P(1)$ is true
Let $P(m)$ is true, $$\sum_{k=1}^mk^2 = \frac{m(m+1)(2m+1)}{6}$$
For $P(m+1),$
$$ \frac{m(m+1)(2m+1)}{6}+(m+1)^2$$
$$=\frac{m(m+1)(2m+1)+6(m+1)^2}6$$
$$=\frac{(m+1)\{m(2m+1)+6(m+1)\}}6$$
$$=\frac{(m+1)(m+2)\{2(m+1)+1\}}6$$ as $m(2m+1)+6(m+1)=2m^2+7m+6=(m+2)(2m+3)$
So, $P(m+1)$ is true if $P(m)$ is true
| {
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"timestamp": "2023-03-29T00:00:00",
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Covergence of $\frac{1}{4} + \frac{1\cdot 9}{4 \cdot 16} + \frac{1\cdot9\cdot25}{4\cdot16\cdot36} + \dotsb$ I am investigating the convergence of the following series: $$\frac{1}{4} + \frac{1\cdot 9}{4 \cdot 16} + \frac{1\cdot9\cdot25}{4\cdot16\cdot36} + \frac{1\cdot9\cdot25\cdot36}{4\cdot16\cdot36\cdot64} + \dotsb$$
This is similar to the series $$\frac{1}{4} + \frac{1\cdot 3}{4 \cdot 6} + \frac{1\cdot3\cdot5}{4\cdot6\cdot8} + \dotsb,$$which one can show is convergent by Raabe's test, as follows: $$\frac{a_{n+1}}{a_n}= \frac{2n-1}{2n+2}= \frac{2n + 2 - 3}{2n+2}=1 - \frac{3}{2(n+1)}.$$
However, in the series I am looking at, $\frac{a_{n+1}}{a_n}=\frac{(2n-1)^2}{(2n)^2}= 1 - \frac{4n-1}{4n^2}$, so Raabe's test doesn't work (this calculation also happens to show that the ratio and root tests will not work).
Any ideas for this one? It seems the only thing left is comparison...
| I just read that there is a partial converse to Raabe's test that if $\frac{a_{n+1}}{a_n}\geq 1-p/n$ for $p\leq 1$, then the series diverges. So I think that settles that it diverges.
| {
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Using trig substitution to evaluate $\int \frac{dt}{( t^2 + 9)^2}$ $$\int \frac{\mathrm{d}t}{( t^2 + 9)^2} = \frac {1}{81} \int \frac{\mathrm{d}t}{\left( \frac{t^2}{9} + 1\right)^2}$$
$t = 3\tan\theta\;\implies \; dt = 3 \sec^2 \theta \, \mathrm{d}\theta$
$$\frac {1}{81} \int \frac{3\sec^2\theta \mathrm{ d}\theta}{ \sec^4\theta} = \frac {1}{27} \int \frac{ \mathrm{ d}\theta}{ \sec^2\theta} = \dfrac 1{27}\int \cos^2 \theta\mathrm{ d}\theta $$
$$ =\frac 1{27}\left( \frac{1}{2} \theta + 2(\cos\theta \sin\theta)\right) + C$$
$\arctan \frac{t}{3} = \theta \;\implies$
$$\frac{1}{27}\left(\frac{1}{2} \arctan \frac{t}{3} + 2 \left(\frac{\sqrt{9 - x^2}}{3} \frac{t}{3}\right)\right) + C$$
This is a mess, and it is also the wrong answer.
I have done it four times, where am I going wrong?
| Try returning to your integral: $$\int \cos^2\theta = \frac{\theta}{2} + \frac{\sin\theta\cos\theta}{2} + C$$
We get a factor of $\frac 12,$ and not $2$, multiplying the second term in the sum.
Note, more importantly, that your substitutions for $\cos\theta\sin\theta$ in terms of $\theta = \arctan(t/3)$ are also incorrect.
We have that $\theta = \arctan(t/3) \implies \tan \theta = \dfrac t3.\;$ Corresponding to this is $\;\cos\theta = \dfrac{3}{\sqrt{t^2 + 9}}$ and $\;\sin\theta = \dfrac t{\sqrt{t^2 + 9}}.$
That gives us $$\frac{1}{ 27}\left( \frac 12\cdot \arctan \frac{t}{3} + \frac 12 \underbrace{\frac{t}{\sqrt{t^2 + 9}}}_{\sin \theta}\cdot \underbrace{\frac{3}{\sqrt{t^2 + 9}}}_{\cos\theta}\right) + C$$ $$ = \frac{1}{54}\left( \arctan \frac{t}{3} + \frac{3t}{t^2+9}\right) + C$$
| {
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Prove that: $a^2+b^2+(1-a-b)^2\ge \frac {1}{3}$ Where $a$ and $b$ are any given real number. I have tried solving it using partial derivative.
$$
s=a^2+b^2+(1-a-b)^2$$
$$\frac{\partial s}{\partial a}=2a-2(1-a-b) \tag{1}$$
$$\frac{\partial s}{\partial b}=2b-2(1-a-b) \tag{2}$$
for maxima both (1) and (2) are 0..from here we get two equations from where we get values of $a$ and $b$
$$2a-2(1-a-b)=0 \tag{3}$$
$$2b-2(1-a-b)=0 \tag{4}$$
putting the values of $a$ and $b$ we find from (3) and (4) in the function the maximum value of the function should be $\frac {1}{3}$.which in this case is not
| Using Titu Andreescu's Lemma, we have
$\dfrac{a^2}{1}+\dfrac{b^2}{1}+\dfrac{(1-a-b)^2}{1} \ge \dfrac{(a+b+1-a-b)^2}{3}=\dfrac{1}{3}$
| {
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Does $y_n=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n}$ converge or diverge? I have to show whether $y_n=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n}$ is convergent or divergent. I tried using the squeeze theorem to prove it was convergent. So what I did was
bound ${y_n}$ in between $\frac{-1}{n}$ and $\frac{1}{n}$. That is $\frac{-1}{n} \leq \frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n} \leq \frac{1}{n}$ Since we also proved earlier that $lim\frac{1}{n}=0$ and $lim\frac{-1}{n}= -lim\frac{-1}{n}=0$ It follows by the squeeze theorem that $lim(y_n)=0$. Would this be correct?
Edit: This what I have now.
Well this what I get so far that $ \sum_{k=n+1}^{2n}\frac{1}{k}≥ \sum_{k=n+1}^{2n}\frac{1}{k+1}$ Hence its divergent. I think it works since $ \sum_{k=n+1}^{2n}\frac{1}{k+1}=\frac{1}{(n+1)+1}+\frac{1}{(n+2)+1}+...+\frac{1}{2n+1}$ ≤ $ \sum_{k=n+1}^{2n}\frac{1}{k}= \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} $.
| Your reasoning would be correct if $y_n<\frac{1}{n}$, which is not true. The sequence actually diverges. To prove it, we use the fact that for $k\in[\![1,n-1]\!]$,
\begin{align}
\frac{1}{kn+1}+\frac{1}{kn+2}+...+\frac{1}{kn+n} &\geq \frac{1}{kn+n}+\frac{1}{kn+n}+...+\frac{1}{kn+n} \\
&\geq \frac{n}{n(k+1)}=\frac{1}{k+1}
\end{align}
Then, we can rewrite $y_n$ using such partial sums:
\begin{align}
y_n&=\sum_{k=1}^{n-1} \frac{1}{kn+1}+\frac{1}{kn+2}+...+\frac{1}{kn+n}\\
&\geq\sum_{k=1}^{n-1}\frac{1}{k+1}
\end{align}
This sum diverges when $n\to\infty$, thus $y_n$ also diverges.
| {
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Series involving Marcum Q function I would like to have a better form of this series:
$$\sum_{k=0}^{\infty}\,\frac{1}{k!}\,\left(\frac{ab\sin(c)}{\sqrt{2}}\right)^{2k}\,Q_{k+\frac{3}{2}}\left(ab\cos(c),bx\right)$$
where $Q_m(\alpha,\beta)$ is the generalized Marcum Q-function.
Does anyone an idea on how to proceed, if possible to make it better (e.g. getting rid of the series)?
Thanks!
EDIT:
could the result be just proportional to $Q_{\frac{3}{2}}\left(ab,bx\right)$???
| Well, I guess, in some form of. But at the same time you can get a closed form solution.
For simplicity let's set $\alpha=\left(\frac{ab\sin(c)}{\sqrt{2}}\right)^{2}$ and $\beta=ab\cos(c)$
Taking into account the definition of the generalized Marcum Q-function:
$$
\begin{eqnarray}
\sum_{k=0}^{\infty}\,\frac{\alpha^{k}}{k!}\,\,Q_{k+\frac{3}{2}}\left(\beta,bx\right)&=&
\sum_{k=0}^{\infty}\,\frac{\alpha^{k}}{k!}\,\,\int_{bx}^{\infty} t \left( \frac{t}{\beta}\right)^{k+\frac{1}{2}} \exp \left( -\frac{t^2 + \beta^2}{2} \right) I_{k+\frac{1}{2}} \left( \beta t \right) \ \mathrm dt\\
&=&\frac{\exp \left(\!-\!\frac{\beta^2}{2}\! \right)}{\sqrt{\beta}}\int_{bx}^{\infty} t^{\frac{3}{2}}e^{-\frac{t^2}{2}}\sum_{k=0}^{\infty}\frac{\left(\frac{\alpha t}{\beta}\right)^k}{k!}I_{k+\frac{1}{2}}(\beta t) \ \mathrm dt\\
\end{eqnarray}
$$
Then one can use Gradshteyn and Ryzhik's Table of Integrals, Series, and Products (Second Part):
$$\sum_{k=0}^{\infty}\frac{\xi^k}{k!}I_{k+\nu}(\eta)=\left(\frac{2\xi}{\eta}+1\right)^{-\frac{\nu}{2}}I_{\nu}\left(\sqrt{\eta^2+2\eta\xi}\right), \qquad |2\xi|<|\eta|$$
So
$$
\begin{eqnarray}
\sum_{k=0}^{\infty}\,\frac{\alpha^{k}}{k!}\,\,Q_{k+\frac{3}{2}}\left(\beta,bx\right)
=\frac{\exp \left(\!-\!\frac{\beta^2}{2}\! \right)}{\sqrt{\beta}}\left(1+\frac{2\alpha}{\beta^2}\right)^{-\frac{1}{4}}
\int_{bx}^{\infty} t^{\frac{3}{2}}e^{-\frac{t^2}{2}}I_{\frac{1}{2}}(\sqrt{2\alpha+\beta^2}|t|) \ \mathrm dt=\\
=\text{A} \ \left[\sqrt{\pi } \sqrt{2 \alpha +\beta ^2} e^{\alpha +\frac{1}{2} \left(b^2 x^2+\beta ^2\right)+b x \sqrt{2 \alpha +\beta ^2}}\left(\text{erfc}\left(\frac{ \sqrt{2 \alpha +\beta ^2}+b x}{\sqrt{2}}\right)+\text{erfc} \left(\frac{b x-\sqrt{2 \alpha +\beta ^2}}{\sqrt{2}}\right) \right)+\sqrt{2} \left(e^{2 b x \sqrt{2 \alpha +\beta ^2}}-1\right)\right]
\end{eqnarray}
$$
where $\text{A}=\frac{\exp \left(-\frac{\beta ^2}{2}-\frac{1}{2} b x \left(2 \sqrt{2 \alpha +\beta ^2}+b x\right)\right)}{2 \sqrt{\pi } \sqrt{\left(2 \alpha +\beta ^2\right) \text{sgn}(\beta)}}$
The last equation was obtained via Mathematica (or you can turn to Gradshteyn and Ryzhik ;)). The equation was simplified assuming that $\alpha>0, \ x>0, \ \beta\in\mathbb{R}$. If not the result gets nastier (with lots of square roots).
I guess you can turn the last integral into some form of the generalized Marcum Q-function, if you'd like to:
$$\int_{bx}^{\infty} t^{\frac{3}{2}}e^{-\frac{t^2}{2}}I_{\frac{1}{2}}(\sqrt{2\alpha+\beta^2}|t|) \ \mathrm dt=(2\alpha+\beta^2)^{\frac{1}{4}}e^{\frac{|2\alpha+\beta^2|}{2}}Q_{\frac{3}{2}}\left(\sqrt{2\alpha+\beta^2},bx\right)$$
So
$$\sum_{k=0}^{\infty}\,\frac{\alpha^{k}}{k!}\,\,Q_{k+\frac{3}{2}}\left(\beta,bx\right)=e^{\alpha}Q_{\frac{3}{2}}\left(\sqrt{2\alpha+\beta^2},bx\right)$$
Noting that $\sqrt{2\alpha+\beta^2}=a \ b $ the result turns out to be even simplier.
| {
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What is the limit of $\frac {x^4 +y^4}{x^3 +y^3}$ as $(x,y) \to(0,0)$ What is the limit of $$\lim_{(x,y)\to(0,0)} \dfrac {x^4 +y^4}{x^3 +y^3}$$
if it exists?
I have tried to solve it by converting it to polar system $(x,y)=(r\cos a,r\sin a)$ and another settings. However I could not find the limit and not to show that there is no limit.
thanks in advance.
| The idea to prove that the limit doesn't exist is to choose a direction such that $x^3+y^3$ is convergent to $0$ faster than $x^4+y^4$ so let $y=-x+x^3$ hence we find
$$x^3+y^3=x^3+(-x+x^3)^3=3x^5+o(x^5)\ \text{and}\ x^4+y^4=x^4+(-x+x^3)^4=2x^4+o(x^4)$$
hence
$$\lim_{(x,y=-x+x^3)\to(0,0)} \dfrac {x^4 +y^4}{x^3 +y^3}=\lim_{x\to0}\dfrac {2x^4 +o(x^4)}{3x^5 +o(x^5)}=\infty $$
| {
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Prove that, if $0 < x < 1$, then $(1+\frac{x}{n})^n < \frac1{1-x}$ More fully,
if $n\ge 2$ is an integer
and
$0 < x < 1$,
prove that
$(1+\frac{x}{n})^n < \frac1{1-x}$.
In addition,
if $c > 1$ and
$0 < x \le \frac{c-1}{c}$,
prove that
$(1+\frac{x}{n})^n < 1+cx$.
Proofs by elementary means
(no calculus or limits)
are particularly sought.
As an example of the utility of this result,
set $x = \frac12$.
Then
this shows that
$2 > (1+\frac1{2n})^n$
or
$2^{1/n} > 1+\frac1{2n}$
.
This is an example
of what I call a
contra-Bernoulli inequality
(CBI)
which gives an upper bound to
$(1+y)^n$
as opposed to Bernoulli's inequality,
which gives a lower bound to
$(1+y)^n$
of $1+ny$.
Note that
any CBI of the form
$(1+y)^n < 1+c y$
for $n \ge 2$
requires that $y$ is bounded,
since
$(1+y)^n > 1+y^n$
so
$1+cy > 1+y^n$
or
$cy > y^n$
or
$y < c^{1/(n-1)}$.
| Here's my elementary proof by induction that
$(1+x/n)^n < 1/(1-x)$.
Let $y = x/n$.
This becomes
$(1+y)^n < 1/(1-ny)$
when $0 < y < 1/n$.
For $n=1$,
this is
$1+y < 1/(1-y)$
or
$1-y^2 < 1$
which is true.
Suppose it it true for $n$,
so that
$(1+y)^n < \frac1{1-ny}$.
Then,
if $0 < y < \frac1{n+1}$,
$\begin{align}
(1-(n+1)y)(1+y)^{n+1}
&=(1-(n+1)y)(1+y)(1+y)^{n}\\
&<\frac{(1-(n+1)y)(1+y)}{1-ny}\\
&=\frac{1-ny-(n+1)y^2}{1-ny}\\
&=\frac{1-ny}{1-ny}\\
&=1
\end{align}
$
so
$(1+y)^{n+1} < \frac1{1-(n+1)y}$.
| {
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Prove that $\cot(A+B)=\frac{\cot A\cot B-1}{\cot A+\cot B}$ The question is:
Prove that:
$$ \cot(A+B)=\frac{\cot A\cot B-1}{\cot A+\cot B} $$
I have tried expanding it as $\dfrac{\cos(A+B)}{\sin(A+B)}$ and $\dfrac{1}{\tan(A+B)}$.
| It is probably easier to start from the RHS:
$$\frac{\cot A\cot B-1}{\cot A+\cot B}= \frac{\frac{\cos A}{\sin A}\frac{\cos B}{\sin B} -1}{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B} }$$
Now, all the following computations are natural, and pretty authomatic:
$$=\frac{\frac{\cos A \cos B}{\sin A \sin B}-\frac{\sin A\sin B }{\sin A\sin B} }{\frac{\cos A \sin B+ \sin A \cos B}{\sin A \sin B}}=\frac{\cos A \cos B-\sin A\sin B }{\cos A \sin B+ \sin A \cos B}=\frac{\cos(A+B)}{\sin(A+B)}$$
| {
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Solve in integers $ (y^3+xy-1)(x^2+x-y)=(x^3-xy+1)(y^2+x-y)$ Solve in integers: $$ (y^3+xy-1)(x^2+x-y)=(x^3-xy+1)(y^2+x-y)$$
My idea:
$$\Longleftrightarrow (y^3+xy-1)(x^2+x-y)-(x^3-xy+1)(y^2+x-y)=0$$
$$\Longleftrightarrow -x^4-x^3y^2+2x^3y+x^2y^3+2x^2y-x^2+2xy^3-2xy^2-2x-y^4-y^2+2y=0$$
$$\Longleftrightarrow 2xy(x^2+y^2)+2y(x^2+1)+x^2y^3+2y=x^4+y^4+x^3y^2+x^2+y^2+2x$$
following I can't work.
| We first note that if $(x,y)$ is a solution then so is $(-y,-x)$.Consider the term
$$
\frac{x^3-xy+1}{x^2+x-y}
$$
This can be expanded as
$$
x-1+\frac{x+1-y}{x^2+x-y}
$$
For the latter term to be integral implies that $x^2+x-y <= x+1-y$ which implies that $|x|<=1$. This gives us 3 possible values for $x$ namely $0,\pm1$.
Case 1. $x=0$ .
The original equation becomes $$(y^3−1)(−y)=(+1)(y^2−y)$$ So $y=0$ is a solution. Reducing the equation further gives us $$ y^3-1=-(y-1)$$ So $y=1$ is also a solution. This can be reduced further to $$y^2+y+1=-1 \Rightarrow y^2+y+2=0$$ This does not have an real integral solutions so we can ignore.
Case 2. $x=1$ The original equation reduces to $$(y^3+y-1)(2-y)=(2-y)(y^2+1-y)$$
So $y=2$ is a solution. Reducing further gives us $$y^3-y^2+2y-2=0\Rightarrow(y^2+1)(y-2)=0$$ No other new solution can be got from this
Case 3. $x=-1$ The original equation reduces to $$(y^3-y-1)(-y)=(y)(y^2-1-y)$$
So $y=0$ is a solution. Reducing further gives us $$y^3+y^2-2y-2=0\Rightarrow(y^2-2)(y+1)=0$$ So $y=-1$ is a solution
Combining the solutions from above 3 cases we get
*
*$(0,0)$
*$(0,1)$
*$(1,2)$
*$(-1,-1)$
Using the original observation that $(x,y)\implies(-y,-x)$ is a solution too we obtain the following solutions for this
*
*$(0,0)$
*$(0,1)$, $(-1,0)$
*$(1,2)$, $(-2,-1)$
*$(-1,-1)$, $(1,1)$
Now consider $(x,y)$ such that $x^2+x-y=0$. Solving the quadratic and substituting for discriminant we can show that $(k,k(k+1))$ and $(-(k+1),k(k+1))$ satisfies this for all values of $k$. We then see if any of the terms in RHS also vanish for the same $(x,y)$ for some $k$ that might give us additional solutions.
case 1. $(k,k(k+1))$. The RHS term becomes
$$(k^3-k^2(k+1)+1)(k^2(k+1)^2+k-k(k+1)) \to (-k^2+1)k(k^3+k^2+1).
$$
which vanishes for $k=0,\pm1$. We can verify that these values of $k$ give known existing solutions above
case 2. $(-(k+1),k(k+1))$. The RHS term becomes
$$(-(k+1)^3+k(k+1)^2+1)(k^2(k+1)^2-(k+1)-k(k+1)) \to -(k^2+2k)(k+1)(k^3+k^2-k-1) \to -k(k+2)(k+1)^2(k^2-1).
$$
which vanishes for $k=0,\pm1,-2$. We can verify that these values of $k$ give known existing solutions above including a new solution for $k=1$ i.e $(-2,2)$
Hence entire solution set can be give by
*
*$(0,0)$
*$(0,1)$, $(-1,0)$
*$(1,2)$, $(-2,-1)$
*$(-1,-1)$, $(1,1)$
*$(-2,2)$
Some notes:
*
*This is not as rigorous as I'd like it to be. For example $x^2+x-y$ needn't divide $x^3-xy+1$ it should divide the product $(x^3-xy+1)(y^2+x-y)$
*It is curious that all obtained solutions involve atleast one term on each side vanishing. Is it possible that for certain values of $(x,y)$ none of the terms vanish but LHS = RHS ? Certainly @chubakueno tests show that this is not the case
*Where did you obtain this equation from ? The equation is non-homogenous and hence cannot be factored into quadratic and cubic forms.
| {
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For what $x\in[0,2\pi]$ is $\sin x < \cos 2x$ What's the set of all solutions to the inequality $\sin x < \cos 2x$ for $x \in [0, 2\pi]$? I know the answer is $[0, \frac{\pi}{6}) \cup (\frac{5\pi}{6}, \frac{3\pi}{2}) \cup (\frac{3\pi}{2}, 2\pi]$, but I'm not quite sure how to get there.
This is what I have so far: $\\
\sin x - \cos 2x < 0\\
\sin x - \cos^2 x + \sin^2 x < 0\\
2\sin^2 x + \sin x - 1 < 0\\
-1 < \sin x < \frac{1}{2}$
Any help will be much appreciated.
| You are doing fine, now just need to find the values of $x$ that make $\sin x$ come out to those values. One way to see $\sin \frac \pi6=\frac 12$ is to consider an equilateral triangle cut in half vertically. The base (adjacent) is $\frac 12$ and the hypotenuse is $1$. To see $\sin \frac {3\pi}2=-1$ just consider the unit circle-the $\sin$ is $-1$ only straight down.
| {
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Proof by induction that the sum of terms is integer I'm having some trouble in order to solve this induction proof.
Proof that $\forall{n} \in \mathbb{N}$ the number $\frac{1}{5}n^5+\frac{1}{3}n^3 + \frac{7}{15}n$ is an integer.
I've tried proving this by induction, but I've not succeed so far. What I did was:
Let's assume that $\frac{1}{5}n^5+\frac{1}{3}n^3 + \frac{7}{15}n = k$, with $k \in \mathbb{Z}$. So, by induction,
$n=1$
$\frac{1}{5} + \frac{1}{3} + \frac{7}{15} = 1$ and $1 \in \mathbb{Z}$.
Inductive step
I want to prove that if $\frac{1}{5}n^5+\frac{1}{3}n^3 + \frac{7}{15}n = k$, with $k \in \mathbb{Z}$ then $\frac{1}{5}(n+1)^5+\frac{1}{3}(n+1)^3 + \frac{7}{15}(n+1) = l$, with $l \in \mathbb{Z}$
And then I'm stuck. I've tried developing every binomial but, for example, for $\frac{1}{5}(n+1)^5$ I get a $\frac{1}{5}n^4$ that is not in the inductive hypothesis, so therefore, I cannot get rid of it and it's not an integer, so I cannot conclude that $l \in \mathbb{Z}$.
Any help or ideas from where I can follow? Thanks in advance!
| Let $f(n)=\frac{n^5}5+\frac{n^3}3+\frac{7n}{15}$
$$f(n+1)-f(n)=\frac{(n+1)^5}5+\frac{(n+1)^3}3+\frac{7(n+1)}{15}-\left(\frac{n^5}5+\frac{n^3}3+\frac{7n}{15}\right)$$
$$=\frac{(n+1)^5-n^5}5+\frac{(n+1)^3-n^3}3+\frac{7(n+1)-7n}{15}$$
$$=\frac{5n^4+10n^3+10n^2+5n+1}5+\frac{3n^2+3n+1}3+\frac{7}{15}$$
$$=(n^4+2n^3+2n^2+n)+(n^2+n)+\left(\frac15+\frac13+\frac7{15}\right)$$ (using Binomial Theorem)
Now, as you have shown $\frac15+\frac13+\frac7{15}=1$
$$\text{So, if }f(n)=\frac{n^5}5+\frac{n^3}3+\frac{7n}{15} \text{ is an integer for integer } n$$
$$\text{ so will be }f(n+1)=\frac{(n+1)^5}5+\frac{(n+1)^3}3+\frac{7(n+1)}{15}$$
This problem can be extended to more primes and proved using the fact that prime $p$ divides $\binom pr$ for $1\le r\le p-1$
I can not keep back a non-inductive proof:
$$\frac{n^5}5+\frac{n^3}3+\frac{7n}{15}=\frac{n^5-n}5+\frac{n^3-n}3+\frac{7n}{15}+\frac n3+\frac n5=\frac{n^5-n}5+\frac{n^3-n}3+n$$
Now, using Fermat's Little Theorem, $n^p-n\equiv0\pmod p$ for any integer $n$ and prime $p$
Similarly, $$\frac{n^7-n}7+\frac{n^5-n}5+\frac{n^3-n}3+n-\left(\frac n7+\frac n3+\frac n5\right)=\frac{n^7}7+\frac{n^5}5+\frac{n^3}3+\frac{34n}{105}$$ will be an integer
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Calculating $\int \dfrac{\cot^3x}{\sqrt{1+\csc^4x}}\;dx$
Evaluate:$$\int \dfrac{\cot^3x}{\sqrt{1+\csc^4x}}\;dx$$
| Rewrite the integral as
$$\int dx \frac{\cot{x} (\csc^2{x}-1)}{\sqrt{1+\csc^4{x}}}$$
Split the integral along the numerator. The first piece is
$$\int dx \frac{\csc^2{x} \cot{x}}{\sqrt{1+\csc^4{x}}} = -\int \frac{d(\cot{x}) \cot{x}}{\sqrt{1+(1+\cot^2{x})^2}} = -\frac12 \int \frac{du}{\sqrt{(u+1)^2+1}}$$
where $u=\cot^2{x}$. The second piece is
$$-\int dx \frac{\cot{x}}{\sqrt{1+\csc^4{x}}} = -\int dx \frac{\cos{x}\sin{x}}{\sqrt{1+\sin^4{x}}} = -\frac12 \int \frac{dv}{\sqrt{1+v^2}}$$
where $v=\sin^2{x}$. Use
$$\int \frac{dw}{\sqrt{w^2+1}} = \log{(w+\sqrt{w^2+1})}+C$$
and you should be able to finish this off.
| {
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Radius of convergence and interval of convergence of $\sum_{n=1}^\infty\frac{(-3)^n}{n \sqrt n}x^n$ $$\sum_{n=1}^\infty\frac{(-3)^n}{n \sqrt n}x^n$$
I tried the ratio test on it but got stuck.
| I'm going to throw this in, since it looks more like what is found in calculus texts:
$$\lim_{n \rightarrow \infty} \ \left| \frac{a_{n+1}}{a_n} \right| \ = \ \lim_{n \rightarrow \infty} \ \left| \frac{(-3)^{n+1} x^{n+1}/ (n+1)^{3/2}}{(-3)^n x^n / n^{3/2}} \ \right| $$
$$= \ \lim_{n \rightarrow \infty} \ \left| \ \frac{(-3)^{n+1}}{(-3)^n} \ \cdot \ \frac{x^{n+1} }{x^n} \ \cdot \ \left(\frac{ n}{n+1} \right)^{3/2} \ \right| \ = \ \left| \ 3 \ \cdot \ x \ \cdot \ 1 \ \right| \ < \ 1 $$
$$\Rightarrow \ \vert x \vert \ < \ \frac{1}{3} , $$
so we need to look at the interval centered on $ \ x = 0 \ $ with a "radius" of $ \ \frac{1}{3} \ $ .
The "endpoints" must be checked separately. Using $ \ x = \frac{1}{3} \ \ \text{and} \ \ x = -\frac{1}{3} \ $ produces series
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^{3/2}} \ \text{and} \ \ \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \ . $$
We have a " $p-$series" which is absolutely convergent, so our series converges at both endpoints, making the interval of convergence $ \ \left[ -\frac{1}{3} , \frac{1}{3} \right] \ . $
| {
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Help with $\int\frac{1}{1+x^8}dx$ I'm not sure how to proceed. I tried factoring like you do to evaluate $\int\frac{1}{1+x^4}dx$, and since it came out nasty I checked out WolframAlpha to see if I was on the right track. In fact, Wolfram doesn't have a step-by-step solution for the integral in question, but the primitive is full of trig expressions like $\csc\frac{\pi}{8}$ and I'm not sure where these would come from.
| Decompose the integrand as
\begin{align}\frac{1}{1+x^8}
&=\frac1{2\sqrt2}\left(\frac{x^2+\sqrt2}{x^4-\sqrt2x^2+1}
- \frac{x^2-\sqrt2}{x^4+\sqrt2x^2+1}\right)\\
&=\frac{2+\sqrt2}8 \left( \frac{1+x^2}{x^4+\sqrt2x^2+1}
+\frac{1-x^2}{x^4-\sqrt2x^2+1} \right)\\
&\hspace{5mm}+\frac{2-\sqrt2}8 \left( \frac{1+x^2}{x^4-\sqrt2x^2+1}
+\frac{1-x^2}{x^4+\sqrt2x^2+1} \right)
\end{align}
Then, integrate
$$\begin{align}
I_+(a)&=\int\frac{1+x^2}{x^4+ax^2+1}dx
= \int\frac{d(x-\frac1x)}{(x-\frac1x)^2 +2+a}=\frac1{\sqrt{2+a}}\tan^{-1}\frac{x^2-1}{x\sqrt{2+a}} \\
I_-(a)&=\int\frac{1-x^2}{x^4+ax^2+1}dx
= \int\frac{d(x+\frac1x)}{2-a-(x+\frac1x)^2}=\frac1{\sqrt{2-a}}\coth^{-1}\frac{x^2+1}{x\sqrt{2-a}} \\
\end{align}$$
Thus
\begin{align}
\int \frac{1}{1+x^8}dx
&= \frac{2+\sqrt2}8 \left[ J_+(\sqrt2)+ J_-(-\sqrt2) \right]
+\frac{2-\sqrt2}8 \left[ J_+(-\sqrt2)+ J_-(\sqrt2) \right]\\
& =\frac{\sqrt{2+\sqrt2}}8 \left( \tan^{-1}\frac{x^2-1}{x\sqrt{2+\sqrt2}}
+ \coth^{-1}\frac{x^2+1}{x\sqrt{2+\sqrt2}}\right)\\
&\hspace{5mm}+\frac{\sqrt{2-\sqrt2}}8 \left( \tan^{-1}\frac{x^2-1}{x\sqrt{2-\sqrt2}}
+ \coth^{-1}\frac{x^2+1}{x\sqrt{2-\sqrt2}}\right)+C\\
\end{align}
| {
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If $x,y,z \in \Bbb{R}$ such that $x+y+z=4$ and $x^2+y^2+z^2=6$, then show that $x,y,z \in [2/3,2]$. If $x,y,z\in\mathbb{R}$ such that $$x+y+z=4,\quad x^2+y^2+z^2=6;$$then show that the each of $x,y,z$ lie in the closed interval $[2/3,2]$.
I have been able to solve using $2(y^2+z^2)\geq(y+z)^2$.
Is there any another method to solve it.
| We have $(x-1)^2+(y-1)^2+(z-1)^2=x^2+y^2+z^2-2(x+y+z)+3=6-8+3=2$, i.e.,
$$(x-1)^2+(y-1)^2+(z-1)^2=1.$$
This implies $(x-1)^2\le1$ and $x\in[0,2]$. Similarly for the other two variables.
Similarly we get
$$\left(x-\frac53\right)^2+\left(y-\frac53\right)^2+\left(z-\frac53\right)^2=1$$
which implies $(x-\frac53)^2\le1$ and $x\in[2/3,8/3]$.
| {
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Taylor Polynomial $\;y = e^{\sin x},\,$ $a = \frac{\pi}{2},\;$ $x = 1.5$ $y = e^{\sin x};\;$ $a = \frac{\pi}{2};\;$ $x = 1.5$
So first things get the derivative. I need to calculate the error and the $T_2$
$$f'(x) = \cos x e^{\sin x}$$
$$f''(x) = \cos ^2x e^{\sin x} - \sin x e^{\sin x}$$
So that now gives me
$$f(a) + f'(a)/1 (x-a) + f''(a)/2 (x-a)^2$$
$$e^ {\sin \frac{\pi}{2}} + \cos (\frac{\pi}{2}) e^{\sin {\frac{\pi}{2}}} * (1.5 - \frac{\pi}{2})^2 + \cos ^2 (\frac{\pi}{2}) e^{\sin {\frac{\pi}{2}}} - \sin {\frac{\pi}{2}} e^{\sin {\frac{\pi}{2}}} $$
This turns into $e$ I think.
So for the error
$$|f(x) - T_n (x)| < K \frac{|x - a|^{n+1}}{(n+1)!}$$
$$e^{\sin(1.5)} - e < K \frac{(1.5 - \frac{\pi}{2})^3}{6}$$
That is a mess...where did I go wrong?
| Your derivatives $f'(x), f''(x)$ are fine. And the following is correct:
$$T_2(x) = f(a) + f'(a)(x-a) + \dfrac{f''(a)(x-a)^2}{2}\tag{1}$$
Where the problem seems to be is in your evaluation of $(1)$, given $f(a), f'(a), f''(a)$ with $a = \pi/2, x = 1.5$:
You wrote: $$e^ {\sin \frac{\pi}{2}} + \cos (\frac{\pi}{2}) e^{\sin {\frac{\pi}{2}}} * (1.5 - \frac{\pi}{2})^2 + \cos ^2 (\frac{\pi}{2}) e^{\sin {\frac{\pi}{2}}} - \sin {\frac{\pi}{2}} e^{\sin {\frac{\pi}{2}}} $$
which does not follow from $(1)$.
What you should obtain is:
$$T_2(1.5) = e^{\sin\frac{\pi}{2}} + \cos \left(\frac{\pi}{2}\right) e^{\sin {\frac{\pi}{2}}}\left(1.5 - \frac{\pi}{2}\right) + \dfrac{\left(\cos ^2 \left(\frac{\pi}{2}\right) e^{\sin {\frac{\pi}{2}}} - \sin \left({\frac{\pi}{2}}\right) e^{\sin {\frac{\pi}{2}}}\right)\left(1.5 - \frac{\pi}2\right)^2 }{2}$$
$$\begin{align}T_2(1.5) &= e^{1} + {(0)e^{1}}\left (1.5-\frac{\pi}{2}\right)+ \dfrac{\left((0)e^{1}-1\cdot e^{1}\right)\left(1.5-\frac{\pi}{2}\right)^2}{2} \\ \\
& = e -{e}\cdot \dfrac{(1.5-\frac{\pi}{2})^2}{2}\\ \\
& = e\left(1 - \dfrac{\left(1.5 -\frac{\pi}{2}\right)^2}{2}\right)\end{align}$$
| {
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How to solve problems involving roots. $\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$ How to solve problems involving roots. If we square them they may go to fourth degree.There must be some technique to solve this.
$$\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$$
| Straight, you can get the following equation :
$$
\sqrt{(2-\sqrt{x-1})^2} + \sqrt{(3-\sqrt{x-1})^2} =1
$$
which leads to the following equation :
$$
|2-\sqrt{x-1}| + |3-\sqrt{x-1}| =1
$$
Then you will have three cases to discuss :
*
*case : $\sqrt{x-1} \leq2$ (equivalent to $x\leq5$) :
$\sqrt{x-1} = 2$ then $x = 5$
*
*case : $\sqrt{x-1} >2$ and $\sqrt{x-1} <3$ (equivalent to $5<x<10$) :
The equation below can be written :
$$
\sqrt{x-1}-2 + 3-\sqrt{x-1} =1
$$
equivalent to :
$
1=1
$
The solutions belongs to $]5,10[$
*
*case : $\sqrt{x-1} \geq3$ (equivalent to $x\geq10$) :
$\sqrt{x-1} = 3$ then $x = 10$
The solutions belongs to $[5,10]$
| {
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$p\nmid 2n-1,$ then $\sum_{k=1}^{p-1}\frac{1}{k^{2n-1}}\equiv 0 \pmod{p^3} \Leftrightarrow \sum_{k=1}^{p-1}\frac{1}{k^{2n}}\equiv 0 \pmod{p^2} $ Is it true that if $p$ is a prime and $p\nmid 2n-1,$ then
$$\sum_{k=1}^{p-1}\frac{1}{k^{2n-1}}\equiv 0 \pmod{p^3}
\hspace{12pt}\Leftrightarrow \hspace{12pt}
\sum_{k=1}^{p-1}\frac{1}{k^{2n}}\equiv 0 \pmod{p^2} $$
I find some examples: $(2n-1,p)=(3,37)(7,67)(7,877)(9,5)(13,7)(13,59)(13,607)$
| We show something stronger, namely that
$$
2\sum_{k=1}^{p-1}\frac{1}{k^{2n-1}}+(2n-1)\cdot p\cdot\sum_{k=1}^{p-1}\frac{1}{k^{2n}}\equiv 0\mod{p^3}.
$$
Let $n\geq 1$ be a fixed integer. If we compute the difference
$$
\frac{1}{(1+x)^{2n-1}}-\left(1-(2n-1)x+n(2n-1)x^2\right),
$$
we see that it is of the form
$$
\frac{f(x)}{(1+x)^{2n-1}},
$$
where $f(x)\in\mathbb{Z}[x]$ is some polynomial which is divisible by $x^3$. It follows that for any $x\in p\mathbb{Z}_{(p)}$,
$$
(\star)\,\,\,\,\,\,\,\,\,\,\frac{1}{(1+x)^{2n-1}}\equiv 1-(2n-1)x+n(2n-1)x^2\mod{p^3}.
$$
Now, we have
$$
\begin{eqnarray*}
\sum_{k=1}^{p-1}\frac{1}{k^{2n-1}}&=&\sum_{k=1}^{p-1}\frac{1}{(p-k)^{2n-1}}\\
&=&-\sum_{k=1}^{p-1}\frac{1}{k^{2n-1}}\frac{1}{\left(1-\frac{p}{k}\right)^{2n-1}}\\
&\equiv&-\sum_{k=1}^{p-1}\frac{1}{k^{2n-1}}-(2n-1)p\sum_{k=1}^{p-1}\frac{1}{k^{2n}}-n(2n-1)p^2\sum_{k=1}^{p-1}\frac{1}{k^{2n+1}}\mod{p^3},
\end{eqnarray*}
$$
where in the third line we have used $(\star)$ with $x=-\frac{p}{k}$.
For $p\geq 2n+3$,
$$
\sum_{k=1}^{p-1}\frac{1}{k^{2n+1}}\equiv 0\mod{p}
$$
(perhaps you are familiar with this fact?), so in this case we have
$$
2\sum_{k=1}^{p-1}\frac{1}{k^{2n-1}}+(2n-1)\cdot p\cdot\sum_{k=1}^{p-1}\frac{1}{k^{2n}}\equiv 0\mod{p^3},
$$
implying the fact you want.
| {
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How to find the sum of the series $1+\frac{1}{2}+ \frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}$? How to find the sum of the following series?
$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{n}$$
This is a harmonic progression. So, is the following formula correct?
$\frac{(number ~of ~terms)^2}{sum~ of~ all ~the~ denominators}$
$\Rightarrow $ if $\frac{1}{A} + \frac{1}{B} +\frac{1}{C}$ are in H.P.
Therefore the sum of the series can be written as :
$\Rightarrow \frac{(3)^3}{(A+B+C)}$
Is this correct? Please suggest.
| There are other ways to represent the harmonic series, for example, recall the geometric sum:
$$\frac{1-r^n}{1-r}=1+r+r^2+\dots+r^{n-1}$$
And integrate both sides with respect to $r$ from $0$ to $1$:
$$\begin{align}\int_0^1\frac{1-r^n}{1-r}dr&=\int_0^11+r+r^2+\dots+r^{n-1}dr\\&=\left.\frac11r+\frac12r^2+\frac13r^3+\dots+\frac1nr^n\right|_0^1\\&=\frac11+\frac12+\frac13+\dots+\frac1n\end{align}$$
So we can rewrite the harmonic series as
$$1+\frac12+\frac13+\dots+\frac1n=\int_0^1\frac{1-r^n}{1-r}dr$$
Which is most useful for deriving many formulas. I used in deriving a representation of the gamma function, for example.
Note that:
$$\ln(\Gamma(n+1))=\ln(n\Gamma(n))=\ln(\Gamma(n))+\ln(n)$$
Take the derivative of both sides to get
$$\psi(n+1)=\psi(n)+\frac1n$$
Repeated application of this formula gives:
$$\begin{align}\psi(n+1)&=\psi(n)+\frac1n\\&=\psi(n-1)+\frac1{n-1}+\frac1n\\&=\psi(1)+1+\frac12+\frac13+\dots+\frac1n\\\psi(n+1)+\gamma&=1+\frac12+\frac13+\dots+\frac1n\end{align}$$
| {
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show that $\int_{-\infty}^{+\infty} \frac{dx}{(x^2+1)^{n+1}}=\frac {(2n)!\pi}{2^{2n}(n!)^2}$ show that:
$$\int_{-\infty}^{+\infty} \frac{dx}{(x^2+1)^{n+1}}=\frac {(2n)!\pi}{2^{2n}(n!)^2}$$
where $n=0,1,2,3,\ldots$.
is there any help?
thanks for all
| It is easier to find the integral in a more general form below:
$$I_{n}(a)=\int_{-\infty}^{+\infty} \frac{d x}{\left(x^{2}+a\right)^{n+1}} \quad \text {, where } n=0,1,2,3, \cdots \textrm{ and }a>0.$$
We first start with the easiest one $I_0(a)$, $$\int_{-\infty}^{+\infty} \frac{d x}{x^{2}+a}=\frac{1}{\sqrt{a}} \left[ \tan ^{-1} \frac{x}{\sqrt{a}}\right]_{-\infty}^{\infty}=\frac{\pi}{\sqrt{a}}
$$
In order to arrive at the power of (n+1), I plan to differentiate $I_{0}(a)$ w.r.t $a$ by $n$ times. $$
\frac{d^{n}}{da^{n}}\int_{-\infty}^{+\infty} \frac{d x}{x^{2}+a}=\pi \frac{d^{n}}{da^{n}}\left(a^{-\frac{1}{2}}\right)
$$
$$
\int_{-\infty}^{\infty} \frac{(-1)^{n} \cdot n !}{\left(x^{2}+a\right)^{n+1}} d x=\pi\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \cdots\left(-\frac{2 n-1}{2}\right) a^{-\frac{2 n+1}{2}}
$$
$$
\boxed{\int_{-\infty}^{\infty} \frac{d x}{\left(x^{2}+a\right)^{n+1}}=\frac{(2n-1)!!\pi}{2^{n} n !}a^{-\frac{2 n+1}{2}}}
$$
Putting $a=1$ yields the result $$
\int_{-\infty}^{\infty} \frac{dx }{\left(x^{2}+1\right)^{n+1}}=\frac{(2 n-1) ! ! \pi}{2^{n} n !} \cdot \frac{2\cdot4 \cdot 6 \cdots(2 n)}{2 \cdot 4 \cdot 6 \cdots(2 n)}= \frac{(2 n) ! \pi}{2^{2 n}(n !)^{2}}. \quad \blacksquare
$$
| {
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If $(a+1)(b+1)(c+1)=8,$ then prove $abc \le 1$? My proof:
$\sqrt[3]{(a+1)(b+1)(c+1)} = 2 \le \frac{a+b+c}{3} + 1 $according to AM-GM.
Thus, $1 \le \frac {a+b+c}{3}$.
Also, $ \sqrt[3]{abc} \le \frac{a+b+c}{3}$.
Then, either $\frac {a+b+c}{3} \le 1$ or $1 \le \frac {a+b+c}{3}$.
Suppose the latter equation is true. Then $a+b+c \ge 3$ meaning $a,b,c \ge 1$. If that is the case, then $(a+1)(b+1)(c+1) \ge 8$. Therefore, this does not satisfy the original equation so $1 \le \frac {a+b+c}{3}$ must be true.
Is this a sufficient proof? If not, can someone solve this question?
| Your proof doesn't work, as pointed out by vadim
Hint: Apply AM-GM directly to each term, we get that
$$ 8 = (a+1)(b+1)(c+1) \geq 2\sqrt{a} 2 \sqrt{b} 2 \sqrt{c} $$
| {
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Approximation of $ \frac{n!}{(n-2x)!}(n-1)^{-2x} $ I would like to find an approximation when $ n \rightarrow\infty$ of $ \frac{n!}{(n-2x)!}(n-1)^{-2x} $. Using Stirling formula, I obtain $$e^{\frac{-4x^2+x}{n}}. $$ The result doesn't seem right!
Below is how I derive my approximation. I use mainly Stirling Approximation and $e^x =(1+\frac{x}{n})^n $.
$$\frac{n!}{(n-1)^{2 x} (n-2 x)!}\approx \frac{\left(\sqrt{2 \pi } n^{n+\frac{1}{2}} e^{-n}\right) (n-1)^{-2 x}}{\sqrt{2 \pi } e^{2 x-n} (n-2 x)^{-2 f+n+\frac{1}{2}}}\approx \frac{n^{n+\frac{1}{2}} e^{-2 x} n^{-2 x} \left(1-\frac{1}{n}\right)^{-2 x}}{(n-2 x)^{n-2 x+\frac{1}{2}}}\approx e^{-2 x} \left(\frac{n}{n-2 x}\right)^{n-2 x+\frac{1}{2}} \left(\left(1-\frac{1}{n}\right)^n\right)^{-\frac{2 x}{n}}\approx e^{-2 x} \left(\frac{n}{n-2 x}\right)^{n-2 x+\frac{1}{2}} e^{\frac{2 x}{n}}\approx \left(\frac{n-2 x}{n}\right)^{-n+2 x-\frac{1}{2}} \left(\frac{n-2 x}{n}\right)^n e^{\frac{2 x}{n}}\approx \left(\frac{n-2 x}{n}\right)^{2 x-\frac{1}{2}} e^{\frac{2 x}{n}}\approx \left(\left(1-\frac{2 x}{n}\right)^n\right)^{\frac{2 x}{n}-\frac{1}{2 n}} e^{\frac{2 x}{n}}\approx \left(e^{-2 x}\right)^{\frac{2 x}{n}-\frac{1}{2 n}}=e^{\frac{x-4 x^2}{n}}$$
I would appreciate your input!
| The easiest way to get the correct approximation is to use Stirling in the logarithmic form:
$$\ln n!=n(\ln n-1)+\frac12 \ln \left(2\pi n\right)+\frac{1}{12n}+O(n^{-3}),$$
which also implies
$$\ln\frac{n!}{(n-2x)!}=2x \ln n+\frac{x-2x^2}{n}+O(n^{-2}).$$
Now using that
$$\ln (n-1)^{-2x}=-2x\ln(n-1)=-2x\ln n+\frac{2x}{n}+O(n^{-2}),$$
we get
$$\ln\left[\frac{n!}{(n-2x)!}(n-1)^{-2x}\right]=\frac{3x-2x^2}{n}+O(n^{-2}).$$
| {
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Prove that 360 divides (a-2)(a-1)a.a.(a+1)(a+2) there's a question which asks to prove that
360 | a2(a2-1)(a2-4)
I attempted it in the following manner.
a2(a2-1)(a2-4) = (a-2)(a-1)(a)(a+1)(a+2)(a)
The first 5 terms represent the product of 5 consecutive terms. Hence,
One of them will have a factor of 5
One of them will have a factor of 4
One of them will have a factor of 3
One of them will have a factor of 2
=> 5 x 4 x 3 x 2 = 120 will divide the given expression for sure.
Now, how do i bring a factor of 3. I know that i have missed one power of a. But, i think i am missing something. Help will be appreciated
Thank you
| $$
\begin{align}
(a-2)(a-1)a\cdot a(a+1)(a+2)
&=5!\,a\binom{a+2}{5}\\
&=120[(a+3)-3]\binom{a+2}{5}\\
&=120(a+3)\binom{a+2}{5}-360\binom{a+2}{5}\\
&=120\cdot6\binom{a+3}{6}-360\binom{a+2}{5}\\
&=360\left[2\binom{a+3}{6}-\binom{a+2}{5}\right]
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/459334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
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A problem of divisibility ... If $7\nmid n$, $7\nmid n-1$ and $7\nmid n^{3}+1$ then $7\mid n^{2}+n+1$ $7\nmid n$, $7\nmid n-1\;\;$ and $\;\;7\nmid n^{3}+1$ $\Longrightarrow 7\mid n^{2}+n+1$$$$$Do not quite understand how to do, but do not want to use modular arithmetic, thought of using consecutive numbers, but did not get the idea ..
Just know that$$n^{3}-1=(n-1)(n^{2}+n+1)$$
| $$n^7-n=n(n^6-1)=n(n^3-1)(n^3+1)=n(n-1)(n^2+n+1)(n^3+1).$$
Ask Little Fermat to help.
If you are really into consecutive numbers, then you may also observe that
$$
(n-2)(n-4)=n^2-6n+8=n^2+n+1-7(n-1),
$$
and
$$
(n+2)(n+4)=n^2+6n+8=n^2-n+1+7(n+1).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/459491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Inductive demonstration I have this:
$$1^2+2^2+3^2+\dots+n^2=\frac{n(n+1)(2n+1)}6$$
So I was suggested of doing this:
$$\begin{align}
(1^2+2^2+3^2+\dots+k^2)+(k+1)^2
&= \frac{k(k+1)(2k+1)}6+(k+1)^2 \\
&= \frac{k(k+1)(2k+1)+\color{blue}{6(k+1)^2}}6 \\
&= \frac{(k+1)[k(2k+1)+6(k+1)]}6 \\
&= \frac{(k+1)[2k^2+7k+6]}6 \\
&= \frac{(k+1)(k+2)(2k+3)}6 \\
\end{align}$$
For making the prof, we suppose that the statement is true for any number. (n=k) Then we can see that it is also true for any number k+1, adding this in both terms. Every thing is ok here but then, like we can see in the right term (marked in blue) appears some coefficient 6 of (k+1)^2 and I don't know it's origin.
Can any one help me here?
The prof after say:
"Thus the left-hand side of (2) is equal to the right-hand side of (2). This proves the inductive step. Therefore, by the principle of mathematical induction, the given statement is true for every positive integer n".
And also in this statement it is not clear to me why is true for n.
| \begin{align*}
\frac{k(k+1)(2k+1)}6+(k+1)^2&=\frac{k(k+1)(2k+1)}6+\frac{6(k+1)^2}{6}\\
&=\frac{k(k+1)(2k+1)+6(k+1)^2}6
\end{align*}
For the last line,
$$\frac{(k+1)(k+2)(2k+3)}6=\frac{(k+1)[(k+1)+1][2(k+1)+1]}6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/459568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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LU factorization with pivot to solve linear system I read that LUP matrix exist for any square matrix such that it is not a singular one.
But I came across a matrix that when I get the LUP and calculate Lz=Pb -> Ux = z the answer is wrong, I don't know why.
The linear equation was like that
x+y+z=2
z=2
2x+3y+z=0
the matrix is like that
1 1 1
0 0 1
2 3 1
calculation with LUP gave out
x=-4, y=2, z=2
Though manual substitution calculations showed that the correct answer is
x=2,y=-2,z=2
Can any one tell me why this happens is it a calculation mistake or something or it is a limitation with the LU factorization?!
[Edit]
Attempt to solve:
$$A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 1 \\ 2 & 3 & 1\\ \end{bmatrix}\rightarrow swap(r1,r3)$$
$$A = \begin{bmatrix} 2 & 3 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1\\ \end{bmatrix}\rightarrow R_3 - \frac12 R_1$$
$$A = \begin{bmatrix} 2 & 3 & 1 \\ 0 & 0 & 1 \\ 0 & - \frac12 & \frac12\\ \end{bmatrix}\rightarrow swap(r2,r3)$$
$$A = \begin{bmatrix} 2 & 3 & 1 \\ 0 & - \frac12 & \frac12 \\ 0 & 0 & 1 \\ \end{bmatrix} = U$$
$$L = \begin{bmatrix} 1 & 0 & 0 \\ \frac12 & 1 & 0 \\ 0 & 0 & 1\\ \end{bmatrix}$$
$$b = \begin{bmatrix} 0 \\2\\2 \end{bmatrix}$$
| As noted in the comments, we have:
$$A = P L U x = b \rightarrow LUx = P^{-1}b \rightarrow Ly = P^{-1}b, Ux = y$$
Note: you can see my $L$ and $U$ below.
$$P = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{bmatrix}$$
Using this, we have:
$Ly = P^{-1}\cdot b \rightarrow \begin{bmatrix}1 & 0 & 0\\2 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} \cdot \begin{bmatrix}y_1 \\ y_2\\ y_3\end{bmatrix} = \begin{bmatrix}2 \\ 0\\ 2\end{bmatrix}$
This will lead to $y = \begin{bmatrix}2 \\ -4 \\ 2\end{bmatrix}$
Next we solve:
$Ux = y \rightarrow \begin{bmatrix}1 & 1 & 1\\0 & 1 & -1 \\ 0 & 0 & 1\end{bmatrix} \cdot \begin{bmatrix}x_1 \\ x_2\\ x_3\end{bmatrix} = \begin{bmatrix}2 \\ -4\\ 2\end{bmatrix}$
This will lead to $x= \begin{bmatrix}2 \\ -2 \\ 2\end{bmatrix}$.
| {
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"url": "https://math.stackexchange.com/questions/461149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to compute $\prod\limits^{\infty}_{n=2} \frac{n^3-1}{n^3+1}$
How to compute
$$\prod^{\infty}_{n=2} \frac{n^3-1}{n^3+1}\ ?$$
My Working :
$$\prod^{\infty}_{n=2} \frac{n^3-1}{n^3+1}= 1 - \prod^{\infty}_{n=2}\frac{2}{n^3+1}
= 1-0 = 1$$
Is it correct
| $$\dfrac{r^3-a^3}{r^3+a^3}=\dfrac{r-a}{r+a}\cdot\dfrac{r^2+ra+a^2}{r^2-ra+a^2}$$
Let $f(r)=r^2-ra+a^2$
$$f(r+h)=(r+h)^2-(r+h)a+a^2=r^2+r(2h-a)+a^2-ah$$
If we set $r^2+ra+a^2=f(r+h),$
$$r^2+ra+a^2=r^2+r(2h-a)+a^2-ah$$
Comparing the coefficients of $r,$ $$a=2h-a\iff h=a$$
Compare the constants $a^2=a^2-ah+h^2\iff h(h-a)=0$
$\implies h=a,$ $$r^2+ra+a^2=f(r+a)$$
$$\dfrac{r^3-a^3}{r^3+a^3}=\dfrac{f(r+a)}{f(r)}\cdot\dfrac{g(r-2a)}{g(r)}\text{ where } g(m)=m+a$$
Clearly
$$\prod_{r=1}^n\dfrac{f(r+a)}{f(r)}\cdot\dfrac{g(r-2a)}{g(r)}$$ telescopes for finite integer $a$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/462082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
$2 \cos^{-1}x =\sin^{-1}(2x \sqrt {1-x^2})$ is valid for which values of $x $
Problem: $2 \cos^{-1}x =\sin^{-1}(2x \sqrt {1-x^2})$ is valid for which values of $x $
Solution: $2 \cos^{-1}x =\sin^{-1}(2x \sqrt {1-x^2})$
$2 \cos^{-1}x =2 \sin^{-1}x$
$ \cos^{-1}x = \sin^{-1}x$
Am I doing right ?
| As the principal value of $\arccos x$ lies in $\in[0,\pi]$
$\implies 2\arccos x$ will lie in $\in[0,2\pi]$
Again as the principal value of $\arcsin x$ $\in[-\frac\pi2,\frac\pi2]$
Observe that the intersection of region is $[0,\frac\pi2]$
So, $\sin^{-1}(2x \sqrt {1-x^2})=2\arccos x$
if $0\le 2\arccos x\le \frac\pi2\iff 0\le \arccos x\le\frac\pi4\iff 1\ge x\ge\cos\frac\pi4=\frac1{\sqrt2} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/463604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Closed form of $\operatorname{Li}_2(\varphi)$ and $\operatorname{Li}_2(\varphi-1)$ I am trying to calculate the dilogarithm of the golden ratio and its conjugate $\Phi = \varphi-1$. Eg the solutions of the equation $u^2 - u = 1$.
From Wikipdia one has the following
\begin{align*}
\operatorname{Li}_2\left( \frac{1 + \sqrt{5}}{2} \right)
& = -\int_0^\varphi \frac{\log(1-t)}{t}\,\mathrm{d}t
= \phantom{-}\frac{\pi^2}{10} - \log^2\left( \Phi\right) \\
\operatorname{Li}_2\left( \frac{1 - \sqrt{5}}{2} \right)
& = -\int_0^\Phi \frac{\log(1-t)}{t}\,\mathrm{d}t
= -\frac{\pi^2}{15} - \log^2\left( -\Phi\right) \\
\end{align*}
I am quite certain that these two special values can be shown by combining the identites for the dilogarithm, and forming a system of equations. But I am having some problems obtaining a set of equations only involving $\operatorname{Li}_2(\varphi)$ and $\operatorname{Li}_2(\Phi)$. Can anyone show me how to set up the system of equations from the identites, or perhaps a different path in showing these two values?
| This solution is very similar to @Start wearing purple's solution above.
Using the following identities (check Eq $3$,$4$ and $5$).
$$\text{Li}_2(1-x)+\text{Li}_2\left(\frac{x-1}{x}\right)=-\frac12\ln^2\left(\frac{1}{x}\right)\tag1$$
$$\text{Li}_2(x)+\text{Li}_2(-x)-\frac12\text{Li}_2(x^2)=0\tag2$$
$$\text{Li}_2(x)+\text{Li}_2(1-x)=\zeta(2)-\ln(x)\ln(1-x)\tag3$$
If we set $x=\frac{\sqrt{5}-1}{2}$ we notice that
$$1-x=\frac{3-\sqrt{5}}{2}$$
$$\frac{x-1}{x}=\frac{1-\sqrt{5}}{2}$$
$$\frac1x=\frac{1+\sqrt{5}}{2}$$
$$x^2=\frac{3-\sqrt{5}}{2}$$
$$\ln(x)\ln(1-x)=2\ln^2\left(\frac{1+\sqrt{5}}{2}\right)$$
Plugging these values in $(1)$, $(2)$ and $(3)$ we get
$$\color{blue}{\text{Li}_2\left(\frac{3-\sqrt{5}}{2}\right)}+\color{red}{\text{Li}_2\left(\frac{1-\sqrt{5}}{2}\right)}=-\frac12\ln^2\left(\frac{1+\sqrt{5}}{2}\right)$$
$$\text{Li}_2\left(\frac{\sqrt{5}-1}{2}\right)+\color{red}{\text{Li}_2\left(\frac{1-\sqrt{5}}{2}\right)}-\frac12\color{blue}{\text{Li}_2\left(\frac{3-\sqrt{5}}{2}\right)}=0$$
$$\text{Li}_2\left(\frac{\sqrt{5}-1}{2}\right)+\color{blue}{\text{Li}_2\left(\frac{3-\sqrt{5}}{2}\right)}=\zeta(2)-2\ln^2\left(\frac{1+\sqrt{5}}{2}\right)$$
Solving this system of equations, we have
$$\color{blue}{\text{Li}_2\left(\frac{3-\sqrt{5}}{2}\right)}=\frac{\pi^2}{15}-\ln^2\left(\frac{1+\sqrt{5}}{2}\right)$$
$$\color{red}{\text{Li}_2\left(\frac{1-\sqrt{5}}{2}\right)}=-\frac{\pi^2}{15}+\frac12\ln^2\left(\frac{1+\sqrt{5}}{2}\right)$$
$$\text{Li}_2\left(\frac{\sqrt{5}-1}{2}\right)=\frac{\pi^2}{10}-\ln^2\left(\frac{1+\sqrt{5}}{2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/463682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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Finding the $n^{\text{th}}$ term of $\frac{1}{4}+\frac{1\cdot 3}{4\cdot 6}+\frac{1\cdot 3\cdot 5}{4\cdot 6\cdot 8}+\ldots$ I need help on finding the $n^{\text{th}}$ term of this infinite series?
$$
s=\frac{1}{4}+\frac{1\cdot 3}{4\cdot 6}+\frac{1\cdot 3\cdot 5}{4\cdot 6\cdot 8}+\ldots
$$
Could you help me in writing the general term/solving?
| The first thing you can do is start with $a_1=\frac{1}{4}$ and then realize that $$a_{n+1} =a_n\frac{2n-1}{2n+2}$$
that doesn't seem to get you anywhere, however.
As commentator Tenali notes, you can write the numerator as $$1\cdot 3\cdots (2n-1) = \frac{1\cdot 2 \cdot 3 \cdots (2n)}{2\cdot 4\cdots (2n)}=\frac{(2n)!}{2^nn!}$$
The denominator, on the other hand, is $$ 4\cdot 6\cdots \left(2(n+1)\right) = 2^n\left(2\cdot 3\cdots (n+1)\right) = 2^n(n+1)!$$
So this gives the result:
$$a_n = \frac{(2n)!}{2^{2n} n!(n+1)!} = \frac{1}{2^{2n}}\frac{1}{n+1}\binom{2n}{n} = \frac{1}{2^{2n}}C_n$$
where $C_n$ is the $n^{\text{th}}$ Catalan number.
If all you want is then $n^{\text{tn}}$ term, that might be enough - you can even skip the part about Catalan numbers and just write it as $a_n=\frac{1}{4^n(n+1)}\binom{2n}n$.
As it turns out, the Catalan numbers have a generating function (see the link above:)
$$\frac{2}{1+\sqrt{1-4x}} = \sum_{n=0}^\infty C_nx^n$$
So, if the series converges when $x=\frac{1}{4}$, it converges to $2$.
(It does converge, since $C_n \sim \frac{2^{2n}}{n^{3/2}\sqrt{\pi}}$.)
| {
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"url": "https://math.stackexchange.com/questions/464285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\cos^4 \theta −\sin^4 \theta = x$. Find $\cos^6 \theta − \sin^6 \theta $ in terms of $x$. Given $\cos^4 \theta −\sin^4 \theta = x$ , I've to find the value of $\cos^6 \theta − \sin^6 \theta $ .
Here is what I did:
$\cos^4 \theta −\sin^4 \theta = x$.
($\cos^2 \theta −\sin^2 \theta)(\cos^2 \theta +\sin^2 \theta) = x$
Thus
($\cos^2 \theta −\sin^2 \theta)=x$ ,
so $\cos 2\theta=x$ .
Now $x^3=(\cos^2 \theta −\sin^2 \theta)^3=\cos^6 \theta-\sin^6 \theta +3 \sin^4 \theta \cos^2 \theta -3 \sin^2 \theta \cos^4 \theta $
So if I can find the value of $3 \sin^4 \theta \cos^2 \theta -3 \sin^2 \theta \cos^4 \theta $ in terms of $x$ , the question is solved. But how to do that ?
| For typing ease, let $a=\cos^2\theta$ and $b=\sin^2\theta$. Thus $a+b=1$. We are told that $a^2-b^2=x$, or equivalently that $a-b=x$.
We want $a^3-b^3$. It will be enough to find $a^2+ab+b^2$. Note the identity
$$4(a^2+ab+b^2)=3(a+b)^2+(a-b)^2.$$
Remark: For other problems of a similar character, it might be preferable to extract $ab$ from the identity $4ab=(a+b)^2-(a-b)^2$, and use standard techniques for expressing symmetric functions of $a$ and $b$ in terms of the elementary symmetric functions $a+b$ and $ab$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/464504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
Show that $19\mid 12^{2013}+7^{2013}$ Show that $$19\mid 12^{2013}+7^{2013}$$$$$$No use of modular arithmetic, have not come this content.
| First we prove two theorems that we will use
If $a,\;b,\;c\;\in \mathbb{N}$ with $a\neq0$ and $x,y\in\mathbb{N} $such that $a\mid b$ and $a\mid c$, then, $a\mid bx+cy$;
Show:$a\mid b$ and $a\mid c$ implies that there $m,n\in\mathbb{N}$ such that $b=a\cdot m$ and $c=a\cdot n$;$$bx+cy=am\cdot x+an\cdot y=a(mx+ny)\Longrightarrow a\mid bx+cy \;\;\;\Box$$Now we have to prove the theorem we use to solve this issue, and in this test we use the theorem proved above
Are $a,b,n\in\mathbb{N}$ with $a+b\neq0$ we have $a+b\mid a^{2n+1}+b^{2n+1}$
Show: Have demonstrated induction $n=0$ the statement is true because $a+b\mid a^{2\cdot 0+1}+b^{2\cdot0+1}\Longrightarrow a+b\mid a+b$; hypothesis: $a+b\mid a^{2n+1}+b^{2n+1}$;$$a^{2(n+1)+1}+b^{2(n+1)+1}=a^{2n+3}+b^{2n+3}=a^{2n+1+2}+b^{2n+1+2}=a^2a^{2n+1}+b^2b^{2n+1}=$$$$=a^2a^{2n+1}+b^2b^{2n+1}\underbrace{-b^2a^{2n+1}+b^2a^{2n+1}}_{=0}=a^2a^{2n+1}-b^2a^{2n+1}+b^2b^{2n+1}+b^2a^{2n+1}=$$$$=a^{2n+1}(a^2-b^2)+b^2(a^{2n+1}+b^{2n+1})$$we know that $a+b\mid a^2-b^2$ because $a^2-b^2=(a+b)(a-b)$; our hypothesis gives us that $a+b\mid a^{2n+1}+b^{2n+1}$ and the theorem shown above assures us that $a+b\mid a^{2n+1}(a^2-b^2)+b^2(a^{2n+1}+b^{2n+1})$ thus showing that the theorem is valid for $n +1$, then, goes for any $n\in\mathbb{N}\;\;\Box$
Question: Show that $19\mid 12^{2013}+7^{2013}$
Using the theorem proved above, we have easily that $19=12+7$ and $$12+7\mid 12^{2013}+7^{2013}$$because $2013=2\cdot1006+1$ and $1006\in\mathbb{N}$
| {
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"url": "https://math.stackexchange.com/questions/464767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the equation : $x^2 − 6 |x − 2| − 28 = 0$ The following is an absolute value quadratic equation that I want to solve:
$$x^2 − 6 |x − 2| − 28 = 0$$
Here is what I did :
$x^2 − 6 |x − 2| − 28 = 0$
$x^2 − 6 |x − 2| − 28 = 0$
$-6|x-2|=28-x^2$
$6|x-2|=x^2-28$
$6x-12=x^2-28$ or $28-x^2$
(Is this step correct ?)
Solving this two quadratic equations I get the answers $x=-2,8,-10,4$
But when I actually substitute these in the original equation I see that only $x=8,-10 $ satisfy and other two solutions are invalid . Why so?
| Once you get to $6|x-2|=x^2-28$, you need to consider two cases.
Case $x \ge 2:$
If $x \ge 2$, then $6x-12 = x^2-28 \implies x^2-6x-16=0 \implies (x+2)(x-8)=0
\implies x \in \{-2, 8\}$ But we assumed $x \ge 2$, so we have to reject $x=-2$ as an extraneous root. Hence if $x \ge 2$, then $x=8$.
Case $x < 2:$
If $x < 2$, then $12-6x = x^2-28 \implies x^2+6x-40=0 \implies (x-4)(x+10)=0
\implies x \in \{4, -10\}$ But we assumed $x < 2$, so we have to reject $x=4$ as an extraneous root. Hence if $x < 2$, then $x=-10$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/464917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Marginal Density - Probability (X,Y) is a two-dimensional abolute continuous random vector with the density function fx,y given by:
(1)
$f_{X,Y}(x,y) = \begin{cases}
\frac{1}{2} & 0 \le x \le 1, 0 \le y \le 4x \\
0 & \text{otherwise} \\
\end{cases}$
Show the density functions fx and fy for X and Y are:
$f_{X}(x) = \begin{cases}
2x & 0 \le x \le 1\\
0 & \text{otherwise} \\
\end{cases}$
$f_{Y}(y) = \begin{cases}
\frac{1}{2}(1-\frac{y}{4}) & 0 \le y \le 4\\
0 & \text{otherwise} \\
\end{cases}$
I'm having a hard time figuring out how this works, i have been looking for a while for some similar problems without any luck.
I can make the step with $$ f_{X}(x) =\int_{0}^{4x} {\frac{1}{2} dy} = 2x$$
But $f_{Y}$ i can't figure out and makes me wonder if the first one is done correctly.
Another example i can't solve is:
(2)
$f_{X,Y}(x,y) = \begin{cases}
\frac{3}{4}x & 0 \le x \le y \le 2\\
0 & \text{otherwise} \\
\end{cases}$
$f_{X}(x) = \begin{cases}
\frac{3}{4}x(2-x) & 0 \le x \le 2\\
0 & \text{otherwise} \\
\end{cases}$
$f_{Y}(y) = \begin{cases}
\frac{3}{8}y^2 & 0 \le y \le 2\\
0 & \text{otherwise} \\
\end{cases}$
From what i gathered i should be able to solve it with $\int{f_{X,Y}(x,y)dx}$, but can't seem to do it right. Any help is appreciated.
| What you did for $f_{X}$ is OK ! For $f_{Y}$, if you just write $f_{Y}(y) = \int_{0}^{1} \frac{1}{2} \: dx$, you're missing that the condition on $y$ (which is $0 \leq y \leq 4x$) also depends on $x$.
To do this right, use indicator functions. $f_{(X,Y)}$ can be re-written as follows :
$$ f_{(X,Y)}(x,y) = \frac{1}{2} \textbf{1}_{\lbrace 0 \leq x \leq 1 \, ; \, 0 \leq y \leq 4x \rbrace} $$
Which also writes :
$$ f_{(X,Y)}(x,y) = \frac{1}{2} \textbf{1}_{\lbrace 0 \leq x \leq 1 \, ; \, 0 \leq \frac{y}{4} \leq x \rbrace} $$
So, we have :
$$ f_{(X,Y)}(x,y) = \frac{1}{2} \textbf{1}_{\lbrace \frac{y}{4} \leq x \leq 1 \rbrace} $$
Now, let's use our favourite formula : $f_{Y}(y) = \int f_{(X,Y)}(x,y) \: dx$. It leads to the expected result :
$$ f_{Y}(y) =
\begin{cases}
\frac{1}{2} \left( 1 - \frac{y}{4} \right) & \text{if } 0 \leq y \leq 4 \\
0 & \text{otherwise} \\
\end{cases}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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For $a$, $b$, $c$, $d$ the sides of a quadrilateral, show $ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0$. (A generalization of IMO 1983 problem 6)
Let $a$, $b$, $c$, and $d$ be the lengths of the sides of a quadrilateral. Show that
$$ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0 \tag{$\star$}$$
Background: The well known 1983 IMO Problem 6 is the following:
IMO 1983 #6. Let $a$, $b$ and $c$ be the lengths of the sides of a triangle.
Prove that $$a^{2}b(a - b) + b^{2}c(b - c) +c^{2}a(c - a)\ge 0. $$
See: here.
A lot of people have discussed this problem. So this problem (IMO 1983) has a lot of nice methods.
Now I found for quadrilaterals a similar inequality. Are there some nice methods for inequality $(\star)$?
| WLOG, assume that $a = \max(a, b, c, d)$.
Let $x = a- b, \ y = a-c, \ z = a-d$. Then $x, y, z \ge 0$.
Let $w = b+c+d - a$. Then $w > 0$ since $a, b, c, d$ are the sides of a quadrilateral.
The inequality is written as $$\frac{1}{4}Aw^2 + \frac{1}{4}Bw + \frac{1}{4}C \ge 0$$ where
\begin{align}
A &= 2\, x^2 - x\, y - 2\, x\, z + 2\, y^2 - y\, z + 2\, z^2,\\
B &= 2\, x^3 + 4\, x^2\, y - 2\, x\, y^2 - 4\, x\, y\, z + 2\, y^3 + 4\, y^2\, z - 2\, y\, z^2 + 2\, z^3,\\
C &= (x+3z)y^3 - 4xzy^2 + (3x^3+3x^2z-3xz^2+z^3)y.
\end{align}
It suffices to prove that $A, B, C \ge 0$.
We have
\begin{align}
A = (x^2 - xy + y^2) + (y^2 - yz + z^2) + (z^2 - 2zx + x^2) \ge 0
\end{align}
and
\begin{align}
B &= 2x^3 + 3x^2y + (x^2y - 2xy^2 + y^3) - 2(2xz)y + y^3 + 3y^2z + z^3 + (y^2z - 2yz^2 + z^3)\nonumber\\
&\ge 2x^3 + 3x^2y - 2(x^2+z^2)y + y^3 + 3y^2z + z^3\nonumber\\
&= 2x^3 + x^2y + y^3 + 2y^2z + (y^2z + z^3 - 2yz^2)\nonumber\\
&\ge 0.
\end{align}
And $C\ge 0$ follows from
\begin{align}
&4(x+3z)(3x^3+3x^2z-3xz^2+z^3) - (4zx)^2\nonumber\\
=\ & 12x^4+48x^3z+8x^2z^2-32xz^3+12z^4\nonumber\\
=\ & (12x^4 - 16x^3z + 24x^2z^2) + 64x^3z - 16x^2z^2 - 32xz^3 + 12z^4\\
\ge \ & 64x^3z - 16x^2z^2 - 32xz^3 + 12z^4 \nonumber \\
= \ & 4z(4x+3z)(2x-z)^2\nonumber\\
\ge \ &0.
\end{align}
We are done.
| {
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} |
How to find the second derivative of $2x^3 + y^3 = 5$? I'm finding it difficult to find the second derivative of the following equation: $2x^3 + y^3 = 5$.
My answer is $\dfrac{-4xy^3 - 8x^4}{y^5}$, but the answer key says $\dfrac{-20x}{y^5}$.
My friends and I have the same answer, but I don't know where I went wrong.
Here is my work:
I got the first derivative as $\dfrac{-2x^2}{y^2}$.
I then used the quotient rule to get
$-2 \left(\dfrac{2xy^2 - 2x^2y\frac{dy}{dx}}{y^4}\right)$
After factoring out $2y$ from the numerator, I got $-4 \left(\dfrac{xy - x^2\left(\frac{-2x}{y^2}\right)}{y^3}\right)$.
After splitting the fraction into $\dfrac{x}{y^2} + \dfrac{2x^4}{y^5}$, I ended up getting
$\dfrac{-4x}{y^2}-\dfrac{-8x^4}{y^5}$.
Could somebody point out my mistake? This isn't homework by the way. I'm preparing for a placement test tomorrow, but I'm not really sure where I went wrong.
| Your answer only requires some re-arrangement:
$$
2x^3 + y^3 = 5
$$
multiply by $x$ on both sides to get:
$$
2x^4 + xy^3 = 5 x
$$
or:
$$
xy^3 = 5x - 2x^4
$$
...(substitute this in your equation to get):
$$
\frac{-4(5x - 2x^4 )- 8x^4}{y^5}\to y''=\frac{-20x}{y^5}
$$
| {
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Inequality with square roots: $\sqrt{x^2+1}+\sqrt{y^2+1}\ge \sqrt{5}$
Let $x$ and $y$ be nonnegative real numbers such that $x+y=1$. How do I show that $\sqrt{x^2+1}+\sqrt{y^2+1}\ge \sqrt{5}$?
How do I deal with square roots inside the inequality?
| If you are familiar with Minkowski's inequality, this is straightforward.
$$\sqrt{x^2+1}+\sqrt{y^2+1} \ge \sqrt{(x+y)^2 + (1+1)^2} = \sqrt5$$
| {
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} |
equilateral triangle; $3(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.$ In equilateral triangle ABC of side length d, if P is an internal point with PA = a, PB = b, and PC = c, the following pleasingly symmetrical relationship holds:
$3(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.$
Please prove this identity.
source: http://www.qbyte.org/puzzles/p117s.html
| Here's a brute-force approach:
Abbreviating the measures of angles surrounding point $P$ thusly,
$$\alpha := |\angle BPC| \qquad \beta := |\angle CPA| \qquad \gamma := |\angle APB|$$
we have
$$\cos\gamma = \cos\left(2\pi - \alpha - \beta\right) = \cos\left(\alpha+\beta\right) = \cos\alpha \cos \beta - \sin\alpha \sin\beta $$
so that
$$\left( \cos\alpha \cos\beta - \cos\gamma \right)^2 = \left(1-\cos^2\alpha\right)\left(1-\cos^2\beta\right)$$
Now, substitute-in these relations from the Law of Cosines
$$\cos\alpha = \frac{b^2+c^2-d^2}{2bc} \qquad
\cos\beta = \frac{c^2+a^2-d^2}{2ca} \qquad
\cos\gamma = \frac{a^2+b^2-d^2}{2ab}$$
and do a bit of algebra.
| {
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proof for the Ramanujan's formula ? I found this formula in a textbook in which the proof to the formula was not given
Ramanujam's formula
$$\sqrt{1 +n\sqrt{1 +(n+1)\sqrt{1 + (n+2)\sqrt{1 + (n+3)\sqrt{1 +....\infty}}}}} = n+1$$
Its a great equation andhow do you prove this. its a bit difficult for me and tried different methods to solve this. Iam an undergraduate and I want you to elaborate the method of solving if its complex.
| Here is this answer adapted to this question.
Define
$$
f(x)=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\dots}}}}
$$
then $f(x)^2=1+xf(x+1)$. This indicates we should look at $f(x)=x+1$.
Considering $f(x)=x+1$, we are lead to show inductively that
$$
\hspace{-18pt}x+1=\small\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+\dots+\sqrt{1+(x+k-2)\sqrt{1+(x+k-1)(x+k+1)}}}}}\tag{1}
$$
Define
$$
f_{k,x}(y)=\left\{\begin{array}{}
y&\text{if }k=0\\
\sqrt{1+xf_{k-1,x+1}(y)}&\text{if }k>0
\end{array}\right.
$$
unrolled, that is
$$
f_{k,x}(y)=\small \sqrt{1+x\sqrt{1+(x+1)\sqrt{1+\dots+\sqrt{1+(x+k-2)\sqrt{1+(x+k-1)y}}}}}
$$
Note that $f_{0,x}(x+1)=x+1$. Suppose that, for some $k\ge0$, $f_{k,x}(x+k+1)=x+1$. Then
$$
\begin{align}
f_{k+1,x}(x+k+2)
&=\sqrt{1+xf_{k,x+1}(x+k+2)}\\
&=\sqrt{1+x(x+2)}\\
&=x+1
\end{align}
$$
Therefore, for all $k\ge0$, we have $f_{k,x}(x+k+1)=x+1$.
Thus $(1)$ is confirmed.
Our sequence is bounded by $f_{k,x}(1)$ and $f_{k,x}(x+k+1)$. To show that our sequence converges to $x+1$, we want to show that
$$
\lim_{k\to\infty}f_{k,x}(1)=x+1\tag{2}
$$
Note that
$$
\frac{f_{0,x+k}(x+k+1)}{f_{0,x+k}(1)}=\frac{x+k+1}{1}
$$
Suppose that, for some $j\ge0$,
$$
\frac{f_{j,x+k-j}(x+k+1)}{f_{j,x+k-j}(1)}\le(x+k+1)^{1/2^j}
$$
Then
$$
\begin{align}
\frac{f_{j+1,x+k-j-1}(x+k+1)}{f_{j+1,x+k-j-1}(1)}
&=\sqrt{\frac{1+(x+k-j-1)f_{j,x+k-j}(x+k+1)}
{1+(x+k-j-1)f_{j,x+k-j}(1)}}\\
&\le\sqrt{\frac{f_{j,x+k-j}(x+k+1)}{f_{j,x+k-j}(1)}}\\
&\le(x+k+1)^{1/2^{j+1}}
\end{align}
$$
Therefore, for all $j\ge0$,
$$
\frac{f_{j,x+k-j}(x+k+1)}{f_{j,x+k-j}(1)}\le(x+k+1)^{1/2^j}
$$
in particular, for $j=k$,
$$
\frac{f_{k,x}(x+k+1)}{f_{k,x}(1)}\le(x+k+1)^{1/2^k}
$$
Since $f_{k,x}(x+k+1)=x+1$, we have
$$
(x+1)(x+k+1)^{-1/2^k}\le f_{k,x}(1)\le (x+1)
$$
and by the squeeze theorem, we have
$$
\lim_{k\to\infty}f_{k,x}(1)=x+1
$$
confirming $(2)$, as desired.
| {
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Finding $x^4 + y^4 + z^4$ using geometric series This is a problem from the 2001 Stanford Math Tournament Algebra section.
$$$$Given that $$x+y+z=3$$ $$x^2 + y^2 + z^2 = 5$$$$x^3+y^3+z^3=7$$Find $x^4+y^4+z^4$. $$$$My friend claimed that he was able to solve this question using geometric series, i.e. adding the above three equations together and forming geometric series with three terms, but I can't see how he solved it using that method. I solved it using the method described in the solutions document, but I haven't seen a solution using geometric series. Can anyone find a way to solve the problem using geometric series? Any help with this problem would be great. Thanks for the help.
| This is a rather encumbered solution for a not-as-hard problem, and it is the only way I see to make sense of what your friend is saying. On the other hand, it is cool that it let us know about Newton's identities and their proof using generating functions.
Let us form the formal series \begin{align}F(x,y,z,t)&:=(x+y+z)t+(z^2+y^2+z^2)t^2+(x^3+y^3+z^3)t^3+\ldots\\&=3t+5t^2+7t^3+at^4+\ldots,\end{align}
where $a:=x^4+y^4+z^4$ is the number we want to find.
Forming this power series is kind of adding the given equations (as your friend suggested), but we multiply them first by $t,t^2,t^3,...$, respectively, before adding.
Consider the polynomial \begin{align}G(x,y,z,t)&:=-1+(x+y+z)t+(xy+xz+yz)t^2-(xyz)t^3\\&=-1+3t+(xy+xz+yz)t^2-(xyz)t^3\end{align}
We multiply $F$ times $G$. It happens that we get a polynomial:
\begin{align}
FG&=\left(\sum_{k=1}^{\infty}(xt)^k+\sum_{k=1}^{\infty}(yt)^k+\sum_{k=1}^{\infty}(zt)^k\right)G\\
&=-\left(\frac{xt}{1-xt}+\frac{yt}{1-yt}+\frac{yt}{1-xt}\right)(1-xt)(1-yt)(1-zt)\\
&=t\left((-x)(1-yt)(1-zt)+(-y)(1-xt)(1-zt)+(-z)(1-xt)(1-yt)\right)\\
&=-(x+y+z)t+(xy+xz+yz)t^2-(xyz)t^3\\
&=-3t+(xy+xz+yz)t^2-(xyz)t^3
\end{align}
In the second equality we used the sum of the geometric series more or less like your friend said.
Comparing coefficients of the multiplication $FG$ with the result in the last line we get (the Newton's identities for the values in this problem)
\begin{align}
2(xy+xz+yz)&=9-5\\
3xyz&=3(xy+xz+yz)-15+7\\
0&=3(xyz)-5(xy+xz+yz)+21-(x^4+y^4+z^4)
\end{align}
From these, substituting from top to bottom, we find the value of $x^4+y^4+z^4$. Notice that the method allows us to continue using the equalities between coefficients of higher degree in $t$ to compute $x^5+y^5+z^5$, $x^6+y^6+z^6$, and so on ...
| {
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Limit of a sequence of integrals Determine
$$\lim _{n \rightarrow \infty} \int_{-\infty}^\infty \frac{dx}{n(e^{x^2}-1) +1/n}$$
Since $e^{x} \geq t+1$ we know that $e^{x^2}-1 \geq 2t +t^2 \geq t^2$ if $t\geq 0$
So due to symmetry, we should be able to use DCT?
But my solution is $$\lim _{n \rightarrow \infty} \int_{-\infty}^\infty \frac{dx}{n(e^{x^2}-1) +1/n} = \int_{-\infty}^\infty \lim _{n \rightarrow \infty} \frac{dx}{n(e^{x^2}-1) +1/n} = 0$$. Which is the wrong answere.
Can someone help me out here?
We have not startet with variable change in measure theory, so please do not use that.
| We start from the the inequality $ e^y \le 1+y +\frac {3y^2} {2}$ if $ y \ge 0,\, y \le 1$. This follows from the Taylor's theorem
in the Lagrange form. Next, we obtain with help of Maple
$$J_n:=\int_{-\frac 1 {\sqrt{n}}}^{\frac 1 {\sqrt{n}}} \frac{dx}{n(e^{x^2}-1) +1/n} \ge \int_{-\frac 1 {\sqrt{n}}}^{\frac 1 {\sqrt{n}}} \frac{dx}{nx^2+3nx^4/2 +1/n}=$$ $$ \frac {1}{\sqrt {{n}^{2}-6}\sqrt { \left( \sqrt {{n}^{2}-6}-n
\right) n}\sqrt { \left( \sqrt {{n}^{2}-6}+n \right) n}}$$ $$-2\,n\sqrt {3} \left( \mathop{\rm arctanh} \left( {\frac {\sqrt {3n}}{\sqrt { \left( \sqrt {{n}^{2}-6}-n \right) n}}} \right) \sqrt {
\left( \sqrt {{n}^{2}-6}+n \right) n}+$$
$$ \left.\sqrt { \left( \sqrt{{n}^{2}-6
}-n \right) n}\arctan \left( {\frac {\sqrt {n}\sqrt {3}}{\sqrt {
\left( \sqrt {{n}^{2}-6}+n \right) n}}} \right) \right)\to \pi,\, n \to \infty.(1)
$$
On the other hand,
$$J_n= \int_{-\frac 1 {\sqrt{n}}}^{\frac 1 {\sqrt{n}}} \frac{dx}{n(e^{x^2}-1) +1/n} \le \int_{-\frac 1 {\sqrt{n}}}^{\frac 1 {\sqrt{n}}} \frac{dx}{nx^2 +1/n} =\frac 1 n \int_{-\frac 1 {\sqrt{n}}}^{\frac 1 {\sqrt{n}}} \frac{dx}{x^2 +1/n^2}=2\arctan(\sqrt{n}) \to \pi,\,n \to \infty .(2)$$
It remains to estimate the tails: $$\int_{-\infty}^{-\frac 1 {\sqrt{n}}} \frac{dx}{n(e^{x^2}-1) +1/n}+ \int_{\frac 1 {\sqrt{n}}}^{\infty} \frac{dx}{n(e^{x^2}-1) +1/n}\le 2\int_{\frac 1 {\sqrt{n}}}^{\infty} \frac{dx}{nx^2} =\frac 2 {\sqrt{n}} \to 0,\, n \to \infty .(3)$$ Combining (1)-(3), we deduce that the limit under consideration equals $\pi$.
| {
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The Converse of the Pythagoras Theorem Let $ABC$ be a triangle in the plane. Suppose that $AB^2+AC^2=BC^2.$ Prove that:
$\angle BAC$ is right angle.
Remark: I believe this to be true. Now I have the following difficulty. I'm looking for a proof which a 10th grade student can understand, but any effort of mine is via proof by contradiction. I mean assuming that the given angle is not a right angle and then arrive on a contradiction. So I'm looking for a solution which could avoid the indirect method but if not then a solution by the indirect method.
| Well, what always worked for me as a tenth grader was to cut a triangle into right triangles:
Construct a point $D$ on $BC$ so that $DA$ is perpendicular to $BC$ so $ADC$ is a right triangle and $AD^2 + DC^2 = AC^2$.
THere are three possibilities.
i) $\angle ABC$ is acute.
Then $D$ is between $B$ and $C$ and $BC = BD + DC$.
Then $AB^2 = BD^2 + AD^2; AC^2 = AD^2 + DC^2; BC^2 = BD^2 + 2BD*DC + DC^2$
And $AB^2 + BC^2 = AD^2 + DC^2 + 2(BD^2+ BD*DC) = AC^2 + 2(BD^2+ BD*DC) > AC^2$
(It helps if you draw a picture and note $AB > AD; BC > BD$ so of course $AB^2 + BC^2 > AD^2 + DC^2 = AC$. But the calculation tells you exactly how much bigger.)
ii) $\angle ABC$ is obtuse.
Then $B$ is between $D$ and $C$ and $DC = DB + BC$.
Then $AB^2 = BD^2 + AD^2; AC^2 = AD^2 + DC^2; DC^2 = DB^2 + 2DB*BC + BC^2$
And $AB^2 + BC^2 = AD^2 + DC^2 - 2(BD^2+ BD*DC) = AC^2 - 2(BD^2+ BD*DC) < AC^2$
iii) $\angle ABC$ is right.
Then $B=D$ and $C$ and $DC = BC $.
Then $AB^2 = AD^2; AC^2 = AD^2 + DC^2=AB^2 + BC^2; BC^2 = DC^2 $
So i) $AB^2 + BC^2 > AC^2 \implies \triangle ABC$ is acute and $AB^2 + BC^2 = AC + 2(BD^2 + BD*DC)$.
ii) $AB^2 + BC^2 < AC^2 \implies \triangle ABC$ is abtuse and $AB^2 + BC^2 = AC - 2(BD^2 + BD*DC)$.
iii) $AB^2 + BC^2 = AC^2 \implies \triangle ABC$ is right.
.... which gives the tenth grader a hint of the law of cosines to come.
[i.e. $\pm(BD^2 + DB*DC) = AB*BC*\cos \angle ABC$ which is $0$ if $\angle ABC$ is right; positive if obtuse and negative if acute.]
| {
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How do I factorise this polynomial Please help factorise this:
$$6x^2+x+4=0$$
In my attempts, I assumed $a=6$, $b=1$ and $c=4$.
I multiplied $c$ by $a$ and attempted to get the factors that give us the sum of $b$.
The only factor pairs of $24$ are $1\cdot 24$, $2\cdot 12$, $3\cdot 8$ and $4\cdot 6$.
None of the above can give me a sum of $1$.
| Let us complete the square:
\begin{align*}
6 x^2 + x + 4 &= 6 \left(x^2 + \frac{1}{6} x + \frac{3}{2}\right) \\
&= 6 \left(x^2 + \frac{1}{6} x + \frac{1}{144} + \frac{3}{2} - \frac{1}{144}\right) \\
&= 6 \left(x + \frac{1}{12}\right)^2 + 6 \left(\frac{3}{2} - \frac{1}{144}\right) \\
&= 6 \left(x + \frac{1}{12}\right)^2 + \left(9 - \frac{1}{24}\right)
\end{align*}
Setting this to zero gives
$$6 \left(x + \frac{1}{12}\right)^2 = - \left(9 - \frac{1}{24}\right) < 0$$
which has no real solutions.
| {
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"source": "stackexchange",
"question_score": "1",
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Maximum of a trigonometric Polynomial
Given
$$x+y+z=\pi$$
$$3\sin(x)+4\sin(y)+18 \sin(z)=A$$
Question：find maximum of $A$.
I spend so many time on this question.
answer is $ 35\sqrt{7} /4$, but why?
| Assuming that $x,y,z>0$, you can write this as:
$$3\sin(x)+4\sin(y)+18 \sin(\pi - x - y)=A$$
Now find when the derivatives:
$$\frac{\partial A}{\partial x}, \frac{\partial A}{\partial y}$$
are equal to zero. You'll get:
$$3 \cos(x) + 18 \cos(x + y)= 0,\ 4 \cos(y) + 18 \cos(x + y) = 0$$
Which leads to:
$$y=\arccos\left(\frac{3 \cos x}{4}\right)$$
Plugging back in to the $\cos(x+y)$ term, and using $\cos(\arctan x)=1/\sqrt{1+x^2}$, you'll get a quadratic in $t = \cos x$:
$$3 t + \frac{27 t^2}{2} - 18 \sqrt{1 - t^2} \sqrt{1 - 9 t^2/16} = 0$$
Or:
$$t^2 (229 + 36 t) = 144 \ \to \ t=\cos x = 3/4$$
Finally, plug in the values $(x = \arccos 3/4, y = \arccos 9/16)$ into the original function. The trick here is to note that:
$$\arccos\alpha \pm \arccos\beta = \arccos\left(\alpha\beta \mp \sqrt{(1-\alpha^2)(1-\beta^2)}
\right)$$
$$\sin(\arccos x) = \sqrt{1-x^2}$$
So that:
$$18\sin\left(\arccos 3/4 + \arccos 9/16\right) = \frac{27 \sqrt{7}}{4}$$
And:
$$A_{max} = \frac{3 \sqrt{7}}{4} + \frac{5 \sqrt{7}}{4} + \frac{27 \sqrt{7}}{4} = \frac{35 \sqrt{7}}{4}$$
| {
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If this is a telescoping series then how does it collapse? $\frac{3r+1}{r(r-1)(r+1)}$ Express $$\frac{3r+1}{r(r-1)(r+1)}$$ in partial fractions. Hence, or otherwise, show
$$\sum_{r=2}^n\frac{3r+1}{r(r-1)(r+1)}=\frac52-\frac2n-\frac{1}{n+1}$$
So, I have obtained the partial fractions $$\frac{3r+1}{r(r-1)(r+1)}=\frac{2}{r-1}-\frac{1}{r}-\frac{1}{r+1}$$
then for $r=(2,3,4,5...(n-1),n)$
$$
\begin{align}
& {}+2-\frac12-\frac13 \\[8pt]
& {}+1-\frac13-\frac14 \\[8pt]
& {}+\frac23-\frac14-\frac15 \\[8pt]
& {}+\frac12-\frac15-\frac16 \\[8pt]
& {}+\cdots \\[8pt]
& {}+\frac{2}{n-2}-\frac{1}{n-1}-\frac{1}{n} \\[8pt]
& {}+\frac{2}{n-1}-\frac{1}{n}-\frac{1}{n+1}
\end{align}
$$
If this is a telescopic series how does it collapse?
| HINT:
There are at least two methods for $$\frac{3r+1}{r(r-1)(r+1)}=\frac{2}{r-1}-\frac{1}{r}-\frac{1}{r+1}$$
Method $1:$
$$\frac{2}{r-1}-\frac{1}{r}-\frac{1}{r+1}=-\left(\frac1{r+1}-\frac1r\right)-2\left(\frac1r-\frac1{r-1}\right)$$
The survivor of the summation $2\le r\le n$ after cancellation,
for the first part will be $\displaystyle-\left(\frac1{n+1}-\frac12\right)=\frac12-\frac1{n+1}$
for the second part will be $\displaystyle-2\left(\frac1n-\frac11\right)=2-\frac2n$
Method $2:$
$$\frac{2}{r-1}-\frac{1}{r}-\frac{1}{r+1}=\left(\frac1{r-1}-\frac1r\right)-\left(\frac1{r+1}-\frac1{r-1}\right)$$
Put a few values of $r$ like $2,3,4,5$
and $n-3,n-2,n-1,n$ to recognize the Telescopic nature & the surviving terms after cancellation
| {
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Given that $T=\frac{x-iy}{x+iy}$, where $x, y, T \in\Bbb R$, show that $\frac{1+T^2}{2T}=\frac{x^2-y^2}{x^2+y^2}$ Given that $T=\frac{x-iy}{x+iy}$, where $x, y, T \in\Bbb R$, show that $$\frac{1+T^2}{2T}=\frac{x^2-y^2}{x^2+y^2}$$
I have got rid of $i$ in the denominator, leaving $$\frac{x^2-y^2-2ixy}{x^2+y^2}$$
but now I am stuck. What are the next steps please?
| You made an error somewhere.
Writing $T = \frac{a}{b}$, the expression you are trying to simplify is
$$\frac{1+\frac{a^2}{b^2}}{2\frac{a}{b}}$$
which after clearing out the denominators is
$$\frac{b^2+a^2}{2ab}.$$
Plugging in $a$ and $b$ and using the fact that they are complex conjugates gives
$$\frac{x^2 +2ixy - y^2 + x^2 -2ixy - y^2}{2(x^2+y^2)}$$
which simplifies to the expression you want.
| {
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Partial Fraction decomposition when denominator is in $x^2 + a$ form Why isn't the expansion of $$ \frac{s^3 - 2s^2 + 16s - 2}{(s^2+1)(s^2+16)} $$
in the form of
$$ \frac{As+B}{s^2+1} + \frac{Cs+D}{s^2+16} $$ since (as I recall) the denominator is of square power and you should decompose it (in the numerator) until the s diminishes.
Apparently wolframalpha is saying the correct form should be
$$ \frac{A}{s^2+1} + \frac{B}{s^2+16} $$
Please clarify.
Thank you in advance.
| We have:
$$ \dfrac{s^3 - 2s^2 + 16s - 2}{(s^2+1)(s^2+16)} = \dfrac{as+b}{s^2+1} + \dfrac{cs+d}{s^2+16} $$
When we multiply things out and equate sides, we have:
$$ s^3 - 2s^2 + 16s - 2 = 16 b + d + 16 a s + c s + b s^2 + d s^2 + a s^3 + c s^3$$
Equating like powers leads to:
$$a+c = 1, b+d = -2, 16a+c = 16, 16b+d=-2$$
When you solve for the constants, you get:
$$a = 1, b = 0, c = 0, d = -2$$
So,
$$ \dfrac{s^3 - 2s^2 + 16s - 2}{(s^2+1)(s^2+16)} = \dfrac{s}{s^2+1} - \dfrac{2}{s^2+16} $$
| {
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Sum of the series $1+\frac{1\cdot 3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$ Decide if the sum of the series
$$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+ \cdots$$
is: (i) $\infty$, (ii) $1$, (iii) $2$, (iv) $4$.
| As suggested in another answer, one can write:
$$\prod_{k=1}^n \frac{2k+1}{2k+4}=\frac{8}{\pi}\cdot \frac{\Gamma (n+\frac{3}{2})\Gamma (\frac{3}{2})}{\Gamma (n+3)}=\frac{8}{\pi}\text{B}\left(n+\frac{3}{2},\frac{3}{2}\right)$$
Now, using the definition of the beta function, our sum is:
$$1+\frac{8}{\pi}\sum_{n\geq 1}\int_0^1 t^{n+\frac{1}{2}}(1-t)^{\frac{1}{2}}\,dt=1+\frac{8}{\pi}\int_0^1 t^{\frac{3}{2}}(1-t)^{-\frac{1}{2}}\,dt$$
Letting $t=\sin^2 w,$ this gives
$$1+\frac{16}{\pi}\int_0^{\frac{\pi}{2}}\sin^4 w\,dw=1+\frac{16}{\pi}\left(\frac{3\pi}{16}\right)=4$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving Inequality with Square Roots and Algebraic Manipulation Question:
Edit: Original question is: Estimate $\sqrt{1}+\sqrt{2}+\cdots+\sqrt{10000}$ to nearest hundred.
I used $\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n}\geq \int_{0}^{n}\sqrt{x}dx$ and $\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n}\leq \frac{4n+3}{6}\sqrt{n}$, which I need to prove.
So prove that: $$\left ( \frac{4n+3}{6} \right )\sqrt{n}+\sqrt{n+1}\leq \left ( \frac{4n+7}{6} \right )\sqrt{n+1}$$
Attempt:
$$\left ( \frac{4n+3}{6} \right )\sqrt{n}+\sqrt{n+1}= \frac{(4n+3)\sqrt{n}+6\sqrt{n+1}}{6}\leq \frac{(4n+3)\sqrt{n+1}+6\sqrt{n+1}}{6}= \frac{(4n+9)\sqrt{n+1}}{6}$$
But $$\frac{(4n+9)\sqrt{n+1}}{6}\geq \frac{(4n+7)\sqrt{n+1}}{6}$$
How can I prove that $$\frac{(4n+3)\sqrt{n}+6\sqrt{n+1}}{6}\leq \frac{(4n+7)\sqrt{n+1}}{6}$$
| Your desired inequality
$$\frac{(4n+3)\sqrt{n}+6\sqrt{n+1}}{6}\leq \frac{(4n+7)\sqrt{n+1}}{6}$$
is equivalent to
$$(4n+3)\sqrt{n}\le(4n+1)\sqrt{n+1}\;.$$
Since we’re dealing with non-negative numbers here, this is equivalent to
$$n(16n^2+24n+9)\le(n+1)(16n^2+8n+1)\;.$$
If you expand that, you’ll see that it’s true.
| {
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Can't simplify this fraction: $ \frac{1+x^6}{1+x^2}$ I've been having trouble simplifying this fraction :
$$
\frac{1+x^6}{1+x^2}
$$
Can anyone explain step by step on how to solve this?
Thank you.
| $1+x^6=1+({x^2})^3$ which is equivalent to
$(1+x^2)(x^4+1-x^2)$ using $a^3+b^3=(a+b)(a^2+b^2-ab)$
Hence the expression becomes $$\frac{(1+x^2)(x^4+1-x^2)}{1+x^2} = (x^4+1-x^2).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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IMO 1988, problem 6 In 1988, IMO presented a problem, to prove that $k$ must be a square if $a^2+b^2=k(1+ab)$, for positive integers $a$, $b$ and $k$.
I am wondering about the solutions, not obvious from the proof. Beside the trivial solutions a or $b=0$ or 1 with $k=0$ or $1$, an obvious solution is $a=b^3$ so that the equation becomes $b^6+b^2=b^2(1+b^4)$ . Are there any other solutions?
| For given positive integers a and b, and applying simple division of $(a^2 + b^2)$ by $(ab + 1)$ we have quotient as a/b and remainder as $(b^2 - a/b)$.
For the result to be a perfect division, the remainder must be zero. Therefore $(b^2 - a/b) = 0$ which gives the relation $a=b^3$.
Applying $a=b^3$ in the equation $x=(a^2 + b^2)/(ab + 1)$, we have $x=b^2$ which is a perfect square of an integer.
All the number pairs $(a,b)$ of the form $(b^3,b)$ are solutions of the above equation and the number which is a perfect square is of the form $x = b^2$ applicable for all $b$ belonging to a set of positive integers.
| {
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"source": "stackexchange",
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Solving a system of equations using modular arithmetic modulo 5 Give the solution to the following system of equations using modular arithmetic modulo 5:
$4x + 3y = 0 \pmod{5}$
$2x + y \equiv 3 \pmod{5}$
I multiplied $2x + y \equiv 3 \pmod 5$ by $-2$, getting $-4x - 2y \equiv -6 \pmod{5}$.
$-6 \pmod{5} \equiv 4 \pmod 5$
Then I added the two equations:
$4x + 3y \equiv 0 \pmod{5}$
$-4x - 2y \equiv 4 \pmod{5}$
This simplifies to $y \equiv 4 \pmod{5}$.
I then plug this into the first equation: $4x + 3(4) = 0 \pmod{5}$
Wrong work:
Thus, $x = 3$.
But when I plug the values into the first equation, I get $2(3) + 4 \not\equiv 3 \pmod{5}$.
What am I doing wrong?
EDIT:
Revised work:
$x = -3 \pmod{5} = 2 \pmod{5}$.
Now when I plug the values into the first equation, I get $2(2) + 4 \equiv 8 \pmod{5} \equiv 3 \pmod{5}$.
| It's the last step, where you're solving for $x$.
$4x + 12 \equiv 0 \Rightarrow 4x\equiv 3 \Rightarrow x\equiv 2$.
| {
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Sum of Fibonacci-Numbers Is there a closed formula for the sum of the first $n$ even (or odd) Fibonacci numbers, like there is one for the $n$th number (Moivre-Binet)?
| Let $\varphi=\frac12\left(1+\sqrt5\right)$ and $\widehat\varphi=\frac12\left(1-\sqrt5\right)$, so that $F_n=\frac1{\sqrt5}\left(\varphi^n-\widehat\varphi^n\right)$. Then
$$\varphi^3=\left(\frac{1+\sqrt5}2\right)^3=2+\sqrt5\qquad\text{and}\qquad\widehat\varphi^3=\left(\frac{1-\sqrt5}2\right)^3=2-\sqrt5\;,$$
so the sum of the first $n$ even Fibonacci numbers is
$$\begin{align*}
\sum_{k=0}^{n-1}F_{3k}&=\frac1{\sqrt5}\sum_{k=0}^{n-1}\left(\varphi^{3k}-\widehat\varphi^{3k}\right)\\\\
&=\frac1{\sqrt5}\left(\sum_{k=0}^{n-1}\varphi^{3k}-\sum_{k=0}^{n-1}\widehat\varphi^{3k}\right)\\\\
&=\frac1{\sqrt5}\left(\frac{\varphi^{3n}-1}{\varphi^3-1}-\frac{\widehat\varphi^{3n}-1}{\widehat\varphi^3-1}\right)\\\\
&=\frac{(2+\sqrt5)^n-1}{5+\sqrt5}-\frac{1-(2-\sqrt5)^n}{5-\sqrt5}\\\\
&=\frac{(2+\sqrt5)^n}{5+\sqrt5}+\frac{(2-\sqrt5)^n}{5-\sqrt5}-\left(\frac1{5+\sqrt5}+\frac1{5-\sqrt5}\right)\\\\
&=\frac{(2+\sqrt5)^n}{5+\sqrt5}+\frac{(2-\sqrt5)^n}{5-\sqrt5}-\frac12\;.
\end{align*}$$
Moreover, $2-\sqrt5\approx-0.23607$, and $5-\sqrt5\approx2.76393$, so even for $n=1$ the middle term is only about $-0.08541$, and it decreases rapidly as $n$ increases. Thus, you want the integer nearest
$$\frac{(2+\sqrt5)^n}{5+\sqrt5}-\frac12\;,$$
which is
$$\left\lfloor\frac{(2+\sqrt5)^n}{5+\sqrt5}\right\rfloor\;.$$
| {
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Olympiad number theory problem I found this problem in previous problems of the olympiads of my country
If $t^2+n^2=r^2$, where $t$ has $3$ positive divisors, $n$ has $30$ positive divisors and $t,n,r$ are natural numbers, find the sum of all the possible values of $t$
I did gave it a try but only solved it partially, and I really doubt this is the best way to do it.
My progress:
First, $t$ must be of the form $p^2$ where $p$ is prime
Then, we use pythagorean triplets
Case 1:
$$t=p^2=2ab$$
$$n=a^2-b^2$$
Therfore, $p$ must be $2$, therefore $a=2, b=1$, but this does not satisfy $n$ having $30$ divisors.
Case 2:
$$t=p^2=a^2-b^2=(a-b)(a+b)$$
$$n=2ab$$
Since $p$ is prime, we get $a=b+1$, therefore
$$p^2=2b+1$$
$$n=2(b+1)b$$
If $n$ has $30$ divisors, it must be of one of the forms
$$q^{29},r^{14}q,r^{9}q^2,r^4q^5,r^4q^2s$$
Where $r,q,s$ are different primes
Since $p$ is odd, $p^2=8k+1=2b+1$, $b=4k$ for some integer $k$
Case 2.1
$$8k(4k+1)=r^{29}\implies 4k(4k+1)=2^{28}$$
That is clearly impossible
Case 2.2
$$8k(8k+1)=r^{14}q\implies r=2$$
$$4k(4k+1)=2^{13}q$$
$$\implies 4k=2^{13} \implies q=4k+1=2^{13}+1=3×2731$$
Case 2.3
$$8k(4k+1)=r^9q^2\implies r=2$$
$$8k(4k+1)=2^9q^2$$
$$4k(4k+1)=2^8q^2$$
$$\implies 4k=2^8\implies q^2=4k+1=256+1=257$$
Case 2.4
$$8k(4k+1)=r^4q^5$$
Case 2.4.1($r=2$)
$$4k(4k+1)=2^3q^5$$
$$\implies 4k=8\implies q^5=4k+1=8+1=9$$
Case 2.4.2($q=2$)
$$4k(4k+1)=r^42^4$$
$$\implies 4k=16\implies r^4=4k+1=16+1=17$$
Case 2.5
$$8k(4k+1)=r^4q^2s\implies r=2$$
$$8k(4k+1)=16q^2s$$
$$8k(8k+2)=32q^2s$$
$$(p-1)(p+1)=32q^2s$$
$$p^2-1=32q^2s$$
Where all $p,q,s$ are odd primes
The question now is: how do I proceed from here? Was there an easier way to solve this?
| We add a little to the analysis by Yury. It could be that $a^2-b^2=p$, the odd leg of our triple is $p(a^2-b^2)$, and the even leg $p(2ab)$, where $(a^2-b^2,2ab,a^2+b^2)$ is a primitive triple.
Then $2ab$ must have $15$ divisors. That leaves very few cases, since the power of $2$ that divides the even one of $a$ and $b$ must be $2^1$, $2^3$, or $2^{13}$. The odd one then must be $q^4$, $q^2$, or $1$, where $q$ is prime. A little fooling around shows that we need $b=8$, and therefore $a=9$.
So we get the solution $t=17^2$, $n=(17)(144)$
| {
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Binomial expansion of modulus? How one can perform binomial expansion of modulus of a quantity.
For example, can we expand |-1+x| ? If, yes how?
Thanks
| Two cases it depends on (-1+x) value
Case 1: (-1+x) is positive then write it as $(x-1)^n$ and use normal binomial theorem
Case 2: if (-1+x) is negative your answer is $-(x-1)^n$
Again if n is not a positive integer then you have to use
$$(1+x)^n=1+nx+ \frac{n(n-1)}{1 \cdot 2} x^2 + \frac{n(n-1)(n-2)}{1\cdot2\cdot3} x^3+\cdots$$
So then case 1 will be $$(-1)^n \bigg[\frac{n(n-1)}{1 \cdot 2}x^2 + \frac{n(n-1)(n-2)}{1\cdot2\cdot3} x^3+ \cdots \bigg]$$
Case 2 will be $$(-1)^{n+1}\bigg[1+nx+ \frac{n(n-1)}{1 \cdot 2}x^2 + \frac{n(n-1)(n-2)}{1\cdot2\cdot3} x^3+ \cdots\bigg]$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving inequality $(a+\frac{1}{a})^2 + (b+\frac{1}{b})^2 \geq \frac{25}{2}$ for $a+b=1$ If $a, b$ are positive real numbers and $a+b = 1$, prove that :
$$\left(a+\frac{1}{a}\right)^2 + \left(b+\frac{1}{b}\right)^2 \geq \frac{25}{2}$$
I can see that the value $\frac{25}2$ is attained for $a=b=\frac12$. But I do not know how to show that this is the minimal possible value.
Thank you.
| Let $f(x)=\left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2$. Note that $f(x)$ is convex on the $(0;+\infty)$ because its second derivative is equal to
$$
f''(x)=2+\frac{6}{x^4}>0,~\text{for all}~x\in(0;+\infty).
$$
Hence, by Jensen's inequality ($p+q=1$)
$$
f(p)+f(q)\geqslant 2f\left(\frac{p+q}{2}\right)=2f\left(\frac{1}{2}\right)=\frac{25}{2}.
$$
Hence,
$$
\left(p+\frac{1}{p}\right)^2+\left(q+\frac{1}{q}\right)^2\geq \frac{25}{2}.
$$
Thus, the correct answer is C.
Here is an elementary approach. We will prove that for $p\in(0;1)$ the following inequality holds
$$
\left(p+\frac{1}{p}\right)^2+\left(1-p+\frac{1}{1-p}\right)^2\geq \frac{25}{2},
$$
or
$$
p^2+(1-p)^2+\frac{1}{p^2}+\frac{1}{(1-p)^2}\geq \frac{17}{2},
$$
or
$$
(p^2+(1-p)^2)\left(\frac{p^2(1-p)^2+1}{p^2(1-p)^2}\right)\geq \frac{17}{2}.
$$
Now, note that
$$
p^2+(1-p)^2=(p+(1-p))^2-2p(1-p)=1-2p(1-p).
$$
Denote $s=p(1-p)$. Thus, we need to prove that if $p\in(0;1)$ then
$$
(1-2s)\frac{s^2+1}{s^2}\geq\frac{17}{2},
$$
or
$$
2(1-2s)(s^2+1)\geq 17s^2.
$$
The last inequality is equivalent to
$$
4s^3+15s^2+4s-1\leq 0.
$$
However, $4s^3+15s^2+4s-2=(4s-1)(s^2+4s+2)\leq 0$ because $s=p(1-p)\leq\frac{1}{4}$.
| {
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Proving $\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab} \le\frac{1}{2}(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac})$
Prove $$\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\le\frac{1}{2}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right),$$
where $a,b,c > 0$ and $a,b,c \in \mathbb{R}$
Well, I've been trying for 3 good hours, nothing worked at all.
I already applied HM < AM but I'm still stuck. It gave the following:
$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{a+b+c}{abc}\le2(a+b+c)$$
| Hint: $$\frac{1}{a^2+bc} \le \frac{1}{2a\sqrt{bc}} = \frac{1}{2\sqrt{ab}\sqrt{ac}}.$$
Now apply AM-GM inequality.
$$\frac{1}{a^2+bc} + \frac{1}{b^2+ac} + \frac{1}{c^2+ab} \le \frac{1}{2\sqrt{ab}\sqrt{ac}} + \frac{1}{2\sqrt{ab}\sqrt{bc}} + \frac{1}{2\sqrt{ac}\sqrt{bc}} \le $$
$$\le \frac14 \left(\frac{1}{ab}+\frac{1}{ac} + \frac{1}{ab}+\frac{1}{bc} + \frac{1}{ac}+\frac{1}{bc}\right)=\frac12\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that a sequence is between certain values at certain n I'm given that $a_1=1$, and for every $n \gt1, a_{n+1} = a_n + \frac{1}{a_{n}}$. I need to prove that $20 < a_{200} < 24$. I tried finding a limit at infinity setting both limits to $L$ ( for $a_n$ and $a_{n+1}$ but that gave me nothing useful. All I figured out was that this sequence is increasing..
| Lower bound:
Let's prove that $a_n>\sqrt{2n}$ for $n\geqslant 3$, other words:
$$
(a_n)^2>2n, \qquad (n \geqslant 3).
\tag{1}
$$
$a_2 = 1+1=2$.
First, $(a_3)^2 = (2+\frac{1}{2})^2 = 4+2+\frac{1}{4} > 2\cdot 3$.
Now, using math. induction, we will show, that
$$(a_n)^2>2n \implies (a_{n+1})^2>2(n+1).$$
Yes, if $(a_n)^2>2n$, then
$$
(a_{n+1})^2 = \left(a_n+\frac{1}{a_n}\right)^2 = (a_n)^2 + 2 + \left(\frac{1}{a_n}\right)^2>2n+2=2(n+1).
$$
Statement $(1)$ is proved.
So, $a_{200}>\sqrt{2\cdot 200} = 20$.
Upper bound:
Let's prove that $a_n<\sqrt{\frac{13}{6}n}$ for $n\geqslant 3$, other words:
$$
(a_n)^2 < \frac{13}{6}n, \qquad (n \geqslant 3).
\tag{2}
$$
First, $(a_3)^2 = \left(2+\frac{1}{2}\right)^2 = 4+2+\frac{1}{4}<6+\frac{1}{2} = \frac{13}{6}\cdot 3$.
Using math. induction and $(1)$, we will show, that
$$(a_n)^2<\frac{13}{6}n \implies (a_{n+1})^2<\frac{13}{6}(n+1).$$
Yes, if $(a_n)^2<13n/6$, then
$$
(a_{n+1})^2 = \left(a_n+\frac{1}{a_n}\right)^2 = (a_n)^2 + 2 + \left(\frac{1}{a_n}\right)^2<
\frac{13}{6}n + 2 + \frac{1}{2n} \leqslant \frac{13}{6}n + 2+\frac{1}{6} = \frac{13}{6}(n+1),
$$
since $(a_n)^2>2n\geqslant 6$, when $n\geqslant 3$.
Statement $(2)$ is proved.
So, $a_{200}<\sqrt{\frac{13}{6}\cdot 200} \approx 20.81666 <24$.
| {
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Bessel function to $\sin(kr)$ $J_{\frac{1}{2}}(kr)=\frac{\sqrt{\frac{2}{\pi }} \text{Sin}[\text{kr}]}{\sqrt{\text{kr}}})$ This can be easily obtained by Mathematica,
How to do the details?
| Here's the quick and dirty way to do it for $kr\ge0$. The case of $kr<0$ involves a little trick with Bessel functions of negative integers. I will show this below. The following is often taken as the definition for $J_{\nu}$.
$$J_{\nu}(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!\Gamma(n+\nu+1)}\left(\frac{x}{2}\right)^{2n+\nu}$$
Letting $\nu = \frac{1}{2}$ we have
$$J_{\frac{1}{2}} = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!\Gamma(n+\frac{1}{2}+1)}\left(\frac{x}{2}\right)^{2n+\frac{1}{2}} = \left(\frac{2}{x}\right)^\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^n}{n!\Gamma(n+\frac{1}{2}+1)}\left(\frac{x}{2}\right)^{2n+1}$$
Here I've just multiplied by a clever form of $1$ to get something we're used to seeing in the power series for $\sin$ (this is where your term in the denominator comes from). Now all that is really left is to evaluate $\Gamma(m+\frac{1}{2})$ for $m$ a natural number (plus 0).
From the properties of the gamma function, we have that $\Gamma(m+\frac{1}{2}) = \frac{(2n)!\sqrt{\pi}}{4^nn!}$. Substituting this into our expression we have
$$J_{\frac{1}{2}} = \left(\frac{2}{x}\right)^\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^n}{n!\frac{(2(n+1))!\sqrt{\pi}}{4^{n+1}(n+1)!}}\left(\frac{x}{2}\right)^{2n+1}$$
This quickly reduces to $\sin$ after a little manipulation:
$$J_{\frac{1}{2}} = \left(\frac{2}{x}\right)^\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^n4^{n+1}(n+1)}{(2n+2)!\sqrt{\pi}}\left(\frac{x}{2}\right)^{2n+1} = \left(\frac{2}{\pi x}\right)^\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^n2^{2n+2}(n+1)}{2(n+1)(2n+1)!}\left(\frac{x}{2}\right)^{2n+1}$$
$$J_{\frac{1}{2}} = \left(\frac{2}{\pi x}\right)^\frac{1}{2}\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1} = \sqrt{\frac{2}{\pi x}}\sin x$$
The case of $x<0$ is pretty simple once you have this done. We have for $x>0$ that $J_{\nu}(-x) = e^{\nu\pi i}J_{\nu}(x)$. In our case $\nu = \frac{1}{2}$ and as we know, $e^{i\pi/2} = i$ so $J_{\nu}(-x) = i J_{\nu}(x)$. This is verified with Wolfram Alpha.
| {
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"url": "https://math.stackexchange.com/questions/489582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all positive integers $n$ for which $1 + 5a_n.a_{n + 1}$ is a perfect square. The sequence $a_1, a_2, \ldots $ is defined by the initial conditions $$a_1 = 20; \quad a_2 = 30$$ and the recursion
$$a_{n+2} = 3a_{n+1} - a_n$$ and
for $n \geq 1$. Find all positive integers $n$ for which $1 + 5a_n * a_{n+1}$ is a perfect square.
I could only find the $n$-th term and don't know how to proceed further.pls help
| Note:Just a try. If I'm wrong feel free to comment. Maybe this approach will give someone an idea
I'll post a solution, or rather attempt to solve this problem.
Let $1 + 5 \cdot a_n \cdot a_{n+1} = x^2$
First obviously every number in the sequence $a$ is multiple of $10$. So:
$$a_n \cdot a_{n+1} \equiv 0 \pmod {100}$$
$$5 \cdot a_n \cdot a_{n+1} \equiv 0 \pmod {100}$$
$$1 + 5 \cdot a_n \cdot a_{n+1} \equiv 1 \pmod {100}$$
$$x^2 \equiv 1 \pmod {100} \implies x \equiv \pm 1, \pm49 \pmod {100}$$
This means that $x$ can be written as: $ x = 50k \pm 1$
$$1 + 5 \cdot a_n \cdot a_{n+1} = (50k \pm 1)^2$$
$$1 + 5 \cdot a_n \cdot a_{n+1} = 2500k^2 \pm 100k + 1$$
$$5 \cdot a_n \cdot a_{n+1} = 100k(25k \pm 1)$$
Now we can introduce another sequence $b$, where $b_n = \frac{a_n}{10}$
$$5 \cdot b_n \cdot b_{n+1} = k(25k \pm 1)$$
The RHS need to be divisible by $5$, but that's only possible if $k$ is multiple of $5$. So we write $k=5l$
$$5 \cdot b_n \cdot b_{n+1} = 5l(25k \pm 1)$$
$$b_n \cdot b_{n+1} = l(125l \pm 1)$$
Now using the formula for the sequence we write $b_{n+1} = 3b_n - b_{n-1}$
$$b_n (3b_n - b_{n-1}) = l(125l \pm 1)$$
$$3b_n^2 - b_n \cdot b_{n-1} - l(125l \pm 1) = 0$$
We are now solving a quadratic equation for $b_n$
$$b_n = \frac{b_{n-1} \pm \sqrt{b_{n-1}^2 + 12l(125l \pm 1)}}{6}$$
But because $b_n$ can have one unique value it means that this equation has double root, implying that:
$$b_{n-1}^2 + 12l(125l \pm 1) = 0$$
But $12l(125l \pm 1) > 0$, which means that $b_{n-1}^2 < 0$, which is impossible. This leads to conclusion that an $1 + 5 \cdot a_n \cdot a_{n+1}$ can't be a perfect square
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/489663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Prove that if $a_{n+1} = a_n^2$, the last $n$ digits of $a_{n+1}$ are the same as the last $n$ digits of $a_n$. I have been working on this problem for a while. I know that I have to prove it using induction, but I'm unsure of the next step. The formula for the terms is: $a_{n+1} = 5^{2n}$ with $a_1 = 5$. The sequence is 5, 25, 625, etc. This is what I have thus far:
$a_{n+1} = a_n^2 = 5^{2n}$
$5^{2(n+1)} - 5^{2n} \equiv 0 \pmod {10^{n+1}}$
$5^{2n}5^2 - 5^{2n} \equiv 0 \pmod {10^{n+1}}$
$5^{2n}(24) \equiv 0 \pmod {10^{n+1}}$
Any hints? Thanks
| Assuming that the formula is actually $a_{n+1} = 5^{2^n}$ (which matches up with $a_{n+1} = a_n^2$ and $a_1 = 5$), then we have as follows:
$a_{n+1} - a_n = 5^{2^n} - 5^{2^{n-1}} = (5^{2^{n-1}})^2 - 5^{2^{n-1}} = 5^{2^{n-1}}(5^{2^{n-1}} - 1)$, and we want to show that this number is divisible by $10^n$.
Since $5$ and $2^n$ are coprime, we know from Euler's theorem that
$5^{2^{n-1}} \equiv 1 \pmod {2^n}$, since $\phi(2^n) = 2^{n-1}$.
Thus, $(5^{2^{n-1}} - 1)$ is divisible by $2^n$.
For $n > 1$, we know that $2^{n-1} \geq n$, so $5^{2^{n-1}}$ is divisible by $5^n$.
Therefore, $5^{2^{n-1}}(5^{2^{n-1}} - 1)$ is divisible by both $5^n$ and $2^n$, and is therefore divisible by $10^n$ and we're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/490025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Definite integral involving square root of polynomial How can I solve the integral below?
$$\int_0^1 (x^4+x^3+x^2+x^1+1)^{1/2} \mathrm dx $$
| You can use the horrible result of calculating $\int(x^4+x^3+x^2+x+1)^\frac{1}{2}~dx$ by Wolfram Integrator, but I have the following relatively simpler approach:
$\int_0^1(x^4+x^3+x^2+x+1)^\frac{1}{2}~dx$
$=\int_0^1\left(\dfrac{1-x^5}{1-x}\right)^{\frac{1}{2}}~dx$
$=\int_0^1\dfrac{(1-x^5)^\frac{1}{2}}{(1-x)^\frac{1}{2}}~dx$
$=\int_0^1\sum\limits_{n=0}^\infty\dfrac{(2n)!x^{5n}}{4^n(n!)^2(1-2n)(1-x)^\frac{1}{2}}dx$
$=\sum\limits_{n=0}^\infty\dfrac{(2n)!B\left(5n+1,\dfrac{1}{2}\right)}{4^n(n!)^2(1-2n)}$
$=\sum\limits_{n=0}^\infty\dfrac{(2n)!(5n)!\left(-\dfrac{1}{2}\right)!}{4^n(n!)^2\left(5n+\dfrac{1}{2}\right)!(1-2n)}$
$=\sum\limits_{n=0}^\infty\dfrac{(2n)!(5n)!\sqrt\pi}{4^n(n!)^2\dfrac{(10n)!\sqrt\pi}{4^{5n}(5n)!}(1-2n)}$
$=\sum\limits_{n=0}^\infty\dfrac{256^n(2n)!((5n)!)^2}{(n!)^2(10n)!(1-2n)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/490790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to integrate $\int_{3\sqrt{2}}^6 1/\big(t^3\sqrt{t^2-9}\big)\;dt$ Here's the full problem
$$\int_{3\sqrt{2}}^6\frac{1}{t^3\sqrt{t^2-9}}\;dt.$$
The problem for me is what to do with the $-9$. I know that $1 - \sec^2(t) = \tan^2(t)$, so I'm trying to substitute $9\sec(x)$ for $t$ to get rid of the square root. So that works, but when I recalculate the limits, I end up with numbers that don't make any sense. I'm certain that if I were doing this correctly I would end up with nice numbers for the new limits, so something must be wrong.
$$
\begin{align}
t &= 9\sec(x)\\
a &= 9\sec^{-1}(3\sqrt{2})\\
b &= 9\sec^{-1}(6)
\end{align}
$$
These answers should be $\displaystyle\frac{x}{\pi}$ or something to that effect. Sorry if this makes no sense. Is there a way for me to evaluate the arcsec function without using a calculator so I get values I can understand? Would I use the unit circle? How can these limits be in radians but not have $\pi$?
| $$
\int_{3\sqrt{2}}^6\frac{1}{t^3\sqrt{t^2-9}}\;dt = \frac12 \int_{3\sqrt{2}}^6\frac{2t\,dt}{t^4\sqrt{t^2-9}} = \frac12 \int_{18}^{36} \frac{du}{u^2\sqrt{u-9}}.
$$
Now let $w=\sqrt{u-9}$, so that $w^2 = u-9$ and $2w\,dw=du$, and $u = w^2+9$. Then the integral is
$$
\frac12 \int_3^{3\sqrt{3}} \frac{2w\,dw}{(w^2+9)^2 w} = \int_3^{3\sqrt{3}} \frac{dw}{(w^2+9)^2} = \int_3^{3\sqrt{3}} \frac{Aw+B}{w^2+9} + \frac{Cw+D}{(w^2+9)^2} \,dw
$$
You need to do some algebra to find $A$, $B$, $C$, and $D$. Then you have
$$
\int \frac{w}{w^2+9}\,dw = \frac12\log(w^2+9)+\text{constant}
$$
and
$$
\int \frac{1}{w^2+9}\,dw = \frac13 \arctan\frac w3 + \text{constant}.
$$
The term involving $C$ can be done by the same method as the one that yields a logarithm above, but you don't get a logarithm or anything transcendental.
Finally, we have
$$
\int \frac{dw}{(w^2+9)^2}.
$$
At this point I'd start with $w=3\tan\theta$.
| {
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"url": "https://math.stackexchange.com/questions/491175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Integration to minimize area I have the following problem:
Given a line $f(x) = ax + b$ which passes trough the point $(1, 2)$, find the values for $a$ and $b$ which results in the minimum value for $\int_{-1}^{1}(ax + b)^2\mathrm{d}x$. I already know the result of the area will be $\dfrac{2a^3}{3} + 2b$. But I don't even know how to start cause there are infinite infinite number of different $a, b$ which passes though this point.
How to start ?
| Firstly, the result of the integration is incorrect:
\begin{align*}
\int_{-1}^1 (ax+b)^2\ dx &= \int_{-1}^1 a^2x^2+2abx+b^2 \ dx \\
&=\left[ \frac{a^2}{3}x^3+abx^2+b^2x \right]_{-1}^1 \\
&=\frac{a^2}{3}1^3+ab \cdot 1^2+b^2 1 - \left( \frac{a^2}{3}(-1)^3+ab(-1)^2+b^2 (-1) \right) \\
&=\tfrac{2}{3}(a^2+3b^2).
\end{align*}
Now, using the point $(1,2)$, we can find an equation for $b$ in terms of $a$, namely $b=2-a$ (since $f(x)=ax+b$). Thus the area is $$\tfrac{2}{3}(a^2+3(2-a)^2).$$
Now the minimum area is when $$g(x):=\tfrac{2}{3}(x^2+3(2-x)^2)$$ is minimized. We can find this minimum by using the usual process of differentiating and equating to zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/491681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Digit Root of $2^{m−1}(2^m−1)$ is $1$ for odd $m$. Why? The Wiki page on Perfect numbers says:
[A]dding the digits of any even perfect number (except $6$), then adding the digits of the resulting number, and repeating this process until a single digit (called the digital root) is obtained, always produces the number $1$. For example, the digital root of $8128$ is $1$, because $8 + 1 + 2 + 8 = 19$, $1 + 9 = 10$, and $1 + 0 = 1$. This works ... with all numbers of the form $2^{m−1}(2^m−1)$ for odd integer (not necessarily prime) $m$.
How does that work?
Ok, this means for example perfect numbers (except $6$) never have a factor of $3$, but does this help...?
| Let $e = m-1$ be even. Then $2^e$ can be congruent to one of $1, 4, 7$ modulo $9$. Then $2^{e+1}-1$ is congruent to $2\cdot 1-1 = 1,\, 2\cdot 4 - 1 = 7,\, 2\cdot 7 - 1 = 13 \equiv 4$ modulo $9$, hence $2^e(2^{e+1}-1)$ is congruent to $1\cdot 1$ or $4\cdot 7 = 28 \equiv 1$ modulo $9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/495778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factor $(x+y)^7-(x^7+y^7)$ I encountered the following problem while preparing for upcoming math contests.
Factor $(x+y)^7-(x^7+y^7)$.
I got zero for $(x+y)^7-(x^7+y^7)$, however, the solutions said it's
$$7xy(x+y)(x^2+xy+y^2)(x^2+xy+y^2)$$
Can someone explain how this is possible?
| Expand $(x+y)^7$, you would get:
\begin{align*}
(x+y)^7-(x^7+y^7)&=x^7+7x^6y+21x^5y^2+35x^4y^3+35x^3y^4+21x^2y^5+7xy^6+y^7-x^7-y^7
\\ &=7x^6y+21x^5y^2+35x^4y^3+35x^3y^4+21x^2y^5+7xy^6
\\ &=7xy(x^5+3x^4y+5x^3y^2+5x^2y^3+3xy^4+y^5)
\\ &=7xy(x+y)(x^4+2x^3y+3x^2y^2+2x^3+y^4)
\\ &=7xy(x+y)(x^2+xy+y^2)^2,\text{which is your given answer.}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/498584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 1
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