Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Examine the continuity of function $f(x)=\frac{2x^2-4x}{|x+1|+|x-3|-2}$ Using the definition of absolute value for
$$|x+1|=\begin{cases}
x+1, & x\ge -1\\
-x-1, & x>-1
\end{cases}$$
and
$$|x-3|=\begin{cases}
x-3, & x\ge 3\\
-x+3, & x>3
\end{cases}$$
and checking cases when $$x<0,x=0,x>0$$
I get this definition of a function $f(x)$:
$$f(x)=\begin{cases}
3, & x=3,x=-1\\
x(x-2), & x>0\\
0, & x=0\\
2-x, & x<0
\end{cases}$$
Is this definition of a function $f(x)$ correct, and if not what is wrong?
| $$|x+1|=\begin{cases}
x+1& x\ge -1\\
-x-1, & x\color{red}<-1
\end{cases}$$
and
$$|x-3|=\begin{cases}
x-3, & x\ge 3\\
-x+3, & x\color{red}<3
\end{cases}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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integer solution of $(x-y)(x+y)xy=z^2$ By wolfram alpha, integer solution of $(x-y)(x+y)xy=z^2$ is $x=y=z=0$.
How to show that there are not another solutions with $z \neq 0$.
Thanks.
| If $\gcd(a,b) = d$, and $(x,y,z) = (a,b,c)$ is a nontrivial solution, then $(\frac{a}{d}, \frac{b}{d},\frac{c}{d^2})$ is also a solution, where $\gcd(\frac{a}{d},\frac{b}{d}) = 1$. So if there is no solution $(a,b,c)$ having $\gcd(a,b) = 1$, then there are no nontrivial solutions.
Assume $\gcd(a,b) = 1$. We want to find $a$ and $b$ where the expression $(a+b)(a-b)ab$ would be a perfect square. But $(a+b)$, $(a-b)$, and their product $(a^2-b^2)$ are all coprime to both $a$ and $b$. So it is necessary that $a$ and $b$ are both squares, because any unsquared factors they may have will not be able to have their squares completed by the other terms.
$a = a'^2$ and $b = b'^2$. $(a+b)(a-b) = a^2 - b^2 = a'^4 - b'^4$. If there is a solution, then there must be some $c$ where $c^2 = a'^4 - b'^4$, or $c^2 + b'^4 = a'^4$. Fermat proved this was impossible, so there cannot be any nontrivial solutions.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to simplify a diabolical expression involving radicals A friend and I have been working on this problem for hours - how can the following expression be simplified analytically?
It equals $\frac{1}{2},$ and we have tried the following to no avail:
*
*Substitution of $x = \sqrt{5}$
*Substitution of $x = 2\sqrt{5}$
*Substitution of $x = 5+\sqrt{5}$
*Substitution of $x = \sqrt{5 + \sqrt{5}}$
*Manipulations by substituting the golden ratio
Here goes:
$$\dfrac{\dfrac{\sqrt{5 + 2\sqrt{5}}}{2} + \dfrac{\sqrt{5(5 + 2\sqrt{5})}}{4} - \dfrac{\sqrt{10 + 2\sqrt{5}}}{8}}{\dfrac{\sqrt{5(5 + 2\sqrt{5})}}{4} + 5 \cdot \dfrac{\sqrt{5 + 2\sqrt{5}}}{4}}$$
Thanks in advance for any help.
| Hint $\ $ Let $\ a = \sqrt{5+2\sqrt 5},\ b = \sqrt{10+2\sqrt 5}.\,$ Show $\,\color{#c00}{b = (\sqrt5 -1) a},\,$ so scaling the top and bottom of the fraction by $\,8\,$ yields $\ \dfrac{4a+ 2\sqrt 5 a - \color{#c00}{(\sqrt5 -1) a}}{10a + 2\sqrt 5 a}^{\phantom{1^{1^1}}}\!\!\!\!\!\! =\, \dfrac{1}2$
| {
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"url": "https://math.stackexchange.com/questions/1317880",
"timestamp": "2023-03-29T00:00:00",
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Derivative of Function with Rational Exponents $f(x)= \sqrt[3]{2x^3-5x^2+x}$ I have a question following:
$$f(x)=\sqrt[3]{2x^3-5x^2+x}$$
Here's what I did,
$$f(x)=\sqrt[3]{2x^3-5x^2+x}
\\ = (2x^3-5x^2+x)^{3\over2}
\\\\f'(x) = {3\over 2}(2x^3-5x^2+x)^{3\over2}(6x^2-10x+1)$$
Did I do this correctly?? Because I have different answer on the answer page. Can I reduce or factor any?? Or was there any mistakes?
Thanks
| No, you didn't. We can write $\sqrt[3]{t}=t^{1/3}$, not $t^{3/2}$. So
$$
f(x)=(2x^3-5x^2+x)^{1/3}
$$
and
$$
f'(x)=\frac{1}{3}(2x^3-5x^2+x)^{-2/3}(6x^2-10x+1)=
\frac{1}{3}\frac{6x^2-10x+1}{\sqrt[3]{(2x^3-5x^2+x)^2}}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$x^2-y^2=196$, can we find the value of $x^2+y^2$? $x$ and $y$ are positive integers.
If $x^2-y^2=196$, can we know what the value of $x^2+y^2$ is?
Can anyone explain this to me? Thanks in advance.
| Hint 1: $x^2-y^2 = (x+y)(x-y)$.
Hint 2: Divisors of $196$: $1$, $2$, $4$, $7$, $14$, $28$, $49$, $98$, $196$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the number of integer solutons to $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 60$. Find the number of integer solutons to $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 60$ if $x_1 \geq2, x_2\geq 5, 2\leq x_3\leq 7, x_4\geq 1, x_5\geq 3, x_6\geq 2$.
Here is my approach: If you let $M = \{\infty x_1, \infty x_2, \infty x_3, \infty x_4, \infty x_5, \infty x_6\}$, then a multisubset of $M$ of size $60$ will represent such a solution, for example $\{20x_1, 15x_2, 15x_3, 10x_4, 0x_5, 0x_6\}$. It can generally be proven that there are ${n+r-1\choose r}$ multisubsets of size $r$ of a multiset with $n$ types. Since we know that $x_1\geq 2, x_2\geq 5...$, we need to count the multisubsets of size $60 - (2+5+2+1+3+2) = 45$,but there musnt be greater than 7 $x_3$'s, so we split it into cases. Case 1: $2$ $x_3$'s ,so there are $49\choose45$ such solutions . Case 2: $3$ $x_3$'s so $48\choose 45$ such solutions and so on to get a total of ${49\choose 45} + {48\choose 44} + {47\choose 43} + {46\choose 42} + {45\choose 41} = 897001$ solutions. But the answer in the book is $1032752$, so there must be some mistake.
The question is where is that mistake?
| $$S(x)=\underbrace{(x^2+x^3+\cdots)}_{x_1}\underbrace{(x^5+x^6+\cdots)}_{x_2}\underbrace{(x^2+x^3+\cdots+x^7)}_{x_3}\underbrace{(x+x^2+\cdots)}_{x_4}\\\times\underbrace{(x^3+x^4+\cdots)}_{x_5}\underbrace{{(x^2+x^3+\cdots)}}_{x_6}\\
={x^2\over1-x}{x^5\over1-x}{x^2(1-x^6)\over 1-x}{x\over1-x}{x^3\over1-x}{x^2\over1-x}={x^{15}(1-x^6)\over(1-x)^6}\\
(1-x)^{-6}=\sum_{r=0}^{\infty}{6\cdot7\cdots(6+r-1)\over r!}x^r$$
Every solution of the equation contributes $1$ to the coefficient of $x^{60}=x^{x_1+x_2+\cdots x_6}$ in $S(x)$. For example if we take $x_1=2,x_2=5,x_3=2,x_4=4,x_5=3,x_6=44$, we can pick $x^2,x^5,x^2,x^4,x^3,x^{44}$ from the $1$-st, $2$-nd,...and $6$-th term respectively to construct $x^{60}$.
Coefficient of $x^{60}$ in $S(x)$ is
$${{6\cdot7\cdots(6+45-1)\over 45!}}-{{6\cdot7\cdots(6+39-1)\over 39!}}=1032752$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Easiest way to calculate the determinant of this 4x4 matrix I have this 4x4 matrix:
$$A=
\begin{pmatrix}
2 & 3 & 1 & 0 \\
4 & -2 & 0 & -3\\
8 & -1 & 2 & 1\\
1 & 0 & 3 & 4\\
\end{pmatrix}
$$
I read that it's easy to calculate it by converting the matrix to upper diagonal. I tried that using line operations but I couldn't make it upper diagonal. Is this the best/easiest method? If so can anyone help me with the process?
Then I have to calculate the eigenvalues and eigenvectors. Any suggestion on how to find them? Do I have to calculate the det(A -λI) to find the characteristic equation? Is there an easy way to find it for such a matrix like A?
Any help will be much appreciated.
Thanks in advance!
| As suggested in the comments, Gauss elimination is usually the way to go, and the fastest in this case, too:
$$\det A=
\det\begin{pmatrix}
2 & 3 & 1 & 0 \\
4 & -2 & 0 & -3\\
8 & -1 & 2 & 1\\
1 & 0 & 3 & 4\\
\end{pmatrix}
=
\det\begin{pmatrix}
0 & 3 & -5 & -8 \\
0 & -2 & -12 & -19\\
0 & -1 & -22 & -31\\
1 & 0 & 3 & 4\\
\end{pmatrix}
=
(-1)^{4+1}\cdot 1\cdot\det\begin{pmatrix}
3 & -5 & -8 \\
-2 & -12 & -19\\
-1 & -22 & -31\\
\end{pmatrix}
=
-\det\begin{pmatrix}
0 & -71 & -101 \\
0 & 32 & 43\\
-1 & -22 & -31\\
\end{pmatrix}
=
-1\cdot(-1)^{3+1}\cdot(-1)\cdot\det\begin{pmatrix}
-71 & -101 \\
32 & 43\\
\end{pmatrix}
= (-71)\cdot 43-(-101)\cdot 32=179
$$
(Wolfram Alpha-verified result; I never could remember the 3x3-formula, so I don't use it)
If you absolutely want an upper diagonal matrix, you can do this, but it's only a restriction of the normal algorithm:
$$\det A=
\det\begin{pmatrix}
2 & 3 & 1 & 0 \\
4 & -2 & 0 & -3\\
8 & -1 & 2 & 1\\
1 & 0 & 3 & 4\\
\end{pmatrix}
=
\det\begin{pmatrix}
2 & 3 & 1 & 0 \\
0 & -8 & -2 & -3\\
0 & -13 & -2 & 1\\
0 & -\frac12 & \frac52 & 4\\
\end{pmatrix}
=
\det\begin{pmatrix}
2 & 3 & 1 & 0 \\
0 & -8 & -2 & -3\\
0 & 0 & ? & ?\\
0 & 0 & ? & ?\\
\end{pmatrix}
=
\det\begin{pmatrix}
2 & 3 & 1 & 0 \\
0 & -8 & -2 & -3\\
0 & 0 & ? & ?\\
0 & 0 & 0 & ?\\
\end{pmatrix}
$$
(I'm too lazy to calculate the $?$ now, just continue with the Gaussian Elimination. The determinant will then be the product of the entries on the diagonal.)
For the eigenvalues: yes, you have to calculate the characteristic polynomial.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate $\int \frac { \sin x+\cos x }{ \sin^4 x+\cos^4x}\, dx$? How can one find $$\int \frac { \sin x+\cos x }{ \sin^4 x+\cos^4x}\, dx?$$
| Let
\begin{align}
I &= \int {\frac{{\sin x + \cos x}}{{\sin ^4 x + \cos ^4 x}}dx} \\
&= \int {\frac{{\sin x}}{{\sin ^4 x + \cos ^4 x}}dx} + \int {\frac{{\cos x}}{{\sin ^4 x + \cos ^4 x}}dx} \\
&= \int {\frac{{\sin x}}{{\sin ^4 x + \cos ^4 x}}dx} + \int {\frac{{\sin \left( {{\textstyle{\pi \over 2}} - x} \right)}}{{\sin ^4 x + \sin ^4 \left( {{\textstyle{\pi \over 2}} - x} \right)}}dx} ,\,\,\,\,\,\, (\text{since}\,\,
{\sin \left( {{\textstyle{\pi \over 2}} - x} \right)}=\cos x)
\end{align}
Substituting $u=\frac{\pi}{2}-x$ in the second integral we get
\begin{align}
I&= \int {\frac{{\sin x}}{{\sin ^4 x + \cos ^4 x}}dx} - \int {\frac{{\sin u}}{{\sin ^4 \left( {{\textstyle{\pi \over 2}} + u} \right) + \sin ^4 u}}du}
\\
&= \int {\frac{{\sin x}}{{\sin ^4 x + \cos ^4 x}}dx} - \int {\frac{{\sin u}}{{\cos ^4 u + \sin ^4 u}}du} ,\,\,\,\,\,\,\,\,\,\,\,\, (\text{since}\,\,
{\sin \left( {{\textstyle{\pi \over 2}} + u} \right)}=\cos u)
\\
&=0
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the number of equations having real roots.
If both $a$ and $b$ belong to the set $\{1,2,3,4\}$ ,
then number of equations of the form $ ax^2+bx+1=0$ having real roots is
$a.)\ 10\\
\color{green}{b.)\ 7}\\
c.)\ 6\\
d.)\ 12\\ $
To solve this I had to make a table and check each of the $16$ cases.
$$\begin{array}{|c|c|c|} \hline
a & b & b^2-4a\geq 0 \\ \hline
1 & 1 & \\ \hline
1 & 2 & \checkmark \\ \hline
1 & 3 & \checkmark \\ \hline
1 & 4 & \checkmark \\ \hline
2 & 1 & \\ \hline
2 & 2 & \\ \hline
2 & 3 & \checkmark \\ \hline
2 & 4 & \checkmark \\ \hline
3 & 1 & \\ \hline
3 & 2 & \\ \hline
3 & 3 & \\ \hline
3 & 4 & \checkmark \\ \hline
4 & 1 & \\ \hline
4 & 2 & \\ \hline
4 & 3 & \\ \hline
4 & 4 & \checkmark \\ \hline
\end{array}$$
But I would like to know if there is any short method for it.
I have studied maths up to $12$th grade, thanks.
| For real roots,
$b^2\ge{4a}$
Put $a=1$.
$b^2\ge4$
$b$ can be $\{2,3,4\}$.
Now put $a=2$
$b$ can take two values.
Put $a=3$, $b$ can take one value and finally put $a=4$. Here $b$ can take one value.
Hence there are $7$ ordered pairs $(a,b)$ so there are $7$ such equations
| {
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"timestamp": "2023-03-29T00:00:00",
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Nesbitt's Inequality for 4 Variables I'm reading Pham Kim Hung's 'Secrets in Inequalities - Volume 1', and I have to say from the first few examples, that it is not a very good book. Definitely not beginner friendly.
Anyway, it is proven by the author, that for four variables $a, b, c$, and $d$, each being a non-negative real number, the following inequality holds:
$$\frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b}\ge 2$$
I have no idea how the author proves this. It comes under the very first section, AM-GM. I get the original Nesbitt's inequality in 3 variables that the author proves (which is also cryptic, but I was able to decipher it).
My effort: I understood how the author defines the variables $M, N$ and $S$.
$$S = \frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b}$$
$$M = \frac{b}{b+c} + \frac{c}{c+d} + \frac{d}{d+a} + \frac{a}{a+b}$$
$$N = \frac{c}{b+c} + \frac{d}{c+d} + \frac{a}{d+a} + \frac{b}{a+b}$$
$M + N = 4$, pretty straightforward. The numerators and denominators cross out to give four 1s.
Then the author, without any expansion/explanation, says
$$M + S = \frac{a+b}{b+c} + \frac{b+c}{c+d} + \frac{c+d}{d+a} + \frac{d+a}{a+b}\ge 4$$
Which is also true, since the AM-GM inequality says
$$\frac{M+S}{4}\ge \left(\frac{a+b}{b+c}\cdot\frac{b+c}{c+d}\cdot\frac{c+d}{d+a}\cdot\frac{d+a}{a+b}\right)^{1/4}$$
The RHS above evaluates to $1^{1/4}$ since all the numerators and denominators cancel out.
The next part is the crux of my question.
The author claims,
$$N + S =\frac{a+c}{b+c}+\frac{a+c}{a+d}+\frac{b+d}{c+d} + \frac{b+d}{a+b}\ge\frac{4(a+c)}{a+b+c+d}+\frac{4(b+d)}{a+b+c+d}$$
This is completely bizarre for me! Where did the author manage to get a sum of $(a+b+c+d)$??
As a side note, I'd definitely not recommend this book for any beginner in basic algebraic inequalities (even though the title of the book promotes that it's a treatment of basic inequalities). The author takes certain 'leaps of faith', just assuming that the student reading the book would be able to follow.
| You can indeed use AM-GM, as the book says, to derive the result. Admittedly, it would have been more transparent had the book added one line of derivation as shown below.
For positive $x$ and $y$,
$$\frac{x+y}2\,\frac12\Big(\frac1x+\frac1y\Big)\ge\sqrt{xy}\sqrt{\frac1{xy}}=1,$$
thus
$$\frac1x+\frac1y\ge \frac4{x+y}.$$
Substituing $x=b+c$ and $y=a+d$, you get the desired inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
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$1+\sqrt[3]{e^{2a}}\sqrt[5]{e^{b}}\sqrt[15]{e^{2c}} \leq \sqrt[3]{(1+e^{a})^2}\sqrt[5]{1+e^{b}}\sqrt[15]{(1+e^{c})^2}$ The inequality $1+\sqrt[3]{e^{2a}}\sqrt[5]{e^{b}}\sqrt[15]{e^{2c}} \leq \sqrt[3]{(1+e^{a})^2}\sqrt[5]{1+e^{b}}\sqrt[15]{(1+e^{c})^2}$ is true for all $a,b,c\in\mathbb{R}$?
I've tried to use the Bernoulli inequality $(1+k)^n \geq 1+kn$ but my inequality don't satisfy the conditions.
My other apporach was by the binomial theorem $(1+x)^n = \sum_{k=0}^n {n \choose k}x^k$ but with non integer power I don't see a way here to solve it.
| Hint: use Holder inequality we have
$$(1+e^a)^{10}\cdot (1+e^b)^3\cdot(1+e^c)^{2}\ge (1+\sqrt[15]{e^{10a}\cdot e^{3b}\cdot e^{2c}})^{15}$$
I add $n=3$ Holder inequality proof,ie:
$$(a^3+b^3+c^3)(p^3+q^3+r^3)(u^3+v^3+w^3)\ge (apu+bqv+crw)^3,a,b,c,p,q,r,u,v,w>0$$
since
$$\sum_{cyc}\left(\dfrac{a^3}{a^3+b^3+c^3}+\dfrac{p^3}{p^3+q^3+r^3}+\dfrac{u^3}{u^3+v^3+w^3}\right)=3$$
use AM-GM inequality we have
$$3\ge 3\sum_{cyc}\dfrac{apu}{\sqrt[3]{(a^3+b^3+c^3)(u^3+v^3+w^3)(p^3+q^3+r^3)}}$$
so
$$(a^3+b^3+c^3)(p^3+q^3+r^3)(u^3+v^3+w^3)\ge (apu+bqv+crw)^3$$
then you understand How to prove general Holder inequality?
| {
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Limit involving Zeta and Gamma function Can someone help me evaluate this limit?
$$\lim_{x\to +\infty}\frac {\zeta(1+\frac 1x)}{\Gamma(x)}$$
I never came across this kind of limit so I don't even know where to start.
| Hint: Note that for $x>1$, we have $$\zeta \left(1+\frac{1}{x}\right) = \frac{1}{1^{1+\frac{1}{x}}}+\frac{1}{2^{1+\frac{1}{x}}}+\frac{1}{3^{1+\frac{1}{x}}}+\frac{1}{4^{1+\frac{1}{x}}}+\frac{1}{5^{1+\frac{1}{x}}}+\cdots< \\ \frac{1}{1^{1+\frac{1}{x}}}+\frac{1}{2^{1+\frac{1}{x}}}+\frac{1}{2^{1+\frac{1}{x}}}+\frac{1}{4^{1+\frac{1}{x}}}+\frac{1}{4^{1+\frac{1}{x}}}+\cdots = \\ 1+\frac{2}{2^{1+\frac{1}{x}}}+\frac{4}{4^{1+\frac{1}{x}}}+\frac{8}{8^{1+\frac{1}{x}}}+\cdots = \\ 1+\frac{1}{2^{\frac{1}{x}}}+\frac{1}{4^{\frac{1}{x}}}+\frac{1}{8^{\frac{1}{x}}}+\cdots= \\ (2^{-\frac{1}{x}})^0+(2^{-\frac{1}{x}})^1+(2^{-\frac{1}{x}})^2+(2^{-\frac{1}{x}})^3+\cdots = \\ \frac{1}{1-2^{-\frac{1}{x}}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why does $\cos(x) + \cos(y) - \cos(x + y) = 0$ look like an ellipse? The solution set of $\cos(x) + \cos(y) - \cos(x + y) = 0$ looks like an ellipse. Is it actually an ellipse, and if so, is there a way of writing down its equation (without any trig functions)?
What motivates this is the following example. The solution set of $\cos(x) - \cos(3x + 2y) = 0$ looks like two straight lines, and indeed we can determine the equations of those lines.
$$
\begin{align}
\cos(x) &= \cos(3x + 2y) \\
\implies x &= \pm (3x + 2y) \\
\implies x + y &= 0 \text{ or } 2x + y = 0
\end{align}
$$
Can we do a similar thing for the first equation?
| @Mr Spock: You can go from $\cos(x)+\cos(y)=\cos(x+y)$ to a rational function (you ask about it!) via identities $$\tan (\frac x2)=t$$$$\tan (\frac y2)=s$$ $$\tan (\frac {x+y}{2})=\frac{t+s}{1-ts}$$ from which you have
$$\cos (x)=\frac{1-t^2}{1+t^2}$$ $$\cos (y)=\frac{1-s^2}{1+s^2}$$
Since $$\tan (\frac{x+y}{2})=\frac{t+s}{1-ts}$$ it follows
$$\cos (x+y)=\frac {(1-ts)^2-(t+s)^2}{(1-ts)^2+(t+s)^2}=\frac{(1+t+s-ts)(1-t-s-ts)}{1+t^2s^2+t^2+s^2}$$ Thus
$$2(1-t^2s^2)(1+t^2s^2+t^2+s^2)= (1+t^2)(1+s^2)[(1-ts)^2-(t+s)^2]$$
In contour plots below you can see the silhouetted “ellipses”.
| {
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Find the sum of first $20$ terms of a sequence Define a sequence $$a_n=\sqrt{1+\left(1-\frac{1}{n}\right)^2}+\sqrt{1+\left(1+\frac{1}{n}\right)^2}$$ for $n \geq 1$. Find $$\sum_{i=1}^{20}\frac{1}{a_i}$$ Some insight on the approach is highly appreciated. What is the general way of solving such problems? Thanks.
| First you calculate $\frac{1}{a_i}$ and write it a bit differently:
$\frac{1}{a_n} =
\frac{\sqrt{1+(1-\frac{1}{n})^2}-\sqrt{1+(1+\frac{1}{n})^2}}{(1+(1-\frac{1}{n})^2)-(1+(1+\frac{1}{n})^2)}
=\frac{\sqrt{1+(1-\frac{1}{n})^2}-\sqrt{1+(1+\frac{1}{n})^2}}{-\frac{4}{n}}
=\frac{1}{4} \left ( \sqrt{n^2+(n+1)^2}-\sqrt{n^2+(n-1)^2} \right )$.
Now, if you look at $\sum_{n=1}^{20} \frac{1}{4} \left ( \sqrt{n^2+(n+1)^2}-\sqrt{n^2+(n-1)^2} \right )$, you can see that it is a telescopic sum and thus can be simplified (because most of the terms cancel each other out):
$\sum_{n=1}^{20} \frac{1}{4} \left ( \sqrt{n^2+(n+1)^2}-\sqrt{n^2+(n-1)^2} \right )=\frac{1}{4} (\sqrt{20^2+21^2}-\sqrt{1^2-0^2})= 7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Two different trigonometric identities giving two different solutions Using two different sum-difference trigonometric identities gives two different results in a task where the choice of identity seemed unimportant. The task goes as following:
Given $\cos 2x =-\frac {63}{65} $ , $\cos y=\frac {7} {\sqrt{130}}$, under the condition $0<x,y<\frac {\pi}{2}$, calculate $x+y$.
The first couple of steps are the same: finding $\sin$ and $\cos$ values for both $x$ and $y$.
From $\cos 2x$ we have:
\begin{align*}
\cos 2x &=-\frac {63}{65} \\
\cos2x &= \cos^2x-\sin^2x = \cos^2x - (1-\cos^2x)=2\cos^2x -1 \\
2\cos^2x -1 &= -\frac {63}{65} \\
2\cos^2x &=\frac {-63+65}{65} \\
\cos^2x &=\frac {1}{65} \\
\cos x &=\frac {1} {\sqrt{65}}
\end{align*}
(taking only the positive value of $\cos x$ because $\cos x$ is always positive under the given domain)
\begin{align*}
\sin^2x &=1-\frac {1}{65} \\
\sin^2x &=\frac {64}{65} \\
\sin x &=\frac {8} {\sqrt{65}}
\end{align*}
(again, only positive value)
From $\cos y$ we have:
\begin{align*}
\cos y &=\frac {7} {\sqrt{130}} \\
\cos^2y &=\frac {49} {130} \\
\sin^2y &=1-\frac {49} {130} \\
\sin^2y &=\frac {81} {130} \\
\sin y &=\frac {9} {\sqrt{130}}
\end{align*}
Now that we've gathered necessary information, we proceed to calculate value of some trigonometric function of $x+y$, hoping we will get some basic angle:
sin(x+y):
\begin{align*}
\sin(x+y) &=\sin x \cos y + \sin y \cos x =\frac {8} {\sqrt{65}} \frac {7} {\sqrt{130}} + \frac {9} {\sqrt{130}}\frac {1} {\sqrt{65}} \\
\sin(x+y) &=\frac {65} {\sqrt{65}\sqrt{130}} \\
\sin(x+y) &=\frac {\sqrt{2}}{2}
\end{align*}
Thus,
$x+y =\frac {\pi}{4}+2k{\pi}$ OR $x+y =\frac {3\pi}{4}+2k{\pi}$
Since $x$ and $y$ are in the first quadrant, their sum must lie in first or second quadrant.
Solutions are:
$x+y= \{\frac {\pi}{4}, \frac {3\pi}{4} \}$
cos(x+y):
\begin{align*}
\cos(x+y) &= \cos x \cos y - \sin x \sin y =\frac {1} {\sqrt{65}} \frac {7} {\sqrt{130}} - \frac {8} {\sqrt{65}}\frac {9} {\sqrt{130}} \\
\cos(x+y) &=-\frac {65} {\sqrt{65}\sqrt{130}} \\
\cos(x+y) &=-\frac {\sqrt{2}}{2}
\end{align*}
Thus,
$x+y =\frac {3\pi}{4}+2k{\pi} $ OR $x+y =\frac {5\pi}{4}+2k{\pi}$
Now we can only have one solution: $x+y=\{\frac {3\pi}{4}\}$
Similar happens with $\cos(x-y)$.
My question is: why do these two formulas give two different solutions? General insight would be great, since I found a lot of examples with similar problems. Thank you in advance.
| Simply: Solving for an angle using its sine value is ambiguous. (This is one reason that the Law of Cosines is preferred over the Law of Sines for determining angles from sides.)
Here's an analogous situation:
$$\begin{align}x^2 + 2 x = 35 \quad(1)\\
x^2 - 2 x = 15 \quad(2)
\end{align}$$
Adding $(1)+(2)$, we see that $2 x^2 = 50$, so that $x = \pm 5$. On the other hand, $(1)-(2)$ gives $4x = 20$, so that $x = 5$. Two valid approaches seem to give different solution sets. But note: in the first case, one of the proposed values of $x$ is extraneous. You can check this by substitution: $-5$ is not a solution to $(1)$ or $(2)$.
In your problem, you can likewise determine that one of the values from the sine-based solution is extraneous.
We're given that $\cos 2x$ is negative, while $x$ itself is strictly positive and acute; therefore, $2x$ must be strictly larger than a right angle, which implies that $x > \pi/4$. Since $y$ is also assumed positive, it cannot be that $x+y = \pi/4$. Thus, $3\pi/4$ is the only valid solution.
The lesson here is that some approaches to a problem introduce extraneous solutions. You should always double-check that proposed solutions actually work in the original equations.
| {
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"url": "https://math.stackexchange.com/questions/1337704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Partial fraction of $\frac 1{x^6+1}$ Can someone please help me find the partial fraction of $$1\over{x^6+1}$$
?
I know the general method of how to find the partial fraction of functions but this seems a special case to me..
| We have
$$\begin{align}x^6+1&=(x^2)^3+1\\&=(x^2+1)(x^4-x^2+1)\\&=(x^2+1)(x^4+2x^2+1-3x^2)\\&=(x^2+1)((x^2+1)^2-(\sqrt 3\ x)^2)\\&=(x^2+1)(x^2-\sqrt 3\ x+1)(x^2+\sqrt 3\ x+1)\end{align}$$
Now setting $$\frac{1}{x^6+1}=\frac{a}{x^2+1}+\frac{bx^2+c}{x^4-x^2+1}$$
will give you $a=\frac 13,b=-\frac 13,c=\frac 23.$
Then, set
$$\frac{-\frac 13x^2+\frac 23}{x^4-x^2+1}=\frac{dx+e}{x^2-\sqrt 3\ x+1}+\frac{fx+g}{x^2+\sqrt 3\ x+1}$$
to get $d,e,f,g$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1338080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Roots addition.
$$\sqrt{\frac{a+x^2}{x}-2\sqrt{a}}+\sqrt{\frac{a+x^2}{x}+2\sqrt{a}}=Q $$
One is expected to find $Q$ respecting $a>0$, $x>\sqrt{a}$ .
I'd like to have my solution checked; namely the correct answer is $2\sqrt{x}$ but I simply fail to see where I made a mistake.
And it'd be nice to hear if there is any smoother solution, or any other way to solve this.
I introduced two subsequent substitutions.
*
*$\frac{a+x^2}{x}=A, \;\;\;2\sqrt{a}=B \; \Rightarrow \; \sqrt{A-B}+\sqrt{A+B}=Q$
*$\sqrt{A-B}=k, \;\;\; \sqrt{A+B}=n$
Now I have: $ \;\;\;$
$k+n=Q, \;\;\; k^2+n^2=2A, \;\;\; kn=\sqrt{A^2-B^2}$
$$k^2+n^2+2kn=\left ( k+n \right )^2=Q^2$$
$$Q^2=2A+2\sqrt{A^2-B^2}=\frac{2(a+x^2)}{x}+2\sqrt{\frac{\left (a+x^2 \right )^2}{x^2}-\left ( 2\sqrt{a} \right )^2}=$$
$$=\frac{2(a+x^2)}{x}+2\sqrt{\frac{a^2+2ax^2+x^4-4ax^2}{x^2}}=$$
$$=\frac{2(a+x^2)}{x}+2\sqrt{\frac{a^2-2ax^2+x^4}{x^2}}=$$
$$=\frac{2(a+x^2)}{x}+2\sqrt{\frac{\left ( a-x^2 \right )^2}{x^2}}=\frac{2(a+x^2)}{x}+\frac{2\left (a-x^2 \right)}{x}=\frac{4a}{x}$$
And as a result: $$Q=2\sqrt{\frac{a}{x}}$$
| Set $\sqrt{a}=b$ so the first summand is
$$
\sqrt{\frac{b^2+x^2}{x}-2b}=
\sqrt{\frac{(x-b)^2}{x}}=\frac{x-b}{\sqrt{x}}
$$
because, by assumption, $x>b$ (and so also positive).
Similarly
$$
\sqrt{\frac{b^2+x^2}{x}+2b}=
\sqrt{\frac{(x+b)^2}{x}}=\frac{x+b}{\sqrt{x}}
$$
Thus your expression is
$$
Q=\frac{x-b}{\sqrt{x}}+\frac{x+b}{\sqrt{x}}=
\frac{2x}{\sqrt{x}}=2\sqrt{x}
$$
Your error is in
$$
\sqrt{\frac{(a-x^2)^2}{x^2}}=\frac{a-x^2}{x}
$$
because the right-hand side is positive, while the right-hand side is negative.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving that $\frac{1}{a^2}+\frac{1}{b^2} \geq \frac{8}{(a+b)^2}$ for $a,b>0$ I found something that I'm not quite sure about when trying to prove this inequality.
I've proven that
$$\dfrac{1}{a}+\dfrac{1}{b}\geq \dfrac{4}{a+b}$$
already. My idea now is to replace $a$ with $a^2$ and $b^2$, so we now have
$$\dfrac{1}{a^2}+\dfrac{1}{b^2}\geq \dfrac{4}{a^2+b^2}$$
So to prove the required result, we just need to show that
$$\dfrac{4}{a^2+b^2} \geq \dfrac{8}{(a+b)^2}$$
or equivalently
$$(a+b)^2 \geq 2(a^2+b^2)$$
But this inequality cannot be true since the pair $(a,b)=(1,2)$ doesn't work. If anything, the reverse is always true!
What have I done wrong here?
| The problem is simple enough that you can just be mechanical about it
$$
\frac{1}{a^2}+\frac{1}{b^2}\geq\frac{8}{(a+b)^2}\iff(a^2+b^2)(a+b)^2-8a^2b^2\geq0.
$$
The claim now follows because
$$
(a^2+b^2)(a+b)^2-8a^2b^2=(a-b)^2(a^2+4ab+b^2)\geq0,\quad a,b>0.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate the indefinite integral $\int \frac{\cos \theta}{ \sqrt{2 - 9 \sin^2 \!\theta}} \mathrm{d}\theta$ I want to evaluate $$\int \dfrac{\cos \theta \, \mathrm{d}\theta}{ \sqrt{2 - 9 \sin^2 \theta}}$$
but I can't seem to get the answer, my working is as below:
| Hint
$$\int\dfrac{\cos{\theta}}{\sqrt{2-9\sin^2{\theta}}}d\theta=\int\dfrac{d\sin{\theta}}{\sqrt{2-9\sin^2{\theta}}}$$
and
$$\int\dfrac{1}{\sqrt{2-9t^2}}dt=\dfrac{1}{\sqrt{2}}\cdot\int\dfrac{1}{\sqrt{1-\left(\frac{3}{\sqrt{2}}t\right)^2}}dt=\dfrac{1}{3}\cdot\int\dfrac{1}{\sqrt{1-\left(\frac{3}{\sqrt{2}}t\right)^2}}d\left(\dfrac{3}{\sqrt{2}}t\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find remainder when $777^{777}$ is divided by $16$
Find remainder when $777^{777}$ is divided by $16$.
$777=48\times 16+9$. Then $777\equiv 9 \pmod{16}$.
Also by Fermat's theorem, $777^{16-1}\equiv 1 \pmod{16}$ i.e $777^{15}\equiv 1 \pmod{16}$.
Also $777=51\times 15+4$. Therefore,
$777^{777}=777^{51\times 15+4}={(777^{15})}^{51}\cdot777^4\equiv 1^{15}\cdot 9^4 \pmod{16}$ leading to $ 81\cdot81 \pmod{16} \equiv 1 \pmod{16}$.
But answer given for this question is $9$. Please suggest.
| Alternative to other answers: Note that $9^2=81\equiv 1\pmod{16}$, so $777^{777}\equiv 9^{777}=9\cdot(9^2)^{388}\equiv 9\pmod{16}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Expand $\binom{xy}{n}$ in terms of $\binom{x}{k}$'s and $\binom{y}{k}$'s Motivated by this question, I want to find a complete set of relations for the ring of integer-valued polynomials, where the generators are the polynomials $\binom{x}{n}$ for $n\in \mathbb{N}$. The best way to do this is would be to describe how to decompose $\binom{x+y}{n}$ and $\binom{xy}{n}$ as a sum of products of $\binom{x}{0},\dots, \binom{x}{n}$ and $\binom{y}{0},\dots,\binom{y}{n}$. This can be done in principle by peeling off the binomials starting with the highest degree and working one's way down. Playing around with Sage, one soon guesses that
$\binom{x+y}{n} = \sum_{k=0}^n \binom{x}{k}\binom{y}{n-k}$
and in fact, I think this has a combinatorial proof which straightforwardly generalizes that of the identity $\binom{m}{n} = \binom{m}{n-1} + \binom{m-1}{n-1}$.
But $\binom{xy}{n}$ seems to be not so straightforward. The first few expansions are:
$\binom{xy}{2} = 2\binom{x}{2}\binom{y}{2} + x\binom{y}{2} + y \binom{x}{2}$
$\binom{xy}{3} = 6\binom{x}{3} \binom{y}{3} + $
$\qquad ~ ~ 6 \binom{x}{3}\binom{y}{2} + 6\binom{x}{2}\binom{y}{3} + $
$\qquad ~ ~ x \binom{y}{3} + 4 \binom{x}{2} \binom{y}{2} + y \binom{x}{3} $
$\binom{xy}{4} = 24\binom{x}{4}\binom{y}{4} + $
$\qquad ~ ~ 36\binom{x}{3}\binom{y}{4} + 36 \binom{x}{4}\binom{y}{3} + $
$ \qquad ~ ~ 14 \binom{x}{2}\binom{y}{4} + 45\binom{x}{3}\binom{y}{3} + 14 \binom{x}{4}\binom{y}{2} + $
$\qquad ~ ~ 12 \binom{x}{2}\binom{y}{3} + 12 \binom{x}{3}\binom{y}{2} + $
$\qquad ~ ~ \binom{x}{2}\binom{y}{2}$
and it is not so easy to discern a pattern.
This must be well-known: what is a closed-form expression for the expansion of $\binom{xy}{n}$?
| There's another formula for the expression in equation (3) on the bottom of page 183 of my paper here: http://www.sciencedirect.com/science/article/pii/S0022404905002161
The formula is ${{xy} \choose n} = \underset{l_1 + 2 l_2 + \cdots + nl_n = n}{\sum} {{\sum_{i=1}^n l_i} \choose {l_1, l_2, \ldots, l_n}} {y \choose {\sum_{i=1}^n l_i}} \prod_{i=1}^n {x \choose l_i}$, proved by expanding $(1+T)^{xy} = ((1+T)^x)^y$ using the binomial and multinomial theorems.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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Prove that $\frac{8}{5}\le 2a+b\le 8$ Let $a,b,c,d,e$ be real numbers such that
$$\begin{cases}
a+b+c+d+e=8\\
a^2+b^2+c^2+d^2+e^2=16
\end{cases}$$
Prove that:
$$\dfrac{8}{5}\le 2a+b\le 8$$
| Use Cauchy-Schwarz inequality we have
$$3(c^2+d^2+e^2)\ge (c+d+e)^2\Longrightarrow 48-3a^2-3b^2\ge (8-a-b)^2$$
then let $2a+b=t$, you have
$$8a^2+(8-7t)a+2t^2-8t+8\le 0$$
$$\Longrightarrow \Delta_{a}=(8-7t)^2-32(2t^2-8t+8)\ge 0\Longrightarrow \dfrac{8}{5}\le t\le 8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A general method for integration of rational function. $\int\frac {x^3}{1+x^5}$
ATTEMPT:
I did the following substitution:
Let $x=\frac{1}{t}.$
$dx=\frac{-1}{t^2}dt.$
substituting back:
$I=\int\frac{-1}{1+t^5}dt$ which doesn't seems a simpler integration.
Next i substituted $x=p^\frac{2}{5}.$
$dx=\frac{2}{5}\frac{p^{3/5}}{1+p^2}.$
Now let $p=tan\theta.$
$dp=sec^2\theta$$d\theta.$
substituting back:
$I=\frac{2}{5}\int tan^{3/5}\theta$$d\theta$ which i couldn't integrate further even by trying By parts method.
Is there a general approach for problems of the form (not the algebraic twins):
$\int \frac{x^m}{1+x^n}dx$ where $n-m \ne1 $
| What you already made is $$I=\int\frac {x^3}{1+x^5}dx=-\int\frac {dt}{1+t^5}$$ Now $$1+t^5=\prod_{i=1}^5 (t-\alpha_i)$$ where $\alpha_i=-(-1)^{\frac i5}$ which means that, using partial fraction decomposition, $$\frac {1}{1+t^5}=\sum_{i=1}^5 \frac{\beta_i}{t-\alpha_i}$$ $$\int\frac {dt}{1+t^5}=\sum_{i=1}^5 \beta_i \log({t-\alpha_i})$$ which is simple, except that the $\alpha_i$'s and $\beta_i$'s are complex numbers.
But the real part of $$(a+i b)\log(c+id)=\frac a2\, \log(c^2+d^2)-b\, \tan^{-1} \big(\frac d c\big)$$ and this has to be applied to each term (noticing that four of the roots $\alpha_i$ are conjugated by pairs). So, isolating the case of the real root, we need to combine by pairs the logarithms and the inverse tangents.
This is how, in "Table of Integrals, Series, and Products" by I.S. Gradshteyn and I.M. Ryzhik, they arrive to equation $2.142$.
$$\frac{1}{20} \left(4 \log (t+1)+\left(\sqrt{5}-1\right) \log \left(t^2+\frac{1}{2}
\left(\sqrt{5}-1\right) t+1\right)-\left(1+\sqrt{5}\right) \log
\left(t^2-\frac{1}{2} \left(1+\sqrt{5}\right) t+1\right)-2 \sqrt{10-2
\sqrt{5}} \tan ^{-1}\left(\frac{-4 t+\sqrt{5}+1}{\sqrt{10-2 \sqrt{5}}}\right)+2
\sqrt{ \left(10+2\sqrt{5}\right)} \tan ^{-1}\left(\frac{4 t+\sqrt{5}-1}{\sqrt{
\left(10+2\sqrt{5}\right)}}\right)\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Substitutions that transform Fermat Equations to Elliptic Curves I was reading Chapter 1 of Elliptic Curves - Number Theory and Cryptography by Lawrence C Washington. He was considering Fermat equations $$a^4+b^4=c^4\text{ and }a^3+b^3=c^3.$$ For the 1st equation, let $$x = 2\frac{b^2+c^2}{a^2},\quad y=\frac{4b(b^2+c^2)}{a^3};$$ and for the 2nd equation, let $$x=\frac{12c}{a+b},\quad y=\frac{36(a-b)}{a+b}.$$ Straightforward calculation shows that under these substitutions we transform Fermat equations to elliptic curves: $$y^2=x^3-4x\text{ and }y^2=x^3-432.$$ Before I looked at the substitutions, I tried and failed to find the solutions myself. I am wondering if there is a natural way to deduce those substitutions and the motivation behind.
| @Zilin J.:I have a method that sends all binomial addition $a+b=c$ (so, for example, $a^4+b^4=c^4$) to an elliptical curve, noted $V_A$, of the shape $$X^3+Y^3=AZ^3$$, where $A$ is a cube-free number, and this curve is birational equivalent to a Wierstrass one, the curve $$y^2z=x^3-432A^2z^2$$ (one can obviously work with affine curve making $z=1$), via $$(X,Y)\to ({\frac {12A}{X+Y}, \frac {36A(X-Y)}{X+Y}})$$. We have $$a+b=c\iff [9ab^2+(b-a)^3]^3+[9a^2b-(b-a)^3]^3=abc[3(a^2+ab+b^2]^3 (*)$$ where $A$ can be taken as the cube-free part of $abc$.
This equivalence can be verified straightforward way, simplifying the right side (a more theoretical way is as follow:
$a+b=c\iff a^2b+ab^2=abc\iff P=(\sqrt[3]{a^2b},\sqrt[3]{ab^2},1)\in V_{abc}$;
one can verify that the irrational point $P\in V_{abc}$ becomes rational and gives (*) via the isogeny $P\to3P$).
I will post about this topic with more information soon.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to solve the integral $\int\tan^{3}x \sec^{3/2}x\; dx$? How to solve the following indefinite integral
$$\int \tan^{3}x \sec^{3/2}x \; dx$$
to get the solution in the form of
$$\large\frac{2}{7}\sec^{7/2}x - \frac{2}{3}\sec^{3/2}x +c$$
I tried taking
$$u = \sec^{2}x \implies du = \tan x \; dx$$
$$\large \int \tan^{3} x \sec^{\frac{3}{2}}x\;dx = \int (u-1) u ^{\frac{3}{4}}du = \int u^{\frac{7}{4}}- u^{\frac{3}{4}}du = \frac{4}{11} (\sec x)^{\frac{11}{2}} \frac{4}{7} (\sec x)^{\frac{7}{2}} +c$$
Where did I go wrong?
| While the OP is interested in understanding the source of an error from a specific substitution, I thought that it would be of some interest to show another way to evaluate this indefinite integral.
To that end, we can simply write
$$\tan^3 x\sec^{3/2}x=(1-\cos^2x)\sin x\cos^{-9/2}x$$
so that we have
$$\begin{align}
\int\tan^3x\sec^{3/2}x\,dx&=\int\left(\cos^{-9/2}x-\cos^{-5/2}x\right)\sin x\,dx\\\\
&=\frac27 \cos^{-7/2}x-\frac23 \cos^{-3/2}x+C
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of elements in discrete $n$-dimensional simplex such that $x_1 \leq \ldots \leq x_n$ For positive integers $n,d$, how many elements are there in the set $S = \{(x_1,\ldots,x_n) \in \mathbb{Z}^n\ |\ 0 \leq x_1 \leq \ldots \leq x_n \wedge \sum_i x_i = d \}$?
I'm hoping that the order constraints on the $x_1,\ldots,x_n$ can be accounted for somehow by "adjusting" the figurate number, which gives the number of elements for an unconstrained discrete simplex. But I'm a bit out of my depth combinatorically.
| Introduce change of variable
$$\begin{cases}
x_1 &= y_1\\
x_2 &= y_1 + y_2\\
x_3 &= y_1 + y_2 + y_3\\
&\;\vdots\\
x_n &= y_1 + y_2 + y_3 + \cdots + y_n
\end{cases}$$
We have
$$\begin{array}{c}
0 \le x_1 \le x_2 \le x_n\\
\text{ and }\\
x_1 + x_2 + \cdots + x_n = d
\end{array}
\quad\iff\quad
\begin{array}{c}
y_1, y_2, \ldots, y_n \ge 0\\
\text{ and }\\
ny_1 + (n-1)y_2 + \cdots + y_n = d
\end{array}
$$
This means the number of solutions of $(x_k)$ satisfying LHS is the same as the number of solutions of $(y_k)$ satisfying RHS. Let us denote this number as $p(d,n)$.
For each solution of $(y_k)$ for the RHS, there is a corresponding partition of integer $d$ into parts whose part not exceeding $n$. i.e.
$$d = \overbrace{n + n + \cdots + n}^{y_1}
+ \overbrace{(n-1) + (n-1) + \cdots + (n-1)}^{y_2}
+ \cdots
+ \overbrace{1 + 1 + \cdots + 1 }^{y_n}$$
This correspondence is one to one. This means $p(d,n)$ equals to the number of ways of expressing integer $d$ as a sum of integers from the set $\{\; 1, 2, \ldots, n \;\}$. For fixed $n$, the generating function of latter is given by:
$$
\text{OCF}_n(t) \stackrel{def}{=} \sum_{d=0}^\infty p(d,n) t^d
= \prod_{k=1}^n (1 + t^k + t^{2k} + t^{3k} + \cdots )
= \prod_{k=1}^n \frac{1}{1 - t^k}
$$
I'm not aware of any formula of $p(d,n)$ for general $d$ and $n$.
However, for small $n$, you can use these OGFs to derive formula for $p(d,n)$.
For example,
*
*$n = 1$,
$$\text{OCF}_1(t) = \frac{1}{1-t}\quad\implies\quad
p(d,1) = 1.$$
*
*$n = 2$,
$$\begin{align}\text{OCF}_2(t)
&= \frac{1}{(1-t)(1-t^2)}
= \frac{1}{2(1-t)^2} + \frac{1}{4(1-t)} + \frac{1}{4(1+t)}\\
&= \frac12 \sum_{d=0}^\infty (d+1)t^d
+ \frac14 \sum_{d=0}^\infty t^d
+ \frac14 \sum_{d=0}^\infty (-t)^d\\
\implies
p(d,2) &= \frac{d+1}{2} + \frac{1+(-1)^d}{4} = \left\lfloor \frac{d}{2} \right\rfloor + 1
\end{align}
$$
*$n = 3$,
$$\begin{align}
\text{OCF}_3(t)
&= \frac{1}{(1-t)(1-t^2)(1-t^3)}
= \frac{1/6}{(1-t)^3} + \frac{1/4}{(1-t)^2} + \frac{1/4}{1-t^2} + \frac{1/3}{1-t^3}
\\
&= \frac16\sum_{d=0}^\infty \binom{d+2}{2} t^d + \frac14\sum_{d=0}^\infty (d+1) t^d + \frac14 \sum_{d=0}^\infty t^{2d} + \frac13 \sum_{d=0}^\infty t^{3d}\\
&= \sum_{d=0}^\infty\left( \frac{(d+3)^2}{12} t^d -\frac13 t^d + \frac14 d^{2d} + \frac13 d^{3d}\right)\\
&= \sum_{d=0}^\infty \left(\frac{(d+3)^2}{12} + \epsilon(d)\right) t^d\\
\implies
p(d,3) &= \frac{(d+3)^2}{12} + \epsilon(d)
\end{align}
$$
where $\epsilon(d)$ takes only the values $-\frac13, -\frac{1}{12}, 0, \frac14$.
Since $p(d,3)$ is an integer and $|\epsilon(d)| < \frac12$, we can further simplify and get
$$p(d,3) = \left\{ \frac{(d+3)^2}{12} \right\}
\quad\text{ where }
\{ x \} \text{ is the nearest integer to } x.
$$
There are formulas of $p(d,n)$ for other $n$. A good reference should be the book
*
*Integer Partitions, by George E. Andrews, Kimmo Eriksson.
Some of the formula here is copied from this book.
| {
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} |
Integral of $\frac{x^2+1}{(1-x^2)\sqrt{1+x^4}}$ So we have to evaluate $\int\frac{x^2+1}{(1-x^2)\sqrt{1+x^4}}dx$.
My work-
We can write the integrand as $\frac{(x+1)^2-2x}{(1-x)(1+x)\sqrt{1+x^4}}dx$.
So we wish to deduce $\int\frac{(x+1)}{(1-x)\sqrt{1+x^4}}dx-\int\frac{2x}{(1-x^2)\sqrt{1+x^4}}dx$
So lets write it as $I_1+I_2$.
I tried and I can't evaluate $I_1$.
For $I_2$, I took $x^2=t;\ 2xdx=dt$.
The integral becomes $\int\frac{dt}{(1-t)\sqrt{1+t^2}}dt$.
Okay, now we do $t=\tan\theta; \ dt=\sec^2\theta d\theta.$
$\int\frac{\sec\theta}{1-\tan\theta} d\theta$
Now what to do? Please guide me.?
| My method :
Am guiding you in different way which is too easy to understand
Make it a try ...
First step : remove $x^2$ from numerator and $x$ from first factor and $x^2$ from under square root function
Then it becomes $\int\frac{1+\frac1{x^2}}{(\frac1{x} -x)\sqrt{x^2+\frac{1}{x^2}}}$
Next
Why don't you take $\frac1{x}-x=t; -(\frac{1}{x^2}+1)dx=dt)$
Then it changes
$\int \frac{-dt}{t \sqrt{t^2+2}}$
Again take $t^2 +2=u^2$
$2t dt =2udu$
$t dt= u du$
Now use these then problem becomes $-\int \frac{du}{u^2-2}$
Using standard results we get
$\frac{-1}{2\sqrt{2}}.In|{\frac{u-\sqrt2}{u+\sqrt2}}|$
Change $u$ to $t$ and change $t$ to $x$ that's ur answer
May i think you like it.
| {
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Show that $p \in \left[\frac{4^m}{2\sqrt{m}},\frac{4^m}{\sqrt{2m+1}}\right]$
If the number of ways in which $m$ identical apples can be put in $2m$ boxes, so that no box contains more than one apple, is $p$, prove that $$p \in \left[\frac{4^m}{2\sqrt{m}},\frac{4^m}{\sqrt{2m+1}}\right]$$
I did this as follows :
Let the number of apples in $i^{th}$ box be $x_{i}$, then,
$$\sum_{i=1}^{2m}x_{i}=m$$
where $x_{i} \in \{0,1\}$ and the number of ways would be the number of solutions to this equation, which is equal to the coefficient of $x^m$ in $(1+x)^{2m}$ i.e, $p=\dbinom{2m}{m}$
However, I can't prove that $p \in \left[\frac{4^m}{2\sqrt{m}},\frac{4^m}{\sqrt{2m+1}}\right]$. Can we prove this inequality without induction? Also, is my method correct ?
Any help will be appreciated.
Thanks
| Here is another rather elementary answer.
Let $a_m=\frac{1}{4m}\binom{2m}{m}$. We show the following is valid
\begin{align*}
\frac{1}{2\sqrt{m}}\leq a_m \leq \frac{1}{\sqrt{2m+1}}\qquad\qquad m\geq 1\tag{1}
\end{align*}
We start similarly to @CuriousGuest
\begin{align*}
a_m&=\frac{1}{4^m}\binom{2m}{m}=\frac{(2m)!}{4^mm!m!}=\frac{(2m)!!(2m-1)!!}{4^mm!m!}\\
&=\frac{(2m-1)!!}{2^mm!}=\frac{(2m-1)!!}{(2m)!!}\\
&=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots\frac{2m-1}{2m}
\end{align*}
Now we consider $a_m^2$ and obtain
\begin{align*}
a_m^2&=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{5}{6}\cdots\frac{2n-1}{2n}\cdot\frac{2n-1}{2n}\\
&>\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdots\frac{2m-2}{2m-1}\cdot\frac{2m-1}{2m}\\
&=\frac{1}{4m}
\end{align*}
Thus
\begin{align*}
a_m>\frac{1}{2\sqrt{m}}
\end{align*}
which proves the left inequality of (1)
$$$$
In order to show the right inequality we observe
\begin{align*}
a_m&=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots \frac{2m-1}{2m}\\
&<\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdots \frac{2m}{2m+1}\\
&=\frac{2}{1}\cdot\frac{4}{3}\cdot\frac{6}{5}\cdots \frac{2m}{2m-1}\,\cdot\,\frac{1}{2m+1}\\
&=\frac{1}{a_m}\cdot\frac{1}{2m+1}
\end{align*}
We conclude
\begin{align*}
a_m^2&<\frac{1}{2m+1}\\
a_m&<\frac{1}{\sqrt{2m+1}}
\end{align*}
which proves the right inequality of (1).
| {
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Straight line is tangent to the curve. The straight line $y=mx+1$ is tangent to the curve $x^2+y^2-2x+4y=0$. Find the possible values of $m$.
My attempt
Substitute the $y=mx+1$ into the equation $x^2+y^2-2x+4y=0$. $$x^2+(mx+1)^2-2x+4(mx+1)=0$$ $$x^2+m^2x^2+2mx+1-2x+4mx+4=0$$ $$(1+m^2)x^2+6mx-2x+5=0$$ $$(1+m^2)x^2+(6m-2)x+5=0$$
I think what I did is wrong as I don't know how to continue from my steps. Can anyone explains it? Thanks
| The general equation of a tangent to a circle $X^2 + Y^2 = a^2$ is given by $Y = mX \pm a\sqrt{1 + m^2}$.
For the purpose of your question, $X = x - 1$, $Y = y + 2$, $a = \sqrt{5}$. Also,
$$
y = mx + 1 \\
\implies y + 2 = m(x - 1) + (m + 3) \\
\implies Y = mX + (m + 3).
$$
Hence,
$$
(m + 3) = \sqrt{5} \sqrt{1 + m^2}.
$$
The rest is simple.
| {
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If $x+y+z=3$, then $\sum_{\text{cyc}}\frac{x^2}{2y^2-y+3}\ge\frac{3}{4}$
Let $x,y,z>0$, be such that $x+y+z=3$. Show that
$$\dfrac{x^2}{2y^2-y+3}+\dfrac{y^2}{2z^2-z+3}+\dfrac{z^2}{2x^2-x+3}\ge\dfrac{3}{4}.$$
I've tried many things but all have failed.
$$\left(\sum_{\text{cyc}}\dfrac{x^2}{2y^2-y+3}\right)\left(\sum_{\text{cyc}}(2y^2-y+3)\right)\ge (x+y+z)^2=9.$$
But
$$\sum_{\text{cyc}}(2y^2-y+3)=2\sum_{\text{cyc}}x^2+6\ge 12.$$
| By C-S $\sum\limits_{cyc}\frac{x^2}{2y^2-y+3}\geq\frac{(x^2+y^2+z^2)^2}{\sum\limits_{cyc}(2x^2y^2-x^2y+3x^2)}$. Thus, it remains to prove that $\frac{(x^2+y^2+z^2)^2}{\sum\limits_{cyc}(2x^2y^2-x^2y+3x^2)}\geq\frac{3}{4}$, which is $\sum\limits_{cyc}(3x^4-x^3y-2x^3z+x^2y^2-x^2yz)\geq0$, which is obvious.
| {
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Ratio of $\frac{\sin x_1 }{\sin x_2}$ where $f(x_1)=f(x_2)$ for a trigonometric sum $f(x)$ If $f(x) = \cos(x+a_1)+\frac12\cos(x+a_2)+\frac14\cos(x+a_3)+\cdots+\frac1{2^{n-1}}\cos(x+a_n)$, where $a_1, a_2, ... a_n$ are some constants and $f(x_1)-f(x_2)=0$, where $x_2 \neq m\pi$, find
$$\frac{\sin x_1}{\sin x_2}+\frac{\left|\sin x_1\right|}{\sin x_2}+\frac{\sin x_1}{\left|\sin x_2\right|}+\left|\frac{\sin x_1}{\sin x_2}\right|$$
A friend gave me this one and I tried using
$$0=f(x_1)-f(x_2) = 2\sin\left(\frac{x_2-x_1}2 \right)\left[\sin\left(\frac{x_1+x_2}2+a_1 \right)+\frac12\sin\left(\frac{x_1+x_2}2+a_2 \right) +\dots\right]$$
which gives one possibility $x_2=x_1+2k\pi$ so the desired value can be $0$ or $4$. However I also see a possibility of $a_i=-\frac{x_1+x_2}2+2k_i\pi$ which leads to a lot of possibilities for the value. Am I doing this incorrectly or is there something missing in the question (my friend swears no)?
| Use the addition formula for cosine
$\cos(A+B) = \cos A \cos B - \sin A \sin B$
to write
$f(x) = \cos{x} \displaystyle\sum\limits_{i=1}^{n}{\dfrac{2}{2^{n-k}}\cos a_k} - \sin{x} \displaystyle\sum\limits_{i=1}^{n}{\dfrac{2}{2^{n-k}}\sin a_k}$
Regard the summations as fixed since you are given $a_1,a_2,\dots,a_n$
Let $\sigma_c = \displaystyle\sum\limits_{i=1}^{n}{\dfrac{2}{2^{n-k}}\cos a_k} \tag{1}$ and $\sigma_s = \displaystyle\sum\limits_{i=1}^{n}{\dfrac{2}{2^{n-k}}\sin a_k} \tag{2}$
Then you have:
$f(x) = \sigma_c \cos{x} - \sigma_s \sin{x} \tag{3}$
Since $f(x_1)-f(x_2)=0$, you must have
$\sigma_c \cos{x_1} - \sigma_s \sin{x_1} = \sigma_c \cos{x_2} - \sigma_s \sin{x_2} \tag{4}$
and so finding a $\theta$ such that $\tan\theta = \dfrac{\sigma_s}{\sigma_c} \tag{5}$, you can divide (4) through by $\sqrt{{\sigma_c}^2 + {\sigma_c}^2}$ and set $\sigma_s = \sin\theta, \sigma_c = \cos\theta$ to get
$\cos\theta \cos x_1 - \sin\theta \sin x_1 = \cos\theta \cos x_2 - \sin\theta \sin x_2$
which is really
$\cos(\theta + x_1) = \cos(\theta + x_2) \tag{6}$
and which has two sets of solutions.
The first is (as you had it)
$\boxed{x_2 = x_1 + 2\pi n, n\in\mathbb{Z}} \tag{a}$
Now from (5) without loss of generality, set
$\theta = \arctan\Big(\dfrac{\sigma_s}{\sigma_c}\Big) \tag{6}$
The other set of solutions is given by setting $\theta + x_2 = -(\theta + x_1)$ [from the identity $\cos(-\alpha) = \cos \alpha$] and then allowing for multiples of $2\pi$:
$x_2 = -x_1 - 2\theta + 2\pi n, n\in\mathbb{Z}$
so finally by (6):
$\boxed{x_2 = -x_1 - 2\arctan\Big(\dfrac{\sigma_s}{\sigma_c}\Big) + 2\pi n, n\in\mathbb{Z}} \tag{b}$
with ${\sigma_c},{\sigma_s}$ given by (1) and (2)
| {
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Proving $\sin 20^\circ \, \sin 40^\circ \, \sin 60^\circ \, \sin 80^\circ =\frac{3}{16}$ The problem is to prove that
$$\sin 20^\circ \, \sin 40^\circ \, \sin 60^\circ \, \sin 80^\circ =\frac{3}{16}$$
All my attempts were to get them in $\sin (2A)$ form after eliminating $\sin 60^\circ$ in both sides. Unfortunately, all these attempts were futile.
Any hints are welcomed.
| HINT:
Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,
$$\sin(60^\circ-x)\cdot\sin x\sin(60^\circ+x)=\sin x[\sin^2120^\circ-\sin^2x]=\dfrac{3\sin x-4\sin^3x}4$$
$$\implies4\sin(60^\circ-x)\cdot\sin x\sin(60^\circ+x)=\sin3x$$
Set $x=20^\circ$
| {
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How find $\sum_{0 \leq k: \leq 100 \ gcd \left( k, 100 \right) =1} f \left( \frac{k}{100} \right)$ if $f \left( x \right) = \frac{9^x}{3+9^x}$ ? How find of $\sum_{0 \leq k: \leq 100 \ gcd \left( k, 100 \right) =1} f \left( \frac{k}{100} \right)$ if $f \left( x \right) = \frac{9^x}{3+9^x}$ ?
| We can prove $(k,100)=(100-k,100)$
Now prove $f(x)+f(1-x)=1$
Set $x=\dfrac k{100}$
Now $\phi(100)=\phi(25)\cdot\phi(4)=40$
If $\displaystyle S=\sum_{0\le k\le 100;\left( k, 100 \right) =1} f \left( \frac k{100} \right),$
$\displaystyle S+S=\sum_{0\le k\le 100;\left( k, 100 \right) =1}\left[f \left( \frac k{100} \right)+f \left( \frac{100-k}{100} \right)\right]=\sum_{0\le k\le 100;\left( k, 100 \right) =1}1=40$
| {
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Proof - for all integers $y$, there is integer $x$ so that $x^3 + x = y$ For all integers $y$, there is an integer $x$ so that $$x^3 + x = y.$$
This is what I have done so far:
Proof: Suppose $y$ is some integer. We want to prove that $$x^3 + x = y$$ for some integer $x$.
I am thinking this is false. For example, if $y = 1$, there is no integer $x$ so that $x^3 + x = 1$. But I am not sure how to start about expressing this as a formal proof. Any help please?
| This is indeed not true. Two proofs:
Using basic calculus:
Note that $x^3+x$ is strictly increasing. (Its derivative is $3x^2+1$ and squares are positive)
Now note that $1^3+1=2$ and $0^3+0=0$. Therefore the $x$ such that $x^3+x=1$ must statistify $0<x<1$, so its not an integer.
Using number theory only:
Note that $x \mid x^3+x$, so $x^3+x=0$ or $|x^3+x|\geq |x|$. Thus it can only be 1, 0 or -1 if it is equal to one. But $x=-1$ gives $x^3+x=-2$, $x=0$ gives $x^3+x=0$ and $x=1$ gives $x^3+x=2$. So it is not possible for an integer.
| {
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Converting a matrix from one base base to another. I have this basis $B = ((1,0,1),(0,1,-1),(1,-1,0))$
That is represented by:
$$[T]_B = \begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 2 \\ 1 & 1 & 2 \end{pmatrix}$$
I want to convert this representing matrix into standard basis, but I somehow misunderstood it since I have mistakes.
I know I can always do: $M_E^B[T]_BM_B^E$ to get $[T]_E$ but I'm looking for more of a manual way of doing this in order to fully understand it.
Can someone please elaborate on this process?
Thanks.
| First you find $ [i]_{E^B} $ and $ [i]_{B^E} $, where $i$ is the identity map. Now I assume you are calling $ E $ as the standard basis vectors. To find $ [i]_{B^E} $, the $j^{th} $ column is given by the linear combination of $E_j$ in terms of the vectors from $B$. This is easy because this is already how $B$ is written. So $ [i]_{B^E} = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 1 & -1 & 0 \end{bmatrix} $. Now to find $ [i]_{E^B} $, you must solve for the standard basis vectors in terms of the vectors from $B$. Note that
$$ E_1 = (1,0,0) = \frac{1}{2}(1,0,1) + \frac{1}{2}(0,1,-1) + \frac{1}{2}(1,-1,0) $$
$$ E_2 = (0,1,0) = \frac{1}{2}(1,0,1) + \frac{1}{2}(0,1,-1) - \frac{1}{2}(1,-1,0) $$
$$ E_3 = (0,0,1) = \frac{1}{2}(1,0,1) - \frac{1}{2}(0,1,-1) - \frac{1}{2}(1,-1,0) $$
This means
$$ [i]_{E^B} = \frac{1}{2} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & -1 \\ 1& -1 & -1\end{bmatrix}$$
Now multiplying these guys all together will give you just what you want:
$$ [T]_{E} = [i]_{B^E}[T]_{B}[i]_{E^B} = \frac{1}{2} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & -1 \\ 1& -1 & -1\end{bmatrix} \begin{bmatrix} 1 & 0& 1 \\ 1 & 1 & 2 \\ 1 & 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 1 & -1 & 0 \end{bmatrix} = \frac{1}{2}\begin{bmatrix} 8 & -3 & 1 \\ 2 & -1 & 1 \\ -4 & 1 & 1\end{bmatrix}$$
| {
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Minimizing a summation? I have absolutely no idea how to approach this problem. I've been looking through notes, and I think I missed this when my professor discussed this in class.
$$
\text{Consider the data}\\
i\: x_i\: y_i\\
1\:2\:1\\
2\:3\:2\\
3\:3\:3\\
4\:4\:6\\
5\:5\:5\\
\text{As discussed in class, compute the line } y=p(x)mx+b \text{ that minimizes}\\
F(m,b) = \sum_{i=0}^{5}(y_i-(mx_i+b))^2 \\
\text{You find that}\\
m=\text{____}\\
b=\text{____}\\
F(m,b)=\text{____}
$$
We've been working on maximizing and minimizing functions by both partial derivatives testing for critical points and using lagrange multipliers to find max and min values, but I have no idea what to do with this to get my initial equations to work with. Can someone give me some kind of hint or instructions on what to do?
| Let our best fit line be described as $mx + b$
We would hope that we could solve for $m$ and $b$ without any contradictions using the system of equations:
$\begin{cases} 2m+b = 1\\
3m + b = 2\\
3m + b = 3\\
4m + b = 6\\
5m + b = 5\end{cases}$
Which can be expressed as the following matrix equation:
$\begin{bmatrix} 2 & 1\\3&1\\3&1\\4&1\\5&1\end{bmatrix}\begin{bmatrix}m\\b\end{bmatrix} = \begin{bmatrix}1\\2\\3\\6\\5\end{bmatrix}$
You will unfortunately find that this system is inconsistent and no exact solution can be found ($3m+b=2$ and $3m+b=3$ simultaneously is impossible). However, a least squares fit can be found. To do so, multiply on the left by the transpose of the first matrix.
$\begin{bmatrix}2&3&3&4&5\\1&1&1&1&1\end{bmatrix}\begin{bmatrix}2&1\\3&1\\3&1\\4&1\\5&1\end{bmatrix}\begin{bmatrix}m^*\\b^*\end{bmatrix} = \begin{bmatrix}2&3&3&4&5\\1&1&1&1&1\end{bmatrix}\begin{bmatrix}1\\2\\3\\6\\5\end{bmatrix}$
This system, in the form $A^TAx=A^Tb$, will be consistent and the solution (or solution set in the case of infinitely many solutions) will be the best fit line. Solve the system either by row-reduction, or if $A^TA$ is invertible as $x=(A^TA)^{-1}A^Tb$, or using your other favorite technique.
(Note: it is as mentioned possible for infinitely many possibilities to occur, in particular when $(A^TA)$ is non-invertible.)
In this case it will be invertible, so we take the easy route and complete the problem as:
$\begin{bmatrix}m^*\\b^*\end{bmatrix}=\left(\begin{bmatrix}2&3&3&4&5\\1&1&1&1&1\end{bmatrix}\begin{bmatrix}2&1\\3&1\\3&1\\4&1\\5&1\end{bmatrix}\right)^{-1}\begin{bmatrix}2&3&3&4&5\\1&1&1&1&1\end{bmatrix}\begin{bmatrix}1\\2\\3\\6\\5\end{bmatrix}$
$=\begin{bmatrix}\frac{5}{26}&-\frac{17}{26}\\ -\frac{17}{26}&\frac{63}{26}\end{bmatrix}\begin{bmatrix}66\\17\end{bmatrix}=\begin{bmatrix}\frac{41}{26}\\-\frac{51}{26}\end{bmatrix}$
Giving us that our best fitting line will be of the form $\frac{41}{26}x-\frac{51}{26}$
Alternatively, approaching via another method, expand $F(m,b)$ and try to calculate its minimum value using derivatives.
$F(m,b)=(1-2m-b)^2 + (2-3m-b)^2 + (3-3m-b)^2 + (6-4m-b)^2 + (5-5m-b)^2$
$=5b^2+34bm-34b+63m^2-132m+75$
We try to calculate critical points where $F_{m}=F_{b}=0$, and test $D=F_{m,m}F_{b,b} - F_{m,b}^2$ to see if they are maximums, minimums, or saddle points.
$F_m = 34b + 126m - 132$
$F_b = 10b + 34m - 34$
Setting each equal to zero, we see that each forms a line in $\mathbb{R}^2$ (and they are not parallel) and so there will be only one critical point. Find the intersection of the line and verify that it is indeed a minimum.
These intersect at the point $m=\frac{41}{26}, b=-\frac{51}{26}$. After some tedious arithmetic, we confirm that $D>0$ and $F_{mm}>0$, implying that this point is indeed a minimum for the multivariable function $F(m,b)$.
Giving us that our best fitting line will be of the form $\frac{41}{26}x-\frac{51}{26}$
(double checked, both methods did in fact yield the same answer)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\int \frac{(x+1)}{(x^2+2)^2}dx$ This is the question I want to ask
Integrate $$\int \frac{(x+1)}{(x^2+2)^2}\,dx.$$
I tried it using algebric manipulation
Integrate $x/(x^2+2)^2+1/(x^2+2)^2$.
Then the latter part would not be solved.
| We split the integral into two components as
$$\int\frac{x+1}{(x^2+2)^2}dx=\int\frac{x}{(x^2+2)^2}dx+\int\frac{1}{(x^2+2)^2}dx\tag 1$$
The first integral on the right-hand side of $(1)$ is trivial to evaluate using the substitution $x^2=u\implies x\,dx\to \frac12 du$. Then,
$$\begin{align}
\int\frac{x}{(x^2+2)^2}dx&=\frac12\int \frac{1}{(u+2)^2}\,du\\\\
&=-\frac12 \frac{1}{u+2}\\\\
&=-\frac12 \frac{1}{x^2+2} \tag 2
\end{align}$$
For the second integral on the right-hand side of $(1)$ , we use the classical trigonometric substitution $x=\sqrt{2}\tan \theta\implies dx=\sqrt{2}\sec^2 \theta d\theta$. Then,
$$\begin{align}
\int\frac{1}{(x^2+2)^2}dx&=\frac{\sqrt{2}}{4}\int \frac{\sec^2 \theta}{\sec^4 \theta}\,d\theta\\\\
&=\frac{\sqrt{2}}{4}\int \cos^2 \theta \,d\theta\\\\
&=\frac{\sqrt{2}}{4}\int \left(\frac{1+\cos 2x}{2}\right) \,d\theta\\\\
&=\frac{\sqrt{2}}{8} \left(\theta +\frac12 \sin 2\theta\right)\\\\
&=\frac{\sqrt{2}}{8} \left(\theta + \sin \theta \cos \theta \right)\\\\
&=\frac{\sqrt{2}}{8} \left(\arctan(x/\sqrt{2}) + \frac{\sqrt{2}x}{x^2+2} \right) \tag 3\\\\
\end{align}$$
Putting $(2)$ and $(3)$ together, we have
$$\bbox[5px,border:2px solid #C0A000]{\int\frac{x+1}{(x^2+2)^2}dx=\frac{\sqrt{2}}{8} \arctan(x/\sqrt{2})+\frac14 \frac{x-2}{x^2+2}+C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
The converges of $ \sqrt 2 +\sqrt { 2-\sqrt 2} +\sqrt { 2-\sqrt { 2+\sqrt 2} } + \cdots =$ I would like to know wheather this series converge or diverge?
$\sqrt 2 +\sqrt { 2-\sqrt 2} +\sqrt { 2-\sqrt { 2+\sqrt 2} } +\sqrt { 2-\sqrt { 2+\sqrt { 2+\sqrt 2} } } +\cdots$
My work:
multiplyIing both demominator and numerator to $\sqrt { 2+\sqrt { 2 } } $,$\sqrt { 2+\sqrt { 2+\sqrt { 2 } } } $,$\sqrt { 2+\sqrt { 2+\sqrt { 2+\sqrt { 2 } } } } $ respectively and so on..
$$\sqrt { 2 } +\frac { \sqrt { 2 } }{ \sqrt { 2+\sqrt { 2 } } } +\frac { \sqrt { 2-\sqrt { 2 } } }{ \sqrt { 2+\sqrt { 2+\sqrt { 2 } } } } +\frac { \sqrt { 2-\sqrt { 2+\sqrt { 2 } } } }{ \sqrt { 2+\sqrt { 2+\sqrt { 2+\sqrt { 2 } } } } } +...=\\ =\sqrt { 2 } +\frac { \sqrt { 2 } }{ \sqrt { 2+\sqrt { 2 } } } +\frac { \sqrt { 2 } }{ \left( \sqrt { 2+\sqrt { 2 } } \right) \left( \sqrt { 2+\sqrt { 2+\sqrt { 2 } } } \right) } +\frac { \sqrt { 2 } }{ \left( \sqrt { 2+\sqrt { 2 } } \right) \left( \sqrt { 2+\sqrt { 2+\sqrt { 2 } } } \right) \left( \sqrt { 2+\sqrt { 2+\sqrt { 2+\sqrt { 2 } } } } \right) } +...=\\ =\sqrt { 2 } \left[ 1+\frac { 1 }{ \sqrt { 2+\sqrt { 2 } } } +\frac { 1 }{ \left( \sqrt { 2+\sqrt { 2 } } \right) \left( \sqrt { 2+\sqrt { 2+\sqrt { 2 } } } \right) } +\frac { 1 }{ \left( \sqrt { 2+\sqrt { 2 } } \right) \left( \sqrt { 2+\sqrt { 2+\sqrt { 2 } } } \right) \left( \sqrt { 2+\sqrt { 2+\sqrt { 2+\sqrt { 2 } } } } \right) } +... \right] $$
we can write every term as :${ a }_{ n+1 }=\frac { 1 }{ \sqrt { 2+{ a }_{ n } } } ,{ a }_{ 0 }=\sqrt { 2 } $
it is obvious that the sequences are monotone decreasing so ${ a }_{ n }>{ a }_{ n+1 }$ and I wrote sum as:$$S_{ \infty }=\sqrt { 2 } \sum _{ i=0 }^{ \infty }{ \frac { 1 }{ { a }_{ i+1 }\sqrt { 2+{ a }_{ i } } } } $$(i am not sure about it)
I am stuck here,I suspect series converges but how to show,which convergence tests can i use i don't know?Any hints,help will be appriceated?
P.S.I apologize for my english
| First, the sum we want can be rewritten as $\displaystyle\;\sum_{n=0}^\infty \sqrt{\epsilon_n}\;$
where
$$\epsilon_n = 2 - \overbrace{\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}^{n\; \text{terms}}$$
Notice for $n \ge 0$,
$$\epsilon_{n+1} = 2 - \sqrt{4 - \epsilon_n} = \frac{\epsilon_n}{2+\sqrt{4-\epsilon_n}}$$
It is easy to see $\epsilon_n > 0 \implies \epsilon_{n+1} > 0$. Since $\epsilon_0 = 2 > 0$, we have $\epsilon_n > 0$ for all $n$. Furthermore,
$$\epsilon_{n+1} \le \frac{\epsilon_n}{2}, \forall n \ge 0
\quad\implies\quad \epsilon_n \le \frac{\epsilon_0}{2^n} = 2^{1-n}, \forall n \ge 0 $$
This leads to an upper bound for the sum at hand
$$\sum_{n=0}^\infty \sqrt{\epsilon_n} \le \sum_{n=0}^\infty 2^{\frac{1-n}{2}} = \frac{\sqrt{2}}{1-\frac{1}{\sqrt{2}}} = 2(\sqrt{2}+1) < \infty$$
Since the sum is over non-negative numbers, this implies the sum converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to solve $z^3 + \overline z = 0$ I need to solve this:
$$z^3 + \overline z = 0$$
how should I manage the 0?
I know that a complex number is in this form: z = a + ib so:
$$z^3 = \rho^3\lbrace \cos(3\theta) + i \sin (3\theta)\rbrace$$
$$\overline z = \rho\lbrace \cos(-\theta) + i \sin (-\theta)\rbrace$$
but how about the 0?
EDIT:
ok, following some of your comments/answers this is what I have done:
$$z^3 = - \overline z$$
$$\rho^3\lbrace \cos(3\theta) + i \sin (3\theta)\rbrace = \rho\lbrace \cos(-\theta) + i \sin (-\theta)\rbrace$$
So
$$
\begin{Bmatrix}
\rho^3 = \rho\\
3\theta = -\theta + 2k\pi
\end{Bmatrix}$$
$$
\begin{Bmatrix}
\rho^3 = \rho\\
2\theta = 2k\pi
\end{Bmatrix}$$
$$
\begin{Bmatrix}
\rho = 0 or \rho = 1\\
\theta = k\frac{\pi}{2}
\end{Bmatrix}$$
is this the right way?
| Given
$$
z^3 = - \bar{z}.
$$
First we note that
$$
|z^3| = |-\bar{z}| \Longrightarrow |z|^3 = |z|.
$$
Therefore
$$
z = 0 \vee z = \exp(\zeta \pi \mathbf{i}).
$$
The case $z \ne 0$
We obtain
$$
\exp(3 \zeta \pi \mathbf{i}) = \exp( \pi \mathbf{i} - \zeta \pi \mathbf{i}).
$$
Whence
$$
3 \zeta = 2 k + 1 - \zeta \Longrightarrow \zeta = \frac{1}{4} + k.
$$
The general solution can be written as
$$
z = 0 \vee z = \exp\Big( \big[ \tfrac{1}{4} + k \big] \pi \mathbf{i} \Big)
= \exp\big( \pi \mathbf{i} / 4 \big) \exp\big( k \pi \mathbf{i} \big).
$$
Or as
$$
z_0 = 0 \vee k \in {1,2,3,4} : z_k = \exp\big( \pi \mathbf{i} / 4 \big) \mathbf{i}^k.
$$
Note that
$$
\exp\big( \pi \mathbf{i} / 4 \big) = \frac{1+\mathbf{i}}{\sqrt{2}},
$$
so
$$
z_1 = \frac{-1+\mathbf{i}}{\sqrt{2}}\\
z_2 = \frac{-1-\mathbf{i}}{\sqrt{2}}\\
z_3 = \frac{1-\mathbf{i}}{\sqrt{2}}\\
z_4 = \frac{1+\mathbf{i}}{\sqrt{2}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 8
} |
Solve $\log_3(3x + 2) = \log_9(4x + 5)$ for $x$
Solve for $x$
$$
\log_3(3x + 2) = \log_9(4x + 5)
$$
I changed the bases of the logs
$$
\frac {\log_{10}(3x + 2)} {\log_{10}(3)} = \frac {\log_{10}(4x + 5)} {\log_{10}(9)}
$$
Now I'm stuck, I don't know how to eliminate the logs.
On WolframAlpha I've seen that $\dfrac {\log_{10}(3x + 2)} {\log_{10}(3)} = 0$ gives $ x = -\frac{1}{3}$.
Do you know how does this equation and the equation above can be solved?
Thanks in advance, first question here :D
EDIT: You all solved the first equation, but I still don't understand how to solve the second one (the one solved by WolframAlpha).
| Notice, formula $$\color{blue}{log_{b^n}(a)=\frac{1}{n}log_{b}(a)}$$ Now, we have $$log_{3}(3x+2)=log_{9}(4x+5)$$ $$\implies log_{3}(3x+2)=log_{3^2}(4x+5)$$
$$\implies log_{3}(3x+2)=\frac{1}{2}log_{3}(4x+5)$$ $$\implies 2 log_{3}(3x+2)=log_{3}(4x+5)$$ $$\implies log_{3}(3x+2)^2=log_{3}(4x+5)$$ $$\implies (3x+2)^2=4x+5$$ $$\implies 9x^2+12x+4=4x+5$$ $$\implies 9x^2+8x-1=0$$ $$\implies (9x-1)(x+1)=0$$ $$\implies x=\frac{1}{9} \ \text{&} \ x=-1$$ Now, substituitng $x=-1$ , we get $LHS=\log(3(-1)+2)=\log(-1)$
But log is not defined for negative number hence the correct solution is $\color{blue}{x=\frac{1}{9}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
First principle derivative of a square root and conjugates I'm trying to find the derivative of the equation: $$g(x)=\sqrt {x+2}-3x^2$$.
I can find the solution just fine using the power rule but am finding trouble with First Principles.
Essentially, I understand getting as far as $$\displaystyle\lim_{h\to 0}\frac{\sqrt {x+h+2}-3(x+h)^2 -\sqrt {x+2}+3x^2}{h}.$$ From here I can expand out to $$\lim_{h\to 0}\frac{\sqrt {x+h+2}-3x^2-6xh-3h^2 -\sqrt {x+2}+3x^2}{h}.$$ But then I get stuck.
I'm not sure if I should use the conjugate rule now (but then how would I even apply that?) or if I'm supposed to try and simplify.
The answer is $\dfrac{1}{2\sqrt {x+2}}-6x$ that I got using the power rule.
Any help and guidance is appreciated.
| I am taking from the step you got stuck $$\lim_{h\to 0}\frac{\sqrt {x+h+2}-3x^2-6xh-3h^2 -\sqrt {x+2}+3x^2}{h}$$ $$=\lim_{h\to 0}\frac{\sqrt {x+2}\left(1+\frac{h}{x+2}\right)^{1/2}-\sqrt {x+2}-6xh-3h^2 }{h}$$ Use binomial expansion $(1+x)^n=1+nx+ \ldots $ where, $|x|<1$ $$=\lim_{h\to 0}\frac{\sqrt {x+2}\left(1+\frac{1}{2}\left(\frac{h}{x+2}\right)+o(h^2) \right)-\sqrt {x+2}-6xh-3h^2 }{h}$$ $$=\lim_{h\to 0}\frac{\sqrt {x+2}+\sqrt {x+2}\left(\frac{1}{2}\left(\frac{h}{x+2}\right)+o(h^2) \right)-\sqrt {x+2}-6xh-3h^2 }{h}$$ $$=\lim_{h\to 0}\frac{\sqrt {x+2}\left(\frac{1}{2}\left(\frac{h}{x+2}\right)+o(h^2) \right)-6xh-3h^2 }{h}$$ $$=\lim_{h\to 0}\sqrt {x+2}\left(\frac{1}{2}\left(\frac{1}{x+2}\right)+o(h) \right)-6x-3h $$ $$=\sqrt {x+2}\left(\frac{1}{2}\left(\frac{1}{x+2}\right)+0 \right)-6x-3(0) $$ $$=\frac{1}{2\sqrt{x+2}}-6x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving that the sum of the first $2n$ terms of the series $1^2 - 3^2 + 5^2 - \cdots$ is $-8n^2$ by induction
Use mathematical induction to prove the following for the first $2n$ terms of the series $$1^2 - 3^2 + 5^2 - 7^2 + \cdots = -8n^2.$$
As we have odd numbers that are squared we could use $n = 2k-1$.
But the $2$ sides do not equate for $n=1$ or $2k-1$ (if you set $k=1$).
Also need to find the sum to $2n+1$ terms.
| $1^2−3^2+5^2−7^2+⋯=−8n^2$
Last term of this series is $(4n-3)^2-(4n-1)^2$
$\Rightarrow P(n):1^2−3^2+5^2−7^2+⋯+(4n-3)^2-(4n-1)^2=−8n^2$
For n=1
$\Rightarrow P(1):1^2−3^2=−8(1)^2$
$\Rightarrow P(1):-8=−8$ which is true
Let P(k)be true,
$P(k):1^2−3^2+5^2−7^2+⋯+(4k-3)^2-(4k-1)^2=−8k^2$.......(1)
We have to prove P(k+1) as true.
$P(k+1):1^2−3^2+5^2−7^2+⋯+(4k-3)^2-(4k-1)^2+(4(k+1)-3)^2-(4(k+1)-1)^2=−8(k+1)^2$
$LHS=1^2−3^2+5^2−7^2+⋯+(4k-3)^2-(4k-1)^2+(4k+1)^2-(4k+3)^2$
$=-8k^2+(4k+1)^2-(4k+3)^2$
$=-8k^2+16k^2+8k+1-(16k^2+24k+9)$
$=-8k^2-16k-8=-8(k^2+2k+1)=-8(k+1)^2$=RHS
Therefore P(k+1) is true.So by Induction principle,P(n) is true for all $n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Given the differential equation, how to solve the y function with x as the independent variable? $y\frac{dy}{dx} = x(y^4 + 2y^2 + 1)$
$y = 1$ when $x = 4$
I tired to integrate by substitution, but it doesn't seem to work out.
| To solve this, you can use separation of variables:
$$ y \frac{dy}{dx} = x(y^4+2y^2+1)$$
$$ \frac{y}{y^4+2y^2+1}dy = x\ dx$$
$$ \frac{y}{(y^2+1)^2}= x \ dx$$
Using the substitution $u = y^2+1$, $\frac{du}{dy}= 2y $
$$ \frac{1}{2}\int u^{-2} du= \int x \ dx $$
$$ \frac{-1}{2(y^2+1)} = \frac{x^2}{2} + c$$
where c is a constant of integration.
You can then obtain a value for $c$ using $x =4 \ \text{and}\ y =1$
This should give you $c =\frac{-33}{4}$
Hence, solution:
$$ \frac{2}{y^2+1} =33- 2x^2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Proving $\frac1{4ab}\left(\frac{(b+1)^{b+1}}{b^b}\right)^a<\binom{a(b+1)}a<\left(\frac{(b+1)^{b+1}}{b^b}\right)^a$ Let $a\in\mathbb N$, and $b\in\mathbb R, b\geq 1$. How to prove that $$\frac{1}{4ab}\left(\frac{(b+1)^{b+1}}{b^b}\right)^a<\binom{a(b+1)}{a}<\left(\frac{(b+1)^{b+1}}{b^b}\right)^{a}?$$
| $\def\e{\mathrm{e}}$Lemma 1: If $$f(x) = \left( 1 + \dfrac{1}{x} \right)^x,\ g(x) = \left( 1 + \dfrac{1}{x} \right)^{x + 1}, \quad x \geqslant 1$$
then $f$ is strictly increasing but $g$ is strictly decreasing, and$$
\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} g(x) = \e.
$$
(This is a well-known result.)
Lemma 2: For $m\in \mathbb{N}_+$, $n \in \mathbb{R}$, $n \geqslant 1$,$$
\binom{m + n}{m} < \frac{(m + n)^{m + n}}{m^m n^n}. \tag{2.1}
$$
Proof of lemma: (2.1) will be proved by induction on $m$. For $m = 1$,$$
(1) \Longleftrightarrow n + 1 < \frac{(n + 1)^{n + 1}}{n^n} \Longleftrightarrow n^n < (n + 1)^n.
$$
Suppose (2.1) holds for $m$. To prove (2.1) for $m + 1$, it suffices to prove that$$
\left. \binom{m + n + 1}{m + 1} \middle/ \binom{m + n}{m} \right. \leqslant \left. \frac{(m + n + 1)^{m + n + 1}}{(m + 1)^{m + 1} n^n} \middle/ \frac{(m + n)^{m + n}}{m^m n^n} \right.. \tag{2.2}
$$\begin{align*}
(2.2) &\Longleftrightarrow \frac{m + n + 1}{m + 1} \leqslant \frac{m^m (m + n + 1)^{m + n + 1}}{(m + 1)^{m + 1} (m + n)^{m + n}}\\
&\Longleftrightarrow \frac{(m + 1)^m}{m^m} \leqslant \frac{(m + n + 1)^{m + n}}{(m + n)^{m + n}}\\
&\Longleftrightarrow \left( 1 + \frac{1}{m} \right)^m \leqslant \left( 1 + \frac{1}{m + n} \right)^{m + n},
\end{align*}
where the last inequality holds by Lemma 1. End of induction.
Lemma 3: For $m\in \mathbb{N}_+$, $n \in \mathbb{R}$, $n \geqslant 1$,$$
\binom{m + n}{m} > \frac{1}{\e m} \frac{(m + n)^{m + n}}{m^m n^n}. \tag{3.1}
$$
Proof of lemma: Again, (3.1) will be proved by induction on $m$. For $m = 1$,$$
(3.1) \Longleftrightarrow n + 1 > \frac{1}{\e}·\frac{(n + 1)^{n + 1}}{n^n} \Longleftrightarrow \left( 1 + \frac{1}{n} \right)^n < \e,
$$
where the last inequality is true by Lemma 1.
Suppose (3.1) holds for $m$. To prove (3.1) for $m + 1$, it suffices to prove that$$
\left. \binom{m + n + 1}{m + 1} \middle/ \binom{m + n}{m} \right. \geqslant \left. \frac{(m + n + 1)^{m + n + 1}}{(m + 1)^{m + 2} n^n} \middle/ \frac{(m + n)^{m + n}}{m^{m + 1} n^n} \right.. \tag{3.2}
$$\begin{align*}
(3.2) &\Longleftrightarrow \frac{m + n + 1}{m + 1} \geqslant \frac{m^{m + 1} (m + n + 1)^{m + n + 1}}{(m + 1)^{m + 2} (m + n)^{m + n}}\\
&\Longleftrightarrow \frac{(m + 1)^{m + 1}}{m^{m + 1}} \geqslant \frac{(m + n + 1)^{m + n}}{(m + n)^{m + n}}\\
&\Longleftrightarrow \left( 1 + \frac{1}{m} \right)^{m + 1} \geqslant \left( 1 + \frac{1}{m + n} \right)^{m + n},
\end{align*}
where the last inequality holds because by Lemma 1,$$
\left( 1 + \frac{1}{m} \right)^{m + 1} \geqslant \e \geqslant \left( 1 + \frac{1}{m + n} \right)^{m + n}.
$$
End of induction.
Now for $a \in \mathbb{N}_+$, $b \geqslant 1$, take $(m, n) = (a, ab)$ in Lemma 2, then$$
\binom{a(b + 1)}{a} < \frac{(ab + a)^{ab + a}}{a^a (ab)^{ab}} = \frac{a^{ab + a} (b + 1)^{ab + a}}{a^{ab + a} b^{ab}} = \left( \frac{(b + 1)^{b + 1}}{b^b} \right)^a.
$$
Take $(m, n) = (a, ab)$ in Lemma 3, then$$
\binom{a(b + 1)}{a} > \frac{1}{\e a} \frac{(ab + a)^{ab + a}}{a^a (ab)^{ab}} = \frac{1}{\e a} \left( \frac{(b + 1)^{b + 1}}{b^b} \right)^a \geqslant \frac{1}{4ab} \left( \frac{(b + 1)^{b + 1}}{b^b} \right)^a.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Differentiate the Function: $y=\sqrt{x^x}$ $y=\sqrt{x^x}$
How do I convert this into a form that is workable and what indicates that I should do so?
Anyway, I tried this method of logging both sides of the equation but I don't know if I am right.
$\ln\ y=\sqrt{x} \ln\ x$
$\frac{dy}{dx}\cdot \frac{1}{y}=\sqrt{x}\ \frac{1}{x} +\ln\ x\ \frac{1}{2}x^{-\frac{1}{2}}$
$\sqrt{x}\cdot (\sqrt{x}\ \frac{1}{x} +\ln\ x \ \frac{1}{2}x^{-\frac{1}{2}})$
| That way could work though you made some mistakes, but an easier way shifts the square root to a fractional exponent.
$$\begin{align}
y&=\sqrt{x^x} \\[2ex]
&= \left(x^x\right)^{1/2} \\[2ex]
&= x^{x/2} \\[2ex]
\ln y&= \ln x^{x/2} \\[2ex]
&= \frac x2\ln x \\[2ex]
\frac{dy}{dx}\frac 1y &=\frac 12\ln x+\frac x2\frac 1x \\[2ex]
&= \frac 12\ln x+\frac 12 \\[2ex]
\frac{dy}{dx} &= y\left(\frac 12\ln x+\frac 12 \right) \\[2ex]
&= \sqrt{x^x}\left(\frac 12\ln x+\frac 12 \right) \\[2ex]
&= \frac 12\sqrt{x^x}\left(\ln x+1 \right)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Mathematical Induction proof for a cubic equation. If $ x^3 = x +1$, prove by induction that $ x^{3n} = a_{n}x + b_n + \frac {c_n}{x}$, where $a_1=1, b_1=1, c_1=0$ and
$a_n = a_{n-1} + b_{n-1}, b_n = a_{n-1} + b_{n-1} + c_{n-1}, c_n = a_{n-1} + c_{n-1}$ for $n=2,3,\dots $
For $n=1$ we have $x_3 = a_1x + b_1 + \frac {c_1}{x}$ = x + 1 + $\frac{0}{x}$ = x + 1, which is true.
Assume the case for $n=k$ is true, so $x^{3k} = a_kx + b_k + \frac{c_k}{x}$
So for $n = k+1$ (and using $a_n = a_{n-1} + b_{n-1}, b_n = a_{n-1} + b_{n-1} + c_{n-1}, c_n = a_{n-1} + c_{n-1}$ for $n=2,3,...), $ we have:
$ \begin{align}
x^{3(k+1)} &= (a_{k+1-1}x + b_{k+1-1}){x} + (a_{k+1-1} + b_{k+1-1} + c_{k+1-1}) +
( \frac{(a_{k+1-1} + c_{k+1-1})}{x}) \\
&= (a_k + b_k){x} + ( a_k + b_k + c_k) + \frac{(a_k + c_k)}{x} \\
&= a_{k+1}{x} + b_{k+1} + \frac{c_{k+1}}{x} \\\end{align} $
This is the same form as $ x^{3k} $ but for n=k+1.
Therefore, if the result is true for k, it is also true for (k+1).
Would you proceed like this or would you add the $(k+1)$th term and show that it is equal to the $k$th term with $n$ (or $k$) replaced by $k+1$?
| You should put "$=$" between things ALREADY KNOWN to be equal, not with things you are trying to prove to be equal. You should not start a chain of equalities with the very equality you're trying to prove.
Thus
\begin{align}
x^{3(k+1)} & = x^3 x^{3k} & & \text{(This known, by routine algebra.)} \\[10pt]
& = x^3 ( a_kx + b_k + c_k x^{-1} ) & & \text{(This comes from the induction hypothesis.)} \\[10pt]
& = (x+1)( a_kx + b_k + c_k x^{-1} ) & & \text{(This is true by hypothesis.)} \\[10pt]
& = a_k x^2 + (a_k+b_k)x + (c_k+b_k) + c_k x^{-1} & & \text{(true by routine algebra)} \\[10pt]
& = \frac{a_k x^3} x + (a_k+b_k)x + (c_k+b_k) + c_k x^{-1} & & \text{(ditto)} \\[10pt]
& = \frac{a_k (x+1)} x + (a_k+b_k)x + (c_k+b_k) + c_k x^{-1} & & \text{(by hypothesis)} \\[10pt]
& = (a_k+b_k)x + (a_k + c_k+b_k) + (a_k+c_k) x^{-1} & & \text{(routine algebra again)}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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} |
Trigonometric equation with sine and cosine So the equation is $3\cos ^2t + 5\sin t = 1$
Now I have simplified this to $$3(1-\sin ^2t) + 5\sin t -1 = 0$$
which leads to $$-3\sin ^2t + 5\sin t + 2 = 0$$
Then I get $$-3t^2 + 5 t +2 = 0$$
Is this the correct way to go with this equation then use $t = t/2 \pm \sqrt {(t/2)^ 2 + y}$ where $y$ in this case will be $2/3$ ?
| you have $$-3\sin(t)^2+5\sin(t)+2=0$$ let $$\sin(t)=u$$ then we have $$-3u^2+5u+2=0$$ divided by $-3$ gives $$u^2-\frac{5}{3}u-\frac{2}{3}=0$$
solving this equation we obtain
$$u_{1,2}=\frac{5}{6}\pm\sqrt{\frac{25}{36}+\frac{24}{36}}$$
from here you will come to the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Asymptotic expansion of integral with hyperbolic functions Consider the integral given by
$$f(r)=\int_{0}^{\tanh(r)} \arccos\left(\frac{\sigma}{\sinh(r)\sqrt{1-\sigma^2}}\right)\cdot \frac{1}{\sqrt{\sigma^2+a^2}}d\sigma,$$
where $a>0$. I am wondering how one can derive the asymptotic power series for $r\downarrow 0$?
I have tried the following
$$\arccos\left(\frac{\sigma}{\sinh(r)\sqrt{1-\sigma^2}}\right)=\pi/2-\arcsin\left(\frac{\sigma}{\sinh(r)\sqrt{1-\sigma^2}}\right)\\
=\pi/2+\sum_{k\geq 0}\binom{2k}{k}\frac{\sigma^{2k+1}}{(2k+1)4^k \sinh(r)^{2k+1}\cdot (1-\sigma^2)^{k+1/2}}.$$
Hence we get intuitively (?)
$$f(r)=\frac{\pi}{2}\left( \log\left( \sqrt{a^2+\tanh(r)^2}+\tanh(r)\right)-\log(a)\right)+ \sum_{k\geq 0}\binom{2k}{k}\frac{1}{(2k+1)4^k \sinh(r)^{2k+1}}\cdot \int_{0}^{\tanh(r)}\frac{\sigma^{2k+1}}{ (1-\sigma^2)^{k+1/2}\sqrt{\sigma^2+a^2}}d\sigma.$$
But the expressions become very complicated. Maybe I have not seen a simplification which could be made at the beginning. Does anyone know how one can solve this problem?
Best wishes
| Consider $\;\sigma:=\tanh(u)\,$ then :
\begin{align}
f(r)&:=\int_{0}^{\tanh(r)} \arccos\left(\frac{\sigma}{\sinh(r)\sqrt{1-\sigma^2}}\right)\cdot \frac{1}{\sqrt{\sigma^2+a^2}}\,d\sigma\\
&=\int_{0}^{r} \arccos\left(\frac{\sinh(u)}{\sinh(r)}\right) \frac{1}{\sqrt{\tanh(u)^2+a^2}}\,\frac{du}{\cosh(u)^2}\\
\end{align}
Set $\;t:=\dfrac{\sinh(u)}{\sinh(r)}\;$ (i.e. $\;u:=\operatorname{argsinh}(\sinh(r)\;t)\,$) then :
\begin{align}
&\sinh(u)=t\,\sinh(r),\\
&\cosh(u)^2=1+(t\,\sinh(r))^2\\
\\
&\text{and}\\
f(r)&=\int_{0}^{1} \arccos\left(t\right) \frac{\sqrt{1+t^2\sinh(r)^2}}{\sqrt{t^2\sinh(r)^2+a^2(1+t^2\sinh(r)^2)}}\,\frac{\sinh(r)\;dt}{\sqrt{1+t^2\sinh(r)^2}^3}\\
&=\sinh(r)\int_{0}^{1} \frac{\arccos\left(t\right) }{\sqrt{a^2+(1+a^2)t^2\sinh(r)^2}}\,\frac{dt}{1+t^2\sinh(r)^2}\\
\end{align}
At this point you may get an asymptotic expansion as $r\to 0$ especially by considering the expansion in powers of $s:=\sinh(r)$ and observing that :
$$a_n:=\int_{0}^{1} \arccos\left(t\right) t^{2n}\,dt=\frac{4^n}{(2n+1)^2\binom{2n}{n}}$$
(the indefinite integral of $\arccos\left(t\right) t^{2n}$ admits a closed form)
For $\,s=\sinh(r)\,$ and using the expansion of $\dfrac 1{\sqrt{1-x}}$ in central binomial terms I get :
$$f(r)=\sum_{n=0}^\infty \frac{(-4)^n}{(2n+1)^2\binom{2n}{n}}\left(\frac sa\right)^{2n+1}\sum_{k=0}^n a^{2n-2k}(a^2+1)^k\frac{\binom{2k}{k}}{4^k}$$
The first terms of this expansion are :
$$f(r)=\frac sa-\frac 29\left(\frac sa\right)^3\frac{3a^2+1}{2}+\frac 8{75}\left(\frac sa\right)^5\frac{15a^4+10a^2+3}{8}-+\frac {16}{245}\left(\frac sa\right)^7\frac{35a^6+35a^4+21a^2+5}{16}+O\left(s^9\right) $$
(as confirmed by numerical evaluation)
Using the Maclaurin series for $\sinh(r)$ this becomes (with an error of $\;O\left(r^9\right)\;$) :
$$\frac ra-\left(\frac ra\right)^3\frac{3a^2+2}{18}+\left(\frac ra\right)^5\frac{75a^4+140a^2+72}{1800}-\left(\frac ra\right)^7\frac{1281a^6+4634a^4+5544a^2+2160}{105840}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Determine if this series $ \sum_{n=0}^\infty\frac{n^6+13n^5+n+1}{n^7+13n^4+9n+2}$ converges Determine if the following series converges:
$$
\sum_{n=0}^\infty\frac{n^6+13n^5+n+1}{n^7+13n^4+9n+2}.
$$
(http://i.stack.imgur.com/qWiuy.png)
I don't know how to start.
| Use the limit comparison test.
Let $\displaystyle a_n=\frac{n^6+13n^5+n+1}{n^7+13n^4+9n+2}$ and $\displaystyle b_n=\frac{n^6}{n^7}=\frac{1}{n}$.
Since
$$\begin{align}\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}&=\lim_{n\to\infty}\frac{n^6+13n^5+n+1}{n^7+13n^4+9n+2}\cdot\frac{n}{1}\\
&=\lim_{n\to\infty}\frac{n^7\left(1+13\frac{1}{n}+\frac{1}{n^5}+\frac{1}{n^7}\right)}{n^7\left(1+13\frac{1}{n^3}+9\frac{1}{n^6}+\frac{2}{n^7}\right)} \\
&=1 > 0 \end{align}$$
Both $a_n$ and $b_n$ converges or diverges. Since $b_n$ diverges, so too must $a_n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving an exponential equation that includes division and multiplication The question is simplify the expression $\left(\dfrac{a^2}{27}\right)^{1/3}\left(\dfrac{64}a\right)^{2/3}$
1: Multiply on both sides equals $\dfrac{a^{2/3}}{27^{1/3}}\cdot \dfrac{64^{2/3}}{a^{2/3}}$
Does this give me $\dfrac{a^{2/3}}{3} \cdot \dfrac{16}{a^{2/3}}$ ?
| Assuming that $a\neq 0$; then
$$\Big(\frac{64}{a}\Big)^{\frac{2}{3}}=\Big(\frac{4^6}{a^2}\Big)^{\frac{1}{3}};$$
then $$\Big(\frac{a^2}{3^3}\Big)^{\frac{1}{3}}\Big(\frac{16^3}{a^2}\Big)^{\frac{1}{3}}=\frac{16}{3}.$$
So your answer is correct. Clearly if $a=0$ this expression is not define.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving two diophantine equations. Find at least one 5-tuple of positive integers which satisfy the following two equations
$$a^2-d^2=3(b^2-c^2)$$
$$e^2-b^2=3(d^2-c^2)$$
such that no three of the 5 positive integers $a, b, c, d, e$ form an Arithmetic progression and $a<b<c<d<e$.
Please help.
| Here is a nice context to this seemingly random system. If there is a $5$-tuple $a,b,c,d,e$ such that,
$$\begin{aligned}
a^2-d^2&=3(b^2-c^2)\\[1.5mm]
e^2-b^2&=3(d^2-c^2)
\end{aligned}\tag1$$
then for $k = 2,4$,
$$(a + c)^k + (-a + c)^k + (2 d)^k = (c + e)^k + (c - e)^k + (2 b)^k\tag2$$
and for $k = 1,2,3,5$,
$$(a + 2 c)^k + (-a + 2 c)^k + (-2 c + e)^k + (-2 c - e)^k =\\
(2 b + c)^k + (-2 b + c)^k + (-c + 2 d)^k + (-c - 2 d)^k\tag3$$
There is an infinite number of co-prime solutions to $(1)$. An example is,
$$a,b,c,d,e = x - 3 y,\; x + 2 y,\; x + 3 y,\; z,\; -x + 7 y$$
where $x,y,z$ satisfy,
$$x^2+24y^2 = z^2\tag4$$
The condition $(4)$ can be easily solved as,
$$x,y,z = 2mn,\; m^2-24n^2,\; m^2+24n^2$$
and if we wish that $a<b<c<d<e$, then $m \approx 6n$ for $n$ above a bound. For example, if $m,n = 38,\,6$, then after removing common factors,
$$a,b,c,d,e = -321,\; 404,\; 549,\;577,\;901$$
and since $(1)$ involves squares, then signs don't matter.
P.S. Using Lucian's small solution $a,b,c,d,e = 4,\,41,\,54,\,61,\,64$ on $(2),(3)$, one also gets,
$$58^k+50^k+122^k = 118^k+10^k+82^k,\;\;k=2,4$$
$$112^k+104^k+(-44)^k+(-172)^k = 136^k+(-28)^k+68^k+(-176)^k,\;\; k=1,2,3,5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove the relation for cos inverse Prove the relation $\cos^{-1}x_0=\dfrac{\sqrt {1-x^2_0}}{x_1\cdot x_2\cdot x_3\cdots \text{ ad inf.}}$ where the successive quantities $x_r$ are connected by the relation $x_{r+1}=\sqrt{\frac{1}{2}(1+x_r)}$
My attempt:
$$x_1=\sqrt{\frac{1}{2}(1+x_0)}$$
$$x_2=\sqrt{\frac{1}{2}(1+x_1)}$$
$$x_3=\sqrt{\frac{1}{2}(1+x_2)}$$
$$x_4=\sqrt{\frac{1}{2}(1+x_3)}$$
Multiplying all these, we get
$$x_1\cdot x_2\cdot x_3\cdots\text{ ad inf.}=\sqrt{\frac{1}{2}(1+x_0)\frac{1}{2}(1+x_1)\frac{1}{2}(1+x_2)\frac{1}{2}(1+x_3)\cdots \text{ ad inf.}}$$
Putting in equation,
$$\cos^{-1}x_0=\frac{\sqrt {1-x^2_0}}{\sqrt{\frac{1}{2}(1+x_0)\frac{1}{2}(1+x_1)\frac{1}{2}(1+x_2)\frac{1}{2}(1+x_3)\cdots \text{ ad inf.}}}$$
$$\cos^{-1}x_0=\frac{\sqrt {1-x_0}}{\sqrt{\frac{1}{2}(1+x_1)\frac{1}{2}(1+x_2)\frac{1}{2}(1+x_3)\cdots \text{ ad inf}}}$$
but i could not solve further. Can someone guide me in this question?
| Let $x_0:=\cos y$. Want to prove that $y = RHS$. Also we know, that in case of convergence, $x_r \to 1$ because $\bar x = \sqrt{\frac12(1+\bar x)} \Rightarrow \bar x^2 - \frac12 \bar x - \frac12 = 0$ and $\bar x \ge 0$. The roots of the quadratic are $1$ and $-\frac12$.
We now need to show that
$$y = \frac{\sqrt{1-\cos^2 y}}{\prod_{r=1}^\infty x_r}$$
By restricting $y$ to $[0, \pi]$ we can thus write
$$y = \frac{\sin y}{\prod_{r=1}^\infty x_r}$$
Or equivalently
$$\frac{\sin y}y = \prod_{r=1}^\infty x_r$$
This identity might be related to some other product identity for the cardinal sine. Specifically looking at
$$\frac{\sin y}y = \prod_{k=1}^\infty \cos \left(\frac y{2^k}\right)$$
looks promising.
In fact we can prove that
$$x_r = \cos\left(\frac y{2^k}\right)$$
by induction.
Note that
$$\frac12(1+\cos x) = \frac12 (1+\cos(\frac x2 + \frac x2)) = \frac12(1 + \cos^2 \frac x2 - \sin^2 \frac x2) = \frac12 (2\cos^2 \frac x2) = \cos^2 \frac x2$$
Thus
$$x_1 = \sqrt{\frac12(1+\cos y)} = \cos \frac y2$$
Now by induction
$$x_r = \sqrt{\frac12(1+\cos \frac y{2^{r-1}})} = \cos \frac y{2^r}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378179",
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"source": "stackexchange",
"question_score": "3",
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Why does $\sqrt{6} + \sqrt{10} + \sqrt{15}$ have four conjugates? I am having trouble understanding how algebraic number $\sqrt{6} + \sqrt{10} + \sqrt{15}$ has four conjugates.
Minimal polynomial is $x^4-62 x^2-240 x-239$ according to Wolfram Alpha.
Factorized:
$$\left(x-2\sqrt{15 (4-\sqrt{15})}-8\sqrt{4-\sqrt{15}}-\sqrt{15}\right)\cdot
\left(x-2\sqrt{4-\sqrt{15}}+\sqrt{15}\right) \\
\cdot \left(x+2\sqrt{4-\sqrt{15}}+\sqrt{15}\right)
\cdot\left(x+2\sqrt{15 (4-\sqrt{15})}+8\sqrt{4-\sqrt{15}}-\sqrt{15}\right)$$
| The conjugates of an algebraic number are (by definition) the roots of its minimal polynomial. The number of (distinct) roots of an irreducible polynomial over the rationals is equal to its degree, that is four.
Thus once you know the minimal polynomial "it is clear."
There is some wiggling room as one might or might not count the number itself among its conjugates. But the former is more common.
(This answer leaves open the question "why" the minimal polynomial has degree four or how it could be found as this is covered in other answers.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that the angle between $OP$ and the normal to the curve at $P$ satisfies the following I'm struggling to answer the following question below
I've already worked out the gradient to the curve at $P$, but I'm having difficulty answering the second part of the question. MY attempt is as follows:
I focus on $\triangle OPQ$, where $Q$ is the intersection of the normal at $P$ to the x-axis
Re-writing $P$ as $(x', y')$ to avoid confusion.
The angle $\theta_2$ between $OP$ and the x-axis satisfies $\tan(\theta_2)=\frac{y'}{x'}$
The equation of the normal to the curve at $P$ is $(y-y')=(\frac{ax+y}{x+ay})(x-x')$, substituting $y=0$ into the equation gives the x co-ordinate of $Q$, which gives $x=(x'(\frac{ax+y}{x+ay})-y')(\frac{x+ay}{ax+y})=x'-y'(\frac{x+ay}{ax+y})$
So the length of $OQ$ is $x'-y'(\frac{x+ay}{ax+y})$
The length of $OP$ is $\sqrt{x'^2+y'^2}$
And $\angle QOP$ satisfies $tan(\theta_2)=\frac{y'}{x'}$
But by using the cosine rule to get $\angle OPQ$, I get nothing close to the desired result. Any hints or different approaches would be greatly appreciated.
| Notice, gradient of the curve: $x^2+y^2+2axy=1$ $$\frac{d}{dx}(x^2+y^2+2axy)=\frac{d}{dx}(1)$$ $$2x+2y\frac{dy}{dx}+2ax\frac{dy}{dx}+2ay=0$$ $$2(ax+y)\frac{dy}{dx}=-2(x+ay)$$ $$\frac{dy}{dx}=\frac{-2(x+ay)}{2(ax+y)}$$ $$\implies \color{blue}{\text{slope of tangent}\,, \frac{dy}{dx}=\frac{-(x+ay)}{ax+y}}$$
Above is the gradient of the curve (i.e. gradient of tangent) at the point $(x, y)$.
Now, the slope of line OP joining origin $O(0, 0)$ & $P(x, y)$ $$=\frac{y-0}{x-0}=\frac{y}{x}$$ & the slope of normal at $P(x, y)$ $$=\frac{-1}{\text{slope of tangent at P}}=\frac{-1}{\frac{-(x+ay)}{ax+y}}=\frac{ax+y}{x+ay}$$ Hence, the angle $\theta$ between OP, slope, $m_1=\frac{y}{x}$ & normal at P, slope, $m_2=\frac{ax+y}{x+ay}$ is given by the following formula $$\tan \theta=\frac{|m_1-m_2|}{1+m_1m_2}$$ $$\implies \tan \theta=\frac{\left|\frac{y}{x}-\frac{ax+y}{x+ay}\right|}{1+\frac{y}{x}\times \frac{ax+y}{x+ay} }$$ $$=\frac{\left|ay^2-ax^2\right|}{x^2+axy+axy+y^2}$$ $$\tan \theta=\frac{a\left|y^2-x^2\right|}{x^2+y^2+2axy}\quad \text{since},\ 0<a<1$$ Setting the value of $x^2+y^2+2axy$, we get $$\tan \theta=\frac{a\left|y^2-x^2\right|}{1}$$
$$\implies \color{blue}{\tan \theta=a\left|y^2-x^2\right|}$$
Hence, $\theta$ is duly satisfying above relation
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate derivative of integral I tried to calculate the derivative of this integral:
$$\int_{2}^{3+\sqrt{r}} (3 + \sqrt{r}-c) \frac{1}{2}\,{\rm d}c $$
First I took the anti-derivative of the integral:
$$\frac{1}{2}\left(\frac{-c^2}{2}+c\sqrt{r}+3c\right)$$
Then I evaluated the integral:
$$-\frac{1}{2}\left(\frac{3+\sqrt{r})^2}{2} + (3 + \sqrt{r})\sqrt{r}+3(3+\sqrt{r})\right)-\frac{1}{2}\left(-\frac{(2^2)}{2}+2\sqrt{r}+3\cdot 2\right)$$
After I simplified I got:
$$\frac{1 + 2\sqrt{r}+r}{4}$$
I should get:
$$\frac{1 + \sqrt{r}}{4\sqrt{r}}$$
But I cannot get this result.
Can someone help? what am I missing?
| $$\int_{2}^{3+\sqrt{r}} \left(\frac{1}{2}\left(3+\sqrt{r}-c\right)\right)dc=$$
$$\frac{1}{2}\cdot\int_{2}^{3+\sqrt{r}} \left(3+\sqrt{r}-c\right)dc=$$
$$\frac{1}{2}\left(\int_{2}^{3+\sqrt{r}} (3)dc+\int_{2}^{3+\sqrt{r}} (\sqrt{r}) dc-\int_{2}^{3+\sqrt{r}} (c)dc\right)=$$
$$\frac{1}{2}\left(\left[3c\right]_{2}^{3+\sqrt{r}}+\left[c\sqrt{r}\right]_{2}^{3+\sqrt{r}}-\left[\frac{c^2}{2}\right]_{2}^{3+\sqrt{r}}\right)=$$
$$\frac{1}{2}\left(\left(3\sqrt{r}+3\right)+\left(r+\sqrt{r}\right)-\left(\frac{r}{2}+3\sqrt{r}+\frac{5}{2}\right)\right)=$$
$$\frac{1}{2}\left(\frac{1}{2}\left(\sqrt{r}+1\right)^2\right)=$$
$$\frac{1}{4}\left(\sqrt{r}+1\right)^2=\frac{\left(\sqrt{r}+1\right)^2}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Trigonometric substitution and triangles I'm learning trigonometric substitutions and am having a bit of trouble understanding the intuition behind the conversions (why do most use secant?). If you could explain the conversions geometrically using a triangle, that would be very helpful. For example, if we have $$\int \frac{\sqrt{x^2-4}}{x}\,dx$$ I tried to construct a triangle like so:
To get $$\sin(\theta)=\frac{\sqrt{x^2-4}}{x}$$
But this is incorrect for use as a substitution. Why?
| It's not that it's wrong to take on the integral the way you propose -- it's just not awfully convenient. If we implicitly differentiate your substitution equation, we obtain
$$ \cos \theta \ d\theta \ = \ \frac{x \ \cdot \frac{1}{2} \ (x^2 - 4)^{-1/2} \ \cdot 2x \ - \ (x^2 - 4)^{1/2} \ \cdot \ 1}{x^2} \ dx $$
$$ = \ \frac{4}{x^2 \ (x^2 - 4)^{1/2}} \ dx \ = \ \left(\frac{2}{x} \right)^2 \cdot \ \frac{2}{ \ (x^2 - 4)^{1/2}} \ \cdot \ \frac{1}{2} \ \ dx $$
$$ = \frac{1}{2} \ \cdot \ \cos^2 \theta \ \cdot \ \frac{1}{\tan \theta} \ \ dx $$
[using your right triangle]
$$ \Rightarrow \ \ dx \ = \ 2 \ \tan \theta \ \cdot \ \frac{\cos \theta}{\cos^2 \theta} \ \ d\theta \ = \ 2 \ \tan \theta \ \sec \theta \ \ d\theta \ \ , $$
which is just the result Mike shows. Following your lead, the integral becomes
$$ \int \ \sin \theta \ ( \ 2 \ \tan \theta \ \sec \theta \ \ d\theta \ ) \ = \ 2 \int \ \sin \theta \ \tan \theta \ \frac{1}{\cos \theta} \ \ d\theta \ = \ 2 \int \ \tan^2 \theta \ \ d\theta \ \ . $$
There are easier ways to get there, though.
| {
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"url": "https://math.stackexchange.com/questions/1381257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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New idea to solve this equation I was teaching $\left \lfloor x \right \rfloor$ function properties and equation . I solved this equation in my class . My works are show below. Some students ask me for new Idea...,and now I am looking for various method to solve this (like this) equation .$$\left \lfloor x\right \rfloor+2x=1$$
1st: $$x=n+p \\n \in\mathbb{Z} , 0 \leq p<1 \to \left \lfloor x \right \rfloor=n ,p=x-n\\ $$
and
$$ \left \lfloor x\right \rfloor+2x=1\\n+2(n+p)=1 \to 3n+2p=1 \\3n=0 ,\pm3 ,\pm 6,\pm9,... $$in this case $$3n=0 \to 2p=1 \\n=0 , p=\frac{1}{2} \to x=n+p=0+\frac{1}{2} $$
2nd:
$$\left \lfloor x\right \rfloor+2x=1 \to \left \lfloor x\right \rfloor=1-2x$$ like $f(x)=g(x)$ by drawing both of them obtain the answer
With respect to the picture ,it suffice to solve $0=1-2x$s o the answer is $x=\frac{1}{2}$
3rd :
we know $$\left \lfloor x\right \rfloor =k \in \mathbb{Z} \to k \leq x <k+1 $$ so $$\left \lfloor x\right \rfloor=1-2x \to 1-2x=k \in \mathbb{Z}\\x=\frac{1-k}{2} \to \left \lfloor \frac{1-k}{2}\right \rfloor =k$$ so we have
$$k \leq \frac{1-k}{2} <k+1 \to \\\left\{\begin{matrix}
k\leq \frac{1-k}{2} \to & 2k \leq 1-k \to & k \leq \frac{1}{3} \to k=\left \{ 0,-1,-2,-3,... \right \}\\
\frac{1-k}{2}<k \to & 1-k<2k \to &k> -\frac{1}{3} \to k=\left \{ 0,1,2,3,... \right \}
\end{matrix}\right.\\
\left \{ ...,-3,-2,-1,0 \right \}\bigcap \left \{ 0,1,2,3,... \right \}=\left \{ 0 \right \}\rightarrow k=0 \\\rightarrow x=\frac{1-k}{2}=\frac{1}{2}$$
| Note that for any integer $n$, $\;\lfloor x+n\rfloor=\lfloor x\rfloor+n $,and that $$\lfloor 2x\rfloor=\begin{cases}2\lfloor x\rfloor&\text{if}\enspace 0\le x-\lfloor x\rfloor<\dfrac12,\\2\lfloor x\rfloor+1&\text{if}\enspace \dfrac12\le x-\lfloor x\rfloor<1.\end{cases} $$
The given equation implies $\;1=\bigl\lfloor\lfloor x\rfloor+ 2x\bigr\rfloor=\lfloor x\rfloor+\lfloor 2x\rfloor=\begin{cases}3\lfloor x\rfloor&\text{if}\enspace 0\le x-\lfloor x\rfloor<\dfrac12,\\3\lfloor x\rfloor+1 &\text{if}\enspace\dfrac12\le x-\lfloor x\rfloor<1 \end{cases}.$
The first case cannot happen, so necessarily $\lfloor x\rfloor=0\;$ and $\;\dfrac12\le x-\lfloor x\rfloor<1$. The equation becomes $\;2x=1$, whence
$$x=\smash{\frac12}.$$
| {
"language": "en",
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Find, in terms of $s$,the coordinates of the point where this normal cuts the curve again.
a) Find the equation of the normal at the point $(2s,\frac{2}{s})$ to the curve whose parametric equations are $x=2s,y=\frac{2}{s}$
b) Find, in terms of $s$,the coordinates of the point where this normal cuts the curve again.
I didn't have any problems with the part a) using implicit differentiation and the fact that $y-y_1=m(x-x_1)$ to get the equation of the normal:
$y-\frac{2}{s}=s^2(x-2s)$
which is correct according to the answers at the back of the book. I also know that the normal will cut the curve again at a point which satisfies the equation of the normal and $x=2s$ and $y=\frac{2}{s}$
I'm just having a bit of problem substituting these values of x and y into the equation of the normal and basically getting zero. I must be going about this the wrong way here - the textbook gives the answer as $(-\frac{2}{s^3},-2s^3)$
Any hints/explanations much appreciated.
| The parametric equation of the curve is $x=2s$ & $y=\frac{2}{s}$
Now, the equation of the curve in cartesian coordinates is obtained by eliminating $s$ as follows $$y=\frac{2}{\frac{x}{2}}=\frac{4}{x}$$ $$\implies \color{blue}{xy=4}$$ Above, equation represents a rectangular hyperbola.
Now, differentiating the equation w.r.t. $x$ to obtain the slope of tangent
$$\frac{d}{dx}(xy)=\frac{d}{dx}(4)$$ $$x\frac{dy}{dx}+y\frac{d}{dx}(x)=0$$ $$\frac{dy}{dx}=\frac{-y}{x}$$ Hence, the slope of normal at the point $\left(2s, \frac{2}{s}\right)$ $$=\left[\frac{-1}{\frac{dy}{dx}}\right]_{x=2s, y=\frac{2}{s}}=\left[\frac{-1}{\frac{-y}{x}}\right]_{x=2s, y=\frac{2}{s}}=\frac{2s}{\frac{2}{s}}=s^2$$ Hence, the equation of the normal pasing through the point $\left(2s, \frac{2}{s}\right)$ & having slope $m=s^2$ $$y-\frac{2}{s}=s^2(x-2s)$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{s^3x-sy+2(1-s^4)=0}}$$
Now, solving the equations of the curve & the normal as follows $$s^3x-s\frac{4}{x}+2(1-s^4)=0$$ $$s^3x^2+2(1-s^4)x-4s=0$$ Now, solving above quadratic equation for $x$ as follows $$x=\frac{-2(1-s^4)\pm\sqrt{4(1-s^4)^2-4(s^3)(-4s)}}{2s^3}$$ $$x=\frac{-2(1-s^4)\pm 2(1+s^4)}{2s^3}\iff x=2s, x=\frac{-2}{s^3}$$ Now, substituting the values of $x$ we get corresponding y-coordinates $$x=2s\implies y=\frac{4}{2s}=\frac{2}{s}$$
$$x=\frac{-2}{s^3}\implies y=\frac{4}{\frac{-2}{s^3}}=-2s^3$$ Hence the points of intersection of normal & curve are $\left(2s, \frac{2}{s}\right)$ & $\left( \frac{-2}{s^3}, -2s^3\right)$
Hence it is clear that the normal at point $\left(2s, \frac{2}{s}\right)$ cuts the curve at another point given as
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\left(\frac{-2}{s^3}, -2s^3\right)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$A+B+C=2149$, Find $A$ In the following form of odd numbers
If the numbers
taken from the form where $A+B+C=2149$
Find $A$
any help will be appreciate it, thanks.
| $A$ is going to be the $k$th entry in the $n$th row, where $1 \leq k \leq n$. That will make $B$ the $k$th entry in the $n+1$st row, and $C$ the $k+1$st entry in the $n+1$st row.
So the first question is, can we write a formula for $A$ in terms of $k$ and $n$? If we consider the first entry in the $n$th row, there are $1 + 2 + \cdots + (n-1)$ odd numbers before it, which equals $\frac{n(n-1)}{2}$. So the first entry in the $n$th row must be $2\left(\frac{n(n-1)}{2}\right) + 1 = n^2 - n + 1$. Now going over to the $k$th entry will add $2(k - 1)$ to this, so $A = n^2 - n + 1 + 2(k-1) = n^2 - n + 2k - 1$
This formula also implies that $B = (n+1)^2 - (n+1) + 2k - 1$ and $C = (n+1)^2 - (n+1) + 2(k+1) - 1$.
If we put this all together, $A + B + C = (n^2 - n + 2k - 1) + (n^2 + n + 2k - 1) + (n^2 + n + 2k + 1) = 3n^2 + n + 6k - 1$.
So can we solve:
$$3n^2 + n + 6k - 1 = 2149$$
Or:
$$3n^2 + n + 6k = 2150$$
Solving for $k$:
$$k = \frac{2150 - 3n^2 - n}{6}$$
Now we need $k$ to be an integer, and $1 \leq k \leq n$. Since $k$ is an integer we know $n$ should be $2$ mod 3, and we know $3n^2 \leq 2150$, so $n^2 \leq 717$. So $n < 27$. Now we guess, starting with the highest value of $n$ which is less than $27$ and $2$ mod 3, so $n = 26$. Plugging this in we get $k = 16$, which works!
This gives $A = 26^2 - 26 + 32 - 1 = 681$.
| {
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"url": "https://math.stackexchange.com/questions/1384690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How do you solve for θ in the equation $\tan \frac{\theta}{5} + \sqrt{3} = 0$ $$\tan \frac{\theta}{5} + \sqrt{3} = 0$$
Alright so the $\frac{\theta}{5}$ is confusing me.
Would it be wrong to do
\begin{eqnarray}
\tan \frac{\theta}{5}&=&-\sqrt{3}\\
\frac{\theta}{5}&=&\tan^{-1}(-\sqrt{3})\\
\theta&=& 5\tan^{-1}(-\sqrt{3})\\
\theta&=& -\frac{5\pi}{3} + 5\pi n
\end{eqnarray}
| Notice, $$\tan \frac{\theta}{5}+\sqrt 3=0$$ $$\tan \frac{\theta}{5}=-\sqrt 3=-\tan \frac{\pi}{3}$$ Writing the general solution for $\theta$ as follows $$\frac{\theta}{5}=n\pi-\frac{\pi}{3}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\theta=5n\pi-\frac{5\pi}{3}}}$$
Where, $n$ is any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Closed form for an infinite product The following fascinating formula appears in the paper "On gamma quotients and infinite products" by M.Chamberland and A.Straub (see page 9):
$$\prod_{n=2}^\infty\left(1-\frac{1}{n(n-1)}\right)=-\frac1\pi\cos\left(\frac{\sqrt5}{2}\pi\right).$$
Unfortunately, it's given without any reference or explanation. Could anyone give one?
| This formula is an application of the theorem $1.1$ for $n=2$ from your Chamberland & Straub paper (when $a+b=c+d$) :
$$\tag{1}\prod_{k\ge 0}\frac {(k+a)(k+b)}{(k+c)(k+d)}=\frac {\Gamma(c)\Gamma(d)}{\Gamma(a)\Gamma(b)}$$
Observe first that : $\quad\displaystyle n^2-n-1=\left(n-\frac{\sqrt{5}+1}2\right)\left(n+\frac{\sqrt{5}-1}2\right)\;$
An idea is to use $\quad a:=-\dfrac{\sqrt{5}+1}2,\;b:=\dfrac{\sqrt{5}-1}2$.
note that $\;a+b=-1\;$ and take $c=0,\;d=-1$ but we would have $\;0\cdot (-1)\;$ at the denominator so let's shift all these values with an offset $2$ (i.e. set $n:=k+2$) then :
$$a:=2-\frac{\sqrt{5}+1}2,\;b:=2+\frac{\sqrt{5}-1}2,\;c:=2,\;d:=1$$
we then get :
\begin{align}
\prod_{k\ge 0}\frac {(k+2)^2-(k+2)-1}{(k+2)(k+1)}&=\frac {\Gamma(2)\Gamma(1)}{\Gamma\left(2-\frac{\sqrt{5}+1}2\right)\Gamma\left(2+\frac{\sqrt{5}-1}2\right)}\\
&=\frac 1{\left(1-\frac{\sqrt{5}+1}2\right)\left(1+\frac{\sqrt{5}-1}2\right)\Gamma\left(1-\frac{\sqrt{5}+1}2\right)\Gamma\left(1+\frac{\sqrt{5}-1}2\right)}\\
\tag{2}&=-\frac 1{\Gamma\left(1-\frac{\sqrt{5}+1}2\right)\Gamma\left(\frac{\sqrt{5}+1}2\right)}
\end{align}
Using the Euler reflexion formula $\displaystyle \Gamma(z)\Gamma(1-z)=\dfrac{\pi}{\sin(\pi z)}$ for $z=\dfrac {\sqrt{5}+1}2$ allows to get :
$$\prod_{n\ge 1}\frac {(n+1)^2-(n+1)-1}{n(n+1)}=-\frac {\sin\left(\pi(\sqrt{5}+1)/2\right)}{\pi}$$
and conclude that indeed :
$$\tag{3}\prod_{n\ge 1}\left(1-\frac 1{n(n+1)}\right)=-\frac {\sin\left(\pi(\sqrt{5}+1)/2\right)}{\pi}=-\frac1\pi\cos\left(\frac{\sqrt5}{2}\pi\right)$$
| {
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"source": "stackexchange",
"question_score": "15",
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Show that $\sqrt{4+2\sqrt{3}}-\sqrt{3}$ is rational using the rational zeros theorem What I've done so far:
Let $$r = \sqrt{4+2\sqrt{3}}-\sqrt{3}.$$
Thus, $$r^2 = 2\sqrt{3}-2\sqrt{3}\sqrt{4+2\sqrt{3}}+7$$
and $$r^4=52\sqrt{3}-28\sqrt{3}\sqrt{4+2\sqrt{3}}-24\sqrt{4+2\sqrt{3}}+109.$$
I did this because in a similar example in class, we related $r^2$ and $r^4$ to find a polynomial such that $mr^4+nr^2 = 0$ for some integers $m,n$. However, I cannot find such relation here. Am I doing this right or is there another approach to these type of problems.
|
is there another approach to these type of problems
Since
$$4=1+3,3=1\times 3$$
we can have
$$4+2\sqrt 3=1+3+2\sqrt{1\times 3}=(1+\sqrt 3)^2$$
| {
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"source": "stackexchange",
"question_score": "11",
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How to find eigenvalues of matrix $\begin{bmatrix} 3& a+1\\a+1&3 \end{bmatrix}$ I want to find the eigenvalues of the following matrix
$\begin{bmatrix} 3& a+1\\a+1&3 \end{bmatrix}$ expressed in $a$ using $\begin{bmatrix} \lambda - 3& a+1\\a+1&\lambda-3 \end{bmatrix}$.
But the $a$-term makes it difficult for me to find it. I hope someone can show me how to do this. Thanks in advance !
EDIT:
Determinant = $4a^2+8a+4$
Using the abc rule I get
$(6\overset{+}{-} \sqrt{4a^2+8a+4})/2$
| \begin{align}
\left| \begin{matrix} \lambda -3 & a+1 \\
a+1 &\lambda-3 \end{matrix} \right| & = (\lambda-3)(\lambda-3) - (a+1)(a+1) \\
&= \lambda^2-6\lambda+9-(a^2+2a+1) \\
&= \lambda^2-6\lambda+9-a^2-2a-1 \\
&= \lambda^2-6\lambda-a^2-2a+8 \\
&= \lambda^2-6\lambda-(a^2+2a-8)\\
&= \lambda^2-6\lambda-(a+4)(a-2) \\
&= (\lambda -a-4)(\lambda +a-2) =0\\
\end{align}
Thus $\lambda = a+4$ and $\lambda = -a+2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integrating Factor by Inspection $(x^3+xy^2+y)dx + (y^3+xy^2+x)dy=0$ $$(x^3+xy^2+y) \hspace{.1cm} dx + (y^3+xy^2+x)\hspace{.1cm} dy=0$$ So I tried to solve this problem but can't figure out my integrating factor all I can see here is if I distribute first I can get a $y \hspace{.1cm} dx + x dy$ so that would be $d(xy)$ however I can't integrate the $xy^2dx$ and $xy^2dy$. Help please and can you also give me techniques on how to know the integrating Factors? My professors just told me that it's a trial and error process, is that true?
| Hint:
$(x^3+xy^2+y)~dx+(y^3+xy^2+x)~dy=0$
$\dfrac{dx}{dy}=-\dfrac{y^3+xy^2+x}{x^3+xy^2+y}$
With reference to http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=180:
Let $u=\dfrac{x}{y}$ ,
Then $x=yu$
$\dfrac{dx}{dy}=y\dfrac{du}{dy}+u$
$\therefore y\dfrac{du}{dy}+u=-\dfrac{y^3+y^3u+yu}{y^3u^3+y^3u+y}$
$y\dfrac{du}{dy}=-\dfrac{y^3(u+1)+yu}{y^3(u^3+u)+y}-u$
$y\dfrac{du}{dy}=-\dfrac{y^3(u^4+u^2+u+1)+2yu}{y^3(u^3+u)+y}$
$\dfrac{du}{dy}=-\dfrac{y^2(u^4+u^2+u+1)+2u}{y^3(u^3+u)+y}$
Let $v=\dfrac{1}{y^2}$ ,
Then $\dfrac{du}{dy}=\dfrac{du}{dv}\dfrac{dv}{dy}=-\dfrac{2}{y^3}\dfrac{du}{dv}$
$\therefore-\dfrac{2}{y^3}\dfrac{du}{dv}=-\dfrac{y^2(u^4+u^2+u+1)+2u}{y^3(u^3+u)+y}$
$\left(\dfrac{2}{y^2}+u^3+u+1+\dfrac{1}{u}\right)\dfrac{dv}{du}=\dfrac{2}{uy^4}+\dfrac{2(u^2+1)}{y^2}$
$\left(2v+u^3+u+1+\dfrac{1}{u}\right)\dfrac{dv}{du}=\dfrac{2v^2}{u}+2(u^2+1)v$
This belongs to an Abel equation of the second kind.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the integral $\int_0^{\frac{\pi}{2}} \frac{1}{a + \sin^2 \theta}$ using the residue theorem Can someone show me how to compute this integral using the residue theorem:
$$\int_0^{\frac{\pi}{2}} \frac{1}{a + \sin^2 \theta}d\theta$$
| We can simplify the problem by (i) exploiting the even symmetry of the integrand, (ii) using the identity
$$\sin^2 \theta =\frac{1-\cos 2\theta}{2}$$
and (iii) enforcing the substitution $2\theta \to \theta$. We can write then write the integral of interest $I(a)$ as
$$I(a)=\frac12 \int_0^{2\pi}\frac{1}{(2a+1)-\cos \theta}d\theta \tag 1$$
Then, we move to the complex plane and letting $z=e^{i\theta}$ find that
$$I(a) = i\oint_{|z|=1} \frac{1}{z^2-2(2a+1)z+1}dz$$
For $a>0$ ($a<-1$), the only pole enclosed in the unit circle is at
$$z=(2a+1)-2\sqrt{a(a+1)}\,\,\,\,\left(z=(2a+1)+2\sqrt{a(a+1)}\right)$$
Therefore the residue of $\frac{1}{z^2-2(2a+1)z+1}$ is for $a>1$
$$\begin{align}
\lim_{z\to (2a+1)-2\sqrt{a(a+1)}}\left(\frac{z-(2a+1)+2\sqrt{a(a+1)}}{z^2-2(2a+1)z+1}\right)=\frac{1}{-4\sqrt{a(a+1)}}
\end{align}$$
and for $a<-1$
$$\begin{align}
\lim_{z\to (2a+1)+2\sqrt{a(a+1)}}\left(\frac{z-(2a+1)-2\sqrt{a(a+1)}}{z^2-2(2a+1)z+1}\right)=\frac{1}{4\sqrt{a(a+1)}}
\end{align}$$
Putting it all together yields for $a>0$ or $a<-1$
$$\bbox[5px,border:2px solid #C0A000]{I(a)=(2\pi i) i\,\text{sgn(a)}\,\frac{1}{-4\sqrt{a(a+1)}}=\text{sgn}(a)\,\frac{\pi}{2\sqrt{a(a+1)}}}$$
NOTE:
For $-1\le a\le 0$, the integral of interest is divergent.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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How can I prove remainder side of this inequality? Let $x$, $y$ be two positive numbers such that $x^4+y^4=x^2+y^2$. Prove that
$$1\leqslant x+y\leqslant 2.$$
With $x+y\leqslant 2$. I tried
We have $$x^4+y^4\geqslant \dfrac{(x^2+y^2)^2}{2}.$$
Therefore, $$x^2+y^2\geqslant \dfrac{(x^2+y^2)^2}{2}$$
or
$$(x^2+y^2)^2-2(x^2+y^2)\leqslant 0$$ Implies $x^2+y^2\leqslant 2.$
Another way
$$\dfrac{x+y}{2} \leqslant \sqrt{\dfrac{x^2+y^2}{2}}=1.$$
Thus $x+y\leqslant 2.$
How can I prove $x+y\geqslant 1$?
Is this true?
Let $x$, $y$ be two positive numbers such that $x^m+y^m=x^n+y^n$, where $m$, $n$, ($m \neq n$) be two positive integer numbers, we have
$$ x+y\leqslant 2.$$
| let $x+y=u,xy=v^2 \implies u^2 \ge 4v^2 \\ x^2+y^2=u^2-2v^2,x^4+y^4=(x^2+y^2)^2-2(xy)^2 =(u^2-2v^2)^2-2v^4=u^4-4u^2v^2+2v^4 \implies \\
u^4-(4v^2+1)u^2+2v^4+2v^2=0 \\ u^2=\dfrac{4v^2+1+ \sqrt{8v^4+1}}{2}$
$u^2=\dfrac{4v^2+1-\sqrt{8v^4+1}}{2} \le 2v^2$
$v^2\ge 0 \implies
u^2=\dfrac{4v^2+1+ \sqrt{8v^4+1}}{2} \ge \dfrac{1+1}{2}=1$ when $v^2=0$
| {
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If 2 roots of the equation $(p-1)(x^2+x+1)^2-(p+1)(x^4+x^2+1)$ are real and distinct and $f(x)=\frac{1-x}{1+x}$... Question: If 2 roots of the equation $(p-1)(x^2+x+1)^2-(p+1)(x^4+x^2+1)$ are real and distinct and $f(x)=\frac{1-x}{1+x}$, then $f(f(x))+f(f(\frac{1}{x})) = ?$
(a)p
(b)2p
(c)-p
(d)-2p
Attempt: the give equation can be written as $(x^2+x+1)[(p-1)(x^2+x+1)-(p+1)(x^2-x+1)] = 2(x^2+x+1)(-x^2+px-1)$
$f(f(x))=x$ and $f(f(\frac{1}{x}))=-\frac{1}{x}$
So the required expression $x-\frac{1}{x}$
Now, $2(x^2+x+1)(x^2-px+1)=0$ (taking $-$ outside of second brackett) and $(x^2+x+1)\neq0$, so $(x^2-px+1)=0$. For this equation to have distinct real roots, $p^2-4>0$
But I am getting nowhere near the given options.
| $\bf{My\; Solution::}$ Given
$$(p-1)(x^2+x+1)^2-(p+1)(x^4+x^2+1) = 0$$
$$\displaystyle (p-1)(x^2+x+1)^2=(p+1)(x^4+x^2+1)$$
We can write it as $$\displaystyle \frac{p+1}{p-1}=\frac{(x^2+x+1)^2}{(x^4+x^2+1)}=\frac{(x^2+x+1)^2}{(x^2-x+1)(x^2+x+1)}=\frac{x^2+x+1}{x^2-x+1}$$
Now We can Write it as $$\displaystyle \frac{\left(x+\frac{1}{x}+1\right)}{\left(x+\frac{1}{x}-1\right)}=\frac{p+1}{p-1}$$
Now Using Componendo-Dividendo, We get
$$\displaystyle \frac{\left(x+\frac{1}{x}+1\right)+\left(x+\frac{1}{x}-1\right)}{\left(x+\frac{1}{x}+1\right)-\left(x+\frac{1}{x}-1\right)}=\frac{(p+1)+(p-1)}{(p+1)-(p-1)}$$
So we get $$\displaystyle x+\frac{1}{x} = p..................(1)$$
Now Given $$\displaystyle f(x) = \frac{1-x}{1+x}$$. So $\displaystyle f(f(x)) = x$ and $\displaystyle f\left(f\left(\frac{1}{x}\right)\right)=\frac{1}{x}$
So $$\displaystyle f(x)+f\left(f\left(\frac{1}{x}\right)\right)=x+\frac{1}{x}=p$$ From $(1)$ equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Method of partial fractions when denumerator cannot be factorized? Suppose I'm given an expression: \begin{align*} P(x) = \frac{1+x}{1-2x-x^2}. \end{align*} The denumerator cannot be readily factorized, so I found the zeros, which are \begin{align*} \lambda_1 = -1 - \sqrt{2} \qquad \lambda_2 = -1 + \sqrt{2}. \end{align*}
I want to apply the method of partial fraction to $P(x)$ now. I wanted to write \begin{align*} P(x) = \frac{1+x}{1-2x-x^2} = \frac{A}{(x- \lambda_1)} + \frac{B}{(x- \lambda_2)} \end{align*} but this equality does not hold. Since $(x-\lambda_1) (x- \lambda_2) = x^2 + 2x -1 \neq 1-2x-x^2$. So what denumerators should I choose to go with $A$ and $B$?
| $$\frac{1+x}{1-2x-x^2}=\frac{-1-x}{x^2+2x-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
$n^{th}$ derivative of $y=x^2\cos x$ I am stuck with Leibniz formula
$$D^{n}y = \sum_{k=0}^{n} \binom{n}{k} \, x^{(2k)}\cos^{(n-k)}x$$
Could someone show how to do it?
| I would use the fact that:
$$\cos^{(n)}(x)=\cos\Bigl(x+\frac{n\pi}2\Bigr)$$
and that $$(x^2)'=2x,\quad (x^2)''=2,\quad (x^2)^{(n)}=0\enspace\text{si}\enspace n>2.$$
Thus Leibniz's rule gives:
$$\bigl(x^2\cos x\bigr)^{(n)} =x^2\cos^{(n)}(x)+2nx\cos^{(n-1)}(x)+n(n-1)\cos^{(n-2)}(x)$$
Also, note that $\;\cos^{(n-2)}(x)=-\cos^{(n)}(x)$, hence
\begin{align*}\bigl(x^2\cos x\bigr)^{(n)}&=\bigl(x^2-n(n-1)\bigr)\cos^{(n)}(x)+2nx\cos^{(n-1)}(x)\\
&=\bigl(x^2-n(n-1)\bigr)\cos\Bigl(x+\frac{n\pi}2\Bigr)+2nx\cos\Bigl(x+\frac{(n-1)\pi}2\Bigr)
\end{align*}
There remains to consider the different cases according to the values of $n$ modulo $4$:
$$\text{if }\begin{cases}
n\equiv 0&\bigl(x^2\cos x\bigr)^{(n)} =\bigl(x^2-n(n-1)\bigr)\cos x+2nx\sin x\\
n\equiv 1&\bigl(x^2\cos x\bigr)^{(n)} =2nx\cos x-\bigl(x^2-n(n-1)\bigr)\sin x\\
n\equiv 2&\bigl(x^2\cos x\bigr)^{(n)} =\bigl(n(n-1)-x^2\bigr)\cos x-2nx\sin x\\
n\equiv 3&\bigl(x^2\cos x\bigr)^{(n)} = -2nx\cos x+\bigl(x^2-n(n-1)\bigr)\sin x\\
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\left (1+\frac ba \right )\left (1+\frac ac \right )\left ( 1+ \frac cb\right )\ge 4+3\sqrt2$ Given $a,b,c>0$ and $a^2\ge b^2+c^2$. Prove that $$\left (1+\frac ba \right )\left (1+\frac ac \right )\left ( 1+ \frac cb\right )\ge 4+3\sqrt2$$
This is my try:
I expanded the LHS, and I have to show that $\displaystyle\frac a b +\frac a c +\frac b c +\frac b a +\frac c a +\frac c b \geq 2+3\sqrt2$
I used AM-GM: $\displaystyle \frac{a}{\sqrt2b}+\frac{\sqrt2b}{a}\ge 2, \quad \color{red}{\frac{\left (2-\sqrt2 \right )a}{2b}+\frac{\left ( 1-\sqrt2 \right )b}{a}}$
But it's not good. Then I don't know how.
P/s: I have read this but they are different.
| We have to show $\displaystyle\frac a b +\frac a c +\frac b c +\frac b a +\frac c a +\frac c b \geq 2+3\sqrt2\quad\color{red}{(1)}$
$(1)\iff \displaystyle \left ( b+c \right )\left ( \frac{a}{bc}+\frac{1}a \right )+\left (\frac b c +\frac c b \right ) \ge2+3\sqrt2$
We know that $\displaystyle {b+c\ge 2\sqrt{bc},\quad \frac{a}{2bc}+\frac{a}{2bc}+\frac{1}a\ge3\sqrt[3]{\frac{a}{4b^2c^2}}\ge 3\sqrt[3]{\frac{\sqrt{2bc}}{4b^2c^2}}=\frac{3}{\sqrt{2bc}}\\\implies \left ( b+c \right )\left ( \frac{a}{bc}+\frac{1}a \right )\ge 3\sqrt2,\quad \left (\frac b c +\frac c b \right ) \ge 2}$
Remember that $a^2\ge b^2+c^2\ge 2bc$.
So we have the Q.E.D
The equality happen $\iff a^2=b^2+c^2$ and $b=c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve trigonometric equation $ 3 \cos x + 2\sin x=1 $ Solve trigonometric equation: $$ 3 \cos x + 2\sin x=1 $$
I tried to substitue $\cos x = \dfrac{1-t^2}{1+t^2}, \sin x = \dfrac{2t}{1+t^2}$. Yet with no results.
| Let $u = \cos x$ and $v = \sin x$. Thus, $3u + 2v = 1$ and $u^2 + v^2 = 1$. Isolating one of the variables , we have
$$
u^2 + \dfrac{(1 - 3u)^2}{4} = 1 \quad \Rightarrow \quad 13u^2 - 6u - 3 = 0
$$
and so on...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Solve the binomial equation Solve the binomial equation $$z^4 = -8$$
Below is the steps i have done
1: I have taken |-8| that is 8 and then done 8^(1/4) which is 2^(1/4).
2: Since $z=r(cos\alpha+isin\alpha)$ leads me to
$r^4(cos4\alpha+isin4\alpha)=-8(cos\pi/2+isin\pi/2)$
Divide by 4 since the z term is raised by four gives $2^{1/4}(cos\pi/8 + k * \pi/2 +sin\pi/8 + k * \pi/2) $
Is this the correct way to solve this problem ? I am asking since i just started with binomic equations and been stuck for some hours with the question.
| Notice, $$z^4=-8$$
$$z^4=8i^2$$ $$z=\sqrt[4]{8i^2}$$
$$z=(2)^{3/4}\sqrt{i}$$ $$z=(2)^{3/4}\sqrt{\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}}$$
$$=(2)^{3/4}\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)^{1/2}$$
$$=(2)^{3/4}\left(\cos\left(2k\pi+\frac{\pi}{2}\right)+i\sin\left(2k\pi+\frac{\pi}{2}\right)\right)^{1/2}$$
$$=(2)^{3/4}\left(\cos\left(\frac{(4k+1)\pi}{2}\right)+i\sin\left(\frac{(4k+1)\pi}{2}\right)\right)^{1/2}$$
$$=(2)^{3/4}\left(\cos\left(\frac{(4k+1)\pi}{4}\right)+i\sin\left(\frac{(4k+1)\pi}{4}\right)\right)$$
Setting $k=0$, we get $$z=(2)^{3/4}\left(\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\right)$$
$$=(2)^{3/4}\left(\frac{1}{\sqrt 2}+i\frac{1}{\sqrt 2}\right)$$
$$z=(2)^{1/4}\left(1+i\right)$$
Setting $k=1$, we get $$z=(2)^{3/4}\left(\cos\left(\frac{5\pi}{4}\right)+i\sin\left(\frac{5\pi}{4}\right)\right)$$
$$=(2)^{3/4}\left(\frac{-1}{\sqrt 2}+i\frac{-1}{\sqrt 2}\right)$$
$$z=-(2)^{1/4}\left(1+i\right)$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{z=(2)^{1/4}\left(1+i\right),\ z=-(2)^{1/4}\left(1+i\right)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Sum of an infinite series. I have found a series whose $n$-th term is denoted by $\dfrac{n}{a^n}$. Here $a$ is a constant. I tried to find the sum but failed. Is there any formula for its infinite sum or just an approximation?
| notice, we have $$T_n=\frac{n}{a^n}$$ $$S_n=\sum_{n=1}^{n} T_n=\sum_{n=1}^{n}\frac{n}{a^n}$$ $$S_n=\frac{1}{a}+\frac{2}{a^2}+\frac{3}{a^3}+\frac{4}{a^4}+\ldots +\frac{n}{a^n}\tag 1$$ Multiplying by $\frac{1}{a}$ & rewriting as follows
$$\frac{1}{a}S_n= \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{a^2}+\frac{2}{a^3}+\frac{3}{a^4}+\frac{4}{a^5}+\ldots +\frac{n-1}{a^{n}}+\frac{n}{a^{n+1}}\tag 2$$ Now, subtracting the corresponding terms of (2) from (1) column wise, we get
$$S_n-\frac{1}{a}S_n=\underbrace{\frac{1}{a}+\frac{1}{a^2}+\frac{1}{a^3}+\ldots+\frac{1}{a^n}}_{\text{n-terms in G.P.}}-\frac{n}{a^{n+1}}$$
$$\left(\frac{a-1}{a}\right)S_n=\frac{\frac{1}{a}\left(\frac{1}{a^n}-1\right)}{\frac{1}{a}-1}-\frac{n}{a^{n+1}}$$
$$S_n=\left(\frac{a}{a-1}\right)\left\{\frac{a^n-1}{a^n(a-1)}-\frac{n}{a^{n+1}}\right\}$$
Hence, taking the limit as $n\to \infty$, the sum of infinite terms $S_{\infty}$ is given as follows $$S_{\infty}=\lim_{n\to \infty}S_n$$
$$=\lim_{n\to \infty}\left(\frac{a}{a-1}\right)\left\{\frac{a^n-1}{a^n(a-1)}-\frac{n}{a^{n+1}}\right\}$$
$$=\left(\frac{a}{a-1}\right)\lim_{n\to \infty}\left\{\frac{a^n-1}{a^n(a-1)}-\frac{n}{a^{n+1}}\right\}$$
$$S_{\infty}=\frac{1}{(a-1)^2}\lim_{n\to \infty}\left\{\frac{a^{n+1}-n(a-1)-a}{a^n}\right\}$$
It is obvious that the series converses for $|a|\leq 1$ otherwise diverges
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Area between the curve and x-axis Find the area between the curve $y={2\over x-1}-1$ and the $x$-axis over the interval $[2,4]$
Would this solve the question $\int_{2}^4( {2\over x-1}-1)dx$ ?
So...
$[2\ln(x-1)-x+c]_2^4$ would be incorrect because the curve would be below the x-axis for $3$ to $4$
| As I mentioned in the comment, you should divide the interval $[2, 4]$ to two parts,
\begin{align*}S = &\int_2^3\left(\frac{2}{x - 1} - 1\right) dx \;{\color{red} -}\int_3^4\left(\frac{2}{x - 1} - 1\right)dx \\
= & \left[2\log(3 - 1) - 2\log(2 - 1) - 1\right] - \left[2\log(4 - 1) - 2\log(3 - 1) - 1\right] \\
= & 4\log 2 - 2\log 3.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve trigonometric inequality $\cos x \geq \sin^2 x - \cos^2 x $ Solve trigonometric inequality $$\cos x \geq \sin^2 x - \cos^2 x $$
My incorrect solution:
$$\cos^2 x-\sin^2 x \geq -\cos x $$
$$\cos 2x \geq \cos (\pi - x) $$
which means:
$$ 2x \geq -(\pi + x)$$
$$ x \geq -\pi $$
Which is wrong.
And
$$ 2x \leq 2\pi + (\pi - x)$$
$$ x \leq \pi$$
| $$0\le\cos x+\cos2x=2\cos\dfrac{3x}2\cos\dfrac x2=2\cos^2\dfrac x2\left(4\cos^2\dfrac x2-3\right)$$
$$\implies0\le4\cos^2\dfrac x2-3=2(1+\cos x)-3=2\cos x-1\iff\cos x\ge\dfrac12$$
$2n\pi-\dfrac\pi3\le x\le2n\pi+\dfrac\pi3$ where $n$ is any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
Advanced Algebra Manipulation/Inequality Proof: $\frac{4x^3(x^2+y^2)-2x(x^4+y^4)}{(x^2+y^2)^2} \leq 6|x|$ I need to show that
$$\frac{4x^3(x^2+y^2)-2x(x^4+y^4)}{(x^2+y^2)^2} \leq 6|x|$$
by starting with the left side of the inequality and working from there.
Hints from the textbook said to work from these inequality "tricks,"
$$2|ab| \leq a^2 + b^2$$
$$|a| + |b| \leq \sqrt{2} \sqrt{a^2+b^2}$$
And the triangle inequality,
$$|a+b| \leq |a| + |b|$$
Expanding the numerator gives us
$$\frac{2x^5 + 4x^3y^2 - 2xy^4}{(x^2 + y^2)^2}$$
For starters, I use the triangle inequality theorem to make the numerator look like this:
$$|2x^5 + 4x^3y^2 - 2xy^4| \leq |2x^5|+|4x^3y^2|+|2xy^4|$$
Also, in regard to the denominator, since it is even, it is always positive and thus equal to its absolute value. Then, I multiply both sides by the denominator, $(x^2+y^2)^2$. This gives us
$$|2x^5|+|4x^3y^2|+|2xy^4| \leq 6|x||x^4 + 2x^2y^2+y^4|$$
$$|2x^5|+|4x^3y^2|+|2xy^4| \leq 6|x^5+2x^3y^2+xy^4|$$
My question is, since $|a+b| \leq |a|+|b|$, I feel as though I cannot violate that rule and am stuck. What trick(s) can I use from here? If there were no absolute value symbols, it would be as simple as simplifying the inequality since the orders match up. However, it would not be true for all (x,y) as the right side of the inequality is odd.
| Interestingly, this inequality is not tight. In fact, we can show that
$$\left|\frac{4x^3(x^2+y^2)-2x(x^4+y^4))}{(x^2+y^2)^2}\right|\le 2|x|$$
To that end, we write
$$\begin{align}
\frac{4x^3(x^2+y^2)-2x(x^4+y^4))}{(x^2+y^2)^2}&=\frac{2x}{(x^2+y^2)^2}\left(2x^2(x^2+y^2)-(x^4+y^4)\right)\\\\
&=\frac{2x}{(x^2+y^2)^2}\left(2(x^2+y^2-y^2)(x^2+y^2)\\
\,\,\,\,\,\,\,\,\,\,-((x^2+y^2)^2-2x^2y^2)\right)\\\\
&=2x\left(1-\frac{y^4}{(x^2+y^2)^2}\right) \tag 1
\end{align}$$
It is easy to show that the term in parentheses on the right-hand side of $(1)$ satisfies the inequality
$$0\le \left(1-\frac{y^4}{(x^2+y^2)^2}\right)\le 1$$
We simply note that $x^2+y^2\ge y^2$ so that $0\le\dfrac{y^4}{(x^2+y^2)^2}\le 1$ and therefore $1\ge 1-\dfrac{y^4}{(x^2+y^2)^2}\ge 0$.
Thus, we have for $x>0$
$$0<2x\left(1-\frac{y^4}{(x^2+y^2)^2}\right)\le 2x \tag 2$$
while for $x<0$, we have
$$2x\le 2x\left(1-\frac{y^4}{(x^2+y^2)^2}\right)<0 \tag 3$$
Putting $(1)$, $(2)$, and $(3)$ together yields
$$\bbox[5px,border:2px solid #C0A000]{\left|\frac{4x^3(x^2+y^2)-2x(x^4+y^4))}{(x^2+y^2)^2}\right|\le 2|x|}$$
which provided a tighter inequality than that one that was asked to be shown.
| {
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"url": "https://math.stackexchange.com/questions/1394985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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nice classical nonhomogeneous inequality Let $a,b,c$ be positive reals and $abc=1$. Prove that $$10(a^4+b^4+c^4)+21\ge 17(a^3+b^3+c^3).$$
I have found a solution using MV and I'm wondering if there is a nice solution.
| This is just a solution, not necessarily a nice one. I started by simple rearrangement, and I got
$$ 10(a^4+b^4+c^4 - 3abc) \geq 17(a^3+b^3+c^3 - 3abc)$$ but I couldn't think of any well-known inequality that can prove this. So I decided to do straight calculus:
Let $g(a,b,c) = 1-abc$ and $f(a,b,c)=10*(a^4+b^4+c^4)-17(a^3+b^3+c^3)$. By Lagrange Theorem, to find the extremal points we need to
$$
\begin{cases}
g = 0\\
\nabla f = \lambda \nabla g
\end{cases} \Leftrightarrow \begin{cases}
abc = 1\\
40a^3 - 51a^2 = \lambda bc\\
40b^3 - 51b^2 = \lambda ac\\
40c^3 - 51c^2 = \lambda ab
\end{cases}
$$
which can be simplified to
$$
\begin{cases}
40a^4 - 51a^3 = \lambda \\
40b^4 - 51b^3 = \lambda \\
40c^4 - 51c^3 = \lambda
\end{cases}
$$
Subtracting first two equations in the system we get $40a^4-51a^3 = 40b^4-51b^3$. If $a=b=c=1$ we observe that $f(1,1,1)+21=0 \geq 0$. If $a=b\not =c$, then we may rewrite our system as
$$
\begin{cases}
a^2 c = 1\\
40a^3 - 51a^2 - \lambda ac = 0\\
40c^3 - 51c^2 - \lambda a^2 = 0
\end{cases}
$$
After simple manipulations and eliminating $\lambda$, we get $40a^8+51a^7+51a^2-40=0$ which has two real roots $-1$ and $\approx 0.7527$ (and 6 other complex roots). The only one satisfying all of our conditions is $a=0.7527$.
Setting the values back into our function yields $f\left(0.7527,0.7527,1.765\right) + 21 = 16.4973 \gt 0$. Now we only need to check the boundary case $f(0,0,1)+21 = 14 \gt >0$. By the Lagrange Theorem the minimum value of the function $f(a,b,c)$ with the bounding condition $g(a,b,c)$ is $-4.5027$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate the integration : $\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx$ $$\int{\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx}$$
$$\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx=\int \frac{(1-\sin x)(2-\sin x)}{\sqrt{(1-\sin x)(2-\sin x)(1+\sin x)(2+\sin x)}}dx$$
I am stuck. Please help me....
| Hint:
$$\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx$$
$$=\int\frac{\cos x\sqrt{4-\sin^2 x}}{(1+\sin x)(2+\sin x)}\,dx$$
( multiplying numerator & denominator by $(1+\sin x)(2+\sin x)$ under square root sign.)
Now put , $\sin x=z$.
Expand Hint :
Then ,
$$=\int\frac{1}{1+z}\sqrt{\frac{2-z}{2+z}}\,dz$$
$$=\int u\sqrt{\frac{3u-1}{u+1}}\,du\text{ , by putting $1+z=\frac{1}{u}.$ }$$
$$=\int\frac{u(3u-1)}{\sqrt{(u+1)(3u-1)}}\,du$$
$$=\int\sqrt{3u^2+2u-1}\,du-\frac{1}{2}\int \frac{d(3u^2+2u-1)}{\sqrt{3u^2+2u-1}}+2\int\frac{\,du}{\sqrt{3u^2+2u-1}}$$
$$=\cdots \cdots \cdots \cdots \cdots $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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What is the residue obtained from this integral? Consider the following integral in two complex variables $z_1$ and $z_2$:
$$\frac{1}{(2\pi i)^2}\oint_{{|z_1|=\epsilon}\atop{|z_2|=\epsilon}}dz_1 dz_2\frac{1}{z_1 z_2\left(1+a+\frac{z_1}{z_1+z_2}\right)\left(1+b+\frac{z_2}{z_1+z_2}\right)}$$
If I perform the contour integration in $z_1$ first, get the residue in $z_1$, and then perform the contour integration in $z_2$ to get the final residue, I get overall:
$$Res_{z_1=0,z_2=0}=\frac{1}{(1+a)(2+b)}$$
If I reverse the process and integrate the $z_2$ contour first, and then $z_1$, I end up with the final result:
$$Res_{z_1=0,z_2=0}=\frac{1}{(2+a)(1+b)}$$
These two overall residues are not the same, which implies that something must have gone wrong in the above calculations. How do I properly compute the actual residue of the integral above? Thanks for any suggestion.
EDIT
It was suggested by Robert Israel to rewrite
$$ \dfrac{1}{1+a+z_1/(z_1 + z_2)} = \dfrac{z_1 + z_2}{a z_1 + a z_2 + 2 z_1 + z_2}$$
and similar for the other term with $b$. This would give additional poles if these denominators vanish within $|z_1|=\epsilon$ and $|z_2|=\epsilon$. However, I meant these extra factors to be purely test functions that do not contribute any extra poles. Therefore, let us concentrate on the parameter region in $a$ and $b$ where the only poles in the above integrand are at $z_1=0$ and $z_2=0$. In this case the problem still stands unresolved.
| (EDITED)
The poles at $z_1 = 0$ and $z_2=0$ are not the only ones inside the circles $|z|=\epsilon$. Note that
$$ \dfrac{1}{z_1 z_2 (1+a+z_1/(z_1 + z_2)(1+b+z_2/(z_1+z_2))} = \dfrac{(z_1 + z_2)^2}{z_1 z_2 ((a+2) z_1 + (a+1) z_2)((b+1) z_1 + (b+2) z_2}$$
so (for fixed $z_2$ on the circle $|z|=\epsilon$) this has poles at
$z_1 = 0$, $z_1 = -z_2 (a + 1)/(a + 2)$ and $z_1 = -z_2 (b+2)/(b+1)$.
On the other hand, for fixed $z_1$ on the circle, it has poles at
$z_2 = 0$, $z_2 = -z_1 (a+2)/(a+1)$ and $z_2 = -z_1 (b+1)/(b+2)$.
Leaving out the cases $|a+2|=|a+1|$ and $|b+1|=|b+2|$ where there is a pole on the circle itself, one of $z_1 = -z_2 (a + 1)/(a + 2)$ and
$z_2 = -z_1 (a+2)/(a+1)$ is inside the circle and the other is outside,
and similarly for the poles involving $b$. The residues for poles inside the circle must be taken into account when doing the integrals.
For example, let's say $|a+1| > |a+2|$ and $|b+1| < |b+2|$.
Then for the $z_1$ integral with fixed $z_2$ on the circle, we need only the pole at $z_1=0$:
$$ \eqalign{\frac{1}{2\pi i} \oint_{|z_1|=\epsilon} dz_1\; f(z_1, z_2) &= \text{res}(f(z_1,z_2); z_1 = 0) = \dfrac{1}{(b+2)(a+1) z_2}\cr
\dfrac{1}{(2\pi i)^2} \oint_{|z_2|=\epsilon} dz_2 \oint_{|z_1|=\epsilon} dz_1\; f(z_1,z_2) &= \text{res}\left(\dfrac{1}{(b+2)(a+1)z_2}; z_2 = 0\right) = \dfrac{1}{(b+2)(a+1)}}$$
In the other order, for the $z_2$ integral with fixed $z_1$ on the circle we need the residues at $z_2 = 0$, $z_2 = -z_1 (a+2)/(a+1)$ and $z_2 = -z_1 (b+1)/(b+2)$.
$$ \eqalign{\frac{1}{2\pi i} \oint_{|z_2|=\epsilon} dz_2\; f(z_1, z_2) =& \text{res}(f(z_1,z_2); z_2 = 0) + \text{res}\left(f(z_1,z_2); z_2 = -z_1 \dfrac{a+2}{a+1}\right) \cr &+ \text{res}\left(f(z_1,z_2); z_2 = -z_1 \dfrac{b+1}{b+2}\right)\cr =
{\frac {1}{z_{{1}} \left( b+1 \right) \left( a+2 \right) }}&+{\frac {1
}{z_{{1}} \left( a+2 \right) \left( a+1 \right) \left( a+b+3
\right) }}-{\frac {1}{z_{{1}} \left( b+2 \right) \left( b+1 \right)
\left( a+b+3 \right) }}\cr
&= \dfrac{1}{(b+2)(a+1) z_1}\cr
\dfrac{1}{(2\pi i)^2} \oint_{|z_1|=\epsilon} dz_1 \oint_{|z_2|=\epsilon} dz_2\; f(z_1,z_2) &= \text{res}\left(\dfrac{1}{(b+2)(a+1) z_1}; z_1 = 0\right) = \dfrac{1}{(b+2)(a+1)} }$$
and there is no discrepancy.
| {
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"url": "https://math.stackexchange.com/questions/1396342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the other end of the Diameter For a National Board Exam Review
A circle has it center at $(3,-2)$ and one end of a diameter at $(7,2)$. Find the other end of the diameter.
Answer is $(-1,6)$
$$m=\frac{y^2-y^1}{x^2-x^1}=\frac{2-(-2) }{7-3}$$
$$=(3-7)^2+(-2-2)^2=r^2=32$$
$$r =\sqrt{32}$$
Plugin $(7,2)$ into
$$y = mx+b$$
$$b = -5$$
Solve two linear equations:
$$32 = (x-h)^2 + (y-k)^2$$
$$32 = (x-7)^2 + ((x-5)-2)^2$$
$$y=x-5$$
I get $(3,-2)$. What am I doing wrong? Any hint?
| Let $(a, b)$ be other end of the diameter then the center $(3, -2)$ is the mid-point of line joining the ends points of diameter $(a, b)$ & $(7, 2)$
Hence the coordinates of the center $(3, -2)$ are given as $$\left(\frac{a+7}{2}, \frac{b+2}{2}\right)\equiv(3, -2)$$ By comparing the corresponding coordinates, we get $$\frac{a+7}{2}=3\iff a=6-7=-1$$
$$\frac{b+2}{2}=-2\iff a=-4-2=-6$$ Hence the other end of the diameter is $(a, b)\equiv\color{blue}{ (-1, -6)}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Sum of (arithmetic?) infinite series How the heck do I find the sum of a series like $\sum\limits_{n=3}^\infty\frac{5}{36n^{2}-9}$? I can't seem to convert this to a geometric series and I don't have a finite number of partial sums, so I'm stumped.
| Notice, we have $$\sum_{n=3}^{\infty}\frac{5}{36n^2-9}$$ $$=\frac{5}{9}\sum_{n=3}^{\infty}\frac{1}{4n^2-1}$$
$$=\frac{5}{9}\sum_{n=3}^{\infty}T_n$$
$$\sum_{n=3}^{\infty}\frac{5}{36n^2-9}=\frac{5}{9}\lim_{n\to \infty}\sum_{n=3}^{n}T_n\tag 1$$
Where, $T_n$ is nth term of the series
given as
$$T_n=\frac{1}{4n^2-1}= \frac{1}{(2n-1)(2n+1)}$$ $$=\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)$$ Now, setting $n=3, 4, 5, \dots n$, we get $$T_1=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{7}\right)$$ $$T_2=\frac{1}{2}\left(\frac{1}{7}-\frac{1}{9}\right)$$ $$T_3=\frac{1}{2}\left(\frac{1}{9}-\frac{1}{11}\right)$$ $$..................$$
$$...................$$
$$T_{n-1}=\frac{1}{2}\left(\frac{1}{2n-3}-\frac{1}{2n-1}\right)$$
$$T_{n}=\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)$$
Now, adding all the terms column-wise we get sum of all $(n-2)$ terms as follows $$\sum_{n=3}^{n} T_n=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{2n+1}\right)$$
Now, taking limit as $n\to \infty$, we get $$\lim_{n\to \infty}\sum_{n=3}^{n} T_n=\lim_{n\to \infty}\frac{1}{2}\left(\frac{1}{5}-\frac{1}{2n+1}\right)$$ $$=\frac{1}{2}\lim_{n\to \infty}\left(\frac{1}{5}-\frac{1}{2n+1}\right)$$ $$=\frac{1}{2}\left(\frac{1}{5}-0\right)=\frac{1}{10}$$ Now, setting the above value in (1), we get
$$\sum_{n=3}^{\infty}\frac{5}{36n^2-9}=\frac{5}{9}\times \frac{1}{10}$$ $$=\frac{1}{18}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluation of $\int\frac{1}{1-\tan^2 x}dx$ Evaluation of $\displaystyle \int\frac{1}{1-\tan^2 x}dx$
$\bf{My\; Try::}$ We can write $$\displaystyle \int\frac{1}{(1-\tan x)\cdot (1+\tan x)}dx = \frac{1}{2}\int\frac{(1+\tan x)+(1-\tan x)}{(1-\tan x)\cdot (1+\tan x)}dx$$
So We get $$\displaystyle = \frac{1}{2}\int\frac{1}{1-\tan x}dx+\frac{1}{2}\int\frac{1}{1+\tan x}dx = \frac{1}{2}I + \frac{1}{2}J$$
Now Let $$\displaystyle I = \int\frac{1}{1-\tan x}dx = \frac{1}{2}\int\frac{2\cos x}{\cos x-\sin x}dx = \frac{1}{2}\int\frac{(\cos x+\sin x)+(\cos x-\sin x)}{\cos x-\sin x}dx$$
So we get $$\displaystyle = \frac{1}{2}\int\frac{(\cos x+\sin x)}{(\cos x-\sin x)}dx+\frac{1}{2}x = -\frac{1}{2}\ln\left|\cos x-\sin x\right|+\frac{1}{2}x$$
Now Let $$\displaystyle I = \int\frac{1}{1+\tan x}dx = \frac{1}{2}\int\frac{2\cos x}{\cos x+\sin x}dx = \frac{1}{2}\int\frac{(\cos x+\sin x)+(\cos x-\sin x)}{\cos x+\sin x}dx$$
So we get $$\displaystyle = \frac{1}{2}\int\frac{(\cos x+\sin x)}{(\cos x-\sin x)}dx-\frac{1}{2}x = +\frac{1}{2}\ln\left|\cos x+\sin x\right|+\frac{1}{2}x$$
So $$\displaystyle \int\frac{1}{1-\tan^2 x}dx = -\frac{1}{2}\ln\left|\cos x-\sin x\right|+\frac{1}{2}\ln\left|\cos x+\sin x\right|+x+\mathcal{C}$$
Can we solve it any short method, If yes then plz explain here, Thanks
| We can also put $\tan\left(x\right)=u
$ to get $$\int\frac{1}{1-\tan^{2}\left(x\right)}dx=\int\frac{1}{\left(1-u^{2}\right)\left(u^{2}+1\right)}du=$$ $$=\int\frac{1}{2\left(u^{2}+1\right)}du+\int\frac{1}{4\left(u+1\right)}du-\int\frac{1}{4\left(u-1\right)}du.
$$
| {
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Value of $\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}}$ Here is the question:
$$\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}} = \;?$$
(original image)
I think we need to simplify it writing it in summation sign as you can see here:
$$\frac{\sum\limits_{n=1}^{99} \sqrt{10 + \sqrt{n}}}{\sum\limits_{n=1}^{99} \sqrt{10 - \sqrt{n}}}$$
or in Wolfram Alpha input in comments.
I can compute it too! It's easy to write a script for this kind of question.
I need a way to solve it. How would you solve it on a piece of paper?
| Let $f(n)=\frac{\sum_{k=1}^{n} \sqrt{\sqrt{n+1}+\sqrt{k}}}{\sum_{k=1}^{n} \sqrt{\sqrt{n+1}-\sqrt{k}}}$, and the original problem is to calculate $f(99)$.
Now we claim that $f(n)=\sqrt{2}+1$ for $\forall n>0$.
$f(n)=\frac{\sqrt{\sqrt{2}+1}}{\sqrt{\sqrt{2}-1}} \cdot \frac{\sum_{k=1}^{n} \sqrt{(\sqrt{n+1}+\sqrt{k})(\sqrt{2}-1)}}{\sum_{k=1}^{n} \sqrt{(\sqrt{n+1}-\sqrt{k})(\sqrt{2}+1)}}$
$=(\sqrt{2}+1) \cdot \frac{\sum_{k=1}^{n} \sqrt{A(k)+B(k)}}{\sum_{k=1}^{n} \sqrt{A(k)-B(k)}}$, where $A(k)=\sqrt{2(n+1)}-\sqrt{k}$, $B(k)=\sqrt{2k}-\sqrt{n+1}$.
Notice that:
$\left(\sqrt{A(k)+B(k)}-\sqrt{A(k)-B(k)}\right)^2=2A(k)-2\sqrt{A(k)^2-B(k)^2}\\=2\left(\sqrt{2(n+1)}-\sqrt{k}-\sqrt{n+1-k}\right)$
That means
$\left(\sqrt{A(n+1-k)+B(n+1-k)}-\sqrt{A(n+1-k)-B(n+1-k)}\right)^2\\=2\left(\sqrt{2(n+1)}-\sqrt{n+1-k}-\sqrt{k}\right)$
So that
$\sqrt{A(k)+B(k)}-\sqrt{A(k)-B(k)}\\=\sqrt{A(n+1-k)-B(n+1-k)}-\sqrt{A(n+1-k)+B(n+1-k)}$
So that
$\sum_{k=1}^{n} \sqrt{A(k)+B(k)}=\sum_{k=1}^{n} \sqrt{A(k)-B(k)}$
So that $f(n)=\boxed{\sqrt{2}+1}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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If $3x^2 -2x+7=0$ then $\left(x-\frac{1}{3}\right)^2 =$? If $\ 3x^{2}-2x+7=0$ then $$\left(x-\frac{1}{3}\right)^2 =\text{?} $$
I am so confused. It is a self taught algebra book.
The answer is: $ \large -\frac{20}{9}$ but I don't know how it was derived.
Please explain.
Thanks for everyone who commented! I understand it now.
| Starting from, $$3x^2-2x+7=0$$ Dividing by 3:$$x^2-\frac{2x}{3}+\frac{7}{3}=0$$ Completing the square: $$\left(x-\frac 13\right)^2-\frac{1}{9}+\frac{7}{3}=0$$ $$\left(x-\frac 13\right)^2=\frac{1}{9}-\frac{7}{3}$$ $$\left(x-\frac 13\right)^2=\frac{1-21}{9}$$
$$\left(x-\frac 13\right)^2=-\frac{20}{9}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sigma Notations I have troubles understanding the sigma notation. If for example we have $c_i$ as $$c_i=\frac {x_i-x}{\sum(x_i-x)^2}$$ $$\sum c_i=\sum\frac{x_i-x}{\sum(x_i-x)^2}$$ Do we distribute the sigma to both top and bottom? But then the bottom would have double sigmas which dont make sense?
| Your notation is confusing.
Assuming $i$ goes from $0$ to $n$ and $c_i$ is actually
$$c_i=\frac{x_i-x}{\sum^n_{k=0}(x_k-x)^2}.$$
Then, we have
\begin{align}
\sum_{i=0}^n c_i
&= c_0 + c_1 + \dots + c_n\\
&=\frac{x_0-x}{\sum^n_{k=0}(x_k-x)^2} + \frac{x_1-x}{\sum^n_{k=0}(x_k-x)^2} + \dots +\frac{x_n-x}{\sum^n_{k=0}(x_k-x)^2}\\
&=\frac{\sum_{i=0}^n x_i-x}{\sum_{k=0}^n (x_k-x)^2}.
\end{align}
On the other hand, if $c_i$ is actually:
$$c_i=\frac{x_i-x}{\sum^i_{k=0}(x_k-x)^2}$$
Then, we have
$$
\begin{align}
\sum c_i&= c_0 + c_1 + \dots + c_n\\
&=\frac{x_0-x}{\sum^0_{k=0}(x_k-x)^2} + \frac{x_1-x}{\sum^1_{k=0}(x_k-x)^2} + \dots +\frac{x_n-x}{\sum^n_{k=0}(x_k-x)^2}\\
&=\frac{x_0-x}{(x_0-x)^2} + \frac{x_1-x}{(x_0-x)^2 + (x_1-x)^2} + \dots +\frac{x_n-x}{(x_0-x)^2 + (x_1-x)^2 + \dots + (x_n-x)^2}
\end{align}
$$
We see here that the denominators of the terms of the sum are not equal so you can't distribute the sigma
| {
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"timestamp": "2023-03-29T00:00:00",
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percentage of integers such that $n^4 \pmod{16} \equiv 1$? How do I find the percentage of numbers $n$ in the list $1^4, 2^4, ... 1000^4$ such that $n \pmod{16} \equiv 1$? I know that for any $x$, if $x \pmod{16} \equiv 1$, then $x^n \pmod{16} \equiv 1$, so I know for sure that there are at least around $\lfloor \frac{1000}{16} \rfloor = 62$ values of $n$ such that $n \pmod{16} \equiv 1$, because any multiple of $16$ plus $1$ will equal $1 \pmod {16}$. However, I am stuck on how to find the rest? My answer sheet tells me that between $20$ to $50$ percent of the numbers are $1 \pmod {16}$.
| We have
$$
\begin{align}
(2n)^4
&=16n^4\\
&\equiv0&\pmod{16}
\end{align}
$$
and
$$
\begin{align}
(2n+1)^4
&=16n^4+32n^3+24n^2+8n+1\\
&\equiv16\frac{n(n+1)}2+1&\pmod{16}\\
&\equiv1&\pmod{16}
\end{align}
$$
Therefore, $8$ out of $16$ equivalence classes mod $16$ satisfy $n^4\equiv1\pmod{16}$. So, I'd say $50\%$ of the equivalence classes mod $16$.
Since there are $500$ odd numbers from $1$ to $1000$ and $500$ even numbers, exactly $50\%$ of the numbers from $1$ to $1000$ have $n^4\equiv1\pmod{16}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Is $(x^2 + 1) / (x^2-5x+6)$ divisible? I'm learning single variable calculus right now and at current about integration with partial fraction. I'm stuck in a problem from few hours given in my book. The question is to integrate $$\frac{x^2 + 1}{x^2-5x+6}.$$
I know it is improper rational function and to make it proper rational fraction we have to divide
$$\frac{x^2 + 1}{x^2-5x+6}$$
I'm trying from sometime but couldn't find the right solution.
Please help! Thank you in advance.
| You can use long division. Another way is as follows:
\begin{align*}
\frac{x^2+1}{x^2-5x+6}&=\frac{(x^2-5x+6)+(5x-5)}{x^2-5x+6}\\
&=\frac{x^2-5x+6}{x^2-5x+6}+\frac{5x-5}{x^2-5x+6}\\
&=1+\frac{5x-5}{x^2-5x+6}
\end{align*}
As next step notice that $$x^2-5x+6=(x-2)(x-3)$$
and expand in partial fractions:
\begin{align*}
\frac{5x-5}{x^2-5x+6}&=\frac{A}{x-2}+\frac{B}{x-3}&&\text{Where }A\text{ and }B\text{ are constants}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve equation: $ \frac{x+1}{9x^3-4x} + \frac{3x}{27x^3 - 12x + 8} - \frac{1}{(3x-2)^2}=0 $ How to solve this equation?
$$
\frac{x+1}{9x^3-4x} + \frac{3x}{27x^3 - 12x + 8} - \frac{1}{(3x-2)^2}=0
$$
I try
$$
\frac{81x^5 - 81x^4 - 90 x^3 + 36 x^2 + 16x -16}{x(3x-2)^2(3x+2)(27x^3 - 12x + 8)}=0
$$
And then
$$
81x^5 - 81x^4 - 90 x^3 + 36 x^2 + 16x -16 =0
$$
here $( x \neq 0; x \neq \frac{2}{3}; x \neq - \frac{2}{3}; 27x^3 - 12x + 8 \neq 0 )$.
Is there some simple (and different) way to solve it?
| This is really a very unpleasant problem.
If you consider the function $$f(x)=81 x^5-81 x^4-90 x^3+36 x^2+16 x-16$$ its derivative $$f'(x)=405 x^4-324 x^3-270 x^2+72 x+16$$ shows four roots (which are supposed to be expressed with radicals - see here) but they are so complex that I shall not put the solution here. Their approximate values are $$x_1^*\approx -0.6061944$$ $$x_2^*\approx -0.1533643$$ $$x_3^*\approx +0.3518603$$ $$x_4^*\approx +1.2076984$$ You could show (very funny without a calculator !) that, for all $x_i^*$'s, $f(x_i^*)<0$. So, there is only one root which will be greater than $x_4^*$. By inspection, since $f(\frac 32)=-\frac{311}{32}$ and $f(2)=736$, you can start Newton method from the left. For this problem, the iterative scheme will be $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=\frac{324 x^5-243 x^4-180 x^3+36 x^2+16}{405 x^4-324 x^3-270 x^2+72 x+16}$$ So, starting at $x_0=\frac 32$, the successive iterates will be $$x_1=1.520533474$$ $$x_2=1.519495986$$ $$x_3=1.519493219$$ which is the solution for ten significant figures.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\binom{n}{m}+2\binom{n-1}{m}+......+(n-m+1)\binom{m}{m} = \binom{n+2}{m+2}$
For $m,n\in\mathbb{N},\;n\geq m$, prove the following:
$$
\tag{i}\binom{n}{m}+\binom{n-1}{m}+\binom{n-2}{m}+......+\binom{m}{m} = \binom{n+1}{m+1}
$$
$$
\tag{ii}\binom{n}{m}+2\binom{n-1}{m}+3\binom{n-2}{m}+......+(n-m+1)\binom{m}{m} = \binom{n+2}{m+2}
$$
My Attempt:
For $(\mathrm{i})$, We can write $\binom{n}{m}$ as the coefficient of $x^m$ in $(1+x)^n$. Thus we can also write $\binom{n-1}{m}$ as the coefficient of $x^m$ in $(1+x)^{n-1}$ and $\binom{n-2}{m}$ as the coefficient of $x^m$ in $(1+x)^{n-2}$. So we have to find the coefficient of $x^m$ in
$$
(1+x)^n+(1+x)^{n-1}+(1+x)^{n-2}+........+(1+x)^{m}
$$
Using the formula for the sum of a geometric progression, this sum equals
$$\frac{(1+x)^{n+1}-(1+x)^m}{x}$$
So we now need to find the coefficient of $x^m$ in
$$
\frac{(1+x)^{n+1}-(1+x)^m}{x}
$$
or, equivalently, we need to find the coefficient of $x^{m+1}$ in
$$
(1+x)^{n+1}-(1+x)^m = \binom{n+1}{m+1}
$$
We can use a similar method to solve $(\mathrm{ii})$. Can these questions be solved using combinatorial methods instead?
| Maybe a bit late now but here is my answer:
Consider counting the number of possible bit-strings of length $n+i$:
$$ \underbrace{111...\overbrace{\textbf{1}}^{i^{\text{th}}\ 1}...111}_{m+i\ 1\text{s}}\ 000...000 $$
So, here $n$ is the number of digits to the right of the $i^\text{th}\ 1$.
The number of bit strings is given simply by $\binom{n+i}{m+i}$ where $m$ is the number of $1$s to the right of the $i^{\text{th}}\ 1$.
We may also count bit strings for each possible position of the $i^{\text{th}}\ 1$ (i.e. position $i$, position $i+1$ up to position $i+n-m$) and sum these to give the same result.
When the $i^{\text{th}}\ 1$ is in a position $i+r$ there are $i+r-1$ digits to the left including $i-1$ $1$s and $n-r$ digits to the right including $m$ $1$s.
So if we sum over $r$:
$$\dbinom{n+i}{m+i} = \sum_{r=0}^{n-m} \dbinom{i+r-1}{i-1}\dbinom{n-r}{m} $$
Your two examples are special cases of this identity.
If we put $i=1$ and $i=2$ we have the two required results:
$$\dbinom{n+1}{m+1} = \sum_{r=0}^{n-m} \dbinom{r}{0}\dbinom{n-r}{m} = \sum_{r=0}^{n-m} \dbinom{n-r}{m} = \dbinom{n}{m} + \dbinom{n-1}{m} + ... + \dbinom{m}{m} $$
and
$$\dbinom{n+2}{m+2} = \sum_{r=0}^{n-m} \dbinom{r+1}{1}\dbinom{n-r}{m} = \sum_{r=0}^{n-m} \left(r+1\right)\dbinom{n-r}{m} = 1\dbinom{n}{m} + 2\dbinom{n-1}{m} + ... + \left(n-m+1\right)\dbinom{m}{m} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\sum\limits_{cyc}\sqrt{\frac{a+b}{c}}\ge2\sum\limits_{cyc}\sqrt{\frac{c}{a+b}}$ Let $a,b,c$ be positive numbers. Then we need to prove
$\sqrt{\frac{a+b}{c}}+\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}\ge2\left(\sqrt{\frac{c}{a+b}}+\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}\right).$
I have an idea to set $x=\frac a{b+c}$, $y=\frac b{c+a},z=\frac c{a+b}$ then
$\frac1{1+x}+\frac1{1+y}+\frac1{1+z}=2$
and we need to prove
$\frac1{\sqrt x}+\frac1{\sqrt y}+\frac1{\sqrt z}\ge2\left(\sqrt x+\sqrt y+\sqrt z\right)$
But I could not go further.
| It is a consequence of Chebychev's inequality:
$$
\sum_{cyc}\sqrt{\frac{a+b}{c}}≥2\sum_{cyc}\sqrt{\frac{c}{a+b}}\iff\sum_{cyc}\frac{a+b-2c}{\sqrt{c(a+b)}}≥0
$$
Since the $a+b-2c$ and $\frac{1}{\sqrt{c(a+b)}}$ are ordered in the same way, we can apply Chebychev's inequality to obtain:
$$
\sum_{cyc}\frac{a+b-2c}{\sqrt{c(a+b)}}≥\frac{1}{3}\left(\sum_{cyc}a+b-2c\right)\left(\sum_{cyc}\frac{1}{\sqrt{c(a+b)}}\right)=0
$$
Edit:
In case you are not familiar with this approach:
If we consider two real sequences $a_1,a_2,…,a_n$ and $b_1,b_2,…,b_n$ for which $a_1≤a_2≤…≤a_n$ and $b_1≤b_2≤…≤b_n$, then Chebyvhev tells us, that:
$$
\frac{a_1b_1+a_2b_2+…+a_nb_n}{n}≥\frac{a_1+a_2+…+a_n}{n}\cdot\frac{b_1+b_2+…+b_n}{n}
$$
We are allowed to use it in this case, because by symmetry, we can assume $a≥b≥c>0$. This implies:
$$
a+b-2c≥a+c-2b≥b+c-2a
$$
And:
$$
ab≥ac≥bc\iff a(b+c)≥b(a+c)≥c(a+b) \iff\\ \frac{1}{\sqrt{c(a+b)}}≥\frac{1}{\sqrt{b(a+c)}}≥\frac{1}{\sqrt{a(b+c)}}
$$
So the two sequences we have in the above inequality are indeed ordered in the same way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Understanding degrees of freedom in relation to rank for $\sum_{i=1}^{n}(y_i-\bar{y})^2$ So I'm looking at this website which states:
One of the questions an instrutor [sic] dreads most from a mathematically unsophisticated audience is, "What exactly is degrees of freedom?" It's not that there's no answer. The mathematical answer is a single phrase, "The rank of a quadratic form."
And near the end:
Okay, so where's the quadratic form? Let's look at the variance of a single sample. If $y$ is an $n$ by $1$ vector of observations, then
$$\sum(y_i - \bar{y})^2 = y^{\prime}My\text{, where }M = \begin{pmatrix}
1-\frac{1}{n} & -1/n & \cdot & -1/n \\
-1/n & 1-\frac{1}{n} & \cdot & -1/n \\
\cdot & \cdot & \cdot & \cdot \\
-1/n & -1/n & -1/n & 1-\frac{1}{n}
\end{pmatrix}\text{.}$$
The number of degrees of freedom is equal to the rank of the $n$ by $n$ matrix $M$, which is $n-1$.
I will use $r$ to denote the rank and $C(A)$ to denote the column space of a matrix $A$.
From my notes, I have the definition $r(A) = r\left(C(A)\right)$, which is the number of vectors that create a basis for $C(A)$. Sure.
Let's consider column one:
$$\begin{pmatrix}
1-\frac{1}{n}\\
-1/n \\
\cdot \\
-1/n
\end{pmatrix} = \dfrac{-1}{n}
\begin{pmatrix}
1-n \\
1 \\
\cdot \\
1
\end{pmatrix} = \dfrac{-1}{n}\begin{pmatrix}
1 \\
1 \\
\cdot \\
1
\end{pmatrix} + \dfrac{-1}{n}\begin{pmatrix}
-n \\
0 \\
\cdot \\
0
\end{pmatrix} = \dfrac{-1}{n}\begin{pmatrix}
1 \\
1 \\
\cdot \\
1
\end{pmatrix}+\begin{pmatrix}
1 \\
0 \\
\cdot \\
0
\end{pmatrix}\text{.}
$$
So based on what I'm seeing here, I'm guessing that a basis for $C(A)$ would be something like
$$\mathscr{V} = \left\{\begin{pmatrix}
1 \\
1 \\
\cdot \\
1
\end{pmatrix}, \begin{pmatrix}
1 \\
0 \\
\cdot \\
0
\end{pmatrix}, \begin{pmatrix}
0 \\
1 \\
\cdot \\
0
\end{pmatrix}, \cdots, \begin{pmatrix}
0 \\
0 \\
\cdot \\
1
\end{pmatrix}\right\}$$
which consists of $n+1$ vectors. However, clearly $$\begin{pmatrix}
1 \\
1 \\
\cdot \\
1
\end{pmatrix}$$ needs to be removed, since it is the sum of the other $n$ vectors, so I get that there are $n$ vectors in $\mathscr{V}$, and therefore $r(\mathscr{V}) = r(C(M)) = n$.
Where is the $n-1$ coming from?
| Take en example: the matrix
$$
A=\left[\matrix{1 & 1\\2 & 2}\right].
$$
Applying your method: the first column
$$
\left[\matrix{1 \\ 2}\right]=\left[\matrix{1 \\ 1}\right]+\left[\matrix{0 \\ 1}\right],
$$
and the second column
$$
\left[\matrix{1 \\ 2}\right]=2\left[\matrix{1 \\ 1}\right]-\left[\matrix{1 \\ 0}\right],
$$
hence, collecting all "basic" vectors
$$
\left\{\left[\matrix{1 \\ 1}\right],\left[\matrix{1 \\ 0}\right],\left[\matrix{0 \\ 1}\right]\right\}
$$
and removing the first dependent gives the rank being $2$.
Do you see the problem with your argument?
The vectors you take as basic vectors do not all belong to the subspace. They span a larger subspace than $C(A)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
If $ a + b + c \mid a^2 + b^2 + c^2$ then $ a + b + c \mid a^n + b^n + c^n$ for infinitely many $n$
Let $ a,b,c$ positive integer such that $ a + b + c \mid a^2 + b^2 + c^2$.
Show that $ a + b + c \mid a^n + b^n + c^n$ for infinitely many positive integer $ n$.
(problem composed by Laurentiu Panaitopol)
So far no idea.
| It seems that there's a partial solution.
Suppose that $\mathrm{gcd}(a,a+b+c)=\mathrm{gcd}(b,a+b+c)=\mathrm{gcd}(c,a+b+c)=1$.
Then for $n=k\cdot \phi(a+b+c)+1 \, (k=1,2, \ldots )$, where $\phi$ is Euler's function, we have:
$$
(a^n+b^n+c^n)-(a^2+b^2+c^2)=a^2 (a^{n-1}-1) + b^2 (b^{n-1}-1) +
c^2 (c^{n-1}-1),
$$
where all round brackets are divisible by $a+b+c$ according to Euler theorem.
Therefore $(a+b+c) \mid (a^n+b^n+c^n)$ for all these $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1408323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 0
} |
How to find $ab+cd$ given that $a^2+b^2=c^2+d^2=1$ and $ac+bd=0$? It is given that $a^2+b^2=c^2+d^2=1 $
And it is also given that $ac+bd=0$
What then is the value of $ab+cd$ ?
| I assume $a,b \in \mathbb{R}$. Since $a^2+b^2 = 1$, we have $-1 \leq a \leq 1$ and likewise $-1 \leq b \leq 1$. Let us take $a = \cos(\alpha)$ and $b = \sin(\alpha)$ without loss of generality. Similarly, $c = \cos(\beta)$ and $d = \sin(\beta)$.
We have $ac + bd = \sin(\alpha) \sin(\beta) + \cos(\alpha) \cos(\beta) = \cos(\alpha - \beta) = 0$.
You have
$ab+cd = \cos(\alpha) \sin(\alpha) + \cos(\beta) \sin(\beta)\\
= \frac{1}{2} ( \sin(2 \alpha) + \sin(2 \beta) ) \\
= \sin(\alpha+\beta) \cos(\alpha - \beta) \\
= 0 $
The answer should be 0.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 3
} |
How to find the value of $(a+b+c)(a+b+d)(a+c+d)(b+c+d)$ from the following equation? I have a question about polynomial.
Given a polynomial:
$$x^4-7x^3+3x^2-21x+1=0$$
Given too that the roots of this polynomial are $a, b, c,$ and $d$.
Find the value of $(a+b+c)(a+b+d)(a+c+d)(b+c+d)$?
My attempt:
It seems I need to apply Vieta formula to find the relationship between its roots. From there, I can get:
$$a+b+c+d = 7$$
$$ab+ac+ad+bc+bd+cd= 3$$
$$abc+abd+acd+bcd= 21$$
$$abcd= 1$$
Then,
$$(a+b+c)(a+b+d)(a+c+d)(b+c+d)$$
$$= (a²+b²+ab+3)(c²+d²+cd+3)$$
$$= 3(a^2+b^2+c^2+d^2)+abcd+(ac)^2+(ad)^2+(bc)^2+(bd)^2+a^2cd+b^2cd+abc^2+abd^2$$
From there, I don't have any idea how to go further.
Can somebody help me to explain to solve this equation?
Thanks.
| HINT: Let $p(x)$ be the polynomial. Note that
$$(a+b+c)(a+b+d)(a+c+d)(b+c+d)=(7-d)(7-c)(7-b)(7-a)=p(?)\;.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Trying to solve $\sqrt{2\cos^2(x)-\sqrt{3}}+\sqrt2 \sin(x)=0$ The equation is
$$\sqrt{2\cos^2(x)-\sqrt{3}}+\sqrt2 \sin(x)=0$$
I solve it thus:
$$
\begin{cases}
2\cos^2(x)-\sqrt3=2\sin^2(x) \\
-\sqrt2 \sin(x)\ge 0 \iff \sin(x)\le 0
\end{cases}
$$
The first equation boils down to
$$2\cos^2(x)=2(1-\cos^2(x))+\sqrt3$$
$$4cos^2(x)-2=\sqrt3$$
$$2(2\cos^2(x)-1)=\sqrt3$$
$$2\cos(2x)=\sqrt3$$
$$\cos(2x)=\frac{\sqrt3}{2}$$
$$2x=\pm \arccos(\frac{\sqrt3}{2})+2n\pi$$
$$2x=\pm \frac{\pi}{6}+2n\pi$$
$$x=\pm \frac{\pi}{12}+n\pi$$
Considering the condition $\sin(x)\le 0$, we are left with
$$x=- \frac{\pi}{12}+n\pi$$
But the textbook's answer is
$$x=- \frac{\pi}{12}+2n\pi; x=- \frac{11\pi}{12}+2n\pi$$
What did I overlook?
P.S. The problem and the answer from the texbook:
| Using $\cos2y=1-2\sin^2y=2\cos^2y-1$ on
$$2\sin^2x=2\cos^2x-\sqrt3$$
$$\iff1-\cos2x=1+\cos2x-\sqrt3\iff\cos2x=\dfrac{\sqrt3}2=\cos30^\circ$$
$$2x=360^\circ n\pm30^\circ\iff x=180^\circ n\pm15^\circ$$ where $n$ in any integer
Case$\#1:$ $+\implies x=180^\circ n+15^\circ$
But as $\sin x<0,x$ lies in the third or in the fourth quadrant
$\implies n$ must be odd $=2m+1$(say)
Case$\#2:$ $-\implies x=180^\circ n-15^\circ$
For the reason mentioned in Case$\#1,$ here $n$ must be even $=2m$(say)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Inequality with Four Numbers I am trying to prove the following inequality for real numbers $a,b,c,d$ all of which are greater than $1$
$8(abcd+1) > (a+1)(b+1)(c+1)(d+1)$
I tried the following approaches :
Used the AM-GM inequality
Tried to form a polynomial whose roots are a,b,c,d
Tried the trigonometric substitution $a=\sec x_1$....
But still I couldn't get any closer to the answer.
| $(a-1)(b-1)>0$ as $a,b>1 \implies ab+1>a+b $, or $2(ab+1)>ab+1+a+b=(a+1)(b+1)$.
Similarly $(c+1)(d+1)<2(cd+1)$.
Now $ab,cd>1$ then by similar method $(ab+1)(cd+1)<2(abcd+1)$ hence $4(ab+1)(cd+1)<8(abcd+1) $ therefore $(a+1)(b+1)(c+1)(d+1)<8(abcd+1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Show that the function is continuous To show that the function $f: \mathbb{R}^2 \rightarrow\mathbb{R}$ with $f=\left\{\begin{matrix}
\frac{x^3-y^3}{x^2+y^2} & , (x,y) \neq (0,0)\\
0 & , (x,y)=(0,0)
\end{matrix}\right.$ is continuous on $(0,0)$ we have to show that $|f(x,y)-f(x_0,y_0)| \leq L ||(x,y)-(x_0,y_0)||$ so we have to show that $\left |\frac{x^3-y^3}{x^2+y^2}\right | \leq L \sqrt{x^2+y^2}$.
I have done the following:
$$\left |\frac{x^3-y^3}{x^2+y^2}\right |=\frac{|x-y||x^2+xy+y^2|}{x^2+y^2}\leq \dfrac{(|x-y|)(x^2+y^2+|xy|)}{x^2+y^2} \leq 2\dfrac{(|x-y|)(x^2+y^2)}{x^2+y^2} = 2(|x-y|)\leq 2(|x|+|y|)\leq 2\sqrt{(|x|+|y|)^2}=2\sqrt{|x|^2+|y|^2+2|xy|} \leq 2\sqrt{|x|^2+|y|^2+|x|^2+|y|^2}=2\sqrt{2(|x|^2+|y|^2)}=2\sqrt{2}\sqrt{|x|^2+|y|^2}=2\sqrt{2}\sqrt{x^2+y^2}=2\sqrt{2}||(x,y)||$$
Is this correct?
| Continue by observing that for any $\varepsilon > 0$, all values in the origin-less open disc of radius $\delta = \varepsilon/(2\sqrt{2})$ have absolute value $< \varepsilon$; this, in conjunction with $f(0, 0) = 0$, establishes the continuity of $f$ at $(0, 0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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A way to calculate e? Define three sequences:
The first sequence is $$n^n: 1,\ 4,\ 27,\ 256,\ 3125,\ 46656, \ldots$$
The second sequence is that of the ratios between adjacent members of the first series, or $$\frac{(n+1)^{n+1}}{n^n}: 4,\ \frac{27}4,\ \frac{256}{27}, \ \frac{3125}{256},\ \frac{46656}{3125},\ldots.$$
The third sequence is the difference between adjacent members of the second sequence, or $$\frac{(n+2)^{n+2}}{(n+1)^{n+1}} – \frac{(n+1)^{n+1}}{n^n}: \frac{11}{4},\ \frac{295}{108},\ \frac{18839}{6912},\ \frac{2178311}{800000},\ \ldots.$$
The third sequence converges toward e, from above, and rather quickly so. Is there a proof or explanation of why this must be so?
| This is a continuation of @Mark Bennet's answer in which I will show, why the limit is indeed $e$.
We will concentrate on the term $T_n=(n+1)\left(\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^n\right)$.
Firstly, we make a few modifications:
$$
T_n=(n+1)\left(\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^n\right)=\\
(n+1)\left(1+\frac{1}{n+1}\right)\left(1+\frac{1}{n+1}\right)^{n}-(n+1)\left(1+\frac{1}{n}\right)^n=\\
(n+2)\left(1+\frac{1}{n+1}\right)^{n}-(n+1)\left(1+\frac{1}{n}\right)^n=\\
\underbrace{\left(1+\frac{1}{n+1}\right)^{n}}_{u_n}+\underbrace{(n+1)\left(\left(1+\frac{1}{n+1}\right)^{n}-\left(1+\frac{1}{n}\right)^n\right)}_{v_n}
$$
As we can see quite easily, $u_n$ converges to $e$. For $v_n$ we use the well known factorization $a^n-b^n=(a-b)\cdot\sum_{r=0}^{n-1}a^rb^{n-1-r}$:
$$
v_n=(n+1)\left(\left(1+\frac{1}{n+1}\right)^{n}-\left(1+\frac{1}{n}\right)^n\right)=\\
(n+1)\left(\frac{1}{n+1}-\frac{1}{n}\right)\sum_{r=0}^{n-1}\left(1+\frac{1}{n+1}\right)^r\left(1+\frac{1}{n}\right)^{n-1-r}=\\
-\frac{1}{n}\sum_{r=0}^{n-1}\left(1+\frac{1}{n+1}\right)^r\left(1+\frac{1}{n}\right)^{n-1-r}
$$
With the obvious inequalities $\left(1+\frac{1}{n+1}\right)^r≤\left(1+\frac{1}{n}\right)^r$ and $\left(1+\frac{1}{n}\right)^{n-1-r}≥\left(1+\frac{1}{n+1}\right)^{n-1-r}$ we obtain:
$$
\left(1+\frac{1}{n+1}\right)^{n-1}=\frac{1}{n}\sum_{r=0}^{n-1}\left(1+\frac{1}{n+1}\right)^r\left(1+\frac{1}{n+1}\right)^{n-1-r}≤-v_n\\
≤\frac{1}{n}\sum_{r=0}^{n-1}\left(1+\frac{1}{n}\right)^r\left(1+\frac{1}{n}\right)^{n-1-r}=\left(1+\frac{1}{n}\right)^{n-1}
$$
Both the upper and the lower bound of $-v_n$ converge to $e$, so by the squeeze theorem, $v_n$ converges to $-e$. Therefore:
$$
\lim_{n\to\infty}T_n=\lim_{n\to\infty}u_n+v_n=\lim_{n\to\infty}u_n+\lim_{n\to\infty}v_n=e-e=0
$$
This completes the answer of Mark Bennet with sufficient precision and we can conclude that your limit indeed is $e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 6,
"answer_id": 1
} |
Does the limit $\lim\limits_{x\to0}\left(\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}\right)$ exist? Does the limit: $$\lim\limits_{x\to0}\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}$$ exist?
| You may use the Taylor expansion, as $u \to 0$,
$$
\frac1{1-u}=1+u+O(u^2)
$$ and the Taylor expansion, as $x \to 0$,
$$
\arctan x=x-\frac{x^3}3+O(x^5),
$$ giving
$$
\begin{align}
\frac1{x\arctan x}&=\frac1{x(x-\frac{x^3}3+O(x^5))}\\\\
&=\frac1{x^2}\frac1{(1-\frac{x^2}3+O(x^4))}\\\\
&=\frac1{x^2}(1+\frac{x^2}3+O(x^4))\\\\
&=\frac1{x^2}+\frac13+O(x^2)
\end{align}
$$and
$$
\lim_{x \to 0}\left(\frac1{x\arctan x}- \frac1{x^2}\right)=\frac13.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Trigonometric equation $\sin v = -1/\sqrt{2}$ I'm trying to solve the following: $$\sin(v) = -\frac{1}{\sqrt{2}}$$
My attempt:
$$-\sin(v) = \frac{1}{\sqrt{2}}$$
$$\sin(-v) = \frac{1}{\sqrt{2}}$$
$$v_1 = -\frac{\pi}{4} - 2\pi n $$
$$v_2 = -\pi + \frac{\pi}{4} - 2\pi n = \frac{-3\pi}{4} - 2\pi n$$
$v_1 $ is correct, but $v_2 $ should apparently be $ \frac{5\pi}{4} + 2\pi n$
why?
| Observe that $$-\frac{3\pi}{4} + 2\pi = \frac{5\pi}{4}$$ If you substitute $-m - 1$ for $n$ in the expression $$-\frac{3\pi}{4} - 2\pi n, n \in \mathbb{Z}$$ you obtain $$-\frac{3\pi}{4} - 2\pi(-m - 1) = -\frac{3\pi}{4} + 2\pi m + 2\pi = \frac{5\pi}{4} + 2\pi m$$
where $m = -n - 1 \in \mathbb{Z}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.