Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Proving $\tan\frac{4\pi}{11} + 4\sin\frac{\pi}{11} = \sqrt{11}$ In a similar vein as $\tan\frac{3\pi}{11} + 4\sin\frac{2\pi}{11} = \sqrt{11}$ discussed in this question is this identity:
$$\tan\frac{4\pi}{11} + 4\sin\frac{\pi}{11} = \sqrt{11}$$
Trying to adopt a method on the same line however lamentably fails. I wonder if the arguement of $11$th roots of unity can still be effectively employed in this case. Is there a way to adapt it or there could be possibly an easier way out to prove the result?
| Let $a=\frac\pi{11}$ to evaluate
\begin{align}
& 4 (\sin 2a -\sin a)-(\tan 4a - \tan3a )\\
= & \>4\sin a (2\cos a -1)-\frac{\sin a}{\cos 3a\cos4a} \\
=& \> \frac{4\sin a}{\cos 7a+\cos a}\cdot A\tag1
\end{align}
where
\begin{align}
A
= & \> 2\cos a( \cos7a + \cos a) - (\cos7a + \cos a)-\frac12 \\
= & \> \cos10a+ \cos8a+ \cos 6a +\cos 4a +\cos2a +\frac12\\
= & \> \frac12 \sum_{k=0}^{10} e^{i 2ka}=0
\\
\end{align}
Substitute $A=0$ into (1) to obtain
$$\tan\frac{4π}{11} + 4\sin\frac{π}{11} = \tan\frac{3π}{11} + 4\sin\frac{2π}{11}= \sqrt{11}$$
where How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$
| {
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"url": "https://math.stackexchange.com/questions/3709929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Sequence problem : Find $|a_1|$
Given a sequence such that
$$a_1a_2=1\ ; \ a_2a_3=2 \ ; \ a_3a_4=3 \ ; \ \dots$$ and
$$\lim_{n\to \infty} \frac{a_n}{a_{n+1}}=1$$
find $|a_1|$.
My attempts :
We can deduce from this $a_1a_2=1 \ ; \ a_2a_3=2 \ ; \ a_3a_4=3 \ ; \ \dots$ that :
$$\prod_{n=k}^{n+1} a_n=n$$
Thus :
$$\begin{align}a_n \times a_{n+1}&=n \\ a_{n+1}&=\frac{n}{a_n} \end{align}$$
Therefore :
$$\begin{align} \lim_{n\to \infty} \frac{a_n}{a_{n+1}}&=\lim_{n\to \infty} \frac{a_n}{\frac{n}{a_n}} \\ &=\lim_{n\to \infty} \frac{a_n^2}{n} \\&=1 \end{align} $$
Is this can lead us to say that :
$$\begin{align} |a_n^2|&\sim n \\ |a_n|&\sim \sqrt{n} \end{align} $$
If this is true we can say that :
$$|a_n| =1$$
But I don't know if this is true or not. Any tips or hints ?
Thanks in advance !
| We find $a_1/a_3=1/2$, $a_1a_4=1\cdot3/2$, $a_1/a_5=1\cdot3/(2\cdot4)$.
By induction, it one shows that
$$a_1a_{2n}=\frac{1\cdot3\cdot\dots\cdot(2n-1)}{2\cdot4\cdot\dots\cdot(2n-2)}$$
for integer $n\geq2$, whereas
$$a_1/a_{2n+1}=\frac{1\cdot3\cdot\dots\cdot(2n-1)}{2\cdot4\cdot\dots\cdot(2n)}$$
for integers $n\geq1$.
Since $a_{2n}/a_{2n+1}\to1$ by assumption, we have
$$a_1^2=\lim_{n\to\infty} \frac{(1\cdot3\cdot\dots\cdot(2n-1))^2}
{(2\cdot4\cdot\dots\cdot(2n-2))^2(2n)}=
\lim_{n\to\infty}\frac{(2n-1)!^2}{2^{4n-3}(n-1)!^4n}=
\lim_{n\to\infty}\frac{(2n)!^2n}{2^{4n-1}n!^4}.$$
A calculation using Stirling's formula shows that
$a_1^2=2/\pi$ and hence
$$|a_1|=\sqrt{\frac{2}{\pi}}.$$
Edit: a) Let me complete the solution by a calculation of the limit using Stirling's formula twice:
$$\begin{array}{rcl}2\log((2n)!)-4\log(n!)-(4n-1)\log2+\log n&
=&2\left[(2n+\frac12)(\log n+\log2)-2n+\frac12\log(2\pi)\right]\\
&&-4\left[(n+\frac12)\log n-n+\frac12\log(2\pi)\right]\\
&&+\log n-(4n-1)\log2+ O(\frac1n)\\
&=&2\log 2-\log(2\pi)+O(\frac1n)\to\log(2/\pi).\end{array}$$
b) The solution also works in the complex domain. Then we obtain $a_1=\pm\sqrt{\frac2\pi}$ which is more precise in the complex domain.
c) For arbitrary $a_1$, the solution shows that
$$\frac{a_{2n}}{a_{2n+1}}\to\frac2{\pi a_1^2}.$$
As $\frac{a_{2n}}{a_{2n+2}}=\frac{2n}{2n+1}\to1$, this implies that
$\frac{a_{2n+1}}{a_{2n+2}}\to\frac{\pi a_1^2}2.$
Therefore $a_n/a_{n+1}\to1$ if and only if $a_1^2=2/\pi$.
d) One can also write using the first two formulas of the solution and $a_{2n+2}/a_{2n+1}\to1$
$$a_1^2=\lim_{n\to\infty}a_1^2\frac{a_{2n+2}}{a_{2n+1}}=\lim_{n\to\infty}\frac{(1\cdot3\cdot\dots\cdot(2n-1))^2(2n+1)}
{(2\cdot4\cdot\dots\cdot(2n-2)(2n))^2}=\frac{1\cdot3}{2^2}\cdot
\frac{3\cdot5}{4^2}\cdots=\frac2\pi$$
by the Wallis product.
| {
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"url": "https://math.stackexchange.com/questions/3711305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Minimize area of ellipse that passes through given point The ellipse formula is:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
$a$ and $b$ are assumed to be positive.
The point $(6, -7)$ is given as a point on the ellipse, with the area of the ellipse equal to $\pi ab$.
How would one find the values of $a$ and $b$ that minimize the area of the ellipse?
| Express $a$ in terms of $b$ given the equation of an ellipse and the given point:
You should get:
$$a=\frac{6b}{\sqrt{b^2-49}}$$
You want to maximize the area ($\pi a b$):
$$A=6\pi \cdot \frac{b^2}{\sqrt{b^2-49}}$$
Now differentiate $A$ with respect to $b$ and set $A'=0$ to find the critical points:
$$0=6\pi \left( \frac{2b\sqrt{b^2-49}-\frac{b^3}{\sqrt{b^2-49}}}{b^2-49}\right)$$
$$0= b\left(b^2-98\right)$$
Therefore, $b=0$ and $b= \pm 7\sqrt{2}$ are critical points, but $b \neq 0 \; \text{and} \; b \neq -7\sqrt{2}$. Try finishing the rest.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3711473",
"timestamp": "2023-03-29T00:00:00",
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Using the remainder theorem to prove a quadratic is a factor of a polynomial For example, if I have $P(x) = 3x^4 + 5x^3 -17x^2 -13x + 6$ then to show that $x^2 + x - 6$ is a factor I individually show that $x+3$ and $x-2$ are factors using the factor theorem (i.e. $P(-3) = 0$ and $P(2) = 0$).
However, if $x+3$ and $x-2$ are individually factors, why does that conclude their product would be a factor?
Because I have learnt if two individual numbers are factors, they also need to be "coprime" for the product to be a factor.
(For example: $12$ is divisible by both $6$ and $12$ but not their product $72$)
How would you know that $x+3$ and $x-2$ are coprime? E.g. for $x=7$ they do not seem to be?
Thanks in advance!
| $x+3$ and $x-2$ (or any two distinct monic polynomials of degree $1$) are coprime as polynomials. That is, there is no polynomial of positive degree that divides both of them. Indeed, $(x-2)-(x+3)=-5$. If $P(x)$ is a polynomial that is divisible by both $x-2$ and $x+3$, i.e. there are polynomials $q(x)$ and $r(x)$ such that $P(x) = (x-2) q(x) = (x+3) r(x)$, we have
$$ (x+3) r(x) = P(x) = (x-2)(q(x)) = (x+3) q(x) -5 q(x)$$
so $$ q(x) = (x+3)(q(x)-r(x))/5$$
and $$P(x) = (x-2) q(x) = (x-2)(x+3)(q(x)-r(x))/5$$
is divisible by $(x-2)(x+3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3711835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\frac{1}{4} (a^2+ 3 b^2)$ is of the form $(c^2+ 3 d^2)$ If $ 2 \mid (a^2+ 3 b^2)$ and $(a,b)=1$ then $4\mid (a^2+ 3 b^2)$. How can I show $\frac{1}{4} (a^2+ 3 b^2)$ is also of the form $(c^2+ 3 d^2)$?
Here, clearly $ a$ and $b$ are both odd.
Let $a=2m+1$ and $ b=2n+1$
$\implies\frac{1}{4} (a^2+ 3 b^2)= m^2 + m+1 +3n^2 +3n$.
I am stuck here. Can anyone please help how to approach from here. Any help would be appreciated. Thanks in advance.
| $$(x^2+ 3y^2)(z^2 + 3w^2) =(xz - 3wy)^2 + 3(wx+ yz)^2 $$
This means that the set of the numbers which can be represented as the form $x^2 + 3y^2$, is closed under multiplication. (This holds for general $x^2 + n y^2$)
And another observation is, $4 = 1^2 + 3\cdot 1^2$ itself has the representation.
The situation we have is $xz - 3wy = a$, $wx + yz = b$, $z = 1$, $w = \pm1$. This leads to
$$ x \mp 3y = a \qquad \pm x + y = b$$
$$x = \frac{a+3b}{4}\quad y = \frac{-a+b }{4} \qquad \text{ or}\qquad x = \frac{a - 3b}{4}\quad y = \frac{a+b}{4} $$
For any odd pair $(a, b)$, $x, y$ can be choosen to be integer and satisfies $x^2 + 3y^2 = (a^2 + 3b^2)/4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$ Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$
My try:
we have $$\frac{13}{32}=\frac{2^2+3^2}{2^5}=\frac{1}{8}\left(1+(1.5)^2)\right)$$
Let $x=1.5$
Now consider the function $$f(x)=\frac{1+x^2}{8}-\ln x$$
$$f'(x)=\frac{x}{4}-\frac{1}{x}$$ So $f$ is Decreasing in $(0,2)$
any help here?
| \begin{align*}
\exp\left(\frac{13}{64}\right) & = \exp\left(\frac15\right)\exp\left(\frac1{320}\right) \\
& > \left(1 + \frac15 + \frac1{50} + \frac1{750}\right)\left(1 + \frac1{320}\right) \\
& = \left(1 + \frac{166}{750}\right)\left(1 + \frac1{320}\right) \\
& = \frac{458}{375}\times\frac{321}{320} = \frac{229\times107}{125\times160} \\
& = \frac{24{,}503}{20{,}000} > \frac{24{,}500}{20{,}000} = \frac{49}{40} \\
\therefore\
\exp\left(\frac{13}{32}\right) & > \left(\frac{49}{40}\right)^2 = \frac{2{,}401}{1{,}600} > \frac32.
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Can any cyclic polynomial in $a, b, c$ be expressed in terms of $a^2b+b^2c+c^2a$, $a+b+c$, $ab+bc+ca$ and $abc$? Problem. Let $f(a,b,c)$ be a cyclic polynomial in $a, b, c$.
Can $f$ always be expressed as $g(a+b+c, ab+bc+ca, abc, a^2b+b^2c + c^2a)$ for some polynomial $g(p, q, r, Q)$?
Motivation: If we want to prove $g(a+b+c, ab+bc+ca, abc, a^2b+b^2c + c^2a)\ge 0$, and $g(p, q, r, Q)$ is non-increasing with $Q$,
by using the known inequality $a^2b + b^2c + c^2a \le \frac{4}{27}(a+b+c)^3 - abc$(see How to prove this inequality? $a^{2}+b^{2}+c^{2}\leq 3$),
sometimes, we may prove $g(a+b+c, ab+bc+ca, abc, \frac{4}{27}(a+b+c)^3 - abc)\ge 0$.
As a result, we deal with a symmetric inequality rather than the original cyclic inequality.
Is it known? Is it easily to prove?
Let us see some examples. Denote $p = a+b+c, q = ab+bc+ca, r = abc$ and $Q = a^2b+b^2c+c^2a$.
1) $ab^2 + bc^2 + ca^2 = (a+b+c)(a^2+b^2+c^2) - a^3-b^3-c^3 - (a^2b+b^2c+c^2a)$
2) From the known identity $a^3b+b^3c+c^3a +(ab+bc+ca)^2 = (a+b+c)(a^2b+b^2c+c^2a+abc)$, we have
$a^3b+b^3c+c^3a = p(Q + r) - q^2$.
3) $a^3b^2+b^3c^2+c^3a^2 = Qq - r(p^2-2q) - rq$.
4) $(a^2+b)(b^2+c)(c^2+a) = \cdots$
Any comments and solutions are welcome.
| I also can't not prove but I use my tool in Maple so I can write.
$$f(a+b+c,\,ab+bc+ca,abc,a^2b+b^2c+c^2a) \to F(p,q,r,\Delta)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the period of $\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}\cdots$? In calculus, $\cos x$ is defined as $1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}\cdots$. With this definition, how can we find the period of $\cos x$
| Define $\sin(x) = x - x^3/3! + ...$. Note that $d/dx \sin(x) = \cos(x)$ and $d/dx \cos(x) = -\sin(x)$.
Taylor expand $f(x) = \cos(x + \alpha)$ about $x = 0$ to get $\cos(\alpha + x) = f(x) = \cos(\alpha) - \sin(\alpha) - \cos(\alpha) x^2/2 + \sin(\alpha) x^3 / 3! + ... = \cos(\alpha) (1 - x^2 / 2! + ...) - \sin(\alpha) (x - x^3 / 3! + ...) = \cos(\alpha) \cos(x) - \sin(\alpha) \sin(x)$. That is, for all $x, \alpha$, $\cos(x + \alpha) = \cos(x) \cos(\alpha) - \sin(x) \sin(\alpha)$.
Furthermore, note that $\cos^2(x) + \sin^2(x)$ has a derivative of 0 and thus is constant; therefore, $\cos^2(x) + \sin^2(x) = \cos^2(0) + \sin^2(0) = 1$ for all $x$.
We now seek some $\eta > 0$ such that $\cos(\eta) = 0$. We know such an $\eta$ exists since if one didn't, we would have $d^2/dx^2 \cos(x) = -\cos(x)$ always negative for $x > 0$ and therefore $\cos(x)$ would have a negative, decreasing derivative for all $x > 0$ and yet still remain positive, which is clearly impossible. Take the least such $\eta$; then clearly we must have $\cos'(\eta) = -\sin(\eta) \leq 0$; then $\sin(\eta) = 1$. Define $\tau = 4 \eta$. We have $\cos(\tau / 2) = \cos(\eta + \eta) = \cos^2(\eta) - \sin^2(\eta) = -1$ and therefore $\sin(\tau / 2) = 0$. Then $\cos(\tau) = \cos(\tau/2 + \tau/2) = \cos^2(\tau / 2) - \sin^2(\tau / 2) = 1$, and therefore $\sin(\tau) = 0$. Then for all $x$, we have $\cos(x + \tau) = \cos(x) \cos(\tau) - \sin(x) \sin(\tau) = \cos(x)$. A bit more analysis shows that no $\tau'$ such that $0 < \tau' < \tau$ satisfies $\cos(0 + \tau') = \cos(\tau') = 1$; then $\tau$ is the period of the cosine function.
| {
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If $a, b, c, d\in\mathbb R^+, $ then prove that $\displaystyle\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b}\ge 0.$ My approach: We have:
$\displaystyle\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b}$
$\displaystyle=\frac{a+c}{b+c}-1+\frac{b+d}{c+d}-1+\frac{c+a}{d+a}-1+\frac{d+b}{a+b}-1$
$\displaystyle=(c+a)\left(\frac1{b+c}+\frac1{d+a}\right)+(b+d)\left(\frac1{c+d}+\frac1{a+b}\right)-4\ge (c+a) \left(\frac{2\cdot 2}{b+c+d+a}\right)+(b+d)\left(\frac{2\cdot 2}{c+d+a+b}\right)-4$
(on applying AM $\ge$ HM.)
$\displaystyle=\left(\frac4{a+b+c+d}\right) (c+a+b+d)-4=0.$
Hence the inequality.
But I would love to know is there any other elegant method to prove the same? Please mention. Thanks in advance.
| If $a-b\geq b-c \geq c-d \geq d-a$ and $\frac{1}{b+c}\geq \frac{1}{c+d}\geq \frac{1}{d+a}\geq \frac{1}{a+b}$
We can apply Chebyshev's sum inequality to get :
$$\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b}\geq 0.25(a-b+b-c+c-d+d-a)\Big(\frac{1}{b+c}+ \frac{1}{c+d}+\frac{1}{d+a}+\frac{1}{a+b}\Big)=0$$
For the other cases you can apply rearrangement inequality
| {
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Weird arithmetic and fractions. When the wrong method gives right results. In Ian Stewart's book "Cabinet of mathematical curiosities" he mentions a (fictional) student that performs the following operation:
$\frac{1}{4} \times \frac{8}{5} = \frac{18}{45} $
Stewart asks when this method works (given non-zero digits in the first two fractions). We're then trying to solve:
$\frac{a}{b} \times \frac{c}{d} = \frac{10a+c}{10b+d} $
which simplifies to
$\ ac(10b+d) = bd(10a+c)$
How do you use this expression to find all the possible solutions? Any clever tricks or systematic approaches?
| There are a total of $95$ solutions that satisfies $10ab(c-d)=(b-a)cd$.
One obvious case is when $c=d$ and $b=a$.
Also, note that $a=b \iff c=d$.
Suppose $c \ne d$ or $b \ne a$, then we have a total of $14$ solutions.
Also notice that if $(a,b)$ and $(c,d)$ is a valid pair, then $(b,a)$ and $(d,c)$ is another valid pair.
Hence, we can assume that $a > b$, which would imply that $c< d$ in that case, there are $7$ solutions.
$$10ab(c-d)=(b-a)cd$$
Consider a few cases:
*
*$b-a=-5$, then $-2ab(c-d)=cd$. Notice that the parity of $a$ and $b$ must be distinct. LHS is a multiple of $4$. If none of $c$ and $d$ are multiple of $4$, then both of them are even, and the LHS is indeed a multiple of $8$ which is a contradiction. Hence either $c$ or $d$ is a multiple of $4$.
*
*If $c=4$, $-2ab(4-d)=4d$, that is $ab(d-4)=d$ which is equivalent to $(b+5)b(d-4)=d$ where we use $a=b+5$. That is $1 \le b \le 4$. Also if $b \ge 2$, then the LHS is more than $10$, hence $b=1$, and $d$ must be a multiple of $6$. Hence $(a,b,c,d)=(6,1,4,6)$.
*If $c=8$, then $d=9$. $2ab=72$, $ab=36$, $(b+5)(b)=36$ which implies that $b=4$. Hence $(a,b,c,d)=(9,4,8,9)$.
*If $d=4$, then $-2ab(c-4)=4c$, $ab(4-c)=2c$. If $c=1$, LHS is a multiple of $3$, it's a contradiction. If $c=2$, $2ab=4$, $ab=2$, there is no solution. If $c=3$, $ab=6$. Hence $(a,b,c,d)=(6,1,3,4)$.
*If $d=8$, then $-2ab(c-8)=8c$, $b(b+5)(8-c)=4c$. We just have to substitute $b=1,2,3$ to chek that they are not valid.
*If $c=5$, you can check for $d=6.7,8,9$.
*If $d=5$, you can check for $c=1,2,3,4$.
I will just run my Python program.
ans = set()
for a in range(1, 10):
for b in range(1, 10):
for c in range(1,10):
for d in range(1,10):
if 10*a*b*(c-d) == (b-a)*c*d:
ans.add((a,b,c,d))
print(len(ans))
#for i in ans:
# print(i)
ans2 = [i for i in ans if i[0] != i[1] or i[2] != i[3]]
print(len(ans2))
#print(ans2)
#for i in ans2:
# print(i)
ans3 = [i for i in ans2 if i[0] > i[1]]
for i in ans3:
print(i)
which gives me the following output:
95
14
(9, 1, 5, 9)
(2, 1, 4, 5)
(9, 4, 8, 9)
(6, 2, 5, 6)
(4, 1, 5, 8)
(6, 1, 3, 4)
(6, 1, 4, 6)
| {
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Limit $\lim _{x \to 0}\sqrt {x+\sqrt {x+\sqrt{x+\sqrt{x...}}}}=1$ I've been investigating some interesting infinite square roots, and I've arrived at the hypothesis that $$\lim_{x\to 0}\sqrt {x+\sqrt {x+\sqrt{x+\sqrt{x...}}}}=1$$
However, I have tried to prove this but have found myself unable to do so. For example, I've tried re-writing this as $$1=\sqrt{x+1}$$ so $1=x+1$, which leads us to $x=0$, which doesn't exactly work- replacing $x$ with $0$ yields a value of $0$.
That's another side point: obviously the method I just used yields an incorrect result, but where is the maths flawed?
Please could you either prove or disprove my hypothesis? Thank you in advance.
| Take $a_0 = \sqrt{x}$ and $a_{n+1} = \sqrt{x+a_n}$. We need to show that
$1$. $a_{n+1} > a_n$ (the sequence is monotonically increasing)
$2$. There exists an $m$ such that $a_n \leq m$ for all $n$ (the sequence is bounded)
$1$ is easy. We have $a_0 = \sqrt{x}$ and $a_1 = \sqrt{x+\sqrt{x}}$. First, $a_1 > a_0$ as we have
$$\sqrt{x} > 0 \iff x+\sqrt{x} > x \iff \sqrt{x+\sqrt{x}}>\sqrt{x} \iff a_1 > a_0$$
Assume $1$ holds up to $n$. Then $a_{n+1} = \sqrt{x+a_n} > \sqrt{x+a_{n-1}} = a_n$ so $1$ holds for $a_{n+1}$. By induction, $1$ holds such that $a_n$ is monotonically increasing.
Now for $2$, we do the following. You can use the common way to solve this kind of radical, which is to assign a value to it $y$:
$$y = \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}}$$
$$y^2 = x+\sqrt{x+\sqrt{x+\ldots}}$$
$$y^2=x+y$$
$$y^2-y-x=0$$
Using the quadratic equation,
$$y=\frac{1 \pm \sqrt{1+4x}}{2}$$
$y > 0$ so the smaller solution is extraneous.
$$y = \frac{1+\sqrt{1+4x}}{2}$$
this means
$$\lim_{n \to \infty} a_n = \frac{1+\sqrt{1+4x}}{2}$$
and we can prove the bound
$$a_n \leq \frac{1+\sqrt{1+4x}}{2} $$
inductively. First, we have $a_0 = \sqrt{x} < \frac{1}{2} + \sqrt{\frac{1}{4}+x} = \frac{1+\sqrt{1+4x}}{2}$. Now assume the inequality is true for all $a_i$ for $i \leq n$. Then, for $x\geq 0$,
$$\Big(\frac{1+\sqrt{1+4x}}{2}\Big)^2 = \frac{1+2\sqrt{1+4x}+(1+4x)}{4} = x + \frac{1+\sqrt{1+4x}}{2}$$
so
$$\frac{1+\sqrt{1+4x}}{2} = \sqrt{x+\frac{1+\sqrt{1+4x}}{2}}$$
Then
$$a_{n+1} = \sqrt{x+a_n} \leq \sqrt{x+\frac{1+\sqrt{1+4x}}{2}} = \frac{1+\sqrt{1+4x}}{2}$$
Therefore, this sequence is bounded and monotonically increasing, so it converges. Now we can evaluate: at $0$, it comes out to
$$y\vert_0=\lim_{x \to 0}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}}= \frac{1+
\sqrt{1}}{2}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Given $ax^2+bx+c=0$ with two real roots, $x_1>x_2$, find a quadratic equation whose roots are $x_1+1$ and $x_2-1$ without solving the first equation Roots of the equation $ (1): ax^2+bx+c=0$ are $x_{1}$ and $x_{2}$. They are both real.
Without solving first equation, make up new quadratic equation such that one of the its roots is $x_{1} + 1$ and second one is $x_{2}-1$. Note that $x_{2}>x_{1}$.
I started to solve this with Vieta's theorem but I couldn't continue. Deep explanation would be grateful!
Also I don't know what does not solve mean, I can't use a quadratic formula or something more specific?
| $f(x)=ax^2+bx+c=a(x-x_1)(x-x_2)$ is given to you. Consider
\begin{align*}
g(x)&=a(x-(x_1+1))(x-(x_2-1))\\
&=a(x-x_1-1)(x-x_2+1)\\
&=a(\color{red}{(x-x_1)}-1)(\color{blue}{(x-x_2)}+1)\\
&=a(x-x_1)(x-x_2)+a(x-x_1)-a(x-x_2)-a\\
&=f(x)+a\underbrace{(x_2-x_1)}_{\text{given }>0}-a.
\end{align*}
Now $x_2-x_2=\sqrt{(x_1+x_2)^2-4x_1x_2}=\sqrt{\left(\frac{b^2}{a^2}\right)-\frac{4c}{a}}=\frac{\sqrt{b^2-4ac}}{a}$
Thus
$$g(x)=ax^2+bx+c+\sqrt{b^2-4ac}-a$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3721224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find all n-tuples $(a_1, a_2, ...,a_n)$ of positive integers such that $(a_1 ! – 1)(a_2 ! – 1)... (a_n! – 1) – 16$ is a perfect square. Find all n-tuples $(a_1, a_2,...,a_n)$ of positive integers such that $(a_1! – 1)(a_2! – 1)... (a_n! – 1)– 16$ is a perfect square.
https://photos.app.goo.gl/b98MVt1MwmTyecLz7
I've done this:
$N=k^2+4^2$ this means that there is not any $p$ prime , $p|N$ , $p=4m−1$ => $a_n!−1≠4m−1$ => $4$ doesn't divide $a_n!$ => $a_n<4$ and also $a_n≠1$ (then $N=0$ which is wrong) $a_n=2,3$ we'll try some combinations of them
$(2!-1)=1$ it doesn't change anything so all we need to do is find solutions for $a_n=3$ but I don't know can I use same $ a_n$ two or more times?
| Suppose there are $n$ and $k$ such that $$ (a_1!-1)(a_2!-1)...(a_n!-1)=k^2+16=k^2+4^2\quad (*).$$
Notice that $4$ can not divide the LHS, and consequently $\mathrm{gcd}(k,4)=1$. Then by step $4$ here, every factor of $k^2+4^2$ is a sum of two squares. That is, for all $i=1,\ldots,n$, there are integers $a$ and $b$ such that $$a_i!-1=a^2+b^2.$$
If $a_i\geq 4$, then $a_i!-1\equiv -1\pmod 4$, but $a^2+b^2\not\equiv -1 \pmod 4$. Hence we must have $a_i\leq 3$ for all $i=1,\ldots,n$.
It follows that the LHS in $(*)$ is either zero, which is impossible, or equal to $$(3!-1)^m=5^m$$
for some $0\leq m \leq n$.
Now let's solve $$ k^2+4^2=5^m.$$
*
*If $m$ is even, say $m=2c$, then $k^2+4^2=5^{2c}$ yields $2^4=4^2=(5^c-k)(5^c+k)$. Consequently $5^c-k=2^{u}$ and $5^c+k=2^v$ for some $u,v\geq 0$ with $u+v=4$. This is easy to solve, by just discussing cases.
*If $m$ is odd, say $m=2c+1$, then $k^2+4^2=5\cdot 5^{2c}$ yields $$k^2=((2\omega -1) 5^c-4)((2\omega -1) 5^c+4),$$
where $w=\frac{1+\sqrt{5}}{2}$.
The ring $\mathbb{Z}[\omega]$, which is the ring of integers of $\mathbb{Q}(\sqrt{5})$, is a UFD; see oeis.org/A003172. Moreover $(2\omega -1) 5^c-4$ and $(2\omega -1) 5^c+4$ are coprime (because if $p$ is irreducible and divides $(2\omega -1) 5^c-4$ and $(2\omega -1) 5^c+4$, then $p\bar{p}\mid 8$ and so $p\bar{p}=2\mid k^2$, which is impossible). Hence both factors are squares in $\mathbb{Z}[\omega]$, that is, $$(2\omega -1) 5^c-4=(e+\omega f)^2$$
for some $e,f\in \mathbb{Z}$. But using the fact that $\omega^2= 1+\omega$, we can see that the last equation has no solutions in $e,f\in \mathbb{Z}$.
Consequently all the solutions, if any (as I didn't do the computation), come from the case where $m$ is even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3721595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplification of $\sqrt{2\zeta^2-1+2\zeta\sqrt{\zeta^2-1}}+\sqrt{2\zeta^2-1-2\zeta\sqrt{\zeta^2-1}}$ If I try to evaluate $\sqrt{2\zeta^2-1+2\zeta\sqrt{\zeta^2-1}}+\sqrt{2\zeta^2-1-2\zeta\sqrt{\zeta^2-1}}$ numerically for real $\zeta$, it looks like it is just equal to $2|\zeta|$ for $\zeta \ne 0$ and $2j$ for $\zeta=0$, but I can't figure out how to simplify to get there...
It's of the form $\sqrt{b+c} + \sqrt{b-c}$ with $b=2\zeta^2-1$ and $c=2\zeta\sqrt{\zeta^2-1}$. I can write:
$$\sqrt{b+c} + \sqrt{b-c} = \frac{(b+c) - (b-c)}{\sqrt{b+c} - \sqrt{b-c}}$$
but that doesn't seem to help either....
| If $|\zeta|\ge1$, with a substitution $\zeta=\mathrm{sign}(\zeta)\cosh z$, $z>0$ you can find:
$$
\sqrt{2\cosh^2z-1+2\cosh z\sqrt{\cosh^2z-1}}+\sqrt{2\cosh^2z-1-2\cosh z\sqrt{\cosh^2z-1}} =
\\
\sqrt{\cosh^2z+\sinh^2z+2\cosh z\sinh z}+\sqrt{\cosh^2z+\sinh^2z-2\cosh z\sinh z} = \\
(\cosh z+\sinh z)+(\cosh z-\sinh z) = 2\cosh z=2|\zeta|.
$$
If $|\zeta|<1$, the answer depends on how you define a complex square root. But a substitution $\zeta=\sin x$ might help anyway.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
An $n\times n$ matrix that has exactly one $1$ and one $-1$ in each row and column and others are $0$ I came across the following Question
Assume an $n\times n$ matrix that has exactly one $1$ and one $-1$ in
each row and column and others are $0$. Prove that there is a way that
we can change the places of rows and columns in which it gives the
negative of the matrix.
MY TRY-
Call such matrix A.
All we need to do is to find some permutation matrices $P_{1}$ and $P_{2}$ such that $$P_{1}AP_{2} = -A$$
$A$ can be written as a difference of two permutation matrices i.e. $$A = P-Q$$
where P and Q are some permutation matrices
Example of one such matrix of order $3\times3$
$$ \begin{pmatrix}
1 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 1
\end{pmatrix} = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}-\begin{pmatrix}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{pmatrix}.$$
We could first turn every such matrix $A$ by multiplying
by appropriate permutation matrices to the form $I-R$ :-$$P^{T}A = P^{T}(P-Q) = I-R$$ Clearly the permutation matrix R shouldn't have $1$ at the same position as in $I$. R lies in the class of traceless permutation matrices. Now If we are able to find matrices permutation $P_{1}$ and $P_{2}$ such that $$P_{1}(I-R)P_{2} = (R-I) = -(I-R)$$ we'll have
$$P_{1}P^{T}AP_{2} = -P^{T}A \implies PP_{1}P^{T}AP_{2} = -A $$
and we would be done.
But how could I proceed now to find $P_{1}$ and $P_{2}$?
Would we need some extra equation from the fact that $R$ is a traceless permutation matrix?
It was great to see other approaches to solve the problem by Michael Hoppe and user1551.
But I am curious to see how would it be if we go this way?
| Here's an algorithm to transform the matrices. I'll explain by an example.
We want to transform
$$
\begin{pmatrix}
-1 & 0 & 1 & 0\\
0 & -1 & 0 & 1\\
0 & 1 & -1 & 0\\
1 & 0 & 0 & -1
\end{pmatrix}\quad\text{to}\quad
\begin{pmatrix}
1 & 0 & -1 & 0\\
0 & 1 & 0 & -1\\
0 & -1 & 1 & 0\\
-1 & 0 & 0 & 1
\end{pmatrix}.
$$
We may define the companion of the first matrix as
$$\begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 1 & 2
\end{pmatrix}$$
where the companion's first column gives the position of $-1$ and $1$ in the first column of the matrix resp., that is $\left(\begin{smallmatrix}1\\4\end{smallmatrix}\right)$ and so on.
Now changing two columns in the matrix changes the correspondent columns in the companion; exchanging two rows $j$ and $k$ in the matrix exchanges all values of $j$ and $k$ in the companion.
We want to go from
$$\begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 1 & 2
\end{pmatrix}\quad\text{to}\quad\begin{pmatrix}
4 & 3 & 1 & 2\\
1 & 2 & 3 & 4
\end{pmatrix}.$$
Start with exchanging the first and last column of the companion:
$$\begin{pmatrix}
4 & 2 & 3 & 1\\
2 & 3 & 1 & 4
\end{pmatrix}.$$
Now the first column should be $\left(\begin{smallmatrix}4\\1\end{smallmatrix}\right)$, hence we swap row $1$ and $2$ to obtain
$$\begin{pmatrix}
4 & 1 & 3 & 2\\
1 & 3 & 2 & 4
\end{pmatrix}.$$
Repeat the process with the companion's second column, that is switch columns $2$ and $3$:
$$\begin{pmatrix}
4 & 3 & 1 & 2\\
1 & 2 & 3 & 4
\end{pmatrix}$$
and we're done already. That was easily done, now another example for the systematic way
Take the companion
$$\begin{pmatrix}
2 & 3 & 1 & 5 & 4\\
4 & 5 & 3 & 1 & 2
\end{pmatrix}.
$$
Look for cycles in the permutation, there are two, namely $(3,5,1)$ and $(2,4)$. Now first change $3$ to $5$, that is, exchange row $3$ with row $5$ in the corresponding matrix to get
$$\begin{pmatrix}
2 & 5 & 1 & 3 & 4\\
4 & 3 & 5 & 1 & 2
\end{pmatrix},
$$
then exchange $5$ and $1$:
$$\begin{pmatrix}
2 & 1 & 5 & 3 & 4\\
4 & 3 & 1 & 5 & 2
\end{pmatrix}
$$
and for the first cycle finally $1$ and $3$:
$$\begin{pmatrix}
2 & 3 & 5 & 1 & 4\\
4 & 1 & 3 & 5 & 2
\end{pmatrix}.
$$
For the second cycle exchange $2$ and $4$
$$\begin{pmatrix}
4 & 3 & 5 & 1 & 2\\
2 & 1 & 3 & 5 & 4
\end{pmatrix}.$$
Now change the columns for the correct order:
$$\begin{pmatrix}
4 & 5 & 3 & 1 & 2\\
2 & 3 & 1 & 5 & 4
\end{pmatrix}.$$
Done!
For the first example we could have perform the changes $1\leftrightarrow4$, $4\leftrightarrow2$, $2\leftrightarrow3$, and $3\leftrightarrow1$ and then switch the columns accordingly, but there obviously was an easier way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3725338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Find the value of $a_{0}^{2}-a_{1}^{2}+a_{2}^{2}- \dots+a_{2 n}^{2}$ Let n be a positive integer and $$\left(1+x+x^{2}\right)^{n}=a_{0}+a_{1} x+\cdots+a_{2 n} x^{2 n}$$
then the value of $a_{0}^{2}-a_{1}^{2}+a_{2}^{2} - \dots+a_{2 n}^{2}$ is
My approach:-
Replacing $x$ by $(-1 / x),$ we get
$$
\begin{array}{r}
\left(1-\frac{1}{x}+\frac{1}{x^{2}}\right)^{n}=a_{0}-\frac{a_{1}}{x}+\frac{a_{2}}{x^{2}}+\cdots-a_{2 n-1} \cdot \frac{1}{x^{2 n-1}}+\frac{a_{2 n}}{x^{2 n}} \\
\text { or, }\left(1-x+x^{2}\right)^{n}=a_{0} x^{2 n}-a_{1} x^{2 n-1}+a_{2} x^{2 n-2}+\cdots+a_{2 n}..... \tag{1}
\end{array}
$$
And given $\left(1+x+x^{2}\right)^{n}=a_{0}+a_{1} x+\cdots+a_{2 n} x^{2 n} \ldots \ldots \ldots \ldots \ldots . \tag{2}.$
Multiplying corresponding sides of (1) and $(2),$ we have
$$
\left(1+x^{2}+x^{4}\right)^{n}=\left(a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{2 n} x^{2 n}\right) \times\left(a_{0} x^{2 n}-a_{1} x^{2 n-1}+a_{2} x^{2 n-2}+\cdots+\right.
$$
$\left.a_{2 n}\right) \ldots \ldots...\tag{3}$
$$
\left(1+x^{2}+x^{4}\right)^{n}=\left(a_{0}+a_{1} x^{2}+a_{2} x^{4}+\cdots+a_{n} x^{2n}+\cdots+a_{2 n} x^{4 n}\right) \ldots \ldots\tag{4}
$$
Equating coefficient of $x^{2 n}$ on both sides of (3) and (4)
$$
a_{0}^{2}-a_{1}^{2}+a_{2}^{2} -\cdots +a_{2 n}^{2}=a_{n}
$$
But this method seems very tedious to me.
Any other approach would be greatly appreciated
| We have
$$
S_{\,n} (x) = \left( {1 + x + x^{\,2} } \right)^{\,n} = \sum\limits_{k = 0}^{2n} {a_{\,n,\;k} x^{\,k} }
$$
and
$$
S_{\,n} (x) = x^{\,2n} \left( {1 + x^{\, - 1} + x^{\, - 2} } \right)^{\,n} = x^{\,2n} S_{\,n} (1/x)\quad \Rightarrow \,\quad a_{\,n,\;k} = a_{\,n,\;2n - k}
$$
therefore the coefficients are symmetric wrt $k=n$
Then
$$
\eqalign{
& S_{\,n} (x)S_{\,n} ( - x) = \left( {\left( {1 + x + x^{\,2} } \right)\left( {1 - x + x^{\,2} } \right)} \right)^{\,n} = \cr
& = \left( {\left( {1 + x^{\,2} } \right)^{\,2} - x^{\,2} } \right)^{\,n} = \left( {1 + x^{\,2} + x^{\,4} } \right)^{\,n} = S_{\,n} (x^{\,2} ) \cr}
$$
which implies
$$
S_{\,n} (x)S_{\,n} ( - x) = \sum\limits_{k = 0}^{2n} {\left( {\sum\limits_{j = 0}^k {\left( { - 1} \right)^{\,k - j} a_{\,n,\;j} \,a_{\,n,\;k - j} } } \right)x^{\,k} }
= \sum\limits_{k = 0}^{2n} {a_{\,n,\;k} x^{\,2k} } = S_{\,n} (x^{\,2} )
$$
and thus
$$
\sum\limits_{j = 0}^k {\left( { - 1} \right)^{\,k - j} a_{\,n,\;j} \,a_{\,n,\;k - j} } = \sum\limits_{j = 0}^k {\left( { - 1} \right)^{\,k - j} a_{\,n,\;j} \,a_{\,n,\;2n - \left( {k - j} \right)} }
= \left\{ {\matrix{ 0 & {k\,{\rm odd}} \cr {a_{\,n,\;k/2} } & {k\,{\rm even}} \cr } } \right.
$$
and in particular for $k=2n$
$$
\sum\limits_{j = 0}^{2n} {\left( { - 1} \right)^{\,j} a_{\,n,\;j} \,a_{\,n,\;j} } = a_{\,n,\;n}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3726741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Proving that $0I need to prove that $0<e-S_{n}<\frac{4}{(n+1)!}$ $\quad\forall n\in\mathbb{N}\quad$ for $\quad S_{n}=1+1+\frac{1}{2!}+\ldots+\frac{1}{n!}$
Previously, I proved that $2<e<4$, and that $S_n$ is taylor polynomial of $e$.
I thought on proving it using induction:
Base: $S_1=1<2<e<4$
Step - for $n+1$:
$e-S_{n+1}=e-S_{n}-\frac{1}{(n+1)!}$
What am I missing here? It's supposed to be simple.
Thank you!
| You have
$$\begin{aligned}0 < e- S_n &= \sum_{k=n+1}^\infty \frac{1}{(k+1)!}\\
&= \frac{1}{(n+1)!}\left(1 + \frac{1}{n+2} + \frac{1}{(n+2)(n+3) }+ \dots\right )\\
&< \frac{1}{(n+1)!}\left(1 + \frac{1}{n+2} + \frac{1}{(n+2)^2 }+ \dots\right )\\
&= \frac{1}{(n+1)!} \frac{1}{1-\frac{1}{n+2} } = \frac{1}{(n+1)!} \frac{n+2}{n+1} \le \frac{2}{(n+1)!}
\end{aligned}$$
Or using Taylor's theorem:
$$e - S_n = \frac{e^c}{(n+1)!}$$ with $c \in (0,1)$ and with what you already proved
$$0 < e- S_n < \frac{4}{(n+1)!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3728591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Sum of distances of a variable point from two fixed points. Given that $A\equiv(4,2)$ and $B\equiv(2,4)$, find a point $P$ on the line $3x+2y+10=0$ such that $PA+PB$ is minimum.
My attempt:
In $\triangle PAB$,
$$PA+PB\ge AB$$ Hence, the minimum value should be $AB=2\sqrt 2$. But in the solution point $P$ is given as $\displaystyle(-\frac{14}{5}, -\frac{4}{5})$ from which $PA+PB$ comes out to be $10\sqrt 2$.
Where am I going wrong? Help is requested!
| If $A(4,2), B(2,4)$ are on the same side of the line not on the same line $L=3x+2y+10=0.$
Let the image of $A$ about $L$ be $A'$, then $A'P=AP$ then $AP+BP=A'P+BP$ is minimum
when $A',P,B$ are collinear and $PA;+PB=A'B$
Image $A'(x',y')$ is obtained as
$$\frac{x'-4}{3}=\frac{y'-2}{2}=-2\frac{12+4+10}{9+4}\implies x'=-8, y'=-6$$
The equation of $A'B$ is $x-y=-2$ and the required point $P$ is thr intersection of this line with $L$, we get $P(-14/5,-4/5)$ and $AP+BP \ge A'B=10\sqrt{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3732926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
For what value of $m$ the is sum $\sum_{i = 0}^{m} {10 \choose i}{20 \choose m - i}$ where ${p\choose q} = 0$, if $pFor what value of $m$ the is sum
$$\sum_{i = 0}^{m} {10 \choose i}{20 \choose m - i}\text{where ${p\choose q}$} = 0\text{, if $p<q$, a maximum}$$
My approach
$$\sum_i^{m} {10 \choose i}{20 \choose m - i} = {10 \choose 0}{20 \choose m} + {10 \choose 1}{20 \choose m - 1} + \dots + {10 \choose m}{20 \choose 0}$$
$$(1 +x)^{20} = {20 \choose 0} + {20 \choose 1}x + \dots + {20 \choose m-1}x^{m-1} + {20 \choose m}x^{m} + \dots + {20 \choose 20}x^{20}$$
$$(1 +x)^{10} = {10 \choose 0} + {10 \choose 1}x + \dots + {10 \choose 10}x^{10}$$
Later what to do??
Any other method or hint will be greatly welcomed.
| Write the (3) expansion as $$(1+1/x)^{10}=~^{10}C_0 + ~^{10}C_1 (1/x)+ ~^{10}C_2(1/x^2)+......+ ~^{10}C_n(1/x^{10})~~~~~(3)$$
Now multiply (2) and (3) mainly collecting the terms independent of $x$ in the product $$S=\text{coefficient of $x^0$ in} (1+x)^{20}(1+1/x)^{10}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
where is this function with exponential differentiable? I need to find out where the function
$$f(x,y)=\left\{
\begin{array}{cc}
e^{\tfrac{1}{x^2+y^2-1}}, & x^2+y^2<1 \\
0, & x^2+y^2\geq1
\end{array}
\right.$$
is differentiable. Here is my progress
For $x^2+y^2<1$, I can use the chain rule
$$\frac{\partial f}{\partial x}=\frac{-2xe^{\tfrac{1}{x^2+y^2-1}}}{(x^2+y^2-1)^2}, \ \ \frac{\partial f}{\partial y}=\frac{-2ye^{\tfrac{1}{x^2+y^2-1}}}{(x^2+y^2-1)^2}.$$
This partial derivatives are continiuos, so $f$ is differentiable.
For $x^2+y^2>1$, $f$ is constant, then its partial derivatives are zero and $f$ is differenciable in this case too.
Finally, for $x^2+y^2=1$, the fraction $\dfrac{1}{x^2+y^2-1}$ goes to $0$ and $f(x,y)=1$ and we have
$$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0.$$
Therefore, $f$ is differentiable in the whole $\mathbb{R}^2$.
Is it correct?
| Let's consider $$f(t)=\left\{
\begin{array}{cc}
e^{\dfrac{1}{t-1}}, & t<1 \\
0, & t\geq1
\end{array}
\right.$$
Which is differentiable and its composition with differentiable $g(x,y)=x^2+y^2$
For partial derivative let's consider any $(x,y)$ for which $x^2+y^2=1$. Then
$$\lim_{\Delta x \to 0-}\frac{e^{\frac{1}{(x+ \Delta x)^2+y^2-1}}}{\Delta x}=\lim_{\Delta x \to 0-}\frac{1}{\Delta x}e^{\frac{1}{(x +\Delta x)^2- x^2}}=\lim_{\Delta x \to 0-}\frac{1}{\Delta x} e^{\frac{1}{2 x \Delta x+\Delta x^2}}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The quadratic polynomial $P(x)$ has a zero at $x=2$. The polynomial $P(P(x))$ has only one real zero at $x=5.$ Compute $P(0).$
The quadratic polynomial $P(x)$ has a zero at $x=2$. The polynomial $P(P(x))$ has only one real zero at $x=5.$ Compute $P(0).$
If we have that $P(x) = ax^2 +bx+ c$, we get from the first condition that $P(x) = (x-2)(bx+c).$
From here $P(P(x)) = (ax+bx+c -2)(b(ax+bx+c)+c)$, but this just looks very messy and doesn't seem to be helpful at all. Is there some other trick here I'm missing?
| Write $$P(x)=a(x-2)(x-b)\implies P(P(x))=a(P(x)-2)(P(x)-b)=a(a(x-2)(x-b)-2)(a(x-2)(x-b)-b)=a^3(x-5)^2(x-u)(x-\bar{u})$$ since $\deg(P(P(x))=4$, which means it must have at least a double root of 5, while the other two roots are complex conjugates (not necessarily $\ne 5$). So we have $(3a(5-b)-2)(3a(5-b)-b)=0$.
Since 5 is a double root of $P(P(x))$, it is also a root of its derivative function. Hence $(2x-2-b)(a(x-2)(x-b)-b)+(a(x-2)(x-b)-2)(2x-2-b)=0$ when $x=5$, i.e. $(8-b)(3a(5-b)-b)+(3a(5-b)-2)(8-b)=0$. Clearly, $b=8$ solves this second equation. If we plug this value into the first equation, we get
$$(-9a-2)(-9a-8)=0\implies a=-\dfrac 29 \;\text{or}\;-\dfrac 89$$
However by plugging into the original equation we see that both cases doesn't produce the desired $P(x)$. Note that there's no need to solve the quartic equation, it suffices to check that 5 doesn't solve the equation we get here.
Thus, $6a(5-b)-b-2=0\implies 3a(5-b)=\dfrac{b+2}2$. This gives $(\dfrac{b+2}2-2)(\dfrac{b+2}2-b)=0\implies b=2\;\text{or}\; -1$, corresponding to $a=\dfrac 29$ and $a=\dfrac 1{36}$ respectively.
By checking in the same way as before, we see the only fitting solution is $a=\dfrac 29, b=2$, and therefore $$P(x)=\dfrac 29(x-2)^2$$
which means that $$P(0)=\dfrac 89$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3737712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I simplify this sum of arccosines? After trying to solve a geometry problem, represented by the following image
I've arrived at this expression:
$\alpha=\arccos\left(\frac{d+r \cos \left(\varphi+\frac{\vartheta}2\right)}{\sqrt{d^2+r^2+2 dr\cos \left(\varphi+\frac{\vartheta}2\right)}}\right)+\arccos\left(\frac{d+r\cos \left(\varphi-\frac{\vartheta}2\right)}{\sqrt{d^2+r^2+2dr\cos\left(\varphi-\frac{\vartheta}2\right)}}\right)$
Is there a way to simplify such expression?
| $$\arccos \beta +\arccos \gamma=\arccos \left(\beta\cdot\gamma-\sqrt{1-\beta^2} \cdot \sqrt{1-\gamma^2}\right)$$
On Wikipedia there are also these identities:
$$\arccos x_1+\arccos x_2=
\begin{cases}
\arccos\left(x_1x_2-\sqrt{1-x_1^2}\sqrt{1-x_2^2}\right)&
x_1+x_2\ge0\\
2\pi-\arccos\left(x_1x_2-\sqrt{1-x_1^2}\sqrt{1-x_2^2}\right)&
x_1+x_2<0
\end{cases}
$$
$$\arccos x_1-\arccos x_2=
\begin{cases}
-\arccos\left(x_1x_2+\sqrt{1-x_1^2}\sqrt{1-x_2^2}\right)&
x_1\ge x_2\\
\arccos\left(x_1x_2+\sqrt{1-x_1^2}\sqrt{1-x_2^2}\right)&
x_1<x_2
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3737847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Three boxes with balls Box $U_1$ contains $1$ white ball and $2$ black balls. Box $U_2$ contains $2$ white balls and $2$ black balls. We extract without reinsertion two balls from every boxes. The four balls are put in a third box $U_3$ initially empty. We randomly extract a ball from $U_3$. Find the probability that the ball is white.
Well, I reasoned in this way. The possible combinations that ensure that $U_3$ contains at least one white ball are BNBB, NBBB, NNBB, BNBN, BNNB, BNNN, NBBN, NBNB, NBNN, NNBN, NNNB. Thus:
*
*$\mathbb{P}$($U_3$ contains $3$ white balls)$=\mathbb{P}($(BNBB)$\cap$(NBBB)$)=(\frac{1}{3}\cdot1 \cdot\frac{1}{2}\cdot\frac{1}{3})+(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2}\cdot \frac{1}{3})=0,11$
*$\mathbb{P}(U_3$ contains $2$ white balls)$=\mathbb{P}($(NNBB)$\cap$(BNBN)$\cap$(BNNB)$\cap$(NBBN)$\cap$(NBNB)$)=(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{3})+(\frac{1}{3}\cdot 1\cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{1}{3}\cdot 1 \cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{2}{3}\cdot \frac{1}{3}\cdot \frac{1}{2} \cdot \frac{2}{3})=0,46$
*$\mathbb{P}(U_3$ contains $1$ white ball)$=\mathbb{P}($(BNNN)$\cap$(NBNN)$\cap$(NNBN)$\cap$(NNNB)$)=(\frac{1}{3}\cdot 1 \cdot \frac{1}{2}\cdot \frac{1}{3})+(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{3})+(\frac{2}{3} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{2}{3}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{2}{3})=0,33$
*$\mathbb{P}(U_3$ doesn't contain any white balls)$=2\mathbb{P}($(NNNN)$)=2(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{3})=0,1$
Thus $\mathbb{P}($one white ball from $U_3)=\frac{3}{4}\cdot 0,11+\frac{2}{4}\cdot 0,46+\frac{1}{4}\cdot 0,33+\frac{0}{4}\cdot 0,11=0,395$
Is it correct? Particularly I'm interested in reasoning. Thanks in advance.
| The question can be considered as two events' intersection, that a). $X$ white balls drawn from $U_1$ and $U_2$; b). the ball drawn from $U_3$ is white.
For the probability of event (a), we can consider it as a hypergeometric case. We’ve already known that there’re 3 white balls totally, so X∈(0,1,2,3). Applying the PMF of hypergeometric distribution:
$P(X)=\left(\frac {{3 \choose X}{4 \choose 4-X}}{{7\choose 4}}\right)$
In terms of event b, it’s nothing more than $ P(white)$=$X/4$
So what you need to do is:
$P=\sum_{X=0}^{4} P(X)P(white)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to prove that a sequence is Cauchy How do I show if $x_n = \frac{n^2}{n^2 -1/2}$ is a Cauchy sequence? (using the definition of Cauchy sequence)
My attempt: A sequence is Cauchy if $ \forall \epsilon>0$ $ \exists N \in \mathbb N$ $\forall m,n \geq N$ :|$x_n -x_m$|$\leq \epsilon$
|$x_n -x_m|=|\frac{n^2}{n^2 -1/2} -\frac{m^2}{m^2 -1/2}$| $\leq|\frac{n^2}{n^2 -1/2}-1|+|1-\frac{m^2}{m^2 -1/2}|= 1/2|\frac{1}{n^2 -1/2}|+1/2|\frac{1}{m^2 -1/2}|$
$1/2|\frac{1}{n^2 -1/2}| \leq \epsilon$ and also $1/2|\frac{1}{m^2 -1/2}| \leq \epsilon$
So $1/2|\frac{1}{n^2 -1/2}|+1/2|\frac{1}{m^2 -1/2}| \leq 2 \epsilon$
if we choose $N \in \mathbb N$ such that $N >\sqrt{\frac{\epsilon}{2}+\frac{1}{2}}$
| The definition you posted is correct. Note though that
$$
x_n
= \frac{n^2}{n^2-1/2}
= 1 + \frac{1/2}{n^2-1/2},
$$
and it's clear this is a sequence which decreases to $1$, so $x_{n+1} < x_n$ for all $n$. (You could at this point claim that since the sequence is convergent, it is also Cauchy, but if you need a proof from the fundamentals, read on).
Assuming $N<m<n$,
$$
\begin{split}
\left|x_m - x_n\right|
&= x_m - x_n \\
&= \left(1 + \frac{1/2}{m^2-1/2} - 1 - \frac{1/2}{n^2-1/2}\right) \\
&= \frac12 \left(\frac{1}{m^2-1/2} - \frac{1}{n^2-1/2}\right) \\
&= \frac12
\left(\frac{n^2 - m^2}
{\left(m^2-1/2\right)\left(n^2-1/2\right)}\right) \\
&\le \frac12
\left(\frac{n^2}
{\left(m^2-1/2\right)\left(n^2-1/2\right)}\right) \\
&\le \frac{1/2}{m^2-1/2} \left(\frac{n^2}{n^2-1/2}\right)\\
&\le \frac{1/2}{m^2-1/2} \left(1 + \frac{1/2}{n^2-1/2}\right)\\
&\le \frac{1}{m^2-1/2}\\
&\le \frac{1}{N^2}.
\end{split}
$$
Can you find what $N$ you need to pick in terms of $\epsilon$ to have that expression $< \epsilon$ in the end?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluating $\int_{0}^{1}\frac{x-1}{(x+1)\ln x} dx $ Evaluate the following integral $$\displaystyle I=\int_{0}^{1}\frac{x-1}{(x+1)(\ln x)} \mathrm{d}x $$
My work: I tried it by letting $\displaystyle I(a)=\int_{0}^{1}\frac{(x-1)x^a}{(x+1)(\ln x)} \mathrm{d}x$ and then $\displaystyle I'(a)=\int_{0}^{1}\frac{(x-1)x^a}{x+1} \mathrm{d}x$.
Now $\displaystyle I'(a)=\int_{0}^{1}x^a \mathrm{d}x-\int_{0}^{1}\frac{2x^a}{x+1} \mathrm{d}x$
Now if $\displaystyle J(a)=\int_{0}^{1}\frac{x^a}{x+1}\mathrm{d}x$ , then by applying integration by parts, we get the reccurence relation $J(a)+J(a-1)=\dfrac{1}{a}$ and we can solve it then, but the thing is, we neet to find $I(0)$ ,so even if we compute $J(a)$, it wouldn't be defined at $0$ and so would $I(a)$, then how do I find $I'(a)$ by other method?
I also tried the substitution $x \to \frac{1}{x}$, which yields $\displaystyle I=\int_{1}^{\infty}\frac{(x-1)}{(x+1)(\ln x)} \mathrm{d}x$ and when I saw their graphs, it clearly doesn't seem that the area under the graph of this function from $0$ to $1$ and from $1$ to $\infty$ are equal.
I would appreciate if someone could continue from my method and other solutions are also welcomed...
| Here, we will more focus on answering to OP's specific questions:
1. The substitution $x\mapsto1/x$ yields
$$ I = \int_{1}^{\infty} \frac{x-1}{x^2(x+1)\log x} \, \mathrm{d}x.
$$
So it seems that OP made a mistake when applying the substitution.
2. Continuing from OP's approach, the recurrence relation and $J(\infty)=0$ together imply
$$ J(a)
= \frac{1}{a+1}-\frac{1}{a+2}+\frac{1}{a+3}-\dots,$$
and hence
\begin{align*}
-I'(a)
&= -\frac{1}{a+1} + 2 \biggl( \frac{1}{a+1}-\frac{1}{a+2}+\frac{1}{a+3}-\dots \biggr) \\
&= \biggl( \frac{1}{a+1} - \frac{2}{a+2} + \frac{1}{a+3} \biggr) + \biggl( \frac{1}{a+3} - \frac{2}{a+4} + \frac{1}{a+5} \biggr) + \dots \\
&= \sum_{n=1}^{\infty} \biggl( \frac{1}{a+2n-1} - \frac{2}{a+2n} + \frac{1}{a+2n+1} \biggr).
\end{align*}
Now we integrate both sides from $0$ to $\infty$. Then the left-hand side becomes $I(0)$ by $I(\infty) = 0$. On the other hand, each term of the summation is non-negative, so we can apply the Fubini-Tonelli Theorem to interchange the order of integration and summation to get
\begin{align*}
I(0)
&= \int_{0}^{\infty} (-I'(a)) \, \mathrm{d}a \\
&= \sum_{n=1}^{\infty} \int_{0}^{\infty} \biggl( \frac{1}{a+2n-1} - \frac{2}{a+2n} + \frac{1}{a+2n+1} \biggr) \, \mathrm{d}a \\
&= \sum_{n=1}^{\infty} (-\log(2n-1) + 2 \log(2n) - \log(2n+1)) \\
&= \log \Biggl( \prod_{n=1}^{\infty} \frac{2n}{2n-1} \cdot \frac{2n}{2n+1} \Biggr).
\end{align*}
By using the Wallis product formula, the product term reduces to $\frac{\pi}{2}$, proving
$$ I(0) = \log\left(\frac{\pi}{2}\right). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
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How to show that $J_{n+1} = \frac{3n-1}{3n} J_n$? Let $$J_n := \int_{0}^{\infty} \frac{1}{(x^3 + 1)^n} \, {\rm d} x$$
where $n > 2$ is integer. How to show that $J_{n+1} = \frac{3n-1}{3n} J_n$?
| Beware: overkill. By letting $\frac{1}{1+x^3}=u$ and exploiting Euler's Beta function we have
$$ J_n = \frac{1}{3}\int_{0}^{1} u^{n-4/3}(1-u)^{-2/3}\,du = \frac{1}{3}\,B\left(n-\frac{1}{3},\frac{1}{3}\right)=\frac{\Gamma\left(n-\frac{1}{3}\right)\Gamma\left(\frac{1}{3}\right)}{3\,\Gamma(n)}$$
hence
$$ \frac{J_{n+1}}{J_n} = \frac{n-\frac{1}{3}}{n}=\frac{3n-1}{3n}=1-\frac{1}{3n}.$$
Since $J_1=\frac{2\pi}{3\sqrt{3}}$ by partial fraction decomposition we also have
$$ J_n=\frac{2\pi}{3\sqrt{3}}\prod_{k=1}^{n-1}\left(1-\frac{1}{3k}\right) $$
for any $n\geq 2$. $J_n$ is a log-convex function by the Cauchy-Schwarz inequality and $J_n$ behaves like $\frac{\Gamma(1/3)}{3\sqrt[3]{n}}$ for large values of $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Problem with summation by method of difference Question: What would be the result of: $$\sum_{k=1}^{n}\frac{1}{n(n+2)}$$
My Approach:
Let $T_n$ denote the $n^{th}$ term of the given series. Then we have
$$T_1=\frac12 \left(\frac11-\frac13\right)$$
$$T_2=\frac12 \left(\frac12-\frac14\right)$$
$$T_3=\frac12 \left(\frac13-\frac15\right)$$
And so on up till
$$T_n=\frac12 \left(\frac1n-\frac1{n+2}\right)$$
I can see that the series telescopes and the terms start to cut each other after an interval of one. My only problem is, how do I find the terms that remain in the end?
| For all $k \geq 1$ you have
$$
\frac{1}{k(k+2)} = \frac{1}{2} \left( \frac{1}{k} - \frac{1}{k+2} \right)
$$
so
$$
\begin{aligned}
\sum_{k=1}^n \frac{1}{k(k+2)} &= \frac{1}{2}\sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k+2} \right) \\
&= \frac{1}{2}\left( \sum_{k=1}^n \frac{1}{k} - \sum_{k=1}^n \frac{1}{k+2} \right) \\
&= \frac{1}{2}\left( \sum_{k=1}^n \frac{1}{k} - \sum_{k=3}^{n+2} \frac{1}{k} \right) \\
&= \frac{1}{2}\left( 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2}\right) \\
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3748524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Area of a Sector of a Circle Question
In the figure, $AB$ and $CD$ are two arcs subtended at center $O$. $r$ is the radius of the sector $AOB$. I was told to find the radius, $x$ (the angle), and the shaded area. I know $2\pi r\cdot\dfrac x{360} = 13$. And $\pi(r+4)^2 \cdot \frac{x}{360} - \pi r^2 \cdot \frac{x}{360}$ = shaded area
| Hint:
$$\begin{align}2\pi r\dfrac{x}{360} &= 13\tag{1}\\ 2\pi (r + 4)\dfrac x{360} &= 17\tag{2}\end{align}$$
What happens if you divide $(1)$ with $(2)$?
$$\begin{align}\dfrac{r}{r + 4} &= \dfrac{13}{17}\implies r = 13 \\ x = \dfrac{13\cdot360}{2\pi\cdot 13} &= \dfrac{180}{\pi}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3749391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Does $\sum\limits_{k \geq 0} \frac{1}{(4k+1)(4k+2)} = \frac{\log(2)}{4} + \frac{\pi}{8}$ hold? Context: I'm interested in trying to find as many explicit formulae as possible for:
$$ S_n(s) = \sum_{k \geq 0} \frac{1}{\prod_{i \in s} (nk+i)} $$
for all $ n \geq 2 $ and for all $ s $ subset of $ \{ 1, \dots, n \} $ such that $ \mbox{Card}(s) = 2 $.
So far, I can establish
$$ S_2(\{ 1, 2 \}) = \log(2) $$
$$ S_3(\{ 1, 2 \}) = \frac{\pi \sqrt{3}}{9},
S_3(\{ 2, 3 \}) = \frac{\log\left(3\right)}{2}-\frac{\pi\sqrt{3}}{18},
S_3(\{ 1, 3 \}) = \frac{\log\left(3\right)}{4}+\frac{\pi\sqrt{3}}{36} $$
$$ S_4(\{ 1, 3 \}) = \frac{\pi}{8}, S_4(\{ 2, 4 \}) = \frac{\log(2)}{4} $$
But I just guess that
$$ S_4(\{ 1, 2 \}) = \frac{\log(2)}{4} + \frac{\pi}{8} $$
Is there a generic method to establish all formulae?
| $$\begin{aligned} S&= \sum_{k\geq 0}\frac{1}{(4k+1)(4k+2)} \\& =\sum_{k\geq 0}\left(\frac{1}{4k+1}-\frac{1}{4k+2}\right)\\& =\frac{1}{4}\sum_{k\geq 0} \left(\frac{1}{k+\frac{1}{4}}-\frac{1}{k+\frac{1}{2}}\right)=\frac{1}{4}\left(\psi^0\left(\frac{1}{2}\right)-\psi^0\left(\frac{1}{4}\right)\right)\\& =\frac{1}{4}\left(-\gamma -\ln(4)+\gamma+\frac{\pi}{2}+\ln(8)\right)=\frac{\ln2}{4} +\frac{\pi}{8}\end{aligned} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3751246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Evaluate $\int_{1}^{\sqrt{2}} \frac{x^4}{(x^2-1)^2+1}\,dx$ Evaluate the integral:
$$\int_{1}^{\sqrt{2}} \frac{x^4}{(x^2-1)^2+1}\,dx$$
The denominator is irreducible, if I want to factorize and use partial fractions, it has to be in complex numbers and then as an indefinite integral, we get
$$x + \frac{\tan^{-1}\left(\displaystyle\frac{x}{\sqrt{-1 - i}}\right)}{\sqrt{-1 - i}} + \frac{\tan^{-1}\left(\displaystyle\frac{x}{\sqrt{-1 + i}}\right)}{\sqrt{-1 + i}}+C$$
But evaluating this from $1$ to $\sqrt{2}$ is another mess, keeping in mind the principal values. I also tried the substitution $x \mapsto \sqrt{x+1}$, which then becomes
$$\frac{1}{2}\int_{0}^1 \frac{(x+1)^{3/2}}{x^2+1}\,dx$$
I don't see where I can go from here. Another substitution of $x\mapsto \tan x$ also leads me nowhere.
Should I approach the problem in some other way?
| Note
\begin{align}
I=&\int_{1}^{\sqrt{2}} \frac{x^4}{(x^2-1)^2+1}\,dx\\
= &\int_{1}^{\sqrt{2}} \left(1+\frac{2x^2-2}{x^4-2x^2+2}\right)\,dx\\
= &\sqrt2-1+\int_{1}^{\sqrt{2}} \frac{2-\frac2{x^2}}{x^2+\frac2{x^2}-2}dx\\
=& \sqrt2-1 + (1+\frac1{\sqrt2})I_1 + (1-\frac1{\sqrt2})I_2\tag1\\
\end{align}
where
\begin{align}
I_1= \int_{1}^{\sqrt{2}} \frac{1-\frac{\sqrt2}{x^2}}{x^2+\frac2{x^2}-2}dx
&=\int_{1}^{\sqrt{2}} \frac{d(1+\frac{\sqrt2}{x})}{(x+\frac{\sqrt2}x)^2-2(1+\sqrt2)}=0 \\
I_2= \int_{1}^{\sqrt{2}} \frac{1+\frac{\sqrt2}{x^2}}{x^2+\frac2{x^2}-2}dx
&=\int_{1}^{\sqrt{2}} \frac{d(1-\frac{\sqrt2}{x})}{(x-\frac{\sqrt2}x)^2+2(\sqrt2-1)}\\
&=\sqrt{\frac2{\sqrt2-1}} \tan^{-1}\sqrt{\frac{\sqrt2-1}2}
\end{align}
Plug $I_1$ and $I_2$ into (1) to obtain
$$I = \sqrt2-1 + \sqrt{\sqrt2-1}\tan^{-1}\sqrt{\frac{\sqrt2-1}2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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How to integrate $\int \sqrt{1-\dfrac{1}{25x^2}}\ dx$? How to integrate the following:
$$\int \sqrt{1-\dfrac{1}{25x^2}}\ dx$$
What I did is:
$$\int \sqrt{1-\dfrac{1}{25x^2}}\ dx=\int \dfrac{\sqrt{25x^2-1}}{5x}\ dx$$
I substituted $5x=\sec\theta$, $dx=\dfrac{1}{5}\sec\theta\tan\theta\ d\theta $
$$=\int \dfrac{\sqrt{\sec^2\theta-1}}{\sec\theta}\ \dfrac{1}{5}\sec\theta\tan\theta\ d\theta$$
$$=\frac15\int \tan^2\theta\ d\theta$$
used $\tan^2\theta=\sec^2\theta-1$
$$=\frac15\int( \sec^2\theta-1)\ d\theta$$
$$=\dfrac15\tan\theta-\frac15\theta+c$$
back to $x$
$$=\dfrac15\sqrt{25x^2-1}-\frac15\sec^{-1}(5x)+c$$
I am not sure whether my answer is correct.
My question: Can I integrate this with other substitutions? If yes, please help me. Thank you
| Integrate over all domain $x$ as follows\begin{align}
\int \sqrt{1-\frac{1}{25x^2}}\ dx
&\overset{ibp}= x \sqrt{1-\frac{1}{25x^2}} + \frac1{5} \int \frac1{\sqrt{1-\frac{1}{25x^2}}}d(\frac1{5x})\\
&= x \sqrt{1-\frac{1}{25x^2}} +\frac15 \sin^{-1}\frac1{5x}+C
\end{align}
| {
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How can i evaluate $\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x\right)\ln \left(1+x\right)}{x}\:dx$ I want to evaluate $$\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x\right)\ln \left(1+x\right)}{x}\:dx$$
I tried integration by parts and shape it in a way that i could expand either $\ln$ terms.
$$-\int _0^1\frac{\ln ^3\left(x\right)\ln \left(1-x\right)}{1+x}dx+\int _0^1\frac{\ln ^3\left(x\right)\ln \left(1+x\right)}{1-x}\:dx$$
After this i tried expand the terms but i still couldnt go through, any different approaches are welcome.
| To evaluate this you can make use of the following identity
$$\ln \left(1-x\right)\ln \left(1+x\right)=-\sum _{k=1}^{\infty }x^{2k}\frac{H_{2k}-H_k}{k}-\frac{1}{2}\sum _{k=1}^{\infty }\frac{x^{2k}}{k^2}$$
Resuming on your integral,
$$\int _0^1\frac{\ln \left(1-x\right)\ln ^2\left(x\right)\ln \left(1+x\right)}{x}\:dx$$
$$=-\sum _{k=1}^{\infty }\frac{H_{2k}-H_k}{k}\int _0^1x^{2k-1}\ln ^2\left(x\right)\:dx-\frac{1}{2}\sum _{k=1}^{\infty }\frac{1}{k^2}\int _0^1x^{2k-1}\ln ^2\left(x\right)\:dx$$
$$=-\frac{1}{4}\sum _{k=1}^{\infty }\frac{H_{2k}}{k^4}+\frac{1}{4}\sum _{k=1}^{\infty }\frac{H_k}{k^4}-\frac{1}{8}\sum _{k=1}^{\infty }\frac{1}{k^5}$$
$$=-\frac{7}{4}\sum _{k=1}^{\infty }\frac{H_k}{k^4}-2\sum _{k=1}^{\infty }\frac{\left(-1\right)^kH_k}{k^4}-\frac{1}{8}\zeta \left(5\right)$$
$$=-\frac{21}{4}\zeta \left(5\right)+\frac{7}{4}\zeta \left(2\right)\zeta \left(3\right)-\zeta \left(2\right)\zeta \left(3\right)+\frac{59}{16}\zeta \left(5\right)-\frac{1}{8}\zeta \left(5\right)$$
$$=\frac{3}{4}\zeta \left(2\right)\zeta \left(3\right)-\frac{27}{16}\zeta \left(5\right)$$
Those sums are evaluated here.
| {
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Let $n$ be an odd positive integer. Show that $24 \vert(n^3-n).$
Let $n$ be an odd positive integer. Show that $24 \vert(n^3-n).$
So since $n^3-n=(n-1)n(n+1)$, we have that $3\vert(n^3-n).$
Also since we have that $n$ is odd we can say that $n=2k+1$, for some $k \in \mathbb{Z^+}.$
This implies that $(n+1) = 2k +2 = 2(k+1)$, hence $2\vert(n+1).$ Similar argument can be made for $(n-1)$.
Now $24 = 2 \cdot 3 \cdot 4$ so essentially I'm only missing the part where I would have to show that $4$ divides some of the terms. How can I find that?
| If $n$ is odd, then both $n-1$ and $n+1$ are even.
Either $n-1$ or $n+1$ must be a multiple of $4$ since for every pair of consecutive even numbers, one of them is a multiple of $4$.
Hence $8$ divides $n^3-n$.
| {
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Evaluate limit integrate: $\lim_{n \rightarrow +\infty}{\int_{0}^{\frac {\pi}{2}} \sin^{n} x dx}$ I know,
when n is even,
$$\int_{0}^{\frac {\pi}{2}} \sin^{n} x dx = \frac {(2m-1)!!}{(2m)!!}\frac {\pi}{2} $$
and n is odd,
$$\int_{0}^{\frac {\pi}{2}} \sin^{n} x dx = \frac {(2m)!!}{(2m+1)!!} $$
But when let limit n to infinite, how can I get the limit, or if the limit exists.
--
I find a good way to solve the limit:$$\lim_{x \to +\infty} \frac{(2n-1)!!}{(2n)!!}
$$
as we know,
$2 =\frac{1+3}{2} > \sqrt{1 \cdot 3}$ , so $2n = \frac{(2n-1)+(2n+1)}{2} > \sqrt{(2n-1)(2n+1)}$
so,
$$\lim_{x \to +\infty} \frac{(2n-1)!!}{(2n)!!} = \frac{\sqrt{1 \cdot 3}\cdots \sqrt{(2n-3)(2n-1)}\sqrt{2n-1}}{\sqrt{1\cdot 3} \cdots \sqrt{(2n-3)(2n-1)} \sqrt{(2n-1)(2n+1)}} = \frac{1}{\sqrt{2n+1}} \to 0 (n \to \infty)$$
| More generally, for any real $n>-1$ the formula is: $$\int_0^{\pi/2}\sin^n(x)\mathrm d x =\dfrac{\surd\pi}{2}\cdotp\dfrac{\Gamma(\tfrac{n+1}{2})}{\Gamma(\tfrac{n+2}{2})}$$
Now, consider whether the ratio of these Gamma functions will converge--and if so where to--as $n$ increases, ....
| {
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System of Equations $ 2x ^ 3 + x + 2y - y ^ 3 = 0$, $4x ^ 3 - 2xy + 3y ^ 2 = 10$ I have found a system of equations which I cannot solve.
$$\begin {cases}
2x ^ 3 + x + 2y - y ^ 3 = 0 \\
4 x ^ 2 - 2xy + 3y ^ 2 = 10 \\
\end {cases}
$$
I noticed that the second equation can be written as $\frac{8x ^ 3 + y ^ 3}{2x + y} + 2y ^ 2 = 10$, so I tried to write the first equation similarly and ended up with $8x ^ 3 + y ^ 3 = 5y ^ 3 - 4x - 8y$, but, in the end, it did not really help.
| Homogenization helps.
We obtain:
$$2x^3+(x+2y)\cdot\frac{4x^2-2xy+3y^2}{10}-y^3=0$$ or
$$(y-2x)(12x^2+9xy+4y^2)=0$$ or $$y=2x.$$
Can you end it now?
| {
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Distance between set and point, confused of partial derivatives. Let $H = \{(x,y,z)\ \in \mathbb{R}^{3}: x^2+y^2 - z^2 + 4 = 0$
Compute the shortest distance between H and point $p=(2,4,0)$.
I am a bit confused because I tried a direct approach.
$$ x^2+y^2 + 4 = z^2$$
Let $D(H,p) = \sqrt{(2-x)^{2}+(4-y)^{2} + x^{2} + y^{2}+4}$
So I tried compute $$\frac{\partial D}{\partial x} = (\sqrt{2} (-1 + x))/\sqrt{12 - 2 x + x^2 - 4 y + y^2}$$
$$\frac{\partial D}{\partial y} = (\sqrt{2} (-2 + y))/\sqrt{12 - 2 x + x^2 - 4 y + y^2}$$
It seems not nice to compare with zero.
Do you have another idea?
| You can find the minimum using convexity:
$$D^2(H,p) =(2-x)^{2}+(4-y)^{2} + x^{2} + y^{2}+4$$ $$= 2\left(\frac 12 (2-x)^{2} + \frac 12 x^{2}\right) + 2\left(\frac 12(4-y)^{2}+ \frac 12y^{2}\right) + 4$$
$$\stackrel{convexity\: of \: t^2}{\geq}2\left(\frac{2-x+x}2\right)^2 + 2\left(\frac{4-y+y}2\right)^2+ 4 = 14$$
Equality is reached for $2-x=x$ and $4-y=y$ or $x=1$ and $y=2$, hence, $z=\pm 3$.
| {
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Proving $\binom{2n}{n} \geq \frac{2^{2n}}{n+1}$ How can I prove this?
$$\binom{2n}{n} \geq \frac{2^{2n}}{n+1}$$
I tried using AM-GM but it didn't help.
| This smells like induction to me...
First, let's rewrite this inequality as
$${(2n)!\over (n!)^2}\geq {2^{2n}\over n+1}$$
This inequality is true for $n=0$, which will be our base case. Now suppose this inequality is true for an arbitrary $n$. Then, consider the $n+1$ case.
$$\frac{(2n+2)!}{((n+1)!)^2}=\frac{(2n)!}{(n!)^2}\times\frac{(2n+2)(2n+1)}{(n+1)^2}=\frac{(2n)!}{(n!)^2}\times\frac{2(2n+1)}{n+1}$$
$$\frac{2^{2n+2}}{n+2}=\frac{2^{2n}}{n+1}\times\frac{4(n+1)}{n+2}$$
Notice that
$$\frac{2(2n+1)}{n+1}\geq\frac{4(n+1)}{n+2}$$
Because
$$(4n+2)(n+2)\geq 4(n+1)^2$$
$$4n^2+10n+4\geq 4n^2+8n+4$$
for all $n\geq 0$. So, since $\frac{(2n)!}{(n!)^2}\geq\frac{2^{2n}}{n+1}$ and $\frac{2(2n+1)}{n+1}\geq\frac{4(n+1)}{n+2}$, then the inductive step holds, i.e.
$$\frac{(2n+2)!}{((n+1)!)^2}\geq\frac{2^{2n+2}}{n+2}$$
This completes the proof by induction.
| {
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Find $\lim_{x\to 0} \frac{\sqrt{ax+b}-1}{x}=1$ My answer
Let $\sqrt{ax+b}=y$
Then
$$\lim_{y\to \sqrt b} \frac{(y-1)a}{y^2-b}$$
Let $b=1$
Then $$\lim _{y\to 1} \frac{a}{\frac{y^2-1}{y-1}}$$
$$=\frac a2 =1$$
$$a=2$$
The answer is correct, but this relies on assuming $b=1$, and that doesn’t seem appropriate. What is the correct answer for this?
| Since
$$\lim_{x\to 0}(\sqrt{ax+b}-1) = \lim_{x\to 0}\left(x\frac{\sqrt{ax+b}-1}{x}\right) =0$$
we get
$$\lim_{x\to 0}\sqrt{ax+b}= 1\Rightarrow b=1$$
The limit itself is now the first derivative of $\sqrt{ax+1}$ at $x=0$:
$$\left(\sqrt{ax+1}\right)'(0) = \left. \frac a{2\sqrt{ax+1}}\right|_{x=0} = \frac a{2}\stackrel{!}{=}1 \Rightarrow a=2$$
| {
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Solving $\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{4-2x}$ Can we find the solutions for this equation?
$$\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{4-2x}, \quad x \in \mathbb{R}$$
I tried to amplify the second square root in the $LHS$ with the conjugate and then use AM-GM in order to find where $x$ can be.
Also, the existence conditions imply $x \leq 2$. I obtained $x \leq \frac{4}{3}$.
| The square roots are all well defined real numbers if and only if $0\le x\le2$. In this range, the numbers on both sides are positive, so we can say that
$$\begin{align}
LHS=RHS
&\iff\left(\sqrt{1+\sqrt{2x-x^2}}+\sqrt{1-\sqrt{2x-x^2}}\right)^2=\left(\sqrt{4-2x}\right)^2\\
&\iff2+2\sqrt{1-(2x-x^2)}=4-2x\\
&\iff1+\sqrt{(1-x)^2}=2-x\\
&\iff|1-x|=1-x\quad\text{(note the absolute values!)}\\
&\iff1-x\ge0\\
&\iff x\le1
\end{align}$$
Together with the restriction $0\le x\le2$, we have
$$\sqrt{1+\sqrt{2x-x^2}}+\sqrt{1-\sqrt{2x-x^2}}=\sqrt{4-2x}\iff 0\le x\le1$$
Remark: If it seems odd that the identity suddenly stops working at $x=1$, note that what we really have is the two-piece identity
$$\sqrt{4-2x}=
\begin{cases}
\sqrt{1+\sqrt{2x-x^2}}+\sqrt{1-\sqrt{2x-x^2}}\quad\text{for }0\le x\le1\\
\sqrt{1+\sqrt{2x-x^2}}-\sqrt{1-\sqrt{2x-x^2}}\quad\text{for }1\le x\le2
\end{cases}$$
| {
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Evaluating an integral with a division by $0$ issue Problem:
Evaluate the following integral.
$$ \int_{0}^{4} \frac{dx}{\sqrt{4-x}} $$
Answer:
Let $u = 4 - x$. We have:
\begin{align*}
du &= -dx \\
\int_{0}^{4} \frac{dx}{\sqrt{4-x}} &= \int_{4}^{0} \frac{- du}{\sqrt{u}} \\
\int_{0}^{4} \frac{dx}{\sqrt{4-x}} &= \int_{0}^{4} \frac{- du}{\sqrt{u}} \\
\int_{0}^{4} \frac{dx}{\sqrt{4-x}} &= \int_{0}^{4} u^{-\frac{1}{2}} \,\, du \\
\int_{0}^{4} \frac{dx}{\sqrt{4-x}} &= 2u^{ \frac{1}{2} } \Big|_0^4 = 2(4)^{ \frac{1}{2} } - 2(0)^{ \frac{1}{2} } \\
\int_{0}^{4} \frac{dx}{\sqrt{4-x}} &= 2(2) - 0 \\
\int_{0}^{4} \frac{dx}{\sqrt{4-x}} &= 4
\end{align*}
My answer matches the back of the book. However, the original function was not defined
at $x = 4$ (due to a divide by $0$ issue ) and I did not do anything special for this case. Should I have? Is my solution correct?
| Your solution is fine, if a little informal.
If you want to evaluate it more formally, you need to recognize the integral as an improper integral, so the first step would become $$\int_0^4 \frac{dx}{\sqrt{4-x}} := \lim_{t\to4 {}-} \int_0^t \frac{dx}{\sqrt{4-x}}$$ but then the rest of the solution is almost identical to yours (with an extra limit hanging around until the end), so it doesn't add much.
| {
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Partial Fraction Decomposition of $\frac{1}{x^2(x^2+25)}$ I have been reviewing some integration techniques and have been searching for tough integrals with solutions online. When I was going through the solution, however, I found a discrepancy between my solution and theirs and think what I did was correct instead.
I am trying to solve the indefinite integral: $\int\frac{dx}{x^2(x^2+25)}$. My first step was to break it into the fractions $$\frac{1}{x^2(x^2+25)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+25}$$ Then multiplying both sides by $x^2(x^2+25)$, we find our basic equation to be$$1=A*x(x^2+25)+B*(x^2+25)+(Cx+D)*x^2$$ Solving the system of linear equations, I found that $B=\frac{1}{25}$, $D=\frac{-1}{25}$, and $A=C=0$.
This is where I found the discepancy. The online solution has the basic equation as $$1=A*x(x^2+25)+B*(x^2+25)+(Cx+D)*x$$ so when they solve for coefficients they find that $B=\frac{1}{25}$, $C=\frac{-1}{25}$, and $A=D=0$.
Am I correct or are they? And if my answer is incorrect how does one of the $x$'s cancel out from the $(Cx+D)$ term? Thanks for any help!
| Since the central theme is to assist in the evaluation of an integral another method is to consider the following.
$$ \frac{1}{x^2 \, (x^2 + 5^2)} = \frac{1}{25} \, \left( \frac{1}{x^2} - \frac{1}{x^2 + 5^2} \right) $$
and
\begin{align}
\frac{1}{x^2 + 5^2} &= \frac{A}{x - 5 \, i} - \frac{B}{x + 5 \, i} \\
&= \frac{(A-B) \, x + (A + B) \, 5 \, i}{x^2 + 5^2} \\
&= \frac{1}{10 \, i} \, \left( \frac{1}{x - 5 \, i} - \frac{1}{x + 5 \, i} \right).
\end{align}
Now,
$$ \frac{1}{x^2 \, (x^2 + 5^2)} = \frac{1}{25 \, x^2} - \frac{1}{250 \, i} \, \left( \frac{1}{x - 5 \, i} - \frac{1}{x + 5 \, i} \right) $$
and
\begin{align}
\int \frac{dx}{x^2 \, (x^2 + 5^2)} &= \int \left( \frac{1}{25 \, x^2} - \frac{1}{250 \, i} \, \left( \frac{1}{x - 5 \, i} - \frac{1}{x + 5 \, i} \right) \right) \, dx \\
&= - \frac{1}{25 \, x} + \frac{i}{250} \, \ln\left(\frac{x - 5 \, i}{x + 5 \, i} \right) \\
&= \frac{1}{5^3} \, \left( \tan^{-1}\left(\frac{5}{x}\right) - \frac{5}{x} \right) + c_{0} \\
&= \frac{1}{5^3} \, \left( \cot^{-1}\left(\frac{x}{5}\right) - \frac{5}{x} \right) + c_{0}
\end{align}
| {
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One number is removed from the set of integers from $1$ to $n.$ The average of the remaining numbers is $163/4.$ Which integer was removed?
One number is removed from the set of integers from $1$ to $n.$ The average of the remaining numbers is $\dfrac{163}4$. Which integer was removed?
Source. British Mathematical Olympiad 2010/11, Round 1, Problem 1
I was hoping if someone could spot the flaw in my working for this question.
Attempt.
I began by letting the integer that was removed be $x$.
Then: $$\frac{1 + 2 + \cdots + (x-1) + (x+1) +\cdots + n} {n-1} = \frac{163}{4}$$
There is two arithmetic sums in the denominator, the first from 1 to $x$ and the second from $x+1$ to $n$.
These are equal to $\frac{x(x-1)}{2}$ and $\frac{(n-x)(n+x+1)}{2}$, and subbing in to first equation this gives:
$$\frac{x(x-1) + (n-x)(n+x+1)}{2(n-1)} = \frac{163}{4}$$
which reduces to:
$$\frac{n^2 + n - 2x}{2(n-1)} = \frac {163}{4}$$
And then:
$$2(n^2 + n -2x) = 163(n-1)$$
At first I thought you could consider factors, as 163 was prime then:
$n-1 = 2$ giving $n = 3$ and $n^2 + n - 2x = 163$, which using $n=3$ gives $x= -75.5$ which isn't our positive integer.
I then tried considering a quadratic in $n$ and using the discriminant but again that just looked to give a negative value of $x.$ I would be grateful for any help
| We know that $n$ is odd. Notice that
$$ n^2 - n \leq n^2 + n -2x \leq n^2 +n - 2 $$
$$ \implies n^2 - n \leq \frac{163}{2}(n-1) \leq n^2 +n - 2 $$
which gives us $n \geq 79.5$ and $n\leq 81.5$, so $n=81$
| {
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If $\lim_{x\to 0} \frac{1+a\cos 2x + b\cos 4x}{x^4}$ exists for all $x\in\mathbb R$ and is equal to $c$, find $\lfloor a^{-1} +b^{-1} + c^{-1}\rfloor$ $$\lim_{x\to 0} \frac{1+a(1-\frac{4x^2}{2!} + \frac{16x^4}{4!}-\cdots)+b(1-\frac{16x^2}{2!} + \frac{256x^4}{4!}-\cdots)}{x^4}$$
$$=\lim_{x\to 0} \frac{ (1+a+b) -\frac{x^2}{2!} (4a+16b) + \frac{x^4}{4!} (16a+256b)}{x^4}$$
For limit to exist, $1+a+b=0$ and $4a+16b=0$
So $a=-\frac 43$ and $b=\frac 13$
Now $$c=\lim_{x\to 0} \frac{\frac{x^4}{4!} (16a+256b)\cdots}{x^4}$$
$$c=\frac{16a+256b}{24}$$
$$c=\frac{2a}{3}+\frac{32}{3}b=\frac{24}{9}$$
Then $$\lfloor a^{-1} + b^{-1} + c^{-1} \rfloor = \left\lfloor 3-\frac 34 + \frac{9}{24}\right\rfloor = \lfloor 2.625\rfloor=2$$
The given answer is $8.$
What is wrong in my solution?
| $$L=\lim_{x\to 0} \frac{1+a\cos 2x + b\cos 4x}{x^4}$$
$$L=\lim_{x\to 0} \frac{1+a(1-\frac{4x^2}{2!} + \frac{16x^{4}}{4!}\cdots)+b(1-\frac{16x^2}{2!} + \frac{256x^4}{4!}\cdots)}{x^4}$$
$$\implies L=\lim_{x\to 0} \frac{(1+a+b)+(-2a-8b)x^2+(2a/3+32b/3)x^4}{x^4}$$
for the limit to exist $(1+a+b)=0, (-2a-8b)=0, (2a/3+32b/3=c$ then
$a=-4/3,b=1/3, c=8/3; [1/a+1/b+1/c]=[21/8]=2.
OP's answer is right.
| {
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Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively. So here is the Question :-
Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively. Find $f(-3)$ .
What I tried:- Since it's given that $f(x)$ is a polynomial of degree $3$ , I can assume $f(x) = ax^3 + bx^2 + cx + d$ for some integers $a,b,c,d$ and $a\neq 0$. Then we have :-
$$ax^3 + bx^2 + cx + d = (x^2 - 1)y + (2x - 5)$$
$$ax^3 + bx^2 + cx + d = (x^2 - 4)z + (-3x + 4)$$
This gives that $(x^2 - 1)y + (2x - 5) = (x^2 - 4)z + (-3x + 4)$ . But I am not sure how to proceed further since we have $3$ variables to deal with , and I am stuck here.
Any hints or explanations for this problem will be greatly appreciated !!
| $f(x)$ leaves remainder $2x-5$ when divided by $x^2-1.$
Then,
$$f(1)=2(1)-5=-3$$
$$f(-1)=2(-1)-5=-7$$
$f(x)$ leaves remainder $-3x+4$ when divided by $x^2-4.$
So we have,
$$f(x)=(x^2-4)\,q(x)-3x+4$$
Since $f(x)$ is of degree $3$, the degree of $q(x)$ should be $1.$
Let $q(x)=ax+b.$ Then,
$$f(x)=(x^2-4)(ax+b)-3x+4$$
$$f(x)=ax^3+bx^2-(4a+3)x-(4b-4)$$
Let $x=1$ then,
$$f(1)=a(1)^3+b(1)^2-(4a+3)(1)-(4b-4)=-3$$
$$a+b=\dfrac{4}{3}\quad\quad\dots(1)$$
Let $x=-1$ then,
$$f(-1)=a(-1)^3+b(-1)^2-(4a+3)(-1)-(4b-4)=-7$$
$$a-b=-\dfrac{14}{3}\quad\quad\dots(2)$$
Solving (1) and (2) we get,
$$a=-\dfrac{5}{3}$$
$$b=3$$
Therefore,
$$f(x)=-\dfrac{5}{3}x^3+3x^2+\dfrac{11}{3}x-8$$
Now,
$$\underline{\underline{f(-3)=53}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3770874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Hint to an Inequality I have the following problem in my book-
Prove that for all non-negative real numbers $x, y, z$- $$ 6(x+y-z)(x^2+y^2+z^2)+27xyz \le 10(x^2+y^2+z^2)^{3\over 2} $$
And the solution:
$$\color{red}{10(x^2+y^2+z^2)^{3\over 2}-6(x+y-z)(x^2+y^2+z^2)}=(x^2+y^2+z^2)(10\sqrt{x^2+y^2+z^2}-6(x+y-z))$$ $$ =(x^2+y^2+z^2)\left(\color{red}{{10\over 3}\sqrt{(x^2+y^2+z^2)(2^2+2^2+1^2)}}-6(x+y-z)\right) $$ By C-S,
$$ \ge(x^2+y^2+z^2)\left({10(2x+2y+z)\over 3}-6(x+y-z)\right)={(x^2+y^2+z^2)(2x+2y+28z)\over 3}$$By AM-GM,
$ x^2+y^2+z^2 \ge 9\sqrt[9]{x^8y^8z^2\over 4^8} $ and $ 2x+2y+28z\ge 9\sqrt[9]{4^8xyz^7} $
and multiplication accompanied by division with 3 ends the proof.
The validity of the proof is undoubted, but I want to know that how the author comes up with such an idea?
What has the question reflected that hints the major steps to the proof in red?
Note:- I want just the idea that leads to the proof, not a solution.
| Because the equality occurs for $(x,y,z)=(2,2,1)$ and we want to delete a radical.
We can try to make it by using C-S:
$$\sqrt{(2^2+2^2+1^2)(x^2+y^2+z^2)}\geq2x+2y+z$$ and it gives a right inequality.
The rest is smooth enough.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3770960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
If $p$ and $q$ are primes such that $q \mid {\frac{x^p-1}{x-1}}$ then prove that $q\equiv 1 \pmod{p}$ or $q\equiv 0 \pmod{p}$.
QUESTION: If $p$ and $q$ are primes such that $$q \mid {\frac{x^p-1}{x-1}} , (x\in\Bbb{N}, x>1)$$ then prove that $q\equiv 1 \pmod{p}$ or $q\equiv 0 \pmod{p}$.
MY ANSWER: I came across this lemma, but couldn't prove the second part properly. Here's what I did -
By Fermat's Little Theorem we know that $x^p\equiv{x}\pmod{p}$. Therefore, $$\frac{x^7-1}{x-1}\equiv\frac{x-1}{x-1}=1\pmod{7}$$ Therefore, $q\equiv{1}\pmod{7}$.
Now, I cannot prove that $q\equiv{0}\pmod{7}$. Not simultaneously ofcourse, I know that ..
Here's my try -
We can write the above equation as $$x^6+x^5+x^4+x^3+x^2+x+1$$ But what after this? Even if I chose $q=7$, it does not divide the above equation for all values of $x$. Say $x=7$, then the equation can be rewritten as $$7^6+7^5+7^4+7^3+7^2+7+1$$ and $$7\nmid{7^6+7^5+7^4+7^3+7^2+7+1}$$ So, how do I rigorously prove that $q\equiv{0}\pmod{p}$ ? Or, for which cases is this true?
Any help will be much appreciated. Thank you :)
EDIT: Terrible Mistake :P The first proof of $q\equiv{1}\pmod{7}$ is wrong. So, now I am left with a full question to be proved °_°
| Generalizing this answer first suppose that $q \not \equiv 1\pmod{p}$ so that $\gcd(q-1, p) =1$. Then there exists integers $n,m$ such that $pn + (q-1)m =1$ by Bezout's lemma. Then we have $$x \equiv x^{pn +(q-1)m} \equiv (x^p)^n(x^{q-1})^m \equiv 1 \pmod{q} $$
by Fermat's little theorem and the fact that $q|x^p-1.$
Hence $$\frac{x^p-1}{x-1} = 1+x+\dots+x^{p-1} \equiv p \pmod{q}.$$ This shows that $q=p$. The contrapositive says $q \ne p$ implies $q \equiv 1 \pmod{p}$, so we're done since either $q=p$ or $q \ne p$.
| {
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"url": "https://math.stackexchange.com/questions/3771781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solving : $ \left(1+x^{2}\right) \frac{d y}{d x}+2 x y=4 x^{2} $ Solve the following ODE : $$
\left(1+x^{2}\right) \frac{d y}{d x}+2 x y=4 x^{2}
$$
After rearrangement I get : $$\frac{dy}{dx}=\frac{4x^{2}-2xy}{1+x^{2}}$$ please help me after this step.
| This is a linear DE. Rearranging this we have that
\begin{align*}
&\frac{dy}{dx} + \frac{2x}{1 + x^2}\cdot y = \frac{4x^2}{1 + x^2}\\
\implies&\frac{dy}{dx} + P(x)\cdot y = Q(x)
\end{align*}
Now the integrating factor is $$ I = e^{\int{{2x}/{1 + x^2}}dx} = e^{{{\ln(1 + x^2)}}} = 1 + x^2$$
On multiplication by the integrating factor and on integrating we obtain this
$$y \cdot I = \int Q(x) \cdot I dx + C $$
And on substitution of $I$, we have that
\begin{align*}
&~y (1 + x^2) = \int 4x^2 + C\\
\implies &~ y \cdot (1 + x^2) = \frac{4}{3} \cdot x^3 + C
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to obtain the sum of the series $\sum_{n=0}^{\infty}\frac{1}{2^{n}(3n+1)}$? How to prove what follows?
$$\sum_{n=0}^{\infty}\frac{1}{2^{n}(3n+1)}=\frac{2^{\frac{1}{3}}}{3}\ln\left(\frac{\sqrt{2^{\frac{2}{3}}+2^{\frac{1}{3}}+1}}{2^{\frac{1}{3}}-1}\right)+\frac{\sqrt[3]{2}}{3}\arctan\left(\frac{2^{\frac{2}{3}}+1}{\sqrt{3}}\right)-\frac{2^{\frac{1}{3}}\pi}{6\sqrt{3}}$$
My attempt:
$$\sum_{n=0}^{\infty}\frac{1}{2^n(3n+1)}=\sum_{n=0}^{\infty}\frac{x^{3n+1}}{2^n(3n+1)}|_{x=1}$$
We put $$S(x)=\sum_{n=0}^{\infty}\frac{x^{3n+1}}{2^n(3n+1)}\implies S^{'}(x)=\sum_{n=0}^{\infty}\frac{x^{3n}}{2^n(3n+1)}$$
$$S^{'}(x)=\sum_{n=0}^{\infty}\frac{(\frac{x^3}{2})^n}{3n+1}=\sum_{n=0}^{\infty}(\frac{x^3}{2})^n(1-\frac{3n}{3n+1})=\sum_{n=0}^{\infty}(\frac{x^3}{2})^n-\sum_{n=0}^{\infty}(\frac{x^3}{2})^n\frac{3n}{3n+1}=\frac{1}{2-\frac{x^3}{2}}-\sum_{n=0}^{\infty}(\frac{x^3}{2})^n(\frac{3n}{3n+1})=\alpha-\beta$$
Where
$$\beta=\sum_{n=0}^{\infty}(\frac{x^3}{2})^n(\frac{3n}{3n+1})$$
So $$\beta=?$$
Waiting for your help to find a beta or prove equal above.
| For all $|x|<1$ we have the elementary geometric series $$\sum_{n\geq 0}x^n=\frac{1}{1-x} \\\underbrace{=}^{x\to x/2}\sum_{n\geq 0}\frac{x^n}{2^n} =\frac{2}{2-x}$$ now replacing $x$ by $x^3$ and then on integration from $0$ to $1$ we have $$\sum_{n\geq 0}\frac{1}{2^n(3n+1)}=\int_0^1\frac{2}{2-x^3}dx$$ since $x^3-2=(x-\sqrt[3]2)(x^2-\sqrt[3]{2}x+\sqrt[3]{8})$ by partial fraction of the integrand we write the last expression as $$-\int_0^1\frac{2}{x^3-2} dx=-\frac{2}{3\sqrt[3]4}\int_0^1\left(\color{red}{\frac{1}{x-\sqrt[3]{2}}}-\color{blue}{\frac{x+\sqrt[3]{16}}{x^2+\sqrt[3]{2}x +\sqrt[3]4}}\right)dx$$ It's is easy to see that red integral $$\int_0^1\color{red}{\frac{1}{x-\sqrt[3]2}}dx =\ln(|x- \sqrt[3]2|)\bigg|_0^1=\ln(|1-\sqrt[3]2|)-\ln\sqrt[3]2\cdots(1)$$
Further note that $$\int_0^1\color{blue}{\frac{x+\sqrt[3]{16}}{x^2+\sqrt[3]{2}x+\sqrt[3]{4}}}dx=\frac{1}{2}\int_0^1\left(\frac{2x+\sqrt[3]2}{x^2+\sqrt[3]2 x+\sqrt[3]4}+\frac{3\cdot
\sqrt[3]{2}}{x^2+\sqrt[3]2x+\sqrt[4]{4}}\right)dx$$ The last two integral are standard and elementary logarithm and arctangent integrals and integrating them we have $$\frac{1}{2}\ln(x^2+\sqrt[3]2x+\sqrt[3]4)+\sqrt[3]{2}\tan^{-1}\left(\frac{\sqrt[3]{4}x+1}{\sqrt 3}\right)\bigg|_0^1=\frac{\ln(1+\sqrt[3]{2}+\sqrt[3]{4})-\ln(\sqrt[3]{4})}{2}+\sqrt{3}\tan^{-1}\left(\frac{\sqrt[3]{4}+1}{\sqrt{3}}\right)-\frac{\pi}{2\sqrt{3}}\cdots(2)$$ subtract $(1)$ from $(2)$ and multiply by the factor $-\frac{2}{3\sqrt[3]{2}}$ and simplification gives us the desired result of the series.
$$\frac{2}{3\sqrt[3]{4}}\left(\sqrt{3}\tan^{-1}\left(\frac{\sqrt[3]{4}+1}{\sqrt 3}\right)+\frac{1}{2}\ln(1+\sqrt[3]{2}+\sqrt[3]{4})-\ln|\sqrt[3]{2}-1| -\frac{\pi}{2\sqrt{3}}\right)\approx 1.18143\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Question on determinant Given $x \in \Bbb R$ and
$$P = \begin {bmatrix}1&1&1\\0&2&2\\0&0&3\end {bmatrix}, \qquad Q=\begin {bmatrix}2&x&x\\0&4&0\\x&x&6\end {bmatrix}, \qquad R=PQP^{-1}$$
show that
$$\det R = \det \begin {bmatrix}2&x&x\\0&4&0\\x&x&5\end {bmatrix}+8$$
for all $x \in \Bbb R$.
My attempt: $|R|=\frac{|P||Q|}{|P|}=|Q|$
$$\left|\begin {array}&2&x&x\\0&4&0\\x&x&6\end {array}\right|=\left|\begin {array}&2&x&x\\0&4&0\\x&x&5\end {array}\right|+\left|\begin {array}&2&x&x\\0&4&0\\x&x&1\end {array}\right|=\left|\begin {array}&2&x&x\\0&4&0\\x&x&6\end {array}\right|+8-4x^2$$
What's my mistake?
| You used multilinearity incorrectly. It should be
$$\left|\begin {array}&2&x&x\\0&4&0\\x&x&6\end {array}\right|=\left|\begin {array}&2&x&x\\0&4&0\\x&x&5\end {array}\right|+\left|\begin {array}&2&x&x\\0&4&0\\0&0&1\end {array}\right|=\left|\begin {array}&2&x&x\\0&4&0\\x&x&5\end {array}\right|+8.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Are there any elementary functions $\beta(x)$ that follows this integral $\int_{y-1}^{y} \beta(x) dx =\cos(y)$ Are there any simple functions $\beta(x)$ that follows this integral $$\int_{y-1}^{y} \beta(x) dx =\cos(y)$$
I think there is an infinite amount of solutions that are continuous everywhere but how can I find one that only uses elementary functions?
| Try this:
$$
\beta(x) = \frac{\sin(x+1)}{2} - \frac{\sin(1)\;\cos(x+1)}{2\cos(1)-2}
$$
How to find this? If $N$ is a positive integer, then
$$
\int_0^N \beta(x)\;dx = \sum_{n=1}^N\int_{n-1}^n\beta(x)\;dx
=\sum_{n=1}^N \cos n
=-\frac{\cos(N+1)}{2}-\frac{\sin(2)\;\sin(N+1)}{2(\cos(1)-1)} - \frac{1}{2}
$$
Now make a guess that the same formula will also work when we
replace $N$ with $y$ which is not
a positive integer. Differentiate to get $\beta(y)$.
Then check that it works.
Check:
\begin{align}
\beta(x) &= \frac{\sin(x+1)}{2} - \frac{\sin(1)\;\cos(x+1)}{2\cos(1)-2}
\\
\int\beta(x)\;dx &=
\frac{-\cos(x+1)}{2} - \frac{\sin(1)\;\sin(x+1)}{2\cos(1)-2}
\\
\int_{y-1}^y\beta(x)\;dx &=
\frac{-\cos(y+1)+\cos(y)}{2}
+ \frac{\sin(1)\;(-\sin(y+1)+\sin(y))}{2\cos(1)-2}
\\ &=
\frac{-\cos(1)\cos(y)+\sin(1)\sin(y)+\cos(y)}{2}
+ \frac{\sin(1)\;(-\cos(1)\sin(y)-\sin(1)\cos(y)+\sin(y))}{2\cos(1)-2}
\\ &=
\left[\frac{-\cos(1)+1}{2}
+\frac{-\sin^2(1)}{2(\cos(1)-1)}\right]\cos(y)
+\left[\frac{\sin(1)}{2}
+\frac{-\cos(1)\sin(1)-\sin(1)}{2(\cos(1)-1)}\right]\sin(y)
\\ &=
\frac{-\cos^2(1)+2\cos(1)-1-\sin^2(1)}{2(\cos(1)-1)}\cos(y)
+\frac{\cos(1)\sin(1)-\sin(1)-\sin(1)\cos(1)+\sin(1)}{2(\cos(1)-1)}\sin(y)
\\ &=
\frac{2\cos(1)-2}{2(\cos(1)-1)}\cos(y)
+\frac{\cos(1)\sin(1)-\sin(1)-\sin(1)\cos(1)+\sin(1)}{2(\cos(1)-1)}\sin(y)
\\ &=
\cos(y)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3778088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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Product of sines when the angles form a sequence I am wondering how to find the following value of $x$
$$x=\prod_{n=1}^{9}\sin\left(\frac{n\pi}{10}\right)$$
I notice that it has something to do with the de moivre's theorem as the angles are root angles of $1^\frac{1}{10}$
To my surprise, the value of the above product seems to be a rational number.
The solution is given as
$$x=\frac{10}{2^9}$$
I attempted to solve this product in different ways but none seems to give out such an elegant solution. Thank you for any help!
| $x = (\prod\limits_{n = 1}^4 \sin(\frac{n \pi}{10}))^2$
$\prod\limits_{n = 1}^4 \sin(\frac{n \pi}{10}) = \prod\limits_{n = 1}^2 \sin(\frac{n \pi}{10}) \prod\limits_{n = 1}^2 \cos(\frac{n \pi}{10}) = \prod\limits_{n = 1}^2 \sin(\frac{n \pi}{10}) \cos(\frac{n \pi}{10}) = \prod\limits_{n = 1}^2 \frac{\sin(\frac{n \pi}{5})}{2} = \frac{1}{4} \sin(\frac{\pi}{5}) \sin(\frac{2 \pi}{5})$
$x = \frac{1}{16} \sin^2(\pi/5) \sin^2(2\pi/5) = \frac{1}{64} (1 - \cos(2\pi / 5))(1 - \cos(4\pi / 5))$
Now consider the fact that $\sum\limits_{j = 0}^4 \cos(2j \pi / 5) = 0$. In particular, we have $2 \cos(2 \pi / 5) + 2 \cos(4 \pi / 5) = -1$. Then, letting $w = \cos(2\pi / 5)$, we have $2 w + 4 w^2 - 2 = -1$. Then $2w + 4w^2 = 1$.
We have $x = \frac{1}{64} (1 - w)(2 - 2w^2)$. Now we have $4(1 - w)(2 - 2 w^2) = 2(1 - w)(4 - 4w^2) = 2(1 - w)(3 + 2w) = 2 (3 - 2w^2 - w) = 6 - 4w^2 - 2w = 6 - 1 = 5$. Then $x = \frac{1}{64} \frac{5}{4} = \frac{5}{2^8}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I solve $(\cos2x+1)^2=1/2$? $x$ belongs to $$[0,\pi/2[$$
$$(\cos(2x+1))^2 = \frac{1}{2}$$
I tried to find $x$ using
$$2x+1=\frac{\pi}{4} +\frac{n\pi}{2}
\qquad\text{or}\qquad
2x+1=\frac{3\pi}{4} + n\pi$$
but I didn't find my answer in MCQ which is
$$\frac{3\pi-4}{8}
\qquad\text{and}\qquad
\frac{5\pi-4}{8}.$$
I just need a hint on how to solve such equation.
| Your inital step is right, indeed we have that
$$(\cos(2x+1))^2 =\frac12 \iff \cos(2x+1)=\pm\frac{\sqrt 2}2 \iff 2x+1=\frac \pi 4+k\frac \pi 2$$
then
$$x=\frac \pi 8+k\frac \pi 4-\frac12$$
and since $x\in [0,\pi/2[$ we obtain
$$x_1=\frac 3 8\pi-\frac12, x_2=\frac 5 8\pi-\frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Why am I getting derivative of $y = 1/x$ function as $0$? I was finding the derivative of the function: $y = 1/x$. I did the followed steps:
\begin{align*}
\frac{\frac{1}{x+dx} - \frac{1}{x}}{dx} &=\left(\frac{1}{x+dx} - \frac{1}{x} \right) \frac{1}{dx} \\
&= \frac{1}{x dx + (dx)^2} - \frac{1}{x dx}.
\end{align*}
Since, $(dx)^2$ would be extremely small, I removed it, so
$$\frac{1}{x dx} - \frac{1}{x dx}$$
which is equal to zero.
why am I getting the derivative of $y = 1/x$ as $0$?
| Do not remove $(dx)^2$:
$$\frac{\frac{1}{x+dx} - \frac{1}{x}}{dx}=\frac{x-(x+dx)}{(x+dx)xdx}=-\frac{1}{x(x+dx)} \to -\frac{1}{x^2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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We have $a,b,c$ and $d$ are real numbers such that $\frac{b + c + d}{a} = \frac{a + c + d}{b} = \frac{a + b + d}{c} = \frac{a + b + c}{d} = r$.
If $a,b,c$ and $d$ are real numbers such that $\frac{b + c + d}{a} = \frac{a + c + d}{b} = \frac{a + b + d}{c} = \frac{a + b + c}{d} = r$ , find the sum of all the possible values of $r$ .
What I Tried :- First of all, when $a = b = c = d$ , the $4$ equations hold and we get $r = 3$ as $1$ solution .
For other solutions I simplified to get $6$ expressions as :-
$b^2 + bc + bd = a^2 + ac + ad$
$bc + c^2 + cd = a^2 + ab + ad$
$bd + cd + d^2 = a^2 + ab + ac$
$ac + c^2 + cd = ab + b^2 + bd$
$ad + cd + d^2 = ab + b^2 + bc$
$ad + bd + d^2 = ac + bc + c^2$
Now I have no idea how to start finding solutions for $r$ from here . Can anyone help?
| First of all $$\frac{b + c + d}{a} = \frac{a + c + d}{b} = \frac{a + b + d}{c} = \frac{a + b + c}{d} = r$$ gives you $$b + c + d=ar\\a + c + d=br\\a + b + d=cr\\a + b + c=dr$$ summing all these we get $$3(a+b+c+d)=r(a+b+c+d)$$ Now if $(a+b+c+d)\neq0$ then we must have $r=3$.
Now if $(a+b+c+d)=0$, then $r=\frac{a+b+c}{d}\implies r=\frac{a+b+c+d-d}{d}=\frac{-d}{d}=-1$ (For clarity: since we are writing $a,b,c,d$ in the denominators, we must need $a,b,c,d$ to be non-zero.). Therefore the sum of all possible values of $r$ is $3-1=2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given matrices $A$ and $B$, solve for $Z$ in the following equation: $A^{-4}BZ^{3}A^{5}B^{2}A=-3A^{-6}$ $
A=
\begin{pmatrix}
2 & -4 & 3\\
4 & -1 & 2\\
-1 & 1 & -1\\
\end{pmatrix}
$
$
B=
\begin{pmatrix}
3 & -4 & 5\\
1 & -2 & 3\\
-2 & 3 & -4\\
\end{pmatrix}
$
Solve Z in the following equation: $A^{-4}BZ^{3}A^{5}B^{2}A=-3A^{-6}$.
I tried this way:
$A^{-4}BZ^{3}A^{5}B^{2}A=-3A^{-6}$
$\Leftrightarrow BZ^{3}A^{5}B^{2}A=-3A^{-2}$
$\Leftrightarrow Z^{3}A^{5}B^{2}A=-3B^{-1}A^{-2}$
$\Leftrightarrow Z^{3}A^{5}B^{2}=-3B^{-1}A^{-3}$
$\Leftrightarrow Z^{3}A^{5}=-3B^{-1}A^{-3}B^{-2}$
$\Leftrightarrow Z^{3}=-3B^{-1}A^{-3}B^{-2}A^{-5}$
But in the final equation, $B^{-1}$ does not exist so how can I solve this?
| I think this problem wasn't meant to be solved. Whoever set this question probably wanted you to point out that the equation is insolvable, because the LHS is singular (as $B$ is singular) but the RHS is not.
| {
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Show that the series $\frac{1} {\sqrt{1}} -\frac{1} {\sqrt{2}} +\frac{1} {\sqrt{3}} +\dots$ converges, and its square (formed by Abel's rule) doesn't. Show that the series $\frac{1} {\sqrt{1}} -\frac{1} {\sqrt{2}} +\frac{1} {\sqrt{3}} +\dots$ (say $\sum u_n$) converges, and its square (say $\sum v_n$)(formed by Abel's rule) doesn't.
Abel's rule: given $\sum a_n, \sum b_n$, $\sum_{n=0} ^\infty c_n=\sum_{n=0} ^\infty [\sum_{i=0} ^n a_{n-i}b_i]$ is the infinite series gotten from multiplication of two series.
According to this rule,
$(\frac{1} {\sqrt{1}} -\frac{1} {\sqrt{2}} +\frac{1} {\sqrt{3}} +\dots)^2
=\frac{1} {\sqrt{1}}\frac{1} {\sqrt{1}}-[\frac{1} {\sqrt{1}} \frac{1} {\sqrt{2}}+\frac{1} {\sqrt{2}} \frac{1} {\sqrt{1}}]+\dots
-[\frac{1} {\sqrt{1}} \frac{1} {\sqrt{2k}}+\frac{1} {\sqrt{2}} \frac{1} {\sqrt{2k-1}}+\dots
+\frac{1} {\sqrt{k}}\frac{1} {\sqrt{k+1}}
+\frac{1} {\sqrt{k+1}}\frac{1} {\sqrt{k}}\dots+ \frac{1} {\sqrt{2k}}\frac{1} {\sqrt{1}}]
+[\frac{1} {\sqrt{1}} \frac{1} {\sqrt{2k+1}}+\frac{1} {\sqrt{2}} \frac{1} {\sqrt{2k}}+\dots
+\frac{1} {\sqrt{k}}\frac{1} {\sqrt{k+2}}+\frac{1} {\sqrt{k+1}} \frac{1} {\sqrt{k+1}}
+\frac{1} {\sqrt{k+2}}\frac{1} {\sqrt{k}}\dots+ \frac{1} {\sqrt{2k+1}}\frac{1} {\sqrt{1}}]\dots,$
which, if we sum by adding nearby items first, equals
$\frac{1} {\sqrt{1}}\frac{1} {\sqrt{1}}+\dots
+[(-\frac{1} {\sqrt{1}} \frac{1} {\sqrt{2k}}+\frac{1} {\sqrt{1}} \frac{1} {\sqrt{2k+1}}
-\frac{1} {\sqrt{2}} \frac{1} {\sqrt{2k-1}}+\frac{1} {\sqrt{2}} \frac{1} {\sqrt{2k}}+\dots
-\frac{1} {\sqrt{k}}\frac{1} {\sqrt{k+1}}+\frac{1} {\sqrt{k}}\frac{1} {\sqrt{k+2}}
-\frac{1} {\sqrt{k+1}}\frac{1} {\sqrt{k}}+\frac{1} {\sqrt{k+2}}\frac{1} {\sqrt{k}}
\dots-\frac{1} {\sqrt{2k}}\frac{1} {\sqrt{1}}+ \frac{1} {\sqrt{2k+1}}\frac{1} {\sqrt{1}})
+\frac{1} {\sqrt{k+1}} \frac{1} {\sqrt{k+1}}]<\frac{1} {\sqrt{1}}\frac{1} {\sqrt{1}}+\dots
+[
\frac{1} {\sqrt{k+1}} \frac{1} {\sqrt{k+1}}]=\sum_{k=0}^\infty \frac{1}{k+1}$ which diverges.
But in order to prove the series diverges we probably need to prove it's no less than another divergent series.
It's possibly not complicated. I will see how I can modify the proof to make it work. Perhaps I need to use the ratio test $\frac{c_{n+1}}{c_n}=1+\frac{A}{n}+O(\frac{1}{n^2})$, considering the series approximates $\sum_{k=0}^\infty \frac{1}{k+1}$
Context: the significance of the statement is that if it is true, then $\lim_{x\to 1}v_n x^n \neq c_n$, and so (though, for $\sum u_n x^n$ converges absolutely, we have $(\sum u_n x^n)^2=\sum v_n x^n$, letting $x\to 1$,) we don't have $(\sum u_n)^2$ (i.e. the limit of the left side) equals $\sum v_n$.
| I'd call that the Cauchy product.
To show that your series diverges, it suffices to use what might be the bluntest tool available, the $n$th term test.
The $n$th term of the "product" is given by
$$\pm \sum_{i=1}^{n-1} \frac{1}{\sqrt{n-i}}\frac{1}{\sqrt{i}} = \pm \sum_{i=1}^{n-1} \frac{1}{\sqrt{i(n-i)}}.$$
We want to show that the series diverges by showing that these don't converge to $0$, which we can do by finding a lower bound on the absolute value.The terms in the sum are minimized when $i(n-i)$ is maximized, and we know that the function $f(x) = x(n-x)$ is maximized at $x=n/2$ (it is a downard pointing parabola so the maximum is at the critical point, which can be identified with the derivative).
All the terms in the sum are nonnegative, so the absolute value won't cause trouble,
$$\left|\pm \sum_{i=1}^{n-1} \frac{1}{\sqrt{i(n-i)}}\right| = \left|\sum_{i=1}^{n-1} \frac{1}{\sqrt{i(n-i)}}\right| = \sum_{i=1}^{n-1} \frac{1}{\sqrt{i(n-i)}},$$
and the bound is then easy to apply,
$$\geq \sum_{i=1}^{n-1} \frac 1{\sqrt{(n/2)^2}} = \sum_{i=1}^{n-1} \frac 2 n = \frac{2(n-1)}{n}.$$
The latter is greater than or equal to $1$ for all $n\geq 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3786298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding a closed form for $\sum_{n=1}^{\infty}\frac{1}{F(n)F(n+2k)}$, where $F(n)$ is the $n$-th Fibonacci number So, I want to find the closed value for the sequence
$$\sum_{n=1}^{\infty} \frac{1}{F(n)F(n+2k)}$$
Here by $F(n)$ I mean the $n$-th term of Fibonacci Sequence.
I got motivated for this by the YouTube video "Using partial fractions to evaluate two Fibonacci reciprocal sums" by Prof. Michael Penn. So in the end of video he says for finding this for the general case : k. So i tried it and here is what i did:-
Let
$$\frac{1}{F(n)F(n+2k)} = \frac{A}{F(n)} + \frac{B}{F(n+2k)}$$
so by rearranging we can say that: $AF(n+2k) + BF(n) = 1$
Now let $C_0 = F(n+2k)$
then $C_1 = F(n+2k-1) + F(n+2k-2)$
so $C_2 = 2F(n+2k-2) + F(n+2k-3)$
then by induction we can prove that if i repeat this process j times then $C_j = F(j+1)F(n+2k-j) + F(j)F(n+2k-j-1)$
So now let $j = 2k - 1$
F
then $C_{2k-1} = F(2k)F(n+1) + F(2k-1)F(n)$
now putting this in to the Eq, of A and B then $AF(2k)F(n+1) + AF(2k-1)F(n) + BF(n) = 1$
now by putting $F(n+1) = F(n+2) - F(n)$
then I will get $AF(2k)F(n+2) + (B+AF(2K-1) - AF(2k))F(n = 1)$
there can be infinite values of A and B satisfying this relation so if I let $A = \frac{1}{F(2k)F(n+2)}$ becaues the first term in the Eq. will become 1 then $B = \frac{F(2k-2)}{F(2k)F(n+2)}$
By putting this values in the summation I will get $\sum_{n=1}^{\infty}\frac{1}{F(2k)F(n)F(n+2)} + \frac{F(2k-2)}{F(n+2)F(n+2k)}$
now in the video It is showned that $\sum_{n=1}^{\infty}\frac{1}{F(n)F(n+2)} = 1$ so I am not going to derive it.
Now putting it we get $\frac{1 + F(2k-2)\sum_{n=1}^{\infty}\frac{1}{F(n+2)F(n+2k)}}{F(2k)}$
$\frac{1 + F(2k-2)\sum_{n=3}^{\infty}\frac{1}{F(n)F(n+2(k-1))}}{F(2k)}$ in this I have changed the limit.
Now by adding and subtracting some termes I get $Sum(k) = \frac{1 + F(2k-2)\sum_{n=1}^{\infty}\frac{1}{F(n)F(n+2(k-1))} - \frac{F(2k-2)}{F(2k-1)} - \frac{F(2k-2)}{F(2k)}}{F(2k)}$
Now finnaly I ended up with this $Sum(k) = \frac{1 + F(2k-2)Sum(k-1) - \frac{F(2k-2)}{F(2k-1)} - \frac{F(2k-2)}{F(2k)}}{F(2k)}$
But this is not a direct formulae. Can someone help?
| After revising your work, you should find
$$\sum_{n=1}^{\infty} \frac{1}{F_nF_{n+2k}}=\frac{1}{F_{2k}}\sum_{j=1}^k\frac{1}{F_{2j-1}F_{2j}}.$$
The first terms for $k\geq 1$ are $1$, $\frac{7}{18}$, $\frac{143}{960}$, $\frac{4351}{76440}$, $\frac{814001}{37437400}$ (see also OEIS sequence A333088).
Proof. By induction with respect to $k\geq 1$, we verify that
$$\frac{F_{2k}}{F_nF_{n+2k}}=\sum_{j=1}^k\frac{1}{F_{n+2(j-1)}F_{n+2j}}.$$
Indeed, it holds for $k=1$ and, for the inductive step, we have
$$\sum_{j=1}^{k+1}\frac{1}{F_{n+2(j-1)}F_{n+2j}}
=\frac{F_{2k}}{F_nF_{n+2k}}+\frac{1}{F_{n+2k}F_{n+2k+2}}=\frac{F_{2k+2}}{F_nF_{n+2k+2}}.$$
where at the last step we applied Vajda's identity: $$F_{2k+2}F_{n+2k}-F_{2k}F_{n+2k+2}=(-1)^{2k}F_2F_n=F_n.$$
Hence, since $\sum_{n=1}^{\infty} \frac{1}{F_nF_{n+2}}=1$ (watch the video or see Sum $\frac{1}{1\times2}+\frac{1}{1\times3}+\frac{1}{2\times5}+\frac{1}{3\times8}+\cdots$), it follows that
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{F_nF_{n+2k}}&=
\frac{1}{F_{2k}}\sum_{n=1}^{\infty}\sum_{j=1}^k\frac{1}{F_{n+2(j-1)}F_{n+2j}}\\
&=\frac{1}{F_{2k}}\sum_{j=1}^k\sum_{n=1}^{\infty}\frac{1}{F_{n+2(j-1)}F_{n+2j}}=\frac{1}{F_{2k}}\sum_{j=1}^k\left(1-\sum_{n=1}^{2(j-1)}\frac{1}{F_{n}F_{n+2}}\right)\\
&=\frac{1}{F_{2k}}\sum_{j=1}^k\left(1-\sum_{n=1}^{2(j-1)}\left(\frac{1}{F_{n}F_{n+1}}-\frac{1}{F_{n+1}F_{n+2}}\right)\right)\\
&=\frac{1}{F_{2k}}\sum_{j=1}^k\left(1-1+\frac{1}{F_{2j-1}F_{2j}}\right)
=\frac{1}{F_{2k}}\sum_{j=1}^k\frac{1}{F_{2j-1}F_{2j}}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3786380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
if $a^5+b^5<1$ and $c^5+d^5<1$ then prove that ${a^2}{c^3}+{b^2}{d^3}<1$ if $a^5+b^5<1$ and $c^5+d^5<1$ then prove that ${a^2}{c^3}+{b^2}{d^3}<1$ given $a,b,c,d$ are non negative real numbers
My try:
It is easy to deduce that $a,b,c,d<1$ ,thus
$$a^2c^3+b^2d^3<a^5c^5+b^5d^5<(a^5+b^5)(c^5+d^5)<1$$
I want to know if there is a more cleaner method to solve this problem and if possible with calculus. I would also want to know if my proof has a mistake anywhere.
| Using the AM-GM inequality, we have
$$2a^5 + 3c^5 \geqslant 5\sqrt[5]{(a^5)^2 \cdot (c^5)^3} = 5a^2 c^3,$$
similar
$$ 2b^5 + 3d^5 \geqslant 5b^2 d^3.$$
Therefore
$$5(a^2c^3+b^2d^3) \leqslant 2(a^5+b^5)+3(c^5+d^5).$$
But $$2(a^5+b^5)+3(c^5+d^5) < 2 \cdot 1 +3\cdot 1 = 5.$$
So
$$a^2c^3+b^2d^3 \leqslant 1.$$
Done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Maclaurin/Laurent series of a complex log I've been asked to find the Maclaurin series of
\begin{equation}
\text{Log}(1+2z)
\end{equation}
Which I've done, and found to be $2z-2z^2+\frac{8z^3}{3}...$,
I'm now told 'hence find the first three terms about $z=0$ of the function
\begin{equation}
\frac{1}{z^2\text{Log}(1+2z)}
\end{equation}
I see the second function is the reciprocal of the first multiplied by $\frac{1}{z^2}$ but I'm not sure how their series are related?
| Hint:
\begin{align*}
\frac{1}{{z^2 \log (1 + 2z)}} & = \frac{1}{{2z^3 }}\frac{{2z}}{{\log (1 + 2z)}} = \frac{1}{{2z^3 }}\frac{{2z}}{{2z - 2z^2 + \frac{{8z^3 }}{3} - \cdots }} = \frac{1}{{2z^3 }}\frac{1}{{1 - z + \frac{{4z^2 }}{3} - \cdots }} \\ & = \frac{1}{{2z^3 }}\left( {1 - \left( { - z + \frac{{4z^2 }}{3} - \cdots } \right) + \left( { - z + \frac{{4z^2 }}{3} - \cdots } \right)^2 - \cdots } \right) = \cdots .
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
how to solve $\int _0^1\frac{\ln \left(1+x\right)}{a^2+x^2}\:\mathrm{d}x$
how to solve $$\mathcal{J(a)}=\int _0^1\frac{\ln \left(1+x\right)}{a^2+x^2}\:\mathrm{d}x$$
i used the differentiation under the integral and got
\begin{align}
\mathcal{J(b)}&=\int _0^1\frac{\ln \left(1+bx\right)}{a^2+x^2}\:\mathrm{d}x
\\[3mm]
\mathcal{J'(b)}&=\int _0^1\frac{x}{\left(a^2+x^2\right)\left(1+bx\right)}\:\mathrm{d}x
\\[3mm]
&=\frac{a^2b}{1+a^2b^2}\int _0^1\frac{1}{a^2+x^2}\:\mathrm{d}x+\frac{1}{1+a^2b^2}\int _0^1\frac{x}{a^2+x^2}\:\mathrm{d}x-\frac{b}{1+a^2b^2}\int _0^1\frac{1}{1+bx}\:\mathrm{d}x
\\[3mm]
&=\frac{ab}{1+a^2b^2}\operatorname{atan} \left(\frac{1}{a}\right)+\frac{1}{2}\frac{\ln \left(1+a^2\right)}{1+a^2b^2}-\frac{\ln \left(a\right)}{1+a^2b^2}-\frac{\ln \left(1+b\right)}{1+a^2b^2}
\end{align}
But we know that $\mathcal{J}(1)=\mathcal{J(a)}$ and $\mathcal{J}(0)=0$
\begin{align}
\int_0^1\mathcal{J'(b)}\:\mathrm{d}b&=a\:\operatorname{atan} \left(\frac{1}{a}\right)\int _0^1\frac{b}{1+a^2b^2}\:\mathrm{d}b+\frac{\ln \left(1+a^2\right)}{2}\int _0^1\frac{1}{1+a^2b^2}\:\mathrm{d}b-\ln \left(a\right)\int _0^1\frac{1}{1+a^2b^2}\:\mathrm{d}b
\\
&-\int _0^1\frac{\ln \left(1+b\right)}{1+a^2b^2}\:\mathrm{d}b
\\[3mm]
\mathcal{J(a)}&=\frac{1}{2a}\operatorname{atan} \left(\frac{1}{a}\right)\ln \left(1+a^2\right)+\frac{1}{2a}\ln \left(1+a^2\right)\operatorname{atan} \left(a\right)-\frac{1}{a}\ln \left(a\right)\:\operatorname{atan} \left(a\right)-\underbrace{\int _0^1\frac{\ln \left(1+b\right)}{1+a^2b^2}\:\mathrm{d}b}_{\mathcal{I}}
\end{align}
but how to calculate ${\mathcal{I}}$, i tried using the same technique but it didnt work
| The integral $I$ has been evaluated here by Felix Marin.
Using it yields the closed form of the integral
\begin{align}
J\left(a\right)&=\int_0^1\frac{\ln\left(1+x\right)}{a^2+x^2}\:dx=\frac{1}{2a}\operatorname{arctan} \left(\frac{1}{a}\right)\ln \left(1+a^2\right)+\frac{1}{2a}\ln \left(1+a^2\right)\operatorname{arctan} \left(a\right)
\\
&-\frac{1}{a}\ln \left(a\right)\:\operatorname{arctan} \left(a\right)-2\ln \left(2\right)\frac{\operatorname{arctan} \left(a\right)}{a}-\frac{1}{a}\mathfrak{I}\left(\operatorname{Li}_2\left(\frac{2a}{i+a}\right)-\operatorname{Li}_2\left(\frac{a}{i+a}\right)\right)
\end{align}
Complex analysis seems like the only way to go with this integral because software also gives closed forms featuring the imaginary unit with polylogs, i also checked if this worked for particular values and it does, though i'd still like to see if anyone can come up with an approach without any complex methods.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3789171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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How to get started on the problem $\text{Var}(fg) \le 2\text{Var}(f)M(g)^2 + 2\text{Var}(g)M(f)^2$ where $M(h) = \max\limits_{x \in [0,1]} |h(x)|.$
$f, g: [0,1] \to \mathbb{R}$ are continuous functions. Show that $\text{Var}(fg) \le 2\text{Var}(f)M(g)^2 + 2\text{Var}(g)M(f)^2$ where $M(h) = \max\limits_{x \in [0,1]} |h(x)|, \text{Var}(h) = \int_0^1 (h(x)-\mu_h)^2 \, dx, \mu_h = \int_0^1 h(x) \, dx.$
How do you get started on this problem? If one of $f, g$ is constant, the result is true. I also noticed that $$M(f)^2 \ge \mu_f^2, M(f) \ge \text{Var}(f)$$ and $$\mu_f^2 \text{Var}(g) + \mu_g^2 \text{Var}(f) + \text{Var}(f)\text{Var}(g) = (\int_0^1 f^2)(\int_0^1 g^2) - \mu_f^2 \mu_g^2$$
If you assume $\mu_f = \mu_g = 0,$ the inequality reduces to $\int f^2 g^2 - \mu_{fg}^2 \le \int f^2g^2 \le 2(\int f^2)M(g)^2 + 2(\int g^2)M(f)^2,$ which I have solved below.
I have no idea how to tie all of these observations together. Does anyone have an idea about how you can make progress on this problem? I think that the Cauchy-Schwarz inequality $(\int fg)^2 \le (\int f^2)(\int g^2)$ will show up in some way but I'm not sure where's the best point to use it. Every time I see a spot where it applies, the inequality I'm investigating is pointing in the wrong direction for it to be useful.
Here's another idea: Replace $f, g$ with $fM(f), gM(g)$ so that $|f|, |g| \le 1$ now and $1$ is achieved at some point. In the case $\mu_f = \mu_g = 0,$ the inequality reduces to $\int f^2 g^2 \le 2\int (f^2+g^2),$ or $\int (2f^2 - f^2 g^2 + 2g^2) \ge 0.$ With $a=f^2, b=g^2,$ we have $2(a+b) \ge 4\sqrt{ab} \ge ab$ since $ab \le 16,$ so this is true. However, I am very suspicious about just how much freedom there is given that we actually have $ab \le 1.$ The general case will not be this easy.
Using this trick for the general case and expanding everything out gives $\int f^2 g^2 + 2\mu_f^2 + 2\mu_g^2 \le \mu_{fg}^2 + 2\int (f^2+g^2) \, \, (*)$ as the inequality to prove.
Update: I just recalled the tactic of turning squares of integrals into double integrals. We may rewrite $(*)$ as $$\int\limits_{[0,1]^2} [2f(x)^2+2g(x)^2+f(x)g(x)f(y)g(y) - 2f(x)f(y) - 2g(x)g(y) - f(x)^2g(x)^2] \, dx dy\ge 0$$ and we still have the condition that $|f|, |g| \le 1.$
If you symmetrize the integrand by replacing $2f(x)^2, 2g(x)^2, f(x)^2g(x)^2$ with $f(x)^2+f(y)^2, g(x)^2+g(y)^2, 0.5(f(x)^2g(x)^2+f(y)^2g(y)^2),$ then let $a=f(x), b=f(y), c=g(x), d=g(y),$ the integrand becomes $a^2+b^2+c^2+d^2+abcd-2ab-2cd-0.5(a^2c^2+b^2d^2) = (a-b)^2 + (c-d)^2 - 0.5(ac-bd)^2.$ Thus, we just need to show $(a-b)^2 + (c-d)^2 \ge 0.5(ac-bd)^2$ given $a,b,c,d \in [-1,1].$ Let $A = (0,0), B = u = (a, c), C = v = (b, d),$ and this becomes $BC^2 = ||u-v||^2 \ge 0.5 \det [u,v]^2 = 2 [ABC]^2,$ or $BC \ge \sqrt{2} \cdot [ABC].$ If $h$ is the length of the altitude from $A,$ then this finally reduces to $h \le \sqrt{2}.$ The altitude is shorter than the side lengths. In particular, $h \le AB = ||u|| = \sqrt{a^2+c^2} \le \sqrt{2}.$ I'll add this as an answer and look out for simpler solution.
| If $M(f) = 0$ or $M(g) = 0,$ then $f=0$ or $g=0$ and the inequality trivially holds. Else, let $h_1 = f/M(f), h_2 = g/M(g)$ so that the inequality becomes $\text{Var}(h_1 h_2) \le 2\text{Var}(h_1) + 2\text{Var}(h_2).$ After expanding everything out, this becomes $$2 \int (h_1^2 + h_2^2) + \left(\int h_1 h_2\right)^2 - 2\left(\int h_1\right)^2 - 2\left(\int h_2\right)^2 - \int h_1^2 h_2^2 \ge 0.$$
We can express this as an inequality for a double integral:
$$\int\limits_{[0,1]^2} \left[ h_1(x)^2 + h_1(y)^2 + h_2(x)^2 + h_2(y)^2 + h_1(x)h_2(x)h_1(y)h_2(y) - 2h_1(x)h_1(y) - 2h_2(x)h_2(y) - 0.5(h_1(x)^2h_2(x)^2+h_1(y)^2h_2(y)^2)\right] \ge 0.$$
Let $a = h_1(x), b = h_1(y), c = h_2(x), d = h_2(y).$ The integrand is $(a-b)^2 + (c-d)^2 - 0.5(ac-bd)^2,$ so it suffices to show $(a-b)^2 + (c-d)^2 \ge 0.5(ac-bd)^2.$ Let $A = (0,0), B = (a,c), C = (b,d).$ Then the inequality becomes $BC^2 \ge 2[ABC]^2 \Leftrightarrow BC \ge \sqrt{2} [ABC].$ Letting $h$ be the altitude from $A$ to $BC,$ this reduces to $h \le \sqrt{2}.$ Since the altitude is the shortest distance to BC, we have $h \le AB = \sqrt{a^2+c^2} \le \sqrt{2}$ as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What is a condition on my parameter for this matrix to be diagonalizable? I am looking for a condition on $b$ so that the matrix $B$ below can be diagonalized:
$$B=\begin{pmatrix}2b+1 &-2b-2 &b-2\\ 2b-1&-2b&b-2\\ -6&6&-1\end{pmatrix}.$$
I know that if $B$ has $3$ linearly independent eigenvectors, $B$ is diagonalizable, but I want to avoid going all the way to get eigenvalues and eigenvectors of $B$.
Do we have any other easier ways?
| Observe that the first two rows of $B$ are nearly identical and the first two columns of $B$ are almost the negative of each other. This motivates one to inspect the following matrix that is similar to $B$:
$$
C=\pmatrix{1&-1&0\\ 0&1&0\\ 0&0&1}
\pmatrix{2b+1&-2b-2&b-2\\ 2b-1&-2b&b-2\\ -6&6&-1}
\pmatrix{1&1&0\\ 0&1&0\\ 0&0&1}
=\left(\begin{array}{c|cc}2&0&0\\ \hline2b-1&-1&b-2\\ -6&0&-1\end{array}\right).
$$
The only repeated eigenvalue of $C$ is the eigenvalue $-1$ of multiplicity $2$. Hence $B$ and $C$ are diagonalisable if and only if $\operatorname{nullity}(C+I)=2$, i.e. if and only if $b=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3795852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using complex numbers to prove Napoleon's Theorem
Let $ABC$ be a triangle and erect equilateral triangles on sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ outside of $ABC$ with centers $O_A$, $O_B$, $O_C$. Prove that $\bigtriangleup O_AO_BO_C$ is equilateral and that its center coincides with the centroid of triangle $ABC$
I have already seen this answer Proving Napoleon's Theorem with complex numbers
but my doubt is different ,
Now,
In this answer https://artofproblemsolving.com/community/c618937h1650553_proposition_634_napoleons_theorem ($5$th post)
they wrote -
$O_AC$ is a $\frac\pi6$ rotation of $BC$ followed by a dilation with ratio $\frac1{\sqrt3}$ at $C,$ so we have
$\begin{align*}
\frac{o_A-c}{b-c}&=\frac1{\sqrt3}\cdot\frac{\sqrt3+i}{2}\end{align*}$ but i am not able to understand this ,can anybody explain this step please ?
Note-I have solved this problem using simple angle chasing ,but i want to understand properly that how they got co-ordinates of $O_A$
thankyou
| First of all, for any equilateral triangle $XYZ$ with side $x$ and centroid $G$, then
$$XG= \frac{2}{3} \cdot \frac{\sqrt{3}}{2}x=\frac{x}{\sqrt3} $$
Now locate the triangle $ABC$ with its centroid $O$ on the pole of the complex plane.
Let $z_{1}, z_{2},z_{3}$ be the complex numbers representing respectively the points $A, B $ and $C$, then $$z_{1}+z_{2}+z_{3}=0\cdots (*)$$.
Also denoting the three centroids $G_i$‘s of the equilateral triangles so-formed by the complex numbers $g_1, g_2$ and $g_3$, then $g_i$’s can be found by rotating the sides of the triangle $ABC$ by $\frac{\pi}{6}$ in clockwise direction as below:
$$
\begin{array}{l}
g_{1}=z_{2}+\frac{1}{\sqrt{3}}\left(z_{3}-z_{2}\right) e^{\frac{\pi}{6} i} \cdots(1) \\
g_{2}=z_{3}+\frac{1}{\sqrt{3}}\left(z_{3}-z_{1}\right) e^{\frac{\pi}{6} i} \cdots(2) \\
g_{3}=z_{1}+\frac{1}{\sqrt{3}}\left(z_{2}-z_{1}\right) e^{\frac{\pi}{6} i} \cdots(3)
\end{array}
$$
Adding them together yields
$$
\begin{aligned}
g_{1}+g_{2}+g_{3} &=z_{1}+z_{2}+z_{3}+\frac{1}{\sqrt{3}} e^{\frac{\pi}{6} i}\left(z_{3}-z_{2}+z_{3}-z_{1}+z_{2}-z_{1}\right) \\
&=0+0 \\
&=0
\end{aligned}
$$
Then centroid of the triangle of centroids $=\dfrac{g_{1}+g_{2}+g_{3}}{3}=0$ which coincide with the centroid of the original triangle $ABC$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limit without de l'Hôpital: $\lim _{x\to 0\color{red}{\boldsymbol -}}\left(1+x^3\right)^{\frac{1}{\left(x^2+1\right)^4-1}}$ I have this limit of this form
$$f(x)^{g(x)}=e^{g(x)\ln(f(x))}$$
$$\lim _{x\to 0\color{red}{\boldsymbol -}}\left(1+x^3\right)^{1/\left((x^2+1)^4-1\right)}$$
In our case I can write in the exponent:
$${g(x)\ln(f(x))}=\frac{\ln(f(x))}{\frac1{g(x)}}$$
and I have an indeterminate form $(0/0)$ and I can apply de l'Hôpital rule. Right now I just thought to write
$$(1+x^3)=\left(1+\frac{1}{\frac1{x^3}}\right)$$ and I call $x^3=t$ but I think to obtain the exponent too long and it will be more complicated.
| $$A=\left(1+x^3\right)^{\frac{1}{\left(x^2+1\right)^4-1}}\implies \log(A)=\frac{1}{\left(x^2+1\right)^4-1}\log(1+x^3)$$
$$\log(A)=\frac{x^3-\frac{1}{2}x^6+\frac{1}{3}x^9+O\left(x^{12}\right)}{4 x^2+6 x^4+4 x^6+x^8}$$ Long division
$$\log(A)=\frac{x}{4}-\frac{3 x^3}{8}+O\left(x^{4}\right)$$
$$A=e^{\log(A)}=1+\frac{x}{4}+\frac{x^2}{32}+O\left(x^3\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the maximum value of $x^2y$ given constraints Find the maximum value of $${ x }^{ 2 }y$$ subject to the constraint $$x+y+\sqrt { 2{ x }^{ 2 }+2xy+3{ y }^{ 2 } } =k$$ where k is a constant.
I tried it by substituting value of x and then differentiating w.r.t $x$ but not able to proceed further.
| Let $y=a x$ and from the constraint (assuming $x>0$
$$x=\frac{k}{\sqrt{3 a^2+2 a+2}+a+1}\qquad y=\frac{ak}{\sqrt{3 a^2+2 a+2}+a+1}$$
$$x^2y=\frac{a k^3}{\left(\sqrt{3 a^2+2 a+2}+a+1\right)^3}$$ Differentiate with respect to $a$ to get
$$-\frac{(2 a-1) \left(3 a+2+\sqrt{a (3 a+2)+2}\right) k^3}{\sqrt{a (3 a+2)+2}
\left(a+\sqrt{a (3 a+2)+2}+1\right)^4}=0$$
So, either $a=\frac 12$ which would give
$$x^2 y=\frac{4 k^3}{\left(3+\sqrt{15}\right)^3}$$ or
$a=-\frac{1}{6} \left(5+\sqrt{13}\right)$ which would give
$$x^2y=\frac{1}{54} \left(35-13 \sqrt{13}\right) k^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3798526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all the $\bf{x}\in\mathbb{R}^3$, which has a property that $\displaystyle \lim_{ n \to \infty } A^n\bf{x} = \bf{x}$ $A$ is a matrix with elements
$$\begin{pmatrix}
\frac{2}{3} & -\frac{2}{3}& -\frac{1}{3} \\
-\frac{2}{3}& 1 & \frac{2}{3} \\
\frac{4}{3} & 0 & -\frac{1}{3}
\end{pmatrix}$$
Find all the $\bf{x}\in\mathbb{R}^3$, which has a property that $$\displaystyle \lim_{ n \to \infty } A^n\bf{x} = \bf{x}$$
A's eigenvalues are $-\frac{1}{3}$,1 and $\frac{2}{3}$ and eigenvectors are $\left(
\begin{array}{c}
0 \\
-\frac{1}{2} \\
1
\end{array}
\right),\left(
\begin{array}{c}
1 \\
-1 \\
1
\end{array}
\right),\left(
\begin{array}{c}
\frac{3}{4} \\
-\frac{1}{2} \\
1
\end{array}
\right)$ for each.
I tried to get $A^n$ at first then got $$\begin{pmatrix}
(\frac{2}{3})^n & -2+\frac{2^{n+1}}{3}& 0 \\
-\frac{2+2^{n+1}}{3^{n+1}}& \frac{1-2^{n+2}}{3^{n+1}}-2 & \frac{2}{3^n} \\
\frac{4+2^{n+2}}{3^{n+1}} & \frac{2-2^{n+3}}{3^{n+1}} & -\frac{1}{3^n}
\end{pmatrix}$$
Then I tried find $$\displaystyle \lim_{ n \to \infty }(\frac{2}{3})^nx_1+(-2+\frac{2^{n+1}}{3})x_2=x_1 \\\displaystyle \lim_{ n \to \infty }-\frac{2+2^{n+1}}{3^{n+1}}x_1+(\frac{1-2^{n+2}}{3^{n+1}}-2)x_2+\frac{2}{3^n}x_3=x_2\\\ \displaystyle \lim_{ n \to \infty }\frac{4+2^{n+2}}{3^{n+1}}x_1+\frac{2-2^{n+3}}{3^{n+1}}x_2-\frac{1}{3^n}x_3=x_3$$
But I could not lead meaningful conclusion from these equations but $(0,0,0)$.
Where did I get wrong?
| As the Eigenvalues of $A$ are $-\dfrac13,1,\dfrac23$, when you take the $n^{th}$ power of any vector, only the Eigenvector associated to $\lambda=1$ will remain.
Hence $\mathbb x$ are the multiple of the second Eigenvector.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find $r$ such that the equation $x^4+x^2(1-2r)-2x+1=0$ has only one real solution I'm looking for $r$ such that the equation $$x^4+x^2(1-2r)-2x+1=0$$
has only one real solution. I've made some attempts to this problem, but it seems that I even didn't get close to the solution. The approximation for $r$ is 0.3347498
Is it possible to find analytic solution for $r$ and if yes then how?
Thanks for all the help.
| Equation $$x^4 + x^2 (1 - 2 r) - 2 x + 1 = 0$$
has only one real solution if the curve $y=x^4 + x^2 (1 - 2 r) - 2 x + 1$ is tangent to $x$-axis, that is: the curve has a absolute minimum, provided that $y''=2 (1-2 r)+12 x^2$ is positive, which means $1-2r>0\to r<\frac{1}{2}$
Derivative is $y'=4 x^3 +2 (1-2 r) x-2$. We have $y'=0$ for $r=\frac{2 x^3+x-1}{2 x}$.
Substitute in the given equation
$$x^4 + x^2 \left(1 - 2\cdot \frac{2 x^3+x-1}{2 x}\right) - 2 x + 1 = 0$$
which gives
$$x^4+x-1=0$$
Real solutions are
$$x_1=-1.22074;\;x_2= 0.724492$$
Correspondent $r$ values are
$$r_1=2.3998;\; r_2=0.33475$$
$r_1$ must be discarded because it is larger than $1/2$ so the unique solution is
$$r = 0.33475$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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How to determine the value of $x$ such that the parallelogram has a given area?
Determine the value(s) of $x$ such that the area of the parallelogram
formed by the vectors $a = (x+1, 1, -2)$ and $b = (x, 3, 0)$ is
$\sqrt{41}$.
My work (using cross product)
\begin{align}
|a \times b| &= ((1)(0) - (-2)(3), (-2)(x) - (x+1)(0), (x+1)(3) - (1)(x))\\
|a \times b| &= (6, -2x, 3x + 3 - x)\\
|a \times b| &= (6, -2x, 2x +3)\\
\sqrt{41} &= (6, -2x, 2x + 3)
\end{align}
I don't know what to do next and how I should isolate for $x$. If anyone can help me out, I'd appreciate it.
| You have to be careful with your equality signs. What you already showed is that
$$a \times b = (6, -2x, 2x + 3).$$
In order to get the area to be $\sqrt{41}$, we need
$$\| a \times b \|^2 = 6^2 + (-2x)^2 + (2x + 3)^2 = 41.$$
This is a quadratic equation, which you can solve for $x$.
Spoiler:
$x_1 = -1/2, \quad x_2 = -1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the sum of all the roots of the equation $f^{[5]}(x)=0$?
Suppose $f(x)=x^2-3x+2$, what is the sum of all the roots of the equation $f^{[5]}(x)=0$?
I think one of the common ways of solving questions about $f^{[n]}(x)$ is fixed-point iteration.
I used this method, solving $x^2-3x+2=x$ and got $x=2 \pm \sqrt{2}$.
Therefore $$f(x)-(2+\sqrt{2})=[x-(2+\sqrt{2})][x+(\sqrt{2}-1)]......(1)$$
And $$f(x)-(2-\sqrt{2})=[x-(2-\sqrt{2})][x-(\sqrt{2}+1)]......(2)$$
I tried $\frac{(1)}{(2)}$ and immediately got trapped since I can’t see the next step. The reason why I wanted to know what $f^{[5]}(x)$ is is that if I can know the coefficient of $x^{31}$, I can easily obtain the answer by Vieta’s theorem.
Maybe I’m missing something obvious here. Any suggestions or hints would be appreciated.
Edit: I just realized different regions may have different expressions on this one.
Basically, $f^{[2]}(x)=f(f(x))$, $f^{[3]}(x)=f(f(f(x)))$, and so on. I am really sorry for not considering this trouble in the first place.
| It seems obvious that, as you apply $f$, the degree of the polynomial doubles. (Proof: induction.)
Let $f^{[5]}(x)=x^{32}+ax^{31}+\ldots$.
As $f^{[4]}(x)=x^{16}+bx^{15}+\ldots$, this means that $f^{[5]}(x)=(x^{16}+bx^{15}+\ldots)^2-3(x^{16}+bx^{15}+\ldots)+2=x^{32}+2bx^{31}+\ldots$, so $a=2b$.
Similarly, as $f^{[3]}(x)=x^8+cx^7+\ldots$, this means that $f^{[4]}(x)=(x^8+cx^7+\ldots)^2-3(x^8+cx^7+\ldots)+2=x^{16}+2cx^{15}+\ldots$, so $b=2c$.
Again, $f^{[2]}(x)=x^4+dx^3+\ldots$, so as $f^{[3]}(x)=(x^4+dx^3+\ldots)^2-3(x^4+dx^3+\ldots)+2=x^8+2dx^7+\ldots$, we have $c=2d$.
Finally, $f(x)=x^2-3x+2$, so $f^{[2]}(x)=(x^2-3x+2)^2-3(x^2-3x+2)+2=x^4-6x^3+\ldots$ implies $d=-6$
Altogether, we have $d=-6, c=-12, b=-24, a=-48$. So, the sum of all zeros is $48$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that an inequality holds for all positive real numbers $a, b$ such that $ab \geq 1$ I found the following question on a past international competition:
Show that:
$\left(a+2b+\frac{2}{a+1}\right)\left(b+2a+\frac{2}{b+1}\right) \geq 16$
for all positive real numbers $a, b$ such that $ab\geq 1$.
I solved it in the following way:
$\left(a+2b+\frac{2}{a+1}\right)\left(b+2a+\frac{2}{b+1}\right)$
$\displaystyle =ab+2a^2+\frac{2a}{b+1}+2b^2+4ab+\frac{4b}{b+1}+\frac{2b}{a+1}+\frac{4a}{a+1}+\frac{4}{(a+1)(b+1)}$
$\displaystyle \ge 5ab+2(a^2+b^2)+\frac{2(a+2b)}{b+1}+\frac{2(b+2a)}{a+1}+\frac{4}{(a+1)(b+1)}$
$\displaystyle \ge 5+2(a^2+b^2)+\frac{2(a+2b)}{b+1}+\frac{2(b+2a)}{a+1}+\frac{4}{(a+1)(b+1)}$
$\displaystyle \ge 9+\frac{2(a^2+2ab+a+2b)+2(b^2+b+2ab+2a)+4}{(a+1)(b+1)}$ (from AM-GM we have that $a^2+b^2\ge 2ab \ge 2$)
$\displaystyle \ge 9+4(a+1)(b+1)+\frac{2b^2+2a^2+4ab+2a+2b}{(a+1)(b+1)}$
$\displaystyle \ge 13+\frac{2b^2+2a^2+4ab+2a+2b}{(a+1)(b+1)}$
However we have that $a^2+b^2\ge (a+b)*\sqrt{a^2b^2} \ge a+b$ (this is true for the well known inequality that $x1^2+x2^2+...+xn^2\ge (x1+x2+...+xn)*\sqrt[n]{x1x2...xn}$), $a^2+b^2\ge 2ab\ge 2$. Hence:
$2b^2+2a^2+ab\ge a+b+3$ so $2b^2+2a^2+4ab+2a+2b\ge 3ab+3a+3b+3$
So, we have that $\displaystyle \left(a+2b+\frac{2}{a+1}\right)\left(b+2a+\frac{2}{b+1}\right)\ge 13+\frac{3ab+3a+3b+3}{(a+1)(b+1)}\ge 13+\frac{3(a+1)(b+1)}{(a+1)(b+1)} \ge 16$
I believe that my solution is correct, however I'm not completely certain, so could you please have a look at it and also share if there is an easier and simpler way of solving the problem?
| From the calculation, you have done, we have:
$f(a, b) = 2(a + b)^{2} + ab + \left[\frac{2a}{1 + b} + \frac{4b}{1 + b} + \frac{2b}{1 + a} + \frac{4a}{1 + a} + \frac{4}{(1 + a)(1 + b)}\right]$
$ = 2(a + b)^{2} + ab + \frac{\left [(1 + a) + (1 + b) \right]^{2}}{(1 + a)(1 + b)} + \frac{\left [a^{2} +b^{2} + 6ab + 2a + 2b \right]}{(1 + a)(1 + b)}
$
We know from the AM-GM inequality we have $2(a + b)^{2} \geq 8$ and $\frac{\left [(1 + a) + (1 + b) \right]^{2}}{(1 + a)(1 + b)} \geq 4$ and by the hypothesis $ab \geq 1$.
It follows that $f(a, b) \geq 8 + 1 + 4 + \frac{\left [a^{2} +b^{2} + 6ab + 2a + 2b \right]}{(1 + a)(1 + b)}$ and if we can bound from above the last term by $3$ we would be done.
We have: $a^{2} +b^{2} + 6ab + 2a + 2b - 3(1 + a)(1 + b) = a^{2} + b^{2} + 3ab - 3 - a - b$
$ \geq a(a - 1) + b(b -1)\geq b^{-1}(b^{-1} -1) + b(b-1)$
$= b^{-2}[(1-b) + b^{3}(b-1)] = b^{-2}(b-1)[b^{3} - 1] $
$= b^{-2}(b-1)^{2} (b^{2} + b + 1) > 0$.
The desired bound is reached, as it was promised.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $x, y \in \mathbb{R}^+$ and $ x^3 + y^3 = x-y$. Prove $x^2 + 4y^2 < 1$ For $x, y \in \mathbb{R}^+$ , I am given that $ x^3 + y^3 = x-y$. I have to prove that $x^2 + 4y^2 < 1$. Now I have $ y + y^3 = x(1-x^2) $. Since $y + y^3 > 0 $ and $ x > 0$, we have $1 - x^2 > 0$. Which means that $ 0 < x < 1$. Also, since $x^3 + y^3 > 0$, we have $x - y > 0$. So, $x > y$. So, we get $0 < y < x < 1$. Using AM-GM inequality, I also get
$$ (x^3 + y^3 + 1) \geqslant 3xy $$
What else can be dome here ?
Thanks
| We need to prove that $$x^2+4y^2<\frac{x^3+y^3}{x-y}$$ or since $x-y>0$, $$y(5y^2-4xy+x^2)>0,$$ which is obvious because $$5y^2-4xy+x^2=y^2+(2y-x)^2>0.$$
| {
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Can all Lissajous curves be described by an implicit equation? I was trying to find the implicit form of the parametric curve $L(x(t),y(t))$ where $x(t)=5\cos\left(3 \pi t +\frac{\pi}{4}\right)$ and $y(t)=3cos\left(6 \pi t -\frac{\pi}{4}\right)$. At the end I wasted a lot of time and paper because I couldn't figure out the implicit expression (maybe I just didn't manipulate it properly). On the other hand, I saw a paper where these curves where described using Chebyshev's polynomials. I just want to know if the implicit form can be found without recurring to these polynomials.
| I'd break it down like:
$$
\begin{align}
\frac{x}{5}
&=\cos(3\pi t + \pi/4)\\
&=\cos(3\pi t)\cos(\pi/4)-\sin(3\pi t)\sin(\pi/4)\\
&=kc-ks&k=1/\sqrt{2},c=\cos(3\pi t),s=\sin(3\pi t)
\end{align}
$$
and
$$
\begin{align}
\frac{y}{3}
&=\cos(6\pi t - \pi/4)\\
&=\cos(6\pi t)\cos(\pi/4)+\sin(6\pi t)\sin(\pi/4)\\
&=\left(\cos^2(3\pi t)-\sin^2(3\pi t)\right)\cos(\pi/4)+2\sin(3\pi t)\cos(3\pi t)\sin(\pi/4)\\
&=kc^2-ks^2+2kcs
\end{align}
$$
So now given these two relations along with $c^2+s^2=1$, you'd like to eliminate $c$ and $s$. It seems helpful to square sides of the first relation so it can be quadratic in $c$ and $s$ like the other two. Normalize by $k^2$, and by $k$ in the second relation.
$$
\begin{align}
\left(\frac{x}{5k}\right)^2&= c^2 +s^2 - 2cs
&\frac{y}{3k}&= c^2 -s^2 + 2cs
&1&= c^2 +s^2
\end{align}
$$
This is now three equations with three things to eliminate ($c^2,s^2,cs$). We'd like to have more equations than things to eliminate. To that end, square the sides of each equation. Also take each pairwise product of two left sides, equal to the corresponding product of right sides. This makes six equations:
$$
\begin{align}
\left(\frac{x}{5k}\right)^4 &= c^4 - 4c^3s + 6c^2s^2 - 4cs^3 + s^4 \\
\left(\frac{y}{3k}\right)^2 &= c^4 + 4c^3s + 2c^2s^2 - 4cs^3 + s^4 \\
1 &= c^4 + 2c^2s^2 + s^4 \\
\left(\frac{x}{5k}\right)^2\frac{y}{3k}
&= c^4 - 4c^2s^2 + 4cs^3 - s^4 \\
\left(\frac{x}{5k}\right)^2 &= c^4 - 2c^3s + 2c^2s^2 - 2cs^3 + s^4 \\
\frac{y}{3k} &= c^4 + 2c^3s + 2cs^3 - s^4
\end{align}
$$
Now we can try to use linear algebra to find a relation on the six vectors from $\mathbb{R}^5$ suggested by the right sides. Row reduction leads to:
$$-2(\text{right}_1)-(\text{right}_2)-(\text{right}_3)-2(\text{right}_4)+4(\text{right}_5)+2(\text{right}_6)=0$$
Therefore
$$-2\left(\frac{x}{5k}\right)^4-\left(\frac{y}{3k}\right)^2-1-2\left(\frac{x}{5k}\right)^2\frac{y}{3k}+4\left(\frac{x}{5k}\right)^2+2\frac{y}{3k}=0$$
Here is that implicit plot and here is the original parametric plot, confirming a match.
| {
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Product of $3$ consecutive triangular numbers is a perfect square When is the product of $3$ consecutive triangular numbers a perfect square
My try:
$k^2 = n(n+1)^2(n+2)^2\frac{n+3}{ 8}$ where $n$ and $k$ are integers, then
$\frac{n(n+3)}{2}$ must be a square number
$n(n+3) =2x^2$ , $n^2+3n-2x^2=0$
then the discriminant must also be a perfect square.
$9+8x^2 = y^2$ but I don't know how to solve this in integers
| You have a Pell-type equation
$$y^2-8x^2=9.$$
This implies that $y$ and $x$ are multiples of $3$, so
$$(y/3)^2-8(x/3)^2=1$$
which is a genuine Pell equation. Its solution is
$$(y+2x\sqrt2)/3=\pm(3+2\sqrt2)^n$$
for $n\in\Bbb Z$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Eigenvalues of $A^{2018}$ Find the eigenvalues and eigenvectors of $A^{2018}$.
$$
A=\begin{bmatrix} 1 & 3 & 4\\ 3 & 1 & 4\\ 0 & 0 & 4\end{bmatrix}
$$
My solution:
First, by substracting first row times three from second row we get:
$$
A\approx \begin{bmatrix} 1 & 3 & 4\\ 0 & -8 & -8\\ 0 & 0 & 4\end{bmatrix}
$$
We achieved the upper triangular matrix so the characteristic polynomial is:
$$
\chi_{A^{2018}}(\lambda)=det (\begin{bmatrix} 1 & 3 & 4\\ 0 & -8 & -8\\ 0 & 0 & 4\end{bmatrix}^{2018}-\lambda I)=(1^{2018}-\lambda)((-8)^{2018}-\lambda)(4^{2018}-\lambda)
$$
Therefore the set of eigevalues is $\{1,4^{2018},8^{2018},\}$.
Please verify if this the correct solution, and in case it isn't, help me find the correct one.
| Your matirx is the sum of a relatively easy one (symmetric) and a nilpotent matrix, and these commute. $N$ is zero except for the pair of $4$ in positions $(1,3)$ and $(2,3).$ Call the symmetric one $S$ which can be nicely diagonalized. because they commute,
$$ (S + N)^{2018} = S^{2018} + 2018 S^{2017} N. $$
It stops there since $N^2 = 0.$
I just guessed this, all that is needed is explicit $P^{-1} S P = D.$ It is not really necessary that $P$ be orthogonal. In fact, what I do is
$$ P^T S P = D $$
$$\left(
\begin{array}{rrr}
\frac{1}{\sqrt 2} & \frac{1}{\sqrt 2} & 0 \\
-\frac{1}{\sqrt 2} & \frac{1}{\sqrt 2} & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 3 & 0 \\
3 & 1 & 0 \\
0 & 0 & 4 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
\frac{1}{\sqrt 2} & -\frac{1}{\sqrt 2} & 0 \\
\frac{1}{\sqrt 2} & \frac{1}{\sqrt 2} & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
4 & 0 & 0 \\
0 & - 2 & 0 \\
0 & 0 & 4 \\
\end{array}
\right)
$$
where $PP^T = P^TP = I$
$ P^T S P = D $ so that $S = PDP^T.$ Thus $S^{n} = P D^n P^T$ for all $n \geq 1$
the $S+N$ is called a Jordan Chevalley decomposition, sometimes much easier than Jordan canonical form. This time, not much different
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the maximum value of $\sum_{cyc}\frac{1}{a^2-4a+9}$ s.t. $a+b+c =1$ Problem: Let $a$, $b$, and $c$ be non-negative real numbers such that $a+b+c =1$. Find the maximum value of
\begin{align}
\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9}.
\end{align}
By using the Karush–Kuhn–Tucker conditions, I know the maximum occurs at $a=0$, $b=0$, and $c=1$. But I want to know how to solve this problem without using calculus, because it is a math olympiad problem.
| Because$:$
$$\dfrac{a}{18} + \dfrac{1}{9} - \dfrac{1}{a^2-4a+9} = \dfrac{a(a-1)^2}{18(a^2-4a+9)} \geq 0,$$
$$\therefore \dfrac{1}{a^2-4a+9} \leq \dfrac{a}{18} + \dfrac{1}{9}.$$
So$:$ $$ \sum \dfrac{1}{a^2-4a+9} \leq \dfrac{a+b+c}{18} + \dfrac{1}{3} = \dfrac{7}{18}.$$
Equality occur when $a:b:c=1:0:0$ or any permution.
Update. Let
$$f(a) = (\,ma + n\,)(\,a^2-4a+9\,) - 1.$$
We will try to find $m,\,n$ such that $f(a) \geqslant 0$ for all $a \in [\,0,1\,].$
Proof 1. Let $a = 0$ and $a =1$ we get
$$\{9n = 1,\; 6(m+n) = 1\}.$$
Solve equation we get $m = \dfrac 1 {18},$ $n = \dfrac 1 9.$
Proof 2. Write inequality as
$$f(a) = a \Big[\,a^2m-(4m-n)a+9m-4n\,\Big] +9n-1.$$
It's easy choose $n = \dfrac{1}{9},$ we get
$$f(a) = a \Big[\,ma^2-(4m-n)a+9m-4n\,\Big].$$
For a quadratic polynomial $ma^2-(4m-n)a+9m-4n,$ we have$:$
$$\Delta = -(10m+n)(2m-n) = -\left(10m+\dfrac 1 9\right)\left(2m-\dfrac 1 9\right).$$
We need $m\geqslant 0.$
So choose $2m = \dfrac{1}{9}$ which means $m = \dfrac{1}{18}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplifying $ 1^5+2^5+3^5+\dots+14^5+15^5 \pmod{13} $
$$ 1^5+2^5+3^5+\dots+14^5+15^5 \pmod{13} $$
I found this question on my old textbook, it seems very trivial but my answer and the answer of book are different. My solution is that:
$$ [1^5+2^5+3^5+4^5+5^5+6^5+(-6)^5+(-5)^5+(-4)^5+(-3)^5+(-2)^5+(-1)^5+0^5+1^5+2^5] \pmod{13} $$
So, it gives us $33\pmod{13}=7$, but the answer is $8$. Where am I missing?
Moreover, if you know any trick or shortcut for these types of problem, can you share your knowledge?
| Since $5$ is prime to $\phi(13)=12$, each residue $\bmod 13$ is the fifth power of one residue and then
$1^5+2^5+...13^5\equiv1+2+...+13\equiv 78\equiv0\bmod 13$.
So the given sum reduces to $14^5+15^5\equiv1^5+2^5\equiv33\equiv7$.
And $7$ is correct.
Either the book is wrong or, as sometimes happens, you mistakenly read an answer to an adjacent problem. That, of course, cannot be resolved here.
| {
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Is this alternative proof of the inequality $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$ correct? Prove that for all positive real numbers:
$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq\dfrac{3}{2}$$
This is same as this question but a different approach is used there whereas I want to verify my approach to this problem.
My Approach:
$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\Big(\dfrac{a}{b+c}+1\Big)+\Big(\dfrac{b}{c+a}+1\Big)+\Big(\dfrac{c}{a+b}+1\Big)-3$$
$$=(a+b+c)\Big[\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\Big]-3$$
By AM-HM inequality:
$$\dfrac{3}{\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}}\leq\dfrac{2(a+b+c)}{3}\Rightarrow (a+b+c)\Big[\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\Big]\geq \dfrac{9}{2}$$
$$(a+b+c)\Big[\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\Big]-3\geq \dfrac{3}{2}$$
$\therefore \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq\dfrac{3}{2}\space \forall\ a,b,c\in \mathbb R$ and $a,b,c>0$
Please check this approach and provide suggestions. Also please provide alternative solutions if available.
THANKS
| Solution by the Tangent Line method.
Since our inequality is homogeneous, we can assume that $a+b+c=3$ and we onbtain: $$\sum_{cyc}\frac{a}{b+c}-\frac{3}{2}=\sum_{cyc}\frac{a}{3-a}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{3-a}-\frac{1}{2}\right)=$$
$$\sum_{cyc}\frac{3(a-1)}{2(3-a)}=\frac{3}{2}\sum_{cyc}\left(\frac{a-1}{3-a}-\frac{1}{2}(a-1)\right)=\frac{9}{4}\sum_{cyc}\frac{(a-1)^2}{3-a}\geq0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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SOS proof for $\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$ I need an SOS(sum of squares) proof for $$\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$$
if $a,b,c>0$
I already have a am-gm proof but is there a way to use SOS.
Am-gm proof :
$\frac{a^3}{bc}+b+c\ge 3a$ .....by(AM-GM ineq.)
thus $$\sum \frac{a^3}{bc}+2\sum a \ge 3\sum a$$
or $$\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$$
| Using SOS ... as requested.
\begin{eqnarray*}
(a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2+2(a^2-bc)^2+2(b^2-ca)^2+2(c^2-ab)^2 \geq 0.
\end{eqnarray*}
Now divide by $4$ and we have
\begin{eqnarray*}
a^4+b^4+c^4 \geq abc(a+b+c).
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that this sequence is convergent by using the limit definition Suppose I have the following sequence:
$ a_n = \frac{2n -\sqrt (n^2+1)}{ n+cos(n)}$
The definition says that for each $\epsilon > 0 $ there exists a positive integer $n_0 $ such that for all n > $n_0 : |a_n - L | < \epsilon$
Before starting to prove this I try to find a value for $ n_0 $ in function of $ \epsilon $.
We can write that:
$ a_n = \frac{2n - \sqrt (n^2+1)}{ n+cos(n)} \leq \frac{2n}{ n+cos(n)} $
But then I am stuck with the $ cos(n) $ in the denominator.
My idea was to use the fact that the cosine is always between -1 and 1.
Thus, $ cos(n) \leq 1 $ and $ a_n = \frac{2n - \sqrt (n^2+1)}{ n+cos(n)} \leq \frac{2n}{ n+cos(n)} \leq \frac{2n}{ n+1} \leq \frac{2n}{ n} \leq 2 $
But then again I am stuck because this doesn't help me to find an $ n_0 $ in function of $ \epsilon $
EDIT:
@TheSilverDoe reminded me that I forgot the '-L' part in the definition.
I have to calculate the limit first:
Write:
$ \frac{2n - \sqrt (n^2+1)}{ n +1} \leq \frac{2n - \sqrt (n^2)}{ n+1} \leq \frac{2n - \sqrt (n^2+1)}{ n+cos(n)} \leq\frac{2n - \sqrt (n^2)}{ n-1} \leq\frac{2n - \sqrt (n^2+1)}{ n-1} $
or
$ \frac{n}{ n+1} \leq \frac{2n - \sqrt (n^2+1)}{ n+cos(n)} \leq\frac{n}{ n-1} $
From that we can easily see that the limit is equal to 1.
Back to my original question we have:
$ a_n - L = \frac{2n - \sqrt (n^2+1)}{ n+cos(n)} - 1 \leq \frac{2n}{ n+cos(n)} - 1 \leq \frac{2n}{ n+1} - 1 \leq \frac{2n}{ n} - 1 \leq 2 - 1 = 1$
This brings me back to the same question I had originally
| It doesn't help you to show that $a_n \le 2$ (or $\le 3$ or $\le 35436$) unless you can get close to the $2$. And as $\sqrt{n^2 + 1} > n$ then $2n -\sqrt{n^2 + 1} < 2n - n =n$ you cant really get close to $2$.
But we could do $\frac {n-1}{n+1}=\frac {2n-(n+1)}{n+1} = \frac{2n-\sqrt{n^2 + 2n+1}}{n+1}<\frac {2n-\sqrt{n^2+1}}{n+\cos n}= a_n < \frac{2n-\sqrt{n^2}}{n-1} = \frac {n}{n-1}$.
We should just be able to look and $\frac {n-1}{n+1} \to 1$ and $\frac {n}{n-1}\to 1$ and use the squeeze theorem but... where's the fun in that? Which is my tongue in check way of saying we'll never learn how to not fear delta epsilon proofs if we keep avoid them.
$\frac {n-1}{n+1} = 1-\frac 2{n+1} < 1 < 1 + \frac1{n-1} = \frac n{n-1}$.
So we have both $a_n$ and $1$ between the extremes of $\frac {n-1}{n+1}$ and $\frac n{n-1}$.
So $|1-a_n| < |\frac n{n-1} - \frac {n-1}{n+1}| = \frac 1{n-1} + \frac 2{n+1}< \frac 1{n-1} +\frac 2{n-1} = \frac 3{n-1}$.
So if we want $\frac 3{n-1} < \epsilon$ it is sufficient (more than sufficient) to have $\frac 3\epsilon < n-1$ so let $n_0 = \frac 3\epsilon +1$.
| {
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A necklace is made up of 3 beads of one color and 6 beads of another color A necklace is made up of 3 beads of one color and 6 beads of another color beads of same color are identical the number of necklace that are possible ?
I have attempted the question in this way :
there are total 9 places and in which 3 objects of one kind and 6 objects of other kind are repeating so number of ways would be $\frac{9!}{3!.6!}$ = 84 which turns out to be a wrong answer when i inspected so please help me where I am wrong.
answer given is 7 possible necklaces
| Consulting the following fact sheet on necklaces and
bracelets
we get for the cycle index of the cyclic group
$$Z(C_n) = \frac{1}{n} \sum_{d|n} \varphi(d) a_d^{n/d}.$$
We will apply PET so we need
$$Z(C_9) = \frac{1}{9} (a_1^9 + 2 a_3^3 + 6 a_9).$$
Our answer is then given by
$$[X^3 Y^6] Z(C_9)(X+Y) \\ =
\frac{1}{9} [X^3 Y^6] (X+Y)^9
+ \frac{2}{9} [X^3 Y^6] (X^3+Y^3)^3
+ \frac{2}{3} [X^3 Y^6] (X^9+Y^9).$$
This yields
$$\frac{1}{9} {9\choose 3} + \frac{2}{9} [X Y^2] (X+Y)^3
= \frac{28}{3} + \frac{2}{9} {3\choose 1} =
\bbox[5px,border:2px solid #00A000]{10.}$$
With reflections allowed we get bracelets and the cycle index
$$Z(D_9) = \frac{1}{2} Z(C_9) + \frac{1}{2} a_1 a_2^4$$
which now yields
$$\frac{1}{2} 10 + \frac{1}{2} [X^3 Y^6] (X+Y) (X^2+Y^2)^4
\\ = 5 + \frac{1}{2} [X^3 Y^6] X (X^2 + Y^2)^4
\\ = 5 + \frac{1}{2} [X^2 Y^6] (X^2+Y^2)^4
= 5 + \frac{1}{2} [X Y^3] (X+Y)^4
= 5 + \frac{1}{2} {4\choose 1} =
\bbox[5px,border:2px solid #00A000]{7.}$$
| {
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Proving $\frac{a}{b^3}+\frac{b}{c^3}+\frac{c}{a^3}\geqslant \frac{a+b}{b^3+c^3}+\frac{b+c}{c^3+a^3}+\frac{c+a}{a^3+b^3}$ For $a,b,c>0.$ Prove$:$ $$\dfrac{a}{b^3}+\dfrac{b}{c^3}+\dfrac{c}{a^3}\geqslant \dfrac{a+b}{b^3+c^3}+\dfrac{b+c}{c^3+a^3}+\dfrac{c+a}{a^3+b^3}\quad (\text{Tran Quoc Thinh}) $$
It's easy with Buffalo Way and computer so I will not post it.
(Please don't post solution by Buffalo Way, thanks for a real a lot!)
So$,$ we try to find a solution by hand.
I get this SOS$:$
$$\sum {\dfrac { \left( 6\,{a}^{5}+5\,{a}^{4}b+2\,{a}^{4}c+4\,{a}^{3}{b}^{2}+4
\,{a}^{3}bc+8\,{a}^{2}{b}^{3}+6\,{a}^{2}{b}^{2}c+3\,a{b}^{4}+4\,a{b}^{
3}c-2\,{b}^{5}+2\,{b}^{4}c \right) \left( a-b \right) ^{2}}{{a}^{3}{b
}^{3} \left( a+b \right) \left( {a}^{2}-ab+{b}^{2} \right) }} \geqslant 0,$$
By SOS theorem$,$ if $$S_a+S_b+S_c \geqslant 0 ; S_a S_b +S_b S_c +S_cS_a\geqslant 0.$$
Then $$S_a (b-c)^2 +S_b (c-a)^2 +S_c(a-b)^2\geqslant 0.$$
Here$,$ we can prove$:$ $$S_a+S_b+S_c \geqslant 0,$$
but $$S_a S_b +S_b S_c +S_cS_a\geqslant 0$$ is not true!
pqr or $uvw$ technique give a very high degree, I think it is impossble.
| We need to prove that:
$$\sum_{cyc}(a^{10}c^6+a^9b^7-a^7b^6c^3-a^6b^6c^4)\geq0,$$ which is true by AM-GM:
$$\sum_{cyc}a^{10}c^6=\frac{1}{38}\sum_{cyc}\left(14a^{10}c^6+21b^{10}a^6+3c^{10}b^6\right)\geq$$
$$\geq\sum_{cyc}\sqrt[38]{a^{14\cdot10+21\cdot6}b^{21\cdot10+3\cdot6}c^{16\cdot6+3\cdot10}}=\sum_{cyc}a^7b^6c^3$$ and
$$\sum_{cyc}a^9b^7=\frac{1}{67}\sum_{cyc}\left(33a^9b^7+19b^9c^7+15c^9a^7\right)\geq$$
$$\geq\sum_{cyc}\sqrt[67]{a^{33\cdot9+15\cdot7}b^{33\cdot7+19\cdot9}c^{19\cdot7+15\cdot9}}=\sum_{cyc}a^6b^6c^4$$ and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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A hint on finding all solutions of $S = \frac{a}{a+b+d} + \frac{b}{a+b+c} + \frac{c}{b+c+d} + \frac{d}{a+c+d}$? I've been working at this for awhile but I haven't been able to figure out the right approach. The question is to find all values of $S = \frac{a}{a+b+d} + \frac{b}{a+b+c} + \frac{c}{b+c+d} + \frac{d}{a+c+d}$ for $a,b,c,d > 0$. Anyone have a hint (no solutions, please) at a more principled way to approach the problem?
I've been just throwing stuff at the wall and trying to find something that sticks.
What I've tried so far:
Plugging in some quick values. Putting $a=b=c=d=1$ yields $S = 4/3$. Taking $a=c=1, b=d=2$ yields $7/5$ so $4/3$ is not the only answer. Note: if $(a,b,c,d)$ yields $k$ then $(sa,sb,sc,sd)$ yields $k$ for any $s > 0$.
If I fix $a=b=c=1$ then $S = 4/3$ regardless of the value of $d$.
If I fix $a=c, b=d$, I can write $b = ka$, which gives $\frac{2}{2k+1} + \frac{2k}{k+2} = S$. This is a quadratic in $k$, with positive discriminant for any value of $S$. But $k$ is not necessarily positive, so I can't claim that all values of $S$ are valid.
Getting some quick bounds: $S > \frac{a}{a+b+c+d} + \frac{b}{a+b+c+d} + \frac{c}{a+b+c+d} + \frac{d}{a+b+c+d} = 1$, and $S < a/a + b/b + c/c + d/d = 4$, so I know $1 < S < 4$. The thought is to either try and tighten these bounds somehow (not sure how to approach this) or figure out how to represent an arbitrary $x$ in this range with some choice of $a,b,c,d$.
Some substitutions: Imposing the constraint $a+b+c+d = 1$ or $u = a+b+d, v = a+b+c, w = b+c+d, x = a+c+d$ come to mind, both attempts at removing the sums from the denominator. Respectively those give $S = \frac{a}{1-c} + \frac{b}{1-d} + \frac{c}{1-a} + \frac{d}{1-b}$ and $S = \frac{u+v+w-2x}{3u} + \frac{u+v+w-2x}{3v} + \frac{v+x+w-2u}{3w} + \frac{u+w+x-2v}{3x}$.
Another way of looking at things: $\frac{a}{a+b+d} + \frac{b}{a+b+c} + \frac{c}{b+c+d} + \frac{d}{a+c+d}$ is continuous in $(a,b,c,d)$. And I think we can get arbitrarily close to 1 by setting $a$ very large, and then $c$ very small relative to $b + d$. So 1 is probably the $\inf$ of $S$. If we do the opposite and set $a$ large and $c$ large, we can get close to 2. My suspicion is that $2$ is the $\sup$ but I'm not sure how to prove it.
| We have
$$\frac{a}{a+b+d} < \frac{a}{a+b}, \frac{b}{a+b+c} < \frac{b}{a+b}, \frac{c}{b+c+d} < \frac{c}{c+d}, \frac{d}{a+c+d} < \frac{d}{c+d}$$
So $S < \frac{a+b}{a+b} + \frac{c+d}{c+d} = 2$, which gives the upper bound that I was missing. Then as mentioned in the question, send $a$, $c$ to be large, (or $b,d\to 0$ as in Michael's answer) to get arbitrarily close to 2, which gives the missing piece.
The case of $S < 1$ is also covered in the question.
| {
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Evaluating $\int_{-1/2}^{1} \cos^{-1} \frac{1-x^2}{1+x^2} dx$ Let us do the integration by parts $$I=\int_{-1/2}^{1} ~\cos^{-1} \frac{1-x^2}{1+x^2}. 1~ dx =\left.x \cos^{-1} \frac{1-x^2}{1+x^2}\right|_{-1/2}^{1}-\int_{-1/2}^{1} \frac{2x}{1+x^2} dx$$ $$=\frac{\pi}{2}+\frac{1}{2}\cos^{-1} \frac{3}{5}-\left .\ln(1+x^2)\right|_{-1/2}^{1}=\frac{\pi}{2}+\frac{1}{2}\cos^{-1} \frac{3}{5}-\ln\frac{8}{5}.$$
The question is whether something is amiss here and whether the answer is right.
EDit
The correct answer is $$I=\frac{\pi}{2}+\frac{1}{2}\cos^{-1} \frac{3}{5}-\ln\frac{5}{2}.$$
| Set $\tan^{-1}x=y, -\dfrac\pi2<y<\dfrac\pi2\iff-\pi<2y\le\pi $
$$\cos^{-1}\dfrac{1-x^2}{1+x^2}=\cos^{-1}(\cos2y)=\begin{cases}2y &\mbox{if }0\le2y\le\pi\iff x\ge0 \\
-2y & \mbox{if } x<0\end{cases} $$
$$\implies\int_{-1/2}^1\cos^{-1}\dfrac{1-x^2}{1+x^2}dx=-2\int_{-1/2}^0\tan^{-1}x\ dx+2\int_0^1\tan^{-1}x\ dx$$
Now integrate by parts $$\displaystyle\int\tan^{-1}x\ dx=\tan^{-1}x\int\ dx-\int\left(\dfrac{d(\tan^{-1}x)}{dx}\int\ dx\right)dx$$
| {
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Evaluation of $\int_{-2}^{-1} \tan^{-1}\sqrt{\frac{x+1}{x-1}} dx$ Evaluation of $$\int_{-2}^{-1} \tan^{-1}\sqrt{\frac{x+1}{x-1}} dx$$
requires a special care when done by hand. The question is: How to do it and what is the right answer.
| Note that when $x\in (-\infty, -1] \cup (1,\infty)$,
$$\frac{d}{dx} \tan^{-1} \sqrt{\frac{x+1}{x-1}}=-\frac{1}{2|x|} \frac{1}{\sqrt{x^2-1}}.$$
Next the integration by parts leads to $$I=\int_{-2}^{-1} \tan^{-1} \sqrt{\frac{x+1}{x-1}}.~1 ~dx=\left . x \tan^{-1} \sqrt{\frac{x+1}{x-1}}\right|_{-2}^{-1}-\frac{1}{2}\int_{-2}^{-1} \frac{dx}{\sqrt{x^2-1}}.$$
$$\implies I=\frac{\pi}{3}-\frac{1}{2} \int_{1}^{2} \frac{dz}{\sqrt{z^2-1}}=\frac{\pi}{3}-\ln\sqrt{2+\sqrt{3}}.$$
| {
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Doubt: $f\bigl(f(x)f(y)\bigr) + f(x+y) = f(xy) $ over $\mathbb R$
Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that for any real numbers $x$ and $y$, $$f\bigl(f(x)f(y)\bigr) + f(x+y) = f(xy) \text.$$
This is from Evan Chen's IMO 2017 Solution Notes:
The only solutions are $f(x) = 0$, $f(x) = x − 1$ and $f(x) = 1 − x$, which clearly work. Note that If $f$ is a solution, so is $−f$.
I understood how $f(x) = 0$, $f(x) = x − 1$ and $f(x) = 1 − x$ work.
When $f(x)=0$, we get
$$f\bigl(f(x)f(y)\bigr) + f(x+y) = f(0)+0=0=f(xy) \text.$$
When $f(x)=x-1$, we get
$$f\bigl(f(x)f(y)\bigr) + f(x+y) = f\bigl((x-1)(y-1)\bigr)+ x+y-1 \\
= xy-x-y+x+y-1=xy-1= f(xy) \text.$$
When $f(x)=1-x$, we get
$$f\bigl(f(x)f(y)\bigr) + f(x+y)= f\bigl((1-x)(1-y)\bigr) + 1 -x-y \\
= -xy +x+y +1 -x-y= 1-xy f(xy) \text.$$
Can someone explain this line "If $f$ is a solution, so is $−f$".
Thanks in advance!
| Let $g=-f$ with $f$ a solution to the equation. Then $$g(xy)=-f(xy)=-f(f(x)f(y))-f(x+y)\\ =g(-g(x)(-g(y)))+g(x+y) = g(g(x)g(y)) +g(x+y)$$
and so $g$ also satisfied the functional equation.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How many ways $5$ identical green balls and $6$ identical red balls can be arranged into $3$ distinct boxes such that no box is empty?
How many ways $5$ identical green balls and $6$ identical red balls can be arranged into $3$ distinct boxes such that no box is empty?
My attempt :
Finding coefficient of $x^{11}$ in the expansion of $$( x + x^2 + x^3 + x^4 + x^5+x^6 )^3 ( x + x^2 + x^3 + x^4 + x^5 )^ 3$$ and arranging them which was wrong when inspected
Please help me out
| When we neglect the nonempty condition we can distribute the $5$ green balls in ${5+2\choose2}=21$ ways into the three distinct boxes, and independently of this the $6$ red balls in ${6+2\choose2}=28$ ways. This gives $21\cdot28=588$ possible distributions.
In the same way we can compute the number of distributions where the third box has to remain empty. This gives a total of ${5+1\choose1}\cdot{6+1\choose1}=42$ distributions, and the same number arises when another box has to remain empty.
The total number $N$ of admissible distributions therefore is given by
$$N=588-3\cdot 42+3=465\ .$$
At the end we have added $3$, because the three distributions where two boxes remain empty have been subtracted twice in the $3\cdot42$ term.
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of real roots $x^8-x^5+x^2-x+1=0$ Find the number of real roots of $x^8-x^5+x^2-x+1$.
My attempt: $f(x)$ has $4$ sign changes and $f(-x)$ has no sign changes, so the possibility of having real roots is $4+0=4$. Since this is a polynomial of degree $8$ it should have $8-4=4$ imaginary roots.
It is quite impossible to see that this equation does not have any real roots by observing the factor $x^8-x^5+x^2-x+1= (x-1)(x^7+x^6+x^5+x)+1>0$, hence it can't have any real roots. My question is how to prove using this equation doesn't have any real roots using Descartes' Rule? Please give some useful hints.
| We will prove that $f(x)=x^8-x^5+x^2-x+1$ has no real roots using Descartes' Rule only.
Since $f(-x)=x^8+x^5+x^2+x+1$ has no non-negative coefficients we know $f(x)$ has no real roots in $(-\infty,0]$.
Since $f(x+1)=x^8+8x^7+28x^6+55x^5+65x^4+46x^3+19x^2+4x+1$ has no non-negative coefficients we know $f(x+1)$ has no real roots in $[0,\infty)$. This means $f(x)$ has no real roots in $[1,\infty)$.
It suffices to show that $f(x)$ has no real roots in $(0,1)$. Perform the substitution $t=1/x$ so it suffices to show that $g(t)=t^8f(t)=t^8-t^7+t^6-t^3+1$ has no real roots in $(1,\infty)$.
Since $g(t+1)=t^8+7t^7+22t^6+41t^5+50t^4+40t^3+19t^2+4t+1$ has no non-negative coefficients we know $g(t+1)$ has no real roots in $(0,\infty)$. This means that $g(t)$ has no real roots in $(1,\infty)$, so $f(x)$ has no real roots in $(0,1)$. This completes the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\frac{{a}^3+{b}^3+ac(a-c)+bc(b-c)-5abc}{{(a+b+c)}^3} \geqslant -\frac{1}{5}$ For $a,b,c\geqslant 0.$ Prove$:$
$$\dfrac{{a}^3+{b}^3+ac(a-c)+bc(b-c)-5abc}{{(a+b+c)}^3} \geqslant -\dfrac{1}{5}$$
I found an AM-GM proof.
Since $$P+\frac{1}{5}\geqslant 0\Leftrightarrow 6\,{a}^{3}+6\,{b}^{3}+8\,{a}^{2}c-2\,a{c}^{2}+8\,{b}^{2}c-2\,b{c}^{2}-19\,abc+3\,{a}^{2}b+3\,a{b}^{2}+{c}^{3} \geqslant 0$$
And by AM-GM$:$
$$2\,a{c}^{2}\leqslant 6{a}^{3}+\frac49{c}^{3},$$
$$2\,b{c}^{2}\leqslant 6{b }^{3}+\frac49{c}^{3},$$
$$19\,abc\leqslant \frac19{c}^{3}+3a{b}^{2}+3{a}^{2}b+8 \,{a}^{2}c+8\,{b}^{2}c.$$
So we are done!
Is there another nice proof$?$ Thanks for a real lot!
| We write inequality as
$$a^3+8(b+c)a^2+(8b^2-19bc-2c^2)a+b^3+8b^2c-2bc^2+c^3 \geqslant 0.$$
Because $a^3-ab(2a-b) = a(a-b)^2 \geqslant 0,$ so we will show that
$$ab(2a-b)+8(b+c)a^2+(8b^2-19bc-2c^2)a+b^3+8b^2c-2bc^2+c^3 \geqslant 0,$$
or
$$f(a) = 2(5b+4c)a^2+(7b^2-19bc-2c^2)a+b^3+8b^2c-2bc^2+c^3 \geqslant 0.$$
Because $b^3+8b^2c-2bc^2+c^3 \geqslant 0,$ therefore
If $7b^2-19bc-2c^2 \geqslant 0$ then $f(a) \geqslant 0.$
If $7b^2-19bc-2c^2 \leqslant 0,$ we have
$$\Delta_a = (b^2-68bc-28c^2)(3b-c)^2 \leqslant 0.$$
So $f(a) \geqslant 0.$
| {
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Base $7$ is backwards base $16$ Find the sum of all positive integers such that their expression in base $7$ digits is the reverse of their expression in base $16$ digits. Express your answer in base $10$.
I tried expressing the digits as $d_1$ $d_2$... and making separate equations but that didn't work. Help!
| In particular, these numbers have at most as many digits in base seven as they have in base sixteen. So, if $16^{k-1} \leq n < 16^{k}$ is such an integer, then $n$ has at most $k$ digits in base $7$, so $n < 7^k$. Thus $16^{k-1} < 7^k$. This forces $k \leq 3$. By assumption, you know that $n$ has as base-16 digits only $0,1,2,3,4,5,6$.
The one-digit numbers contribute $21$.
For the two-digit numbers: write $n$ as $ab$ in base $16$. Then $16a+b=7b+a$, so $15a=6b$ ie $5a=2b$, so $a=2,b=5$ is the only solution, ie $n=37$.
For the three-digit numbers: write $n$ as $abc$, then $256a+16b+c=49c+7b+a$, thus $255a+9b=48c$, hence $85a+3b=16c$. As $a > 0$, $c > 5$ so $c=6$ and $3|85a$, so $16c=85a+3b > 255$, so $c > 6$, impossible.
So finally it’s $21+37=58$.
| {
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$A,B,C$ and $D$ are concyclic.$AC$ is the diameter of the circle and $AD=DC$.The area of quadrilateral $ABCD$ is $20cm^2$. $A,B,C$ and $D$ are concyclic . $AC$ is the diameter of the circle and $AD=DC$ . The area of quadrilateral $ABCD$ is $20c$m$^2$.
Draw a line $DE$ such that $E$ is a point on $AB$, and $DE$ $\bot$ $AB$.
Find the length of $DE$.
The answer keys says that it is ${2}\sqrt{5}$ $cm$. How was the length found?
I imagine that the use of similar triangles and using the fact that angle in a semicircle is $90^\circ $may be involved, but I cannot go beyond that.
| Let $AB=x$ and $BC=y$.
Thus, $$AC=\sqrt{x^2+y^2}$$ and by the given:
$$\frac{xy}{2}+\frac{x^2+y^2}{4}=20,$$ which gives $$x+y=4\sqrt5.$$
Id est, $$DE=AD\sin\left(45^{\circ}+\measuredangle BAC\right)=$$
$$=\sqrt{\frac{x^2+y^2}{2}}\cdot\frac{1}{\sqrt2}\left(\sin\arctan\frac{y}{x}+\cos\arctan\frac{y}{x}\right)=$$
$$=\sqrt{\frac{x^2+y^2}{2}}\cdot\frac{1}{\sqrt2}\left(\frac{\frac{y}{x}}{\sqrt{1+\left(\frac{y}{x}\right)^2}}+\frac{1}{\sqrt{1+\left(\frac{y}{x}\right)^2}}\right)=\frac{1}{2}(x+y)=2\sqrt5.$$
| {
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Calculate the integral $\iiint\limits_{D}dxdydz$ over domain D Let $a\in (-1,1)$ and domain $D=\{(x,y,z)\in \Bbb R:x^2+y^2+z^2<1,z>a\}$.
I try to calculate the integral $$\iiint\limits_{D}dxdydz$$ with the use of cylindrical coordinates, but I couldn't find the limits of the integration. Do you have any idea about the limits of the integral ?
| As in our region $D$ we have
$$ x^2+y^2+z^2 < 1 $$
and
$$
z > a,
$$
so we can take $r$, $\theta$, and $z$ as follows:
$$
0 \leq r < 1,
$$
$$
0 \leq \theta \leq 2 \pi,
$$
and
$$
a < z < 1.
$$
Moreover, we note that, since
\begin{align}
x &= r \cos \theta, \\
y &= r \sin \theta, \\
z &= z,
\end{align}
therefore we find that
\begin{align}
\frac{\partial(x, y, z) }{\partial(r, \theta, z)} &= \left| \begin{matrix} \frac{\partial x}{\partial r } & \frac{\partial x}{\partial \theta } & \frac{\partial x}{\partial z } \\
\frac{\partial y}{\partial r } & \frac{\partial y}{\partial \theta } & \frac{\partial y}{\partial z } \\
\frac{\partial z}{\partial r } & \frac{\partial z}{\partial \theta } & \frac{\partial z}{\partial z }
\end{matrix} \right| \\
&= \left| \begin{matrix} \cos \theta & - r\sin \theta & 0 \\
\sin \theta & r \cos \theta & 0 \\
0 & 0 & 1 \end{matrix} \right| \\
&= \left| \begin{matrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{matrix} \right| \\
&= r \cos^2 \theta - \left(-r \sin^2 \theta \right) \\
&= r,
\end{align}
and so
\begin{align}
\int\int\int_{D}\, dx \, dy\, dz &= \int_{z = a}^{z = 1} \int_{\theta = 0}^{\theta = 2 \pi} \int_{r = 0 }^{r = 1} r d r\, d \theta \, dz \\
&= \int_{z = a}^{z = 1} \int_{\theta = 0}^{\theta = 2 \pi} \frac{1}{2} d \theta \, dz \\
&= \int_{z = a}^{z = 1} \frac{1}{2} (2 \pi ) dz \\
&= \int_{z = a}^{z = 1} \pi dz \\
&= \pi (1 - a).
\end{align}
| {
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The sequence $A_n=\prod_{k=1}^n\left(1+\frac{k}{n^2}\right)$ is decreasing Let $A$ be the sequence of real numbers defined by :
$$\forall n\in\mathbb{N}^\star,\,A_n=\prod_{k=1}^n\left(1+\frac{k}{n^2}\right)$$
I know how to prove that this sequence converges to $\sqrt e$, using the following inequalities :
$$\forall t>0,\,t-\frac{t^2}2\leqslant\ln(1+t)\leqslant t$$
I found numerical evidence that $(A_n)$ is decreasing, but wasn't able to prove it. Any help will be appreciated.
| It is easy to verify $A_1 > A_2 > A_3 > A_4$.
It suffices to prove that $A_n > A_{n+1}$ for all $n\ge 4$.
It suffices to prove that, for all $n\ge 4$,
$$\sum_{k=1}^n \ln (1 + k/n^2) > \sum_{k=1}^{n+1} \ln (1 + k/(n+1)^2)$$
or
$$\sum_{k=1}^n \ln \frac{1 + k/n^2}{1 + k/(n+1)^2} > \ln (1 + 1/(n+1))$$
or
$$\sum_{k=1}^n \ln\left(1 + \frac{k(2n+1)}{n^2(n^2+k+2n+1)} \right) > \ln (1 + 1/(n+1)).$$
By using $\ln (1+x) \ge \frac{x}{1+x}$ for $x > 0$, we have
\begin{align}
\ln\left(1 + \frac{k(2n+1)}{n^2(n^2+k+2n+1)} \right)
&\ge \frac{k(2n+1)}{(n^2+k)(n+1)^2}\\
&= \frac{k(2n+1)}{n^2(n+1)^2}\, \frac{1}{1 + k/n^2}\\
&\ge \frac{k(2n+1)}{n^2(n+1)^2}(1 - k/n^2).
\end{align}
Also, by using $\ln(1+x) < \frac{x^2+6x}{6+4x}$ for $x > 0$, we have
$$\ln (1 + 1/(n+1)) < \frac{7+6n}{2(3n+5)(n+1)}.$$
Thus, it suffices to prove that, for all $n\ge 4$,
$$\sum_{k=1}^n \frac{k(2n+1)}{n^2(n+1)^2}(1 - k/n^2) > \frac{7+6n}{2(3n+5)(n+1)}$$
or
$$\frac{(n-1)(2n+1)(3n+1)}{6n^3(n+1)} > \frac{7+6n}{2(3n+5)(n+1)}$$
or
$$\frac{6n^3-17n^2-23n-5}{6n^3(n+1)(3n+5)} > 0.$$
It is true. We are done.
| {
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I am not able to solve the integral $I=\int { \frac { \sqrt { \sqrt { x } +\sqrt { x-1 } } }{ 1+\sqrt { x } } }$ $J=\int { \frac { \sqrt { \sqrt { x } -\sqrt { x-1 } } }{ 1+\sqrt { x } } } $ then I-J
I substituted $x=\sec ^{ 2 }{ \theta } $.
$I-J=\frac { \sqrt { \sec { \theta } +\tan { \theta } } -\sqrt { \sec { \theta } -\tan { \theta } } }{ 1+\sec { \theta } } \times 2\sec ^{ 2 }{ \theta } \tan { \theta d\theta } \\ I-J=\frac { \sqrt { \sec { \theta } +\tan { \theta } } -\frac { 1 }{ \sqrt { \sec { \theta } +\tan { \theta } } } }{ 1+\sec { \theta } } \times 2\sec ^{ 2 }{ \theta } \tan { \theta d\theta } \\ I-J=\frac { \frac { \sec { \theta } +\tan { \theta } -1 }{ \sqrt { \sec { \theta } +\tan { \theta } } } }{ 1+\sec { \theta } } \times 2\sec ^{ 2 }{ \theta } \tan { \theta d\theta } $
I am not able to solve further Please help
| Hint:
$$\begin{align}I-J &=\int \frac{\sqrt{\sqrt{x}+\sqrt{x-1}}-\sqrt{\sqrt{x}-\sqrt{x-1}}}{\sqrt{x}+1}dx \\ &= \int\frac{\left(\left(\sqrt{\sqrt{x}+\sqrt{x-1}}-\sqrt{\sqrt{x}-\sqrt{x-1}}\right)^{2}\right)^{\frac{1}{2}}}{\sqrt{x}+1}dx \\ &=\sqrt{2} \int \frac{\sqrt{\sqrt{x}-1}}{\sqrt{x}+1}dx \end{align}$$
| {
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Show that the equation $x^3+2y^2+4z=n$ has an integer solution $(x,y,z)$ for all integers $n.$
Show that the equation $$x^3+2y^2+4z=n$$ has an integer solution $(x,y,z)$ for all integers $n.$
I tried to use parity in order to get somewhere, but couldn't get quite far.
I was given a hint that I should first show that $n$ can be of some of the following forms $n=4k, n=4k+1, n=4k+2, n=4k+3$. How can I come to this conclusion?
|
I was given a hint that I should first show that n can be of some of the following forms n=4k,n=4k+1,n=4k+2,n=4k+3. How can I come to this conclusion?
If you take an integer $n$ and divide it by $4$ you will get a quotient $k$ and a remainder $r$. And you can say $n = 4k + r$ and that $r = 0,1,2,$ or $3$.
That is how you come to that conclusion.
Now, if we let $z = k$ then we have $x^3 + 2y^2 + 4z = x^3+2y^2+4k = n = 4k +r$ so
$x^3 + 2y^2 = r$ so if we can find integer solutions to $x^3 +2y^2 = 0,1,2,3$ we will be done.
Now $0^3 + 2*0^2 = 0$ is clearly a solution for $r = 0$. So $(0,0,k)$ is a solution to $x^3 + 2y^2 + 4z = n = 4k$.
And $1^3 + 2*0^2=1$ is a solution for $r=1$. So $(1,0,k)$ is a solution to $x^3 + 2y^2 + 4z = n = 4k+1$.
And $0^3 + 2*1^2 = 2$ is a solution for $r=2$. So $(0,1,k)$ is a solution to $x^3 + 2y^2 + 4z = n = 4k +2$.
And $1^3 + 2*1^2 =3$ (I'm embarrassed to say that that it took me a long time to come up with that as my instinct so to use $x=-1$ and $z =k+1$ or other variations first) is a solution for $r=3$ and so $(1,1,k)$ is a solution for $x^3 + 2y^2 + 4z = n = 4k+3$.
.....
Now obviously these aren't the only solutions. (if $(x,y,z)= (3,2,1)$ then $x^3 + 2y^2 + 4z= 27+8 + 4= 39$ is a solution for $n=39$..... but so is $(1,1,9)$ as $39=4*9+3$). But they are enough to show solutions always exist.
| {
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Elementary solution to $ \int \frac{1}{x^5+1} \, dx $ Since the original question is heavily downvoted, I'm not sure if the moderators will delete it before I can post my answer.
Here's my attempt:
Following JG's remarks, we have
$$ x^5 + 1 = \frac14(x+1)(2x^2 + x(\sqrt5 - 1) + 2)(2x^2 - x(\sqrt5 - 1) + 2) $$
Then by partial fractions, we have $$ \dfrac1{x^5 + 1} = \dfrac4{(x+1)(2x^2 + x(\sqrt5 - 1) + 2)(2x^2 - x(\sqrt5 - 1) + 2)} $$ is equal to $$ \dfrac2{5+\sqrt5} \cdot \dfrac1{x+1} -\dfrac{4\sqrt5}{(\sqrt5 - 1)(5+\sqrt5)} \cdot \dfrac x{2x^2 - x(\sqrt5 - 1) + 2} + \dfrac4{(\sqrt5-1)(5+\sqrt5)} \cdot \dfrac{x + \sqrt5 + 1}{2x^2 + x(\sqrt5 - 1) + 2}$$
What's left is to apply $ \int \dfrac1{x^2+a^2} \, dx = \frac1a \tan^{-1} (\frac xa) + C$ and $ \int \dfrac{1}{x^2-a^2}\, dx = -\frac 1a \tanh^{-1} (\frac xa) + C$.
Obviously all the calculations above are fairly tedious. Is there a(another) way to evaluate this indefinite integral?
To clarify, I'm not interested to see ANY roots of unity ($\omega $) in the final result.
Naturally, I'm also interested if there's other approach is applicable for the indefinite integral of $\frac1{x^n + 1}$ for all positive integers $n$.
| Let $g(x,\phi)=\frac{2\phi x-2}{x^2-2\phi x+1}$ and integrate
\begin{align}
I(x,\phi) &= \int g(x,\phi)dx
=\phi\ln\left(x^2-2\phi x+1\right)
-2\sqrt{1-\phi^2}
\tan^{-1}\frac{x-\phi}{\sqrt{1-\phi^2}}
\end{align}
Then, with $\phi_{\pm} = \frac{1\pm\sqrt5}{4}$
\begin{align}
\int \frac{1}{1+x^5}dx
&=\frac15 \int \left(\frac1{x+1}- g(x,\phi_+)
- g(x,\phi_-) \right) dx\\
&=\frac15\left[\ln(x+1)-I(x,\phi_+)-I(x,\phi_-)\right] + C
\end{align}
| {
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How to find the cartesian equation of a plane given the vector equation? The vector equation is given by $r=\begin{pmatrix}-3\\ -5\\ -1\end{pmatrix}+k\begin{pmatrix}6\\ 6\\ -3\end{pmatrix}$
To find the cartesian equation, I have to consider a point on the plane:
$$\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}-3\\ -5\\ -1\end{pmatrix}+k\begin{pmatrix}6\\ 6\\ -3\end{pmatrix}$$
which gives:
$$\frac{x+3}{\:6}=\:\frac{y+5}{6}=\frac{z+1}{-3}$$
Now this is the part where I don't understand. Up till now my working follows with the answers provided by my book, but then the book did something that puzzle me:
$$\frac{x+3}{6}+\frac{y+5}{6}=2(\frac{z+1}{-3})$$
How did they reached here?
| We have that
*
*$x=-3+6k \implies k=\frac{x+3}6$
*$y=-5+6k\implies k=\frac{y+5}6$
*$z=-1-3k\implies k=-\frac{z+1}3$
therefore
$$\frac{x+3}6+\frac{y+5}6=2k=-2\frac{z+1}3$$
moreover
$$\frac{x+3}6-\frac{y+5}6=2k=0$$
and the given line can be represented, in cartesian form, by the intersection of the two planes
*
*$x+y+4z+12=0$
*$x-y-2=0$
| {
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question from South Korean selection exam 1998, about proving that an inequality holds true if $a+b+c=abc$ I was just doing the following question:
If $a,b,c>0$ such that $a+b+c=abc$, prove that:
$\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}\le \frac{3}{2}$
I think that this question can be solved through the use of homogenization, something which I attempted to do in the following way:
We have that $\frac{a+b+c}{abc}=1$ and hence also $\sqrt{\frac{a+b+c}{abc}}=1$. So $\frac{3}{2}*\sqrt{\frac{a+b+c}{abc}}=\frac{3}{2}$.
So all we have to prove now is $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}\le\frac{3}{2}*\sqrt{\frac{a+b+c}{abc}}$ which are homogenized and hence we do not need the original equality any more. This is where I couldn't continue from, and got stuck. Could you please explain to me how I could finish it off like this, or tell me why it can't and how it can be done using homogenization?
| We write inequality as
$$\sum \sqrt{\frac{bc}{(a+b)(a+c)}} \leqslant \frac{3}{2}.$$
Using known inequality
$$(a+b)(b+c)(c+a) \geqslant \frac{8}{9}(a+b+c)(ab+bc+ca),$$
and the Cauchy-Schwarz inequality we have
$$\sum \sqrt{\frac{bc}{(a+b)(a+c)}} \leqslant \sqrt{(ab+bc+ca) \sum \frac{1}{(a+b)(a+c)}} $$
$$= \sqrt{\frac{2(ab+bc+ca)(a+b+c)}{(a+b)(b+c)(c+a)}} \leqslant \frac 32.$$
Done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Alternative approach for proving that for any $x\in\mathbb R^+$, $x^2+3x+\frac{1}{x} \ge \frac{15}{4}$. Let $x \in \mathbb{R^+}$. Prove that:
$$x^2+3x+\frac{1}{x} \ge \frac{15}{4}.$$
While this is indeed easily proven using derivatives, where the minimum is obtained when $x=\frac{1}{2}$, is it possible to prove it through other means, say by AM-GM? Any hints would be much appreciated.
| The AM-GM inequality would not be any better than derivatives at finding your minimum. But if you know (or suspect) that the minimum is at $x = \frac12$, then we can use it to get a proof.
First: at $x = \frac12$, the three terms are equal to $\frac14$, $\frac32$, and $2$. If we want to use a weighted AM-GM inequality to prove that $x^2 + 3x + \frac1x \ge \frac{15}{4}$, this has to be the equality case. So, we rewrite $x^2 + 3x + \frac1x$ as
$$
\frac1{15} (15 x^2) + \frac25 \left(\frac{15}{2}x\right) + \frac{8}{15} \left(\frac{15}{8x}\right)
$$
because the three terms $15x^2$, $\frac{15}{2}x$, and $\frac{15}{8x}$ are equal at $x=\frac12$. (We get the coefficients $\frac1{15}$, $\frac25$, and $\frac{8}{15}$ by normalizing $\frac14$, $\frac32$, and $2$ to add to $1$.)
Now, we apply weighted AM-GM:
$$
\frac1{15} (15 x^2) + \frac25 \left(\frac{15}{2}x\right) + \frac{8}{15} \left(\frac{15}{8x}\right) \ge (15 x^2)^{1/15} \left(\frac{15}{2}x\right)^{2/5} \left(\frac{15}{8x}\right)^{8/15}.
$$
The right-hand side simplifies to $\frac{15}{4}$, as all the powers of $x$ cancel, and we get the lower bound we wanted.
The general procedure at work here is called geometric programming. Usually, we do not have a candidate solution, and we will try arbitrary weights $\delta_1, \delta_2, \delta_3 \ge 0$ with $\delta_1 + \delta_2 + \delta_3 = 1$ (to apply AM-GM) and $2\delta_1 + \delta_2 - \delta_3 = 0$ (so that the powers of $x$ cancel). Then we would get a lower bound on $x^2 +3x + \frac1x$ depending on these weights, and we would optimize our choice of weights to get the best lower bound possible.
In this example, this technique doesn't work any better than setting the derivative equal to $0$. It really shines in multivariable examples.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Apparent loophole in set of coupled partial differential equations I have a set of three coupled differential equations, using coordinates $x$ and $\theta$, and would like to solve for $\xi_x (x, \theta)$, $\xi_{\theta} (x, \theta)$, given by:
\begin{equation}
\partial_x \xi_x = -\frac{x}{1+x^2} \xi_x,
\end{equation}
\begin{equation}
\left( \frac{2}{x} - \partial_x \right)\xi_{\theta} = \partial_{\theta} \xi_x.
\end{equation}
\begin{equation}
\partial_{\theta} \xi_{\theta} = -x (x^2+1) \xi_x,
\end{equation}
I solve the first one by integration and find
\begin{equation}
\xi_x=\frac{A}{\sqrt{1+x^2}} + f(\theta),
\end{equation}
where $A$ is just a constant and $\xi_x$ may depend on $\theta$, which is why I added the function $f(\theta)$. Now I plug this solution into the second equation, i.e.
\begin{equation}
\left( \frac{2}{x} - \partial_x \right)\xi_{\theta} = \partial_{\theta} f(\theta),
\end{equation}
and solve for $\xi_{\theta}$ to find
\begin{equation}
\xi_{\theta} = x \cdot \partial_{\theta} f(\theta) + Bx^2,
\end{equation}
with $B$ another constant. Finally, I plug both of my solutions into the third equation to obtain
\begin{equation}
x \cdot \partial_{\theta}^2 f(\theta) = -x(x^2+1) \left(\frac{A}{\sqrt{x^2+1}} + f(\theta), \right)
\end{equation}
which has a solution such that
\begin{equation}
f(\theta) = -\frac{A}{\sqrt{x^2+1}} + C \cos (\sqrt{x^2+1} \theta) + D \sin (\sqrt{x^2 +1 } \theta).
\end{equation}
Now here's my problem: $f(\theta)$ was assumed to depend only on $\theta$, not on $x$. Plugging this back into my first equation already shows this can not be the solution. Does anyone see what my mistake is? Thank you for your time in advance.
Edit: I tried to write the solution for $\xi_x$ as
\begin{equation}
\xi_x = \frac{A}{\sqrt{1+x^2}} \cdot f(\theta),
\end{equation}
but this makes it such that the second equation has no solutions.
| The first equation gives
$$
\xi_x = \frac{A(\theta)}{\sqrt{1+x^2}} \, .
$$
The other equations then give
$$
\left(\frac2x - \partial_x\right) \xi_\theta = \frac{A'(\theta)}{\sqrt{1+x^2}} ,\qquad
\partial_\theta\xi_\theta = -A(\theta)x{\sqrt{1+x^2}} .
$$
The first one yields
$$
\xi_\theta = A'(\theta)x{\sqrt{1+x^2}} + B(\theta) x^2
$$
so that by substitution in the last equation,
$$
\partial_\theta\xi_\theta = A''(\theta)x{\sqrt{1+x^2}} + B'(\theta) x^2 = -A(\theta)x{\sqrt{1+x^2}} \, .
$$
For this to be satisfied for all $x$, we choose $A(\theta) = a \cos\theta + b\sin\theta$ and $B(\theta) = c$. Finally,
$$
\xi_x = \frac{a \cos\theta + b\sin\theta}{\sqrt{1+x^2}} , \qquad
\xi_\theta = \left(-a \sin\theta + b\cos\theta\right) x\sqrt{1+x^2} + cx^2 \, .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3839777",
"timestamp": "2023-03-29T00:00:00",
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Set of Pythagorean Identities and The Golden Ratio.
Using Degrees
Where A intercepts C the Y coordinate is Phi
The X coordinate is (sin^-1(1/ phi^2))/ 2
Where B intercepts C the Y coordinate is 1/Phi
The X coordinate is (sin^-1(1/phi) )/ 2
I got these equations while messing around with Area-Diagonal relationships.
Basically replaced x and y with sin and cos.
Why does the golden ratio show up here?
| Interesting observation. We need the identities $\sin^2x+\cos^2x=1$ and $\sin2x=2\sin x\cos x$.
To solve $A$ and $C$, we set
$$\begin{align}(\sin x + \cos x)^2&=\frac {2 \sin x \cos x}{(\sin x - \cos x)^2}
\\(\sin^2 x + 2\sin x\cos x+\cos^2 x)(\sin^2 x - 2\sin x\cos x+\cos^2 x)&=2 \sin x \cos x\\(1+\sin 2x)(1 - \sin 2x)&=\sin 2x
\\1-\sin^2 2x&=\sin 2x \\\sin^2 2x+\sin 2x-1&=0\end{align}$$
and this equation leads to the golden ratio, since
$$\sin 2x = \frac {-1\pm\sqrt5}2 \implies (\sin x+\cos x)^2=1+\sin2x=\frac {1\pm\sqrt5}2$$ Similarly for $B$ and $C$:
$$\begin{align}(\sin x - \cos x)^2&=\frac {2 \sin x \cos x}{(\sin x - \cos x)^2}
\\(\sin^2 x - 2\sin x\cos x+\cos^2 x)^2&=2 \sin x \cos x\\(1 - \sin 2x)^2&=\sin 2x
\\1-2\sin 2x + \sin^2 2x&=\sin 2x \\\sin^2 2x-3\sin 2x+1&=0\end{align}$$
and we thus have
$$\sin 2x = \frac {3\pm\sqrt5}2 \implies (\sin x-\cos x)^2=1-\sin2x=\frac {-1\pm\sqrt5}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.