Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Nature of algebraic structure I am given $G = \{x + y \sqrt7 \mid x^2 - 7y^2 = 1; x,y \in \mathbb Q\}$ and the task is to determine the nature of $(G, \cdot)$, where $\cdot$ is multiplication. I'm having trouble finding the inverse element (I have found the neutral and proven the associative rule.
| For $a+b\sqrt{7}$ we seek $x+y\sqrt{7}$ such that $(a+b\sqrt{7})(x+y\sqrt{7})=1$. Expanding these brackets and comparing coefficients gives
$$\begin{align} ax+7by &= 1 \\ bx+ay &= 0 \end{align}$$
It is then just a task of solving for $x$ and $y$.
This can be done using matrices:
$$\begin{align} \begin{pmatrix} a & 7b \\ b & a \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} 1 \\ 0\end{pmatrix} \\ \Rightarrow \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} a & 7b \\ b & a \end{pmatrix}^{-1} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \end{align}$$
Note that $a^2-7b^2 \ne 0$ for any $a,b \in \mathbb{Q}$ and so this matrix is invertible. It comes out from here in a couple of lines; then $x+y\sqrt{7} = (a+b\sqrt{7})^{-1}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Linear Algebra, eigenvalues and eigenvectors For any $m \times m$ matrix, I will get a characteristic polynomial of degree $m$
with $m$ eigenvalues.
But for the matrix
$$A = \pmatrix{2 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1}$$
I got the characteristic polynomial
$$P(A)=t(t+1)(1-t)^3.$$
This means $5$ eigenvalues: $\{1,1,1,-1,0\}$.
Did I do some thing wrong?
| I calculated the same polynomial and I got
$$P(X)= X^2 (X-1)^3 (X-4) \,.$$
Note that $tr(A)=7$ has to be the sum of eigenvalues.
Just to get you started:
$$\det(A-tI)= \det \pmatrix{2-t & 1 & 1 & 1 & 1 & 1 \\ 1 & 1-t & 0 & 1 & 0 & 1 \\ 1 & 0 & 1-t & 0 & 0 & 1 \\ 1 & 0 & 0 & 1-t & 0 & 0 \\ 1 & 0 & 0 & 0 & 1-t & 0 \\ 1 & 0 & 0 & 0 & 0 & 1-t}$$
Subtract the 6th row from 4th and 5th:
$$\det(A-tI)= \det \pmatrix{2-t & 1 & 1 & 1 & 1 & 1 \\ 1 & 1-t & 0 & 1 & 0 & 1 \\ 1 & 0 & 1-t & 0 & 0 & 1 \\ 0 & 0 & 0 & 1-t & 0 & t-1 \\ 0 & 0 & 0 & 0 & 1-t & t-1 \\ 1 & 0 & 0 & 0 & 0 & 1-t}$$
Now, $(1-t)$ common factor on rows 4 and 5.
$$\det(A-tI)= (t-1)^2\det \pmatrix{2-t & 1 & 1 & 1 & 1 & 1 \\ 1 & 1-t & 0 & 1 & 0 & 1 \\ 1 & 0 & 1-t & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 & 1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 1-t}$$
Next add Column 4 and column 5 to Column 6, and you can get a smaller $4 \times 4$ determinant....
| {
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Expressing $\sin^4x-\sin^6x$ in another way I am slightly confused on how one would subtract $\sin^4x-\sin^6x$.
I know that $\sin^2x=(1/2)(1-\cos2x)$,
so $\sin^4x$ would logically be $[(1/2)(1-\cos2x)]^2=(1/4)(1-2\cos(2x)+\cos^2(2x)$
However the value of $\sin^6x$ eludes me. Would it be $(1-\cos2x)^3$? I did that and got $1-\cos2x-2\cos2x-2\cos^2(2x)+\cos^2(2x)-\cos^3(2x)$
I am not sure if that is correct.
| $\sin^4x-\sin^6x=\sin^4x(1-\sin^2x)=\sin^4x \cos^2x=\sin^2x(\sin^2x \cos^2x)$ $$=\frac{\sin^2 x\,\sin^2 2x}{4}=\left(\frac{\sin x \sin 2x}{2}\right)^2=\left(\frac{\cos x-\cos3x}{4}\right)^2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If a number can be expressed as a product of n unique primes............ If a number can be expressed as a product of n unique primes, in how many ways can the number be expressed as a difference of two squares?
| Write $m = p_1 \times \ldots \times p_n$ and for now assume each $p_i$ is odd. Then each factorization of $m = a\times b$ into a pair of odd numbers $(a,b)$ gives a representation $$m = ab = \left(\frac{a+b}{2} + \frac{a-b}{2}\right)\left(\frac{a+b}{2} - \frac{a-b}{2}\right) = \left(\frac{a+b}{2}\right)^2 - \left(\frac{a-b}{2}\right)^2$$ and vice versa. So you need to consider all possible pairs of factors $(a,b)$ (where order doesn't matter). There are $\frac{1}{2}\left(\binom{n}{0} + \binom{n}{1} + ... + \binom{n}{n}\right) = 2^{n-1}$ of these.
If one of the $p_i$ is two, then there are no representations: by the unique primes assumptions, $m \equiv 2 \bmod 4$ and all differences of two squares are one of $0,1,3 \bmod 4$.
| {
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How do I transform the equation based on the condition? If $q$ and $w$ are the roots of the equation $$2x^2-px+7=0$$
Then $q/w$ is a root of ?
P.s:- It is an another question of How do I transform the equation based on this condition?
| Let $y=\frac{q}{w}$.
Clearly, q,w are $\frac{p ±\sqrt{p^2-56}}{4}$
If we take q to be $\frac{p +\sqrt{p^2-56}}{4}$, w is $\frac{p -\sqrt{p^2-56}}{4}$
So, $y=\frac{q}{w}=\frac{p +\sqrt{p^2-56}}{p -\sqrt{p^2-56}}$
Applying componendo dividendo, $\frac{y+1}{y-1}=\frac{p}{\sqrt{p^2-56}}$
On squaring,
$\frac{y^2+1+2y}{y^2+1-2y}=\frac{p^2}{p^2-56}$
Applying componendo dividendo again, $\frac{y^2+1}{2y}=\frac{p^2-28}{28}$
or, $28y^2-2(p^2-28)y+28=0$
$=>14y^2-(p^2-28)y+14=0$ is the desired equation whose one root is $\frac{q}{w}$.
If we take q to be $\frac{p -\sqrt{p^2-56}}{4}$, w is $\frac{p +\sqrt{p^2-56}}{4}$
So, $y=\frac{q}{w}=\frac{p -\sqrt{p^2-56}}{p +\sqrt{p^2-56}}$
Applying componendo dividendo, $\frac{y+1}{y-1}=-\frac{p}{\sqrt{p^2-56}}$
On squaring we shall get the desired quadratic equation whose one root is $\frac{q}{w}$.
Just observe that both the derived equations are same.
This is the minimal equation with rational coefficients for any integer p.
If we allow irrational coefficients, $\frac{q}{w}$ will be a root y=$\frac{p -\sqrt{p^2-56}}{p +\sqrt{p^2-56}}$ or y=$\frac{p + \sqrt{p^2-56}}{p - \sqrt{p^2-56}}$.
| {
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Inequality. $a^2+b^2+c^2 \geq a+b+c$ Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that
$a^2+b^2+c^2 \geq a+b+c$.
Thanks
| Using Cauchy-Schwarz inequality we get
$$
a+b+c=a\cdot 1+b\cdot 1+c\cdot 1+\leq\sqrt{a^2+b^2+c^2}\sqrt{1^2+1^2+1^2}\tag{1}
$$
From AM-GM we obtain $a^2+b^2+c^2\geq 3\sqrt[3]{a^2b^2c^2}=3$, so
$$
\sqrt{3}\leq\sqrt{a^2+b^2+c^2}\tag{2}
$$
From $(1)$ and $(2)$ it follows
$$
a+b+c\leq\sqrt{a^2+b^2+c^2}\sqrt{3}\leq\sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}=a^2+b^2+c^2
$$
| {
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Inequality. $\frac{1}{16}(a+b+c+d)^3 \geq abc+bcd+cda+dab$ I want to prove the following inequality :
$$\frac{1}{16}(a+b+c+d)^3 \geq abc+bcd+cda+dab, $$
$a,b,c,d \in \mathbb{R}_{+} .$
In my book, at the answers chapter the author uses AM $\geq$ GM, but I haven't any idea how I can use that.
Thanks :)
| You need to show that
$$(a + b + c + d)^3 - 16(abc + bcd + cda + dab) \geq 0$$
It suffices to show this on any set of the form $0 \leq a,b,c,d \leq N$. By calculus (the "extreme value theorem") the function $(a + b + c + d)^3 - 16(abc + bcd + cda + dab)$ achieves its minimum at some $(a,b,c,d)$ in the set $0 \leq a,b,c,d \leq N$. I claim that this minimum has to occur when $a = b = c = d$.
Write $$(a + b + c + d)^3 - 16(abc + bcd + cda + dab) = (a + b + c + d)^3 - 16(a + b)cd - 16(c + d)ab$$
Note that by AM-GM, we have $16(c + d)ab \leq 16(c + d)({a + b \over 2})^2$. If we had $a \neq b$, we could replace $a$ and $b$ by ${a + b \over 2}$, leaving $c$ and $d$ constant, and we'd get a smaller value. So since $(a,b,c,d)$ is the minimum, this can't happen and we conclude that $a = b$. For similar reasons $c = d$, reversing the roles of the terms $16(a + b)cd$ and $16(c + d)ab$
Next, write $$(a + b + c + d)^3 - 16(abc + bcd + cda + dab) = (a + b + c + d)^3 - 16(b + d)ac - 16 (a + c)bd$$ Then arguing like above gives $b = d$ and $a = c$. Combining the above gives $a = b = c = d$, whereupon $(a + b + c + d)^3 - 16(abc + bcd + cda + dab) = (4a)^3 - 16(4a^3) = 0$. Since this is the minimum, the expression $(a + b + c + d)^3 - 16(abc + bcd + cda + dab)$ is nonnegative as needed.
| {
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Random point uniform on a sphere If $X=(x,y,z)$ is a random point uniform on the unit sphere in $\mathbb{R}^3$, Are the coordinates $x$, $y$, $z$ uniform in interval $(-1,1)$?
| Let $Z_1$,$Z_2$, $Z_3$ be independent identically distributed standard normal random variables. Then the random vector
$$
\vec{X} = \left( \frac{Z_1}{\sqrt{Z_1^2+Z_2^2+Z_3^2}}, \frac{Z_2}{\sqrt{Z_1^2+Z_2^2+Z_3^2}}, \frac{Z_3}{\sqrt{Z_1^2+Z_2^2+Z_3^2}} \right)
$$
is uniformly distributed on a unit sphere (e.g. sphere point picking on MathWorld).
The distribution of components of $\vec{X}$ is clearly symmetric, so for $-1<x<0$:
$$\begin{eqnarray}
\mathbb{P}\left( \frac{Z_1}{\sqrt{Z_1^2+Z_2^2+Z_3^2}} \leqslant x \right) &=&
\mathbb{P}\left( Z_1^2 \geqslant x^2 (Z_1^2+Z_2^2+Z_3^2) | Z_1 < 0\right) \\ &=&
\frac{1}{2}\mathbb{P}\left( \frac{2 Z_1^2}{Z_2^2+Z_3^2} \geqslant \frac{2 x^2}{1-x^2} \right)
\end{eqnarray}
$$
The ratio $\frac{2 Z_1^2}{Z_2^2+Z_3^2}$ follows $F_{1,2}$-distribution (wiki), hence:
$$
\mathbb{P}\left( \frac{Z_1}{\sqrt{Z_1^2+Z_2^2+Z_3^2}} \leqslant x \right) = \frac{1}{2}\left(1 - \sqrt{\frac{\frac{2x^2}{1-x^2}}{2+ \frac{2x^2}{1-x^2}}}\right) = \frac{1}{2} \left(1- \sqrt{x^2}\right) = \frac{1+x}{2}
$$
which the cumulative distribution function of the uniform distribution indeed.
Added:
As suggested by @did, consider the case of $\mathbb{R}^d$. In that case the cumulative distribution function of the component, for $-1<x<0$ reads:
$$
\mathbb{P}\left( \frac{Z_1}{\sqrt{Z_1^2+Z_2^2+\cdots+Z_d^2}} \leqslant x\right) =
\frac{1}{2}\mathbb{P}\left(\frac{(d-1) Z_1^2}{Z_2^2+\cdots+Z_d^2} \geqslant \frac{(d-1)x^2}{1-x^2} \right)
$$
The ratio $\frac{(d-1) Z_1^2}{Z_2^2+\cdots+Z_d^2}$ follows $F$-distribution with $1$ and $d-1$ degrees of freedom, thus:
$$
\mathbb{P}\left( \frac{Z_1}{\sqrt{Z_1^2+Z_2^2+\cdots+Z_d^2}} \leqslant x\right) = \frac{1}{2} \frac{\operatorname{B}\left(1-x^2 ;\frac{d-1}{2}, \frac{1}{2}\right)}{\operatorname{B}\left(\frac{d-1}{2}, \frac{1}{2}\right)}
$$
where $\operatorname{B}\left(z ;a, b\right)$ is the incomplete Beta function. Differentiating we get the probability density:
$$
f(x) = \frac{1}{\operatorname{B}\left(\frac{d-1}{2}, \frac{1}{2}\right)} \left(1-x^2\right)^{\tfrac{d-3}{2}} [-1<x<1]
$$
This clearly shows that the distribution is not uniform for $d\not=3$.
| {
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How to construct minimal polynomial? This is an exam question from last semester.
We have the finite field
$$ \mathbb F_{81} = \mathbb Z_3 [x]/(x^4+x^2+x+1)$$
(a) Prove that the polynomial $$ x^4+x^2+x+1 $$
is irreducible
(b) Construct the minimal polynomial of the element $$ x^3+x^2+x+1 \space\epsilon\space Z_3 [x]/(x^4+x^2+x+1)$$
Use y as a formal variable in this polynomial. Hint: using $$ x^3+x^2+x+1 = (x^2+1)(x+1) $$ should help with the calculations.
(c) Construct the subfield F9 in $$ Z_3 [x]/(x^4+x^2+x+1)$$
I tried a and I think you can prove it by showing the polynomial has no Zeros? So assuming we call the polynomial g(x). I just filled in {0,1,2} and none of them gave 0 --> You can't split up the polynomial in polynomials of lower orders -> it's irreducible?
I don't know how to do b and c though.
Can someone please tell me how to do it in general and what the solution is here? Really need the answer.
| For (a), a reducible polynomial of degree 4 must have either linear or quadratic factors. Having linear factors is the same as having a root, and it is easy to plug in $0,1,$ and $2$ and see that they are not roots of the polynomial. Therefore, we must verify that
$$x^4+x^2+x+1 \neq (x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+\ldots +bd$$
for any choice of $a,b,c,d\in \mathbb{F}_3$. If we had equality, then the $x^3$ term tells us $a=-c$ and the constant term tells us $b=1/d$ so either $b=d=1$ or $b=d=-1$. Fully expanding out, we have
$$ (x^2+ax+b)(x^2-ax+b)=x^4+(2b-a^2)x^2+b^2 $$
which has zero $x$ term, and therefore cannot equal $x^4+x^2+x+1$.
For (b), we need to find the smallest linear relation over $\mathbb{F}_3$ between powers of $y=(x^2+1)(x+1)$ modulo $x^4+x^2+x+1$. First, we note that because our field is a vector space over the subfield generated by $y$, the minimal polynomial will have degree $1$, $2$, or $4$.
One can calculate that $y^2\equiv x^3+x^2+x-1\pmod{x^4+x^2+x+1}$ over $\mathbb{F}_3$, and therefore $y^2+2=y$ is the minimal polynomial for $y$. Presumably the factorization helps for doing the calculation by hand.
For (c), since the minimal polynomial of $y$ is quadratic, the subfield generated by $y$ is isomorphic to $\mathbb{F}_9$.
| {
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inequality fraction I saw this problem in an Australian maths olympiad:
$6/10 < a/b < 10/15$
The problem asked for the lowest possible value of $b\in \mathbb{Z}$.
I tried manipulating but couldnt derive one of their answers.
| That's how I did it. First of all, note that \begin{equation} \frac{11}{15} - \frac{7}{10}= \frac{1}{30} \end{equation} thus b=31 certainly works, i.e. is a higher bound to the answer.
Then notice that $b$ cannot be any divisor of 10 or 15, as $\frac{11}{15} < \frac{8}{10}$ and $\frac{10}{15}< \frac{7}{10}$.
It remains to check a few values. As $\frac{7}{10}=0.70$ you are looking for a fraction which is 'slightly more' than $0.70$. $b=4$ does not work as $\frac{3}{4} > \frac{11}{15}$ and $b=6$ does not work either because we know $\frac{4}{6} < 0.70$ and $\frac{5}{6}$ is clearly too big. Now check $b=7$ and you find $\frac{5}{7}$ is in the required range, so $b=7$ is your answer.
| {
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Can a partial differential equation have two different solutions? Consider:
$$x^2p+y^2q=(x+y)z$$ where $p=\frac{\partial z}{\partial x}$ and $q=\frac{\partial z}{\partial y}$. Thus by Lagrange's Method
$$\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{dz}{(x+y)z}$$
$$\Rightarrow \frac{dx}{x^2}=\frac{dy}{y^2}
\Rightarrow \frac{1}{x}-\frac{1}{y}=k$$
$$\Rightarrow \frac{dx}{x^2}=\frac{dz}{(x+y)z}
\Rightarrow -\frac{1}{x}-\frac{\ln(z)}{(x+y)}=c$$
$$\Rightarrow \phi (\frac{1}{x}-\frac{1}{y},x+y+x\ln(z))=c$$
Or if we explore further
$$\frac{yzdx+xzdy-xydz}{x^2yz+xy^2z-xyz(x+y)}\Rightarrow \frac{yzdx+xzdy-xydz}{x^2yz+xy^2z-x^2yz-xy^2z}\Rightarrow yzdx+xzdy-xydz=0 \Rightarrow xyz=k$$
So can we even write $f(\frac{1}{x}-\frac{1}{y},xyz)=k$. So which one is right?
Soham
| I think that your work using the characteristics method was right at the start including :
$$\frac{dx}{x^2}=\frac{dy}{y^2} \Rightarrow \frac{1}{x}-\frac{1}{y}=k$$
But the implication here was wrong :
$$\frac{dx}{x^2}=\frac{dz}{(x+y)z} \Rightarrow
-\frac{1}{x}-\frac{\ln(z)}{(x+y)}=c$$
because the $x$ appears at the denominator of $dz$ as well as the numerator of $dx$ so that I think this should be :
$$\frac{dx}{x^2}=\frac{dz}{(x+y)z}
\implies \frac{(x+y)\,dx}{x^2}=\frac{dz}z$$
with Joriki's correction and using $\ y=\dfrac 1{\frac 1x-k}=\dfrac x{1-kx}$ we get :
$$\left(\frac 1x+\frac 1x+\frac k{1-kx}\right)dx=\frac{dz}z\implies 2\ln(x)-\ln(1-kx)-\ln(z)=C_0$$
$$\implies \ln\left(\frac {x^2}{1-kx}\right)-\ln(z)=\ln(xy)-\ln(z)=C_0$$
(we replaced by $y$ again to remove the $k$ constant)
$$\implies \phi\left(\frac{1}{x}-\frac{1}{y},\ \frac {xy}z\right)=C$$
(you had another error here : forgetting the denominator of the second parameter)
After that you wrote :
EDITED:
Or if we explore further
$$\frac{yzdx+xzdy-xydz}{x^2yz+xy^2z-xyz(x+y)}\Rightarrow
\frac{yzdx+xzdy-xydz}{x^2yz+xy^2z-x^2yz-xy^2z}\Rightarrow
yzdx+xzdy-xydz=0 \Rightarrow xyz=k$$
If you divide $\ yzdx+xzdy-xydz=0$ by $xyz\ $ you get :
$\dfrac {dx}x +\dfrac {dy}y -\dfrac {dz}z = 0\implies\dfrac {xy}z=C_1$
So that the other way to write the solution doesn't differ from the first one :
$$\phi\left(\frac 1x -\frac 1y,\ \frac {xy}z\right)=C$$
Because of the arbitrary character of $\phi$ other equivalent parameters could have been obtained/chosen. Like replacing one of the parameters by the product $\dfrac{y-x}z$ or something more elaborate.
| {
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Power series: Is the radius of convergence $\frac{1}{3}$ for $ \frac{x}{1\cdot3} + \frac{x^2}{2\cdot3^2} + \frac{x^3}{3\cdot3^3}+...$ I should calculate the radius of convergenc and would like to know, if the result $\frac{1}{3}$ is correct.
Here the exercise:
$$ \frac{x}{1\cdot3} + \frac{x^2}{2\cdot3^2} + \frac{x^3}{3\cdot3^3} + \frac{x^4}{4\cdot3^4}... $$
This is:
$$
\sum\limits_{n=0}^\infty \frac{x^{n+1}}{(n+1)3^{n+1}} \\
\lim\limits_{n \to \infty} \left| \frac{(n+1)\cdot3^{n+1}}{(n+2)\cdot3^{n+2}} \right| = \left| \frac{1}{3} \right|
$$
I’m right? Thanks.
Summery
I could test with the ratio test if a power series is convergent.
I could use $$\lim\limits_{n \to \infty} \frac{|a_{n+1}|}{\left|a_{n}\right|}$$ and get the $\left|x\right|$ for which the series is convergent. With that test the series is convergent, if the result is $<1$.
| Using the ratio test for absolute convergence.
$$
|a_{n+1}| = \frac{|x|^{n+2}}{(n+2)3^{n+2}} \\
$$
$$
|a_{n}| = \frac{|x|^{n+1}}{(n+1)3^{n+1}} \\
$$
$$
\frac{|a_{n+1}|}{\left|a_{n}\right|} = |x| \left( \frac{n+1}{n+2} \right)\left( \frac{1}{3} \right)
$$
$$
\lim\limits_{n \to \infty} \frac{|a_{n+1}|}{\left|a_{n}\right|} = \frac{|x|}{3}
$$
The series converges absolutely if $\frac{|x|}{3} < 1 $, which is when $|x| < 3$. Absolute convergence implies convergence.
You also need to check for convergence when $|x| = 3$ to determine if those points are in the radius of convergence.
| {
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} |
showing that there is a shape which has three connected black squares We have square lattice with dimensions $n × n$, such that $n \ge 2$. Some of the squares on this lattice are coloured black. How can we show that there are at least 3 connected black squares if there are $$1+\frac{n^{2}}{2}$$ black squares when $n$ is even and $$\frac{n(n+1)}{2}$$ black squares when $n$ is odd?
| Note that any $2\times 2$ tile can contain at most $2$ black squares. In addition, any path of $k$ squares can contain at most $2(k+1)/3$ black squares. For an $n\times n$ square region where $n$ is even, the decomposition into $\frac{1}{4}n^2$ tiles gives an upper bound of $\frac{1}{2}n^2$ black squares. For an $n\times n$ square region where $n$ is odd, the decomposition into $\frac{1}{4}(n-1)^2$ tiles and a path of length $2n-1$ gives an upper bound of $\frac{1}{2}(n-1)^2+\frac{4}{3}n=\frac{1}{2}n^2+\frac{1}{3}n+\frac{1}{2}$ black squares. This latter is a tighter bound than the one in the problem, which is $\frac{1}{2}n^2+\frac{1}{2}n-1$.
Note that the checkerboard pattern with black squares in the four corners, which is an obvious candidate for optimality in the odd-$n$ case, has only $\frac{1}{2}n^2+\frac{1}{2}$ black squares. To see that this is not optimal, consider the $5\times5$ region. The checkerboard pattern has $13$ black squares; but a pattern with $14$ black squares is obtained by placing $4$ black squares in the first, third, and fifth columns and a single black square in the second and fourth columns.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $a$, $a+2$ and $a+4$ are prime numbers then, how can one prove that there is only one solution for $a$? If $a$, $a+2$ and $a+4$ are prime numbers then, how can one prove that there is only one solution for $a$?
when, $a=3$
we have, $a+2=5$ and $a+4=7$
| $a\equiv 0 \mod 3\Rightarrow a=3$
$a\equiv 1\mod 3\Rightarrow a+2\equiv 0\mod 3\Rightarrow a+2=3\Rightarrow a=1$
$a\equiv 2\mod 3\Rightarrow a+4\equiv0\mod 3\Rightarrow a+4=3\Rightarrow a=-1$
So the only possibility is the first one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/189179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 8,
"answer_id": 4
} |
Showing that $ 1<\sin\frac{\alpha}{2}+\sin\frac{\beta}{2}+\sin\frac{\gamma}{2}$ I would like to show that:
$$ 1<\sin\frac{\alpha}{2}+\sin\frac{\beta}{2}+\sin\frac{\gamma}{2}$$
where $\alpha, \beta, \gamma$ are the angles of a triangle.
I know that the inequality $$ 1<\cos \alpha+\cos \beta+\cos \alpha $$
is a direct consequence of the identity $$ \cos \alpha+\cos \beta+\cos \alpha =1+\frac{r}{R}$$
with circumradius $R$ and inradius $r$.
So is there a similar expression for $$ \sin\frac{\alpha}{2}+\sin\frac{\beta}{2}+\sin\frac{\gamma}{2}?$$
| In order to demonstrate the inequality above we may actually prove a more general inequality, which can come in handy another time:
For all natural $n > 1$ and $x_1, x_2, \ldots, x_n \in (0, \pi)$ we have:
\[ |\sin(x_1 + x_2 + ...+x_n)| < \sin x_1 + \sin x_2 + ... + \sin x_n.\]
This can be proved by induction:
*
*for $n = 2$
\[
\begin{split}
|\sin(x_1 + x_2)| &= |\sin x_1 \cos x_2 + \sin x_2 \cos x_1| \leq |\sin x_1 \cos x_2| + |\sin x_2 \cos x_1| \\
&= |\sin x_1| \cdot |\cos x_2| + |\sin x_2| \cdot |\cos x_1| < \sin x_1 + \cos x_2
\end{split}
\]
by the properties of absolute value and since $\cos x < 1$ for $x \in (0, \pi)$
*induction step
\[
\begin{split}
|\sin (x_1 + \ldots + x_{n+1})| &= |\sin(x_1 + \ldots + x_n) \cos x_{n+1} + \sin x_{n+1} \cos(x_1 + \ldots + x_n)| \\
&\leq |\sin(x_1 + \ldots + x_n)| \cdot |\cos x_{n+1}| + |\sin x_{n+1}| \cdot |\cos(x_1 + \ldots + x_n)| \\
&< |\sin(x_1 + \ldots + x_n)| + |\sin x_{n+1}| \\
&< \sin x_1 + \sin x_2 + \ldots + \sin x_{n+1}.
\end{split}
\] by the induction hypothesis and the fact $|\sin x| = \sin x$ for $x \in (0, \pi)$
When plugging in halves of the angles of a triangle we obtain the desired
\[ \sin(\dfrac{\alpha + \beta + \gamma}{2}) = 1 < \sin \dfrac{\alpha}{2} + \sin\dfrac{\beta}{2} + \sin\dfrac{\gamma}{2} .\]
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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} |
Catalan constant is irrational. What is wrong with this proof? Assume $G = p/q$ with $q > 1$ where $p$ and $q$ are coprime. Let $k$ be an arbitrary large odd integer s.t. $2k + 1 > q$ and $p/q$ is not a rational multiple of
$$s_k = \sum_{n = 0}^k \frac{(-1)^n}{(2n + 1)^2} = 1 - \frac{1}{3^2} + \frac{1}{5^2} - \ldots - \frac{1}{(2k + 1)^2}.$$
For odd $k$, $p/q - s_k$ is negative and any positive real number must greater than $p/q - s_k$. Note that $1/(2k + 1)!^2 > 0$, so we have
$$-\ell < G - s_k < \frac{1}{(2k + 1)!^2}$$
or
$$-(2k + 1)!^2\ell < (2k + 1)!^2\left( \frac{p}{q} - s_k \right) < 1$$
where $\ell$ is some integer.
Now $(2k + 1)!^2(p/q - s_k)$ is not an integer because $p/q$ is not a rational multiple of $s_k$ by our choice of $k$. However, note that $q, 3^2, 5^2, \ldots, (2k + 1)^2$ must all be divisors of $(2k + 1)!^2$, i.e.,
\begin{align}
(2k + 1)!^2\left( \frac{p}{q} - s_k \right) &= \frac{(2k + 1)!^2 p}{q} - (2k + 1)!^2 + \frac{(2k + 1)!^2}{3^2}\\
&\quad -\frac{(2k + 1)!^2}{5^2} + \cdots + \frac{(2k + 1)!^2}{(2k + 1)^2}
\end{align}
is an integer, a contradiction.
| You write assume the fraction p/q.
Also p/q is not a rational multiple of a partial sum.
But that partial sum is a rational number.
Hence p/q is a rational multiple since any rational is a rational multiple of any other.
Contradiction and thats whats wrong.
| {
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Find a positive number $\delta$? Find a positive number $\delta$ such that when $|x-1|<\delta$, then $|x^2-1| < 0.45$
Part 2:
Find the LARGEST number $\lambda$ with the property that when $|x-1|< \lambda$, then $|x^2-1| < 0.45$
I don't understand this at all. Help?
| For example $\delta = 1/8 $ solve. in fact,
\begin{eqnarray}
|x^2-1| &=& |x-1||x+1|\\
&\le& |x-1|(|x|+1)\\
&\le & 1/8 \cdot 2 = 1/4 <0,45.
\end{eqnarray}
Because
\begin{equation}
|x| \le |x-1| +|1| \le 2.
\end{equation}
Part 2.
If $x>1, |x^2-1| = x^2-1$ and $x^2-1<0,45 \Rightarrow x< \sqrt{1,45}$. If $x<1, |x^2-1| = 1-x^2$ and $1-x^2<0,45 \Rightarrow x> \sqrt{0,55} $.Then $\delta = \max (|1-\sqrt{0,55}|,|1-\sqrt{1,45}|)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/191112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Inequality. $(n+1)\left(a^{n+1}+b^{n+1}\right) \geq (a+b)\left(a^n+a^{n-1}b+\ldots+b^{n}\right). $
Let $a$ and $b$ be positive numbers, and $n \in \mathbb{N}$. Prove that (using Rearrangement Inequality)
$$(n+1)\left(a^{n+1}+b^{n+1}\right) \geq (a+b)\left(a^n+a^{n-1}b+\ldots+b^{n}\right). $$
Thanks :)
| As the sequences $(a^{k+1},b^{k+1}),(a^{n-k},b^{n-k})$ are similarly sorted, therefore Rearrangement Inequality gives $a^{n+1} + b^{n+1} \ge a^{k+1}b^{n-k} + a^{n-k}b^{k}$.
The same inequality for $k-1$ is $a^{n+1} + b^{n+1} \ge a^{k}b^{n-k+1} + a^{n-k+1}b^{k-1}$, hence adding them gives $$a^{n+1} + b^{n+1} \ge (a+b)(a^{k}b^{n-k} + a^{n-k}b^{k})$$ Adding these inequalities for $k=0,1,2, \cdots, n $ gives the required.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality. $\frac{x^3}{1+9y^2xz}+\frac{y^3}{1+9z^2yx}+\frac{z^3}{1+9x^2yz} \geq \frac{(x+y+z)^3}{18}$ Let $x,y$ and $z$ be positive numbers such that $xy+yz+zx=1$. Prove that (using Hölder's inequality):
$$\frac{x^3}{1+9y^2xz}+\frac{y^3}{1+9z^2yx}+\frac{z^3}{1+9x^2yz} \geq \frac{(x+y+z)^3}{18}$$
Thanks :)
What I try:
$$\left(\frac{x^3}{1+9y^2xz}+\frac{y^3}{1+9z^2yx}+\frac{z^3}{1+9x^2yz} \right)\left(1+9xy^2z+1+9xyz^2+1+9xyz^2\right)\left(1+1+1\right) \geq \left(\sum_{x,y,z}{\left(\sqrt[3]{\frac{x^3}{1+9y^2xz}\cdot\left(1+9y^2xz\right) \cdot 1}\right)}\right)^{3}=\left(\sum_{x,y,z}{x}\right)^{3}.$$ So we have to prove that :
$$\large\frac{\left(\sum_{x,y,z}{x}\right)^{3}}{\left(1+9xy^2z+1+9xyz^2+1+9xyz^2\right)\left(1+1+1\right)} \geq \frac{(x+y+z)^3}{18} $$ or
$$3\cdot \left(3+9xyz\left( x+y+z\right)\right) \leq 18 \Leftrightarrow$$ $$xyz\left(x+y+z\right) \leq \frac{1}{3},$$ but I don't know if this can help me to prove the inequality.
Thanks )
| I'm using the Hölder's inequality
$$\left(\frac{x^3}{1+9y^2xz}+\frac{y^3}{1+9z^2yx}+\frac{z^3}{1+9x^2yz} \right)\left(1+9xy^2z+1+9xyz^2+1+9xyz^2\right)\left(1+1+1\right) \geq \left(\sum_{x,y,z}{\left(\sqrt[3]{\frac{x^3}{1+9y^2xz}\cdot\left(1+9y^2xz\right) \cdot 1}\right)}\right)^{3}=\left(\sum_{x,y,z}{x}\right)^{3}.$$ So we have to prove that :
$$\large\frac{\left(\sum_{x,y,z}{x}\right)^{3}}{\left(1+9xy^2z+1+9xyz^2+1+9xyz^2\right)\left(1+1+1\right)} \geq \frac{(x+y+z)^3}{18} $$ or
$$3\cdot \left(3+9xyz\left( x+y+z\right)\right) \leq 18 \Leftrightarrow 3xyz(x+y+z) \leq \left(xy+yz+zx\right)^2=1^2$$
And this is true because
$$x^2y^2+y^2z^2+z^2x^2+2x^2yz+2xy^2z+2xyz^2\geq 3x^2yz+3xy^2z+3xyz^2$$
$$x^2y^2+y^2z^2+z^2x^2\geq x^2yz+xy^2z+xyz^2$$ and this is the followint inequality:
$$a^2+b^2+c^2 \geq ab+bc+ca$$ for:
\begin{eqnarray}
a&=&xy\\
b&=&yz\\
c&=&zx.
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the sum $\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\dots+\frac{1}{99\cdot 100}$ Please help me calculate the following sum
$$\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\dots+\frac{1}{99\cdot 100}$$
| $$\frac{1}{2 \cdot 3} =\frac{1}{2} - \frac{1}{3}$$
$$\frac{1}{3 \cdot 4}=\frac{1}{3} -\frac{1}{4}$$
$$\ldots$$
$$\frac{1}{99 \cdot 100}= \frac{1}{99}-\frac{1}{100}$$
So:
$$\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{99 \cdot 100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+ \ldots-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}.$$
general case:
$$\frac{1}{n \cdot (n+1)}=\frac{1}{n}-\frac{1}{n+1}.$$
| {
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"url": "https://math.stackexchange.com/questions/194846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the sixth Martin quadruple $\sqrt[n]{x_1^k+x_2^k+x_3^k+x_4^k} =\text{Integer}$ for $k=1,2,3$? Define a Martin quadruple {a,b,c,d} as a solution in non-zero integers to the system,
$a+b+c+d = x^2$
$a^2+b^2+c^2+d^2 = y^2$
$a^3+b^3+c^3+d^3 = z^3$
It can be shown that there are an infinite number of solutions. However, the smallest five that are positive and 6th-power primitive (no common factor that is a 6th power, re cyclochaotic's comment below) have the curious linear sums as smooth numbers,
$\begin{aligned}
&10 + 13 + 14 + 44 = 9^2 = 3^4\\
&54 + 109 + 202 + 260 = 25^2 = 5^4\\
&102 + 130 + 234 + 318 = 28^2 = 2^4\cdot7^2\\
&198 + 630 + 1594 + 1674 = 64^2 = 2^{12}\\
&570 + 742 + 1094 + 1690 = 64^2 = 2^{12}\end{aligned}$
found by James Allen and Seiji Tomita. Of course, the squares and the cubes of the addends also add up to a square and cube, respectively.
What is the sixth such quadruple?
| The sixth primitive positive one is
$$
\{a,b,c,d\}=\{1630,2594, 3562, 3878 \}.
$$
$$
1630+2594+3562+3878=108^2=2^4 3^6;
$$
$$
1630^2+2594^2+3562^2+3878^2=6092^2;
$$
$$
1630^3+2594^3+3562^3+3878^3=5004^3.
$$
(update)
The $7$th such quadruple is
$$
\{a,b,c,d\}=\{259, 1307, 3485, 9349 \}.
$$
$$
259+1307+3485+9349=120^2=2^6 3^2 5^2;
$$
$$
259^2+1307^2+3485^2+9349^2=10066^2;
$$
$$
259^3+1307^3+3485^3+9349^3=9516^3.
$$
| {
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"url": "https://math.stackexchange.com/questions/198002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Find all positive integers $n$ such that $n$ is divisible by all the positive integers less than or equal to $\sqrt{n}$ Find all positive integers $n$ such that $n$ is divisible by all the
positive integers less than or equal to $\sqrt{n}$
My thought:
If n is a positive integer, let d(n) denote the number of positive
divisors of n and I will find n such that $d(n) \ge 2\sqrt{n} - 1$
| By manual testing, $n=1$, $n=2$, $n=4$, $n=6$, $n=8$ have the desired property and are the only numbers $<9$ with it.
Let $n$ be a natural number that is divisible by all natural numbers $\le \sqrt n$. So if we let $a=\lfloor \sqrt n\rfloor$ then $a$ and $a-1$ (if $>0$) are by divisors of $n$. If we additionally assume $n\ge 9$ then $a\ge 3$.
Since $a$ and $a-1$ are relatively prime, $a^2-a=a(a-1)$ divides $n$. Thus $n=k(a^2-a)$ for some $k$. Clearly $k>1$ because $n\ge a^2$. Hence $(a+1)^2>n\ge 2a^2-2a$, i.e., $0\ge a^2-4a=(a-4)a$.
This implies $a\le 4$.
If $a=3$ we find $9\le n<16$ and $n$ must be a multiple of $2$ and $3$. We need only check $n=12$, which gives a solution.
If on the other hand $a=4$, then $16\le n<25$ and $n$ is a multiple of $3$ and $4$. We need only check $n=24$, which gives a solution.
Thus the positive integers with the property are precisely $$1, 2, 3, 4, 6, 8, 12, 24.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof the logarithmic identity $\log_{b+c} a+\log_{c-b} a=2\log_{b+c} a\cdot\log_{c-b} a$ Please help me proof $\log_{b+c} a+\log_{c-b} a=2\log_{b+c} a\cdot\log_{c-b} a$, for $a,b,c>0$ and $a^2+b^2=c^2$. Thanks.
| $$\log_{c+b}a+\log_{c-b}a=\frac{\log a}{\log (b+c)}+\frac{\log a}{\log (c-b)}$$ $$=(\log a)(\frac{1}{\log (b+c)}+\frac{1}{\log (c-b)})$$ $$=(\log a)(\frac{\log(c+b)+\log(c-b)}{\log(c+b)\log(c-b)})$$ $$=(\log a)(\frac{\log(c^2-b^2)}{\log(c+b)\log(c-b)})$$ $$=(\log a)(\frac{\log(a^2)}{\log(c+b)\log(c-b)})$$ $$=2(\frac{\log a}{\log(c+b)})(\frac{\log a}{\log(c-b)})$$ $$=2\log_{c+b}a\cdot\log_{c-b} a$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/200243",
"timestamp": "2023-03-29T00:00:00",
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How I can prove that the sequence $\sqrt{2} , \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}$ converges to 2?
Prove that the sequence $\sqrt{2} , \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}} \ $ converges to $2$.
My attempt
I proved that the sequence is increasing and bounded by $2$, can anyone help me show that the sequence converges to $2$?
Thanks for your help.
| You can also observe that
$$\sqrt{2\sqrt{2\sqrt{2 ...\sqrt{2}}}} \cdot \sqrt{\sqrt{\sqrt{...\sqrt{2}}}}=2$$
thus
$$\sqrt{2\sqrt{2\sqrt{2 ...\sqrt{2}}}} =\frac{2}{\sqrt[2^n]{2}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $0, \frac{1}{2}, 0, \frac{1}{3}, \frac{2}{3}, 0, \frac{1}{4}, \frac{2}{4}, ...$ equidistributed in $[0, 1)$ Prove
$0, \frac{1}{2}, 0, \frac{1}{3}, \frac{2}{3}, 0, \frac{1}{4}, \frac{2}{4}, ...$
equidistributed in $[0, 1)$.
A sequence of numbers $\xi_1, \xi_2, \xi_3, ...$ in $[0, 1)$ is said to be equidistributed if for every
interval $(a, b) \subset [0, 1)$
$$\lim\limits_{N\to\infty} \frac {\bigl|\{1\le n\le N: \xi_n \in (a, b)\}\bigr|} {N}
= b-a$$
It's from Chapter 4 in the book Fourier Analysis: An Introduction
| Let $a < b$ be given. We have for $0 \le p < q$ that $\frac pq \in (a,b)$ iff $qa < p < qb$ iff $\lfloor qa\rfloor < p < \lceil qb \rceil$.
For $N \in \mathbb N$, choose the maximal $q$ with $\frac 12q(q+1) < N$, then all fractions with denominator $\le q$ have apperead under $(\xi_n)_{n\le N}$. We have
\begin{align*}
\left|\{1 \le n \le N: \xi_n \in(a,b)\}\right| &\ge \sum_{r=2}^q \bigl(\lceil rb \rceil - \lfloor ra\rfloor - 1\bigr)\\
&\ge \sum_{r=2}^q \bigl(r(b-a) - 3\bigr)\\
&= \left(\frac 12q(q+1) - 1\right)(b-a) - 3(q-1)
\end{align*}
But now
\[ \frac 12 q(q+1) < N \le \frac 12 (q+1)(q+2) \]
hence
\[ 1 \le \frac N{\frac 12q(q+1)} \le \frac{(q+1)(q+2)}{q(q+1)} \to 1, q \to \infty \]
So, for $N \to \infty$
\begin{align*}
\frac 1N \left|\{1 \le n \le N: \xi_n \in(a,b)\}\right|
&\ge \frac 1N\left(\frac 12q(q+1) - 1\right)(b-a) - \frac 3N(q-1)\\
&\to b-a.
\end{align*}
| {
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Prove the trigonometric identity Please help me prove the identity:
$$\cos^2\alpha-\cos^4\alpha+\sin^4\alpha=\frac{1}{2}-\frac{1}{2}\cos2\alpha$$
| Use the identities, $\sin^2\alpha+\cos^2\alpha=1$ and $\cos2\alpha=1-2\sin^2\alpha$
Since, $\cos^2\alpha-\cos^4\alpha=\cos^2\alpha(1-\cos^2\alpha)=\cos^2\alpha\cdot\sin^2\alpha$
So, $$\cos^2\alpha-\cos^4\alpha+\sin^4\alpha=\cos^2\alpha\cdot\sin^2\alpha+\sin^4\alpha$$ $$=\sin^2\alpha(\cos^2\alpha+\sin^2\alpha)$$ $$=\sin^2\alpha=\frac{1-\cos2\alpha}{2}$$
| {
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"url": "https://math.stackexchange.com/questions/202760",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Should L'Hopital's Rule be used for this limit? Question
$$\lim_{x \to 0} \left(\dfrac{\sin x}{x}\right)^{\dfrac{1}{1 - \cos x}}$$
Attempt
This limit is equal to,
$$\lim_{x \to 0} \exp\left({\dfrac{1}{1 - \cos x}\ln \left(\dfrac{\sin x}{x}\right)}\right).$$
I hope to solve this by focusing on the limit,
$$\lim_{x \to 0} {\dfrac{1}{1 - \cos x}\ln \left(\dfrac{\sin x}{x}\right)}.$$
Which is indeterminate of the form "$\frac{0}{0}$". However, repeated application of L'Hopital's Rule seems to lead to an endless spiral of trigonometric terms, but wolfram Alpha says the limit I am focusing on exists.
Is there another way to solve this?
| One can also simplify as follows:
$\displaystyle \begin{aligned}L &= \lim_{x \to 0}\dfrac{1}{1 - \cos x}\log\left(\dfrac{\sin x}{x}\right)\\
&= \lim_{x \to 0}\dfrac{1}{1 - \cos x}\log\left(1 + \dfrac{\sin x}{x} - 1\right)\\
&= \lim_{x \to 0}\dfrac{1}{1 - \cos x}\left(\dfrac{\sin x}{x} - 1\right)\dfrac{\log\left(1 + \dfrac{\sin x}{x} - 1\right)}{\left(\dfrac{\sin x}{x} - 1\right)}\\
&= \lim_{x \to 0}\dfrac{1}{1 - \cos x}\left(\dfrac{\sin x}{x} - 1\right)\cdot 1\\
&= \lim_{x \to 0}\dfrac{x^{2}}{1 - \cos x}\cdot \dfrac{\sin x - x}{x^{3}}\\
&= \lim_{x \to 0}\dfrac{x^{2}}{1 - \cos x}\cdot\lim_{x \to 0}\dfrac{\sin x - x}{x^{3}}\\
&= \lim_{x \to 0}\dfrac{x^{2}}{1 - \cos x}\cdot\lim_{x \to 0}\dfrac{\cos x - 1}{3x^{2}}\text{ (by L'Hospital's Rule)}\\
&= \lim_{x \to 0}\dfrac{x^{2}}{1 - \cos x}\cdot\dfrac{\cos x - 1}{3x^{2}}\\
&= -\frac{1}{3}\end{aligned}$
This solves the problem with minimal use of L'Hospital Rule. We have made use of the standard limit $\lim_{y \to 0}\dfrac{\log (1 + y)}{y} = 1$ where $y = \dfrac{\sin x}{x} - 1$ which tends to $0$ as $x \to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/206536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
sum of the series
could anyone tell me how to calculate these sums?I am not finding any usual way to calculate them.
| 5.6:
$$\begin{align*}
\frac{1}{2\cdot3}+\frac{1}{4\cdot5}+\frac{1}{6\cdot7}
&=\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\cdots\\
&=\sum_{n=2}^{\infty}\frac{(-1)^{n}}{n}\\
&=\sum_{n=2}^{\infty}\int_{0}^{-1}x^{n-1}dx\\
&=\int_{0}^{-1}\sum_{n=2}^{\infty}x^{n-1}dx\\
&=\int_{0}^{-1}\frac{x}{1-x}dx\\
&=1-\ln2
\end{align*}$$
5.8:
$$\begin{align*}
\frac{1}{3}+\frac{1}{4}\cdot\frac{1}{2!}+\frac{1}{5}\cdot\frac{1}{3!}+\cdots
&=\sum_{n=1}^{\infty}\frac{1}{n+2}\cdot\frac{1}{n!}\\
&=\sum_{n=1}^{\infty}\frac{n+1}{(n+2)!}\\
&=\sum_{n=1}^{\infty}[\frac{1}{(n+1)!}-\frac{1}{(n+2)!}]\\
&=\frac{1}{2}
\end{align*}$$
The sum to $m$ terms is
$$\begin{align*}
\sum_{n=1}^{m}\frac{1}{n+2}\cdot\frac{1}{n!}
&=\sum_{n=1}^{m}\frac{n+1}{(n+2)!}\\
&=\sum_{n=1}^{m}[\frac{1}{(n+1)!}-\frac{1}{(n+2)!}]\\
&=\frac{1}{2}-\frac{1}{(m+2)!}
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/207062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Approximation for $2^r\ln \frac{2^r}{2^r-r}$ I know the function
$$2^r\ln \frac{2^r}{2^r-r}$$
is about linear in $r$, but I need an argument that an undergraduate could follow. Is there a simple way to explain this? I'd be happy with a simple upper-bound.
| Consider the function $f(\theta) = -\log(1-\theta)$. If we take $|\theta| < \frac{1}{2}$, the power series expansion of $f$ is given by $f(\theta) = \theta + \sum_{k=2}^\infty \frac{\theta^k}{k}$, from which we can derive the estimate $|f(\theta) - \theta| \leq K |\theta|^2$, for some constant $K$.
Let $x>2$, then this gives $|f(\frac{1}{x}) - \frac{1}{x}| \leq K \frac{1}{x^2}$, from which we obtain $|x f(\frac{1}{x}) - 1| \leq K \frac{1}{x}$.
Now note that $\frac{2^r}{r} \log(\frac{2^r}{2^r-r}) = \frac{2^r}{r} \log(\frac{1}{1-\frac{1}{2^r/r}}) = -\frac{2^r}{r} \log(1-\frac{1}{2^r/r}) =\frac{2^r}{r} f(\frac{r}{2^r})$. Using the above estimate, we get:
$$|\frac{2^r}{r} \log(\frac{2^r}{2^r-r})-1| \leq K \frac{r}{2^r}$$
which is valid when $2^r > 2 r$. Multiplying through by $r$ gives
$$|2^r \log(\frac{2^r}{2^r-r})-r| \leq K \frac{r^2}{2^r},$$
which shows that $2^r \log(\frac{2^r}{2^r-r})$ is arbitrarily close to $r$ for large values of $r$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/207212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Number of solutions of a linear system --- dependence on parameters For which values of $a$ and $b$ does the system:
$3x+ay=b$
$ax+3y=b$
have:
$a)$ no solutions,
$b)$ 1 solution,
$c)$ infinite solutions.
How to tackle this? Please don't use advanced mathematics..
| If $(x,y)$ is a solution, then by subtracting we get
$$3x-ax+ay-3y=b-b=0.$$
This can be rewritten as
$$(3-a)(x-y)=0.$$
If $a=3$, the equation holds for all $x$ and $y$. So let's check what happens with our original equations when $a=3$.
We get $3x+3y=b$, $3y+3x=b$. For any $b$, there are infinitely many solutions, since we can arrange in infinitely many ways to have $x+y=b/3$.
Thus if $a=3$ and $b$ is anything, there is a unique solution.
We have dealt with the case $a=3$. Now suppose $a\ne 3$. Then from the equation $(3-a)(x-y)=0$ we get $x=y$.
Now go back to the original equations, with $a\ne 3$ and $x=y=t$.
The first equation reads $3t+at=b$, the second reads $at+3t=b$, the same. If $a+3=0$ and $b\ne 0$, there is no solution. So there is no solution if $a=-3$ and $b\ne 0$.
If $a=-3$, and $b=0$, there are infinitely many solutions, since any $t$ works.
If $a\ne -3$, we can comfortably divide, getting the unique soluton $x=y=t=\dfrac{b}{a+3}$.
Summary: (1) No solution if $a=-3$ and $b\ne0$.
(2) Infinitely many solutions if $a=3$ and $b$ is anything; also if $a=-3$ and $b=0$.
(3) Otherwise, unique solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/207288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
at least $a\sqrt{n}$ solutions
Moderator Note: At the time that this question was posted, it was from an ongoing contest. The relevant deadline has now passed.
Consider a number $0<a<2\pi$. How can we show that there is an infinite amount of integers $n$ such that
$x^2 + y^2 + z^2=n$
has at least $a\sqrt{n}$ integer solutions?
I apologize for not editing in the boldened bit earlier.
| Assuming that $a$ and $b$ are fixed and that $x^2+y^2+z^2 = n$ has $\le a \sqrt{n}+b$ integer solutions for all $n$, we get that the number of integer solutions of $x^2+y^2+z^2 < n$ is bounded above by
$$\sum_{k=0}^{n-1} (a \sqrt{k}+b) = bn+ an^{3/2}\sum_{k=0}^{n-1} \sqrt{\frac{k}n} \frac1n \le bn+ a n^{3/2} \int_0^1 \sqrt{x} \, dx = bn+\frac{2an^{3/2}}{3},$$
since the second sum is a lower Riemann sum for the integral of $\sqrt{x}$ over $[0,1]$.
Now since the volume of the ball with radius $\sqrt{n}$ is $\frac{4\pi n^{3/2}}3$, and the number of integer solutions of $x^2+y^2+z^2 <n$ has the same asymptotic growth, we get that $\frac{2a}3 \ge \frac{4\pi}3$, i.e., $a \ge 2\pi$.
Lastly, if there are only finitely many $n$ such that $x^2+y^2+z^2 = n$ has $\ge a\sqrt{n}$ integer solutions, it is straightforward that there exists $b$ such that $x^2+y^2+z^2 = n$ has $\le a\sqrt{n}+b$ integer solutions for all $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/208929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Sum of two absolute values in complex plane I'm trying to find out all $z \in C$ that satisfy the following condition:
$|z+1|+|z-i|=3$
I understand that $|z|=r$ represents a circle with a radius of $r$.
I also understand that $|z+1|=r$ can be written as $|(x+1)+yi|=\sqrt{(x+1)^2+y^2}=r$ which can then be squared to get $(x+1)^2+y^2=r^2$ which represents the circle with a radius of r, with center in $(-1,0)$.
So, back to my problem:
Unlike the example with $|z+1|=r$, where it is easy to square the equation, squaring
$|z+1|+|z-i|=3$ written as $\sqrt{(x+1)^2+y^2}+\sqrt{x^2+(y-1)^2}=3$ equals:
$(x+1)^2+(y-1)^2+x^2+y^2+2\sqrt{(x+1)^2+y^2}\sqrt{x^2+(y-1)^2}=9$ which is a nightmare to solve, if at all possible.
I am sure there must be some elegant way to solve this kind of problem. The way I'm thinking is this:
$|z+1|$ by itself seems to represent a circle centered at $(-1,0)$, with an undefined radius, and $|z-i|$ seems to represent a circle centered at $(0,1)$, also with an undefined radius. However, the sum of those two radii must be 3. But I can't seem to wrap my mind around what this would represent (when drawn on Cartesian plane), or how to solve it analytically. So, how would one approach this problem? Even a hint would (probably) suffice.
| Let $\sqrt{(x-a)^2+(y-b)^2}+\sqrt{(x-c)^2+(y-d)^2}=e$
$\sqrt{(x-a)^2+(y-b)^2}=e-\sqrt{(x-c)^2+(y-d)^2}$
Squaring we get, $(x-a)^2+(y-b)^2=e^2+(x-c)^2+(y-d)^2-2e\sqrt{(x-c)^2+(y-d)^2}$
or, $2e\sqrt{(x-c)^2+(y-d)^2}=e^2+(x-c)^2-(x-a)^2+(y-d)^2-(y-b)^2$
or, $2e\sqrt{(x-c)^2+(y-d)^2}=e^2+(2x-c-a)(a-c)+(2y-d-b)(b-d)$
or, $e\sqrt{(x-c)^2+(y-d)^2}=x(a-c)+y(b-d)+f$ where $2f=e^2-(c+a)(a-c)-(d+b)(b-d)$
Squaring we get, $e^2((x-c)^2+(y-d)^2)=(x(a-c)+y(b-d)+f)^2$
$\implies x^2(e^2-(a-c)^2)-2xy(a-c)(b-d)+y^2(e^2-(b-d)^2)-2x(e^2c+f(a-c))-2y(e^2d+f(b-d))+e^2(c^2+d^2)-f^2=0$
Using this, $4(a-c)^2(b-d)^2-4(e^2-(a-c)^2)(e^2-(b-d)^2)$
$=4e^2((a-c)^2+(b-d)^2-e^2)$
Now $e=\sqrt{(x-a)^2+(y-b)^2}+\sqrt{(x-c)^2+(y-d)^2}$
$=|(x,y)-(a,b)|+|(x,y)-(c,d)|<|(a,b)-(c,d)|$
So, $(a-c)^2+(b-d)^2-e^2<0$, so the locus of $(x,y)$ is an ellipse ,but not a circle unless the coefficient of $xy$ is $0$.
Alternatively,
$\sqrt{(x+1)^2+y^2}+\sqrt{x^2+(y-1)^2}=3$
$\sqrt{(x+1)^2+y^2}=3-\sqrt{x^2+(y-1)^2}$
Squaring we get, $ (x+1)^2+y^2=9+x^2+(y-1)^2-6\sqrt{x^2+(y-1)^2}$
$2x+2y-8=-6\sqrt{x^2+(y-1)^2}$
$x+y-4=-3\sqrt{x^2+(y-1)^2}$
Squaring we get, $(x+y-4)^2=9(x^2+(y-1)^2)$
$8x^2-2xy+8y^2+8x-10y-7=0$
Using this or this, $(-2)^2-4\cdot 8 \cdot 8<0$
So, the curve is an ellipse.
Using this , we can remove $xy$ terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/209794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Analytic Function with positive integers as zeros? Do you know any nontrivial analytic function f(z) with zeros only at positive integer values of the argument z = 1, 2, 3, 4, ... ?
If yes, please give some example.
PS: I already thought of $f(x)=\frac{1}{\Gamma(-x+1)}$. Any other nice options?
EDIT: To avoid trivial solutions due to restriction of definition range, please consider the required function to be defined in the whole complex plane.
| $$ f(x) = 3^x - 2^x $$
has 0 as a zero.
$$ f(x) = 4^x - 2 \cdot 3^x + 2^x$$
has 0 and 1 as a zero.
$$ f(x) = 5^x - 3\cdot4^x + 3\cdot3^x - 2^x $$
has 0, 1, and 2 as a zero.
$$ f(x) = 6^x - 4\cdot5^x + 6\cdot4^x - 4\cdot3^x + 2^x $$
has 0, 1, 2, and 3 as a zero.
See the pattern. Continuing
$$ f(x) = 7^x - 5\cdot6^x+10\cdot5^x-10\cdot4^x+5\cdot3^x-2^x$$
has 0, 1, 2, 3, and 4 as a zero.
$$ f(x) = 8^x-6\cdot7^x+15\cdot6^x-20\cdot5^x+15\cdot4^x-6\cdot3^x+2^x $$
has 0, 1, 2, 3, 4, and 5 as a zero.
The pattern is the base decreases by one. Signs alternate. The coefficients are the numbers of the Pascal's triangle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/211674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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How to evaluate this integral $\int_{-\infty}^{+\infty}\frac{x^2e^x}{(1+e^x)^2}dx$? I need to evaluate $$\int_{-\infty}^{+\infty}\frac{x^2e^x}{(1+e^x)^2}dx$$
I think the answer is $\frac{\pi^2}{3}$, but I'm not able to calculate it.
| $$f(x) = \dfrac{x^2 \exp(x)}{(1+\exp(x))^2} = \dfrac{x^2 \exp(-x)}{\left(1 + \exp(-x) \right)^2}$$
Recall that
$$\dfrac{a}{(1+a)^2} = a -2a^2 + 3a^3 - 4a^4 + 5a^5 \mp \cdots = \sum_{k=1}^{\infty}(-1)^{k+1}k a^k$$
For $x > 0$, $$f(x) = x^2 \sum_{k=1}^{\infty} (-1)^{k+1} k \exp(-kx)$$
Now for $a > 0$, $$\int_0^{\infty} x^2 \exp(-ax) = \dfrac2{a^3}$$
Hence, $$\int_0^{\infty} f(x) dx = \sum_{k=1}^{\infty} (-1)^{k+1} \dfrac{2k}{k^3} = 2 \sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k^2} = \dfrac{\pi^2}6$$
Hence, $$\int_{-\infty}^{\infty} f(x) dx = 2 \int_0^{\infty} f(x) dx = \dfrac{\pi^2}3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/213818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
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How to integrate $\int \frac{2x^2+x}{(x+1)(x^2+1)}dx$? How to integrate $\displaystyle \int \frac{2x^2+x}{(x+1)(x^2+1)}dx$? I Tried using partial fractions but i got lost, thanks.
| The key is to write $(2x^2 + x)$ as $A(x^2+1) + (Bx+C)(x+1)$
$$(2x^2 + x) = A(x^2+1) + (Bx+C)(x+1) = (A+B)x^2 + (B+C)x + (A+C)$$
This gives us $A+B = 2$, $B+C = 1$ and $A+C = 0$ i.e. $A+B = 2$ and $B-A = 1$.
$$A = \dfrac12, B = \dfrac32, C = -\dfrac12$$
Hence, $$\dfrac{2x^2+x}{(x+1)(x^2+1)} = \dfrac1{2(x+1)} + \dfrac{3x-1}{2(x^2+1)} = \dfrac1{2(x+1)} + \dfrac34 \dfrac{2x}{x^2+1} - \dfrac12 \dfrac1{x^2+1}$$
Now you should be able to integrate it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/216742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$f(x)$ is a monotonic function prove the following Prove that:
$ \rightarrow\sum_{k=1}^n f(\frac{k}{n})\sum_{k=1}^n k{f(\frac{k}{n})}^2\le\sum_{k=1}^n kf(\frac{k}{n})\sum_{k=1}^n{f(\frac{k}{n})}^2 $
Given $f(x)$ is a positive function and also monotonic decreasing function.
| Denote $a_k=f\left(\frac{k}{n}\right).$
(Proof by induction). Let $P(n)$ be the statement $$\sum\limits_{k=1}^n f\left(\frac{k}{n}\right)\sum\limits_{k=1}^n k\left(f\left(\frac{k}{n}\right)\right)^2\leqslant\sum\limits_{k=1}^n kf\left(\frac{k}{n}\right)\sum\limits_{k=1}^n\left(f\left(\frac{k}{n}\right)\right)^2$$ or, in shorter form
$$\sum\limits_{k=1}^n a_k\sum\limits_{k=1}^n ka_k^2\leqslant\sum\limits_{k=1}^n ka_k\sum\limits_{k=1}^na_k^2.$$
*
*(Basis) For $n=1$ we have: $a_1\cdot a_1= a_1\cdot a_1$
*(Induction step). We assume that $P(n)$ is true and prove that $P(n+1)$ is true. For LHS of $P(n+1)$:
\begin{equation}
\sum\limits_{k=1}^{n+1} a_k\sum\limits_{k=1}^{n+1} ka_k^2 =
\left(\sum\limits_{k=1}^{n} a_k +a_{n+1} \right) \left(\sum\limits_{k=1}^{n} k a_k^2 + (n+1) a_{n+1}^2 \right)=\\
\sum\limits_{k=1}^{n} a_k\sum\limits_{k=1}^{n} ka_k^2 + (n+1)a_{n+1}^2 \sum\limits_{k=1}^{n} a_k + a_{n+1} \sum\limits_{k=1}^{n} ka_k^2 + (n+1)a_{n+1}^3 \leqslant \\
\sum\limits_{k=1}^n ka_k\sum\limits_{k=1}^na_k^2 + (n+1)a_{n+1}^2 \sum\limits_{k=1}^{n} a_k + a_{n+1} \sum\limits_{k=1}^{n} ka_k^2 + (n+1)a_{n+1}^3\overset{def}{=} A.
\end{equation} RHS of $P(n+1)$ equals:
\begin{equation}\sum\limits_{k=1}^{n+1} ka_k\sum\limits_{k=1}^{n+1} a_k^2 = \left(\sum\limits_{k=1}^{n} ka_k + (n+1)a_{n+1}\right)\left(\sum\limits_{k=1}^{n} a_k^2 + a_{n+1}^2 \right) = \\
\sum\limits_{k=1}^n ka_k\sum\limits_{k=1}^na_k^2 + (n+1)a_{n+1}\sum\limits_{k=1}^{n} a_k^2 +a_{n+1}^2\sum\limits_{k=1}^{n} ka_k + (n+1)a_{n+1}^3\overset{def}{=} B.
\end{equation} Next,
\begin{equation}
A-B= \left((n+1)a_{n+1}^2 \sum\limits_{k=1}^{n} a_k + a_{n+1} \sum\limits_{k=1}^{n} ka_k^2 \right) - \\
\left( (n+1)a_{n+1}\sum\limits_{k=1}^{n} a_k^2 +a_{n+1}^2\sum\limits_{k=1}^{n} ka_k \right) = \\
(n+1)a_{n+1} \left(a_{n+1}\sum\limits_{k=1}^{n} a_k -\sum\limits_{k=1}^{n} a_k^2 \right) + a_{n+1} \left(\sum\limits_{k=1}^{n} ka_k^2 - a_{n+1}\sum\limits_{k=1}^{n} k a_k \right)=\\
a_{n+1}^2 \sum\limits_{k=1}^{n} \left((n+1)a_k -k a_k \right) + a_{n+1}\sum\limits_{k=1}^{n} \left(k a_k^2 - (n+1)a_k^2 \right)=\\
a_{n+1}^2 \sum\limits_{k=1}^{n} {a_k(n+1-k)} - a_{n+1} \sum\limits_{k=1}^{n} {a_k^2(n+1-k)}=\\
a_{n+1} \sum\limits_{k=1}^{n} \left(a_{n+1}a_k (n+1-k) - {a_k^2(n+1-k)} \right)=\\
a_{n+1} \sum\limits_{k=1}^{n}{(n+1-k)(a_{n+1}a_k - a_k^2)}=\\
a_{n+1} \sum\limits_{k=1}^{n}{(n+1-k)a_k(a_{n+1} - a_k)}<0
\end{equation} since $\{a_k \}$ decreases.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the possible values of $a$, $b$ and $c$? Given $(a,\space b,\space c)\in \mathbb Z^3$ and that $$\sqrt[3]{\sqrt{a}+\sqrt{b}} + \sqrt[3]{\sqrt{a}-\sqrt{b}} = c$$
Find the possible values of $a $, $b$, and $c$.
| We try to find all solutions in integers $a,b,c$ for which the square and cube root are unambiguously defined, that is we require $a\ge0$ and $b\ge0$.
Let $u=\sqrt[3]{\sqrt a+\sqrt b}$, $v=\sqrt[3]{\sqrt a-\sqrt b}$, $w=uv=\sqrt[3]{a-b}$. (Note that $u,v,w$ need not be integers).
Note that $u^3+v^3=2\sqrt a\ge0$ implies $u^3\ge -v^3$ and hence $u\ge-v$ and finally $c\ge0$. The case $c=0$ leads to $u=-v$ and hence $a=0$. Then we find the solutions
$$\tag1(0,b,0)\quad\text{with arbitrary }b\ge0.$$
For the rest of the argument we may assume that $c>0$.
Moreover $u^3+v^3=2\sqrt a$ implies
$$\tag22\sqrt a = u^3+v^3=(u+v)^3-3(u+v)uv=c^3-3cw,$$
hence by isolating $-3cw$ and cubing
$$-c^{27}+6c^9\sqrt a-12c^3a+8a\sqrt a=-27c^3(a-b),$$
i.e.
$$\tag33(3c^9+4a)\sqrt a \in\mathbb Z.$$
Since $c>0$ and $a\ge0$, we have $3c^9+4a\ne0$ and conclude that $a=d^2$ is a perfect square with $d\in\mathbb Z$.
Next observe that
$$4b=(u^3-v^3)^2=u^6-3u^3v^3+v^6\\=(u+v)^6-6uv(u+v)^4+9(uv)^2(u+v)^2-4(uv)^3
\\=c^6-6c^4w+9c^2w^2-4(a-b).$$
Thus $w$ is root of a quadratic and of a cubic rational polynomial, hence is rational, i.e. $a-b$ is a perfect cube, say $a=b+e^3$ with $e\in \mathbb Z$.
With this, $(2)$ becomes
$$\tag42d=c^3-3ce.$$
Note that $u,v$ are roots of
$$x^2-cx+e = x^2-(u+v)x+uv= 0,$$
i.e.
$$\tag5u=\frac{c+\sqrt{c^2-4e}}2\quad v=\frac{c-\sqrt{c^2-4e}}2$$
and we require $c^2\ge4e$.
Now let us go backwards:
Select integers $c>0$ and $e\le\frac{c^2}4$ such that $c\equiv0\pmod 2$ or $e\equiv 1\pmod 2$.
Then $a:=\frac{c^2(c^2-3e)^2}4$ is a nonnegative integer.
Set $b:=a-e^3$. Then $b\ge0$ because either $e\le 0$ and then $b\ge a\ge0$; or $e>0$ and then $c^2-3e\ge e>0$, i.e. $b = \frac{c^2(c^2-3e)^2-4e^3}4\ge \frac{c^2e^2-4e^3}4=\frac{(c^2-4e)e^2}4\ge0$.
With these values, $(a,b,c)$ is a solution.
With nice parametrizations depending on the parity of $c$ we thus find for even $c=2m$ and $e=m^2-n$ ($m>0$ ,$n\ge0$):
$$\tag6\begin{matrix} a&=&m^2(m^2+3n)^2,\\
b&=&m^2(m^2+3n)^2-(m^2-n)^3=n(3m^2+n)^2,\\
c&=&2m.\end{matrix}$$
And for odd $c$ and odd $e$ ($c=2m+1$, $e=m(m+1)-2n-1$ with $m,n\ge0$):
$$\tag7\begin{matrix}a&=&(2m+1)^2\left(\frac{m(m+1)}2+2+3n\right)^2,\\
b&=&(2m+1)^2\left(\frac{m(m+1)}2+2+3n\right)^2-\left(m(m+1)-2n-1\right)^3,\\
c&=&2m+1.\end{matrix}$$
The solutions given by $(1),(6),(7)$ are complete.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
How to solve this trigonometric equation? I want to solve the equation $$(3-\cos 4x)(\sin x - \cos x)=2.$$
I solve by putting $t = \sin x - \cos x$, but I can not find all solutions.
| I have just a solution.
\begin{equation*}
(3-\cos 4x)(\sin x - \cos x) = 2.
\end{equation*}
Note that, $3-\cos 4x > 0$, therefore $\sin x - \cos x>0$.
Put $t = \sin x - \cos x = \sqrt{2}\sin\left(x - \dfrac{\pi}{4}\right)$, and then $0<t \leqslant \sqrt{2}$. We have
$$ t^2 = (\sin x - \cos x)^2 = 1 -\sin 2x.$$
Another way,
$$3 - \cos 4x =3 - (1-2\sin^2 2x )= 2[1-(1-t^2)^2] = 2(t^4 -2t^2 + 2)$$
The given equation has the form
$$2(t^4 -2t^2 + 2)t = 2,$$
equavalent to
$$t^5- 2t^3+2t - 1 = 0.$$ Now we prove that this equation has only root $t = 1$.
First way. The function $f(t)=t^5 -2t^3 +2t - 1$ has $$f'(t) = 5t^4-6t^2+2>0, \forall t$$
Therefore, $f$ is an increasing function on the interval $(0; \sqrt{2}]$. And $f(1) = 0$, thus $t = 1$ is only root.
Second way, we have
$$t^5- 2t^3+2t - 1 = 0 \Leftrightarrow (t-1)(t^4 +t^3 -t^2 -t + 1)=0.$$ Note that
$$t^4 +t^3 -t^2 -t + 1 = \biggl(t^2 + \dfrac{t}2 -1\biggr)^2 + \dfrac{3t^2}4 > 0, \forall t.$$
Thirt way, we have
$$t^4 +t^3 -t^2 -t + 1=0 \Leftrightarrow \biggl(t - \dfrac{1}{t}\biggr)^2 + \biggl(t - \dfrac{1}{t}\biggr) + 1 = 0. $$
The last equation has no sulution.
With $t = 1$, we have $\sin x - \cos x = 1$, then $x = \dfrac{\pi}2 + k2\pi$ and $x = \pi + 2m\pi.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Range of a trignonmetric function I came across this in an Engineering entrance book,
What is the range of this: $a^2 \sin^2 x + b \sin x \cos x + c \cos^2 x$
What is the method to find it? I tried the graph approach but didn't know how to proceed.
| Hint: Use the two double angle identities $\sin 2x=2\sin x\cos x$ and $\cos 2x=2\cos^2 x-1=1-2\sin^2 x$.
After a short while you will find that your expression is equal to an expression of the shape $A\sin 2x+B\cos 2x +C$, where $A$, $B$, and $C$ are easy to compute in terms of your parameters $a$, $b$, and $c$.
The $C$ is harmless. As for $A\sin 2x+B\cos 2x$, write it as
$$\sqrt{A^2+B^2}\left((\sin 2x)\frac{A}{\sqrt{A^2+B^2}}+(\cos 2x)\frac{B}{\sqrt{A^2+B^2}}\right).$$
There is an angle $\theta$ whose cosine is $A/\sqrt{A^2+B^2}$ and whose sine is $B/\sqrt{A^2+B^2}$. So we are looking at
$$\sqrt{A^2+B^2}\sin(2x+\theta).$$
The sine has maximum value $1$ and minimum value $-1$. Now we can find the range.
There is only one minor hitch. For some values of the parameters (as currently given, $b=0$ and $c=a^2$) we have $A=B=0$. So the proof breaks down since we cannot divide by $\sqrt{A^2+B^2}$. But that case is easy, the function is constant.
| {
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} |
Prove that $\frac{1}{4-\sec^{2}(2\pi/7)} + \frac{1}{4-\sec^{2}(4\pi/7)} + \frac{1}{4-\sec^{2}(6\pi/7)} = 1$
How can I prove the fact $$\frac{1}{4-\sec^{2}\frac{2\pi}{7}} + \frac{1}{4-\sec^{2}\frac{4\pi}{7}} + \frac{1}{4-\sec^{2}\frac{6\pi}{7}} = 1.$$
When asked somebody told me to use the ideas of Chebyshev polynomial, but I haven't learnt that in school.
I tried doing this way:
Look at $y =\cos\theta + i \sin\theta$ where $\displaystyle\theta \in \Bigl\{\frac{2\pi}{7},\frac{4\pi}{7},\cdots,2\pi\Bigr\}$
Then we have
\begin{align*}
y^{7} &=1 \\ y^{7}-1 &=0 \\ (y-1) \cdot (y^{6}+y^{5}+\cdots + 1) &= 0
\end{align*}
Now the root $y=1$ corresponds to $\theta = 2\pi$, and that $$y^{6} + y^{5}+\cdots + 1 =0$$
have roots $\cos\theta + i \sin\theta$, where $\theta \in \Bigl\{\frac{2\pi}{7},\frac{4\pi}{7} ,\cdots \Bigr\}$. Looking at $y+\frac{1}{y} $ will give me the roots as $\cos\theta$ and then i can put $z=y^{2}$ to get $\cos^{2}$ as the roots and the invert to get $\sec^{2}$, but I have some problems.
Can anyone help me out with a neat solution.
Thanks.
|
Motto: Use as much as possible the set of complex roots of unity as a whole rather than each complex root in isolation.
Sub-motto: The complex exponential is easier to use than the sine and cosine functions hence reducing the latter ones to the former is often fruitful.
Note that, for every $x$ such that this fraction exists,
$$
\frac1{4-\sec^2x}=\frac14+\frac14\frac1{2\cos(2x)+1}=\frac14+\frac14\frac{\mathrm e^{2\mathrm ix}}{\mathrm e^{4\mathrm ix}+\mathrm e^{2\mathrm ix}+1},
$$
hence the identity to be proved is
$$
\sum\limits_z\frac{z^2}{z^4+z^{2}+1}=1,
$$
where the sum is over $z$ in $\{\omega,\omega^2,\omega^3\}$ with $\omega=\mathrm e^{2\mathrm i\pi/7}$.
If $z^7=1$ and $z\ne1$, then $\dfrac{z^2}{z^4+z^2+1}=-z^3-z^{4}$
hence the sum in the LHS is
$$
-\omega^3-\omega^{4}-\omega^6-\omega^{8}-\omega^9-\omega^{12},
$$
that is,
$$
-\omega^3-\omega^{4}-\omega^6-\omega-\omega^2-\omega^5=1-\sum\limits_{k=1}^7\omega^k=1.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\lim\limits_{x\to 0}\frac{x^2-1+\cos^2x}{x^4+x^3\sin x}=\frac{1}{6}$ without De L'Hôpital Rule or Taylor Expansion? Any ideas how to show $\lim\limits_{x\to 0}\dfrac{x^2-1+\cos^2x}{x^4+x^3\sin x}=\lim\limits_{x\to 0}\dfrac{x-\sin(x)}{x^3}=\dfrac{1}{6}$ without using the De L'Hôpital rule (or proving a special case of it?).
How can we reduce this to $\lim\limits_{x\to 0}\dfrac{1-\cos(x)}{x^2}=\dfrac{1}{2}$?
You can suppose that we know the limit in question exists
and therefore use inequalities to bound it
| Use $1-\cos^2(x) = \sin^2(x)$:
$$
\lim_{x\to 0} \frac{x^2-1+\cos^2(x)}{x^4 + x^3 \sin(x)} = \lim_{x\to 0} \frac{x^2 - \sin^2(x)}{x^4(1+\frac{\sin(x)}{x})} = \lim_{x\to 0} \frac{1 - \left(\frac{\sin(x)}{x}\right)^2}{x^2(1+\frac{\sin(x)}{x})} = \lim_{x\to 0} \frac{1 - \frac{\sin(x)}{x}}{x^2} = \frac{1}{6}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/226867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
wave equation with initial values and boundary condititon I have a homogenious 2-dimensional wave equation:
$$ - \frac{\partial^2 u}{\partial x^2} (x, y, t) - \frac{\partial^2 u}{\partial y^2} (x, y, t) + \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2} (x, y, t) = 0$$
With:
$$ 0 < x < a, 0 < y < b, t > 0$$
$$ u(x, y, 0) = 0$$
$$ \dot u(x, y, 0) = x(x-a)(y-b)$$
And boundary condition:
$$ u(0, y, t) = u(a, y, t) = u(x, 0, t) = u(x, b, t) = 0$$
It says that I should solve it using separation of variables with $u(x, y, t) = G(x)H(y)w(t)$ as a starting point. With that approach, I got this solution:
$$ u(x, y, t)
= \sin\left(\frac{n\pi}a x\right)
\sin\left(\frac{n\pi}b y\right)
\sin\left(\left(\frac{n\pi}a - \frac{n\pi}b\right) c^2 t\right)
$$
This solved everything, except the initial value for $\dot u$. A friend of mine proved that if such a seperated function would solve this initial value, the function $w(t)$ would depend on $x$ and $y$.
How do I solve this differential equation so that it satifies the $\dot u$ initial value?
| In a more simple way, take $u(x,y,t) = T(t) f(x,y)$ and substitute in your equation. Then
$$
\frac{1}{c^2 T(t)} \frac{d^2 T(t)}{d t^2} = \frac{1}{f(x,y)} \Delta f(x,y)
$$
Where $\Delta$ is the Laplace operator. The only way for this equality to be satisfied for all $(x,y,t)$ is that both sides of it are a constant, lets say $-\lambda$. Then
\begin{align}
\frac{d^2 T}{d t^2} + \lambda c^2 T &= 0\\
\Delta f + \lambda f &= 0
\end{align}
Now, looking at the spatial part, take $f(x,y) = X(x)Y(y)$, then
$$
\frac{1}{X(x)}\frac{d^2 X(x)}{d x^2} + \frac{1}{Y} \frac{d^2 Y(y)}{d y^2} + \lambda = 0
$$
The same argument as above, leads to the two ode's
\begin{align}
\frac{d^2 X}{d x^2} + \lambda_x X &= 0\\
\frac{d^2 Y}{d y^2} + \lambda_y Y &= 0
\end{align}
where $\lambda_x + \lambda_y = \lambda$. The solutions for both equations are
\begin{align}
X(x) &= A \sin(\sqrt{\lambda_x} x)\\
Y(y) &= B \sin(\sqrt{\lambda_y} y)
\end{align}
where we have made use that $u(0,y,t) = u(x,0,t) = 0$. The conditions on the other side of the domain will either lead to the trivial solutions or to a restriction on the values of $\lambda_{x,y}$. Hence
$$
\lambda_x = \frac{m^2 \pi^2}{a^2}, \qquad \lambda_y = \frac{n^2 \pi^2}{b^2}
$$
where $m,\,n\, \in \mathbb{Z}$. Then
$$
f_{nm}(x,y) = A_{mn} \sin\big(\tfrac{m \pi}{a} x\big) \sin\big(\tfrac{n \pi}{b} y\big)
$$
and
$$
T_{mn}(t) = C \sin\Big(\sqrt{\tfrac{m^2 \pi^2}{a^2} + \tfrac{n^2 \pi^2}{b^2}}\, ct\Big) + D \cos\Big(\sqrt{\tfrac{m^2 \pi^2}{a^2} + \tfrac{n^2 \pi^2}{b^2}}\, ct\Big)
$$
Finally, due linearity of the PDE and homogeneous boundary conditions, the solution is
$$
u(x,y,t) = \sum_{m,n = 0}^\infty B_{mn} \sin\big(\tfrac{m \pi}{a} x\big) \sin\big(\tfrac{n \pi}{b} y\big) \sin\Big(\sqrt{\tfrac{m^2 \pi^2}{a^2} + \tfrac{n^2 \pi^2}{b^2}}\, ct\Big)
$$
given that $u(x,y,0) = 0$.
To obtain the value of $B_{nm}$, one have to use the other initial condition and the orthogonality of $\{\sin(\frac{m \pi}{a} x)\}_m$, $\{\sin(\frac{m \pi}{b} y)\}_n$ in the intervals $(0,a)$ and $(0,b)$ respectively. Multiplying by $\sin(\frac{r \pi}{a}x)$ where $r$ is an integer, and integrating from $0$ to $a$, we have
\begin{multline}
\sum_{m,n = 0}^\infty \left(B_{mn} \sqrt{\tfrac{m^2 \pi^2}{a^2} + \tfrac{n^2 \pi^2}{b^2}}\, c\right) \sin\big(\tfrac{n \pi}{b} y\big) \int_0^a \sin\big(\tfrac{m \pi}{a} x\big) \sin\big(\tfrac{r \pi}{a} x\big) dx = \\
(y-b) \int_0^a x(x-a)\sin\big(\tfrac{r \pi}{a} x\big) dx
\end{multline}
which, given orthogonality, leads to
$$
\sum_{n = 0}^\infty \frac{a}{2}\left(B_{rn} \sqrt{\tfrac{r^2 \pi^2}{a^2} + \tfrac{n^2 \pi^2}{b^2}}\, c\right) \sin\big(\tfrac{n \pi}{b} y\big) =
\frac{2 a^3}{r^3 \pi^3}\left[(-1)^r - 1\right](y-b)
$$
Multiplying by $\sin(\frac{s \pi}{b} y)$, were $s$ is an integer, and integrating from $0$ to $b$, we have
$$
\frac{ab}{4}\left(B_{rs} \sqrt{\tfrac{r^2 \pi^2}{a^2} + \tfrac{s^2 \pi^2}{b^2}}\, c\right) = -\frac{2 a^3 b^2}{r^3 s \pi^4}\left[(-1)^r - 1\right]
$$
hence
$$
B_{mn} = -\frac{8 a^2 b \left[(-1)^m - 1\right]}{m^3 n \pi^4 c \sqrt{\tfrac{m^2 \pi^2}{a^2} + \tfrac{n^2 \pi^2}{b^2}}}
$$
and the solution is
$$
u(x,y,t) = -\sum_{m,n = 0}^\infty \tfrac{8 a^2 b \left[(-1)^m - 1\right]}{m^3 n \pi^4 c \sqrt{\tfrac{m^2 \pi^2}{a^2} + \tfrac{n^2 \pi^2}{b^2}}} \sin\big(\tfrac{m \pi}{a} x\big) \sin\big(\tfrac{n \pi}{b} y\big) \sin\Big(\sqrt{\tfrac{m^2 \pi^2}{a^2} + \tfrac{n^2 \pi^2}{b^2}}\, ct\Big)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/229659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
About regular surfaces I never had seen this exercise, but I'm confused again, I don't know what I have to use.
I have the surface $S=\{(x,y,z)\in \mathbb{R}^3|xy+xz+yz=1,x>0,y>0,z>0\}$, is $S$ regular?. Then, if $S$ is regular, I have to found the higher value of $f$ defined by $f(x,y,z)=xyz$ where $(x,y,z)$ is in $S$.
Can anyone help me?
| Define $F:\mathbb{R}_{x,y,z>0} \to \mathbb{R}$ by $F(x,y,z) = xy + xz + yz.$ Then $dF = (y+z, x+z, x+y)^T$ so $F$ has no critical values, and $1$ is a regular value of $F.$ The preimage of a regular value is a regular surface, so $S= F^{-1}(1)$ is a regular surface.
Then for points on the surface $xy+xz+yz=1$, we have $z = \frac{1-xy}{x+y}$ so we have to maximize $$g(x,y)=\frac{ xy (1-xy)}{x+y}$$ over $\mathbb{R_{x,y>0}}.$
We compute that $g_x = -\frac{y}{(x+y)^2} (x^2+2xy-1)$ and $g_y = -\frac{x}{(x+y)^2} (y^2+2xy-1)$ so $g_x=0$ when $y= \frac{1-x^2}{2x}$ and $g_y=0$ when $x= \frac{1-y^2}{2y}.$ When both these hold we have $1-x^2=2xy=1-y^2$ so $x=y$, and $x^2+2xy-1=0$ leads to $x=y=\frac{1}{\sqrt{3}}$ so the maximum value of $g$ is $\frac{1}{3\sqrt{3}}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/234275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Expressing in rationals I have this question to express it in a specific form
Express $\dfrac{1}{\sqrt{2} + \sqrt{3} + \sqrt{7}}$ in the form $a \sqrt{2} + b \sqrt{3} + c \sqrt{7} + d \sqrt{42}$, for some rationals $a,b,c$ and $d$.
So the solution that I have and am trying to understand is this
So what I don't understand is that why in the world is there a 2sqrt6 here where did it come from?
| Just for the record, elaborating on Berci's suggestion, you could also try solving
$$(a \sqrt{2} + b \sqrt{3} + c \sqrt{7} + d \sqrt{42})(\sqrt{2} + \sqrt{3} + \sqrt{7}) = 1.$$
Writing out the product, you get
$$(2a + 3b + 7c) + \sqrt{6}(a + b + 7d) + \sqrt{14}(a + c + 3d) + \sqrt{21}(b + c + 2d) = 1.$$
With all coefficients rational, this implies that the following matrix equation has to be satisfied:
$$\begin{pmatrix} 2 & 3 & 7 & 0 \\ 1 & 1 & 0 & 7 \\ 1 & 0 & 1 & 3 \\ 0 & 1 & 1 & 2 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}.$$
There are several ways to solve such systems, all of them leading to the solution:
$$\begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} = \frac{1}{10} \begin{pmatrix} 4 \\ 3 \\ -1 \\ -1 \end{pmatrix}.$$
This does not involve any tricks with getting rid of square roots, although one may argue this way to get to the result takes more time.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/234872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Geometric identity, cannot show equivalence using trigonometric identities
clearly
$$(x+a \cos\theta)^2+(y-a \sin\theta)^2=b^2$$
expanding and using the Weierstrass substitution we find that
$$\theta= 2 \arctan \frac{\left( 2ay- \sqrt{ 4a^2y^2 - ( (x-a)^2+y^2-b^2)( (x+a)^2+y^2-b^2) }\right)}{(x-a)^2+y^2-b^2} $$
if we use the law of cosines, with $c^2=x^2+y^2$
$$\theta = \arctan_2(y,x) - \arccos( (c^2+a^2-b^2) / (2ac) )$$
Is there a way to pass from one expression to the other using trigonometric identities?
| Try this:
$$(x+a \cos\theta)^2+(y-a \sin\theta)^2=b^2$$
Working through we get $$x\cos\theta-y\sin\theta=\frac {(b^2-a^2-y^2-x^2)}{2a}$$
Now use $x^2+y^2=c^2$ on both sides and set $\phi=\arctan \frac y x$. Divide through by $c$ to obtain:
$$\cos \phi \cos \theta - \sin\phi\sin\theta =\cos(\theta+\phi)=\frac {(b^2-a^2-c^2)}{2ac}$$ and you can do it from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/236510",
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"source": "stackexchange",
"question_score": "4",
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How to find k given determinant? So I've got this matrix here, and need to solve for $k$
$$\text{det}\;\begin{pmatrix}
3 & 2 & -1 & 4 \\
2 & k & 6 & 5 \\
-3& 2 & 1 & 0 \\
6 & 4 & 2 & 3 \\
\end{pmatrix}=33$$
Doing some row operations $(R3+R1) \to R3\text{ and}\; (R4-2R1)\to R4$), I end up with
$$\text{det}\;\begin{pmatrix}
3 & 2 & -1& 4 \\
2 & k & 6 & 5 \\
0 & 4 & 0 & 4 \\
0 & 0 & 4 & -5\\
\end{pmatrix}=33$$
I expand along the first column and somehow my $k$ value is a decimal. Am I doing this correctly? I've tried making this into an upper and lower diagonal matrix and it just gets messy.
| Your row reductions seem fine, thus far. You can row reduce some more:
try to get an upper triangular matrix; short of that, it will simplify the calculation of the determinant.
$(a)$ You can factor out $4$ from the third row.
$$\text{ det}
\begin{pmatrix}
3 & 2 & -1& 4 \\
2 & k & 6 & 5 \\
0 & 4 & 0 & 4 \\
0 & 0 & 4 & -5\\
\end{pmatrix}
=33\iff 4 \text{ det} \begin{pmatrix}
3 & 2 & -1& 4 \\
2 & k & 6 & 5 \\
0 & 1 & 0 & 1 \\
0 & 0 & 4 & -5\\
\end{pmatrix}
=33$$
$$
\text{Add}\;-2(R_3) \text{ to}\;R_1 \implies 4 \text{ det} \begin{pmatrix}
3 & 0 & -1& 2 \\
2 & k & 6 & 5 \\
0 & 1 & 0 & 1 \\
0 & 0 & 4 & -5\\
\end{pmatrix}
= 33$$
$$
\text{Add }\;R_4 \text{ to}\; R_2\implies 4\text{ det} \begin{pmatrix}
3&0&-1&2\\
2&k&10&0\\
0&1&0&1\\
0&0&4&-5\\
\end{pmatrix}
= 33$$
At this point, I'd suggest simply expanding along the column containing $k$; with the additional row reduction, that may simplify the process. Don't worry if you end up with with an equation in which $k$ evaluates to a fraction (decimal)!
$$
\text {det} \begin{pmatrix}
3 & 0 & -1& 2 \\
2 & k & 10 & 0 \\
0 & 1 & 0 & 1 \\
0 & 0 & 4 & -5\\
\end{pmatrix}
= \frac{33}{4}\tag{1}$$
$$k\text{ det}\; \begin{pmatrix}
3 &-1&2\\
0&0&1\\
0&4&-5\\
\end{pmatrix}
- \text{ det}\; \begin{pmatrix}
3&-1&2\\
2&10&0\\
0&4&-5\\
\end{pmatrix}\tag{2}$$
$$= k\text{ det}\; \begin{pmatrix}
3 &-1&2\\
0&0&1\\
0&4&-5\\
\end{pmatrix}
- \left(3\text{ det}\; \begin{pmatrix}
10&0\\
4&-5\\
\end{pmatrix}
-2\text{ det}\; \begin{pmatrix}
-1&2\\
4&-5\\
\end{pmatrix}\right)= \frac{33}{4}
\tag{3}$$
$$-12k - [3(-50) - 2(-3)] = \frac{33}{4}$$
$$-12k +144 = \frac{33}{4}$$
$$k=\frac{181}{16}$$
Note:In $(1)$, I simply divided both sides of the equation by $4$.
In $(2)$, I expanded along the second column, the column containing $k$. Note the sign of each resulting determinants.
In $(3)$, I expanded along the first column of the second matrix in $2$; again, we need to keep.
The rest is simplification.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/238384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Helping turning this into sum/difference logarithm? Hey guys so i'm trying to turn this equation into it's sum/difference logarithm. However, the part that messes me up is turning the bottom of the fraction
$$
\log\left(\frac{x^2 +2x+1}{x^2 -3x +2}\right)^2\;.
$$
I think it will turn into this:
$$
4\log(x+1) -2\big(\log(x-1) +\log(x-2)\big)\;,
$$
but I don't want it to be $(x-1)^2$ times $(x-2)^2$. How do I solve it so it's $\big((x-1)(x-2)\big)^2$?
| $$\log((x^2 +2x+1) / (x^2 -3x +2))^2$$
$$\log((x+1)^2 / (x^2 -2x-(x-2))^2$$
$$\log((x+1)^2 / (x(x-2)-(x-2))^2$$
$$\log((x+1)^2 / ((x-2)(x-1))^2$$
$$\log\left(\frac{(x+1)^2}{(x-2)(x-1)}\right)^2=2(\log(x+1)^2-\log(x-2)(x-1))$$
$$=2(2\log(x+1)-\log(x-2)-\log(x-1))=$$
$$=4\log(x+1)-2\log(x-2)-2\log(x-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/239729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Tricky integration by substitution $\int_{-1}^{1} \frac{ \sqrt{1-x^2}}{1+x^{2}} dx$ I have to get this integral
$$\int_{-1}^{1} \frac{ \sqrt{1-x^2}}{1+x^{2}} dx$$
into
$$\int_{-\pi }^{\pi } \frac{1}{1+\cos^2\theta } \,d\theta - \pi$$
any tips would be recommended.
| I will use the substitution $x = \cos θ$. And my result is very close to the required.
$dx = - \sinθ d \theta $
When $x = 1, \ θ=\dots = 0$ and when $x = –1, \ θ= \dots = – π.$
We have :
$$ \begin{align}
& \int_{- \pi}^{0} \cfrac{ \sqrt{1- \cos^2 \theta} }{1+ \cos^{2} \theta} (- \sin \theta d \theta) \\
& = – \int_{-\pi}^{0} \cfrac{\sin^2 \theta}{1+ \cos^2 \theta} d \theta \\
& = – \int_{- \pi}^{0} \cfrac{1 – \cos^2 \theta}{1+ \cos^2 \theta} d \theta \\
& = – \int_{- \pi}^{ 0} \left( –1 + \cfrac {2}{1 + \cos^2 \theta} \right) d \theta \\
& = \int_{- \pi}^{0} d \theta – 2\int_{- \pi}^{\theta} \left( \cfrac {1}{1 + cos^{2} \theta} \right) d \theta \\
& = \pi –2\int_{- \pi}^{0} \left( \color{blue}{\cfrac {1}{1 + cos^{2} \theta}} \right) d \theta \\
& = \pi –\int_{-\pi}^{\pi}\left( \cfrac {1}{1 + cos^{2} \theta} \right) d \theta \end{align} $$
Area under [-π, π] = 2 * that under [-π. 0] (in blue)
Which differs from the requested form by a sign only.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to calculate the limit of $\frac{\sin(ax)}{x}$ for $x\to0$ How can I calculate the following limit epsilon-delta definition?
$$\lim_{x \to 0} \left(\frac{\sin(ax)}{x}\right)$$
Edited the equation, sorry...
| This is easy if you know the power series of the sine function.
$$\sin(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} +\cdots$$
Then $$\sin(ax) = a x - \frac{a^3 x^3}{6} + \frac{a^5 x^5}{120} -\cdots$$
and $\frac{\sin(ax)}{x} = a - \frac{a^3 }{6}x^2 + \frac{a^5 }{120}x^4 -\cdots$ for all $x \neq 0$. Since power series are continuous you can find the limit for $x \rightarrow 0$ by simply setting $x=0$ in this expression, so the limit is $a$.
| {
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How to calcualate how many unique set of 6 can i have in a given set. Hello my question is quite simple i would think but i just cant seem to find an answer. I have a set of $\{1,2,3,4,5,6,7,8,9,10\}$ and i would like to calculate how many unique given sets of $6$ can i get from this set.
In other words for the number $1$
i would end up with
$[1,2,3,4,5,6]
[1,3,4,5,6,7]
[1,4,5,6,7,8]
[1,5,6,7,8,9]
[1,6,7,8,9,10]$
I would move down the line with the number $2$ to compare to unique sets of $6$
note:
when moving to two I would no longer do this
$[2,1,3,4,5,6]$ because it repeats my first case above.
its there a formula to figure this sort of thing?
Thanks in advance.
when I work this out on paper i end up with 15 sets
here is how
for 1
[1,2,3,4,5,6]
[1,3,4,5,6,7]
[1,4,5,6,7,8]
[1,5,6,7,8,9]
[1,6,7,8,9,10]
for 2
[2,3,4,5,6,7]
[2,4,5,6,7,8]
[2,5,6,7,8,9]
[2,6,7,8,9,10]
for 3
[3,4,5,6,7,8]
[3,4,6,7,8,9]
[3,5,6,7,8,9,10]
for 4 [4,5,6,7,8,9]
[4,6,7,8,9,10]
for 5 [5,6,7,8,9,10]
after that i cant make any more groups of $6$ thus i end up with $15$ sets.
| Yes, there is, it is called the binomial, written $\binom{n}{k}$, read $n$ choose $k$. The value is $$\binom{n}{k}=\frac{n!}{k!(n-k)!}.$$ So, in your case, you have $$\binom{10}{6}=\frac{10!}{6!4!}=210.$$ I hope you find this helpful!
| {
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"timestamp": "2023-03-29T00:00:00",
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How come $2 \times 3^k + 3^k = 3 \times 3^k$ I get something else
$$\begin{align*}
&2 \times 3^k + 3^k=\\
&2 \times 3^k \times 2=\\
&4 \times 3^k
\end{align*}$$
What does $3^k + 3^k$ give exactly?
| Your calculation appears to be based on the notion that $2\times 3^k+3^k$ means $2\times(3^k+3^k)$, which would indeed be $4\cdot3^k$. However, in the absence of parentheses multiplication is performed before addition, so $2\times 3^k+3^k$ actually means $(2\times 3^k)+3^k$. This is $(2\times 3^k)+(1\times 3^k)$, which is clearly just $3\times 3^k$, or $3^{k+1}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $(a,b)=1$ then prove $(a+b, ab)=1$.
Let $a$ and $b$ be two integers such that $\left(a,b\right) = 1$. Prove that $\left(a+b, ab\right) = 1$.
$(a,b)=1$ means $a$ and $b$ have no prime factors in common
$ab$ is simply the product of factors of $a$ and factors of $b$.
Let's say $k\mid a+b$ where $k$ is some factor of $a$.
Then $ka=a+b$ and $ka-a=b$ and $a(k-l)=b$.
So $a(k-l)=b, \ a\mid a(k-1)$ [$a$ divides the left hand side] therefore $a\mid b$ [the right hand side].
But $(a,b)=1$ so $a$ cannot divide $b$.
We have a similar argument for $b$.
So $a+b$ is not divisible by any factors of $ab$.
Therefore, $(a+b, ab)=1$.
Would this be correct? Am I missing anything?
| By contradiction, let $d>1$ divide both $ab$ and $a^2+b^2$. Then $d$ has a prime divisor $p$, which divides both $ab$ and $a^2+b^2$.
Then, $p|a(a+b) - ab = a^2$
Since $p$ is a prime which divides $a^2$ $\implies$ that $p|a$.
And $p|b(a+b) - ab = b^2$ $\implies$ $p|b$.
Hence $p|a$ and also, $p|b$, so $p$ divides $gcd(a,b) = 1$. However, this forces $p = 1$, which is a contradiction, since 1 isn't a prime.
Thus the only positive common divisor of $ab$ and $a^2+b^2$ is $1$, and hence they are coprime.
| {
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"timestamp": "2023-03-29T00:00:00",
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Do there exist any odd prime powers that can be represented as $n^4+4^n$? Well, I wrote up a solution on it, but according to the place I found the problem, it isn't quite correct. Ah, I'm simply hoping someone will point out where I got wrong.
Now, let, $n^4+4^n = p^k$, where $p$ is an odd prime and $k$ is a positive integer.
Further, $p^k \equiv 1 \pmod 2$. Therefore, $n^4 + 4^n \equiv n^4 \equiv 1 \pmod 2$, and so $n \equiv 1 \pmod 2$. $n$ must be odd.
Okay, now let, $n = 2m +1$. Substituting it in $n^4+4^n$ and using the Sophie Germain inequality, we have,
$$n^4+4\cdot4^{2m} = n^4 + 4(2^m)^4 = (n^2 + 2^n+2^{m+1}\,n)(n^2 + 2^n-2^{m+1}\,n) = p^k$$
Now, as $p^k$ can only be factorized into smaller powers of $p$, let $n^2 + 2^n+2^{m+1}\,n = p^i$, and let $n^2 + 2^n-2^{m+1}\,n = p^j$ where $i+j= k$, obviously, and $i>j$.
Now consider this:
$$\begin{align}
p^i - p^j & \equiv 0\\
2\cdot2^{m+1}\,n = 2^{m+2}\,n &\equiv 0 \pmod p
\end{align}$$
But, as $p$ is odd, $\gcd(p, 2) = 1$, so $n \equiv 0 \pmod p$.
Now look at this:
$$\begin{align}
p^i + p^j &\equiv 0 \\
2(n^2 + 2^n) &\equiv 0 \\
n^2 + 2^n &\equiv 0 \pmod p
\end{align}$$
But we just established that $n \equiv 0 \pmod p$, so $2^n \equiv 0 \pmod p$.
Therefore, let $2^n = jp$ for some integer $j$.
Now, $2^n$ is its own prime factorization, which is unique according to the Fundamental Theorem of Arithmetic and does not include $p$.
Therefore, the above statement is an impossibility!
There exist no such $p$ and $n$, and no odd prime powers can be written as $n^4+4^n$.
Ah, well, that's it.
Sorry for the tediousness of it. I've still no clue how to use $\LaTeX$.
Thank you everybody,
Cheers.
| I think that your argument works as long as the smaller of your two factors
$$n^2+2^n−2^{m+1}n$$
is not equal to $1$. In the case of $n=1$, it does equal $1$, and you get Henry's counterexample. For larger $n$,
$$\begin{align}n^2+2^n−2^{m+1}n&=n^2+2^n-2^{(n+1)/2}n\\
&=n^2+\sqrt{2^n}(\sqrt{2^n}-n\sqrt{2})\end{align}$$
will be larger than $1$ for sure once $n$ is large enough for $\sqrt{2^n}-n\sqrt{2}>0$. This is equivalent to $2^n>2n^2$. This inequality is true for $n=7$. Calculus can confirm that it holds once $n$ is beyond $7$ too:
Let $f(n)=2^n$ and $g(n)=2n^2$. Then $f''(n)=2^n(\ln(2)^2)$ and $g''(n)=4$. So certainly for $n\ge7$, $f$ has the larger second derivative. The first derivative of $f$ at $n=7$ is $128\ln(2)$, which is larger than $g$'s first derivative at $n=7$, which is $28$. So for all $n\ge7$, $f$ has the larger first derivative. The same argument applies to $f'$ and $g'$, and so we conclude that $f(n)>g(n)$ when $n\ge7$.
We can directly check that $n^2+2^n-2^{(n+1)/2}n\neq1$ when $n$ is $3$ or $5$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Analytical form of an integral I'm trying to find the analytical form of the following integral:
$$\int_0^{2\pi}\left|\frac{(1-i\cos(\phi))}{(1+a-i\cos(\phi))}\right|^2d\phi$$
I've tried in Wolfram Alpha, which ran out of free calculation time, and in Mathematica I get the original expression for the indefinite integral and the program stays in the "Running..." state for several minutes, when evaluating the definite integral, and I stopped it.
However, I am reading a paper, which says there is an analytical expression for this integral, and also to me it seems rather straightforward. Any hints?
| We have
\begin{align*}
\int_0^{2\pi} \left| \frac{1- i \cos(\phi)}{1+a-i\cos(\phi)} \right|^2 d\phi &= \int_0^{2\pi} \frac{1+\cos^2\phi}{(1+a)^2 + \cos^2\phi} d\phi\\
& = \int_0^{2\pi} \left[1 - \frac{a^2+2a}{(1+a)^2 + \cos^2\phi}\right]d\phi\\
& = 2\pi - 2(a^2+2a)\int_{-\pi/2}^{\pi/2} \frac{1}{(1+a)^2+\cos^2\phi}d\phi\\
& = 2\pi - 2(a^2+2a)\int_{-\pi/2}^{\pi/2} \frac{1}{(1+a)^2 \sin^2 \phi + [1+(1+a)^2]\cos^2\phi}d\phi\\
& = 2\pi - 2(a^2+2a)\int_{-\pi/2}^{\pi/2} \frac{\sec^2\phi}{(1+a)^2 \tan^2 \phi + [1+(1+a)^2]}d\phi.
\end{align*}
Now let $u=\frac{|1+a|}{\sqrt{a^2+2a+2}}\tan\phi$, getting
$$
2\pi - 2\frac{(a^2+2a)}{|1+a|\sqrt{a^2+2a+2}}\int_{-\infty}^{\infty} \frac{1}{u^2 + 1}du= 2\pi - \frac{2\pi(a^2+2a)}{|1+a|\sqrt{a^2+2a+2}}.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing that: $(\frac{a}{b+c})^2+(\frac{b}{a+c})^2+(\frac{c}{a+b})^2+\frac{10abc}{(a+b)(b+c)(c+a)}\ge 2$ Let a;b;c>0. Prove:
$$\left(\frac{a}{b+c}\right)^2+\left(\frac{b}{a+c}\right)^2+\left(\frac{c}{a+b}\right)^2+\frac{10abc}{(a+b)(b+c)(c+a)}\geq 2$$
I think
$$\frac{2a}{b+c}=x;\frac{2b}{c+a}=y;\frac{2c}{a+b}=z$$
We have: $xy+yz+zx+xyz=4$
$$(\frac{x}{2})^2+(\frac{y}{2})^2+(\frac{z}{2})^2+\frac{10xyz}{8} \ge 2$$
$\Leftrightarrow x^2+y^2+z^2+5xyz \ge 8$
$\Leftrightarrow x^2+y^2+z^2-5(xy+yz+zx) +12 \ge 0$
deadlock
Can you help me? Thank you very much
| Nice problem!
Here is my solution.
Let $x=\frac{a}{b+c},y=\frac{b}{c+a},z=\frac{c}{a+b}$
Then we have to show that
$$ x^2+y^2+z^2+10xyz\geq 2 $$
Fisrtly,we have the identity
$$xy+yz+zx+2xyz=1 $$
Secondly,we have the known inequality
$$ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{3}{2}$$
that implys
$$x+y+z\geq \frac{3}{2}$$
Now Using Schur of the third degree,we have
$$
x^2+y^2+z^2+6xyz+4xyz\geq x^2+y^2+z^2+\frac{9xyz}{x+y+z}+4xyz\geq 2(xy+yz+xz)+4xyz=2
$$
Hence we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1} \leq \frac{9}{10}$ if $a+b+c=1$. Let $a,b,c$ are real number such that $a+b+c=1$. Prove that:
$$\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1} \leq \frac{9}{10}.$$
| Let $S(a,b,c) = \frac{a}{a^2+1}+ \frac{b}{b^2+1}+ \frac{c}{c^2+1}$. Since we have that $a,b$ and $c$ are all represented symmetrically in $S$ (interchanging any two variables has no effect), it must be that case that $S$ has a stiationary value (maximum or minimum) when all parameters are equal, ie. $a=b=c=x$. Using the fact that $a+b+c=1$ gives $x=\frac{1}{3}$, so we must classify the point $(a,b,c)=(\frac{1}{3},\frac{1}{3},\frac{1}{3})$ as a maximum or minimum of the function $S(a,b,c)$. Consider increasing the value of an parameter by a small quantity $\delta$, and thereby decreasing the value of another parameter by $\delta$. In other words, without loss of generality take $a=\frac{1}{3}+\delta$, $b=\frac{1}{3}-\delta$ and $c=1/3$. We then have that:
$$
S(\frac{1}{3}+\delta,\frac{1}{3}-\delta,\frac{1}{3})=\frac{3}{10}+\frac{\frac{1}{3}+\delta}{(\frac{1}{3}+\delta)^2+1} + \frac{\frac{1}{3}-\delta}{(\frac{1}{3}-\delta)^2+1}
$$
$$
S(\frac{1}{3}+\delta,\frac{1}{3}-\delta,\frac{1}{3})\approx\frac{3}{10}+\frac{\frac{1}{3}+\delta}{\frac{10}{9}+\frac{2\delta}{3}} + \frac{\frac{1}{3}-\delta}{\frac{10}{9}-\frac{2\delta}{3}}=\frac{33}{10}+\frac{6}{3\delta-5}-\frac{6}{3\delta+5}
$$
Then, to leading order in $\delta$, we have the following series expansion:
$$
S(\frac{1}{3}+\delta,\frac{1}{3}-\delta,\frac{1}{3})\approx \frac{9}{10} - \frac{108}{125}\delta^2
$$
Thus, $\delta=0$ is a maximum, which tells us that $(a,b,c)=(\frac{1}{3},\frac{1}{3},\frac{1}{3})$ is a maximum! Furthermore, this tells us that $S_{\text{max}} = \frac{9}{10}$ which immediately implies that $S(a,b,c)\leq \frac{9}{10}$ as required.
| {
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Show $\lim\limits_{n\to\infty} \dfrac{n^2+3^{2n}}{(n^3+3^n)^2} = 1$ Could you help me show that
$$\lim\limits_{n\to\infty} \dfrac{n^2+3^{2n}}{(n^3+3^n)^2} = 1$$
| $$\lim\limits_{n\to\infty} \dfrac{n^2+3^{2n}}{(n^3+3^n)^2} =\lim_{n\to\infty}{n^2+3^{2n}\over n^6+2n^33^n+3^{2n}}=\lim_{n\to\infty}{\frac{n^2}{3^{2n}}+1\over\frac{n^6}{3^{2n}}+2\frac{n^3}{3^n}+1}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/267322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Pretty solution to the trigonometric equation Problem
Consider the trigonometric equation:
$$
a\sin x+b\cos x-\cos x\sin x=0\qquad(0\le x<2\pi)\tag{*}
$$
try to analyze the number of solutions to equation (*) with parameters $a,b$, i.e, let $A=a^{2/3}+b^{2/3}-1$, we have:
*
*$A<0$, there are four distinct solutions.
*$A>0$, there are two distinct solutions.
Endeavors
Let $f(x)=a\sin x+b\cos x-\cos x\sin x$, we have $f^\prime(x)=a\cos x-b\sin x-\cos2x$. It seems no advance to calculate the derivative, because $f^\prime$ is as hard as $f$.
Let $u=\cos x$ and $v=\sin x$, we have $u^2+v^2=1$ and $av+bu=uv$. We can work on these equations, but I prefer the trigonometric way, i.e, analyze the properties of $f(x)$.
I want to illustrate some details about $f(x)$, which might be useful. Let $a=r\cos\phi$ and $b=r\sin\phi$, where $r=\sqrt{a^2+b^2}$, we have
$f(x)=r\sin(x+\phi)-\frac12\sin2x$. It's a linear combination of $\sin(x+\phi)$ and $\sin2x$. I don't know whether there's a systematical way to deal with it.
Any idea? Thanks!
| Let us consider only four cases in our analysis. The other missing cases can be analyzed by a similar approach.
Let $A$ a real number such that $A=a^\frac{2}{3}+b^\frac{2}{3}-1$ and $g(x)= a\sin x + b\cos x-\cos x\sin x$ a function with domain $[0, 2\pi[$ , let's analyze the number of zeros of $g(x)$ in function of $A$.
(1) Case ($a=0$ and $b \neq 0$)
$$g(x)=0 \Leftrightarrow b\cos x-\cos x\sin x = 0$$
$$ \Leftrightarrow \cos x(b-\sin x)=0$$
For $b^2>1$ we have $A>0$ and two distinct solutions.
For $b^2<1$ we have $A<0$ and four distinct solutions.
For $b^2=1$ we have $A=0$ and a double root and a simple one.
(2) Case ($a \neq 0$ and $b = 0$)
$$g(x)=0 \Leftrightarrow a\sin x-\cos x\sin x=0$$
$$\Leftrightarrow \sin x(a-\cos x)=0$$
For $a^2>1$ we have $A>0$ and two distinct solutions.
For $a^2<1$ we have $A<0$ and four distinct solutions.
For $a^2=1$ we have $A=0$ and a double root and a simple one.
(3) Case ($a=0$ and $b = 0$)
$$g(x)=0 \Leftrightarrow -\cos x\sin x=0$$
We have $A<0$ and the equation has four distinct solutions.
(4) Case ($a>0$ and $b>0$)
$$g(x)=0 \Leftrightarrow a\sin x + b\cos x-\cos x \sin x=0$$
For the equation above $0$, $\frac{\pi}{2}$, $\pi$, and $\frac{3\pi}{2}$ are never solutions (Verify by yourself).
Let's make the following factorising:
$$g(x)=(a\tan x + b - \sin x)\cos x$$
And let's define $f(x)=a\tan x + b - \sin x$.
Every root of $f(x)$ is a root of $g(x)$.
So let's search for roots of $f(x)$.
$$f(x)=0 \Leftrightarrow a\tan x + b= \sin x$$
If we change the values of $a$ and $b$ ($a>0$ and $b>0$) the graph of $a\tan x+b$ will always cross the graph of $\sin x$ once in the intervals $[\frac{\pi}{2}, \frac{3\pi}{2}]$ and $[\frac{3\pi}{2}, 2\pi[$. So $f(x)$ has at least two distinct roots in $[0, 2\pi[$.
See the graph below:
For $x \in [0, \frac{\pi}{2}]$ we must analyze the values of $a$ and $b$.
The local minimum of $f(x)$ in $[0, \frac{\pi}{2}]$ occurs at $\cos x_0 = a^\frac{1}{3}$ i.e. $x_0= \arccos a^\frac{1}{3}$.
If we substitute $x_0$ in $f(x)$ we get:
$$f(x_0)= \frac{a \sin (\arccos a^\frac{1}{3})}{a^\frac{1}{3}} + b - \sin (\arccos a^\frac{1}{3})$$
After some algebra we get:
$$f(x_0) = -(1-a^\frac{2}{3})^\frac{3}{2} + b$$
Now let's analyze the possibilities.
Note that
$$ f(x_0) >0 \Leftrightarrow -(1-a^\frac{2}{3})^\frac{3}{2} + b>0 \Leftrightarrow b>(1-a^\frac{2}{3})^\frac{3}{2}$$
$$\Leftrightarrow a^\frac{2}{3}+ b^\frac{2}{3} -1 >0 \Leftrightarrow A >0.$$
So
If $f(x_0) > 0$ ($A>0$), then $a\tan x +b$ and $\sin x$ have no intersection ($f(x_0) >0$), $f(x)$ has only two distinct roots in $[0, 2\pi[$ and so do $g(x)$.
If $f(x_0) =0$ ($A = 0$), then $a\tan x +b$ and $\sin x$ are tangent at $x_0$, $f(x)$ has a double root in $[0,\frac{\pi}{2}]$ (a double root and two distinct ones in $[0, 2\pi[$) and so do $g(x)$.
If $f(x_0)<0$ ($ A < 0)$, then $a\tan x +b$ and $\sin x$ have two intersections, $f(x)$ has only two distinct roots in $[0,\frac{\pi}{2}]$ (four distinct roots in $[0, 2\pi[$) and so do $g(x)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $xy=3$ and $4^{x^2}+2^{y^2}=72$ I have a system of equations $xy=3$ and $4^{x^2}+2^{y^2}=72$ whose solution I know is $x=y=\sqrt 3$, but what are the steps to solve it?
| The only solutions are $\pm(\sqrt{3},\sqrt{3})$ and $\pm(\sqrt{1.5},\sqrt{6})$. Proof:
Substitute $a=2x^2, b=y^2$. This becomes $2^a + 2^b = 72$. The relation between $a,b$ becomes $ab=2(xy)^2=18$ and $a,b>0$.
Let $f(x)=2^{x} + 2^{18/x}$. We are interested in positive solutions to $f(x)=72$. Since $f(x)=f(\frac{18}{x})$, we can restrict ourselves to $x \le \sqrt{18}$.
One solution is $f(3)=2^3 + 2^6=72$, which corresponds to $2x^2=3,y^2=6$, i.e. $\pm(\sqrt{1.5},\sqrt{6})$.
Another solution is recovered - $6=\frac{18}{3}$, which corresponds to $2x^2=6,y^2=3$, i.e. $\pm(\sqrt{3},\sqrt{3})$.
I'll show that $f(x)=72$ has no solutions for $0<x<\sqrt{18}$ other than $x=3$. The proof will be by showing that $f$ is decreasing in that interval:
$$f'(x)=\ln 2 ( 2^x -\frac{18}{x^2} 2^{\frac{18}{x}})$$
For $0<x<\sqrt{18}$,
$$1<\frac{18}{x^2}, 2^{x} < 2^{\frac{18}{x}}$$
So $2^x < \frac{18}{x^2}2^{\frac{18}{x}}$, proving $f' <0$.
| {
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Definite integral involving $\sqrt{\log}$ In some old lecture notes I found the following problem:
$$ \int_1^e \frac{1+2x \sqrt{\log x}}{2 x \sqrt{\log x}(x+\sqrt{\log x})} \;dx$$
I just don't seem to get a handle on this rather difficult looking integral. Does anybody have some insights?
| First note that
\begin{align*}
\left(\sqrt{\log{x}}\right)'=\frac{1}{2x\sqrt{\log{x}}}
\end{align*}
Then for the integrand it follows that
\begin{align*}
\frac{1+2x\sqrt{\log x}}{2x\sqrt{\log x}(x+\sqrt{\log x})} &=\frac{\frac{1}{2x\sqrt{\log x}}+1}{x+\sqrt{\log x}}\\
&=\frac{(x+\sqrt{\log x})'}{x+\sqrt{\log x}}.
\end{align*}
This enables us to calculate
\begin{align*}
\int\frac{(x+\sqrt{\log x})'}{x+\sqrt{\log x}}dx=\log\left(x+\sqrt{\log x}\right)
\end{align*}
and therefore
\begin{align*}
\int^e_1\frac{(x+\sqrt{\log x})'}{x+\sqrt{\log x}}dx =\log\left(e+\sqrt{\log e}\right)-\log\left(1+\sqrt{\log 1}\right)=\log\left(e+1\right).
\end{align*}
| {
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Possible Jordan Canonical Forms Given Minimal Polynomial I was supposed to find all possible Jordan canonical forms of a $5\times 5$ complex matrix with minimal polynomial $(x-2)^2(x-1)$ on a qualifying exam last semester. I took the polynomial to mean that there were at least two 2's and one 1 on the main diagonal, and that the largest Jordan block with eigenvalue 2 is $2\times 2$ while the largest Jordan block with eigenvalue 1 is $1\times 1$. Did I miss any matrices or interrupt the minimal polynomial incorrectly?
\begin{pmatrix}
2 &1 &0 &0 &0\\
0 &2 &0 &0 &0\\
0 &0 &2 &1 &0\\
0 &0 &0 &2 &0\\
0 &0 &0 &0 &1
\end{pmatrix}
\begin{pmatrix}
2 &1 &0 &0 &0\\
0 &2 &0 &0 &0\\
0 &0 &2 &0 &0\\
0 &0 &0 &2 &0\\
0 &0 &0 &0 &1
\end{pmatrix}
\begin{pmatrix}
2 &1 &0 &0 &0\\
0 &2 &0 &0 &0\\
0 &0 &2 &0 &0\\
0 &0 &0 &1 &0\\
0 &0 &0 &0 &1
\end{pmatrix}
\begin{pmatrix}
2 &1 &0 &0 &0\\
0 &2 &0 &0 &0\\
0 &0 &1 &0 &0\\
0 &0 &0 &1 &0\\
0 &0 &0 &0 &1
\end{pmatrix}
\begin{pmatrix}
2 &0 &0 &0 &0\\
0 &2 &0 &0 &0\\
0 &0 &1 &0 &0\\
0 &0 &0 &1 &0\\
0 &0 &0 &0 &1
\end{pmatrix}
| Yes, you did.
Based on the minimal polynomial, you must have a two-by-two Jordan block for eigenvalue 2 and a one-by-one block for eigenvalue 1. You can fill in the five-by-five matrix with more of those blocks or with one-by-one blocks for eigenvalue 2. Using those rules yields precisely your first four matrices. Your fifth matrix is not correct.
Furthermore, you can permute the blocks. Thus,
*
*your first matrix yields 3!/2! = 3 Jordan forms,
*your second and third matrices yield 4!/2! = 12 forms each, and
*your four matrix yields 4!/3! = 4 forms
for a total of 3 + 2 · 12 + 4 = 31 forms.
| {
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limits of polynomials A few polynomial limits:
$$\lim_{x\rightarrow 0} \frac{x^2 + 5x}{x^3+6x^2+3x}=\lim_{x\rightarrow 0} \frac{x + 5}{x^2+6x+3}=\frac{5}{3}$$
$$\lim_{x\rightarrow 0} \frac{x^4 + 5x}{x^3+6x^2+3x}=\lim_{x\rightarrow 0} \frac{x^3 + 5}{x^2+6x+3}=\frac{5}{3}$$
$$\lim_{x\rightarrow 0} \frac{x^4}{x^6}=\infty$$
Please verificate if I'm doing it correctly. For which two polynomials $W$, $P$ limit of $W(x)/P(x)$ as x goes to $0$ can be a real number?
EDIT
Why for example
$$\lim_{x\rightarrow0}\frac{x+1}{x^2}=\infty$$
$$\lim_{x\rightarrow0}\frac{x}{x^2+1}=0$$
| Lets denote by $n$ the multiplicity of $0$ as a root of $W$ and by $k$ the multiplicity of $0$ as a root of $P$.Also denote $w(x)=\dfrac{W(x)}{x^n}$ and $p(x)=\dfrac{P(x)}{x^k}$. If $k=0$ then $\lim_{x\to0}\dfrac{W(x)}{P(x)}=\dfrac{W(0)}{P(0)}$.
If $n=k>0$ then $\lim_{x\to0}\dfrac{W(x)}{P(x)}=\dfrac{w(0)}{p(0)}$.
If $n>k>0$ then $\lim_{x\to0}\dfrac{W(x)}{P(x)}=0$.
If $n<k>0$ and $k-n$ is even then $\lim_{x\to0}\dfrac{W(x)}{P(x)}=\text{sign}(w(0))\text{sign}(p(0))\infty$.
If $n<k>0$ and $k-n$ is odd then the $\lim_{x\to0}\dfrac{W(x)}{P(x)}$ does not exist.
| {
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Calculate $\int\frac{1}{(x+1)\sqrt{1+x^2}} dx $ How can I calculate the following integral :
$$\int\frac{1}{(x+1)\sqrt{1+x^2}} dx $$
I try to write the integral like :
$$\int\frac{1+x-x}{(x+1)\sqrt{1+x^2}}=\int\frac{1}{\sqrt{1+x^2}}-\int\frac{x}{(x+1)\sqrt{1+x^2}}=\int\frac{1}{\sqrt{1+x^2}}-\int\frac{(\sqrt{x^2+1})'}{(x+1)}$$
but still nothing .
thanks :)
| Let $x=\tan[y]$, then $dx=\sec^{2}[y]dy$ $$\frac{1}{1+x}=\frac{\cos[y]}{\sin[y]+\cos[y]},\frac{1}{\sqrt{1+x^{2}}}=\frac{1}{\sec[y]}=\cos[y]$$
Multiplying everything out you need to integrate $$\frac{1}{\sin[y]+\cos[y]}dy$$
But you can simplify $$\sin[y]+\cos[y]=\sqrt{2}[\sin[y]\cos[\frac{\pi}{4}]+\cos[y]\sin[\frac{\pi}{4}]]=\sqrt{2}\sin[y+\frac{\pi}{4}]$$
So it suffice to integrate $$\frac{1}{\sqrt{2}}\csc[y+\frac{\pi}{4}]dy$$And we know how to do $\int \csc[x]dx$.
| {
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A multiple choice question on number theory Pick out the true statement(s):
(a) If $n$ is an odd positive integer, then $8$ divides $n^2 – 1$.
(b) If $n$ and $m$ are odd positive integers, then $n^2+m^2$ is not a perfect square.
(c) For every positive integer $n$, $n^5/5+n^3/3+7n/15$ is an integer.
(a) is true by induction. But I can not verify the others. Please help me somebody
| (a) can also be proved without induction.
$$(2c+1)^2=8\frac{c(c+1)}2+1\equiv1\pmod 8$$
(b)$$(2a+1)^2+(2b+1)^2=4\{a^2+a+b^2+b\}+2\equiv2\pmod 4$$
But any number $m\equiv0,1,2,3\pmod 4\implies m^2\equiv0,1\pmod 4$
(c) Using Fermat's Little Theorem, prime $p\mid(n^p-n) $ for all integer $n$.
So, $$5\mid (n^5-n) \text{ and } 3\mid(n^3-n)$$
Hence, $$\frac{n^5}5+\frac{n^3}3+\frac{7n}{15}=\frac{n^5-n}5+\frac{n^3-n}3+\left(\frac n3+\frac n5+\frac{7n}{15}\right)=\frac{n^5-n}5+\frac{n^3-n}3+n$$
| {
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Let $a, b$ and $c$ be the lengths of the sides of an arbitrary triangle. Pick out the true statements. Let $a, b$ and $c$ be the lengths of the sides of an arbitrary triangle. Define
$$x =\frac{ab + bc + ca}{a^2 + b^2 + c^2}.$$
Pick out the true statements.
(a) $1/2 ≤ x ≤ 2$.
(b) $1/2 ≤ x ≤ 1$.
(c) $1/2 < x ≤ 1$.
How can I able to solve this problem
| *
*We first rule out (a) by writing $(a-b)^2\geq 0$, $(b-c)^2\geq 0$, $(c-a)^2\geq 0 $ and adding side by side to get $2a^2+2b^2+2c^2\geq 2ab + 2bc+2ac$. This tells us $\frac{ab+bc+ac}{a^2+b^2+c^2}\leq 1 $. Therefore $x$ cannot be greater than 1. Therefore (a) is not your answer.
*We now rule out (b) by writing: from the Cosine Law $a^2+b^2-2ab\cos\gamma=c^2$, $b^2+c^2-2bc\cos\alpha=a^2$ and $c^2+a^2-2ca\cos\beta=b^2.$ Adding side by side gives $a^2+b^2+c^2=2ab\cos\gamma+2bc\cos\alpha+2ac\cos\beta<2ab+2bc+2ac.$ This implies $x=\frac{ab+bc+ac}{a^2+b^2+c^2}>\frac{1}{2}$. Notice the last inequality follows since the angles in a triangle add up to $180^o$.
| {
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Proving an inequality for positive numbers $a, b, c$
Let be $a,b,c$ positive numbers such that $a+b+c=3$. Prove that
$$\frac{b+c+bc}{a^2+b^3+c^4}+\frac{c+a+ca}{b^2+c^3+a^4}+\frac{a+b+ab}{c^2+a^3+b^4} \le 3$$
| Using Cauchy Schwarz inequality:
$(a^2+b^3+c^4)(a^2+b+1) \ge (a^2+b^2+c^2)^2$. Do this for each denominator, then it suffices to show that
$$\sum (a^2+b+1)(b+c+bc) \leq 3(a^2+b^2+c^2)^2$$
where the sum is cyclic. Expand the left hand side, we get
$$\begin{eqnarray}
&\sum a^2b + b^2 + b + a^2c + bc + c + a^2bc + b^2c + bc \leq 3(a^2+b^2+c^2)^2 \\
\Leftrightarrow & 2 \sum a^2b + \sum a^2 c + \sum b^2 + \sum b + \sum c + 2\sum bc + \sum a^2bc \leq 3(a^2+b^2+c^2)^2 \\
\Leftrightarrow & \sum (a^2b+a^2c) + \sum a^2b + \sum a^2 + 3 + 3 + 2\sum ab + 3abc \leq 3(a^2+b^2+c^2)^2 \cdots (*)
\end{eqnarray}$$
using $a+b+c = 3$. Now note the identity
$$\sum(a^2b+a^2c) +3abc = a^2b+ab^2+b^2c+bc^2+c^2a+ca^2+3abc = (ab+bc+ca)(a+b+c)$$
which is equal to $3(ab+bc+ca)$ for this problem, so
$$(*)\Leftrightarrow \sum a^2 + 5 \sum ab + 6 + \sum a^2b \leq 3(a^2+b^2+c^2)^2$$
to be proved now.
Note the following inequalities:
$$\begin{eqnarray}(1) & \hspace{5mm}(a^2+b^2+c^2)^2 +3 \ge 4 \sum a^2b \\
(2)& \hspace{5mm}(a^2+b^2+c^2) \ge 3 \\
(3)& \hspace{5mm}(a^2+b^2+c^2) \ge ab+bc+ca
\end{eqnarray}$$
Here (1) follows from applying AM-GM to $a^4+a^2b+a^2b+1 \ge 4a^2b$ and sum up cyclicly. The other two are routine applications of Cauchy-Schwarz/AM-GM.
These implies
$$\begin{eqnarray}(1') &\hspace{5mm} \frac{(a^2+b^2+c^2)^2}{3} \ge \frac{1}{4} (a^2+b^2+c^2)^2 + \frac{3}{4} \ge \sum a^2b \\
(2')&\hspace{5mm} \frac{2}{3}(a^2+b^2+c^2)^2 \ge 6 \\
(3')&\hspace{5mm} \frac{5}{3}(a^2+b^2+c^2)^2 \ge 5(a^2+b^2+c^2) \ge 5\sum ab \\
(4')&\hspace{5mm} \frac{1}{3} (a^2+b^2+c^2)^2 \ge \sum a^2 \end{eqnarray}$$
The inequality then follows from $(1') + (2') + (3') + (4')$.
| {
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Evaluate: $3\cdot9^{\frac{1}{2}}\cdot27^{\frac{1}{4}}\cdot81^{\frac{1}{8}} \ldots$
Evaluate: $3\cdot9^{\frac{1}{2}}\cdot27^{\frac{1}{4}}\cdot81^{\frac{1}{8}} \dotsb$
Trial: Let
$$\begin{align}
P &= 3 \cdot 9^{\frac{1}{2}} \cdot 27^{\frac{1}{4}} \cdot 81^{\frac{1}{8}} \dotsb\\
\implies \ln P &=\ln3+\frac{1}{2} \ln 3^2+\frac{1}{2^2} \ln 3^3+\frac{1}{2^3} \ln 3^4 + \dotsb\\
&=\ln 3 \sum_{x=0}^{\infty}\frac{x+1}{2^x}
\end{align}$$
Then how I proceed. Is there any other simpler way to solve. Please help.
| We have:
$$ P = \prod_{k=1}^\infty \big(3^k\big)^{\left(2^{1-k}\right)}$$
Take the $\log_3$ of both sides:
$$ \log_3 P = \sum_{k=1}^\infty \log_3 3^{k2^{1-k}} $$
$$ \log_3 P = \sum_{k=1}^\infty k2^{1-k} $$
$$ \log_3 P = 2\sum_{k=1}^\infty k\left(\frac{1}{2}\right)^k$$
It is known that:
$$ \sum_{k=1}^{\infty} k r^k = \frac{r}{\left(1-r\right)^2} $$
So:
$$ \log_3 P = 2 \cdot \frac{\frac{1}{2}}{(1-1/2)^2} $$
$$ \log_3 P = \frac{1}{1/4} $$
$$ \log_3 P = 4 $$
$$ P = 3^4 = 81 $$
| {
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Induction Proof that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1})$ This question is from [Number Theory George E. Andrews 1-1 #3].
Prove that $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}).$$
This problem is driving me crazy.
$$x^n-y^n = (x-y)(x^{n-1}+x^{n-2}y+\dots +xy^{n-2}+y^{n-1)}$$
$(x^n-y^n)/(x-y) =$ the sum for the first $n$ numbers and then I added $(xy^{(n+1)-2}+y^{(n+1)-1})$ which should equal $(x^{n+1}-y^{n+1})/(x-y)$ but I can't figure it out
This is a similar problem in the book and I tried this method but it wasn't working out
$\quad$Thereom $\bf1$-$\bf2$: $\,\,\,\,$ If $\,x$ is any real number other than $1$, then $$\sum_{j=0}^{n-1}x^j=1+x+x^2+\ldots+x^{n-1}=\dfrac{x^n-1}{x-1}.$$
$\quad$Remark: $\displaystyle\sum_{j=0}^{n-1}A_j$ is shorthand for $A_0+A_1+A_2+\ldots+A_{n-1}.$
$\quad$Proof: Again we proceed by mathematical induction. If $n=1$ then $\displaystyle\sum_{j=0}^{1-1}x^j=x^0=1$ and $(x-1)/(x-1)=1$. Thus the theorem is true for $n=1$. $\quad$ Assuming that $\displaystyle\sum_{j=0}^{k-1}x^j=(x^k-1)/(x-1)$, we find that $$
\eqalign{
\sum^{(k+1)-1}_ {j=0}x^j & = \sum^{k-1}_ {j=0}x^j+x^k=\dfrac{x^k-1}{x-1}+x^k \\
&= \dfrac{x^k-1+x^{k+1}-x^k}{x-1}\\
&= \dfrac{x^{k+1}-1}{x-1}.
}$$
Hence condition $(\rm ii)$ is fulfilled, and we have established the theorem.
$\quad$Corollary $\bf1$-$\bf1$: $\,\,$ If $\,m$ and $n$ are positive integers and if $m>1$, then $n<m^n.$
| Hint: Apply $\textbf{Theorem 1.2}$ to $\displaystyle \frac{x}{y}$.
Edit: It follows from $\textbf{Theorem 1.2}$ that $\displaystyle \Bigl(\frac{x}{y}\Bigr)^n -1=\Bigl(\frac{x}{y}-1\Bigr)\Bigl( \Bigl(\frac{x}{y}\Bigr)^{n-1}+\cdots +\frac{x}{y}+1\Bigr)$.
Now multiply the equation by $y^n$ to get
$$\displaystyle y^n\Bigl(\Bigl(\frac{x}{y}\Bigr)^n -1)\Bigr)=y^n\Bigl(\frac{x}{y}-1\Bigr)\Bigl( \Bigl(\frac{x}{y}\Bigr)^{n-1}+\cdots +\frac{x}{y}+1\Bigr)$$
Simplifying on the left-hand side and rewritting $y^n$ as $yy^{n-1}$ on the right-hand side we get
$$(x^n -y^n)=yy^{n-1}\Bigl(\frac{x}{y}-1\Bigr)\Bigl( \Bigl(\frac{x}{y}\Bigr)^{n-1}+\cdots +\frac{x}{y}+1\Bigr)$$
Because the product is commutative you can rewrite the right-hand side to get
$$(x^n -y^n)=y\Bigl(\frac{x}{y}-1\Bigr)y^{n-1}\Bigl( \Bigl(\frac{x}{y}\Bigr)^{n-1}+\cdots +\frac{x}{y}+1\Bigr)$$
Finally, on the right-hand side, factor in $y$ and $y^{n-1}$ accordingly to get
$$(x^n -y^n)=(x-y)(x^{n-1}+\cdots +xy^{n-2}+y^{n-1})$$
| {
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Limit for a Recurrence Relation How I can find a limit for this recursively defined sequence?
$$a_0>0, a_{n+1}=\frac{a_{n}+2}{3a_{n}+2}$$
I'm particularly interested in answers involving concepts like contractive sequences and fixed points.
Many thanks.
| Use the recurrence relation
$$ a_{n+1} - a_{n} = \frac{a_n +2 }{ 3 a_n +2} - \frac{a_{n-1} +2}{3 a_{n-1} +2 } = \frac{4(a_{n-1} -a_{n})}{(3 a_{n-1} +2)(3 a_n +2)}. $$
Since
$$
3 a_n = 3 \frac{a_{n-1} +2 }{3 a_{n-1} +2} > 1 + \frac{1}{a_{n-1}+1} >1, \quad\forall n\geq 1,
$$
it follows that
$$
|a_{n+1} - a_{n}| < \frac{4}{9} |a_{n} - a_{n-1}|.
$$
Iteration gives
$$
|a_{n+1} - a_{n}| < \left(\frac{4}{9}\right)^n |a_{1} - a_{0}|.
$$
The series $\sum_{n=1}^\infty (a_{n+1}-a_n)$, of positive terms, is dominated by the convergent series $|a_1-a_0| \sum_{n=1}^\infty (4/9)^n$ and so converges. We have $\sum_{n=1}^\infty (a_{n+1}-a_n)= \lim_{n\to \infty} a_n - a_1$ which shows that the limit exists.
Then, to find to fixed points we can pass to the limit in the recurrence relation
$$
a_{\infty}= \frac{a_{\infty}+2}{3 a_{\infty} +2},
$$
which leads to $a_{\infty}= 2/3.$
| {
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Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$
Let be $a,b,c \geq 0$ such that: $a^2+b^2+c^2=3$.
Prove that:
$$(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27.$$
I try to apply $GM \leq AM$ for $x=a^3+a+1$, $y=b^3+b+1,z=c^3+c+1$ and
$$\displaystyle \sqrt[3]{xyz} \leq \frac{x+y+z}{3}$$ but still nothing.
Thanks :-)
| For what it's worth, this answer uses derivatives, but it does use a generally applicable method.
Since $a^2+b^2+c^2=3$, any variations, $(\delta a,\delta b,\delta c)$, of $(a,b,c)$ must satisfy
$$
a\,\delta a+b\,\delta b+c\,\delta c=0\tag{1}
$$
We are interested in finding the maximum of
$$
\log(a^3+a+1)+\log(b^3+b+1)+\log(c^3+c+1)\tag{2}
$$
At a critical point, the variations of $(2)$ must satisfy
$$
\frac{3a^2+1}{a^3+a+1}\delta a+\frac{3b^2+1}{b^3+b+1}\delta b+\frac{3c^2+1}{c^3+c+1}\delta c=0\tag{3}
$$
Standard linearity arguments say that if $(3)$ is true for all $(\delta a,\delta b,\delta c)$ that satisfy $(1)$, we have
$$
\left(\frac{3a^2+1}{a^3+a+1},\frac{3b^2+1}{b^3+b+1},\frac{3c^2+1}{c^3+c+1}\right)=k(a,b,c)\tag{4}
$$
That is,
$$
\frac{3a^2+1}{a^4+a^2+a}=\frac{3b^2+1}{b^4+b^2+b}=\frac{3c^2+1}{c^4+c^2+c}\tag{5}
$$
Note that
$$
\frac{\mathrm{d}}{\mathrm{d}x}\frac{3x^2+1}{x^4+x^2+x}
=-\frac{6 x^5+4 x^3-3 x^2+2 x+1}{(x^4+x^2+x)^2}\tag{6}
$$
If $x\ge1$, then $6x^5+4x^3\ge3x^2$ and if $0\le x\le1$, then $2x+1\ge3x^2$. Therefore, for all $x\ge0$, $(6)$ is negative. That is,
$$
\frac{3x^2+1}{x^4+x^2+x}\tag{7}
$$
is monotonic decreasing which, when combined with $(5)$, says that
$$
a=b=c\tag{8}
$$
$(8)$ says that
$$
(a^3+a+1)(b^3+b+1)(c^3+c+1)=27\tag{9}
$$
Condition $(8)$ assumes that (a,b,c) is not on the boundary, that is none are $0$. Suppose that $c=0$, then the same argument yields that $a=b=\frac12\sqrt6$ and therefore
$$
(a^3+a+1)(b^3+b+1)(c^3+c+1)=\frac{83}{8}+\frac52\sqrt6\tag{10}
$$
Suppose that $b=c=0$, then $a=\sqrt3$ and therefore
$$
(a^3+a+1)(b^3+b+1)(c^3+c+1)=1+4\sqrt3\tag{11}
$$
Comparing $(9)$, $(10)$, and $(11)$, the maximum is $27$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/283895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "43",
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Find the nth term of a recursive sequence I have a the following sequence:
$$\begin{gather}
a_1 = 3 \\
a_{n + 1} = 1 + \frac{a_n}{2}
\end{gather}
$$
How can I find the $a_n$ term?
| Note that $$\begin{align}a_{n+1} &=1+\frac{a_n}{2} =1+\frac{1}{2}(1+\frac{a_{n-1}}{2})\\ &=1+\frac{1}{2}+\frac{1}{2^2}a_{n-1} \\ &=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}a_{n-2}\\ & \vdots \\ &= 1+\frac{1}{2}+\frac{1}{2^2}+ \dots+\frac{1}{2^{n-1}}+\frac{1}{2^n}a_1 \\ &=1+\frac{1}{2}+\frac{1}{2^2}+ \dots+\frac{1}{2^{n-1}}+\frac{3}{2^n}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/285717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to compute the series $\sum\limits_{x=0}^\infty\sum\limits_{y=0}^\infty\sum\limits_{z=0}^\infty\frac{1}{2^x(2^{x+y}+2^{x+z}+2^{z+y})}$ How to compute the series $\displaystyle\sum_{x=0}^\infty\sum_{y=0}^\infty\sum_{z=0}^\infty\frac{1}{2^x(2^{x+y}+2^{x+z}+2^{z+y})}$ ?
| By symmetry, the sum $S$ of this triple series $$
S=\sum_{x,y,z}\frac{1}{2^x\cdot(2^{x+y}+2^{x+z}+2^{z+y})}$$ is also
$$
S=\sum_{x,y,z}\frac{1}{2^\color{red}{y}\cdot(2^{x+y}+2^{x+z}+2^{z+y})}=\sum_{x,y,z}\frac{1}{2^\color{red}{z}\cdot(2^{x+y}+2^{x+z}+2^{z+y})}.
$$
Furthermore,
$$
\frac1{2^x}+\frac1{2^y}+\frac1{2^z}=\frac{2^{x+y}+2^{x+z}+2^{z+y}}{2^{x+y+z}}.
$$
Hence, summing these three equivalent formulas for $S$, one gets
$$
3S=\sum_{x,y,z}\frac1{2^{x+y+z}}=\left(\sum_{x}\frac1{2^x}\right)^3,
$$
and, finally,
$$
S=\frac13\cdot2^3=\frac83.
$$
More generally, for every absolutely convergent series $\sum\limits_x\frac1{a_x}$,
$$
\sum_{x,y,z}\frac{1}{a_x\cdot(a_xa_y+a_xa_z+a_za_y)}=\frac13\left(\sum_x\frac{1}{a_x}\right)^3.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/286072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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"answer_id": 0
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Prove this product: $\prod\limits_{k=2}^ n {\frac{k^2+k+1}{k^2-k+1}}=\frac{n^2+n+1}{3}$ How to prove this product?
$$\prod\limits_{k=2}^ n {\frac{k^2+k+1}{k^2-k+1}}=\frac{n^2+n+1}{3}$$
| $$\require{cancel}\prod_{k=2}^n\frac{k^2+k+1}{k^2-k+1}=\frac{\cancel7}{3}\frac{\cancel{13}}{\cancel7}\frac{\cancel{21}}{\cancel{13}}\frac{31}{\cancel{21}}\cdot\ldots$$
Can you see what the only factors that'll remain are?
| {
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Prove $|\frac{1}{n} - \frac{1}{x}| \le \frac{1}{n^2}$ Prove $|\frac{1}{n} - \frac{1}{x}| \le \frac{1}{n^2}$ for all $n \le x \le n+1$, using the mean value theorem applied to $f(x) = \frac{1}{x}$
Immediately, I can recognize some components of the mean value theorem. $\frac{1}{n^2}$ likely comes from the slope of $f$ at $n$, and the $|\frac{1}{n} - \frac{1}{x}|$ expression likely comes from the secant expression $\frac{\frac{1}{n} - \frac{1}{x}}{x}$. But I cannot figure out how they fit together in the end.
| $n-x\leq1 $ therefore $$|\frac{1}{n} - \frac{1}{x}| =|\frac{n-x}{nx}|\leq|\frac{1}{nx}|$$ i know $n \le x $ then $$\leq|\frac{1}{nx}|\leq\frac{1}{n^2}$$
| {
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Minimum of $\left| \sin x- 1\right| + \left|\sin x- 2\right| + \left| \sin x -3\right| + \left| \sin x+1\right|$ I would like to know the minimum value of $$\left| \sin x- 1\right| + \left|\sin x- 2\right| + \left| \sin x -3\right| + \left| \sin x+1\right|$$
for $x \in \mathbb{R}$.
| As $x$ varies through $\mathbb R$, $\sin x$ varies through $[-1,1]$. The problem then reduces to finding the minimum of $f(y)=|y-1|+|y-2|+|y-3|+|y+1|$ for $y\in [-1,1]$.
But on $[-1,1]$, $f(y)=(1-y)+(2-y)+(3-y)+(y+1)=7-2y$, so the minimum value is $5$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Solutions of $\frac{1}{\cos \theta} = a \sin \theta - b$ One of my math professors and I are working on a physics problem involving spinning a chain, and we decided to go as simple as possible and work out the solution explicitly for that case (a long rod hanging from a hinge rotating in a horizontal circle). Then we could hopefully work up from there. In the end, we boiled it down to the point where we had an equation of this form:
$$\frac{1}{\cos \theta} = a \sin \theta - b$$
Depending on the values of $a$ and $b$, there are $0$, $1$, $2$, $3$, or $4$ solutions for $\theta$ in this equation. What I'm curious about is whether there are formulas in terms of $a$ and $b$ that will give these solutions. As an aside, this situation actually reminds me of quadratics - they have $0$, $1$, or $2$ solutions, the solutions are given by the quadratic formula, and the value of $b^2-4ac$ indicates how many real-valued solutions there are. I'm looking for something similar for the equation I've given above, and WolframAlpha is being no help (gasp!).
| If we put $t=\tan \frac\theta2,\sin\theta =\frac{2t}{1+t^2},\cos\theta=\frac{1-t^2}{1+t^2}$
Then,
$$ \frac1{\cos\theta}=a\sin\theta-b--->(1)$$
becomes $$(b-1)t^4-2at^3-2t^2+2at-(b+1)=0--->(2)$$ which is a quartic equation in $t$ hence will definitely have exactly four finite roots if $b-1\ne0$ .
$(1)$ If $b=1,$ the equation reduces to $2at^3+2t^2-2at+2=0--->(3)$
We can make use of this to identify the number of real roots of $(3)$
Clearly, each real root of $(3)$ will correspond to one real of root of of $(1)$ the reason being:
We know, $\tan A=\tan B\implies A=n\pi+B,$ so $\tan(\pi+\frac\theta2)=\tan\frac\theta2$ i.e., or $\tan\left(\frac{2\pi+\theta}2\right)=\tan\frac\theta2$
So, the periods of $\cos\theta,\sin\theta(=2\pi)$ and $\tan \frac\theta2$ are same.
Hence, in each $\in[2n\pi,2(n+1)\pi)$ there will be one-one correspondence between $\cos\theta,\tan \frac\theta2$ and $\sin\theta,\tan \frac\theta2$
$(2)$ If $b\ne1,$ we can write $$t^4-\frac{2a}{b-1}t^3-\frac2{b-1}t^2+\frac{2a}{b-1}t+\frac{b+1}{b-1}=0--->(4)$$
Now, we eliminate the $t^3$ term by putting $y=x-\lambda\implies x=y+\lambda$
$$(y+\lambda)^4-\frac{2a}{b-1}(y+\lambda)^3-\frac2{b-1}(y+\lambda)^2+\frac{2a}{b-1}(y+\lambda)+\frac{b+1}{b-1}=0--->(5)$$
The coefficient of $y^3$ is $4\lambda-\frac{2a}{b-1}$
If we set this to $0,\lambda=\frac a{2(b-1)}\implies y=x+\frac a{2(b-1)}$
Now, we can utilize this to identify the number of real roots of $(5),$ hence of $(4)$
Clearly, each real root of $(4)$ i.e. of $(2)$ will correspond to one real of root $(1)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Loss of significance error
Give exact ways of avoiding loss-of-significance errors in the
following computations:
a. $log(x+1)-log$, with large $x$
b. $\frac{1-cosx}{x^2}$, with $x\approx 0$
c.$(1+x)^{1/3}-1$, with $x\approx 0$
Am I doing this correctly?
a. I got $$log(x+1)-log(x)= log(\frac{x+1}{x})= log(1+\frac{1}{x}) \approx \frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3} ...$$
c. I got $$\frac{x}{(1+x)^{2/3}+(1+x)^{1/3}+1}$$
b. I am stuck
| a)
$$\log{(x+1)} = \log{x} + \log{\left (1 + \frac{1}{x} \right )}$$
Use the Taylor series approximation for $\log{y}$ for small values of $y$:
$$\log{(x+1)} - \log{x} = \frac{1}{x} - \frac{1}{2 x^2} + \frac{1}{3 x^3} + \dots$$
b) Again, use the Taylor series for $\cos{x}$ about $x=0$:
$$\frac{1-\cos{x}}{x^2} = \frac{1}{2} - \frac{x^2}{24} + \dots$$
c) Taylor expansion or a binomial expansion:
$$\begin{align}(1+x)^{1/3}&= 1 + \frac{1}{1!} \frac{1}{3} x + \frac{1}{2!} \frac{1}{3} \left (-\frac{2}{3} \right ) x^2 + \frac{1}{3!} \frac{1}{3} \left (-\frac{2}{3} \right ) \left (-\frac{5}{3} \right ) x^3 + \ldots \\ \end{align} $$
so that
$$(1+x)^{1/3} - 1 = \frac{1}{3} x - \frac{1}{9} x^2 + \frac{5}{81} x^3 - \ldots$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Trigonometry with multiple angle and exact value of $\tan\pi/5$ By considering the equation $\tan5\theta=0$, show that the exact value of $\tan\pi/5$ is $\sqrt{5-2\sqrt{5}}$.
Do I need to evaluate the multiple angle for $\tan5\theta=0$?
| $$\tan(5x) = \dfrac{\tan(3x) + \tan(2x)}{1-\tan(3x) \tan(2x)} = \dfrac{\dfrac{3 \tan(x) - \tan^3(x)}{1-3 \tan^2(x)} + \dfrac{2 \tan(x)}{1-\tan^2(x)}}{1- \dfrac{3 \tan(x) - \tan^3(x)}{1-3 \tan^2(x)} \cdot \dfrac{2 \tan(x)}{1-\tan^2(x)}}$$
Hence,
$$\tan(5x) = 0 \implies (3t-t^3)(1-t^2) + 2t(1-3t^2) = t \left(t^4-4t^2+3 - 6t^2+2 \right) = 0$$
where $t= \tan(x)$. Since we are interested in $\tan(\pi/5)$, we can rule out $t=0$. Hence, we need to solve a bi-quadratic $t^4-10t^2+5 = 0$. This gives us
$$t^2 = \dfrac{10 \pm \sqrt{100-4 \times 5}}{2} = 5 \pm 2 \sqrt5 \implies t = \pm \sqrt{5 \pm 2\sqrt5}$$ We also know that
$\tan(\pi/5) \in (\tan(0),\tan(\pi/4)) = \left(0,1 \right)$. Hence, $$\tan(\pi/5) = \sqrt{5 - 2\sqrt5}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Induction on binomial Identity: $0\cdot {n\choose 0} + 2\cdot {n\choose 2} + 4\cdot {n\choose4}+\ldots = n\cdot2^{n-2}$ I am having trouble proving the following identity:
$0\cdot {n\choose 0} + 2\cdot {n\choose 2} + 4\cdot {n\choose4}+\ldots = n\cdot2^{n-2}$
Here is what I have so far:
Proof:
Base: Let $n=0$:
LHS: $0\cdot {0\choose 0} = 0\cdot1 = 0$
RHS: $0\cdot 2^{0-2} = 0$
Step: Let $k\in \mathbb{Z} s.t k \geq 0$ and assume the identity is true for k.
Consider the LHS for $k+1$ where $k$ is even (I leave out the odd case because I think it will turn out the same?):
\begin{align}
=& 0\cdot{k+1\choose 0}+2\cdot{k+1\choose 2}+4\cdot{k+1\choose 4}+... +k\cdot{k+1\choose k}
\\=&0\cdot\left[{k\choose 0}+{k\choose -1}\right] + 2\cdot\left[{k\choose 2}+{k\choose 1}\right]+ 4\cdot\left[{k\choose 4}+{k\choose 3}\right]+\ldots+ k\cdot\left[{k\choose k}+{k\choose k-1}\right]
\\=&\left[0\cdot{k\choose 0}+2\cdot {k\choose 2}+4\cdot{k\choose 4}+\ldots+k\cdot{k\choose k}\right] + \left[0\cdot{k\choose -1}+ 2\cdot {k\choose 1}+4\cdot {k\choose 3}+\ldots+k\cdot{k\choose k-1}\right]
\\=& k\cdot2^{k-2} + \left[0\cdot{k\choose -1}+ 2\cdot {k\choose 1}+4\cdot {k\choose 3}+\ldots+k\cdot{k\choose k-1}\right]
\\
\end{align}
I know I need to end up with something like:
\begin{align}
=&k\cdot2^{k-2}+ \left[k\cdot2^{k-2} + 2^{k-1}\right]
\\=&2k\cdot 2^{k-2} + 2^{k-1}
\\=&k\cdot 2^{k-1}+2^{k-1}
\\=&(k+1)\cdot 2^{k-1}\end{align}
But, how can I get what I need from the combinations above? It may not end up exactly like that, but what is the reasoning behind this?
| Here is a proof, which relies on induction indirectly.
Note that
$$2k \dbinom{n}{2k} = 2k \dfrac{n!}{(n-2k)!(2k)!} = n \dfrac{(n-1)!}{(n-2k)!(2k-1!)} = n \dbinom{n-1}{2k-1}$$
Hence,
$$\sum_{k=1}^{\lfloor n/2 \rfloor} 2k \dbinom{n}{2k} = n\sum_{k=1}^{\lfloor n/2 \rfloor} \dbinom{n-1}{2k-1}$$
Now note that
$$(1+1)^{n-1} = \sum_{k=0}^{\lfloor n/2 \rfloor} \dbinom{n-1}{2k} + \sum_{k=0}^{\lfloor n/2 \rfloor} \dbinom{n-1}{2k-1} \,\,\,\,\, (\heartsuit)$$
and
$$(1-1)^{n-1} = \sum_{k=0}^{\lfloor n/2 \rfloor} \dbinom{n-1}{2k} - \sum_{k=0}^{\lfloor n/2 \rfloor} \dbinom{n-1}{2k-1} \,\,\,\,\, (\spadesuit)$$
Hence,
$$(\heartsuit) - (\spadesuit) \implies 2 \sum_{k=0}^{\lfloor n/2 \rfloor} \dbinom{n-1}{2k-1} = 2^{n-1}$$
Hence, $$\sum_{k=0}^{\lfloor n/2 \rfloor} \dbinom{n-1}{2k-1} = 2^{n-2}$$
Therefore, we get
$$\sum_{k=1}^{\lfloor n/2 \rfloor} 2k \dbinom{n}{2k} = n\sum_{k=1}^{\lfloor n/2 \rfloor} \dbinom{n-1}{2k-1} = n2^{n-2}$$
| {
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Why, given a natural number $n$, does $n^6$ always have the remainder of 1 when divided by 7? I am trying to prove why a natural number $n$ (which is not a multiple of $7$) when taken to the power of six ($n^6$) and divided by 7 always have the remainder of 1? I am not supposed to use "Fermats little theorem", but I am given the hint that the only numbers I have to account for are $1, 2, 3, 4, 5, 6$.
I would very much appreciate if someone could explain this to me.
Thank you
| Make use of the fact that if
$$a \equiv b \pmod{k}$$ then
$$a^n \equiv b^n \pmod{k}$$
In your case, for any $n$ that is not a multiple of $7$, we have
$$n \equiv \begin{cases}1 \pmod 7\\ 2 \pmod 7 \\ 3 \pmod 7 \\ 4 \pmod 7 \\ 5 \pmod 7 \\ 6 \pmod 7 \end{cases}$$
Hence,
$$n^6 \equiv \begin{cases}1^6 \pmod 7\\ 2^6 \pmod 7 \\ 3^6 \pmod 7 \\ 4^6 \pmod 7 \\ 5^6 \pmod 7 \\ 6^6 \pmod 7 \end{cases} \equiv \begin{cases}1 & \pmod 7\\ (2^3)^2 & \pmod 7 \\ (3^3)^2 & \pmod 7 \\ (4^3)^2 & \pmod 7 \\ (5^3)^2 & \pmod 7 \\ (-1)^6 & \pmod 7 \end{cases} \equiv \begin{cases}1 & \pmod 7\\ 1^2 & \pmod 7 \\ (-1)^2 & \pmod 7 \\ 1^2 & \pmod 7 \\ (-1)^2 & \pmod 7 \\ (-1)^6 & \pmod 7 \end{cases} \equiv 1 \pmod 7$$
| {
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Are $4ab\pm 1 $ and $(4a^2\pm 1)^2$ coprime? Let $a\ne b$ be two positive integers. Are $4ab+1$ and $(4a^2+1)^2$ coprime always?
Can you find $a$ and $b$ such that they are not coprime?
Edit:
It has been proved that $4ab-1$ is not a divisor of $(4a^2-1)^2$.
Are $4ab-1$ and $(4a^2-1)^2$ always coprime?
| When $a=1$ and $b=6$, $4ab+1=25$ and $(4a^2+1)^2=25$.
When $a=1$ and $b=4$, $4ab-1=15$ and $(4a^2-1)^2=9$.
| {
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When does a system of equations have no solution? I have performed Gaussian elimination on this matrix to reduce it to
$$
\left[
\begin{array}{@{}ccc|c@{}}
-3&-1&2 & 1 \\
0& \frac{-5}{3}& \frac{10}{3} & \frac{8}{3} \\
0&0& a+2 & b + \frac{6}{5} \\
\end{array}
\right]
$$
I thought that setting $a$ equal to $-2$ and having $b$ not equal to $-\frac{6}{5}$ would be the answer to this problem, but it apparently isn't. Could someone please explain why?
| Given a system of linear equations represented by the matrix equation: $\mathbf{A}\vec{x}=\vec{b}$, there is no unique set of solutions for $\det{\mathbf{A}}=0$.
Therefore, in your case:
$$\begin{bmatrix}-3 & -1 & 2 \\ 0 & -\frac{5}{3} & \frac{10}{3} \\ 0 & 0 & a+2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}=\begin{bmatrix}1 \\ \frac{8}{3} \\ b+\frac{6}{5}\end{bmatrix}$$
So we are interested in the case when:
$$\begin{vmatrix}-3 & -1 & 2 \\ 0 & -\frac{5}{3} & \frac{10}{3} \\ 0 & 0 & a+2\end{vmatrix}=0 \implies (5a+10)=0\implies a =-\frac{10}{5}=-2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that for any odd integer: $\lceil \frac{N^2}{4} \rceil = \frac{N^2 + 3}{4}$ I'm trying to construct a proof that for any odd integer: the ceiling of $\large \lceil \frac{N^2}{4} \rceil = \frac{N^2 + 3}{4}$.
Anyone have a second to show me how this is done? Thanks!
| Take $N=2k+1$ then we have,
$(N^2+3)/4=k^2+k+1$
$N^2/4=k^2+k+1/4,\Rightarrow $ its ceiling is $\lceil N^2/4\rceil =k^2+k+1=(N^2+3)/4$
| {
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Finding the Solutions of the two systems by using the inverse. I am having a difficult time understanding where I went wrong with the following:
$$\begin{matrix}4x-y = 1 \\ 2x+3y = 3 \end{matrix} $$
$$\begin{matrix}4x-y = -3 \\ 2x+3y = 3 \end{matrix} $$
I found the inverse of the common coefficient matrix of the systems:
$$A^{-1} \begin{cases} \frac3{14}, \frac1{14} \\ \\ -\frac17, \frac27 \end{cases} $$
The issue is this question:
Find the solutions to the two systems by using the inverse, i.e. by evaluating $A^{-1}B$ where $B$ represents the right hand side (i.e. $B = \begin{bmatrix}1 \\ 3 \end{bmatrix}$ for system (a) and $B = \begin{bmatrix}-3 \\ 3 \end{bmatrix}$ for system (b)).
Now whatever I find for x and y for both solutions keep coming as wrong. I am thinking, I might of read the question wrongly....I am using the negative values to find the x and y. Not too sure how to go at this now...
| $A^{-1} = \frac{1}{14}\begin{bmatrix} 3 & 1 \\ -2 & 4\end{bmatrix}$.
$A^{-1} \binom{1}{3} = \frac{1}{14} \binom{6}{10} = \frac{1}{7} \binom{3}{5}$.
$A^{-1} \binom{-3}{3} = \frac{1}{14} \binom{-6}{18} = \frac{1}{7} \binom{-3}{9}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/300867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding all solutions to $ \tan^5x - 9\tan{x} = 0 $ I am stuck when it comes to finding the end value of a trig function. I have the following question:
$$ \tan^5x - 9\tan{x} = 0 $$
I worked the problem and got:
$$ \tan x = 0\\
\tan^4x-9 = 0\\
x = 0, \pi, \frac {\pi}{3}, \frac {2\pi}{3}, \frac {4\pi}{3}, \frac {5\pi}{3} $$
My book answer is $x = \frac {\pi k}{3}$ how do you get that? I understand that tan uses $ \pi $ and sin, cos use $ 2\pi $ but I'm not sure how they got to that answer.
| $$\tan^4x=9\Longrightarrow \tan^2x= 3\Longrightarrow \tan x=\pm\sqrt 3=\pm\frac{\frac{\sqrt 3}{2}}{\frac{1}{2}}=\pm\frac{\sin\frac{\pi}{3}}{\cos\frac{\pi}{3}}$$
Can you see it now?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Different solutions of trigonometric equations Please take a look at this trigonometric equation,
$\cos9x\cos7x = \cos5x\cos3x$
To solve this equation, we can proceed as,
$2\cos9x\cos7x = 2\cos5x\cos3x$
or, $\cos(9x+7x)+\cos(9x-7x) = \cos(5x+3x)+\cos(5x-3x)$
or, $\cos16x+\cos2x = \cos8x+\cos2x$
or, $\cos16x = \cos8x$
From the above situation we can proceed in two ways,
First Way
$\cos16x = \cos8x$
or, $\cos(2\times8x) = \cos8x$
or, $2\cos^28x -1 = \cos8x$
or, $2\cos^28x-\cos8x -1 = 0$
or, $2\cos^28x-2\cos8x+\cos8x -1 = 0$
or, $2\cos8x(\cos8x-1)+(\cos8x -1) = 0$
or, $2\cos8x(\cos8x-1)+(\cos8x -1) = 0$
or, $(\cos8x -1)(2\cos8x+1) = 0$
Either or both of the above factors are zero.
Taking the first one,
$(\cos8x -1) = 0$
or, $\cos8x = 1$
or, $8x = 2n\pi$, where $n$ is an integer, +ve or -ve.
or, $x = {n\pi \over4}$
Taking the second one,
$(2\cos8x+1) = 0$
or, $2\cos8x = -1$
or, $\cos8x = -\frac1 2$
or, $8x = 2n\pi \pm \frac {2\pi} 3$, where $n$ is an integer, +ve or -ve.
or, $x = \frac{n\pi} 4 \pm \frac {\pi} {12}$
Second Way
$\cos16x = \cos8x$
or, $2\sin{8x-16x\over2}\sin{8x+16x\over2} = 0$
or, $2\sin(-4x)\sin{12x} = 0$
or, $-2\sin(4x)\sin{12x} = 0$
Again, either or both of the above factors are zero.
Taking the first one,
$\sin4x = 0$
or, $4x=n\pi$
or, $x=\frac{n\pi}4$
Taking the second one,
$\sin12x = 0$
or, $12x=n\pi$
or, $x=\frac{n\pi}{12}$
Now, as you must have noticed, we are getting two different sets of solutions, $x = \left\{{n\pi \over4}, \frac{n\pi} 4 \pm \frac {\pi} {12}\right\}$ and $x = \left\{{n\pi \over4}, \frac{n\pi}{12}\right\}$. The member $x=\frac{n\pi}4$ is common to both of the sets. Moreover, all these solutions satisfy the equation under consideration.
Could anybody please tell me why is this happening. In addition to the specific answer, some general insight will be most welcome. We have got a number of similar problems in hand. So, unless we can develop some acumen, life may become difficult.
| They aren't actually different sets of numbers. They are both all the numbers of the form $n\pi/12$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there a nice way to integrate this? I'm trying to integrate:
$$\iiint\limits_{x^2+y^2+z^2\leqslant 1}x^{2n}+y^{2n}+z^{2n}\mathrm{d}V$$
but, I can't seem to find any nice way to do this?
| $$\begin{align}
& \int_{x^2+y^2+z^2\le 1} x^{2n}+y^{2n}+z^{2n} dxdydz\\
= & 3 \int_{x^2+y^2+z^2\le 1}x^{2n} dx dy dz\\
= & 3 \int_{-1}^{1} dx\left( x^{2n} \int_{y^2 + z^2 \le 1 - x^2} dy dz \right)\\
= & 3\pi \int_{-1}^{1} dx \left( x^{2n} (1 - x^2)\right)\\
= & 3 \pi \left( \frac{2}{2n+1} - \frac{2}{2n+3} \right)\\
= & \frac{12 \pi}{(2n+1)(2n+3)}
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Parametrization for intersection of sphere and plane Given is the sphere $x^2 + y^2 + z^2 = 4$ and the plane $x + y = 2$ in $\mathbb R^3 $.
How can I find a parametrization for the intersection of the two?
| You are in luck because one of the equations is linear.
Since $x+y=2$ it follows that $x=2-y$. We can substitute this into the first equation:
$$x^2+y^2+z^2=4 \implies (2-y)^2+y^2+z^2=4 \implies 2y^2-4y+z^2=0\,.
$$
We can "complete the square" on the first part to give:
$$2[(y-1)^2-1]+z^2=0 \implies 2(y-1)^2+z^2=2 \, . $$
Considering only the $yz$-plane then we have an ellipse in the $yz$-plane parametrised by $(y(\theta),z(\theta)) = (1+\cos\theta,\sqrt{2}\sin\theta)$. We also know that $x=2-y$, hence:
$$\gamma(\theta) = (1-\cos\theta,1+\cos\theta,\sqrt{2}\sin\theta)$$
would give a regular parametrisation of the set in question.
| {
"language": "en",
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Finding the derivative of $r = \left(\frac{m^2}{a}\right)\left(\frac{1}{b}-\frac{m}{c}\right)$ I needed some help finding the derivative of the following at $m=15$
$$r = \left(\frac{m^2}{a}\right)\left(\frac{1}{b}-\frac{m}{c}\right)$$
Where $a,b,c$ are constants
I have no idea where to start. Can someone please point me in the correct direction?
Thanks!
| $$r =\frac{m^2}{a}\left(\frac{1}{b}-\frac{m}{c}\right)$$
$$r' =\frac{2m}{a}\left(\frac{1}{b}-\frac{m}{c}\right)+\frac{m^2}{a}\left(-\frac{1}{c}\right)$$
$$r'(15) =\frac{2\cdot 15}{a}\left(\frac{1}{b}-\frac{15}{c}\right)+\frac{15^2}{a}\left(-\frac{1}{c}\right)$$
$$r'(15) =\frac{30}{a}\left(\frac{1}{b}-\frac{15}{c}\right)+\frac{225}{a}\left(-\frac{1}{c}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/310462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Value of $ \sum \limits_{k=1}^{81} \frac{1}{\sqrt{k} + \sqrt{k+1}} = \frac{1}{\sqrt{1} + \sqrt{2}} + \cdots + \frac{1}{\sqrt{80} + \sqrt{81}} $? I tried my best, but I am totally clueless about it. Worse thing is we were supposed to arrive at the answer in approximately $ 2 $ minutes. The correct answer is $ 8 $, right? Can you kindly explain how to arrive at it? I hope it won’t be too much bother. Thank you.
| As \begin{align}
\frac{1}{\sqrt{i}+\sqrt{i+1}} = \frac{\sqrt{i}-\sqrt{i+1}}{\sqrt{i}-\sqrt{i+1}} \cdot \frac{1}{\sqrt{i}+\sqrt{i+1}} = \frac{\sqrt{i}-\sqrt{i+1}}{-1} = {\sqrt{i+1}-\sqrt{i}}
\end{align}
Thus
\begin{align}
\sum_{i=1}^{80} \frac{1}{\sqrt{i}+\sqrt{i+1}} = \sum_{i=1}^{80} {\sqrt{i+1}-\sqrt{i}} = \sqrt{80+1} -\sqrt{1} = 8
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Do matrices $ AB $ and $ BA $ have the same minimal and characteristic polynomials?
Let $ A, B $ be two square matrices of order $n$. Do $ AB $ and $ BA $ have same minimal and characteristic polynomials?
I have a proof only if $ A$ or $ B $ is invertible. Is it true for all cases?
| Yes, $AB$ and $BA$ have the same characteristic polynomial.
Basic facts: $\det(A^T) = \det(A)$, $\det(AB) = \det(A) \det(B)$
*
*$A$ and $A^T$ share the same characteristic polynomial.
\begin{align*}
\det(xI-A) = \det((xI-A)^T) = \det(xI-A^T)
\end{align*}
*Similar matrices have the same characteristic polynomial. If $B = PAP^{-1}$,
\begin{align*}
\det(xI - B) &= \det(xI - PAP^{-1}) \\
&= \det(P(xI - A)P^{-1}) \\
&= \det(P)\det(xI - A)\det(P^{-1}) \\
&= \det(xI - A)
\end{align*}
*Determinant of a block triangular matrix:
\begin{align*}
\det \begin{pmatrix}A & B \\0 & C\end{pmatrix} = \det(A) \det(C)
\end{align*}
Using block multiplication, please verify that $\begin{pmatrix}I & -A \\0 & I\end{pmatrix} \begin{pmatrix}AB & 0 \\B & 0\end{pmatrix} = \begin{pmatrix}0 & 0 \\B & BA\end{pmatrix} \begin{pmatrix}I & -A \\0 & I\end{pmatrix}$.
Therefore, the matrices $\begin{pmatrix}AB & 0 \\B & 0\end{pmatrix}$ and $\begin{pmatrix}0 & 0 \\B & BA\end{pmatrix}$ are similar, and have the same characteristic polynomial.
\begin{align*}
\det\left[x\begin{pmatrix}I & 0 \\0 & I\end{pmatrix} - \begin{pmatrix}AB & 0 \\B & 0\end{pmatrix}\right] &= \det(xI - AB) \det(xI)
\end{align*}
\begin{align*}
\det\left[x\begin{pmatrix}I & 0 \\0 & I\end{pmatrix} - \begin{pmatrix}0 & 0 \\B & BA\end{pmatrix}\right] &= \det(xI) \det(xI - BA)
\end{align*}
And there it is. But $AB$ and $BA$ do not need to have the same minimal polynomial. See Jim's answer for a counterexample.
| {
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Show a,b,c congruences when $a^2+b^2=c^2$ i would like to know how to prove that if $a^2+b^2=c^2$ then
a) $3|a$ or $3|b$
b) $5|a$ or $5|b$ or $5|c$
c) $4|a$ or $4|b$
Also that $a^2+b^2+c^2+1$ can't be divisible by 8
Thank you very much!
| a) $a^2=1 \;\text{or}\; 0 \pmod 3,\quad b^2=1 \;\text{or}\; 0 \pmod 3.\quad c^2=0 \;\text{or}\; 1 \mod(3).$
$ c^2$ can't be $0 \pmod 3$, which means $a^2+b^2=1$ or $0 \pmod 3$. Either $3|a \;\text{or}\; 3|b$..
b) $a^2=0,1 \;\text{or}\; 4 \pmod 5,\quad b^2=0,1 \;\text{or}\; 4 \pmod 5.\quad c^2=0,1 \;\text{or}\; 4 \pmod 5$.
$$a^2+b^2=0,1 \;\text{or} 4 \pmod 5.$$
Case 1: When 5 doesn't divide both 'a' and 'b',
$$a^2=1 \;\text{or}\; 4 \pmod 5 ,\quad b^2=4 \;\text{or}\; 1 \pmod 5$$.
Notice that, both can't $1$ or $4 \pmod 5$ simultaneously as $c^2=0,1 \;\text{or}\; 4 \pmod 5)$.
Therefore, $a^2+b^2=0 \pmod 5, \quad 5|c$.
Case 2: When $5$ doesn't divide $c$
$$c^2=1 \;\text{or}\; 4 \pmod 5$$
$$a^2=1,4 \;\text{or}\; 0 \pmod 5,\quad b^2=1, \quad 4 \;\text{or}\; 0 \pmod 5$$
$$a^2+b^2=1 \;\text{or}\; 4 \pmod 5 \text{, therefore, } 5|a \;\text{or}\; 5|b.$$
c) $a^2=1 \;\text{or}\; 0 \pmod 4, \; b^2=1 \;\text{or}\; 0 \pmod4, c^2=1 \;\text{or}\; 0 \pmod 4.$
Adding two squares is not $2 \pmod 4$.
Therefore, if $$c^2=1 \pmod 4.\quad 4|a \;\text{or}\; 4|b$$
$$c^2=0 \pmod 4. \quad 4|a \;\text{or}\; 4|b$$
Well, this is the most elementary solution to your problem.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $(5-\frac5k )(1+\frac{1}{(k+1)^2}) \le 5 - \frac{5}{k+1}$ Can anyone tell me what I am doing wrong? need to prove for $k\ge2$
$$(5-\frac5k )(1+\frac{1}{(k+1)^2}) \le 5 - \frac{5}{k+1}$$$$(5-\frac5k )(1+\frac{1}{(k+1)^2})= 5(1-\frac1k)(1+\frac1{(k+1)^2})$$
$$=5(1+\frac1{k+1)^2}-\frac1k-\frac1{k(k+1)^2})$$
$$= 5(1-\frac{k^2+k+2}{k(k+1)^2})$$
$$=5(1-\frac{k(k+1)}{k(k+1)^2}+\frac2{k(k+1)^2})$$
$$=5(1-\frac{1}{k+1}+\frac2{k(k+1)^2})$$
$$= 5 - \frac5{k+1}+\frac{10}{k(k+1)^2}\le5-\frac5{k+1}$$
which doesn't look true.
| Let $f(x) = (5-\frac{5}{x})(1+\frac{1}{(1+x)^2})-5+\frac{5}{1+x}$. After a little algebra, this gives $f(x) = -\frac{10}{x(x+1)^2}$. Hence $f(x) \leq 0$ when $x>0$.
Hence $f(k) \leq 0$ for $k \geq 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/316601",
"timestamp": "2023-03-29T00:00:00",
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Limit of $y=\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-\sqrt{5+\ldots}}}}}$ I would appreciate any help with this problem:
If
$$y=\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-\sqrt{5+\ldots}}}}}$$
Then how do I find $y^2 - y$?
I'm not sure whether this is an arithmetic or geometric series.
| Define
$$
a_0=0\quad\text{and}\quad a_{k+1}=\sqrt{5+\sqrt{5-a_k}}
$$
Show that for $k\ge1$, $\sqrt5\le a_k\le\sqrt{5+\sqrt5}$.
Initially, $0\le a_0=0\le5$.
Suppose that $0\le a_k\le 5$. Then $\sqrt5\le a_{k+1}\le\sqrt{5+\sqrt5}$.
$$
\begin{align}
\left|\,a_{k+1}-a_k\,\right|
&=\frac{\left|\,\left(5+\sqrt{5-a_k}\right)-\left(5+\sqrt{5-a_{k-1}}\right)\,\right|}{a_{k+1}+a_k}\\
&=\frac{\left|\,\sqrt{5-a_k}-\sqrt{5-a_{k-1}}\,\right|}{a_{k+1}+a_k}\\
&=\frac{\left|\,a_k-a_{k-1}\,\right|}{(a_{k+1}+a_k)\left(\sqrt{5-a_k}+\sqrt{5-a_{k-1}}\right)}\\
&\le\frac{\left|\,a_k-a_{k-1}\,\right|}{\left(\sqrt5+\sqrt5\right)\left(\sqrt{5-\sqrt{5+\sqrt5}}+\sqrt{5-\sqrt{5+\sqrt5}}\right)}\\
&\le\frac{\left|\,a_k-a_{k-1}\,\right|}{13}
\end{align}
$$
Thus, $a_k$ converges since
$$
\lim_{n\to\infty}\sum_{k=1}^n(a_k-a_{k-1})=\lim_{n\to\infty}a_n-a_0
$$
converges absolutely; that is,
$$
\sum_{k=1}^\infty\left|\,a_k-a_{k-1}\right|\le\frac{13}{12}\sqrt{5+\sqrt5}
$$
Set $a=\lim\limits_{k\to\infty}a_k$. Since $\sqrt{5+\sqrt{5-x}}$ is continuous for $x\le5$, we have that $a=\sqrt{5+\sqrt{5-a}}$, which means
$$
\begin{align}
0
&=a^4-10a^2+a+20\\
&=(a^2-a-4)(a^2+a-5)
\end{align}
$$
The roots of $a^2-a-4$ are $\frac{1\pm\sqrt{17}}{2}$ and the roots of $a^2+a-5$ are $\frac{-1\pm\sqrt{21}}{2}$. The only one that is between $\sqrt5$ and $\sqrt{5+\sqrt5}$ is $\frac{1+\sqrt{17}}{2}$. Therefore,
$$
\lim_{k\to\infty}a_k=\frac{1+\sqrt{17}}{2}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Prove $\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}$ I need to show that
$$
\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}
$$
I have already derived the result $\int_0^\infty \frac{\sin^2x}{x^2} = \frac{\pi}{2}$ using complex analysis, a result which I am supposed to start from. Using a change of variable $ x \mapsto 2x $ :
$$
\int_0^\infty \frac{\sin^2(2x)}{x^2}dx = \pi
$$
Now using the identity $\sin^2(2x) = 4\sin^2x - 4\sin^4x $, we obtain
$$
\int_0^\infty \frac{\sin^2x - \sin^4x}{x^2}dx = \frac{\pi}{4}
$$
$$
\frac{\pi}{2} - \int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4}
$$
$$
\int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4}
$$
But I am now at a loss as to how to make $x^4$ appear at the denominator. Any ideas appreciated.
Important: I must start from $ \int_0^\infty \frac{\sin^2x}{x^2}dx $, and use the change of variable and identity mentioned above
| You mentioned that you used complex analysis to evaluate $\int_{0}^{\infty} \frac{\sin^{2}(x)}{x^{2}} \, \mathrm dx$.
We can also use complex analysis to evaluate $\int_{0}^{\infty} \frac{\sin^{4}(x) }{x^{4}} \, \mathrm dx$.
Using the trigonometric identity $ \displaystyle \sin^{4} x = \frac{1}{8} \Big(\cos 4x - 4 \cos 2x + 3 \Big)$, we get
$$ \begin{align} \int_{0}^{\infty} \frac{\sin^{4} x}{x^{4}} \, \mathrm dx &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin^{4} x}{x^4} \, \mathrm dx \\ &= \frac{1}{16} \int_{-\infty}^{\infty} \Re \ \frac{e^{4ix}-4e^{2ix}+3}{x^{4}} \, \mathrm dx \\ &= \frac{1}{16}\int_{-\infty}^{\infty} \Re \ \frac{e^{4ix}-4e^{2ix}+3+4ix}{x^{4}} \, \mathrm dx \\ &= \frac{1}{16} \, \Re \, \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{4ix}-4e^{2ix}+3+4ix}{x^{4}} \, \mathrm dx. \end{align}$$
So let's integrate the function $$f(z) = \frac{e^{4iz}-4e^{2iz}+3+4iz}{z^{4}}$$ around a contour the consists of the real axis from $-R$ to $R$, $R>0$, and the upper half of the circle $|z|=R$. To avoid the simple pole at the origin, the contour needs to be indented at the origin.
Letting the radius of the indentation go to zero and $R \to \infty$, we get
$$ \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{4ix}-4e^{2ix}+3+4ix}{x^{4}} \, \mathrm dx- i \pi \ \text{Res}[f(z),0] = 0,$$
where
$$ \begin{align} \operatorname{Res}[f(z),0] &= \lim_{z \to 0} \frac{e^{4iz}-4e^{2iz}+3+4iz}{z^{3}} \\ &= \lim_{z \to 0} \frac{-64ie^{4iz}+32ie^{2iz} }{6} \\ &= - \frac{16i}{3}. \end{align}$$
Therefore,
$$ \int_{0}^{\infty} \frac{\sin^{4} x}{x^{4}} \, dx = \frac{1}{16} \left(\frac{16 \pi}{3} \right)=\frac{\pi}{3} .$$
Technically, it wasn't necessary to add $4ix$ to the numerator.
For reasons explained here, the Cauchy principal value of $ \int_{-\infty}^{\infty} \frac{e^{4ix}-4e^{2ix}+3}{x^{4}} \, \mathrm dx $ exists even though $\frac{e^{4iz}-4e^{2iz}+3}{z^{4}}$ has a pole of order $3$ at the origin.
| {
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generating functions, can't seem to get the correct answers. So, I've been having some issues with generating functions and counting problems. An example problem is: $$ a_n = a_{n-1} + 9a_{n-2} - 9a_{n-3} \;\;\; (n \geq 3)$$
Where $a_0 = 0, a_1 = 1, a_2 = 2$
Let's define $g(x)$ as our generating function, in which case,
$$ g(x) = a_0 + a_1x + a_2x^2+... = 0 + x + 2x^2 + (a_2 + 9a_1 - 9a_0)x^3 + (a_3+9a_2-9a_1)x^4+...$$
We can factor this like so...
$$g(x) = x + 2x^2 + x(a_2x^2 + a_3x^3 + ...) + 9x^2(a_1x + a_2x^2+...) - 9x^3(a_0 + a_1x + a_2x^2+...)$$
Now, the infinite series, $a_2x^2 + a_3x^3+...$ can be defined as $g(x) - a_0 - a_1x$, and the series $a_1x + a_2x^2+...$ can be defined as $g(x) - a_0$, and lastly, the series $a_0 + a_1x + a_2x^2+...$ is $g(x)$
So, making substitutions:
$$g(x) = x + 2x^2 + xg(x) - x^2 + 9x^2g(x) - 9x^3g(x)$$
and finally,
$$g(x) = \frac{x^2 + x}{9x^3 - 9x^2 - x + 1}$$
I decided to run this through wolfram alpha for the partial fraction decomposition and I get:
$$g(x) = \frac{1}{3}\left (\frac{1}{1+3x}\right ) + \frac{1}{12}\left (\frac{1}{1+3x}\right ) - \frac{1}{4}\left (\frac{1}{1+x}\right )$$
Each of the terms in parenthesis can be expressed as a series, in summation notation:
$$\frac{1}{3}\sum_{k = 0}^{\infty }(-3)^kx^k + \frac{1}{12}\sum_{k = 0}^{\infty }(-3)^kx^k - \frac{1}{4}\sum_{k=0}^{\infty }(-1)^kx^k$$
And, my final answer should be:
$$a_n = \frac{1}{3}(-3)^n + \frac{1}{12}(-3)^n - \frac{1}{4}(-1)^n$$
Which, is incorrect. Could someone point out what I'm doing wrong?
| A simpler way, due to Wilf:
Your recurrence is $a_{n + 3} = a_{n + 2} + 9 a_{n + 1} - 9 a_n$. If you multiply by $x^n$ and add for $n \ge 0$ it is:
$$
\begin{align*}
\sum_{n \ge 0} a_{n + 3} x^n
&= \sum_{n \ge 0} a_{n + 2} x^n + 9 \sum_{n \ge 0} a_{n + 1} x^n - 9 \sum_{n \ge 0} a_n x^n \\
\frac{g(x) - a_0 - a_1 x - a_2 x^2}{x^3}
&= \frac{g(x) - a_0 - a_2 x}{x^2} + 9 \frac{g(x) - a_0}{x} - 9 g(x)
\end{align*}
$$
As you see, it is almost immediate to write the last equation just by looking at the recurrence. A term $a_{n + k}$ gives rise to
$$
\frac{g(x) - a_0 - a_1 x - \ldots - a_{k - 1} x^{k - 1}}{x^k}
$$
The rest, as they say, is algebra (a computer algebra package like maxima helps a lot) and a bit of series expansion.
| {
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"url": "https://math.stackexchange.com/questions/318213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solve $79x \equiv 1 \pmod {90}$ This is an example in my lecture notes and I am a little confused as to how my lecturer has gone about doing this. In my notes, it says
$90 = 2 \times 3^2 \times 5$. So we can now say that solving $79x \equiv 1 \pmod {90}$ is equivalent to solving
$$79x \equiv 1 \pmod 2$$
$$7x \equiv 1 \pmod 9$$
$$4x \equiv 1 \pmod 5$$
...
How has she got the $79, 7$ and $4$ in the equivalent set of equations?
| to solve the problem $79x \equiv 1 \pmod{90}$ you can use the Chinese Remainder Theorem to obtain the answer. For this way you need to factorize the 90 so the factors are pairwise coprimes (like the modules in the congruences your lecturer made), then you need to transform the congruences to the form:
$$
\left\{
\begin{aligned}
x & \equiv b_1 \pmod{a_1} \\
x & \equiv b_2 \pmod{a_2} \\
\vdots & \\
x & \equiv b_r \pmod{a_r}
\end{aligned}
\right.
$$
For this particular case we have that $90 = 2 \cdot 5 \cdot 9$ is a good factorization because the three factors are coprimes pairwise, so we need to fix the congruences to match the structure above:
The first congruence is in mod 2, so we can simplify it because $79 \equiv 1 \pmod 2$, so:
$ 79 \cdot x \equiv 1 \cdot x \pmod{2} \quad \Rightarrow \quad x \equiv 1 \pmod{2} $
Second congruence is $4x \equiv 1 \pmod{5}$. The multiplicative inverse of 4 in mod 5 is the same number: $ 4 \cdot 4 = 16 \equiv 1 \pmod{5}$ and taking product by 4 on the congruence:
$ 4 \cdot 4 \cdot x \equiv 4 \cdot 1 \pmod{5} \Rightarrow x \equiv 4 \pmod{5} $
In the last congruence $7x \equiv 1 \pmod{9}$ the inverse of 7 is again 4.
Using this inverse we obtain an equivalent congruence:
$ 4 \cdot 7 \cdot x \equiv 4 \cdot 1 \pmod{9} \Rightarrow x \equiv 4 \pmod{9} $
Now resolve the initial problem is equivalent to find x that
$$
\left\{
\begin{aligned}
x & \equiv 1 \pmod 2 \\
x & \equiv 4 \pmod 5 \\
x & \equiv 4 \pmod 9
\end{aligned}
\right.
$$
Now x can be obtained using the theorem (check http://en.wikipedia.org/wiki/Chinese_remainder_theorem for more information about this).
To find x is convenient to express in a linear combination of the $b_i$:
$ x = \sum_{1 \le i \le r} m_i b_i \pmod{n} $
In this case we must search for the coefficients:
$
\begin{align}
m_1 & \quad \text{coefficient for mod 2} \\
m_2 & \quad \text{coefficient for mod 5} \\
m_3 & \quad \text{coefficient for mod 9}
\end{align}
$
This coefficients are obtained using the expression $ m_i = s_i \cdot \frac{n}{a_i} $, where $s_i = \left(\frac{n}{a_i}\right)^{-1}$ in mod $a_i$.
For this problem $n = 90$ and
$
\begin{align}
a_1 & = 2 \\
a_2 & = 5 \\
a_3 & = 9
\end{align}
$
Inverses are:
$
\begin{align}
s_1 = \left(\frac{90}{2}\right)^{-1} \pmod 2 = 1 \\
s_2 = \left(\frac{90}{5}\right)^{-1} \pmod 5 = 2 \\
s_3 = \left(\frac{90}{10}\right)^{-1} \pmod 9 = 1
\end{align}
$
and the coefficients in mod 90:
$
\begin{align}
m_1 = s_1 \cdot \frac{90}{2} = 1 \cdot 45 = 45 \\
m_2 = s_2 \cdot \frac{90}{5} = 2 \cdot 18 = 36 \\
m_3 = s_3 \cdot \frac{90}{10} = 1 \cdot 10 = 10
\end{align}
$
Value of x is obtained evaluating the linear combination:
$
x = \sum_{1 \le i \le 3} m_i \cdot b_i
= 45 \cdot 1 + 36 \cdot 4 + 10 \cdot 4
= 229 \equiv 49 \pmod{90}
$
We can check that $x = 49$ satisfies the three congruences:
$
\begin{align}
49 & \equiv 1 \pmod 2 \\
49 & \equiv 4 \pmod 5 \\
49 & \equiv 4 \pmod 9
\end{align}
$
and the initial problem:
$ 79 \cdot 49 = 3871 \equiv 1 \pmod 90 \quad (\text{because } 3871 = 90 \cdot 43 + 1)$
There is a CRT calculator in this page http://davidwees.com/chineseremaindertheorem/
Hope this could be useful.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/318742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Solve the following homogeneous differential equation Initial value problem: $\displaystyle \frac{dy}{dx}= \frac{y}{x}+2 \frac{x^2}{y^2}$, $y(1)=1$. Can anyone help
| Let, $y=ux$
$\frac{dy}{dx}=u+x\frac{du}{dx}=u+\frac{2}{u^2}$
$x\frac{du}{dx}=\frac{2}{u^2}$
$\frac{dx}{x}=\frac{u^2}{2}du$
$\ln x+\ln c=\frac{u^3}{2*3}$
$6(\ln cx)=\frac{y^3}{x^3}$
$6{x^3}(\ln cx)=y^3$
$y=(6{x^3}(\ln cx))^{1/3}$
$y=(6{x^3}(\ln cx))^{1/3}$ -- (|)
$y(1)=(6(\ln c))^{1/3}=1$
$\ln c=\frac{1}{6}$ -- (||)
From, (|) & (||) we have,
$y=(6{x^3}(\ln c+\ln x))^{1/3}$
$y=(x^3(1+6\ln x))^{1/3}$
$y^3=x^3(1+6\ln x)$
$y^3=x^3(1+1*(3-1)*3\ln x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/322535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
finding the derivative using quotient rule and product rule find dy/dx;
a) $\frac{1-2x}{\sqrt{2+x}}$
b.) $3x(1-x^2)^{1/3}$
My attempt at a)
use the quotient rule:
so $dy/dx = -2 \sqrt{2+x}+ (1-2x)0.5(2+x)^{-1/2}$
but then I get stuck there, cannot simplify it, wolfram gives a nice simplified answer but not sure how to get it.
b.) Product rule:
$dy/dx= 3(1-x^2)^{0.5} + 3 \times 1/3 \times (1-x^2)^{-2/3}$ and again i can't seem to simplify that either to a nice wolfram answer.
| $$(a)\;\;\;\left(\frac{1-2x}{\sqrt{2+x}}\right)'=\frac{-2\sqrt{2+x}-\frac{1}{2\sqrt{2+x}}(1-2x)}{2+x}=\frac{-4(2+x)-(1-2x)}{2(2+x)^{3/2}}=\ldots$$
$${}$$
$$(b)\;\;\;\;\;\; \left(3x(1-x^2)^{1/3}\right)'=3(1-x^2)^{1/3}+3x(-2x)\frac{1}{3}(1-x^2)^{-2/3}=\ldots $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/326405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Solve $A_n=A_{n-1}+\lfloor \sqrt{A_{n-1}}\rfloor$ I am trying to solve the recurrence:
$A_0=1$
$A_n=A_{n-1}+\lfloor \sqrt{A_{n-1}}\rfloor,\text{ for } n > 0$
Its obvious that
$A_n=m^2 \implies A_{n+1}=m^2+m$
however my book's solution states that the key insight is to realize:
$A_n=m^2 \implies A_{n+2k+1}=(m+k)^2+m-k\text{, for }0 \le k \le m$
and
$A_n=m^2 \implies A_{n+2k+2}=(m+k)^2+2m\text{, for }0 \le k \le m$
What method or line of thinking would have lead me to this?
This is exercise 3.28 from Concrete Mathematics.
| I'll use $r(x) = \lfloor \sqrt{x}\rfloor$ throughout.
Warm Up Here's some stuff we'll use freely in the next section. It's motivated by the next section though.
What's the floor of the square root of $m^2+7m+5$? You can read it off this table:
$$\begin{array}{|c|c|c|} \hline
m+k & m^2 + 2km + k^2 \\ \hline
m+1 & m^2 + 2m + 1 \\
m+2 & m^2 + 4m + 4 \\
m+3 & m^2 + 6m + 9 \\
m+4 & m^2 + 8m + 16 \\
m+5 & m^2 + 10m + 25 \\
m+6 & m^2 + 12m + 36 \\
m+7 & m^2 + 14m + 49 \\ \hline
\end{array}$$
It must be $m+3$.. but we do need $2k < m$ to use this table. It's easy to find counterexamples (e.g. $r(3^2+7\cdot 3+5)=5$).
Using this we can push your idea of starting with a generic $A_n = m^2$ much further and see the recurrence:
$$\begin{array}{|l|l|} \hline
A_n & r(A_n) \\ \hline
m^2 & m \\
m^2 + m & m \\
m^2 + 2m & m \\
m^2 + 3m & m + 1 \\
m^2 + 4m + 1 & m + 1 \\
m^2 + 5m + 2 & m + 2 \\
m^2 + 6m + 4 & m + 2 \\
m^2 + 8m + 6 & m + 3 \\
\end{array}$$
of course with the same caveat as before.
To try to formulate the pattern in this table (in order to prove it by induction) we get exactly the key insight you mentioned.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/327584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
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