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$\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a>-\sqrt{2}$
If $\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a>-\sqrt{2}$ is satisfied for all real $x>0$ then obtain the possible values of the parameter $a$.
My attempt is as follows:
$$\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a+\sqrt{2}>0$$
First condition: If $\forall$ $x>0$, inequality is satisfied, it means $x=0$ must have been the root of the equation $\quad \{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a+\sqrt{2}=0\quad$ at which given inequality would not be satisfied.
Second condition: If one root is real, then other root must be real as complex roots occur as conjugates if all quadratic equation coefficients are real. So $D>=0$
Third condition: $a>0$ as $\forall$ $x>0$, $\quad\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a+\sqrt{2}>0\quad$
First condition:
$$x=0$$
$$a+\sqrt{2}=0$$
$$a=-\sqrt{2}$$
Second condition:
$$D>=0$$
$$4(a^2-2)^2-4(a-\sqrt{2})(a^2+a-3)(a+\sqrt{2})>=0$$
$$(a-\sqrt{2})(a+\sqrt{2})(a^2-2-a^2-a+3)>=0$$
$$(a-\sqrt{2})(a+\sqrt{2})(a-1)<=0$$
$$a\in \left(-\infty,-\sqrt{2}\right]\quad \cup \quad[1,\sqrt{2}]$$
Third condition:
$$a>0$$
$$(a-\sqrt{2})(a^2+a-3)>0$$
$$(a-\sqrt{2})\left(a-\left(\frac{-1+\sqrt{13}}{2}\right)\right)\left(a-\left(\frac{-1-\sqrt{13}}{2}\right)\right)>0$$
$$a\in \left(\frac{-1-\sqrt{13}}{2},\frac{-1+\sqrt{13}}{2}\right) \cup \left(\sqrt{2},\infty\right) $$
Taking intersection of all three conditions would give
$$a\in \{-\sqrt{2}\}\quad$$
But answer is $a\in [-\sqrt{2},\frac{-1+\sqrt{13}}{2})\quad\cup\quad\left[\sqrt{2},\infty\right)$
What am I missing here, I tried to think of it a lot but didn't any breakthroughs. Please help me in this.
| Why are you taking the intersection of those three conditions? It doesn't make much sense to me.
For example, the second condition $\Delta\geq 0$ is not necessary. If $a>\sqrt{2}$ then we have $\Delta<0$ and the inequality is satisfied for all real $x$ (not only for $x>0$):
$$\underbrace{(a^2+a-3)(a-\sqrt{2})}_{>0}x^2+2(a^2-2)x+(a+\sqrt{2})>0.$$
Also $a=\sqrt{2}$ works because the inequality boils down to $(\sqrt{2}+\sqrt{2})>0$ which holds.
Hence $[\sqrt{2},+\infty)$ is a subset of the possible values of the parameter $a$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Ramanujan, sum of two cubes - how was it discovered? I'm looking for motivation for, and hopefully a derivation of, Ramanujan's sum of cubes formula
$$\left(x^2+7xy-9y^2\right)^3+\left(2x^2-4xy+12y^2\right)^3=\left(2x^2+10y^2\right)^3+\left(x^2-9xy-y^2\right)^3$$
I can't see where one could start to justify this without actually expanding everything out.
Additionally, does it provide a parametrisation of the surface
$$X^3+Y^3=Z^3+1$$ in $\mathbb{Q}^3$
with
$$X=\frac{x^2+7xy-9y^2}{2x^2+10y^2}$$
$$Y=\frac{2x^2-4xy+12y^2}{2x^2+10y^2}$$
$$Z=\frac{x^2-9xy-y^2}{2x^2+10y^2}$$
| The most general homogeneous quadratic function of two variables is $ax^2+bxy+cy^2$. Cubing this gives $$a^3x^6+3a^2bx^5y+(3a^2c+3ab^2)x^4y^2+(b^3+6abc)x^3y^3+(3ac^2+3b^2c)x^2y^4+3bc^2xy^5+c^3y^6.$$(The fact that this sextic is homogeneous makes the above easy to deduce from combinatorics, without manually expanding out brackets. There's even an $x\leftrightarrow y$ symmetry to almost halve the work.) The sum of two such cubes equals another viz.$$(ax^2+bxy+cy^2)^3+(dx^2+exy+fy^2)^3=(gx^2+hxy+iy^2)^3+(jx^2+kxy+ly^2)^3$$iff coefficients match. The first requirement this gives is$$a^3+d^3=g^3+j^3.$$Ramanujan was famous for noting the least positive integer expressible as the sum of two cubes in two different ways is $1729=1^3+12^3=9^3+10^3$, but in this example we work with much smaller examples, so we just swap variables. The "simplest" option is $a=1,\,d=2$ (so that $d\ne a$), whence $g=d=2,\,j=a=1$. For the $y^6$ coefficients, Ramanujan actually does work with the above result about $1729$, rearranging the equality of two sums of cubes as$$(-9)^3+12^3=10^3+(-1)^3$$to reduce the size of the total to $999$. So we try $c=-9,\,f=12,\,i=10,\,l=-1$. We now only need to find $b,\,e,\,h,\,k$, which in the case at hand turn out to be $7,\,-4,\,0,\,-9$. History probably doesn't record whether Ramaujan found these through trial and error or by working with the other coefficient-matching conditions, but I'm sure he did the latter, and you can try it as an exercise. (Remember to plug in the $8$ coefficients we've already fixed to simplify it to a problem in the other $4$.)
| {
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What is the sum of $k^2(n - k)$ for $k = 1$ to $k = n$? What is the value of $\sum\limits_{k=1}^n k^2(n - k)$?
The problem I had was:
for a square grid of size $n \times n$ how many squares have their corners on the intersecting points of the grid. (There are $n \times n$ points, the square's length is $n - 1$).
For every $k$-sized square there are $n - k$ possible squares with their corners laying on the edges of the bigger square. For every $k$-sized square there are $\frac{k(k + 1)(2k + 1)}{6}$ possible smaller squares.
| Let's start by splitting the sum into two different sums:
$$
\sum_{k=1}^n k^2(n-k)=\sum_{k=1}^n k^2n-\sum_{k=1}^n k^3=n\sum_{k=1}^n k^2-\sum_{k=1}^n k^3
$$
Now we have $\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^n k^3=\left( \frac{n(n+1)}{2}\right) ^2=\frac{n^2(n+1)^2}{4}$ and together
$$
n\frac{n(n+1)(2n+1)}{6}-\frac{n^2(n+1)^2}{4}=\frac{2n^2(n+1)(2n+1)-3n^2(n+1)^2}{12}=n^2(n+1) \frac{2(2n+1)-3(n+1)}{12}=n^2(n+1)\frac{4n+2-3n-3}{12}=\frac{n^2(n+1)(n-1)}{12}
$$
| {
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Dealing with degrees in decomposition into partial fractions I had to decompose $ \frac{2x^2}{x^4-1} $ into partial fractions in order to determine its antiderivative.
So, I said:
$$
\frac{2x^2}{x^4-1} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1}
$$
However, in the answer key, they said:
$$
\frac{2x^2}{x^4-1} = \frac{A}{x^2-1} + \frac{B}{x^2+1}
$$
Although I got the same answer, I had to do unpleasant calculations to finally get a system of four equations and four unknowns, which I had to solve as well.
What I want to know is what made them do that assumption? This isn't what we learned about decomposition into partial fractions.
I thought that maybe because in the starting fraction, the polynomial in the numerator is 2 degrees less than that of the denominator, so we must keep the same degree difference in the partial fractions.
However if we look at this example where the numerator is 5 degrees less than the denominator, this what was written in the answer key:
$$
\frac{1}{x^2(x-1)^3} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1} + \frac{D}{(x-1)^2} + \frac{E}{(x-1)^3}
$$
Which is normal and compatible with what I've learned about decomposition into partial fractions.
Please can anyone help? Also please no very complex calculations because I am a biology student. Thank you.
| There are always alternatives, one does not expect an answer key to give every way that works. For finding an antiderivative,
$$
\frac{2x^2}{x^4-1} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx}{x^2+1} + \frac{D}{x^2+1}
$$
is quite good in terms of recognizing things. $C$ leads to a substitution, while $D$ leads to an arctangent
| {
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Integrate area enclosed by $x^4 + y^4 = x^3 + y^3 $ I want to calculate the area bounded by the region
$$x^4 + y^4 = x^3 + y^3 $$
with integral. And also it's perimeter.
Can somebody please help me with it?!
|
Express the curve $x^4 + y^4 = x^3 + y^3$ in its polar coordinates,
$$r(\theta) = \frac{\cos^3\theta+\sin^3\theta}{\cos^4\theta+\sin^4\theta}\tag{1}$$
Recognizing that a complete loop is formed starting at origin and varying $\theta$ from $-\frac{\pi}{4}$ to $\frac{3\pi}{4}$. the area integral is then,
$$A= \int_{-\frac\pi4}^{\frac{3\pi}{4}}\int_0^{r(\theta)}rdrd\theta=\frac12\int_{-\frac\pi4}^{\frac{3\pi}{4}}r^2(\theta)d\theta$$
Simplify the integrand $r(\theta)$ given by (1),
$$r(\theta)=\frac{(\cos\theta+\sin\theta)(\cos^2 -\cos\theta\sin\theta +\sin^2\theta)}{(\cos^2\theta+\sin^2\theta)^2 - 2\cos^2\theta\sin^2\theta}$$
$$=\frac{\sqrt2\sin(\theta+\frac\pi4)(2-\sin2\theta)}{2- \sin^22\theta}$$
and apply the variable change $t=\theta + \frac{\pi}{4}$ to reduce the integral to,
$$A=\int_0^\pi \sin^2t \left(\frac{2+\cos 2t}{2-\cos^2 2t}\right)^2dt$$The area can then be integrated analytically to yield,
$$A= \frac{3\sqrt2}{8}\pi$$
| {
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A Wallis-like formula for $\pi$: $(\frac21)^2(\frac23)^2(\frac43)^2(\frac45)(\frac65)^2(\frac67)^2(\frac87)(\frac89)^2\cdots$ I don't quite understand this and would appreciate a clear explanation, if possible.
If we begin with the Wallis product for $\frac{\pi}{2}$,
$$\prod_{n=1}^\infty \left( \frac{2n}{2n-1} \cdot \frac{2n}{2n+1} \right) = \Big(\frac{2}{1} \cdot \frac{2}{3}\Big) \cdot \Big(\frac{4}{3} \cdot \frac{4}{5}\Big) \cdot \Big(\frac{6}{5} \cdot \frac{6}{7}\Big) \cdot \Big(\frac{8}{7} \cdot \frac{8}{9}\Big) \cdot \; \cdots \\ = \Big(\frac{2}{1}\Big) \cdot \Big( \frac{2}{3}\Big) \cdot \Big(\frac{4}{3}\Big) \cdot \Big(\frac{4}{5}\Big) \cdot \Big(\frac{6}{5}\Big) \cdot \Big(\frac{6}{7}\Big) \cdot \Big(\frac{8}{7}\Big) \cdot \Big(\frac{8}{9}\Big) \cdot \; \cdots \\ = \frac{\pi}{2}$$
By squaring only factors whose numerator and denominator sum to primes, we obtain
$$\Big(\frac{2}{1}\Big)^2 \cdot \Big( \frac{2}{3}\Big)^2 \cdot \Big(\frac{4}{3}\Big)^2 \cdot \Big(\frac{4}{5}\Big) \cdot \Big(\frac{6}{5}\Big)^2 \cdot \Big(\frac{6}{7}\Big)^2 \cdot \Big(\frac{8}{7}\Big) \cdot \Big(\frac{8}{9}\Big)^2 \cdot \; \cdots = \pi\\$$
Why?
| We have to show that the product of those factors is (convergent and) equal to $2$.
The factors are precisely $\frac{p-\chi(p)}{p+\chi(p)}$, where $p$ runs through odd primes, and $$\chi(n)=\begin{cases}(-1)^{(n-1)/2},&n\text{ is odd}\\\hfill 0,\hfill&n\text{ is even}\end{cases}$$ is known as the non-trivial Dirichlet character modulo $4$, and we know that $$\prod_p\big(1-\chi(p)p^{-s}\big)^{-1}=\sum_{n>0}\chi(n)n^{-s}=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^s}\qquad(\Re s>1)$$ since $\chi$ is completely multiplicative. The (harder!) fact that the equality holds at $s=1$ is a particular case of the result discussed here; you may also want to follow the links out there. The sum at $s=1$ is well-known to be $\pi/4$, thus $\prod_p\big(1-\chi(p)/p\big)=4/\pi$. As $$\frac{1-\chi(p)/p}{1+\chi(p)/p}=\frac{\big(1-\chi(p)/p\big)^2}{1-\big(\chi(p)/p\big)^2}=\frac{\big(1-\chi(p)/p\big)^2}{1-1/p^2},$$ and $\prod_p(1-1/p^2)^{-1}=(1-1/4)$$\zeta(2)$$=\pi^2/8$ since the product is over odd primes, we find $$\prod_p\frac{p-\chi(p)}{p+\chi(p)}=\left(\frac{4}{\pi}\right)^2\frac{\pi^2}{8}=2$$ as expected.
| {
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Then find the sum of all possible values of $abc$. Let $a, b, c$ be positive integers with $0 < a, b, c < 11$. If $a, b, $ and $c$ satisfy
\begin{align*}
3a+b+c&\equiv abc\pmod{11} \\
a+3b+c&\equiv 2abc\pmod{11} \\
a+b+3c&\equiv 4abc\pmod{11} \\
\end{align*}then find the sum of all possible values of $abc$.
What I tried:
I only got this far ...
If the equations from top to bottom were labeled $(1),(2),(3)$,
$(1)+3(2)+(3):$
$7a+11b+7c\equiv 11abc\pmod{11}$
$7(a+c)\equiv 0\pmod{11}$.
Since 7 and 11 are coprime, this means $a+c\equiv0\pmod{11}$ or rather just $a+c=11$ in the given range. Note that because 11 is prime, than modular division by any number not divisible by 11 is allowed henceforth.
Back in $(2)$:
$3b+11\equiv3b\equiv2abc\pmod{11}$
$3\equiv 2ac\pmod{11}$
$2ac=2a(11-a)=22a-2a^2\equiv-2a^2\equiv3\pmod{11}$
$2a^2\equiv-3\equiv8\pmod{11}$
$a^2\equiv 4\pmod{11}$
I just got this far ...
| $3a + b + c \equiv abc \pmod{11}$
$a + 3b + c \equiv 2abc \pmod{11}$
$a + b + 3c \equiv 4abc \pmod {11}$
So
$(3a+b+c)+ 3(a+3b+c) + (a+b+3c)\equiv 11abc$
$7(a+c)\equiv 0\pmod{11}$ As $11$ is prime you can divide by $7$. (This might not be the case if $\gcd(7,11)\ne 1$ but in this case you can.)
$a\equiv -c\pmod {11}$
So we have
$2a+b\equiv -a^2b$
$3b\equiv -2a^2b$
$b-2a \equiv -4a^2b$.
Now $11$ is prime and $b\ne 0$ so $b$ is invertible. (It's not clear in your post if you realize that you CAN"T allways divide by $b$. You can in this case because $b\not \equiv 0$ and $\gcd(b,11) = 1$.)
$3\equiv -2a^2$
$3*5\equiv (-2*5)a^2$
$4 \equiv a^2$ and $a\equiv \pm 2$.
These are the only solutions as $a^2 -4\equiv 0$ and so $(a-2)(a+2)\equiv 0$. Again because $p$ is prime the only $0$ divisors is $0\equiv 11$ so one of $a+2, a-2\equiv 0\pmod{11}$. (This would not be true if $11$ were not prime.)
So $a \equiv \pm 2$ and $c \equiv \mp 2$ so
If $a \equiv 2$ then
$4 + b \equiv -4b\equiv 7b$
$3b\equiv -8b$ (already exhausted)
$b-4\equiv b+7 \equiv -8b\equiv 3b$
ie. $6b \equiv 4$ and $2b\equiv 7$
$2*6b\equiv 2*4$ and $6*2b \equiv 6*7$ so $b\equiv 8$.
SO $a= 2,b= 8, c= 9$ so $abc=144$
or if $a \equiv -2$ then
$-4+b\equiv -4b$ or $5b \equiv 4$ or $9*5b \equiv 9*4$ or $b\equiv 3$
$b+4 \equiv -16b$. or $5b \equiv 4$ or $b\equiv 3$.
So $a=9; b=3;c=2$. And $abc = 54$
So the sum is $144+54 = 198$.
| {
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How to find x from the equation $\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x+\sqrt{x}}}$ For m are real number,
Find x from the equation
$$\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x+\sqrt{x}}}$$
I tried to multiply $\sqrt{x + \sqrt{x}}$ to the both sides and I get
$$x + \sqrt{x} - \sqrt{x^{2} - x} = m \sqrt{x}$$
What should I do now to get x? Can anyone show me a hint please?
| From the simplified form of @gt6989b, we get $$\sqrt{x}- \sqrt{x-1}=m-1.$$
Squaring for rationalization one must demand that $$m \ge 1...(1),~~ x \ge 1....(2),~~~\sqrt{x} \ge m-1....(3)$$.The Eq. (3) gives a very interestiong condition which has been missed out in the answers by @Dinno Koluh and @trancelocation. When we demand (1) and (3), we get
$$\sqrt{x} \ge \frac{m^2-2m+2}{2(m-1)} \ge (m-1) \implies m^2-2m \le 0 \implies m(m-2) \le 0 \implies 1<m \le 2.$$ Finally, we clam that the solution of the priposed equation is given as
$$x=\left(\frac{m^2-2m+2}{2(m-1)}\right)^2 ~~if~~ 1<m \le 2.$$
| {
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Rewriting a binomial formula summation I am trying to obtain an explicit solution for $p$ from the following equation:
$p = 1 - \left[ \sum_{k=0}^{n} \binom{n}{k} p^k (1-p)^{n-k} \frac{1}{k+1} \right]^{\frac{1}{{n}}}$.
Obviously, I know that summing a binomial pdf we have
$\sum_{k=0}^{n} \binom{n}{k} p^k (1-p)^{n-k} = 1$
Perhaps there is a neat way to rewrite
$\sum_{k=0}^{n} \binom{n}{k} p^k (1-p)^{n-k} \frac{1}{k+1}$
so I get an explicit solution for $p$.
| $$
\eqalign{
& \sum\limits_{\left( {0\, \le \,} \right)\,k\,\left( { \le \,n} \right)} {{1 \over {k + 1}}\left( \matrix{
n \cr
k \cr} \right)p^{\,k} q^{\,n - k} } = \sum\limits_{0\, \le \,k\,\left( { \le \,n} \right)} {{1 \over {n + 1}}\left( \matrix{
n + 1 \cr
k + 1 \cr} \right)p^{\,k} q^{\,n - k} } = \cr
& = {1 \over {\left( {n + 1} \right)p}}\sum\limits_{0\, \le \,k\,\left( { \le \,n} \right)} {\left( \matrix{
n + 1 \cr
k + 1 \cr} \right)p^{\,k + 1} q^{\,\left( {n + 1} \right) - \left( {k + 1} \right)} } = \cr
& = {1 \over {\left( {n + 1} \right)p}}\sum\limits_{1\, \le \,j\,\left( { \le \,n + 1} \right)} {\left( \matrix{
n + 1 \cr
j \cr} \right)p^{\,j} q^{\,\left( {n + 1} \right) - j} } = \cr
& = {1 \over {\left( {n + 1} \right)p}}\left( {\sum\limits_{0\, \le \,j\,\left( { \le \,n + 1} \right)} {\left( \matrix{
n + 1 \cr
j \cr} \right)p^{\,j} q^{\,\left( {n + 1} \right) - j} } - q^{\,\left( {n + 1} \right)} } \right) = \cr
& = {{1 - q^{\,n + 1} } \over {\left( {n + 1} \right)p}} = {1 \over {\left( {n + 1} \right)}}{{1 - q^{\,n + 1} } \over {1 - q}} \cr}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Definite integration evaluation of $\int_0^{\pi/2} \frac{\sin^2(x)}{(b^2\cos^2(x)+a^2 \sin^2(x))^2}~dx$. $$\int_0^{\pi/2} \frac{\sin^2(x)}{(b^2\cos^2(x)+a^2 \sin^2(x))^2}~dx$$
how to proceed please help
The answer given is $\dfrac{\pi}{4a^3b}$.
| Change variable to $t = \tan x$, we have
\begin{align}
&\quad\int_0^{\frac{\pi}{2}}\frac{\sin ^2 x}{\left(b^2\cos^2 x+a^2\sin^2 x\right)^2}\, dx\\&=\int_0^{\frac{\pi}{2}}\frac{\tan^2 x}{\left(b^2+a^2\tan^2 x\right)^2}\cdot\frac{1}{\cos^2 x}\,dx\\&=\int_0^\infty\frac{t^2}{\left(b^2+a^2t^2\right)^2}\,dt\\&=\left.-\frac{t}{2a^2}\cdot\frac{1}{b^2+a^2t^2}\,\right|_{\,0}^{+\infty} +\frac{1}{2a^2}\int_0^\infty \frac{1}{b^2+a^2t^2}\,dt\\&=\left.\frac{1}{2a^2}\cdot\frac{a}{b^3}\arctan \frac{a}{b}t\,\right|_{\,0}^{+\infty}\\&=\frac{\pi}{4a^3b}.
\end{align}
| {
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The Calculation of an improper integral For the integral $$\int_{0}^{\infty} \frac{x \ln x}{(x^2+1)^2} dx $$
I want to verify from the convergence then to calculate the integral!
*
*For the convergence, simply we can say $$ \frac{x \ln x}{(x^2+1)^2} \sim \frac{1 }{x^3 \ln^{-1} x}$$
then the integral converge because $\alpha=3 > 1$. Is this true?
*To calculate the integral, using the integration by parts where $u = \ln x$ and $dv = \frac{x \ln x}{(x^2+1)^2} dx$.
So, $$\int_{0}^{\infty} \frac{x \ln x}{(x^2+1)^2} dx = \frac{- \ln x}{2(x^2+1)}- \frac{1}{4x^2} +\frac{\ln |x|}{2} ~\Big|_{0}^{\infty} $$
and this undefined while it should converge to $0$ ! what I missed?
I found an error in the calculation so the integral = So, $$\int_{0}^{\infty} \frac{x \ln x}{(x^2+1)^2} dx = \frac{- \ln x}{2(x^2+1)}+ \frac{1}{8} \Big( \ln x^2 - \ln (x^2+1) \Big) ~\Big|_{0}^{\infty} $$
and it's still undefined!
| Convergence:
First, let $f(x)=\frac{x \ln x}{(x^2+1)^2}$ then $f$ is decrease on $x>t$ for some $t\in\mathbb{R}$
Use $$x>\ln x \Rightarrow f(x)\leq\frac{x^2}{(x^2+1)^2}$$
and we know
$$(x^2+1)^2=x^4+2x^2+1\geq x^4 \Rightarrow \frac{1}{(x^2+1)^2}\leq\frac{1}{x^4}$$
Apply this for our first inequality, then we get
$$0\leq f(x)\leq\frac{1}{x^2}\quad (x\geq1)$$
and we know that improper integral converge.
Computing:
\begin{align}
&\int_0^\infty\frac{x\ln x}{(x^2+1)^2}dx=\int_0^1\frac{x\ln x}
{(x^2+1)^2}dx+\int_1^\infty\frac{x\ln x}{(x^2+1)^2}dx \\
&=\int_0^1\frac{x\ln x}{(x^2+1)^2}dx-\int_0^1\frac{\ln t}{(\frac{1}{t^2}+1)^2t^3}dt \\
&=\int_0^1\frac{x\ln x}{(x^2+1)^2}dx-\int_0^1\frac{t\ln t}{(\frac{1}{t^2}+1)^2t^4}dt \\
&=\int_0^1\frac{x\ln x}{(x^2+1)^2}dx-\int_0^1\frac{x\ln x}{(x+1)^2}dx \\
&=0
\end{align}
Substituted $x=\frac{1}{t}$ the second integral one at first line.
| {
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"url": "https://math.stackexchange.com/questions/3409650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Functions $f: \mathbb{R} \to \mathbb{R}$ satisfying $f \big(x ^ 2 + y ^ 2f (x)\big) = xf (y) ^ 2-f (x) ^ 2$
Find all functions $ f : \mathbb R \to \mathbb R $ satisfying
$$ f \left ( x ^ 2 + y ^ 2 f ( x ) \right ) = x f ( y ) ^ 2 - f ( x ) ^ 2 $$
for all $ x , y \in \mathbb R $.
Let $P(x,y)$ denote the functional equation.
$P(0,0)$ gives $f(0)=-f(0)^2 \implies f(0)=0$.
Now $P(1,0)$ gives $f(1)=-f(1)^2 \implies f(1)=0$.
So, $P(1,y)$ gives $f(y)^2=f(1)=0$.
So $f(x)=0$ is the only solution.
| It' easy to see that the $ f ( x ) = 0 $ and $ f ( x ) = - x $ are both solutions to the functional equation
$$ f \big( x ^ 2 + y ^ 2 f ( x ) \big) = x f ( y ) ^ 2 - f ( x ) ^ 2 \text . \tag 0 \label 0 $$
We show that those are the only solutions. Letting $ x = y = 0 $ in \eqref{0}, we find out that either $ f ( 0 ) = 0 $ or $ f ( 0 ) = - 1 $. But if $ f ( 0 ) = - 1 $, then letting $ x = 0 $ and $ y = 1 $ in \eqref{0}, we have $ f ( - 1 ) = - 1 $, which by letting $ x = y = - 1 $ in \eqref{0}, gives $ f ( 0 ) = - 2 $, and that leads to a contradiction. Thus we must have $ f ( 0 ) = 0 $, and letting $ y = 0 $ in \eqref{0}, we get
$$ f \big( x ^ 2 \big) = - f ( x ) ^ 2 \text . \tag 1 \label 1 $$
In particular, letting $ x = 1 $ in \eqref{1}, we find out that either $ f ( 1 ) = 0 $ or $ f ( 1 ) = - 1 $. If $ f ( 1 ) = 0 $, then letting $ x = 1 $ in \eqref{0}, we get $ f ( y ) = 0 $, which gives one of the solutions mentioned above. From now on, we assume that $ f ( 1 ) = - 1 $. Letting $ x = 1 $ in \eqref{0}, we get
$$ f \big( 1 - y ^ 2 \big) = f ( y ) ^ 2 - 1 \text . \tag 2 \label 2 $$
Substituting $ - y $ for $ y $, we find out that $ f ( - y ) ^ 2 = f ( y ) ^ 2 $. Now, if we let $ y = 1 $ in \eqref{0}, we have $ f \big( x ^ 2 + f ( x ) \big) = x - f ( x ) ^ 2 $, which shows that if $ f ( - x ) = f ( x ) $ then $ x = 0 $. Combining with the previous result, we get
$$ f ( - x ) = - f ( x ) \text . \tag 3 \label 3 $$
Now, \eqref{1}, \eqref{2} and \eqref{3} show that
$$ f \big( y ^ 2 - 1 \big) = - f \big( 1 - y ^ 2 \big) = - f ( y ) ^ 2 + 1 = f \big( y ^ 2 \big) + 1 \text , $$
which means that we have $ f ( x + 1 ) = f ( x ) - 1 $ for $ x \ge - 1 $. Substituting $ - x - 1 $ for $ x $ in the last equation and using \eqref{3}, we get the same equation for $ x \le 0 $, and thus it's true for every $ x $. By induction and \eqref{3}, we get $$ f ( x + n ) = f ( x ) - n \tag 4 \label 4 $$
for every integer $ n $. In particular, $ x = 0 $ gives $ f ( n ) = - n $ for every integer $ n $. Now, we combine \eqref{0}, \eqref{1}, \eqref{3} and \eqref{4} to get
$$ f \big( n y ^ 2 \big) = - f \big( - n y ^ 2 \big) = - n ^ 2 - f \big( n ^ 2 - n y ^ 2 \big) = - n ^ 2 - n f ( y ) ^ 2 + f ( n ) ^ 2 = - n f ( y ) ^ 2 = n f \big( y ^ 2 \big) \text . $$
Together with the previously proven facts, we can generalize this to
$$ f ( n x - m ) = n f ( x ) + m \tag 5 \label 5 $$
for every integers $ m $ and $ n $. Now, \eqref{1} shows that $ f $ takes nonpositive values at nonnegative points. This shows that if for some integers $ m $ and $ n $ with $ n > 0 $ we have $ \frac m n \le x $, then by \eqref{5}, we get $ f ( x ) \le - \frac m n $. By \eqref{3}, using a similar argument we get that if $ \frac m n \ge x $ then $ f ( x ) \ge - \frac m n $. As the set of rational numbers is dense in the set of real numbers, we must have $ f ( x ) = - x $, which is the other solution mentioned above.
| {
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"timestamp": "2023-03-29T00:00:00",
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If A commutes with both of these matrices, then A must be a scalar multiple of the identity matrix I am working on the following problem:
Let $A$ be a $4 \times 4$ matrix with entries in a field of characteristic zero. Suppose that $A$ commutes with both $\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & 3 & 0\\ 0 & 0 & 0 & 4 \end{pmatrix}$ and $\begin{pmatrix} 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \end{pmatrix}$. Prove that $A$ is a scalar multiple of the identity matrix.
I know that $A$ is a scalar multiple of the identity matrix if and only if $AB = BA$ for all other possible $4 \times 4$ matrices $B$ with entries in a field of characteristic $0$. However, I'm struggling with deducing here that $A$ commuting with these specific matrices forces $A$ to be a scalar multiple of the identity matrix. Does commuting with these specific matrices force $A$ to commute with all $4 \times 4$ matrices with entries in a field of characteristic $0$? If so, how can I deduce this?
Thanks!
| My answer can be treated as the supplement to the Loup Blanc's answer, I would like to express similar result in a more elementary way.
Denote both mentioned matrices as $D$ and $P$.
It is easy to check that if $A$ commutes with matrices $D$ and $P$ then it commutes also with any power of matrices $D$ and $P$, any polynomial of $D$ and $P$ and generally with any product or linear combination of these matrices.
For powers of $D$ we have results
$D=\text{diag} ( 1 \ \ 2 \ \ 3 \ \ 4) , \\ D^2= \text{diag} ( 1 \ \ 2^2 \ \ 3^2 \ \ 4 ^2) , \\ D^3=\text{diag} ( 1 \ \ 2^3 \ \ 3^3 \ \ 4^3 ) , \\D^4=\text{diag} ( 1 \ \ 2^4 \ \ 3^4 \ \ 4^4) $
Four vectors formed from diagonal entries are linearly independent (if they are columns of $4 \times 4$ matrix then they form Vandermonde matrix) so linear combination of them can generate any diagonal matrix, denote it generally as ${D_i}$.
On the other hand $P= \begin{pmatrix} 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \end{pmatrix}$ is a permutation matrix with its powers
$P^2= \begin{pmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \end{pmatrix}, P^3= \begin{pmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0 \end{pmatrix}, P^4=I$.
It is visible that with the expression $D_0+P D_1 + P^2D_2+P^3D_3$ we can generate any $4 \times 4$ matrix (assuming first we generate appropriate diagonal matrices $D_i$) and hence the matrix $A$ has to commute with all possible $4 \times 4$ matrices.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3413082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Maximum value of $8v_1 - 6v_2 - v_1^2 - v_2^2$ subject to $v_1^2+v_2^2\leq 1$
Given that $g:\mathbb{R}^2 \to \mathbb{R}$ defined by
$$g(v_1,v_2) = 8v_1 - 6v_2 - v_1^2 - v_2^2$$
find the maximum value of $g$ subject to the constraint $v_1^2+v_2^2\leq 1.$
My attempt:
Note that
$$g(v_1,v_2) = 8v_1 - 6v_2 - v_1^2 - v_2^2 = -(v_1-4)^2 - (v_2+3)^2 + 25.$$
So $g$ is a decreasing function.
So, the maximum value of $g$ lies on the circumference of $v_1^2+v_2^2 = 1.$
It suffices to find the intersection between $(0,0)$ and $(4,-3)$ as $(4,-3)$ is the peak point of $g.$
The intersection point lies on both $v_1^2+v_2^2 = 1$ and $v_2 = -\frac{3}{4}v_1.$
Solving the simultaneous equation gives that
$$v_1 = \frac{4}{5}, \quad v_2 = -\frac{3}{5}.$$
So, maximum value of $g$ is
$$g\left(\frac{4}{5}, -\frac{3}{5}\right) = 9.$$
Is my attempt correct?
| Your result is correct but it is not clear what you mean by $g$ is a decreasing function and "find the intersection between $(0,0)$ etc."
So, here is another way using Cauchy-Schwarz:
$$8v_1 - 6v_2 \leq \sqrt{8^2+6^2}\sqrt{v_1^2+v_2^2} = 10 \sqrt{v_1^2+v_2^2}$$
$$g(v_1,v_2)\leq 10\sqrt{v_1^2+v_2^2}- (v_1^2 + v_2^2)=t(10-t) \mbox{ with } t = \sqrt{v_1^2+v_2^2}$$
$t(10-t)$ is strictly increasing for $0\leq t \leq 5 \Rightarrow$ maximum of $g$ is reached for $t=\sqrt{v_1^2+v_2^2} =1$: $1(10-1)=\boxed{9}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3416414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If equation rep $x^2+2y^2-5z^2+2kyz+2zx+4xy=0$ represent pair of plane. then $k$ is The values of $k$ for which the equation
$x^2+2y^2-5z^2+2kyz+2zx+4xy=0$
represents a pair of plane passing Through origin,is
what i try
$x^2+2y^2-5z^2+2kyz+2zx+4xy=(ax+by+cz)(px+qy+rz)$
and camparing coefficients
but it is very tedious work
How do i solve it some short way Help me please
| We can assume that $a=p=1$, so we have $$x^2+2y^2-5z^2+2kyz+2zx+4xy=(x+by+cz)(x+qy+rz)$$
This should be true for all $y$, so it is true for $y=0$ also, and we get:
$$x^2-5z^2+2zx=(x+cz)(x+rz)$$ So $cr=-5$ and $c+r=-2$. Also it is true for all $z$ and specialy for $z=0$: $$x^2+2y^2+4xy=(x+by)(x+qy)$$ so $bq=2$ and $b+q=4$...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $(11 \cdot 31 \cdot 61) | (20^{15} - 1)$ Prove that
$$ \left( 11 \cdot 31 \cdot 61 \right) | \left( 20^{15} - 1 \right) $$
Attempt:
I have to prove that $20^{15}-1$ is a factor of $11$, $31$, and $61$. First, I will prove
$$ 20^{15} \equiv 1 \bmod11 $$
Notice that
$$ 20^{10} \equiv 1 \bmod 11$$
$$ 20^{5} \equiv 9^{5} \bmod 11 = 9^{4} 9 \bmod 11, \:\: 9^{2} \equiv 4 \bmod 11 $$
$$ \implies 9^{5} \equiv 144 \bmod 11 \implies 20^{5} \equiv 1 \bmod 11 $$
Then the proof is done.
Now I will prove:
$$ 20^{15} \equiv 1 \bmod 31 $$
Notice $20^{2} \equiv 28 \bmod 31$, so
$$20 \times (20^{2})^{7} \equiv 20 \times (28)^{7} \bmod 31 \equiv 20 \times (-3)^{7} \bmod 31 \equiv -60 \times 16 \bmod 31\equiv 32 \bmod 31 $$
then the proof is done.
Also, in similar way to prove the $20^{15} \equiv 1 \bmod 61$.
Are there shorter or more efficient proof?
| A variant, with lil' Fermat:
*
*As $20$ is not divisible by $11$, we have $20^{15}\equiv 1\mod11 $.
*$2^{30}\equiv 1\mod 31$ and $2^{15}\equiv-1\mod 31$, so $(-2)^{15}\equiv 5^{15}\equiv 1\mod 31$.
*$2^6\equiv 3\equiv 5^3$, so $\;2^{30}5^{15}\equiv 9^5\mod 61$. Now $9^2\equiv 20$, so $9^3\equiv 180\equiv-3$ and ultimately $9^5\equiv -60\equiv 1\mod 61$.
| {
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"url": "https://math.stackexchange.com/questions/3421282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Constrained minimum with inequalities Show that for all real positive $x$ and $y$ such that $x+y=1$, the following holds: $$\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\geq \frac{25}{2}.$$
I know this can probably be easily solved with Lagrange multipliers, but I am more interested in proving it using inequalities. I think it could be profitable to use AM-HM or something similar but I am not sure about how to proceed.
How can I do it?
Thanks.
| This problem can be rewritten as:
$$\sqrt{\frac{(x+\frac{1}{x})^2+(y+\frac{1}{y})^2}{2}} \geq\frac{5}{2}$$
We can use inequality between square and arithmetic mean:
$$\sqrt{\frac{(x+\frac{1}{x})^2+(y+\frac{1}{y})^2}{2}} \geq \frac{(x+\frac{1}{x})+(y+\frac{1}{y})}{2}=\frac{1+\frac{1}{x}+\frac{1}{y}}{2}$$
We only have to show, that $$\frac{1}{x}+\frac{1}{y} \geq 4$$
But this comes from $x+y=1$ and inequality between arithmetic and harmonic mean:
$$\frac{1}{2}=\frac{x+y}{2}\geq\frac{2}{\frac{1}{x}+\frac{1}{y}} \Rightarrow \frac{1}{x}+\frac{1}{y} \geq 4$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\left |\frac{1}{\sqrt{n}} \sum_{i=1}^{[tn]}X_i-\frac{\sqrt{t}}{\sqrt{[tn]}} \sum_{i=1}^{[tn]}X_i\right |\overset{n}{\to}0$ in probability
As the title states, I would like to prove that
$$\left | \frac{1}{\sqrt{n}} \sum_{i=1}^{\lfloor tn \rfloor }X_i - \frac{\sqrt{t}}{\sqrt{\lfloor tn \rfloor}} \sum_{i=1}^{\lfloor tn \rfloor}X_i \right |$$
converges to zero, with $n $, in probability. Here $X_i $ are independent random variables with zero mean and second moment equal to one. $\lfloor x \rfloor $ denotes the floor function and $t $ is a positive real number.
My first thought was that
$$\left | \frac{1}{\sqrt{n}} \sum_{i=1}^{\lfloor tn \rfloor}X_i - \frac{\sqrt{t}}{\sqrt{\lfloor tn \rfloor}} \sum_{i=1}^{\lfloor tn \rfloor}X_i \right |=\left |\sum_{i=1}^{\lfloor tn \rfloor}X_i \right |\left| \frac{1}{\sqrt{n}} - \frac{\sqrt{t}}{\sqrt{\lfloor tn \rfloor}} \right |$$
Where $\left| \frac{1}{\sqrt{n}} - \frac{\sqrt{t}}{\sqrt{\lfloor tn \rfloor}} \right | \overset {n } {\to } 0 $. But does it converge sufficiently fast so that the product with $\left |\sum_{i=1}^{\lfloor tn \rfloor}X_i \right | $ converges to zero?
Most grateful for any help provided!
| Here's an attempt to an answer using the points from stochasticboy321's comment above.
We will try to apply Chebyshev's inequality which in our case states, since the variance of $\sum_{i=1 } ^{\lfloor nt \rfloor } X_i$ is $\lfloor nt \rfloor$, that
$$P[|\sum_{i=1 } ^{\lfloor nt \rfloor } X_i | \ge \epsilon \sqrt{\lfloor nt \rfloor}] \le \frac{1}{\epsilon^2}, \ \epsilon>0$$
for which is would be sufficient to show that, for some $n_0 $ onwards,
$$\left | \sqrt {\frac{t}{\lfloor nt \rfloor}} - \frac{1}{\sqrt {n}} \right | \le \left | \frac{1}{tn} \right |$$
since then we would have [for $n \ge n_0 $]
$$P[|\sum_{i=1 } ^{\lfloor nt \rfloor } X_i |\left | \frac{1}{\sqrt{n}} - \frac{\sqrt{t}}{\sqrt{\lfloor nt \rfloor}} \right | \ge \epsilon ] \le P[|\sum_{i=1 } ^{\lfloor nt \rfloor } X_i |\left | \frac{1}{\lfloor tn \rfloor} \right | \ge \epsilon ] \le \frac{1}{\epsilon^2 \lfloor tn \rfloor} \overset {n } {\to} 0$$
Attempt to show that $ \left | \frac{1}{\sqrt{n}} - \frac{\sqrt{t}}{\sqrt{\lfloor nt \rfloor}} \right | = \mathcal{O}(t^{-1 } n^{-3/2 })$:
From the inequality $ nt -1 < \lfloor nt \rfloor \le nt $ we get
$$\frac{1}{nt} \le \frac{1}{\lfloor nt \rfloor} < \frac{1}{nt-1} \implies \frac{1}{n} \le \frac{t}{\lfloor nt \rfloor} < \frac{t}{nt-1}=\frac{1}{n} + \frac{1}{n(nt-1)}$$
and since $0 \le a \le b < c \implies b-a \le c-a $ we get
$$\frac{1}{n} - \frac{t}{\lfloor nt \rfloor} \le \left (\frac{1}{n} + \frac{1}{n(nt-1)}\right ) - \frac{1}{n} = \frac{1}{n(nt-1)}$$
and thus
$$\left |\frac{t}{\lfloor nt \rfloor} - \frac{1}{n} \right | \le \left | \frac{1}{n(nt-1)} \right |$$
Since $|\sqrt{a} - \sqrt{b}| \le \sqrt{|a-b|}$ holds for any two non negative numbers [see eg here] we get the inequality.
$$\left | \sqrt {\frac{t}{\lfloor nt \rfloor}} - \frac{1}{\sqrt {n}} \right | \le \left | \frac{1}{\sqrt {n(nt-1)}} \right |$$
I don't think it is true that
$$\frac{1}{\sqrt {n(nt-1)} } = \mathcal{O}(t^{-1 } n^{-3/2})$$
but I do believe we have that
$$\frac{1}{\sqrt {n(nt-1)} } = \mathcal{O}(t^{-1 } n^{-1})$$
which as noted above would be sufficient.
Any comment would be much appreciated!
| {
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The number of real roots of the equation $5+|2^x-1|=2^x(2^x-2)$ I'm trying to find the number of real roots of the equation $5+|2^x-1|=2^x(2^x-2)$.
Let $2^x=a$
$$|a-1|=a^2-2a-5$$
Then there are two cases
$$a-1=a^2-2a-5$$
And $$a-1=-a^2+2a+5$$
Solving both equations
$$a=1,-4,-2,3$$
Now -4 and -2 can be neglected so there are two values
1 and 3.
Then $$2^x=1$$
$$x=0$$
And $$2^x=3$$
$$x=\log_2 3$$
But the answer doesn’t seem to consider the $\log_2 3$ as a viable root, and the answer is 1. Why is that the case?
| There are two cases:
*
*$x\geq0$. The equation becomes $$2^{2x}-2^{x+1}-2^x-4=0$$$$(2^x-2)(2^x-1)= 6$$Clearly, $2^x-2>0$, which implies $x>1$. Let $y=2^x$. We have $y^2-3y-4=0$ which implies $$(y-4)(y+1)=0$$ So, $y=4$ and $x = 2$.
*$x<0$. The equation becomes $$6-2^x=2^{2x}-2^{x+1}$$$$2^{2x}-2^{x+1}+2^x-6=0$$$$(2^x-2)(2^x+1)=4$$Let $y=2^x$. Hence, $$y^2-y-6=0\to(y-3)(y+2)=0$$Neither solution can equal $2^x$ where $x<0$. So there are no solutions.
So, the only solution is $x=2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve this integral for free WiFi I saw this today, I checked in Mathematica and the integral comes out to $\pi$, but I have no idea how to solve it.
FREE Wi-Fi: The Wi-Fi password is the first $10$ digits of the answer.
$$\int_{-2}^2\left(x^3\cos\frac x2+\frac12\right)\sqrt{4-x^2}\ dx$$
| We have that
$$\int_{-2}^2 x^3 \cos\frac x2 \sqrt{4-x^2} dx =0$$
since the integrand is point symmetric in the origin.
Since $\sqrt{4-x^2}$ on $[-2;2]$ is the formula for the upper part of a circle we find that
$$\int_{-2}^2 \sqrt{4-x^2} dx=\frac 12 \pi r^2=\frac 12 \pi 2^2=2\pi$$
So the whole integral is:
$$\int_{-2}^2(x^3\cos \frac x2 +\frac 12 )\sqrt{4-x^2} dx=\int_{-2}^2 x^3 \cos\frac x2 \sqrt{4-x^2} dx+\frac 12\int_{-2}^2 \sqrt{4-x^2} dx=0+\frac 12 2\pi=\pi$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the range of $f(x)=2|{\sin x}|-3|\cos x|$ Find the range of $f(x)=2|{\sin x}|-3|\cos x|$
My attempt is as follows:-
$$f(x)=\sqrt{13}\left(\dfrac{2}{\sqrt{13}}\cdot|\sin x|-\dfrac{3}{\sqrt{13}}\cdot|\cos x|\right)$$
Let's assume $z=\dfrac{2}{\sqrt{13}}\cdot|\sin x|-\dfrac{3}{\sqrt{13}}\cdot|\cos x|$
$$f(x)=\sqrt{13}z$$
Let's find the range of $z^2$
$$z^2=\dfrac{4}{13}\cdot\sin^2x+\dfrac{9}{13}\cdot\cos^2x-\dfrac{6}
{13}|\sin2x|$$
$$z^2=1-\dfrac{6}{13}\cdot|\sin2x|$$
$$z^2=1-\dfrac{6}{13}\cdot[0,1]$$
$$z^2=1-\left[0,\dfrac{6}{13}\right]$$
$$z^2=\left[\dfrac{7}{13},1\right]$$
As $0$ is not there in the range of $z^2$, so we can say $z\in \left[-1,-\dfrac{\sqrt{7}}{\sqrt{13}}\right]\cup \left[\dfrac{\sqrt{7}}{\sqrt{13}},1\right]$
Hence $y \in \sqrt{13}\left(\left[-1,-\dfrac{\sqrt{7}}{\sqrt{13}}\right]\cup \left[\dfrac{\sqrt{7}}{\sqrt{13}},1\right]\right)$
$$y\in[-\sqrt{13},-\sqrt{7}]\cup[\sqrt{7},\sqrt{13}]$$
But actual answer is $[-3,2]$
| For $0\le x\le\dfrac\pi2,$
$$f(x)=\sqrt{2^2+3^2}\sin\left(x-\arcsin\dfrac3{\sqrt{2^2+3^2}}\right)$$
Now $0\le x\le\dfrac\pi2\implies-\arcsin\dfrac3{\sqrt{2^2+3^2}}\le x-\arcsin\dfrac3{\sqrt{2^2+3^2}}\le\dfrac\pi2-\arcsin\dfrac3{\sqrt{2^2+3^2}}$
Again,
$\dfrac\pi2-\arcsin\dfrac3{\sqrt{2^2+3^2}}=\arccos\dfrac3{\sqrt{2^2+3^2}}=\arcsin\dfrac2{\sqrt{13}}$
As $\sin$ is increasing in $\left[-\dfrac\pi2,\dfrac\pi2\right]$
$$\implies-\dfrac3{\sqrt{2^2+3^2}}\le\sin\left(x-\arcsin\dfrac3{\sqrt{2^2+3^2}}\right)\le\dfrac2{\sqrt{13}}$$
Observe that $$f(\pi+x)=f(x)=f(\pi-x)$$
which cover $\left[\pi,\pi+\dfrac\pi2\right]$ and $\left[\dfrac\pi2,\pi\right]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3428466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the unit digit of $2124^{392}+3143^{394}*7177^{392}-8818^{394}$ What I do here is I find the remainder of each parcel in mod 10 and then do the maths and mod 10 again:
$2124^{392} \equiv 4^{392} \pmod{10}$
$$4^1 \equiv 4 \pmod{10} \\
4^2 \equiv 6 \\
4^3 \equiv 4 \\
4^4 \equiv 6 \\
(...)$$
Then I do $392 \pmod{2}$. Problem is that here, it's zero. If it were 1 or 2, the remainder would be either 4 or 6. How do I solve it this way?
| $4^1\equiv 4$
$4^2\equiv 6$
$4^3\equiv 4$
$\vdots$
When the exponent is odd, the remainder is $4$.
When the exponent is even the remainder is $6$.
$0$ is even and so is $392$, thus $4^{392}\equiv 6\pmod{10}$
As an alternative notice that $6^k\equiv 6\pmod{10}$,
so $4^{392} = 16^{196}\equiv 6\pmod{10}$
| {
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"source": "stackexchange",
"question_score": "1",
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General formula for $e^x+\cos(x)$, $e^x+\sin(x)$, $e^x-\sin(x)$, $e^x-\sin(x)$ I have been able to derive the formal series for these four functions:
$e^x+\sin(x) = 1+2x+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dfrac{2x^5}{5!}+\dfrac{x^6}{6!}+\dfrac{x^8}{8!}+\dfrac{2x^9}{9!}+...$
$e^x+\cos(x) = 2+\dfrac{x^3}{3!}+\dfrac{2x^4}{4!}+\dfrac{x^5}{5!}+\dfrac{x^7}{7!}+\dfrac{2x^8}{8!}+...$
$e^x-\sin(x) = 1+\dfrac{x^2}{2!}+\dfrac{2x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{2x^7}{7!}+\dfrac{x^8}{8!}+...$
$e^x-\cos(x) = x+x^2+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\dfrac{2x^6}{6!}+\dfrac{x^7}{7!}+\dfrac{x^9}{9!}+...$
Due to the missing term and the irregularity, I am unable to write the general formula for these series. I wish to find the general formula with compact sigma notation. Can you help with this?
| If the terms in the series for the two functions you're adding or subtracting had the same numbering scheme, you could easily add or subtract them term by term within the same numbering scheme.
The reason you cannot do this with the usual series for these functions is that the $n$th term in the series for $e^x$ is the term in $x^n$ (numbering the terms starting with $n=0$), whereas for $\sin x$ the $n$th term is the term in $x^{2n+1}$:
$$ \sin (x) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1} . $$
To fix this, we need a function $f$ over integers such that
$$
f(n) = \begin{cases}
\phantom{-}0 &&& n \equiv 0 \pmod 4,\\
\phantom{-}1 &&& n \equiv 1 \pmod 4,\\
\phantom{-}0 &&& n \equiv 2 \pmod 4,\\
-1 &&& n \equiv 3 \pmod 4.
\end{cases}
$$
This function can be written compactly using the formula $\Im(i^n),$ the imaginary part of the $n$th power of the imaginary unit.
Alternatively, $\sin\left(\frac\pi2 n\right)$ will do just as well.
You could even use an expression like
$\frac12(1 - (-1)^n)(-1)^{\lfloor n/2\rfloor},$
thereby avoiding all hints of imaginary numbers or trig functions.
But we could just define $f$ by cases as shown above in order to have less obscure jiggery-pokery in our formulas.
So we can write
\begin{align}
e^x &= \sum_{n=0}^\infty \frac{1}{n!} x^n, \\
\sin(x) &= \sum_{n=0}^\infty \frac{f(n)}{n!} x^n, \\
\cos(x) &= \sum_{n=0}^\infty \frac{f(n+1)}{n!} x^n. \\
\end{align}
Now it should not be hard to add or subtract these series term by term.
For example,
$$
e^x - \sin(x) = \sum_{n=0}^\infty \frac{1 - f(n)}{n!} x^n
$$
| {
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"url": "https://math.stackexchange.com/questions/3430117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Logarithm question - spurious solution So I have the equation that I want to solve
$$\log_2(8x) - \log_2(1+\sqrt{x}) = 3, \hspace{1mm} x>0$$
My solution is $$\log_2\left(\frac{8x}{1+\sqrt{x}}\right) = 3 \\ \Rightarrow x = 1 + \sqrt{x} \\ \Rightarrow x - \sqrt{x} - 1 = 0 \\ \Rightarrow \sqrt{x} = \frac{1}{2} \pm \frac{\sqrt{5}}{2} \\ \Rightarrow x = \frac{3}{2} \pm \frac{\sqrt{5}}{2} $$ Now only the larger solution actually solves the equation.
Why is this? I think it involves the squaring stage, but still a bit not sure
| $$ \sqrt{x} = \frac{1}{2} \pm \frac{\sqrt{5}}{2} $$
At this stage you must ditch the negative solution $\frac{1}{2} - \frac{\sqrt{5}}{2}$ beacause $\sqrt{x}$ is non negative.
| {
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Very tricky triple integral in spherical coordinates Evaluate $$\int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{2-\sqrt{4-x^2-y^2}}^{2+\sqrt{4-x^2-y^2}}(x^2+y^2+z^2)^{3/2} \; dz \; dy \; dx$$ by converting to spherical coordinates.
We know that $(x^2+y^2+z^2)^{3/2} = (\rho^2)^{3/2} = \rho^3$. The range of $y$ tells us that we have disk of radius $2$. How can we use this find the limits of each integral and ultimately find the solution? Thanks you for the help.
| The boundary of our sphere is $x^2 + y^2 + z^2 = 4z$
Plugging
$x = \rho\cos\theta \sin\phi\\
y = \rho\sin\theta\sin\phi\\
z = \rho \cos\phi$
We get:
$\rho^2 = 4\rho\cos\phi\\
\rho = 4\cos\phi$
And the sphere is above the $xy$ plane, or $\phi \le \frac {\pi}{2}$
$\int_0^{2\pi}\int_0^{\frac {\pi}{2}}\int_0^{4\cos\phi} (\rho^3)(\rho\sin\phi) \ d\rho\ d\phi\ d\theta\\
\int_0^{2\pi}\int_0^{\frac {\pi}{2}} (\frac 15) (4^5) (\cos^5 \phi)(\sin\phi) d\phi\ d\theta\\
\int_0^{2\pi}\int_0^{\frac {\pi}{2}} -(\frac 1{30}) (4^5) \cos^6 \phi\ d\theta\\
(2\pi)(\frac {1}{30})4^5$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove the identity $\sin2x + \sin2y = 2\sin(x + y)\cos(x - y)$ I've been trying to prove the identity $$\sin2x + \sin2y = 2\sin(x + y)\cos(x - y).$$
So far I've used the identities based off of the compound angle formulas. I'm not quite sure if those identities would work with proving the above identity.
Thank you in advance.
| \begin{align}
2\sin(x + y)\cos(x - y)
&= 2(\sin x \cos y + \cos x \sin y)\cdot
(\cos x \cos y + \sin x \sin y) \\
&= 2\sin x \cos x(\cos^2 y + \sin^2 y) +
2\sin y \cos y(\cos^2 x + \sin^2 x) \\
&= 2\sin x \cos x + 2\sin y \cos y \\
&= \sin 2x + \sin 2y
\end{align}
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Find $\frac{\partial x}{\partial u}$, if $u=x+y^2$, $v=y+z^2$, $w=z+x^2$. I tried very hard but couldn't make a relation out them. I cross checked question too but in the book it's given as above. Hints will be appreciated too. Thanks.
| $\text{We have}$:
$u=x+y^2\tag 1$ $v=y+z^2\tag 2$ $w=z+x^2\tag 3$
$\text{Therefore,}$
$\begin{align}x &=u-y^2[\text{ from }(1)]\\&=u-(v-z^2)^2[\text{ from }(2)]\\&=u-[v-(w-x^2)^2]^2[\text{ from }(3)]\\&=u-[v-(w^2-2wx^2+x^4)]^2\\&=u-[v-w^2+2wx^2-x^4)]^2\end{align}$
$\begin{align}\implies\dfrac{\partial x}{\partial u}&=1-2(v-w^2+2wx^2-x^4)(2w\cdot 2x-4x^3)\dfrac{\partial x}{\partial u}\\&=1-2(v-w^2+2wx^2-x^4)(4wx-4x^3)\dfrac{\partial x}{\partial u}\end{align}$
$\begin{align}\implies\dfrac{\partial x}{\partial u}&=\dfrac{1}{1+2(v-w^2+2wx^2-x^4)(4wx-4x^3)}\\&=\dfrac{1}{1+8x(v-w^2+2wx^2-x^4)(w-x^2)}\\&=\dfrac{1}{1+8x[v-w^2+wx^2+wx^2-x^4]z}[\text{ from }(3)]\\&=\dfrac{1}{1+8x[v-w^2+x^2(w-x^2)]z}\\&=\dfrac{1}{1+8xz[v-w^2+x^2z]}[\text{ from }(3)]\end{align}$
| {
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Range of $a$ in $x^2-a=\sqrt{x+a}$
The equation $x^2-a=\sqrt{x+a}$ has real or
imaginary roots depending on the values of $a.$
Then range of $a$ for which the equation.
$(a)\;\; $ No real roots
$(b)\;\; $ One real root
$(c)\;\;$ Exactly two real roots
$(d)\;\;$ At least two real roots
what i try
$x^2-a=\sqrt{x+a}\Rightarrow (x^2-a)^2=x+a$
$x^4+a^2-2ax^2=x+a\Rightarrow a^2-(2x^2+1)a+x^4-x=0$
$$a=\frac{(2x^2+1)\pm \sqrt{(2x^2+1)^2-4(x^4-x)}}{2}$$
$$a=\frac{2x^2+1\pm (2x+1)}{2}$$
$$a=x^2-x,a=x^2+x+1$$
$(a)$ For no real roots
$x^2-x-a=0$ and $x^2+x+1-a=0$ has no real roots
So $1+4a>0$ and $\displaystyle 1-4(1-a)>0\Rightarrow a>-\frac{1}{4}.$
How do i solve other parts Help me please
| First of all, a quadratic equation has no real roots when its discriminant is negative. So you have $1+4a<0$ and similarly for the other inequality. However, these are not necessarily all the cases. The equation is equivalent to the system
$$\begin{align}
(x^2-a)^2 &= x+a \\
x^2-a &\geq0 \Leftrightarrow x^2\geq a
\end{align}$$
Notice that we don't need $x+a\geq 0$ because that's guaranteed to be non-negative since it's equal to a square. So we're interested in the number of solutions to the system, not just the equation. It might happen that the equation has real roots but they don't satisfy the inequality which would imply the original equation has no real roots. Continuing your work, the equation has roots (by solving the $2$ quadratics)
$$x_{1,2}= \frac{1\pm\sqrt{1+4a}}{2}\\
x_{3,4}=\frac{-1\pm\sqrt{4a-3}}{2}$$
Now you need to plug in each of those in $x^2\geq a$ and solve for $a$. The values for $a$ tell you when the root you plugged in is actually a real root of the original equation. After that, you can determine the answers for (a) to (d).
| {
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prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0. Prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0.
I have tried induction as follows.
Step 1:
Try n = 0, we get: $5 - 3 - 2 = 0$, which is divisible by 30.
Try n = 1, we get: $5^{3} - 3^{3} - 2^{3} = 90$, which is also divisible by 30.
Step 2:
Assume it is true for n = k.
So we are assuming the following equality is true: $5^{2k+1} - 3^{2k+1} - 2^{2k+1} = 30M$, for some integer M.
Step 3:
Now we look at the next case: n = k + 1.
$5^{2(k+1)+1} - 3^{2(k+1)+1} - 2^{2(k+1)+1}$
= $5^{2k+3} - 3^{2k+3} - 2^{2k+3}$
= $25\times5^{2k+1} - 9\times3^{2k+1} - 4\times2^{2k+1}$
= $21\times5^{2k+1} + 4\times5^{2k+1} - 5\times3^{2k+1} - 4\times3^{2k+1} - 4\times2^{2k+1}$
= $21\times5^{2k+1} - 5\times3^{2k+1} + 4\times[5^{2k+1} - 3^{2k+1} - 2^{2k+1}]$
= $21\times5^{2k+1} - 5\times3^{2k+1} + 4\times30M$. (Assumed in step 2)
The last term is divisible by 30. But I cannot get a factor 30 out of the first two terms. I can show divisibility by 15 as follows:
= $7\times3\times5\times5^{2k} - 5\times3\times3^{2k} + 4\times30M$
= $7\times15\times5^{2k} - 15\times3^{2k} + 4\times15\times2M$
But how do I show divisibility by 30?
| You could show by induction that modulo $30$
$5^{2k+1}\equiv5,$
$3^{2k+1}\equiv3$ if $k$ is even and $-3$ if $k$ is odd, and
$2^{2k+1}\equiv2$ if $k$ is even and $8$ if $k$ is odd.
| {
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Problematic inequality & hint I would like to ask for hint for proving following inequality:
$$x^3(1+x)+y^3(1+y)+z^3(1+z)\geq \frac{3}{4}(1+x)(1+y)(1+z)$$
for all $x>0$, $y>0$, $z>0$ such that $xyz=1$.
Generally, I tried to find some elementary solution, but even with some calculus I didn't solve it.
Edit. My attempt:
Let $f:(0,\infty)\times(0,\infty)\rightarrow \mathbb{R}$ be function defined by equality:
$$ f(x,y)=x^3(1+x)+y^3(1+y)+\frac{1}{x^3y^3}(1+\frac{1}{xy})-\frac{3}{4}(1+x)(1+y)(1+\frac{1}{xy})$$
and I calculated partial derivatives:
$$\frac{\partial f}{\partial x }(x,y)=-\frac{4}{x^5y^4}-\frac{3}{x^4y^3}+4x^3+\frac{3(y+1)}{4x^2y}+3x^2-\frac{3(y+1)}{4}$$
$$\frac{\partial f}{\partial y }(x,y)=-\frac{4}{y^5x^4}-\frac{3}{y^4x^3}+4y^3+\frac{3(x+1)}{4y^2x}+3y^2-\frac{3(x+1)}{4}$$
I noticed that:
$$\frac{\partial f}{\partial x }(1,1)=\frac{\partial f}{\partial x }(1,1)=0.$$
I began to wonder if the function is convex (i.e. by demonstrating that $g(w)=w^3(1+w)$ is convex for positive values and $h(x,y)=(1+x)(1+y)(1+\frac{1}{xy})$ is also concave for positive x and y, with latter I had some calculation trouble).
| (Edited version). By AM-GM we have:
$$ \sum_{cyc}(\frac{x^3}{(1+y)(1+z)}+\frac{1+y}{8}+\frac{1+z}{8})$$
$$ \geq \sum_{cyc}3 \sqrt[3]{\frac{x^3}{(1+y)(1+z)} \cdot \frac{1+y}{8} \cdot \frac{1+z}{8}}$$
$$=\frac{3}{4}(x+y+z)$$
we have:
$$
\sum_{cyc}(\frac{x^3}{(1+y)(1+z)}\geq \frac{3}{4}(x+y+z) - \sum_{cyc}(\frac{1+y}{8} + \frac{1+z}{8})
$$
$$
=\frac{2}{4}(x+y+z)-\frac{3}{4}\geq \frac{1}{2}\cdot 3 \sqrt[3]{xyz}-\frac{3}{4}=\frac{3}{4}.
$$Q.E.D.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$ \int \frac{\cosh(x)}{ (a^2 \sinh^2(x) + b^2 \cosh^2(x))^{\frac{3}{2}} } dx$ I am tring to obtain following formula [actually I obtained this resut via mathematica]
\begin{align}
\int \frac{\cosh(x)}{ (a^2 \sinh^2(x) + b^2 \cosh^2(x))^{\frac{3}{2}} } dx
= \frac{\sqrt{2} \sinh(x)}{a^2 \sqrt{a^2-b^2 + (a^2+b^2) \cosh(2x)}} + C
\end{align}
Since the answer contains $\cosh(2x)$
My first trial was
\begin{align}
a^2 \sinh^2(x) + b^2 \cosh^2(x) = a^2 \cosh(2x) + (b^2 -a^2) \cosh^2(x)
\end{align}
Seems not good..
My second trial is
\begin{align}
a^2 \sinh^2(x) + b^2 \cosh^2(x) = (a^2+b^2) \sinh^2(x) + b^2
\end{align}
and this also does not good for integrand....
How one can obtain this integral?
| Use your 2nd trial:
$$
\newcommand{\abs}[1]{\left\vert #1 \right\vert}
\newcommand\rme{\mathrm e}
\newcommand\imu{\mathrm i}
\newcommand\diff{\,\mathrm d}
\DeclareMathOperator\sgn{sgn}
\renewcommand \epsilon \varepsilon
\newcommand\trans{^{\mathsf T}}
\newcommand\F {\mathbb F}
\newcommand\Z{\mathbb Z}
\newcommand\R{\Bbb R}
\newcommand \N {\Bbb N}
$$
\begin{align*}
I &= \int \frac {\diff (\sinh x)}{((a^2+b^2) \sinh^2 x +b^2 )^{3/2}} \\
&= \frac 1{b^2 \sqrt {a^2 +b^2}} \int \frac {\diff \left(\sqrt{\frac {a^2 + b^2} {b^2}} \sinh x\right)} {\left(\frac {a^2 + b^2}{b^2} \sinh^2 x + 1\right)^{\frac{3}{2}}} \\
&=\frac 1{b \sqrt {a^2 + b^2}} \int \frac {\diff (\sinh y)}{(\sinh^2 y + 1)^{\frac{3}{2}}} \\
&= \frac 1{b \sqrt {a^2 + b^2}} \int \cosh (y)^{-2} \diff y\\
&= \frac 1{b \sqrt {a^2 +b ^2}} \int \diff (\tanh y) \\
&= \frac {\tanh y} {b\sqrt {a^2 + b ^2}},
\end{align*}
where $\sinh y =b^{-1} \sqrt {a^2 +b^2}\sinh x$. Hence
$$
\cosh^2 y = \sinh^2 y + 1 = \frac 1{b^2} ((a^2 + b^2) \sinh^2 x + b^2) = \frac 2{b^2} (2b^2 + (a^2 + b^2 ) (\cosh (2x ) - 1)) = \frac 2{b^2} ((a^2 + b^2 ) \cosh (2x) - (a^2 -b^2)),
$$
and
$$
\tanh^2 y = \frac {\sinh^2 y} {\cosh^2 y} = 1 - \frac 1{\sinh^2 y + 1} \implies \tanh y = \frac {\sqrt {a^2 +b^2} \sinh x} {\sqrt {(a^2+b^2 )\sinh^2 x} + b^2}
$$
hence
$$
I = \frac {\sinh x} {b \sqrt {(a^2 +b^2) \sinh^2 x + b^2}} = \frac {\sqrt 2 \sinh x} {\sqrt {(a^2+b^2) \cosh(2x) + (b^2 - a^2)}}
$$
as we desire, where $2\sinh^2 (x) = \cosh (2x) - 1$.
| {
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Prove that $\sqrt{2a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4$ when $a+b=2$. Suppose that $a,b$ are non-negative numbers such that $a+b=2$. I want to prove $$\sqrt{2 a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4.$$
My attempt: I tried proving $$2a^2+b+1\geq 4$$ but this seems wrong (try $a=0,b=1$).
How can I prove the above result?
| Since: $$\sqrt{2 a^2+b+1}= \sqrt{2 a^2-a+3}\geq {3\over 4}a+{5\over 4}$$
and the same for the other root, we have $$...\geq ({3\over 4}a+{5\over 4})+
({3\over 4}b+{5\over 4})=4$$
Here is a detailed proof for $$\sqrt{2 a^2-a+3}\geq {3\over 4}a+{5\over 4}$$
It is equivalent to $$16(2a^2-a+3)\geq (3a+5)^2$$
and this is equivalent to $$23a^2-46a+23\geq 0$$
or $$23(a-1)^2\geq 0$$
which is obviously true.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Limit with radicals, $\cos$, $\ln$ and powers $\underset{x\rightarrow +\infty}\lim{\frac{\sqrt[6]{1-\cos{\frac{1}{x^3}}}\Big(2^{-\frac{1}{x}}\;-\;3^{-\frac{1}{x}}\Big)}{\ln(x-1)^{\frac{1}{x}}-\ln{x^{\frac{1}{x}}}}}=\underset{x\rightarrow +\infty}\lim{\frac{\sqrt[6]{1-\cos{\frac{1}{x^3}}}\Big(2^{-\frac{1}{x}}\;-\;3^{-\frac{1}{x}}\Big)}{\ln(\frac{x-1}{x})^{\frac{1}{x}}}}=\underset{x\rightarrow +\infty}\lim{\frac{\sqrt[6]{1-\cos{\frac{1}{x^3}}}\Big(\frac{\sqrt[x]{3}-\sqrt[x]{2}}{\sqrt[x]{6}}\Big)}{\frac{1}{x}\ln{\Big(1-\frac{1}{x}\Big)}}}=0$
My answer is rather imprecise:$$\underset{x\rightarrow +\infty}\lim{\cos{\frac{1}{x^3}}}=1\implies\underset{x\rightarrow +\infty}\lim{\sqrt[6]{1-\cos{\frac{1}{x^3}}}}=0$$
$$\underset{x\rightarrow +\infty}\lim{\frac{\sqrt[x]{3}-\sqrt[x]{2}}{\sqrt[x]{6}}}=0?$$
I am aware of the mistake I have made by writting $0$ for an undefined term.
$$\underset{x\rightarrow +\infty}\lim{\Big(1-\frac{1}{x^3}\Big)}=1\implies \ln{\Big(1-\frac{1}{x}\Big)}<0\implies\underset{x\rightarrow +\infty}\lim{\Bigg(\frac{1}{\frac{1}{x}\ln{\Big(1-\frac{1}{x}\Big)}}\Bigg)=-\infty}$$
The limit of the denumerator is $0$ $\&$ the limit of the whole expression is $0$.
How can I prove this concisely?
| Your conclusion is wrong since we find an indeterminate form $\frac 0 0$ and we can't conclude that the limit is zero.
Indeed we have that by standard limits
$$\sqrt[6]{1-\cos{\frac{1}{x^3}}} = \frac1{x\sqrt[6]2}+O\left(\frac1{x^2}\right)$$
$$2^{-\frac{1}{x}}\;-\;3^{-\frac{1}{x}} = \frac1x\log \frac32+O\left(\frac1{x^2}\right)$$
therefore
$${\frac{\sqrt[6]{1-\cos{\frac{1}{x^3}}}\Big(2^{-\frac{1}{x}}\;-\;3^{-\frac{1}{x}}\Big)}{\ln(x-1)^{\frac{1}{x}}-\ln{x^{\frac{1}{x}}}}}= \frac{\frac1{x^2}\left(\frac1{\sqrt[6]2}+O\left(\frac1{x}\right)\right)\left( \log \frac32+O\left(\frac1{x}\right)\right)}{{\frac{1}{x}\ln{\Big(1-\frac{1}{x}\Big)}}}=$$
$$= \frac{\left(\frac1{\sqrt[6]2}+O\left(\frac1{x}\right)\right)\left( \log \frac32+O\left(\frac1{x}\right)\right)}{\frac{\ln{\Big(1-\frac{1}{x}\Big)}}{\frac1x}}\to -\frac1{\sqrt[6]2}\log \frac32$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How can I justify this without determining the determinant? I need to justify the following equation is true:
$$
\begin{vmatrix}
a_1+b_1x & a_1x+b_1 & c_1 \\
a_2+b_2x & a_2x+b_2 & c_2 \\
a_3+b_3x & a_3x+b_3 & c_3 \\
\end{vmatrix} = (1-x^2)\cdot\begin{vmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3 \\
\end{vmatrix}
$$
I tried dividing the determinant of the first matrix in the sum of two, so the first would not have $b's$ and the second wouldn't have $a's$.
Then I'd multiply by $\frac 1x$ in the first column of the second matrix and the first column of the second, so I'd have $x^2$ times the sum of the determinants of the two matrices.
I could then subtract column 1 to column 2 in both matrices, and we'd have a column of zeros in both, hence the determinant is zero on both and times $x^2$ would still be zero, so I didn't prove anything. What did I do wrong?
| \begin{align}
&\phantom {=}\,\ \begin{vmatrix}
a_1+b_1x & a_1x+b_1 & c_1 \\
a_2+b_2x & a_2x+b_2 & c_2 \\
a_3+b_3x & a_3x+b_3 & c_3
\end{vmatrix} \\
&=
\begin{vmatrix}
a_1 & a_1x+b_1 & c_1 \\
a_2 & a_2x+b_2 & c_2 \\
a_3 & a_3x+b_3 & c_3
\end{vmatrix}
+ \begin{vmatrix}
b_1x & a_1x+b_1 & c_1 \\
b_2x & a_2x+b_2 & c_2 \\
b_3x & a_3x+b_3 & c_3
\end{vmatrix} \\&= \begin{vmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{vmatrix} + x \begin{vmatrix}
b_1 & a_1x & c_1 \\
b_2 & a_2x & c_2 \\
b_3 & a_3x & c_3
\end{vmatrix} \\&= \begin{vmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{vmatrix} + x^2 \begin{vmatrix}
b_1 & a_1 & c_1 \\
b_2 & a_2 & c_2 \\
b_3 & a_3 & c_3
\end{vmatrix} \\&= 1\cdot \begin{vmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{vmatrix} + (-1) x^2 \begin{vmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{vmatrix} \\&=
(1-x^2)\cdot\begin{vmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3 \\
\end{vmatrix}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3449350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 7,
"answer_id": 2
} |
Value of $L +\frac{153}{L}=$
If $\displaystyle L = \lim_{x\rightarrow 0}\bigg(\frac{1}{\ln(1+x)}-\frac{1}{\ln(x+\sqrt{x^2+1})}\bigg).$
Then value of $\displaystyle L +\frac{153}{L}=$
what i try
$$L=\lim_{x\rightarrow 0}\frac{\ln(x+\sqrt{x^2+1})-\ln(1+x)}{\ln(1+x)\ln(x+\sqrt{x^2+1})}$$
from D L Hopital rule
$$L=\lim_{x\rightarrow 0}\frac{(x^2+1)^{-\frac{1}{2}}-(1+x)^{-1}}{\ln(x+\sqrt{x^2+1})(1+x)^{-1}+\ln(1+x)(x^2+1)^{-\frac{1}{2}}}$$
How do i solve it help me please
| We have that
$$\ln(x+\sqrt{x^2+1})=\ln (\sqrt{x^2+1})+\ln\left(1+\frac x{\sqrt{1+x^2}}\right)=$$
$$=\frac12\ln (1+x^2)+\ln\left(1+x\left(1-\frac12x^2+O(x^4)\right)\right)=$$
$$=\frac12\ln (1+x^2)+\ln\left(1+x-\frac12x^3+O(x^5)\right)=$$
$$=\frac12x^2+x-\frac12x^2+O(x^3)=x+O(x^3)$$
then
$$\frac1{\ln(x+\sqrt{x^2+1})}=\frac1x+O(x)$$
and
$$\log(1+x)=x-\frac12 x^2+O(x^3) \implies \frac1{\ln(1+x)}=\frac1x+\frac12+O(x)$$
and
$$\frac{1}{\ln(1+x)}-\frac{1}{\ln(x+\sqrt{x^2+1})}=\frac12+O(x) \to \frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving $\sum_{cyc} \sqrt{a^2+ab+b^2}\geq\sqrt 3$ when $a+b+c=3$ Good evening everyone, I want to prove the following:
Let $a,b,c>0$ be real numbers such that $a+b+c=3$. Then $$\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}\geq 3\sqrt 3.$$
My attempt: I try noting that $$\sum_{cyc} \sqrt{a^2+ab+b^2}=\sum_{cyc} \sqrt{(a+b)^2-ab}$$
and now I want to apply Cauchy-Schwarz but it is the wrong direction.
| By C-S
$$\sum_{cyc}\sqrt{a^2+ab+b^2}=\sqrt{\sum_{cyc}(a^2+ab+b^2+2\sqrt{(a^2+ab+b^2)(a^2+ac+c^2)}}=$$
$$=\sqrt{\sum_{cyc}(2a^2+ab++2\sqrt{\left(\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}\right)\left(\left(a+\frac{c}{2}\right)^2+\frac{3c^2}{4}\right)}}\geq$$
$$\geq\sqrt{\sum_{cyc}(2a^2+ab+2\left(\left(a+\frac{b}{2}\right)\left(a+\frac{c}{2}\right)+\frac{3bc}{4}\right)}=$$
$$=\sqrt{\sum_{cyc}(4a^2+5ab)}\geq\sqrt{\sum_{cyc}(3a^2+6ab)}=3\sqrt3>\sqrt3.$$
Also, $$\sum_{cyc}\sqrt{a^2+ab+b^2}\geq\sum_{cyc}\sqrt{\frac{3}{4}(a+b)^2}=3\sqrt3>\sqrt3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
How can I evaluate $\lim_{x\to \infty }\left( x^2 - \frac x2 - (x^3 + x+1 ) \ln \left(1+ \frac 1x \right) \right)$ using L'Hospital's rule? How can I find the following limit using the L'Hospital's rule?
$$\lim_{x\rightarrow \infty }\left( x^2 - \frac x2 - (x^3 + x+1 ) \ln \left(1+ \frac 1x \right) \right)$$
I have already tried to replace $1/x$ with $t \,\,(t\rightarrow 0)$, but the only result I get is infinity, which is incorrect.
| Using Taylor series you get $\log(1+1/x)=\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}+\ldots$, thus $$(x^3+x+1)\log(1+1/x)=(x^3+x+1)(\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}+\ldots)=$$
$$x^2-\frac{x}{2}+\frac{1}{3}+1+\text{other terms that goes to 0 for }x\rightarrow\infty.$$
Hence
$$\lim x^2-\frac{x}{2}-(x^3+x+1)\log(1+1/x)=\lim x^2-\frac{x}{2}-[x^2-\frac{x}{2}+\frac{1}{3}+1+\ldots]=-4/3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3452100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Minimize $\frac{(x^2+1)(y^2+1)(z^2+1)}{ (x+y+z)^2}$, $x,y,z>0$ Minimize $\;\;\displaystyle \frac{(x^2+1)(y^2+1)(z^2+1)}{ (x+y+z)^2}$, if $x,y,z>0$.
By setting gradient to zero I found $x=y=z=\frac{1}{\displaystyle\sqrt{2}}$, which could minimize the function.
Question from Jalil Hajimir
| Let $x=\frac{a}{\sqrt2},$ $y=\frac{b}{\sqrt2}$ and $z=\frac{c}{\sqrt2}.$
Thus, since we can assume that $(a^2-1)(b^2-1)\geq0,$ by C-S we obtain:
$$\frac{(x^2+1)(y^2+1)(z^2+1)}{(x+y+z)^2}=\frac{(a^2+2)(b^2+2)(c^2+2)}{4(a+b+c)^2}\geq$$
$$\geq\frac{3(a^2+b^2+1)(1+1+c^2)}{4(a+b+c)^2}\geq\frac{3(a+b+c)^2}{4(a+b+c)^2}=\frac{3}{4}.$$
The equality occurs for $x=y=z=\frac{1}{\sqrt2},$ which says that we got a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What is the period of the $f(x)=\sin x +\sin3x$?
What is the period of the $f(x)=\sin x +\sin3x$?
$f(x)=\sin x+\sin 3x=2\frac{3x+x}{2}\cos\frac{x-3x}{2}=2\sin2x\cos x=4\sin x\cos^2x\\f(x+T)=4\sin(x+T)\cos^2(x+T)=4\sin x\cos^2 x$
how can I deduct this I have no idea
| $f(x)=f(x+2\pi)$ so it has a period which is a factor of $2\pi$.
$df/dx=4$ only when $x$ is a multiple of $2\pi$ so any period is a multiple of $2\pi$.
So the period is $2\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3455375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove by mathematical induction that $(3n+1)7^n -1$ is divisible by $9$ for integral $n>0$ $7^n(3n+1)-1=9m$
$S_k = 7^k(3k+1)-1=9P$
$\Rightarrow 7^k(3k+1) = 9P+1$
$S_{k+1} = 7\cdot7^k(3(k+1)+1)-1$
$= 7\cdot7^k(3k+1+3)-1$
$= 7\cdot7^k(3k+1) +21\cdot7^k -1$
$= 7(9P+1)+21\cdot7^k -1$
$= 63P+7+21\cdot7^k -1$
$= 63P+6+21\cdot7^k$
$= 9(7P +2/3+21\cdot7^k/9)$
therefore it is divisible by $9$
So I believe I have done this right but I've ended up with non-integers in the answer which im pretty sure isn't right.
Where have I gone wrong?
Thanks
| To address your question directly:
You calculations are correct but - as you realized by yourself - the conclusion at the end is still not justified since there are fractions in "$= 9(7P +2/3+21\cdot7^k/9)$".
So, just go one step back and try to squeeze out factor $9$, for example, as follows:
\begin{eqnarray}63P+6+21\cdot7^k
& = & 63P + 3(2+7^{k+1}) \\
& \stackrel{7^{k+1}=(6+1)^{k+1}=6m+1}{=} & 63P + 3(2+1+6m) \\
& = & 63P + 9(1+2m) \\
& = & 9(7P + 1+2m)
\end{eqnarray}
Now, the conclusion is justified.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Determinant of a 3*3 matrix of cosines I need help in evaluating the following determinant:
$$\begin{vmatrix}
\cos\frac{2\pi}{63} & \cos\frac{3\pi}{90} & \cos\frac{4\pi}{77} \\
\cos\frac{\pi}{72} & \cos\frac{\pi}{40} & \cos\frac{3\pi}{88} \\
1 & \cos\frac{\pi}{90} & \cos\frac{2\pi}{99}
\end{vmatrix}$$
I noticed that there is a pattern in the right row ($\frac{4\pi}{77}$,$\frac{3\pi}{88}$ & $\frac{2\pi}{99}$) but I don't know how to utilize the pattern. I also don't see any other patterns: subtracting or adding two rows seems pointless and would further complicate an already complicated determinant.
Is there any way of evaluating this determinant?
| We assume the second entry to be $\cos(3\pi/70)$.
Multiply the last row by $-\cos(\pi/72)$ and add it the the second row; multiply the last row by $-\cos(2\pi/63)$ and add it to the first row. Then use Laplace expansion to obtain
$$\begin{vmatrix}
-\cos(\frac{\pi}{90}) \cos(\frac{\pi}{72}) +
\cos(\frac{\pi}{40})& -\cos(\frac{\pi}{72}) \cos(\frac{2 \pi}{99}) + \cos(\frac{3 \pi}{88})\\
-\cos(\frac{\pi}{90}) \cos(\frac{2 \pi}{63}) +
\cos(\frac{3 \pi}{70}) & -\cos(\frac{2 \pi}{99})
\cos(\frac{2 \pi}{63}) + \cos(\frac{4 \pi}{77})
\end{vmatrix}.
$$
Key observation is that each entry is of the form $-\cos(x)\cos(y)+\cos(x+y)$. But the latter sum equals $\sin(x)\sin(y)$. Hence we have
$$\begin{vmatrix}
\sin(\pi/90) \sin(\pi/72)& \sin(\pi/72) \sin(2 \pi/99)\\
\sin(\pi/90) \sin(2 \pi/63) & \sin(2 \pi/99) \sin(2 \pi/63)
\end{vmatrix},
$$
which is trivially zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3459899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to prove this $\sum_{i=1}^{n}(x_{i})^{S-x_{i}}>1?$ Question:
Let $x_{i} \in (0,1),i=1,2,\cdots,n$. Show that
$$
x_{1}^{S-x_{1}}+x_{2}^{S-x_{2}}+\cdots+x_{n}^{S-x_{n}}>1
$$
where $S=x_{1}+x_{2}+\cdots+x_{n}$.
I have proved when $n=2$,because it use this Bernoulli's inequality
$$
(1+x)^a\le 1+ax,0<a\le 1,x>-1
$$
so we have
$$
x^y=\dfrac{1}{(1/x)^y}=\dfrac{1}{\left(1+\frac{1-x}{x}\right)^y}\ge\dfrac{1}{1+\frac{1-x}{x}\cdot y}=\dfrac{x}{x+y-xy}>\dfrac{x}{x+y}
$$
and simaler we have
$$
y^x>\dfrac{y}{x+y}
$$
so we have
$$
x^y+y^x>\dfrac{x}{x+y}+\dfrac{y}{x+y}=1
$$
Edit: Now the Mr Michael Rozenberg has prove when $n=3$ and MR Czylabson Asa has prove $n\ge 6$ this inequality can't hold, so how to prove $n=4,5?$
Thanks.
| For $n=3$ we can use your work and the Canhang's idea.
Let $\{a,b,c\}\subset(0,1).$ Prove that:
$$a^{b+c}+b^{a+c}+c^{a+b}>1.$$
Proof.
Let $a+b+c\leq1.$
Thus, by Bernoulli
$$\sum_{cyc}a^{b+c}=\sum_{cyc}\frac{1}{\left(1+\frac{1}{a}-1\right)^{b+c}}\geq\sum_{cyc}\frac{1}{1+\left(\frac{1}{a}-1\right)(b+c)}>\sum_{cyc}\frac{1}{1+\frac{b+c}{a}}=1.$$
Let $a+b+c\geq1.$
Thus, by Bernoulli again and by C-S we obtain:
$$\sum_{cyc}a^{b+c}=\sum_{cyc}\frac{1}{\left(1+\frac{1}{a}-1\right)^c\left(1+\frac{1}{a}-1\right)^b}\geq\sum_{cyc}\frac{1}{\left(1+\left(\frac{1}{a}-1\right)c\right)\left(1+\left(\frac{1}{a}-1\right)b\right)}=$$
$$=\sum_{cyc}\frac{a^2}{(a+b-ab)(a+c-ac)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a+b-ab)(a+c-ac)}>1$$ because the last inequality it's just
$$(ab+ac+bc)(a+b+c-1)+abc(3-a-b-c)>0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Stuck on a probability problem/Expectation of coin toss I'm stuck on the following problem that is due for tomorrow:
We're flipping 1 coin indefinitely. $X$ is a random variable that count the amount of coin tosses.
What is the expectation of the number of coin toss until we have the following sequence: THH (T: tail, H: head)
We know that TTT has an expectation of 14.
Edit: Someone told me that
$E[X]= 3 + P[X>3]E[X].$
$P[X>3]=(1 - (1/8))$
$E[X]=24$
But I have now idea how to get to the last step?
| Generating Function Approach
Duration Until $\boldsymbol{THH}$
Any trial can uniquely be constructed from any number of $H$ atoms, then an arbitrary combination of $TH$ and $T$ atoms, then a $THH$ atom. Put this together in a generating function:
$$
\begin{align}
\overbrace{\ \ \frac1{1-x}\ \ }^\text{$H$ atoms}\overbrace{\frac1{1-x-x^2}}^\text{$TH$, $T$ atoms}\overbrace{\quad\ x^3\quad\ \vphantom{\frac11}}^\text{$THH$ atom}
&=\frac{x^3}{1-2x+x^3}
\end{align}
$$
The expected duration is then
$$
\begin{align}
\left.x\frac{\mathrm{d}}{\mathrm{d}x}\frac{x^3}{1-2x+x^3}\right|_{x=\frac12}
&=\left.\frac{x^3(3-4x)}{\left(1-2x+x^3\right)^2}\right|_{x=\frac12}\\[9pt]
&=8
\end{align}
$$
Duration Until $\boldsymbol{THT}$
Any trial can be uniquely constructed from any number of $H$ atoms, then an arbitrary combination of $T$ and $TH^n$ atoms for $n\ge2$, then a $THT$ atom. Put this together in a generating function:
$$
\overbrace{\ \ \frac1{1-x}\ \ }^\text{$H$ atoms}\overbrace{\frac1{1-x-\frac{x^3}{1-x}}}^\text{$T$, $TH^n$ atoms}\overbrace{\quad\ x^3\quad\ \vphantom{\frac11}}^\text{$THT$ atom}=\frac{x^3}{1-2x+x^2-x^3}
$$
The expected duration is then
$$
\begin{align}
\left.x\frac{\mathrm{d}}{\mathrm{d}x}\frac{x^3}{1-2x+x^2-x^3}\right|_{x=\frac12}
&=\left.\frac{x^3\left(3-4x+x^2\right)}{\left(1-2x+x^2-x^3\right)^2}\right|_{x=\frac12}\\[9pt]
&=10
\end{align}
$$
Duration Until $\boldsymbol{TTT}$
Any trial can be uniquely constructed from any number of $H$, $TH$, and $TTH$ atoms, then an $TTT$ atom. Put this together in a generating function:
$$
\overbrace{\frac1{1-x-x^2-x^3}}^\text{$H$, $TH$, and $TTH$ atoms}\overbrace{\quad\ x^3\quad\ \vphantom{\frac11}}^\text{$TTT$ atom}=\frac{x^3}{1-x-x^2-x^3}
$$
The expected duration is then
$$
\begin{align}
\left.x\frac{\mathrm{d}}{\mathrm{d}x}\frac{x^3}{1-x-x^2-x^3}\right|_{x=\frac12}
&=\left.\frac{x^3\left(3-2x-x^2\right)}{\left(1-x-x^2-x^3\right)^2}\right|_{x=\frac12}\\[9pt]
&=14
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3464206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
particular solution of $(D^2+4)y=4x^2\cos 2x$
Find a particular solution to the equation $$(D^2+4)y=4x^2\cos 2x$$
\begin{align}
y_p&=\frac{1}{D^2+4}(4x^2\cos 2x)\\
&=\frac{1}{D^2+4}\left[2x^2(e^{2ix}+e^{-2ix})\right]\\
&=2\left[e^{2ix}\frac{1}{(D+2i)^2+4}x^2+e^{-2ix}\frac{1}{(D-2i)^2+4}x^2\right]\\
&=2\left[e^{2ix}\frac{1}{D(D+4i)}x^2+e^{-2ix}\frac{1}{D(D-4i)}x^2\right]\\
\end{align}
But I stuck at this point because how to solve $\frac{1}{D(D+4i)}x^2?$ The solution provided this
\begin{align}
&=2\left[e^{2ix}\frac{1}{D(D+4i)}x^2+e^{-2ix}\frac{1}{D(D-4i)}x^2\right]\\&=\frac{1}{4i}\left[e^{2ix}D^{-1}\left(2x^2+ix-\frac{1}{4}\right)-e^{-2ix}D^{-1}\left(2x^2-ix-\frac{1}{4}\right)\right]\\
&=\frac{1}{24}[6x^2\cos 2x+x(8x^2-3)\sin 2x]
\end{align}
I still can't figure out how they do this. Any help will be appreciated.
Update: Using @Isham answer I tired
\begin{align}
&=\frac{1}{4i}\left[e^{2ix}D^{-1}\left(2x^2+ix-\frac{1}{4}\right)-e^{-2ix}D^{-1}\left(2x^2-ix-\frac{1}{4}\right)\right]\\
&=\frac{1}{4i}\left[e^{2ix}\left(2\frac{x^3}{3}+i\frac{x^2}{2}-\frac{x}{4}\right)-e^{-2ix}\left(2\frac{x^3}{3}-i\frac{x^2}{2}-\frac{x}{4}\right)\right]\\
&=\frac{1}{4i}\left[\frac{2x^3}{3}(e^{2ix}-e^{-2ix})+\frac{x^2i}{2}(e^{2ix}-e^{-2ix})-\frac{x}{4}(e^{2ix}-e^{-2ix})\right]\\
&=\frac{1}{4i}\left[\frac{2x^3}{3}2\cos 2x+\frac{x^2i}{2}2\cos 2x-\frac{x}{4}2\cos 2x\right]
\end{align}
But I think I am again Lost.
| Hint: $$\dfrac{1}{1+D} = 1-D+D^2-\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3466016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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evaluation of Trigonometric limit
Evaluation of $$\lim_{n\rightarrow \infty}\frac{1}{n\bigg(\cos^2\frac{n\pi}{2}+n\sin^2\frac{n\pi}{2}\bigg)}$$
What i try
Put $\displaystyle \frac{n\pi}{2}=x,$ when $n\rightarrow\infty,$ Then $x\rightarrow \infty$
$$\frac{\pi^2}{2}\lim_{x\rightarrow \infty}\bigg(\frac{ x^{-1}}{\pi\cos^2(x)+2x\sin^2(x)}\bigg)$$
How do i solve it help me please
| \begin{align*}
\lim_{k\rightarrow\infty}\dfrac{1}{2k[\cos^{2}(2k\pi/2)+2k\sin^{2}(2k\pi/2)]}=\lim_{k\rightarrow\infty}\dfrac{1}{2k}=0,
\end{align*}
while
\begin{align*}
&\lim_{k\rightarrow\infty}\dfrac{1}{(2k+1)[\cos^{2}((2k+1)\pi/2)+(2k+1)\sin^{2}((2k+1)\pi/2)]}\\
&=\lim_{k\rightarrow\infty}\dfrac{1}{(2k+1)(2k+1)}\\
&=0,
\end{align*}
so
\begin{align*}
\lim_{n\rightarrow\infty}\dfrac{1}{n[\cos^{2}(n\pi/2)+n\sin^{2}(n\pi/2)]}=0.
\end{align*}
| {
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Find the recurrence relation solution. $a_n$ = $a_{n-1} + 3n - 5$ $a_n$ = $a_{n-1} + 3n - 5$ $,$ $a_0 = 7$
So far what I got is:
$a_0 = 7$
$a_1 = 7+3(1)-5$
$a_2 = 7+3(1)-5 + 3(2)-5$
$a_3 = 7+3(1)-5 + 3(2)-5 + 3(3)-5$
$a_4 = 7+3(1)-5 + 3(2)-5 + 3(3)-5 + 3(4)-5$
This is where I'm stuck I'm having trouble finding a pattern to the relation.
At first I tried solving this as a quadratic sequence and got $a_n = \frac{3}{2}{n^2} - \frac{7}2n + 7$ which works but on a test I solved a recurrence relation problem as a quadratic sequence got the right answer but got a 0 on the problem because the professor wanted us to use forward or backward substitution which is what I am attempting above (forward substitution).
| Let $A(z) = \sum_{n=0}^\infty a_nz^n$. Multiply both sides of the recurrence by $z^n$ and sum over $n$ to obtain
$$
\sum_{n=1}^\infty a_nz^n = \sum_{n=1}^\infty a_{n-1}z^n + \sum_{n=1}^\infty 3nz^n - \sum_{n=1}^\infty 5z^n.
$$
By some algebra we obtain
$$
A(z) - a_0 = zA(z) +\sum_{n=0}^\infty 3nz^n - \sum_{n=1}^\infty 5z^n.
$$
From the identity $\sum_{n=0}^\infty z^n = \frac1{1-z}$ and differentiating we obtain
$$
\sum_{n=0}^\infty 3nz^n - \sum_{n=1}^\infty 5z^n= \frac{3 z}{(1-z)^2} -\frac{5 z}{1-z},
$$
so we have
$$
A(z) - 7 = zA(z) +\frac{3z}{(1-z)^2} - \frac{5z}{1-z}.
$$
Solving for $A(z)$ yields
$$
A(z) = \frac{12 z^2-16 z+7}{(1-z)^3}.
$$
After a partial fraction decomposition, we have
$$
A(z) = -\frac{8}{(1-z)^2}+\frac{3}{(1-z)^3}+\frac{12}{1-z}.
$$
Writing the terms on the right-hand side as series yields
\begin{align}
A(z) &= \sum_{n=0}^\infty -8(n+1)z^n + \sum_{n=0}^\infty \frac32(n+1)(n+2)z^n + \sum_{n=0}^\infty 12z^n\\
&= \sum_{n=0}^\infty \left(\frac{1}{2} n (3 n-7)+7\right) z^n.
\end{align}
It follows then that
$$
a_n = \frac{1}{2} n (3 n-7)+7.
$$
| {
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"answer_id": 1
} |
Without calculating the square roots, determine which of the numbers:$a=\sqrt{7}+\sqrt{10}\;\;,\;\; b=\sqrt{3}+\sqrt{19}$ is greater.
Without calculating the square roots, determine which of the numbers:
$$a=\sqrt{7}+\sqrt{10}\;\;,\;\; b=\sqrt{3}+\sqrt{19}$$
is greater.
My work (I was wondering if there are other ways to prove this):
$$a^2=17+2\sqrt{70}, \;\;b^2=22+2\sqrt{57}$$
$$\sqrt{64}<\sqrt{70}<\sqrt{81}\implies 8<\sqrt{70}<9\implies a^2<35$$
$$\sqrt{49}<\sqrt{57}<\sqrt{64}\implies 7<\sqrt{57}<8\implies b^2>36$$
$$a^2<35<36<b^2\implies a^2<b^2\implies |a|<|b|$$
| Since $a$ and $b$ are both positive, it follows that $a<b \iff a^2<b^2$; namely $$a<b\iff17+2\sqrt{70}<22+2\sqrt{57};$$ that is, $a<b$ iff $2\left(\sqrt{70}-\sqrt{57}\right)<5.$ Continuing equivalent statements in this way, we get
$$
a < b
\iff 508-8\sqrt{3990}<25
\iff 60 \tfrac38 < \sqrt{3990}
\iff 3600+45+\tfrac9{64} < 3990,
$$
the latter clearly being the case.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 5,
"answer_id": 2
} |
Find the minimal polynomial for $\cos(\frac{2\pi}{5})$ and $\sin(\frac{2\pi}{5})$ Let $\omega$ be the primitive 5th root of $1$, then $\cos(\frac{2\pi}{5}) = \frac{w+w^{-1}}{2}$ and $\sin(\frac{2\pi}{5}) = \frac{w-w^{-1}}{2i}$. How to find the minimal polynomial of $\frac{w+w^{-1}}{2}$ then? (without using the Chebyshev polynomials)
Thanks.
| Let $w=e^{2\pi i/5}$. Then $w^5=1$ and $w\ne1$ so $w^4+w^3+w^2+w+1=0$, $w^4=w^{-1},$ and $w^3=w^{-2}$.
Also, $\cos(2\pi i/5)=\dfrac{w+w^{-1}}2$,
so $\cos^2(2\pi i/5)=\dfrac{w^2+w^{-2}+2}4=\dfrac{w^2+w^3+2}4=\dfrac{-1-w-w^{-1}+2}4=\dfrac{1-2\cos(2\pi i/5)}{4}.$
Can you take it from here?
| {
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Show that $2(\sin y + 1)(\cos y + 1) = (\sin y + \cos y + 1)^2$ The question states:
Show that: $$2(\sin y + 1)(\cos y + 1) = (\sin y + \cos y + 1)^2$$
This is what I have done
$2(\sin y + 1)(\cos y + 1) = 2(\sin y + \cos y + 1)^2$
L. H. S. = R. H. S.
From L. H. S.
$2(\sin y +1)(\cos y + 1) = 2(\sin y\cos y + \sin y + \cos y + 1)$
$= 2(\sin y\cos y + \sin y + \cos y + \sin^2 y + \cos^2 y) (\sin^2 y + \cos^2 y = 1)$
$= 2(\sin^2 y + \sin y\cos y + \sin y + \cos^2 y + \cos y)$
I got stuck here. I do not know what to do from here.
I have tried and tried several days even contacted friends but all to no avail.
| $$RHS - LHS=(\sin y + \cos y + 1)^2-2(\sin y + 1)(\cos y + 1) $$
$$=[(\sin y +1)+( \cos y + 1) -1]^2-2(\sin y + 1)(\cos y + 1) $$
$$=(\sin y +1)^2+( \cos y + 1)^2 + 1 -2(\sin y + 1)-2(\cos y + 1) $$
$$=[(\sin y +1)^2-2(\sin y + 1)+1]+[( \cos y + 1)^2 - 2(\cos y + 1)+1] -1 $$
$$=\sin^2 y+\cos^2 y -1 = 0$$
| {
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} |
Possible Jordan Canonical Forms Suppose I have a matrix $A \in M_{n \times n}(\mathbb{C})$ such that its minimal polynomial is either $x-1$ or $(x-1)^{2}$. What are its possible Jordan Canonical Forms? I was thinking that if its minimal polynomial is $x-1$, then its Jordan canonical form is $I_{n}$, the $n \times n$ identity matrix. But if its minimal polynomial is $(x-1)^{2}$ then the number of its Jordan Canonical Forms depend on $n$. I was thinking that the number of forms is $\lfloor \frac{n}{2} \rfloor$. For example, when $n = 7$, we have that $V \cong \left( \mathbb{C}[x] / (x-1) \right)^{5} \oplus \mathbb{C}[x] / (x-1)^{2}$, or $V \cong \left( \mathbb{C}[x] / (x-1) \right)^{2} \oplus \left( \mathbb{C}[x] / (x-1)^{2} \right)^{2}$ or $V \cong \mathbb{C}[x] / (x-1) \oplus \left( \mathbb{C}[x] / (x-1)^{2} \right)^{3}$, which gives $3$ distinct Jordan forms. Also, are the matrices $\begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \end{bmatrix}$ and $\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\0 & 0 & 0 & 1 \end{bmatrix}$ considered different Jordan canonical forms or the same for $n=4$ and the minimal polynomial $(x-1)^{2}$.
| Consider the special case when $A$ is nilpotent:
If minimal polynomial has degree $m=1$, then largest Jordan block will be of size $1\times 1$, thus JCF is diagonal. This can be generalized to any matrix.
If minimal polynomial has degree $m=2$, then largest Jordan block will be of size $2\times 2$, and the rest of the Jordan blocks will depend on the dimensions of $N(A^i)$ (null space of $A^i$), larger the $n$ the more options we have. For example, when $n=7$, we can have JCF with jordan blocks $(2,2,2,1),$ $(2,2,1,1,1),$ $(2,1,1,1,1,1),$ assuming that the Jordan blocks are used from largest to smallest, otherwise we can change the position of jordan blocks, which is the case in your example.
If $A$ has equal eigenvalues $\lambda$, then $A$ and $A-\lambda I$ have same JCF, since $P^{-1}(A-\lambda I)P=P^{-1}AP-\lambda I$, thus your two matrices are equivalent to $A_1=\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 \end{bmatrix}$ and $A_2=\begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\0 & 0 & 0 & 0 \end{bmatrix}$. Now you can see that they have same JCF, they both consists of one $2\times 2$ and two $1\times 1$ Jordan blocks. Also $\mathrm{dim}N(A_1^i)=\mathrm{dim}N(A_2^i)$ for $i=1,2,3,4$ and that $S^{-1}A_1S=A_2$, where permutation matrix $S=\begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \end{bmatrix}$.
| {
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Having problem to find all solutions using the Chinese Remainder Theorem Find all solutions using the Chinese Remainder Theorem.
x = 2 (mod 4)
x = 3 (mod 5)
x = 9 (mod 13)
My step is here:
a = 2 (mod 4) , a = 0 (mod 5) , a = 0 (mod 13)
b = 0 (mod 4) , b = 3 (mod 5) , b = 0 (mod 13)
c = 0 (mod 4) , c = 0 (mod 5) , c = 9 (mod 13)
The general solution for x is given by x = a + b + c + k*lcm(4,5,13) = a+b+c+260k
since a = 0 (mod 5) and a = 0 (mod 13), so a = 65m for some integer m
then 65m = 2 (mod 4)
What should i do next?
I think the step i write 65m - 2 = 4 is wrong
| We have ,
$$x\equiv2 \mod 4 \implies \color{#d05}{x=4a+2}\\ $$
$$\begin{align}x&\equiv3 \mod 5 \\4a+2&\equiv3\mod5\\ 4a&\equiv 1\mod5 \\ a &\equiv 4 \mod 5\implies\color{#2cd}{a = 5b+4}\end{align} $$
$$\begin{align}x&\equiv9 \mod 13 \\ 4(5b+4)+2&\equiv 9\mod13 \\ 20b+18&\equiv9\mod13 \\ 20b&\equiv 4\mod 13 \implies \color{#2c0}{b = 13c+8}\end{align} $$
So , $x = 4a+2 = 20b+18=260c + 178$
| {
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} |
Find a cubic polynomial satisfying the given conditions Find $p(x) = ax^3 + bx^2 + cx + d $, where a, b, c and d are real constants, satisfying the following conditions:
$1)\quad 4x^3 - 12x^2 + 12x - 3 \le p(x) \le 2019(x^3 - 3x^2 + 3x) - 2018,$ for all values of $x \ge 1$.
$2)\quad p(2) = 2011.$
My idea:
The condition $(1) \iff 4(x - 1)^3 + 1\le p(x) \le 2019(x - 1)^3 + 1 $ , for all $x \ge 1.$
Therefore, I predict $p(x) = a(x - 1)^3 + 1 $, where $4 \le a \le 2019 $ . But, I still don't prove this. Please everyone help me.
| Continue with what you obtained,
$$4(x - 1)^3 + 1\le p(x) \le 2019(x - 1)^3 + 1 $$
and rearrange,
$$4(x - 1)^3 \le p(x) - 1 \le 2019(x - 1)^3 $$
Assume $p(x) - 1 = a(x-1)^3$ and substitute $p(2) = 2011$ to obtain $a = 2010$. Thus,
$$p(x) = 2010(x-1)^3+1$$
Edit:
Assume, instead, a general form $p(x)-1=\sum_{k=0}^{\infty}a_k(x-1)^k$ with $a_k\ge 0$. Note that,
$$\lim_{x\to\infty} a_k(x-1)^k > 2019(x-1)^3,\>\>\> k>3$$
$$\lim_{x\to\infty} a_k(x-1)^k < 4(x-1)^3,\>\>\> k<3$$
which leads to $a_k = 0 $, for $k\ne 3$.
| {
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How to prove $2x^2 + 2y^2 + 2z^2 -2xy -2yz \geq 0$ I know how to prove $2x^2 + 2y^2 + 2z^2 -2xy -2yz -2xz \geq 0$ since I can split it up in the three terms $(x-y)^2, (y-z)^2, (z-x)^2$ which are all greater than or equal to zero, but how do I prove this without the last term: $2x^2 + 2y^2 + 2z^2 -2xy -2yz \geq 0$?
| It is $$x^2+y^2-2xy+y^2+z^2-2yz+z^2+x^2\geq 0$$
so
$$(x-y)^2+(y-z)^2+z^2+x^2\geq0$$
| {
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If $m$ and $n$ are integers, show that $\left|\sqrt{3}-\frac{m}{n}\right| \ge \frac{1}{5n^{2}}$ If $m$ and $n$ are integers, show that $\biggl|\sqrt{3}-\dfrac{m}{n}\biggr| \ge \dfrac{1}{5n^{2}}$.
Since $\biggl|\sqrt{3}-\dfrac{m}{n}\biggr|$ is equivalent to $\biggl|\dfrac{ \sqrt{3}n-m}{n}\biggr|$
So I performed the following operation $\biggl|\dfrac{\sqrt{3}n-m}{n}\biggr|\cdot \biggl|\dfrac{\sqrt{3}n+m}{\sqrt{3}n+m}\biggr|$ to get $$\biggl|\dfrac{3n^{2}-m^{2}}{\sqrt{3}n^{2}+mn}\biggr|$$
Since $n,m \ne 0$, we have that $|3n^{2}-m^{2}| \ge 1$. Now for the denominator, we have $$ |\sqrt{3}n^{2}+mn| \le |\sqrt{3n^{2}}| + |mn| $$
Thus it follows that $$\dfrac{1}{|\sqrt{3}n^{2}+mn|} \ge \dfrac{1}{|\sqrt{3}n^{2}| + |mn|}$$
Would I have to work in cases where $m<n$, for example? Then we have $$|\sqrt{3}n^{2}| + |mn| < |\sqrt{3}n^{2}| + n^{2} < 3n^{2} + n^{2} < 5n^{2}$$ which gives us the desired result. Although, the same method doesn't work when $n >m$.
| One aims to give an optimal lower bound as follows:
Theorem. $|\sqrt{3}-\frac mn |\geq \frac 1{(2+\sqrt{3})n^2}$
Proof. Clearly the lower bound is achieved when $n=1,m=2.$ As in other proofs, one may assume without loss of generality that $m,n$ are positive and $n\geq 2$. Observe first that $$\frac 53<\sqrt{3}<\frac 74.$$ We divide into two cases.
Case 1. $\frac m n<\frac 7 4.$ Then $$|\sqrt{3}-\frac m n|=\frac{|3n^2-m^2|}{n^2(\sqrt{3}+\frac m n)}$$ $$>\frac 1{n^2(\frac 7 4+\frac 7 4)}=\frac 1{(3.5)n^2}>\frac 1{(2+\sqrt{3})n^2}.$$
Case 2. $\frac m n\geq \frac 7 4.$ Here one considers three subcases: $n=2,n=3$ and $n\geq 4$. If $n=2,$ then $\frac m n\geq \frac 7 4\Rightarrow m\geq 4,$ so $\frac m n\geq 2$ and hence $$|\sqrt{3}-\frac m n|=\frac m n-\sqrt{3}\geq 2-\sqrt{3}\geq \frac 1{(2+\sqrt{3})n^2}.$$ If $n=3$, then $\frac m n\geq \frac 7 4\Rightarrow m\geq \frac{21}4,$ so $m\geq 6$ and $\frac m n\geq 2$ and the result follows in exactly the same way as the above subcase. It remains to check the subcase when $n\geq 4$, but then $$ |\sqrt{3}-\frac m n|=\frac m n-\sqrt{3}\geq \frac 7 4-\sqrt{3}>\frac 1{(2+\sqrt{3})\cdot 4^2}\geq \frac 1{(2+\sqrt{3})n^2},$$ where one used $$\frac 7 4-\sqrt{3}>\frac 1{(2+\sqrt{3})\cdot 4^2}$$ $$\Leftrightarrow \frac 7 4>\sqrt{3},$$ which is true.
Combining all cases, the result is proven.
Edit. (More details) $$\frac 7 4-\sqrt{3}>\frac 1{(2+\sqrt{3})\cdot 4^2}$$ $$\Leftrightarrow \frac {7-4\sqrt{3}}4>\frac 1{(2+\sqrt{3})\cdot 4^2}$$ $$\Leftrightarrow 7-4\sqrt{3}>\frac 1{(2+\sqrt{3})\cdot 4}$$
$$\Leftrightarrow (7-4\sqrt{3})(2+\sqrt{3})>\frac 1 4$$
$$\Leftrightarrow 2-\sqrt{3}>\frac 1 4$$
$$\Leftrightarrow \frac 7 4>\sqrt{3},$$ which is true.
| {
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If $\cos(A+B+C)=\cos A\cos B\cos C\neq 0$, then evaluate $\left|\frac{8\sin(A+B)\sin(B+C)\sin(C+A)}{\sin 2A\sin 2B\sin 2C}\right|$
If
$$\cos(A+B+C)=\cos A\cos B\cos C, \quad\text{with}\;A,B,C\neq \frac{k\pi}{2}$$
then
$$\left|\frac{8\sin(A+B)\sin(B+C)\sin(C+A)}{\sin 2A\sin 2B\sin 2C}\right|$$ is what integer?
What I tried:
$$\begin{align}
\cos(A+B+C)
&=\cos A \cos B \cos C - \sin A \sin B \sin C - \sin C \sin A \cos B - \sin B \sin C \cos A \\
&=\cos A\cos B\cos C\left(1-\tan A\tan B-\tan B\tan C-\tan C\tan A\right)
\end{align}$$
How do i solve it Help me please
| So, where you have left off,
$$pq+qr+rp=0$$ writing $\tan A=p,\tan B=q,\tan C=r$
we have $p,q,r$ are non-zero and finite
Now $$\dfrac{2\sin(A+B)}{\sin2C}=\cdots=\dfrac{p+q}{r}$$
$$\implies\dfrac{2\sin(A+B)\cdot2\sin(B+C)\cdot2\sin(C+A)}{\sin2A\sin2B\sin2C}$$
$$=\dfrac{(p+q)(q+r)(r+p)}{pqr}$$
$$=\dfrac{2pqr+p(pq+qr+rp)-pqr+q(pq+qr+rp)-pqr+r(pq+qr+rp)-pqr}{pqr}$$
$$=\dfrac{pqr(2-3)}{pqr}=?$$
| {
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If $a^2>b^2$ prove that $\int\limits_0^{\pi} \frac{dx}{(a+b\cos x)^3}=\frac{\pi (2a^2+b^2)}{2(a^2-b^2)^{5/2}}$. Problem: If $a^2>b^2$ prove that $\int\limits_0^\pi \dfrac{dx}{(a+b\cos x)^3} = \dfrac{\pi (2a^2+b^2)}{2(a^2-b^2)^{5/2}}$.
My effort:
If we choose $$x=\tan\frac{\theta}{2}\Longrightarrow d\theta=\frac{2}{x^2+1} \, dx\;\;,\;\;\cos\theta=\frac{1-x^2}{1+x^2}$$ then the integral becomes critical. What is the simplest way to solve?
| It is well-known that
$$\int_0^\pi\frac{dx}{t+\cos x}=\frac{\pi}{\sqrt{t^2-1}}$$
for $t>1$. Differentiating gives
$$-\int_0^\pi\frac{dx}{(t+\cos x)^2}=-\frac{\pi t}{(t^2-1)^{3/2}}.$$
Differentiating again gives
$$2\int_0^\pi\frac{dx}{(t+\cos x)^3}
%=-\frac{\pi(t^2-1)}{(t^2-1)^{5/2}}+\frac{3\pi t^2}{(t^2-1)^{5/2}}
=\frac{\pi(2t^2+1)}{(t^2-1)^{5/2}}.$$
Homogenising this gives your formula.
| {
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If $PA - PB = \text{constant}$ then the locus of $P$ will be a hyperbola. The sound of a cannon firing is heard one
second later at a position B than at position A.
If the speed of sound is uniform, then
(A) The positions A and B are foci of a
hyperbola, with cannon's position on one
branch of the hyperbola
(B) the position A and B are foci of an ellipse
with cannon's position on the ellipse
(C) One of the positions A,B is focus of a
parabola with cannon's position on the
parabola
(D) It is not possible to describe the positions
of A, B and the cannon with the given
information
My Attempt: If I suppose the position of A and B are respectively $(c,0)$ and $(0,-c)$ and $2a = 343\ $m, then we will get the locus of $P$ as follows $$\frac{x^2}{a^2} - \frac{y^2}{(c^2 - a^2)} = 1~.$$
If $c^2 > a^2$ we are done but if not then what will happen?
Explicitly saying if the speed of the sound is greater than the distance between two points then what will happen?
Can anyone please help me if I have gone wrong anywhere?
| EDIT This was my first answer to the original question. OP change his question afterward, so I'll update my answer.
The geometric definition of an hyperbola is :
the points such that the difference of the distance to two foci is
constant. This difference is equal to the distance between the apex of the hyperbola
If A and B are the foci, the cannon is somewhere on an hyperbola.
The geometric definition I gave is equivalent to the following algebric definition of the horizontal hyperbola
If the foci of the hyperbola are located at $(c, 0)$ and $(-c, 0)$, then the hyperbola is given by the equation $$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2}=1 \qquad \text{where }c^2 = a^2+b^2$$
Proof
Starting with the geometric definition. Let $A(c, 0)$ and $B(-c,0)$, the foci, and $P(x, y)$ a point on the hyperbola. Then, the difference of the distances is equal to $2a$.
$$\vert d(P, A) - d(P, B)\vert = 2a$$
$$\left\vert \sqrt{(x-c)^2+(y-0)^2} - \sqrt{(x-(-c))^2+(y-0)^2}\right\vert = 2a$$
If we square each side, it become
$$(x-c)^2+y^2 - 2\sqrt{(x-c)^2+y^2}\sqrt{(x+c)^2+y^2} + (x+c)^2+y^2 = 4a^2$$
Developping the $(x\pm c)^2$ and moving things around give us
$$x^2+y^2+c^2-2a^2=\sqrt{(x-c)^2+y^2}\sqrt{(x+c)^2+y^2}$$
We square once again to get rid of square roots.
$$(x^2+y^2+c^2-2a^2)^2=((x-c)^2+y^2)((x+c)^2+y^2)$$
Developping both side
$$x^4+y^4+c^4+4a^4 + 2x^2y^2+2x^2c^2-4a^2x^2+2y^2c^2-4a^2y^2-4a^2c^2 = x^4-2x^2c^2+c^4+y^4+2y^2x^2+2y^2c^2$$
Cancelling terms on each side and reorganise remaining terms
$$4a^4+4x^2c^2-4a^2x^2-4a^2y^2-4a^2c^2 = 0$$
$$x^2(c^2-a^2)-a^2y^2=a^2(c^2 - a^2)$$
Since $c^2 = a^2+b^2$
$$x^2b^2-a^2y^2=a^2b^2$$
Dividing everywhere by $a^2b^2$
$$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$$
A similar proof could be done for a vertical hyperbola $\dfrac{y^2}{b^2}-\dfrac{x^2}{a^2}=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3479152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Finding $\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x}$ $\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x}$
I'm looking more into simplifying this than the solution itself (which I know involves using t=tan(x/2)).
I did:
$$\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x} = \int\frac{\cos(2x)dx}{(\cos^2x+\sin^2x)^2-2\sin^2x\cos^2x} = \int\frac{\cos(2x)dx}{1-2\sin^2(2x)}$$
I tried applying the t=tan(y/2) (where y = 2x) but it became a mess so I figure this can be simplified further... help?
| Continues, we have $$ \int{ \frac{1}{2 - t^2} dt } $$.
assume $ t = \sqrt{2} \sin (y) $, we differensial and get $ dt = \sqrt{2} \cos (y) dy $
$$ \int{ \frac{1}{2 - t^2} dt } = \frac{1}{2} \int{ \frac{\sqrt{2} \cos (y) dy}{1 - \sin ^2 (y) } dt } = \frac{\sqrt{2}}{2} \int{ \sec(y) dy } = \frac{\sqrt{2}}{2} \sec y \tan y = \frac{1}{2- t^2}$$ . back to question
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3480289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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How to prove $x^4+2x^2y^2+y^4\geq2xy^3$ Suppose that $x,y$ are real numbers. I want to prove $$x^4+2x^2y^2+y^4\geq2xy^3.$$ I noticed that this is the same as $$(x^2+y^2)^2\geq 2xy^3.$$ Can we proceed from here?
| We know that: $$(x^2+y^2)\ge 2xy$$
Multiplying both sides by $y^2$
$$(x^2+y^2)\cdot y^2\ge 2xy^3$$
And clearly $x^2+y^2 \ge y^2$. So
$$(x^2+y^2)\cdot(x^2+y^2)\ge(x^2+y^2)\cdot y^2\ge 2xy^3$$
implying that $$\boxed{(x^2+y^2)^2 \ge 2xy^3}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Expressing $\cos(x)$ and $\sin(x)$ in terms of $\tan(\frac{1}{2} x)$ May I know is there any quick way to express $\cos(x)$ and $\sin(x)$ in terms of $\tan(\frac{1}{2} x)$. I checked that $\cos(x)=\frac{1 - \tan^2(\frac{1}{2} x)}{1+\tan^2(\frac{1}{2} x)}$. I really want to know how could I figure this out quickly, as well as how can I express $\sin(x)$ in terms of $\tan(\frac{1}{2} x)$ .
Thank you very much for your help.
| For the sine, you have
$$\begin{align}
\sin x&=2\sin\frac x2\cos \frac x2 \\
&= \frac{2\frac{\sin \frac x2}{\cos \frac x2}}{\frac{1}{\cos^2 \frac x2}} \\
&= \frac{2\tan\frac x2}{1+\tan^2\frac x2}\\
\end{align}
$$
In the second line, we divide the numerator and denominator ($=1$) by $\cos^2\frac x2$, and we then use $1+\tan^2\frac x2=\frac{1}{\cos^2\frac x2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3481998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
solutions of $a+b=c^2 , a^2+c^2=b^2$ ; $a,b,c$ are natural numbers So it all started with a fun observation, $12+13=5^2$ and these are Pythagorean triplets($5,12,13$), so I thought are there more such numbers?
with brute force I was able to get $(24,25,7)$ and $(40,41,9)$.
Then I was able to find 3 families of solutions.
$(50k^2+50k+12 , 50k^2+50k+13 , 10k+5)$
$(10k+4 , 10k+5 , \sqrt{20k+9})$
$(10k, 10k+1 , \sqrt{20k+1})$
ps: I found these by using the property of Pythagorean triplets that they have at least one multiple of 5 in it.
My question is are there more sets of solution and how do I know I haven't missed any?
| $a^2 + c^2 = b^2$
$c^2 = b^2 - a^2 = (b-a)(a+b)$ but $a+b = c^2$ so if we assume $a+b \ne 0$, we have $b-a = 1$ and $b = a+1$ and we have
$a^2 + c^2 = (a+1)^2$ and $2a + 1 = c^2$
If we replace $c^2$ with $2a+1$ we have $a^2 + 2a + 1 = (a+1)^2$ which is always true. So $2a+1=c^2$ can be any number that is both; an odd number at least equal to $3$ ($a \ge 1$); and a perfect square, and $c=\sqrt {2a+1}$ is a square root of an odd perfect square greater than $3$ which can be any odd integer greater than $1$.
So for any $k \in \mathbb N$ we can have $c = 2k +1$, $a= \frac {c^2 -1}2= 2k(k+1)$ and $b = \frac {c^2 +1}2 = 2k^2 + 2k + 1$.
Those are all of them.
If $k\equiv 0,4 \pmod 5$ then $a\equiv 0 \pmod 5$. If $k\equiv 1,3\pmod 5$ then $b\equiv 0 \pmod 5$ and if $k \equiv 2\pmod 5$ then $c \equiv 0 \pmod 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3482827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
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Close formula of Yukawa Potential in 3 dimension by integrating $\int_{0}^{\infty}{\frac{1}{(4\pi t)^{\frac{d}{2}}}e^{-\frac{x^2}{4t} - \mu^2t}}dt$ In PDE, Yukawa potential can be calculate as
$$G^\mu(x) = \int_{0}^{\infty} \frac{1}{(4\pi t)^{\frac{d}{2}}}e^{-\frac{x^2}{4t} - \mu^2t}dt$$
When $d = 3$ we can get the close formula $G^\mu(x) = \frac{1}{4\pi|x|}e^{-\mu|x|}$.
How should we do the integral?
| Let
\begin{align}
I(\mu, x) = \int^\infty_0 dt\ \frac{1}{(4\pi t)^{3/2}}\exp\left( -\frac{x^2}{4t}-\mu^2 t\right).
\end{align}
Then we see that
\begin{align}
I(\mu, x) = \int^\infty_0 dt\ \frac{1}{(4\pi t)^{3/2}}\exp\left( -\frac{x^2}{2}\left(\frac{1}{\sqrt{2t}}-\frac{\mu}{|x|} \sqrt{2t}\right)^2\right)\exp\left(-\mu |x|\right).
\end{align}
Set $u = \frac{1}{\sqrt{2t}}$ and $\alpha = \frac{\mu}{|x|}$ then it follows
\begin{align}
I(\mu, x) = \int^\infty_0 du\ \exp\left(-\frac{1}{2}x^2\left(u-\frac{\alpha}{u}\right)^2\right)(2\pi)^{-3/2}\exp(-\mu|x|) = \frac{e^{-\mu|x|}}{(2\pi)^{3/2}}J(\mu, x).
\end{align}
Hence it suffices to evaluate $J(\mu, x)$. Set $z=\alpha/u$ then we have
\begin{align}
J(\mu, x) = \int^\infty_0 \frac{dz}{z^2}\ \alpha\exp\left(-\frac{1}{2}x^2\left(z-\frac{\alpha}{z}\right)^2\right)
\end{align}
which means
\begin{align}
2J(\mu, x) =&\ \int^\infty_0 du\ \left(1+\frac{\alpha}{u^2} \right)\exp\left(-\frac{1}{2}x^2\left(u-\frac{\alpha}{u}\right)^2\right)\\
=&\ \int^\infty_0 d\left(u-\frac{\alpha}{u} \right)\exp\left(-\frac{1}{2}x^2\left(u-\frac{\alpha}{u}\right)^2\right)\\
=& \int^\infty_{-\infty} dw\ \exp\left(-\frac{1}{2}x^2w^2\right)= \sqrt{2\pi}\frac{1}{|x|}.
\end{align}
Hence it follows
\begin{align}
I(\mu, x) = \frac{e^{-\mu|x|}}{(2\pi)^{3/2}}J(\mu, x) = \sqrt{\frac{\pi}{2}}\frac{1}{|x|}\frac{e^{-\mu|x|}}{(2\pi)^{3/2}} = \frac{e^{-\mu |x|}}{4\pi |x|}.
\end{align}
Additional: Observe
\begin{align}
\int^\infty_0 dt\ \frac{1}{(4\pi t)^{5/2}}\exp\left( -\frac{x^2}{4t}-\mu^2 t\right) =&\ \int^\infty_0 dt\ \frac{1}{(4\pi t)^{3/2}}\left(-\frac{1}{2\pi x}\right)\frac{d}{dx}\exp\left( -\frac{x^2}{4t}-\mu^2 t\right)\\
=&\ \left(-\frac{1}{2\pi x}\right)\frac{d}{dx}I(\mu, x) = \frac{e^{-\mu|x|}(\mu|x|+1)}{8\pi^2 |x|^3}.
\end{align}
Using this observation, we can recover all expression of $G^\mu(x)$ for odd dimension $d$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3485618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$\log_{2} (3 \sin \theta) = 2.\log_{2}(-3\cos \theta) + 1$ what's the sum of possible $\theta$ $0 \leq \theta \leq 360$
$\log_{2} (3 \sin \theta) = 2.\log_{2}(-3\cos \theta) + 1$
$\log_{2} 3 + \log_{2} \sin \theta = \log_{2} 9 + \log_{2} \cos ^2 \theta + \log_{2} 2$
$\log_{2} \sin \theta - \log_{2} \cos ^2 \theta = \log_{2} 6$
$\log_{2} \sin ^3 \theta = \log_{2} 6$
How to find the possible angle?
| Clearly we need $\sin\theta>0,\cos\theta<0$
$$\implies90^\circ<\theta<180^\circ$$
Choose $180^\circ-\theta=t$
$$\log_2(3\sin t)=2\log_2(3\cos t)+1=\log_2(2(3\cos t)^2)$$
$$3\sin t=18(1-\sin ^2t)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int \sqrt{\frac{5-x}{x-2}}\,dx$ with two different methods and getting two different results I tried Evaluating $\int \sqrt{\dfrac{5-x}{x-2}}dx$ using two different methods and got two different results.
Getting two different answers when tried using two different methods:-
M-$1$:
$$\int \dfrac{5-x}{\sqrt{\left(5-x\right)\left(x-2\right)}}dx$$
$$\dfrac{1}{2}\int\dfrac{-2x+7}{\sqrt{\left(5-x\right)\left(x-2\right)}}dx+\dfrac{3}{2}\int\dfrac{dx}{\sqrt{\left(5-x\right)\left(x-2\right)}}$$
$$\sqrt{\left(5-x\right)\left(x-2\right)}+\dfrac{3}{2}\int\dfrac{dx}{\sqrt{-x^2+7x-10}}$$
$$\sqrt{\left(5-x\right)\left(x-2\right)}+\dfrac{3}{2}\int\dfrac{dx}{\sqrt{\left(\dfrac{3}{2}\right)^2-\left(x-\dfrac{7}{2}\right)^2}}$$
$$\sqrt{\left(5-x\right)\left(x-2\right)}+\dfrac{3}{2}\sin^{-1}\dfrac{x-\dfrac{7}{2}}{\dfrac{3}{2}}$$
$$\sqrt{\left(5-x\right)\left(x-2\right)}+\dfrac{3}{2}\sin^{-1}\dfrac{2x-7}{3}$$
M-$2$:
$$x=5\sin^2\theta+2\cos^2\theta$$
$$dx=\left(10\sin\theta\cos\theta-4\cos\theta\sin\theta\right) \, d\theta$$
$$\int \sqrt{\dfrac{5\cos^2\theta-2\cos^2\theta}{5\sin^2\theta-2\sin^2\theta}}\cdot6\sin\theta\cos\theta \,d\theta$$
$$6\int \cos^2\theta \,d\theta$$
$$\int 3\left(1+\cos2\theta\right) \,d\theta$$
$$3\left(\theta+\dfrac{\sin2\theta}{2}\right)$$
$$3\theta+\dfrac{3}{2}\sin2\theta$$
$$x=5\sin^2\theta+2-2\sin^2\theta$$
$$\sin^{-1}\sqrt{\dfrac{x-2}{3}}=\theta$$
$$\cos2\theta=1-2\sin^2\theta$$
$$\cos2\theta=1-2\cdot\dfrac{x-2}{3}$$
$$\cos2\theta=\dfrac{7-2x}{3}$$
$$3\sin^{-1}\sqrt{\dfrac{x-2}{3}}+\dfrac{3}{2}\cdot\dfrac{\sqrt{9-(49+4x^2-28x)}}{3}$$
$$3\sin^{-1}\sqrt{\dfrac{x-2}{3}}+\sqrt{\left(5-x\right)\left(x-2\right)}$$
In first and second method I am getting the different results of $\dfrac{3}{2}\sin^{-1}\dfrac{2x-7}{3}$ and $3\sin^{-1}\sqrt{\dfrac{x-2}{3}}$ respectively. I checked that these are not inter-convertible. Why am I getting this difference?
| Shown below is that the two results differ by a constant
$-\frac{3\pi}4$.
Define $f(x)$ as the difference of the two results
$$f(x)=\frac32\sin^{-1}\frac{2x-7}3-3\sin^{-1}\sqrt{\frac{x-2}3}$$
where $2<x\le5$. Then, evaluate
$$f’(x) = \frac3{2\sqrt{(5-x)(x-2)}}- \frac3{2\sqrt{(5-x)(x-2)}}=0$$
Thus, $f(x)$ is a constant over $(2,5]$ and can be evaluated with
$$f(x)=f(5)=\frac32\cdot \sin^{-1} 1 -3\sin^{-1}1=-\frac{3\pi}4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Finding the volume of the tetrahedron with vertices $(0,0,0)$, $(2,0,0)$, $(0,2,0)$, $(0,0,2)$. I get $8$; answer is $4/3$. The following problem is from the 7th edition of the book "Calculus and Analytic Geometry Part II". It can be found in section 13.7. It is
problem number 5.
Find the volume of the tetrahedron whose vertices are the given points:
$$ ( 0, 0, 0 ), ( 2, 0, 0 ), ( 0, 2, 0 ), ( 0, 0, 2 ) $$
Answer:
In this case, the tetrahedron is a parallelepiped object. If the bounds of such an object is given by the vectors $A$, $B$ and $C$ then
the area of the object is $A \cdot (B \times C)$. Let $V$ be the volume we are trying to find.
\begin{align*}
x^2 &= 6 - y^2 - z^2 \\[4pt]
A &= ( 2, 0, 0) - (0,0,0) = ( 2, 0, 0) \\
B &= ( 0, 2, 0) - (0,0,0) = ( 0, 2, 0) \\
C &= ( 0, 0, 2) - (0,0,0) = ( 0, 0, 2) \\[4pt]
V &= \begin{vmatrix}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3 \\
\end{vmatrix} =
\begin{vmatrix}
2 & 0 &0 \\
0 & 2 & 0 \\
0 & 0 & 2\\
\end{vmatrix} \\
&= 2 \begin{vmatrix}
2 & 0 \\
0 & 2\\
\end{vmatrix} = 2(4 - 0) \\
&= 8
\end{align*}
However, the book gets $\frac{4}{3}$.
| Referring to this, now you must divide it by $3!$ to get the final answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate $\int \sqrt{\frac{\sin(x-a)}{\sin(x+a)}}dx$ $$\int \sqrt{\dfrac{\sin(x-a)}{\sin(x+a)}}dx$$
My attempt is as follows:-
$$\int \sqrt{\dfrac{\sin(x-a)\sin(x-a)}{\sin(x+a)\sin(x-a)}}dx$$
$$\int \sin(x-a)\sqrt{\dfrac{1}{\sin^2x-\sin^2a}}dx$$
Integrating by parts-
$$-\cos(x-a)\sqrt{\dfrac{1}{\sin^2x-\sin^2a}}+\int \cos(x-a)\dfrac{d}{dx}\left(\dfrac{1}{\sqrt{\sin^2x-\sin^2a}}\right)dx$$
$$-\cos(x-a)\sqrt{\dfrac{1}{\sin^2x-\sin^2a}}+\int \cos(x-a)\left(-\dfrac{\sin x\cos x}{(\sin^2x-\sin^2a)^\frac{3}{2}}\right)dx$$
$$-\cos(x-a)\sqrt{\dfrac{1}{\sin^2x-\sin^2a}}-\int \cos x\cos a\left(\dfrac{\sin x\cos x}{(\sin^2x-\sin^2a)^\frac{3}{2}}\right)dx-\int \sin x\sin a\left(\dfrac{\sin x\cos x}{(\sin^2x-\sin^2a)^\frac{3}{2}}\right)dx$$
$$-\cos(x-a)\sqrt{\dfrac{1}{\sin^2x-\sin^2a}}-\cos a\int\dfrac{\sin x\cos^2 x}{(\cos^2a-\cos^2x)^\frac{3}{2}}dx-\sin a\int \dfrac{\sin^2x\cos x}{(\sin^2x-\sin^2a)^\frac{3}{2}}dx$$
$$\cos x=t$$
$$-\sin x=\dfrac{dt}{dx}$$
$$-\cos(x-a)\sqrt{\dfrac{1}{\sin^2x-\sin^2a}}+\cos a\int \dfrac{t^2}{\left(\cos^2a-t^2\right)^\frac{3}{2}}dt-\sin a\int \dfrac{y^2}{\left(y^2-\sin^2a\right)^\frac{3}{2}}dy$$
$$I=\cos a\int \dfrac{t^2}{\left(\cos^2a-t^2\right)^\frac{3}{2}}dt$$
$$t=\cos a\cos z$$
$$\dfrac{dt}{dz}=-\cos a\sin z$$
$$I=-\int \dfrac{\cos^4a\cos^2z\sin z}{\left(\cos^2a-\cos^2a\cos^2z\right)^\frac{3}{2}}dz$$
$$I=-\int \dfrac{\cos^4a\cos^2z\sin z}{\cos^3a\sin^3z}dz$$
$$I=-\cos a\int \cot^2z$$
$$I=-\cos a\int (\mathrm{cosec}^2z-1) dz$$
$$I=\cos a\cot z+z\cos a$$
$$I_1=-\sin a\int \dfrac{y^2}{\left(y^2-\sin^2a\right)^\frac{3}{2}}dy$$
$$y=\sin a\sec u$$
$$\dfrac{dy}{du}=\sin a\sec u\tan u$$
$$I_1=-\sin a\int \dfrac{\sin^3a\sec^3u\tan u}{(\sin^2a\tan^2u)^\frac{3}{2}}du$$
$$I_1=-\sin a\int \dfrac{\sec^3u}{\tan^2u}du$$
$$I_1=-\sin a\int \dfrac{\sec u}{\sin^2u}du$$
$$\sin u=v$$
$$\cos u=\dfrac{dv}{du}$$
$$du=\dfrac{dv}{\sqrt{1-v^2}}$$
$$-\sin a\int \dfrac{1}{v^2(1-v^2)}dv$$
$$\sin a\int \dfrac{v^2-(v^2-1)}{v^2(v^2-1)}dv$$
$$\sin a\left(\dfrac{1}{2}\ln\left|\dfrac{v-1}{v+1}\right|+\dfrac{1}{v}\right)$$
So integral would be
$$-\cos(x-a)\sqrt{\dfrac{1}{\sin^2x-\sin^2a}}+\cos a\cot z+z\cos a+\sin a\left(\dfrac{1}{2}\ln\left|\dfrac{v-1}{v+1}\right|+\dfrac{1}{v}\right)+C$$
$$-\cos(x-a)\sqrt{\dfrac{1}{\sin^2x-\sin^2a}}+\cos a\dfrac{t}{\sqrt{\cos^2a-t^2}}+\cos a\cos^{-1}\left(\dfrac{t}{\cos a}\right)+\dfrac{\sin a}{2}\cdot\ln\left|\dfrac{\sin u-1}{\sin u+1}\right|+\dfrac{\sin a}{\sin u}+C$$
$$-\cos(x-a)\sqrt{\dfrac{1}{\sin^2x-\sin^2a}}+\cos a\dfrac{\cos x}{\sqrt{\cos^2a-\cos^2x}}+\cos a\cos^{-1}\left(\dfrac{\cos x}{\cos a}\right)+\dfrac{\sin a}{2}\cdot\ln\left|\dfrac{\sqrt{y^2-\sin^2a}-y}{\sqrt{y^2-\sin^2a}+y}\right|+\dfrac{y\sin a}{\sqrt{y^2-\sin^2a}}+C$$
$$-\cos(x-a)\sqrt{\dfrac{1}{\sin^2x-\sin^2a}}+\cos a\dfrac{\cos x}{\sqrt{\cos^2a-\cos^2x}}+\cos a\cos^{-1}\left(\dfrac{\cos x}{\cos a}\right)+\dfrac{\sin a}{2}\cdot\ln\left|\dfrac{\sqrt{\sin^2x-\sin^2a}-\sin x}{\sqrt{\sin^2x-\sin^2a}+\sin x}\right|+\dfrac{\sin x\sin a}{\sqrt{\sin^2x-\sin^2a}}+C$$
$$\cos a\cos^{-1}\left(\dfrac{\cos x}{\cos a}\right)+\dfrac{\sin a}{2}\cdot\ln\left|\dfrac{\sqrt{\sin^2x-\sin^2a}-\sin x}{\sqrt{\sin^2x-\sin^2a}+\sin x}\right|+C$$
First of all, it went very long, what clever way I am missing here? Please feel free to suggest any other shorter alternatives
| $$\int\frac{\sin(x-a)}{\sqrt{\sin^2x-\sin^2a}}dx$$
$$=\cos a\int\frac{\sin x}{\sqrt{\cos^2a-\cos^2x}}dx-\sin a\int\frac{\cos x}{\sqrt{\sin^2x-\sin^2a}}dx$$
Now, substitute $\sin x$ in second integral and $\cos x$ in first integral to reduce to standard formulas.
| {
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"url": "https://math.stackexchange.com/questions/3489684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the determinant of a matrix given by three parameters.
Show that for $a,b,c\in\mathbb R$ $$\begin{vmatrix}b^2+c^2&ab&ac\\ba&c^2+a^2&bc\\ca&cb&a^2+b^2\end{vmatrix} = \begin{vmatrix}0&c&b\\c&0&a\\b&a&0\end{vmatrix}^2 = 4a^2b^2c^2. $$
There must be some trick, like using elementary row operations, to get the determinant into that form, but I am not seeing it. And directly computing the determinant by the cofactor expansion looks very nasty. So is there a simpler way to compute this determinant?
| Use the rule of Sarrus to show
$\begin{vmatrix}0&c&b\\c&0&a\\b&a&0\end{vmatrix}^2 = (2abc)^2$ and then show that $A^2 :=\begin{pmatrix}0&c&b\\c&0&a\\b&a&0\end{pmatrix}^2=\begin{pmatrix}b^2+c^2&ab&ac\\ba&c^2+a^2&bc\\ca&cb&a^2+b^2\end{pmatrix}$ and then use $\det(A*A)=\det(A)*\det(A)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\int \frac{1}{(1+x^4)\sqrt{\sqrt{1+x^4}-x^2}}dx$ I tried substitution $x=\dfrac{1}{t}$, then $z=t^2$, tried rationalizing the denominator but haven't been able to pull it off. I can't think of any trig substitution either. I prefer an intuitive approach rather than an "plucked-out-of-thin-air" counter-intuitive substitution.
| Rewrite the integral as
$$\int \frac{x\:dx}{x^2(1+x^4)\sqrt{\frac{\sqrt{1+x^2}-x^2}{x^2}}}$$
by pulling out an $x^2$ from the square root then multiplying the top and bottom by $x$.
Now let $x^2 = \sinh( t) \implies xdx = \frac{1}{2}\cosh(t) dt$:
$$\int \frac{\left(\frac{1}{2}\cosh(t)\right)\:dt}{\sinh(t)(1+\sinh^2(t))\sqrt{\frac{\sqrt{1+\sinh^2(t)}-\sinh(t)}{\sinh(t)}}} = \int \frac{\sqrt{e^t\sinh(t)}\:dt}{\sinh(2t)} = \sqrt{2}\int \frac{\sqrt{e^{2t}-1}}{e^{2t}-e^{-2t}}dt$$
Now let $e^t = \sec\theta \implies dt = \tan\theta \: d\theta$:
$$\sqrt{2}\int \frac{\sqrt{\sec^2\theta - 1}\tan\theta}{\sec^2\theta - \cos^2\theta}\:d\theta = \sqrt{2}\int \frac{\sin^2\theta}{1-\cos^4\theta}\: d\theta = \int \frac{\sqrt{2}\:d\theta}{1+\cos^2\theta}$$
$$= \int \frac{\sqrt{2}\:d\theta}{2\cos^2\theta+\sin^2\theta} = \int \frac{\sqrt{2}\sec^2\theta \:d\theta}{2 + \tan^2\theta} = \tan^{-1}\left(\frac{\tan\theta}{\sqrt{2}}\right) + C$$
Now undoing the substitutions,
$$\tan\theta = \sqrt{\sec^2\theta - 1} = \sqrt{e^{2t}-1} = \sqrt{2e^t \sinh(t)}$$
which can go in one of two directions, either
$$\sqrt{2(\cosh(t)+\sinh(t))\sinh(t)} = \sqrt{2(\sqrt{1+\sinh^2(t)}+\sinh(t))\sinh(t)} = x \sqrt{2(\sqrt{1+x^4}+x)}$$
$$\implies I = \tan^{-1}\left(x\sqrt{\sqrt{1+x^4}+x^2} \right) + C$$
or
$$\sqrt{\frac{2\sinh(t)}{\cosh(t)-\sinh(t)}} = \sqrt{\frac{2\sinh(t)}{\sqrt{1+\sinh^2(t)}-\sinh(t)}} = x\sqrt{\frac{2}{\sqrt{1+x^4}-x^2}}$$
$$\implies I = \tan^{-1}\left(\frac{x}{\sqrt{\sqrt{1+x^4}-x^2}} \right) + C$$
the latter of which matches wolfram's output, but both are equivalent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Inequality $\frac{x^3}{x^2+y^2}+\frac{y^3}{y^2+z^2}+\frac{z^3}{z^2+x^2} \geqslant \frac{x+y+z}{2}$ Help to prove this Inequality:
If x,y,z are postive real numbers then:
$\dfrac{x^3}{x^2+y^2}+\dfrac{y^3}{y^2+z^2}+\dfrac{z^3}{z^2+x^2} \geqslant \dfrac{x+y+z}{2}$
I tied to use analytic method with convex function but no result:
Since $f(x)=\frac{1}{x}$ is a convex function, by Jensen we obtain:
$$\frac{1}{x+y+z}\sum_{cyc}\frac{x^3}{x^2+y^2}=\sum_{cyc}\left(\frac{x}{x+y+z}\cdot\frac{1}{\frac{x^2+y^2}{x^2}}\right)\geq$$
$$\geq\frac{1}{\sum\limits_{cyc}\left(\frac{x}{x+y+z}\cdot\frac{x^2+y^2}{x^2}\right)}=\frac{x+y+z}{\sum\limits_{cyc}\left(x+\frac{y^2}{x}\right)}.$$
Thus, it's enough to prove that
$$\frac{x+y+z}{\sum\limits_{cyc}\left(x+\frac{y^2}{x}\right)}\geq\frac{1}{2}$$ or
$$x+y+z\geq\sum_{cyc}\frac{y^2}{x},$$ which is wrong.
thanks
| $$\sum_{cyc}\frac{x^3}{x^2+y^2}-\frac{x+y+z}{2}=\sum_{cyc}\left(\frac{x^3}{x^2+y^2}-\frac{x}{2}\right)=\sum_{cyc}\frac{x^3-xy^2}{2(x^2+y^2)}=$$
$$=\sum_{cyc}\left(\frac{x^3-xy^2}{2(x^2+y^2)}-\frac{x-y}{2}\right)=\sum_{cyc}\frac{y(x-y)^2}{2(x^2+y^2)}\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Reducing an inequality to Schur's inequality Given this,
$a\left(a+b\right)\left(a+c\right)+b\left(a+b\right)\left(b+c\right)+c\left(a+c\right)\left(b+c\right)\ge 2ab\left(a+b\right)+2bc\left(b+c\right)+2ac\left(c+a\right)$
how do you obtain Schur's inequality?
It is suppose to reduce to the nice form of Schur's inequality with $t=1$:
$a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)\ge0$
Seems trivial but I can't figure it out.
| RHS can be rearranged as $$2ab\left(a+b\right)+2bc\left(b+c\right)+2ac\left(c+a\right)$$
$$=2a^2(b+c)+2b^2(c+a)+2c^2(a+b)$$
Now, observe that $$a(a+b)(a+c)-2a^2(b+c)=a(a-b)(a-c)$$
and you are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $\alpha, \beta,\gamma$ be the roots of $2x^3+x^2+x+1=0$, show that..........? If $\alpha, \beta,\gamma$ be the roots of $2x^3+x^2+x+1=0$, show that:
$$(\frac{1}{\beta^3}+\frac{1}{\gamma^3}-\frac{1}{\alpha^3})(\frac{1}{\gamma^3}+\frac{1}{\alpha^3}-\frac{1}{\beta^3})(\frac{1}{\alpha^3}+\frac{1}{\beta^3}-\frac{1}{\gamma^3})=16$$
Here's what I have tried,
By Vieta's rule
$\alpha+\beta+\gamma=\frac{-1}{2}\text{. ...........}(1)$
$\alpha \beta+\beta \gamma+\alpha \gamma=\frac{1}{2}\text{. ...........}(2)$
$\alpha \beta \gamma=\frac{-1}{2}\text{. ...........}(3)$
Squaring $(1)$,
$\alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\beta\gamma+\alpha\gamma)=\frac{1}{4}\text{. ...........}(4)$
From $(2)$,
$\alpha^2+\beta^2+\gamma^2=\frac{-3}{4}\text{. ...........}(5)$
Putting the roots and adding these equations,
$2\alpha^3+\alpha^2+\alpha+1=0$
$2\beta^3+\beta^2+\beta+1=0$
$2\gamma^3+\gamma^2+\gamma+1=0$
We get,
$2(\alpha^3+\beta^3+\gamma^3)+(\alpha^2+\beta^2+\gamma^2)+(\alpha+\beta+\gamma)+3=0$
Putting the values,
$2(\alpha^3+\beta^3+\gamma^3)+\frac{-3}{4}+\frac{-1}{2}+3=0$
$(\alpha^3+\beta^3+\gamma^3)=\frac{-7}{8}$
Then I divided $2x^3+x^2+x+1=0$ by $x$ and I found out $(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma})$ the same way I found out $(\alpha^3+\beta^3+\gamma^3)$. Likewise doing the same, finding $\sum\frac{1}{\alpha^2}$ then $\sum\frac{1}{\alpha^3}$ I found out
$\sum\frac{1}{\alpha^3}=-4$
But still I'm far from the answer, also the $-$ sign is creating problems.
Any help would be highly appreciated
| Hint:
$$-(2x^3+1)^3=(x^2+x)^3$$
$$-8(x^3)^3-12(x^3)^2-6(x^3)-1=(x^3)^2+x^3+3x^3(-(2x^3+1))$$
Replace $x^3$ with $y$ to form a cubic equation in $y$ whose roots are $\alpha^3=a$ etc.
Let $z=\dfrac1a+\dfrac1b+\dfrac1c-\dfrac2c=\dfrac{ab+bc+ca}{abc}-\dfrac2c$
We can find the values of $abc, ab+bc+ca$ from the cubic equation in $y$
$\dfrac2c=?$
As $c$ is a root of the new cubic equation in $y$, replace the value of $c$ to form a cubic equation in $z$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Proving the convergence of an infinite product I’m trying to prove the Taylor series of $e$ using binomial expansion:
$$e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$$
The steps I’ve tried so far are:
$$\left(1+\frac{1}{n}\right)^n \\
= 1 + \binom{n}{1}\frac{1}{n} + \binom{n}{2}\frac{1}{n^2} + \binom{n}{3}\frac{1}{n^3} + \dots + \binom{n}{n}\frac{1}{n^n} \\
= 1+\dfrac {n}{1!}\left( \dfrac {1}{n}\right) +\dfrac {n\left( n-1\right) }{2!}\left( \dfrac {1}{n^{2}}\right) + \dfrac {n\left( n-1\right) \left( n-2\right) }{3!}\left( \dfrac {1}{n^{3}}\right) +\dots +\dfrac {n\left( n-1\right) \left( n-2\right) \dots 1}{n!}\left( \dfrac {1}{n^{n}}\right) \\
= 1+\dfrac {1}{1!}+\dfrac {1}{2!}\left( 1-\dfrac {1}{n}\right) +\dfrac {1}{3!}\left( 1-\dfrac {1}{n}\right) \left( 1-\dfrac {2}{n}\right) +\ldots +\dfrac {1}{n!}\left( 1-\dfrac {1}{n}\right) \left( 1-\dfrac {2}{n}\right) \dots
\left( 1-\dfrac {n-1}{n}\right) $$
As $n \to \infty$, the last term becomes
$$\lim_{n\to\infty} \left(\frac{1}{n!} \prod_{k=1}^n \left(1-\frac{k-1}{n}\right)\right)$$ and the infinite product on the right of the term should be proven to be convergent and equals to $1$. So how do I prove that?
| $$\prod_{k=1}^n\left(1-\frac{k-1}{n}\right)=\prod_{k=1}^n\frac{1}{n}\left(n-k+1\right)=\frac{1}{n^n}\big[n(n-1)\cdots(1)\big]=\frac{n!}{n^n}$$
Hence, we have:
$$\lim_{n\to\infty}\frac{1}{n!}\prod_{k=1}^n\left(1-\frac{k-1}{n}\right)=\lim_{n\to\infty}\frac{n!}{n^n\cdot n!}=\lim_{n\to\infty}\frac{1}{n^n}=0$$
This should be expected, as if the last term approached any other value then the terms of the summation would not approach zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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If both roots of the equation $ax^2-2bx+5=0$ are $\alpha$ and roots of the equation $x^2-2bx-10=0$ are $\alpha$ and $\beta$.
find $\alpha^2+\beta^2$
Both equations have a common root
$$(-10a-5)^2=(-2ab+2b)(20b+10b)$$
$$25+100a^2+100a=60b^2(1-a)$$
Also since the first equation has equal roots
$$4b^2-20a=
0$$
$$b^2=5a$$
I could substitute the value of a in the above equation, but that gives me a biquadractic equation in b, and I don’t think it’s supposed to go that way. What am I doing wrong?
| $b^2=5a$, then $\alpha=b/a$
Next we have $\alpha+ \beta=2b$ and $\alpha \beta =-10$
Then $$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha \beta=4b^2+20=20a+20 ~~~(*)$$
Next we have $$\beta=\frac{-10}{\alpha}=2b-\alpha \implies -\frac{10a}{b}=2b-\frac{b}{a}$$
$$\implies -10a^2=2ab^2-b^2 \implies -20a^2=-5a \implies a=\frac{1}{4},~ as ~a\ne 0$$
Finally from (*) we get $\alpha^2+\beta^2=25.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3504014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the maximum number of draws in a tournament. Consider a round robin tournament between 8 teams. A win gives 3 points, tie gives 1 point (to both teams) and loss gives 0 points.
If the match between two teams, say A & B, is a tie, then A & B can't end up with the same number points after the tournament.
The task is to find the maximum number of ties in the tournament.
I have no idea other than bashing all the cases. Please help me how to do it?
| Start by considering the possible outcomes for a player. All players play seven other players and they can tie, win or lose games.
This table shows the score achieved in different situations:
\begin{array}{|c|c|c|c|c|}
\hline
&Ties & Wins & Loses & Score \\ \hline
A&7& 0& 0&7\\ \hline
B&6& 0 &1 &6\\ \hline
C&6& 1 &0 &9\\ \hline
D&5& 0 &2 &5\\ \hline
E&5& 1 &1 &8\\ \hline
F&5& 2 &0 &11\\ \hline
G&4& 0 &3 &4\\ \hline
H&4& 1 &2 &7\\ \hline
I&4& 2 &1 &10\\ \hline
J&4& 3 &0 &13\\ \hline
K&3& 0 &4 &3\\ \hline
L&3& 1 &3 &6\\ \hline
M&3& 2 &2 &9\\ \hline
N&3& 3 &1 &12\\ \hline
O&3& 4 &0 &15\\ \hline
& & & &\\ \hline
& & & &\\ \hline
\end{array}
I could go further but there is no need because we want to maximise the number of ties.
We now want to choose 7 of these outcomes that yield seven different scores. We also want to pick outcomes from nearer the top of the table so that we maximise the number of ties.
So it's obvious we want outcomes A, B, C, D, E and F four our first six players. As well as having lots of ties, their scores are all different to each other.
Now we need to pick two more outcomes for the remaining players.
We can choose outcome G but not outcome H (because we already have an outcome with score of 7). We can choose outcome I instead.
Now we have eight different outcomes for our eight players... or do we?
Let's list the outcomes we have and sum our columns:
\begin{array}{|c|c|c|c|c|}
\hline
&Ties & Wins & Loses & Score \\ \hline
A&7& 0& 0&7\\ \hline
B&6& 0 &1 &6\\ \hline
C&6& 1 &0 &9\\ \hline
D&5& 0 &2 &5\\ \hline
E&5& 1 &1 &8\\ \hline
F&5& 2 &0 &11\\ \hline
G&4& 0 &3 &4\\ \hline
I&4& 2 &1 &10\\ \hline
& & & &\\ \hline
Total& 42& 6 & 8&\\ \hline
\end{array}
Can you see the problem? There are more players losing than winning, but surely these should be equal!
Choosing outcome J instead of outcome I fixes that problem:
\begin{array}{|c|c|c|c|c|}
\hline
&Ties & Wins & Loses & Score \\ \hline
A&7& 0& 0&7\\ \hline
B&6& 0 &1 &6\\ \hline
C&6& 1 &0 &9\\ \hline
D&5& 0 &2 &5\\ \hline
E&5& 1 &1 &8\\ \hline
F&5& 2 &0 &11\\ \hline
G&4& 0 &3 &4\\ \hline
J&4& 3 &0 &13\\ \hline
& & & &\\ \hline
Total& 42& 7 & 7&\\ \hline
\end{array}
That gives us a feasible set of outcomes and shows that the maximum possible number of ties is 42.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3504877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Help in summing: $\sum_{k=m} ^{2m} ~k\cdot ~2^{-k} {k \choose m}$ While solving an interesting problem I require two sums
$$S_1=\sum_{k=m}^{2m} 2^{-k} {k \choose m} ~ \text{and} ~ S_2=\sum_{k=m}^{2m} k \cdot ~ 2^{-k} {k \choose m}$$
The former is know to be unity see inside the solutions of
How to prove that $\sum_{i=0}^n 2^i\binom{2n-i}{n} = 4^n$.
However, here I require your help in finding $S_2$ by hand.
| Starting from
$$\sum_{k=m}^{2m} k 2^{-k} {k\choose m}$$
we have as in the answer that was first to appear
$$\sum_{k=m}^{2m} (k+1) 2^{-k} {k\choose m}
- \sum_{k=m}^{2m} 2^{-k} {k\choose m}
\\ = (m+1) \sum_{k=m}^{2m} 2^{-k} {k+1\choose m+1}
- \sum_{k=m}^{2m} 2^{-k} {k\choose m}.$$
For the first sum term we get
$$(m+1) [z^{m+1}] (1+z) \sum_{k=m}^{2m} 2^{-k} (1+z)^k
\\ = (m+1) 2^{-m} [z^{m+1}] (1+z)^{m+1}
\sum_{k=0}^{m} 2^{-k} (1+z)^k
\\ = (m+1) 2^{-m} [z^{m+1}] (1+z)^{m+1}
\frac{1-((1+z)/2)^{m+1}}{1-(1+z)/2}
\\ = (m+1) \frac{1}{2^{m-1}} [z^{m+1}] (1+z)^{m+1}
\frac{1-((1+z)/2)^{m+1}}{1-z}.$$
This is
$$(m+1) \frac{1}{2^{m-1}}
\sum_{k=0}^{m+1} {m+1\choose k}
- (m+1) \frac{1}{2^{2m}}
\sum_{k=0}^{m+1} {2m+2\choose k}
\\ = 4(m+1) - (m+1) \frac{1}{2^{2m}}
\left(\frac{1}{2} 2^{2m+2} + \frac{1}{2}{2m+2\choose m+1}\right)
\\ = 2(m+1) - (m+1) \frac{1}{2^{2m+1}} {2m+2\choose m+1}.$$
The second one is very similar:
$$[z^{m}] \sum_{k=m}^{2m} 2^{-k} (1+z)^k
\\ = \frac{1}{2^m} [z^{m}] (1+z)^m \sum_{k=0}^{m} 2^{-k} (1+z)^k
\\ = \frac{1}{2^m} [z^{m}] (1+z)^m
\frac{1-((1+z)/2)^{m+1}}{1-(1+z)/2}
\\ = \frac{1}{2^{m-1}} [z^{m}](1+z)^m
\frac{1-((1+z)/2)^{m+1}}{1-z}$$
This is
$$\frac{1}{2^{m-1}} 2^m
- \frac{1}{2^{2m}} \sum_{k=0}^m {2m+1\choose k}
= 2 - \frac{1}{2^{2m}} \frac{1}{2} 2^{2m+1}
= 2 - 1 = 1.$$
Collecting everything we find
$$\bbox[5px,border:2px solid #00A000]{
2m+1 - \frac{m+1}{2^{2m}} {2m+1\choose m}.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given positives $a, b, c$, prove that $\frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$.
Given positives $a, b, c$, prove that $$\large \frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$$
Let $x = \dfrac{b + c}{2}, y = \dfrac{c + a}{2}, z = \dfrac{a + b}{2}$
It needs to be sufficient to prove that $$\sum_{cyc}\frac{\dfrac{a + b}{2} + \dfrac{b + c}{2} - \dfrac{c + a}{2}}{\left(2 \cdot \dfrac{c + a}{2}\right)^2} \ge \frac{9}{\displaystyle 4 \cdot \sum_{cyc}\dfrac{c + a}{2}} \implies \sum_{cyc}\frac{z + x - y}{y^2} \ge \frac{9}{y + z + x}$$
According to the Cauchy-Schwarz inequality, we have that
$$(y + z + x) \cdot \sum_{cyc}\frac{z + x - y}{y^2} \ge \left(\sum_{cyc}\sqrt{\frac{z + x - y}{y}}\right)^2$$
We need to prove that $$\sum_{cyc}\sqrt{\frac{z + x - y}{y}} \ge 3$$
but I don't know how to.
Thanks to Isaac YIU Math Studio, we additionally have that $$(y + z + x) \cdot \sum_{cyc}\frac{z + x - y}{y^2} = \sum_{cyc}(z + x - y) \cdot \sum_{cyc}\frac{z + x - y}{y^2} \ge \left(\sum_{cyc}\frac{z + x - y}{y}\right)^2$$
We now need to prove that $$\sum_{cyc}\frac{z + x - y}{y} \ge 3$$, which could be followed from Nesbitt's inequality.
I would be greatly appreciated if there are any other solutions than this one.
| Recall Nesbitt's inequality
\begin{eqnarray*}
\frac{a}{b+c}+\frac{b}{c+a} + \frac{c}{a+b} \geq \frac{3}{2}.
\end{eqnarray*}
Using Cauchy-Schwartz and Nesbitt gives
\begin{eqnarray*}
&\left( 2 \cdot \frac{a}{b+c}+1 \right)^2 +\left( 2 \cdot \frac{b}{c+a}+1 \right)^2 + \left(2 \cdot \frac{c}{a+b} +1 \right)^2\\ &\geq
\frac{1}{3}\left( 2 \cdot \frac{a}{b+c}+2 \cdot \frac{b}{c+a} + 2 \cdot \frac{c}{a+b} +3 \right)^2 \geq 12.
\end{eqnarray*}
And this can be rearranged to give the inequality.
Edit: In light of Issac's answer ... By Cauchy-Schwartz,
\begin{eqnarray*}
\left(\dfrac{a}{(b + c)^2} + \dfrac{b}{(c + a)^2} + \dfrac{c}{(a + b)^2}\right)(a+b+c) \geq \left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2
\end{eqnarray*}
and the result now follows by Nesbitt's inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Difference in choosing some generator of principal ideal for solving $x^2+d=y^3$ So in solving $E : x^2 + 19 = y^3$ in the integers I determined that $(x + \sqrt{-19}) = I^3$ for some ideal in $$\mathscr{O}_K = \mathbb{Z}\left[\frac{1 + \sqrt{-19}}{2}\right]$$ and since the class number of $K$ is not $3$, that $I$ must be principal.
So what I did was $$\left(a + b \frac{1 + \sqrt{-19}}{2}\right)^3 = x + \sqrt{-19},$$ but the equation I ended up with doesn't have solutions: $b(3a^2+3ab-4b^2)=1$ doesn't have integral solutions in $a$ and $b$.
However, by solving $$\left(\frac{a + b\sqrt{-19}}{2}\right)^3 = x + \sqrt{-19}$$ I got the equation $b(3a^2 - 19b^2) = 8$ which gave me the desired solutions to $E$, namely $(x, y) = (\pm 18, 7)$.
How could I have known which generator for $I$ I should have chosen to get to the right (and only) solutions?
My confusion is in the fact that $$\mathbb{Z}\left[\frac{1 + \sqrt{-19}}{2}\right] = \mathbb{Z} \oplus \mathbb{Z}\frac{1 + \sqrt{-19}}{2},$$ so I thought that my first try would simply do the trick, but it didn't.
| To solve $$\left(a+b\frac{1+\sqrt{-19}}{2}\right)^3=x+\sqrt{-19}$$ for integers $a,b$ you should express the right-hand side in terms of the integral basis $1, \frac{1+\sqrt{-19}}{2}$ as well: $$x+\sqrt{-19} = (x-1)+2\frac{1+\sqrt{-19}}{2}.$$
(You could also equate coefficients w.r.t the $\mathbb{Q}$-basis $1,\sqrt{-19}$, but then you equate half-integers.)
(To work efficiently with $\frac{1+\sqrt{-19}}{2}$, use its minimal polynomial $x^2 - x + 5$.)
Equating the coefficients w.r.t. the $\mathbb{Z}$-basis $1, \frac{1+\sqrt{-19}}{2}$, you get $$a^3 - 15ab^2 - 5b^3 = x-1 \qquad \text{and}\qquad b(3a^2 + 3ab - 4b^2) = \mathbf{2}.$$
The second equation does have solutions, namely $(a,b)=(1,1)$ and $(a,b)=(-2,1)$; this leads to $x = \pm 18$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3510541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Show: For every $\alpha \in \mathbb{Q}$, the polynomial $X^2+\alpha$ has endless different roots in $\mathbb{Q}^{2\times2}$.
Show that for every $\alpha \in \mathbb{Q}$, the polynomial $X^2+\alpha$ has endless different roots in $\mathbb{Q}^{2\times2}$.
What I did:
Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathbb{Q}^{2\times2}, f=X^2+\alpha \in \mathbb{Q}[X].$
Consider $$f(A) = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix}+\begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix} = \\ \begin{pmatrix} a^2+bc & ab+bd \\ ac+cd & bc+d^2 \end{pmatrix} + \begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix} = \\\begin{pmatrix} a^2+bc + \alpha & ab+bd \\ ac+cd & bc+d^2 + \alpha \end{pmatrix}.$$
How do I get to the claim from here on?
Thanks in advance!
| By definition,
$\alpha \in \Bbb Q \Longrightarrow \exists p, q \in \Bbb Z, \; \alpha = \dfrac{p}{q} = pq^{-1}; \tag 1$
set
$Y = \begin{bmatrix} 0 & -p \\ q^{-1} & 0 \end{bmatrix} ; \tag 2$
then
$Y^2 = \begin{bmatrix} 0 & -p \\ q^{-1} & 0 \end{bmatrix} \begin{bmatrix} 0 & -p \\ q^{-1} & 0 \end{bmatrix}$
$= \begin{bmatrix} -pq^{-1} & 0 \\ 0 & -pq^{-1} \end{bmatrix} = \begin{bmatrix} -\alpha & 0 \\ 0 & -\alpha \end{bmatrix} = -\alpha I, \tag 3$
that is,
$Y^2 + \alpha I = 0; \tag 4$
thus $Y$ satisfies
$X^2 + \alpha \in \Bbb Q[X]; \tag 5$
note further for
$S \in \Bbb Q^{2 \times 2}, \tag 6$
with $S$ invertible, that
$(S^{-1}YS)(S^{-1}YS) = S^{-1}Y(SS^{-1})YS = S^{-1}YIYS$
$= S^{-1}Y^2S = S^{-1}(-\alpha I)S = -\alpha S^{-1}IS = -\alpha I, \tag 7$
or
$(S^{-1}YS)^2 + \alpha I = 0; \tag 8$
thus $S^{-1}YS$ also satisfies (5).
To see that there are an infinite number of such matrices, let
$S = \begin{bmatrix} s & 0 \\ 0 & 1 \end{bmatrix}, \;0 \ne s \in \Bbb Q; \tag 9$
we have
$S^{-1}YS = \begin{bmatrix} s^{-1} & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 0 & -p \\ q^{-1} & 0 \end{bmatrix} \begin{bmatrix} s & 0 \\ 0 & 1 \end{bmatrix}$
$= \begin{bmatrix} s^{-1} & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 0 & -p \\ q^{-1}s & 0 \end{bmatrix} = \begin{bmatrix} 0 & -ps^{-1} \\ q^{-1}s & 0 \end{bmatrix}, \tag{10}$
which are distinct for distinct values of $s$.
By further adjusting the vaues of $p$ and $q$ even more such matrices may be obtained; for example, every
$\forall p, q \in \Bbb Z, \; \alpha = pq^{-1} \tag{11}$
yields such a matrix $S^{-1}YS$.
| {
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"url": "https://math.stackexchange.com/questions/3511852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$a^3 + b^3 + c^3 = 3abc$ , can this be true only when $a+b+c = 0$ or $a=b=c$, or can it be true in any other case? Since
$$
a^3 + b^3 + c ^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc ),
$$ from this it is clear that if $(a+b+c) = 0$ , then $a^3 + b^3 + c ^3 - 3abc = 0$ and
$a^3 + b^3 + c ^3 = 3abc$ . Also if $a=b=c$ , $a^3 + b^3 + c^3 = a^3 + a^3 +a^3 = 3a^3 = 3aaa = 3abc$ , hence $a^3 + b^3 + c^3 = 3abc$.
But I was wondering if $a^3 + b^3 + c^3 = 3abc$ can be true even when none of the above two relations are true. Please guide me toward a solution.
| By your work:
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$
Thus, $$a^3+b^3+c^3-3abc=0$$ for $a+b+c=0$ or for
$$a^2+b^2+c^2-ab-ac-bc=0,$$ which is
$$(a-b)^2+(a-c)^2+(b-c)^2=0,$$ which gives $$a=b=c.$$
Id est, we have no another cases for equality occurring for real values.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given reals $a_1, a_2, \cdots, a_{n - 1}, a_n$ such that $\sum_{i = 1}^na_1^2 = 1$. Calculate the maximum value of $\sum_{cyc}|a_1 - a_2|$.
Given reals $a_1, a_2, \cdots, a_{n - 1}, a_n$ such that $a_1^2 + a_2^2 + \cdots + a_{n - 1}^2 + a_n^2 = 1$ $(n \in \mathbb N, n \ge 3)$. Calculate the maximum value of $$\large |a_1 - a_2| + |a_2 - a_3| + \cdots + |a_{n - 1} - a_n| + |a_n - a_1|$$
There must exist $1 < k < n$ such that $a_{k - 1} \le a_k \le a_{k + 1}$
$ \implies |a_{k + 1} - a_k| + |a_k - a_{k + 1}| = |a_{k + 1} - a_{k + 1}|$.
Repeat the above process for about $n - 1$ times and we have that $$|a_1 - a_2| + |a_2 - a_3| + \cdots + |a_{n - 1} - a_n| + |a_n - a_1| \le 2 \cdot \min(|a_i - a_j|, 1 \le i < j \le n)$$
Now we just have to find the maximum value of $\min(|a_i - a_j|, 1 \le i < j \le n)$ for $$a_1^2 + a_2^2 + \cdots + a_{n - 1}^2 + a_n^2 = 1$$, which I don't know how.
| The maximal value is $2\sqrt{n-1}$ if $n$ is odd, and $2\sqrt{n}$ if $n$ is even. We can prove the following:
Let $a_1, \ldots, a_n$ be real numbers, $n \ge 2$. Then
$$ \tag{*}
|a_1 - a_2| + |a_2 - a_3| + \ldots + |a_{n - 1} - a_n| + |a_n - a_1|
\le c_n \sqrt{a_1^2 + \ldots + a_n^2}
$$
where $c_n = 2\sqrt{n-1}$ if $n$ is odd, and $c_n = 2\sqrt{n}$ if $n$ is even. The bounds are sharp.
Proof: Case 1: $n$ is even. Then
$$
|a_1 - a_2| + |a_2 - a_3| + \ldots + |a_{n - 1} - a_n| + |a_n - a_1| \\
\underset{(1)}{\le} \sum_{k=1}^n (|a_k| + |a_{k+1}|) = 2 \sum_{k=1}^n (1 \cdot |a_k|)
\underset{(2)}{\le} 2 \sqrt{n} \sqrt{\sum_{i=1}^{n} a_i^2 } \, ,
$$
where the last step uses the Cauchy-Schwarz inequality.
Equality holds at $(1)$ if the $a_k$ have alternating signs, and equality holds at $(2)$ if all $|a_k|$ are equal. It follows that equality holds in $(*)$ exactly if
$$
(a_1, \ldots, a_n) = (x, -x, \ldots, x, -x)
$$
for some $x \in \Bbb R$.
Case 2: $n$ is odd. There must be (at least) one index $k$ such that $a_{k-1} - a_k$ and $a_k - a_{k+1}$ have the same sign. Without loss of generality $k=n$, so that
$$
|a_{n-1} - a_n | + |a_n - a_{1}| = |a_{n-1} - a_{1}| \, .
$$
Then, using the already proven estimate for the even number $n-1$,
$$
|a_1 - a_2| + |a_2 - a_3| + \ldots + |a_{n - 1} - a_n| + |a_n - a_1| \\
= |a_1 - a_2| + |a_2 - a_3| + \ldots + |a_{n - 1} - a_1| \\
\underset{(3)}{\le} 2\sqrt{n-1} \sqrt{\sum_{i=1}^{n-1} a_i^2 }
\underset{(4)}{\le} 2\sqrt{n-1} \sqrt{\sum_{i=1}^{n} a_i^2 } \, .
$$
Equality holds at $(3)$ if $(a_1, \ldots, a_{n-1}) = (x, -x, \ldots, x, -x)$, and equality at $(4)$ holds if $a_n = 0$. It follows that equality holds in $(*)$ exactly if
$$
(a_1, \ldots, a_n) = (x, -x, \ldots, x, -x, 0)
$$
for some $x \in \Bbb R$, or a cyclic rotation thereof.
| {
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"url": "https://math.stackexchange.com/questions/3514654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\sum_{i=0}^{k}\binom{k-i}{b}\binom{i}{a-1}=\binom{k+1}{a+b}$ I am trying to prove this identity from an exercise:
Given $k$, $a$, $b$, prove that
$$
\sum_{i=0}^{k}\binom{k-i}{b}\binom{i}{a-1}=\binom{k+1}{a+b}
$$
However, I'm having trouble using existing identities to prove this. Any help would be much appreciated.
| Start from
$$\sum_{q=a-1}^{k-b} {k-q\choose b} {q\choose a-1}
= \sum_{q=0}^{k+1-b-a} {k+1-a-q\choose b} {q+a-1\choose a-1}
\\ = \sum_{q=0}^{k+1-a-b}
{k+1-a-q\choose k+1-a-b-q} {q+a-1\choose a-1}
\\ = [z^{k+1-a-b}] (1+z)^{k+1-a} \sum_{q=0}^{k+1-a-b}
{q+a-1\choose a-1} \frac{z^q}{(1+z)^q}.$$
The coefficient extractor enforces the range and we get
$$[z^{k+1-a-b}] (1+z)^{k+1-a} \sum_{q\ge 0}
{q+a-1\choose a-1} \frac{z^q}{(1+z)^q}
\\ = [z^{k+1-a-b}] (1+z)^{k+1-a}
\frac{1}{(1-z/(1+z))^a}
\\ = [z^{k+1-a-b}] (1+z)^{k+1-a}
\frac{(1+z)^a}{(1+z-z)^a}
\\ = [z^{k+1-a-b}] (1+z)^{k+1}
= {k+1\choose k+1-a-b} = {k+1\choose a+b}.$$
This is the claim. BTW when $k-q\lt b$ or $k-b\lt q$ we have
$(k-q)^\underline{b} = 0.$ (This zero value does not include $q=k$ and
$b=0$ because we required $k-q\lt b$.) Similarly when $0\le q\lt a-1$
we have $q^\underline{a-1} = 0.$ (This zero value does not include
$q=0$ and $a=1$ because we required $q\lt a-1$.) Applies to $k,b$
non-negative integers and $a$ a positive integer. For the sum range
not to be empty we also need $k-b\ge a-1$ or $k+1\ge a+b.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the set of values of x for which $\lvert \frac {x-1}{x+1} \rvert <2 $ How to solve this inequality question involving modulus?
I can’t get the same answer as the book [answer below]
I know the properties of absolute values, that is
If $\lvert x \rvert <k$ , then $-k < x < k$.
So for this question, this is my working:
$-2<\lvert \frac {x-1}{x+1} \rvert <2 $
When $-2<\lvert \frac {x-1}{x+1} \rvert $
$-2x-2 < x-1$
$-1<3x$
$x>- \frac {1}{3}$
And when $\lvert \frac {x-1}{x+1} \rvert <2 $
$x-1<2x+2$
$x>-3$
So, I got $x> - \frac {1}{3}\:$ or $\:x>-3$
However, The **answer given is $\{ {x\mid x<-3\: \text{ or }\: x> -\frac {1}{3}, ∈ℝ} \}$ **
I don't think the answer from the book is wrong, since the following graph confirms that the book answer is correct.
Please show me how to get the answer.
Thank you
| If $x=a+ib$ where $a,b$ are real
We have $$\sqrt{(a-1)^2+b^2}<2\sqrt{(a+1)^2+b^2}$$
Squaring and simplifying we get
$$b^2+a^2+\dfrac{10a}3+1>0$$
$$b^2+\left(a+\dfrac53\right)^2>\dfrac{25}9-1=\left(\dfrac43\right)^2$$
So, $x$ needs to lie outside the circle $$y^2+\left(x+\dfrac53\right)^2=\left(\dfrac43\right)^2$$
If $b=0,x=a$ is real
$$\left(a+\dfrac53\right)^2>\left(\dfrac43\right)^2\iff\left(a+\dfrac13\right)(a+3)>0$$
which happens
if $a>$max$\left(-\dfrac13,-3\right)$
or if $a<$min$\left(-\dfrac13,-3\right)$
In any case, $a\ne-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3519520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Eliminating parameters from polynomial relations Let $p =x^3+y^3+z^3$, $q= x^2y+y^2 z+z^2x$, $r=xy^2+yz^2+zx^2$, and $s=xyz$. I want to find some non-zero polynomial in $\phi$ in $p,q,r,s$ such that $\phi(p,q,r,s)=0$; that is, to eliminate $x,y,z$. I have $p^2-2qr-2ps+6s^2=x^6+y^6+z^6$ and $p^3-3q^3-3r^3-24s^3+18qrs=x^9+y^9+z^9$, but at this point it seems like I'm going in circles. The PDF I'm reading is on ring actions, if that gives any context.
| Multiply $r$ and $q$
\begin{eqnarray*}
rq &=& \sum x^3y^3+ 3(xyz)^2+xyz \sum x^3 \\
&=&\sum x^3y^3+ 3s^2 +sp. \\
\end{eqnarray*}
So
\begin{eqnarray*}
\sum x^3y^3= rq -3s^2 -sp \\
\end{eqnarray*}
Now multiply this by $p$
\begin{eqnarray*}
p(rq-sp-3s^2) = \sum_{perms} x^6y^3+ 3(xyz)^3 \\
\sum_{perms} x^6y^3= p(rq-sp-3s^2) -3s^3 \\
\end{eqnarray*}
Now cube $q$
\begin{eqnarray*}
q^3 = \sum_{cyc} x^6 y^3 + 3xyz \sum x^3 y^3 + 3(xyz)^2 \sum x^3 + 6(xyz)^3 \\
= \sum_{cyc} x^6 y^3 + 3 s(rq-3s^2-sp) +3s^2p+6s^3.
\end{eqnarray*}
Do the same for $r$ add and then eliminate using $\sum_{cyc} x^6y^3 +\sum_{cyc} x^3y^6=\sum_{perms} x^6y^3$.
This gives
\begin{eqnarray*}
q^3 - 3 srq +3s^3 +r^3 - 3 srq +3s^3 = pqr-sp^2 -3s^2p -3s^3 \\
\color{red}{q^3+r^3+9s^3 +sp^2+3s^2p-pqr-6srq=0}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3520237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Can any polynomials in the rational field be decomposed like this I' ve learned that the following examples can be used to decompose a
factor in this way:
x^5 - 5 x + 12 = (x - a) (a^4 - (5 a^3)/8 + (7 a^2)/8 + 1/8 (5 a^2 - 12 a) +
1/8 (5 a^3 - 12 a^2) + ((5 a^4)/16 - a^3/8 + (7 a^2)/16 -
3/16 (5 a^2 - 12 a) + 1/16 (12 a^3 - 5 a^4) +
1/8 (12 a^2 - 5 a^3) - (9 a)/4) x^2 + ((5 a^4)/16 + a^3/4 + (
5 a^2)/16 + 1/16 (12 a - 5 a^2) + 1/16 (12 a^3 - 5 a^4) + a/2 +
1/4 (12 - 5 a) - 3) x + a x^3 + a/4 + 1/4 (5 a - 12) + x^4 - 2)
Can any polynomials f(x) in the field of rational numbers be factorized into the form of (x - a) g (x, a)?
Besides a, other coefficients of g (x, a) should also be in the rational number field.
In the case of $x^5-5 x+12$, we can know the algebraic relations of his five roots (The letter a is a root of equation $x^5−5x+12$):
$x^5-5 x+12=(x-a) \left(x-\frac{1}{8} \left(-a^4-a^3-a^2-2 \sqrt{2}
\sqrt{3 a^3-2 a^2+a+4}-a+4\right)\right) \left(x-\frac{1}{8}
\left(-a^4-a^3-a^2+2 \sqrt{2} \sqrt{3 a^3-2
a^2+a+4}-a+4\right)\right) \left(x-\frac{1}{8} \left(a^4+a^3+a^2-2
\sqrt{2} \sqrt{a^4+a^2+6 a-8}-3 a-4\right)\right)
\left(x-\frac{1}{8} \left(a^4+a^3+a^2+2 \sqrt{2} \sqrt{a^4+a^2+6
a-8}-3 a-4\right)\right)$
I used the function in this link to find the relationship between a polynomial Galois group and a root set.
| Here:
In[5]:= qpoly = x^5 - 5 x + 12;
rt = First[x /. Solve[qpoly == 0, x]]
fax = Factor[qpoly, Extension -> rt] /. rt -> a
(% /. a -> Root[#1^5 - 5 #1 + 12 &, 1]) // Simplify
Out[5]=(1/16)*(x - a)*(-a^4 - a^3 - a^2 + (-a^4 - a^3 - a^2 + 3*a + 4)*x - 5*a + 4*x^2 + 8)*(-(2*a^3) + (a^4 + a^3 + a^2 + a - 4)*x - 2*a + 4*x^2 - 4)
x^5-5 x+12
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\sqrt{a^2+4}+\sqrt{b^2+4}\leq\frac{\sqrt{2}}{4}(a+b)^2$ with $\frac1a+\frac1b=1$. Let $ a, b> 0 $ and $\frac{1}{a}+\frac{1}{b}=1.$ Prove that$$\sqrt{a^2+4}+\sqrt{b^2+4}\leq\frac{\sqrt{2}}{4}(a+b)^2.$$
Obviously $a+b=ab\ge4$. Other than that, I do not know what trick to use to deal with the constraint. The Lagrange multiplier method has not gotten me further either.
| Let $x=\frac{1}{a}$ and $y=\frac{1}{b}$. We want to show that $x+y=1$ and $x,y\geq 0$ imply
$$ y^2x\sqrt{4x^2+1}+x^2y\sqrt{4y^2+1}\leq \frac{\sqrt{2}}{4}. $$
Letting $x=\frac{1+t}{2},y=\frac{1-t}{2}$, this is equivalent to finding the maximum of
$$ f(t) =(1-t^2)\left[ (1-t)\sqrt{1+(1+t)^2}+(1+t)\sqrt{1+(1-t)^2}\right]$$
(which is an even function) over $[-1,1]$. We have
$$ f(t) = (1-t^2)\sqrt{4-2t^2+2t^4+2(1-t^2)\sqrt{1+t^4}} $$
$$ f(\sqrt{t}) = \sqrt{2}(1-t)\sqrt{2-t+t^2+(1-t)\sqrt{1+t^2}}$$
and both $1-t$ and $2-t+t^2+(1-t)\sqrt{1+t^2}$ are positive and decreasing functions over $[0,1]$, so the maximum of $f(t)$ is attained at the origin. Indeed
$$2-\tan\theta+\tan^2\theta+(1-\tan\theta)\sqrt{1+\tan^2\theta} = \frac{2+\sqrt{2}\sin\left(\tfrac{\pi}{4}+\theta\right)}{2\sin^2\left(\frac{\pi}{4}+\frac{\theta}{2}\right)}$$
and the derivative of the RHS is
$$ -\frac{1}{\sin^2\left(\frac{\pi}{4}+\frac{\theta}{2}\right)\left(1+\tan\frac{\theta}{2}\right)}<0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3521160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Minimum without computing the derivative. I am trying to find the minimum of the following function:
$$H(x)=\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})},\hspace{1cm}x>0$$
What I usually do to find extrema of a function is computing the derivative of the function and finding its zeros. However, I think there must be a more efficient approach to this problem, since the derivative of this function is quite long. Any tips?
Thanks.
| $$\displaystyle \bigg[\bigg(x+\frac{1}{x}\bigg)^3\bigg]^2-\bigg[x^3+\frac{1}{x^3}\bigg]^2$$
$$\bigg[\bigg(x+\frac{1}{x}\bigg)^3+x^3+\frac{1}{x^3}\bigg]\bigg[\bigg(x+\frac{1}{x}\bigg)^3-x^3-\frac{1}{x^3}\bigg]$$
$$H(x)=\bigg(x+\frac{1}{x}\bigg)^3-x^3-\frac{1}{x^3}=3\bigg(x+\frac{1}{x}\bigg)\geq 3\cdot 2=6$$
Because arithmetic geometric inequality
$$x+\frac{1}{x}\geq 2\;\forall \;x>0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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For real $a$, $b$, $c$ in $(0,1)$, if $a+b+c=2$, then $\frac{a}{1-a}\times\frac{b}{1-b}\times\frac{c}{1-c}\geq 8$
Let $a$, $b$, $c$ be three real numbers such that $0<a,b,c<1$ and $$a + b + c = 2$$
Prove that
$$\frac a{1-a}×\frac b{1-b}×\frac c {1-c}≥8$$
I tried by substituting $x$ for$(1-a)$ and similarly for others and used AM-GM inequality, but couldn't get to the desired result.
| For the edited question
Let $a$, $b$, $c$ be three real numbers such that $0<a,b,c<1$ and $$a + b + c = 2$$
Prove that
$$\frac a{1-a}×\frac b{1-b}×\frac c {1-c}≥8$$
The inequality is equivalent with
$$abc \geq 8(1-a)(1-b)(1-c)$$
or
$$ abc \geq 8\left(\frac{a+b+c}{2}-a\right)\left(\frac{a+b+c}{2}-b\right)\left(\frac{a+b+c}{2}-c\right)$$
or
$$abc \geq (a+b-c)(b+c-a)(c+a-b)$$
Notice that each bracket on the right hand side is positive, so using AM-GM we have
$$\sqrt{(a+b-c)(b+c-a)} \leq \frac{1}{2}[(a+b-c)+(b+c-a)]=b$$
Now multiply with the other two similar inequalities to arrive at
$$(a+b-c)(b+c-a)(c+a-b) \leq abc$$
Equality occurs when $a=b=c=\dfrac{2}{3}$.
This part of the answer is for initial problem which was
Let$ a,b,c$ be three real numbers such that $0<a,b,c<1$and $$a + b + c= 2$$
Prove that:
$$\frac a{1-a} + \frac b{1-b} + \frac c {1-c}≥8$$
I tried by substituting $x$ for$(1-a)$ and similarly for others and
used AM-GM inequality,but couldn't get to the desired reasult.
The correct lower bound is $6$. From Cauchy-Schwarz:
$$(a+b+c)^2 \leq 3(a^2+b^2+c^2)\Rightarrow a^2+b^2+c^2 \geq \frac{4}{3}$$
So, using Cauchy-Schwarz again:
$$LHS = \sum \frac{a^2}{a-a^2} \geq \frac{(a+b+c)^2}{a+b+c-(a^2+b^2+c^2)} \geq \frac{4}{2-\frac{4}{3}} = 6$$
Equality occurs when $a=b=c=\dfrac{2}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3524619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
How do we integrate the expression $\frac{b^{3}}{1 - \theta(1-b)^{2}}$ where $0\leq b\leq 1$ and $\theta\in(0,1)$? MY ATTEMPT
In order to solve this integral, I have tried using the integration by parts method, as suggested by the expression
\begin{align}
\int_{0}^{1}\frac{b^{3}}{1-\theta(1-b)^{2}}\mathrm{d}b = \int_{0}^{1}\frac{b^{2}}{1-\theta(1-b)^{2}}\mathrm{d}b - \int_{0}^{1}\left[\int\frac{b^{2}}{1-\theta(1-b)^{2}}\mathrm{d}b\right]\mathrm{d}b
\end{align}
Based on it, the first summand is given by
\begin{align}\label{eq8}
\int_{0}^{1}\frac{b^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b & = \int_{0}^{1}\frac{(1 - 2b + b^{2}) - (1 - 2b)}{1 - \theta(1-b)^{2}}\mathrm{d}b\\
& = \int_{0}^{1}\frac{(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b - \int_{0}^{1}\frac{(2-2b) - 1}{1 - \theta(1-b)^{2}}\mathrm{d}b\\
& = \int_{0}^{1}\frac{(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b - 2\int_{0}^{1}\frac{1-b}{1 - \theta(1-b)^{2}}\mathrm{d}b + \int_{0}^{1}\frac{1}{1 - \theta(1-b)^{2}}\mathrm{d}b
\end{align}
The first of the last three integrals is given by
\begin{align}
\int_{0}^{1}\frac{(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b & = \frac{1}{\theta}\int_{0}^{1}\frac{\theta(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b\\
& = -\frac{1}{\theta}\int_{0}^{1}\frac{1 - \theta(1 - b)^{2}}{1 - \theta(1-b)^{2}}\mathrm{d}b + \frac{1}{\theta}\int_{0}^{1}\frac{1}{1 - \theta(1-b)^{2}}\mathrm{d}b\\
& = -\frac{1}{\theta} + \frac{1}{\theta}\int_{0}^{1}\frac{1}{1 - \theta(1-b)^{2}}\mathrm{d}b
\end{align}
According to the substitution $u = \sqrt{\theta}(1-b)$, it results that
\begin{align}\label{eq10}
\int_{0}^{1}\frac{1}{1 - \theta(1-b)^{2}}\mathrm{d}b = \frac{1}{\sqrt{\theta}}\int_{0}^{\sqrt{\theta}}\frac{1}{1 - u^{2}}\mathrm{d}u = \frac{1}{2\sqrt{\theta}}\ln\left|\frac{1+\sqrt{\theta}}{1-\sqrt{\theta}}\right|
\end{align}
Finally, based on the same substitution, we have
\begin{align}\label{eq11}
\int_{0}^{1}\frac{1-b}{1 - \theta(1-b)^{2}}\mathrm{d}b = \frac{1}{\theta}\int_{0}^{\sqrt{\theta}}\frac{u}{1-u^{2}}\mathrm{d}u = -\frac{1}{\theta}\ln|1-\theta^{2}|
\end{align}
Combining all these results, we get the first summand from the integration by parts method. The problem arises when I try to determine the second part.
| You can use partial fractions:
$$
\frac{1}{1-\theta(1-b)^2}=\frac12\,\left(\frac1{1-\sqrt{\theta}(1-b)}+\frac{1}{1+\sqrt{\theta}(1-b)}\right).
$$
Now, using the substitution $v=1-\sqrt{\theta}(1-b)$, we have $dv=\sqrt{\theta}\,db$, so the integral of the first half is
\begin{align}
\int_0^1\frac{b^3}{1-\sqrt{\theta}(1-b)}\,db
&=\theta^{-1/2}\int_{1-\sqrt{\theta}}^1\frac{\left(\frac{v-1+\sqrt{\theta}}{\sqrt{\theta}} \right)^3}{v}\,dv
=\theta^{-2}\int_{1-\sqrt{\theta}}^1\frac{\left({v-1+\sqrt{\theta}} \right)^3}{v}\,dv\\ \ \\
&=\frac1{\theta^{2}}\,\int_{1-\sqrt{\theta}}^1\frac{\left(v^3-3v^2(1-\sqrt{\theta})+3v(1-\sqrt{\theta})^2+(1-\sqrt{\theta})^3 \right)}{v}\,dv\\ \ \\
&=\frac1{\theta^{2}}\,\int_{1-\sqrt{\theta}}^1{\left(v^2-3v(1-\sqrt{\theta})+3(1-\sqrt{\theta})^2+\frac{(1-\sqrt{\theta})^3}v \right)}{}\,dv\\ \ \\
&=\frac1{\theta^{2}}\,\left.\vphantom{\int}{\left(\frac{v^3}3-\frac{3v^2(1-\sqrt{\theta})}2+3v(1-\sqrt{\theta})^2+{(1-\sqrt{\theta})^3}\log v \right)}{}\right|_{1-\sqrt{\theta}}^1\\ \ \\
&=\frac1{\theta^{2}}\,{\left(\frac13-\frac{11(1-\sqrt{\theta})^3}6-\frac{3(1-\sqrt{\theta})}2+3(1-\sqrt{\theta})^2-{(1-\sqrt{\theta})^3}\log (1-\sqrt{\theta}) \right)}{}\\ \ \\
\end{align}
The second half can be calculated similarly.
\begin{align}
\int_0^1\frac{b^3}{1+\sqrt{\theta}(1-b)}\,db
&=\theta^{-1/2}\int_1^{1+\sqrt{\theta}}\frac{\left(\frac{1-v+\sqrt{\theta}}{\sqrt{\theta}} \right)^3}{v}\,dv
=\theta^{-2}\int_1^{1+\sqrt{\theta}}\frac{\left({1-v+\sqrt{\theta}} \right)^3}{v}\,dv\\ \ \\
&=\frac1{\theta^{2}}\,\int_1^{1+\sqrt{\theta}}\frac{\left(-v^3+3v^2(1+\sqrt{\theta})-3v(1+\sqrt{\theta})^2+(1+\sqrt{\theta})^3 \right)}{v}\,dv\\ \ \\
&=\frac1{\theta^{2}}\,\int_1^{1+\sqrt{\theta}}{\left(-v^2+3v(1+\sqrt{\theta})-3(1+\sqrt{\theta})^2+\frac{(1+\sqrt{\theta})^3}v \right)}{}\,dv\\ \ \\
&=\frac1{\theta^{2}}\,\left.\vphantom{\int}{\left(-\frac{v^3}3+\frac{3v^2(1+\sqrt{\theta})}2-3v(1+\sqrt{\theta})^2+{(1+\sqrt{\theta})^3}\log v \right)}{}\right|_1^{1+\sqrt{\theta}}\\ \ \\
&=\frac1{\theta^{2}}\,{\left(-\frac13-\frac{7(1+\sqrt{\theta})^3}6-\frac{3(1+\sqrt{\theta})}2+3(1+\sqrt{\theta})^2+{(1+\sqrt{\theta})^3}\log (1+\sqrt{\theta}) \right)}{}.
\end{align}
Now we need to add the two integrals and divide by two. We get
$$
\frac1{2\theta^2}\,\left(
\frac{2\theta^{3/2}}{3}-3\theta+2\sqrt\theta-{(1-\sqrt{\theta})^3}\log (1-\sqrt{\theta})
+
{(1+\sqrt{\theta})^3}\log (1+\sqrt{\theta})
\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3525288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Solve $\sin x + \cos x = \sin x \cos x.$ I have to solve the equation:
$$\sin x + \cos x = \sin x \cos x$$
This is what I tried:
$$\hspace{1cm} \sin x + \cos x = \sin x \cos x \hspace{1cm} ()^2$$
$$\sin^2 x + 2\sin x \cos x + \cos^2 x = \sin^2 x \cos^2x$$
$$1 + \sin(2x) = \dfrac{4 \sin^2 x \cos^2x}{4}$$
$$1 + \sin(2x) = \dfrac{\sin^2(2x)}{4}$$
$$\sin^2(2x) - 4 \sin(2x) -4 = 0$$
Here we can use the notation $t = \sin(2x)$ with the condition that $t \in [-1,1]$.
$$t^2-4t-4=0$$
Solving this quadratic equation we get the solutions:
$$t_1 = 2+ 2\sqrt{2} \hspace{3cm} t_2 = 2 - 2\sqrt{2}$$
I managed to prove that $t_1 \notin [-1, 1]$ and that $t_2 \in [-1, 1]$. So the only solution is $t_2 = 2 - \sqrt{2}$. So we have:
$$\sin(2x) = 2 - 2\sqrt{2}$$
From this, we get:
$$2x = \arcsin(2-2\sqrt{2}) + 2 k \pi \hspace{3cm} 2x = \pi - \arcsin(2-2\sqrt{2}) + 2 k \pi$$
$$x = \dfrac{1}{2} \arcsin(2-2\sqrt{2}) + k \pi \hspace{3cm} x = \dfrac{\pi}{2} - \dfrac{1}{2}\arcsin(2 - 2\sqrt{2}) + k \pi$$
Is this solution correct? It's such an ungly answer, that I kind of feel like it can't be right. Did I do something wrong?
| Note that when you square the equation
$$(\sin x + \cos x)^2 = (\sin x \cos x)^2$$
which can be factorized as
$$(\sin x + \cos x - \sin x \cos x)(\sin x + \cos x + \sin x \cos x)=0$$
you effectively introduced another equation $\sin x + \cos x =- \sin x \cos x$ in the process beside the original one $\sin x + \cos x = \sin x \cos x$. The solutions obtained include those for the extra equation as well.
Normally, you should plug the solutions into the original equation to check and exclude those that belong to the other equation. However, given the complexity of the solutions, it may not be straightforward to do so. Therefore, the preferred approach is to avoid the square operation.
Here is one such approach. Rewrite the equation $\sin x + \cos x = \sin x \cos x$ as
$$\sqrt2 \cos(x-\frac\pi4 ) = \frac12 \sin 2x = \frac12 \cos (2x-\frac\pi2 ) $$
Use the identity $\cos 2t = 2\cos^2 t -1$ on the RHS to get the quadratic equation below
$$\sqrt2 \cos(x-\frac\pi4) = \cos^2 (x-\frac\pi4 ) -\frac12$$
or
$$\left( \cos(x-\frac\pi4) - \frac{\sqrt2-2}2\right)\left( \cos(x-\frac\pi4) - \frac{\sqrt2+2}2\right)=0$$
Only the first factor yields real roots
$$x = 2n\pi + \frac\pi4 \pm \cos^{-1}\frac{\sqrt2-2}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3530949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 0
} |
$\int_0^{\pi/2} \sec^a(t)\,dt= \frac{\sqrt{\pi}}{2\Gamma\left(1-\frac{a}{2}\right)}\Gamma\left(\dfrac{1-a}{2}\right)$ Inside the Wolfram Documentation page for the secant function, an identity is given which involves the gamma function, polygamma function, and Catalan's constant.
Notes on documentation page:
Some special functions can be used to evaluate more complicated definite integrals. For example, polygamma and gamma functions and the Catalan constant are needed to express the following integral:
$$\int_0^{\pi/2} \sec^a(t)\,dt= \frac{\sqrt{\pi}}{2\Gamma\left(1-\frac{a}{2}\right)}\Gamma\left(\dfrac{1-a}{2}\right),\quad\text{$\operatorname{Re}(a)<1$} $$
I know that the Gamma function is defined as
$$\Gamma(z) = \int_0^\infty x^{z-1} e^{-x}\, dx, \quad\text{$\operatorname{Re}(z)>0$}$$
and Catalan's constant can be written as
$$G = \beta(2) = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^2} = \frac{1}{1^2} - \frac{1}{3^2} + \frac{1}{5^2} - \frac{1}{7^2} + \frac{1}{9^2} - \cdots$$
but I don't see how this helps. I couldn't find a source in the Wolfram documentation page and couldn't find a duplicate question on Math SE. How did the author of the Wolfram page arrive at this identity?
| To derive the identity, it is a straight up application of the Beta function in the form
$$\operatorname{B}(m,n) = 2 \int_0^{\frac{\pi}{2}} \cos^{2m - 1} t \sin^{2n - 1} t \, dt.$$
For the secant integral, we have
\begin{align}
\int_0^{\frac{\pi}{2}} \sec^a t \, dt &= \int_0^{\frac{\pi}{2}} \cos^{-a} t \, dt\\
&= \frac{1}{2} \cdot 2 \int_0^{\frac{\pi}{2}} \cos^{2\left (\frac{1 - a}{2} \right ) - 1} t \sin^{2 \left (\frac{1}{2} \right ) - 1} t \, dt\\
&= \frac{1}{2} \operatorname{B} \left (\frac{1 - a}{2}, \frac{1}{2} \right )\\
&= \frac{1}{2} \frac{\Gamma \left (\frac{1 - a}{2} \right ) \Gamma \left (\frac{1}{2} \right )}{\Gamma \left (\frac{1 - a}{2} + \frac{1}{2} \right )}\\
&= \frac{\sqrt{\pi}}{2 \Gamma \left (1 - \frac{a}{2} \right )} \Gamma \left (\frac{1 - a}{2} \right ),
\end{align}
for $a < 1$, as required. Here the following well-known results of
$$\operatorname{B} (a,b) = \frac{\Gamma (a) \Gamma (b)}{\Gamma (a + b)},$$
together with $\Gamma (\frac{1}{2}) = \sqrt{\pi}$, have been used.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3531201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Bernoulli First Order ODE I want to know if my answer is equivalent to the one in the back of the book. if so what was the algebra? if not then what happened?
$$x^2y'+ 2xy = 5y^3$$
$$y' = -\frac{2y}{x} + \frac{5y^3}{x^2}$$
$n = 3$
$v = y^{-2}$
$-\frac{1}{2}v'=y^{-3}$
$$\frac{-1}{2}v'-\frac{2}{x}v = \frac{5}{x^2}$$
$$v'+\frac{4}{x}v=\frac{-10}{x^2}$$
this is now a first order linear ODE where:
$$y(x)=\frac{1}{u(x)}\int u(x)q(x)$$
$u(x)=e^{4\int\frac{1}{x}}=x^4$
$q(x) = \frac{-10}{x^2}$
$$\frac{1}{x^4}\int x^4 \frac{-10}{x^2}=\frac{1}{x^4}\frac{-10x^{2+1}}{2+1}+C$$
which leaves us with :
$$\frac{1}{y^2} = \frac{-10}{3x}+x^{-4}C$$
naturally
$$y^2= \frac{1}{\frac{-10}{3x}+x^{-4}C}$$
The book states the answer as being:
$$y^2= \frac{x}{2+Cx^5}$$
| Write:$$(x^2y)' = 5y^3$$ Now let $z=x^2y$ then we get $$z' = {5z^3\over x^6}\implies {z'\over z^3} = 5x^{-6}$$
So after integrating both sides we get $$ -{z^{-2}\over 2} = -x^{-5}+c'\implies {1\over 2x^4y^2} = {1-c'x^5\over x^5}$$
So $$ y^2 = {x\over 2+cx^5}$$
where $c=-2c'$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3534566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Hölder continuous implying the rate of convergence of the Cesàro mean Let $f$ be a periodic function which is Hölder continuous of order $0<\alpha<1$
And $\sigma_nf$ be its $n^{th}$ Cesàro mean. Let $ \|f \|=\|f\|_\infty+\sup_{x \neq y}\frac{|f(x)-f(y)|}{|x-y|^{\alpha}}$.
The I have to show that $\|\sigma_nf-f\|_{\infty} \leq C\|f\|n^{-\alpha}$ for all natural number $n$ and a constant $C$.
I tried to use the convolution and the Fejér kernel to show this inequality but failed in every attempt and it is unbearably frustrating…
Could anyone please help me?
| The proof for the line can be applied to the circle using inequality $(4)$.
Preliminary Inequalities for $\boldsymbol{0\lt x\lt\frac\pi2}$
Inequality $\bf{1}$:
$$
\begin{align}
\frac{\sin^2(x)}{x^2}
&=\prod_{k=1}^\infty\cos^2\left(\frac{x}{2^k}\right)\tag{1a}\\
&=\prod_{k=1}^\infty\left(1-\sin^2\left(\frac{x}{2^k}\right)\right)\tag{1b}\\
&\ge\prod_{k=1}^\infty\left(1-\frac{x^2}{4^k}\right)\tag1
\end{align}
$$
Explanation:
$\text{(1a)}$: Induction, $\frac{\sin(x)}{2\sin(x/2)}=\cos(x/2)$, and $\lim\limits_{n\to\infty}2^n\sin\left(x/2^n\right)=x$
$\text{(1b)}$: $\cos^2(x)=1-\sin^2(x)\vphantom{\lim\limits_{n\to\infty}}$
$\phantom{\text{b}}\text{(1)}$: $\sin^2(x)\le x^2$
Inequality $\bf{2}$:
$$
\begin{align}
\frac{\tan(x)}{x}
&=\prod_{k=1}^\infty\frac1{1-\tan^2\left(x/2^k\right)}\tag{2a}\\
&\ge\prod_{k=1}^\infty\frac1{1-\frac{x^2}{4^k}}\tag2
\end{align}
$$
Explanation:
$\text{(2a)}$: Induction, $\frac{\tan(x)}{2\tan(x/2)}=\frac1{1-\tan^2(x/2)}$, and $\lim\limits_{n\to\infty}2^n\tan\left(x/2^n\right)=x$
$\phantom{\text{a}}\text{(2)}$: $\tan^2(x)\ge x^2$
Inequality $\bf{3}$:
Multiplying inequalities $(1)$ and $(2)$ we get
$$
\frac{\sin^2(x)\tan(x)}{x^3}\ge1\tag3
$$
Inequality $\bf{4}$:
Inequality $(3)$ implies that
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac1{\sin^2(x)}-\frac1{x^2}\right)
&=\frac2{x^3}-\frac{2\cos(x)}{\sin^3(x)}\tag{4a}\\
&\ge0\tag{4b}
\end{align}
$$
Since $\frac1{\sin^2(x)}-\frac1{x^2}$ is increasing, it attains its maximum on $\left(0,\frac\pi2\right]$ at $\frac\pi2$. Therefore,
$$
\frac1{\sin^2(x)}-\frac1{x^2}\le1-\frac4{\pi^2}\tag4
$$
Hölder Bound on the Kernel
$$
\begin{align}
\int_{-1/2}^{1/2}\frac{\sin^2(\pi nx)}{n\sin^2(\pi x)}|x|^\alpha\,\mathrm{d}x
&=2\int_0^{1/2}\frac{\sin^2(\pi nx)}{n\sin^2(\pi x)}x^\alpha\,\mathrm{d}x\tag{5a}\\
&=2n^{-\alpha-2}\int_0^{n/2}\frac{\sin^2(\pi x)}{\sin^2(\pi x/n)}x^\alpha\,\mathrm{d}x\tag{5b}\\
&\le\frac2{\pi^2}n^{-\alpha}\int_0^{n/2}\frac{\sin^2(\pi x)}{x^{2-\alpha}}\,\mathrm{d}x+\left(1-\frac4{\pi^2}\right)\frac{2^{-\alpha}}{1+\alpha}n^{-1}\tag{5c}\\
&\le\frac2{\pi^2}n^{-\alpha}\left(\frac{\pi^2}{1+\alpha}+\frac1{1-\alpha}\right)+\left(1-\frac4{\pi^2}\right)\frac{2^{-\alpha}}{1+\alpha}n^{-1}\tag{5d}\\[6pt]
&=O\!\left(n^{-\alpha}\right)\tag5
\end{align}
$$
Explanation:
$\text{(5a)}$: apply symmetry
$\text{(5b)}$: substitute $x\mapsto x/n$
$\text{(5c)}$: apply $(4)$ and evaluate the integral of $x^\alpha$
$\text{(5d)}$: estimate the integral on $[0,1]$, where $\sin^2(\pi x)\le\pi^2x^2$,
$\phantom{\text{(5d):}}$ and on $[1,\infty]$, where $\sin^2(\pi x)\le1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find $a$ such that the minimum and maximum distance from a point to the curve $x^2+y^2=a$ are $\sqrt{5}$ and $3\sqrt{5}$ Context: I have to solve the following problem: find $a\in \mathbb{R},\, a>0\,\, /$ the minimum and maximum distance from $(4,2)$ to the curve $x^2+y^2=a$ are $\sqrt{5}$ and $3\sqrt{5}$ respectively.
I can't use polar coordinates. I decided to used Lagrange multipliers.
Problem: after using Lagrange multipliers I got this system:
$$
\left\{
\begin{array}{c}
\frac{x-4}{x}=\frac{y-2}{y} \\
x^2+y^2=a
\end{array}
\right.
$$
And then I add one more equation to the system: $5=(x-4)^2+(y-2)^2$ and $45=(x-4)^2+(y-2)^2$, so I have to find the $a$ that satisfies both system of equations:
$$
\left\{
\begin{array}{c}
\frac{x-4}{x}=\frac{y-2}{y} \\
x^2+y^2=a \\
5=(x-4)^2+(y-2)^2
\end{array}
\right.
$$
and
$$
\left\{
\begin{array}{c}
\frac{x-4}{x}=\frac{y-2}{y} \\
x^2+y^2=a \\
45=(x-4)^2+(y-2)^2
\end{array}
\right.
$$
The thing is that I can't resolve those systems, I already tried a lot of ways but I never get a result. I would really appreciate some help.
| Perhaps a different approach altogether. If $(4,2)$ is inside the circle of radius $\sqrt{a}$, then the minimum and maximum distances combine into one diameter, hence
$$
2a = d = \sqrt{5} + 3\sqrt{5} = 4\sqrt{5}
$$
and the point of interest is $a = \sqrt{2^2+4^2} = \sqrt{20} = 2\sqrt{5}$ away, which means the point $(4,2)$ cannot be inside the circle.
Then the diameter of the circle is exactly the difference between the distances, i.e.
$$
2a = d = 3\sqrt{5} - \sqrt{5} = 2\sqrt{5},
$$
and so the radius must be $a = \sqrt{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
When does $2n-1$ divide $16(n^2-n-1)^2$? Find all integers $n$ such that
$\dfrac{16(n^2-n-1)^2}{2n-1}$
is an integer.
| Let $k = 2n-1$ then $$\dfrac{16(n^2-n-1)^2}{2n-1} = \dfrac{(4n^2-4n-4)^2}{2n-1}$$
$$ = \dfrac{((k+1)^2-2(k+1)-4)^2}{k} = {(k^2-5)^2\over k}$$
So $$k\mid (k^2-5)^2= k^4-10k^2+25$$ $$\implies k\mid 25 \implies k\in\{\pm 1, \pm 5,\pm 25\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Quadratic function with roots in $[0,1]$. Prove that $f(0) \geq \frac49$ or $f(1) \geq \frac49$ Let $a,b$ in $[0,1]$ be such that the polynomial $f(x) = (x-a)(x-b)$ satisfies $f(\tfrac12) \geq \frac1{36}$.
I have found a quite complicated proof of the following inequality using calculus:
$$f(0) \geq \frac49 \quad\text{or}\quad f(1) \geq \frac49.$$
Can you find a simple argument? (with geometric flavour if possible)
| Note,
$$f(\frac12) =(\frac12-a)(\frac12-b)\ge \frac{1}{36}\implies 2ab-(a+b)+\frac{4}{9} \geq 0$$
Use $a+b\ge 2\sqrt{ab}$ to factorize above inequality as
$$(\sqrt{ab}-\frac{1}{3})(\sqrt{ab}-\frac{2}{3})\ge 0$$
which leads to either $\sqrt{ab}\le\frac{1}{3}$ or $\sqrt{ab}\ge\frac{2}{3}$. Examine the two cases below.
Case 1) $\sqrt{ab}\ge\frac{2}{3}$. Evalute $f(0)$,
$$f(0) = (0-a)(0-b) = ab \ge \frac49$$
Case 2) $\sqrt{ab}\le\frac{1}{3}$ leads to $-ab \ge -\frac19$. Take $f'(x) = 2x-(a+b)$ and evalute $f(1)$ in two ways,
$$f(1) = f(\frac12) + \int_{1/2}^1 f'(x)dx=f(\frac12)+ \frac34 -\frac12(a+b)$$
$$f(1) = 1-(a+b)+ab$$
Eliminate $a+b$ and apply $f(\frac12)\ge \frac1{36}$, $-ab \ge -\frac19$ to obtain
$$f(1)=2f(\frac12)+\frac12-ab\ge \frac2{36}+\frac12 - \frac19 = \frac49
\implies f(1) \ge \frac49$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Rationalizing the denominator of $\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}$
Simplify
$$\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}$$
I think this should be expressed without square roots at the denominator. I tried to multiply by conjugate.
| So, you multiply by the denominator over the denominator to get rid of the first square root (while still just technically multiplying only by 1):
$$
\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}} \times \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{\sqrt{2+\sqrt{2+\sqrt{2}}}} =
\frac{2\sqrt{2+\sqrt{2+\sqrt{2}}}}{2+\sqrt{2+\sqrt{2}}}
$$
Now you have an expression that can be multiplied by a conjugate, so that’s what we do:
$$
\frac{2\sqrt{2+\sqrt{2+\sqrt{2}}}}{2+\sqrt{2+\sqrt{2}}}\times\frac{2-\sqrt{2+\sqrt{2}}}{2-\sqrt{2+\sqrt{2}}}=\frac{\left(2\sqrt{2+\sqrt{2+\sqrt{2}}}\right)\left(2-\sqrt{2+\sqrt{2}}\right)}{(2-\sqrt{2})}
$$
This can also be multiplied by its conjugate:
$$
\frac{\left(2\sqrt{2+\sqrt{2+\sqrt{2}}}\right)\left(2-\sqrt{2+\sqrt{2}}\right)}{(2-\sqrt{2})}=\frac{\left(2\sqrt{2+\sqrt{2+\sqrt{2}}}\right)\left(2-\sqrt{2+\sqrt{2}}\right)\left(2+\sqrt2\right)}{2}
$$
So, we’ve finally got an expression with a rationalized denominator.
I'd just leave the numerator unsimplified, as it's kind of hard to do and I know I'd definitely make a mistake (If your teacher/professor wants you to multiply out the numerator they're crazy;)).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3538395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show that $n= 5 + 5^2+ 5^3+...5^{150}$ is divisible by $930$. Show that $n= 5 + 5^2+ 5^3+...5^{150}$ is divisible by $930$.
I'm thinking to show that $n$ is divisible by each of the prime factors of $930$, is that right? I'm stuck
| $$n= 5(1 +5+ 5^2+ ...5^{149}) = 5 {5^{150}-1\over 4}$$
so $5\mid 4n\implies 5\mid n$. Now:
$$5^{150}-1 = 125^{50}-1 = (125-1)(125^{49}+...+125^2+125+1) $$
so $31\mid 4n \implies 31\mid n$ and $$5^{150}-1 = 25^{75}-1 = (25-1)(25^{74}+...+25+1) $$
so $24\mid 4n \implies 6\mid n$ and you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3540558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
$\sum_{k=1}^{n}\frac{\ln k}{(2k-1)(2k+1)}<\frac{1}{4}$
Show that $$\sum_{k=1}^{n}\frac{\ln k}{(2k-1)(2k+1)}<\frac{1}{4}$$
holds for all $n\in\mathbb{N^+}$.
Since this is a positive series, it suffices to show that $$\sum_{k=1}^{\infty}\frac{\ln k}{(2k-1)(2k+1)}<\frac{1}{4},$$which is true by machine computing. WA gives that
$$\sum_{k=1}^{\infty}\frac{\ln k}{(2k-1)(2k+1)}\approx 0.231051.$$
| Someone gives a solution as follows, which is essentially equivalent to @Robert Z 's except some details.
\begin{align*} s(n):&=\sum_{k=1}^n\frac{\ln k}{(2k-1)(2k+1)}\\ &\leq\sum_{k=2}^{\infty}\frac{\ln k}{(2k-1)(2k+1)}\\ &=\frac{1}{2}\sum_{k=2}^{\infty}\left(\frac{\ln k}{2k-1}-\frac{\ln k}{2k+1}\right)\\ &=\frac{1}{2}\sum_{k=2}^{\infty}\left(\frac{\ln (k-1)}{2k-1}-\frac{\ln k}{2k+1}\right)+\frac{1}{2}\sum_{k=2}^{\infty}\frac{\ln\left(1+\frac{1}{k-1}\right)}{2k-1}\\ &=\frac{1}{2}\sum_{k=2}^{\infty}\frac{\ln\left(1+\frac{1}{k-1}\right)}{2k-1}\\&=\frac{1}{2}\left(\frac{\ln 2}{3}+\frac{\ln\left(\frac 32\right)}{5}+\sum_{k=4}^{\infty}\frac{\ln\left(1+\frac{1}{k-1}\right)}{2k-1}\right)\\ &\leq\frac{1}{2}\left(\frac{\ln 2}{3}+\frac{\ln \left(\frac{3}{2}\right)}{5}+\sum_{k=4}^{\infty}\frac{1}{(k-1)(2k-1)}\right)\\ &=\frac{1}{2}\left(\frac{\ln 2}{3}+\frac{\ln \left(\frac{3}{2}\right)}{5}+\frac{47}{30}-2\ln 2\right)\\ &<\frac{1}{4} \end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3541416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Problem regarding unique solution of differential equation
A unique solution to the differential equation $y = x \frac{dy}{dx} - (\frac{dy}{dx})^2$ passing through $(x_0,y_0)$ doesnot exist
then choose the correct option
$1.$ if $ x_0^2 > 4y_0$
$2.$ if $ x_0^2 = 4y_0$
$3.$ if $ x_0^2 < 4y_0$
$4.$ for any $(x_0 , y_0)$
My attempt : Here $y = x \frac{dy}{dx} - (\frac{dy}{dx})^2$
Now i put $x= e^z$ then $z= \log x$
So $y= Dy- D^2y$
$D^2y-Dy -y=0$
so $(D^2-D-1)y=0$
so auxiliary equation will be $m^2-m-1=0$ ,$m= \frac{1 +_{-}\sqrt - 3}{2}$
so $y= e^{\frac{1}{2}x} (c_1 \cos(\frac{\sqrt - 3}{2} ) + c_2\sin ({\frac{\sqrt - 3}{2}} )x)$
After that im not able to proceed further
| Hint: First treat the differantial equation as a quadratic in $\frac{dy}{dx}$
\begin{eqnarray*}
\left( \frac{dy}{dx} \right)^2 -x \frac{dy}{dx} +y=0.
\end{eqnarray*}
Now plug this into the quadratic formula and ... ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3543086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
What is the closed form of this quickly growing sequence? Is there a pattern in the following sequence:
$$3,19,451,22051,\dots$$
I tried setting some $f(n)$ to this sequence, where $n$ is the term number. Then I basically treated it like a matrix equation: multiplying, adding, subtracting and dividing to hopefully obtain a simple, known result. Like the Fibonacci sequence, or something else of the sort. I've just put some randomly growing numbers together in hopes of finding a representation of some kind. This is not homework.
| If I subtract $1$ from each term, I see $2,18,450,22050,...$. The ratio of successive terms is $9,25,49,...$
And now I think I see it : the ratio of successive terms, minus $1$, gives the list of odd squares, but starting from $3^2 = 9$.
In other words, if $f(n)$ denotes the $n$th number of the sequence, then $(f(n)-1) = (f(n-1) - 1)(2n-1)^2$, along with $f(1) = 3$. For example, $$
f(2) - 1 = (f(1) - 1) \times 9 = 18 \\
f(3) -1 = (f(2) - 1) \times 25 = 450 \\
\vdots
$$
It follows from here that $$(f(n) - 1) = (f(1) - 1) \times 9 \times 25 \times ... \times (2n-1)^2 \\ = 2 \times \prod_{k=1}^n (2k-1)^2$$.
(The $k=1$ just gives $1$ so we include it).
Turns out this can be simplified :
\begin{align}
\prod_{k=1}^n (2k-1) & = 1 \cdot 3 \cdot 5 \cdot ... \cdot (2n-1) \\ & = \frac{1 \cdot 2 \cdot 3 \cdot ... \cdot 2k}{2 \cdot 4 \cdot 6 \cdot ... \cdot 2k} \\ &= \frac{(2k)!}{2^kk!}
\end{align}
So, after squaring and completing the formalities :
$$\bbox[yellow,5px,border:2px solid red]{
\color{brown}{f(n) = 1 + \frac{((2n)!)^2}{2^{2n-1}(n!)^2}}}
$$
Trying this out: at $n=1$, we have $1+\frac{2!^2}{2^11!^2} = 1+2 = 3$.
At $n=2$ we have $1 + \frac{4!^2}{2^32!^2} = 19$.
At $n=3$ we have $1 + \frac{6!^2}{2^53!^2} = 451$.
And so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3544881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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