Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Matrices operations My professors asked to find the inverse of M:
$$ M=
\begin{bmatrix}
I_p&0\\
A&I_q\\
\end{bmatrix}
$$
Therefore, to find $M^{-1}$, we calculate:
$$
\begin{bmatrix}
I_p&0\\
A&I_q\\
\end{bmatrix}
\begin{bmatrix}
C&D\\
E&F\\
\end{bmatrix} =
\begin{bmatrix}
I_p&0\\
-A&I_q\\
\end{bmatrix}
$$
How do I go from
$$
\begin{bmatrix}
C&D\\
AC+E&AD+F\\
\end{bmatrix} =
\begin{bmatrix}
I_p&0\\
0&I_q\\
\end{bmatrix}
$$
to
$$
\begin{bmatrix}
C&D\\
E&F\\
\end{bmatrix} =
\begin{bmatrix}
I_p&0\\
-A&I_q\\
\end{bmatrix}
$$
What operations did my professor use to get rid of $AC$ and $AD$ in the first matrix to just get $C$, $D$, $E$, $F$?
| Assuming the matrix equation
$\begin{bmatrix} C&D\\AC+E&AD+F\\\end{bmatrix} =
\begin{bmatrix} I_p&0\\ 0&I_q\\\end{bmatrix} \tag 1$
is to be interpreted in terms of matrix blocks, we immediately have
$C = I_p, \tag 2$
$D = 0, \tag 3$
$AC + E = 0, \tag 4$
$AD + F = I_q; \tag 5$
then
$A + E = AI_p + E = AC + E = 0, \tag 6$
whence
$E = -A, \tag 7$
and
$F = A(0) + F = AD + F = I_q; \tag 8$
substituting (2), (3), (7), and (8) into
$\begin{bmatrix} C & D \\ E & F \end{bmatrix} \tag 9$
yields
$\begin{bmatrix} C & D \\ E & F \end{bmatrix} = \begin{bmatrix} I_p & 0 \\ -A & I_q \end{bmatrix}. \tag{10}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3407230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Taylor series for $\sqrt{x^0+\sqrt{x^1+\sqrt{x^2+\cdots}}}$ Let
$$f(x) =\sqrt{x^0+\sqrt{x^1+\sqrt{x^2+\cdots}}} = A_0 + A_1(x-1)+A_2(x-1)^2+\cdots$$
I got the following:
*
*$A_0 = (1+\sqrt{5})/2$
*$A_1 = 1/5$
*$A_2=-1/25$
*$A_3 = -1/168$
The values seem correct for $A_0, A_1, A_2$ but only a good approximation for $A_3$. Is it possible to find a simple iterative formula that gives all the coefficient $A_k$'s? What is the exact value for $A_3, A_4$ and $A_5$?
Note
$\lim_{x\rightarrow 0^+} f(x) = \sqrt{2}$. That limit is not equal to $1$, despite what it looks like at first glance. A Taylor approximation around $x=0$ seems much more challenging (if at all possible) than around $x=1$.
| Let $f_n(z) = \sqrt{z^n + \sqrt{z^{n+1} + \sqrt{\ldots}}}$.
Thus $f_n(z)^2 = z^n + f_{n+1}(z)$. Since we're expanding
around $z=1$, it may help to write $z = 1+t$. Letting
$f_n(z) = a_0(n) + a_1(n) t + a_2(n) t^2 + \ldots$ and working formally, we have
$$ (a_0(n) + a_1(n) t + a_2 (n) t^2 + \ldots)^2 = (1+t)^n +
a_0(n+1)+a_1(n+1)t + a_2(n+1) t^2 + \ldots $$
Equating coefficients of each power of $t$:
$$ \eqalign{a_0(n)^2 &= 1 + a_0(n+1)\cr
2 a_0(n) a_1(n) &= n + a_1(n+1)\cr
2 a_0(n) a_2(n) + a_1(n)^2 &= \frac{n^2-n}{2} + a_2(n+1)\cr
2 a_0(n) a_3(n) + 2 a_1(n) a_2(n) &= \frac{n^3}{6} - \frac{n^2}{2}+\frac{n}{3} + a_3(n+1)\cr
etc} $$
The first equation is consistent with $a_0(n)$ all being equal to a root of $z^2 = z+1$, presumably $(1+\sqrt{5})/2$.
Assuming that is the case, the second equation is consistent with
$$a_1(n) = \dfrac{n}{\sqrt{5}} + \dfrac{1}{5}$$
Assuming that is the case, the third equation is consistent with
$$ a_2(n) = \dfrac{3\sqrt{5}}{50} n^2 + \left(\frac{1}{25}-\frac{\sqrt{5}}{10}\right) n - \frac{1}{25} $$
Assuming that is the case, the fourth equation is consistent with
$$ a_3(n) = {\frac {7\,\sqrt {5}{n}^{3}}{750}}+ \left( -{\frac{3}{250}}-{\frac {3
\,\sqrt {5}}{50}} \right) {n}^{2}+ \left( -{\frac{9}{250}}+{\frac {22
\,\sqrt {5}}{375}} \right) n+{\frac{1}{125}}-{\frac {4\,\sqrt {5}}{625
}}
$$
So I think your $A_0$ to $A_2$ are correct, and $A_3$ should be ${\frac{1}{125}}-{\frac {4\,\sqrt {5}}{625
}}
$.
At the next steps, I get $A_4 = {\frac{34}{3125}}+{\frac {7\,\sqrt {5}}{625}}$ and $A_5 = -{\frac{353}{15625}}-{\frac {1138\,\sqrt {5}}{78125}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3407391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
$\lim_{x \to 0} \frac{x^{11}-3x^2+\sin x}{e^x - \cos x}$ Is this solution correct? Just want to double check whether all my operations are legal. The result seems to be correct but I want to make sure I didn't make any mistakes.
$\require{cancel}$
$$\lim_{x \to 0} \frac{x^{11}-3x^2+\sin x}{e^x - \cos x}=$$
$$=\lim_{x \to 0} \frac{x^{11}-3x^2}{e^x - \cos x}+\frac{\sin x}{e^x - \cos x}=$$
$$=\lim_{x \to 0} \frac{x^{11}-3x^2}{e^x - \cos x}+\frac{x}{e^x - \cos x}\cdot\cancelto{1}{\frac{\sin x}{x}}=$$
$$=\lim_{x \to 0} \frac{x^{11}-3x^2+x}{e^x - 1 + 1 - \cos x}=$$
$$=\lim_{x \to 0} (\frac {1-\cos x}{x^{11} - 3x^2 + x} + \frac {e^x-1}{x^{11}-3x^2+x})^{-1}=$$
$$=\lim_{x \to 0} (\frac {1-\cos x}{x^2\cancelto{-\infty}{(x^9-3-\frac {1}{x})}} + \frac {e^x-1}{x\cancelto{1}{(x^{10}-3x+1)}})^{-1}=$$
$$ =\lim_{x \to 0} (\cancelto{\frac{1}{2}}{\frac{1-\cos x}{x^2}}\cdot\cancelto{0}{\frac{1}{-\infty}} + \cancelto{1}{\frac {e^x-1}{x}})^{-1}=$$
$$=\lim_{x \to 0} (\frac {1}{2} \cdot0+ 1)^{-1}=$$
$$=(0+1)^{-1} = 1$$
Is this correct?
| The result is correct but we can simplify some step as follows
$$\frac{x^{11}-3x^2+\sin x}{e^x - \cos x}=\frac{x^{11}-3x^2}{e^x - \cos x}+\frac{\sin x}{e^x - \cos x}$$
and since the first term tends to $0$
$$\frac{x^{11}-3x^2}{e^x - \cos x}=\frac{x^{10}-3x}{\frac{e^x - 1}x+x\frac{1-\cos x}{x^2}}\to 0$$
we obtain
$$\frac{\sin x}{e^x - \cos x}=\frac{\frac{\sin x}x}{\frac{e^x - 1}x+x\frac{1-\cos x}{x^2}}\to 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Recursive function proof using induction We get a $D_0, D_1, D_2, ...$ sequence, for the recursive function $D_n = (n-1)(D_{n-1} + D_{n-2})$ for every n ≥ 2, starting at $D_0 = 1, D_1 = 0$.
Using induction, prove that for every n≥0, $D_n = n! \sum_{k=0}^n = \frac{(-1)^k}{k!}$.
Now, taking for example n=2, the recursive function is $D_2 = (2-1)(D_{1} + D_{0}) = 1\cdot(0+1) = 1$.
Then, $D_2 = 2! \sum_{k=0}^2 = \frac{(-1)^k}{k!}$, and we get $2\cdot(1+(-1)+\frac{1}{2}) = 1$.
So how do we connect these two to prove what is requested?
| We have
$$D_{n+1}=(n+1)!\sum_{k=0}^{n+1}\frac{-1^k}{k!}=(n+1)*n!*(\sum_{k=0}^{n}\frac{-1^k}{k!}+\frac{-1^{n+1}}{(n+1)!})=nD_n+n*n!\frac{-1^{n+1}}{(n+1)!}+n!(\sum_{k=0}^{n-1}\frac{-1^k}{k!}+\frac{-1^n}{n!}+\frac{-1^{n+1}}{(n+1)!})=nD_n+nD_{n-1}+n!(n\frac{-1^{n+1}}{(n+1)!}+\frac{-1^{n+1}}{(n+1)!}+\frac{-1^{n}}{(n)!})=n(D_n+D_{n-1})$$
This, together with $D_0=1=0!*\frac{-1^0}{0!}$ proves the induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals. I need to prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals.
I started by using Bernoulli's inequality:
$(1+\frac{2}{\sqrt{n}}) < (1 + \frac{1}{\sqrt{n}})^2$
I can say that:
$(1+\frac{2}{\sqrt{n}}) = (1+\frac{2\sqrt{n}}{n})$
I can also subtract the one and divide by 2 on the left side without changing the inequality (because it makes it even smaller): $(\frac{\sqrt{n}}{n}) < (1 + \frac{1}{\sqrt{n}})^2$
But now I am stuck...
| It is equivalent to prove
$$ n < \left(1 + \frac{1}{\sqrt{n}}\right)^{2n}.$$
By the binomial identity, the right hand side is at least
\begin{align*}
& 1 + \binom{2n}{1}\frac{1}{\sqrt{n}} + \binom{2n}{2}\frac{1}{n} \\
= & 1 + 2\sqrt{n} + 2n - 1 \\
= & 2n + 2\sqrt{n} \\
> & n.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
} |
$\int \frac{1}{(x^2-4)^2}dx$ Calculate:
$$\int \frac{1}{(x^2-4)^2}dx.$$
I tried Partial Fractions method first I write:
$$\frac{1}{(x^2-4)^2}=\frac{A}{X-2}+\frac{Bx+C}{(x-2)^2}+\frac{D}{x+2}+\frac{Ex+F}{(x+2)^2}.$$
We have:
$$A(x-2)(x+2)^2+(Bx+C)(x+2)^2+D(x+2)(x-2)^2+(Ex+F)(x-2)^2=1.$$
$$(A+B+D+E)x^3+(4A-2A+4B+C-4D+2D-4E+F)x^2+(4A-8+4B+4C+4D-8D+4E-4F)x+(-8A+4C+8D+4F)=1$$
So:
$$A+B+C+D+E=0$$
$$2A+4B+C-2D-4E+F=0$$
$$A+B+C-D+E=2$$
$$-8A+4C+8D+4F=1.$$
But how to find $A$, $B$, $C$, $D$, $E$, $F$?
I also tried substitution $$x=2\sec t$,$
but It caused some difficulty.
| Hint:
Another way
Write numerator as $$\left(\dfrac{x+2-(x-2)}4\right)^2$$
Now $\dfrac1{(x+2)(x-2)}=\dfrac{x+2-(x-2)}{4(x+2)(x-2)}=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Problem in the properties of limit: $\lim\limits_{x \to\frac{\pi}{3}}\frac{\sin\left(x-\frac{\pi}{3}\right)}{1-2\cos\left(x\right)}$
$$\lim\limits_{x \to\frac{\pi}{3}}\frac{\sin\left(x-\frac{\pi}{3}\right)}{1-2\cos\left(x\right)}$$
I used the following property:
if $$\lim\limits_{\large x \to\frac{\pi}{3}}f(x)=L$$
then $$\lim\limits_{x \to\frac{\pi}{3}}\frac{1}{f\left(x\right)}=\frac{1}{L}$$
where $L$ is a real number and nonzero,hence we have:
$$\lim\limits_{\large x \to\frac{\pi}{3}}\frac{1-2\cos\left(x\right)}{\sin\left(x-\frac{\pi}{3}\right)}$$
substititute $x-\frac{\pi}{3}=u$:
$$\lim\limits_{\large u \to 0}\frac{1-2\cos\left(u+\frac{\pi}{3}\right)}{\sin\left(u\right)}$$$$=\lim\limits_{\large u \to 0}\frac{1-\cos\left(u\right)+\sqrt{2}\sin\left(u\right)}{\sin\left(u\right)}=\lim\limits_{\large u \to 0}\frac{1-\cos\left(u\right)}{\sin\left(u\right)}+\sqrt{2}$$$$=\lim\limits_{\large u \to 0}\frac{\sin\left(u\right)}{1+\cos\left(u\right)}+\sqrt{2}=\sqrt{2}$$
hence the main limit should be $\frac{1}{\sqrt{2}}$which is wrong, but I don't know why, also is there any way to solve the problem without using Taylor series or L'hopital's rule?
| Your derivation is absolutely fine and right, but we have that
$$1-2\cos\left(u+\frac{\pi}{3}\right)=1-2\frac12\cos u+2\frac {\sqrt 3} 2\sin u=1-\cos u+\color{red}{\sqrt 3}\sin u$$
therefore
$$\lim\limits_{x \to\frac{\pi}{3}}\frac{\sin\left(x-\frac{\pi}{3}\right)}{1-2\cos\left(x\right)}=\frac1{\sqrt 3}$$
Note also that we don't need to invert the expression, indeed in the same way we have
$$\lim\limits_{\large u \to 0}\frac{\sin\left(u\right)}{1-2\cos\left(u+\frac{\pi}{3}\right)}=\lim\limits_{\large u \to 0}\frac{\sin\left(u\right)}{1-\cos u+\sqrt 3\sin u}=\lim\limits_{\large u \to 0}\frac{1}{\frac{1-\cos u}{\sin u}+\sqrt 3}=\frac1{\sqrt 3}$$
since
$$\frac{1-\cos u}{\sin u}=u\frac{1-\cos u}{u^2}\frac{u}{\sin u}\to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3410852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Square-root equation
Solve square-root equation: $\left (\sqrt{5+2\sqrt{6}} \right )^x+\left (\sqrt{5-2\sqrt{6}} \right )^x=10$
$\left (\sqrt{5+2\sqrt{6}} \right )^x+\left (\sqrt{5-2\sqrt{6}} \right )^x=10\\
\left (\sqrt{\left (\sqrt{3}+\sqrt{2} \right )^2} \right )^x+\left (\sqrt{\left (\sqrt{3}-\sqrt{2} \right )^2} \right )^x=10\\
\left (\sqrt{3}+\sqrt{2} \right )^x+\left (\sqrt{3}-\sqrt{2} \right )^x=10$
at the moment I don't know what to do
| Expanding on @DanielFischer's comment, let $y=(\sqrt{3}+\sqrt{2})^x$ so $y+1/y=10$ and $y=5\pm2\sqrt{6}$, so $x=\pm2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Last step in evaluating: $\int {dx\over a\cos x + b\sin x}$ Seems like I've made a mistake somewhere in my calculation and am now struggling to find it without any success. Here is the problem statement:
Evaluate:
$$
\int {dx\over a\cos x + b\sin x}
$$
I've decided to use a $t = \tan{x\over 2}$ substitution:
$$
dx = {2\,dt\over 1+t^2}\\
\sin x = {2t\over 1+t^2}\\
\cos x = {1-t^2\over 1+t^2}\\
$$
That substitution yields:
$$
\begin{align}
I &= \int {2\,dt \over (1+t^2)\left({a(1-t^2)\over 1+t^2} + {2bt\over 1 + t^2}\right)} \\
&=\int {2\,dt \over a(1-t^2) + 2bt} \\
&=-2 \int {dt \over at^2 - 2bt - a}
\end{align}
$$
Factor the denominator:
$$
D = 4b^2 + 4a^2\\
R_{1,2} = {b \pm \sqrt{a^2 + b^2}\over a}\\
{1 \over at^2 - 2bt - a} = {1\over a(t-R_1)(t-R_2)}
$$
By that the integral becomes:
$$
I = -{2\over a}\int {dt\over (t-R_1)(t-R_2)}
$$
Using partial fraction decomposition:
$$
\begin{align}
I &= -{2\over a}\int \left({1\over (R_1 - R_2)(t-R_1)} - {1\over (R_1 - R_2)(t-R_2) }\right)\,dt \\
&= -{2\over a(R_1 - R_2)}\left(\ln\left|t -R_1\right|- \ln\left|t -R_2\right|\right) \\
&= - {2\over a(R_1 - R_2)}\ln\left|{t -R_1 \over t -R_2}\right| + C\\
&= \boxed{ - {2\over a(R_1 - R_2)}\ln\left|{\tan{x\over 2} -R_1 \over \tan{x\over 2} -R_2}\right| + C }
\end{align}
$$
My answer seems legit, here is a simulation in Desmos. The problem that the answer section suggests that:
$$
I = {1\over \sqrt{a^2 + b^2}}\ln\left|\tan \left({x\over 2} + {\phi \over 2}\right)\right| + C
$$
Where $\phi$ satisfies the following conditions:
$$
\sin \phi = {a\over \sqrt{a^2 + b^2}}\\
\cos \phi = {b\over \sqrt{a^2 + b^2}}
$$
I've tried manipulating my answer in different ways, but I still couldn't express it in the way it is expressed in the keys section. I would appreciate it if someone could explain to me the way to do it. Thank you!
| $\tan(\frac x2+\frac\phi 2) = \dfrac{\tan\frac x2+\tan\frac\phi 2}{1 - \tan\frac x2\tan\frac\phi 2}$
but
$$\tan\frac\phi 2 = \dfrac{\sin\phi}{1+\cos\phi} = \dfrac{a}{b+\sqrt{a^2+b^2}} = \dfrac{\sqrt{a^2+b^2} - b}{a} = -R_2.$$
Thus,
$$\ln\tan\left(\frac x2+\frac\phi 2\right) = \ln\dfrac{\tan\frac x2 - R_2}{1+R_2\tan\frac x2} = -\ln\dfrac{1 - \frac {1}{R_1}\tan\frac x2}{\tan\frac x2 - R_2}.$$
Can you see it now? The only nuisance was to use $\ln a = -\ln\frac 1a$, on top of $R_1R_2 = -1$ and when you multiply the last numerator by $R_1$ because it can be absorbed to the constant $C.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3421982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Determine the value of $k$ such that the line determined by the points $(3,2)$ and $(1,-2)$ is tangent to the graph of $y=\frac{k}{x+1}$. I have done the following work but I am stuck on solving for k. Am I doing this right? If so what do I do next?
| We have
$1.$ $k/(x+1)=2x-4$
$2.$ $-k/(x+1)^2=2$
First equation
Multiplying it by $(x+1)$
$k=(x+1)\cdot (2x-4)=2x^2-4x+2x-4=2x^2-2x-4 \quad (1)$
Second equation
Multiply it by $(x+1)^2$
$-k=2(x+1)^2=2x^2+4x+2 \Rightarrow k=-2x^2-4x-2 \quad (2)$
Set $(1)$ equal to $(2)$
$2x^2-2x-4=-2x^2-4x-2$
$4x^2+2x-2=0$
$2x^2+1x-1=0$
Solve for $x$. This gives $x_1=-1$ and $x_2=0.5$
$x_1$ is not in the domain of $y=k/(x+1)$.
Now insert $x=0.5$ in $k/(x+1)=2x-4$ to obtain the value of $k$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3423556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find all $n≥1$ natural numbers such that : $n^{2}=1+(n-1)!$ Problem :
Find all $n≥1$ natural numbers such that : $n^{2}=1+(n-1)!$
My try :
$n=1$ we find : $1=1+1$ $×$
$n=2$ we find : $4=1+1$ $×$
$n=3$ we find : $9=1+2$ $×$
$n=4$ we find : $16=1+6$ $×$
$n=5$ we find : $25=1+24$ $√$
Now how I prove $n=5$ only the solution ?
| Hint If $n \geq 6$ then
$$1+(n-1)! \geq 1+2 \cdot 3 \cdot (n-2) \cdot (n-1)$$
Show that
$$2(n-1) \geq n \\
3(n-2) \geq n$$
for $n \geq 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3427795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Find the units digit of $572^{42}$ The idea of this exercise is that you use the modulus to get the right answer.
What I did was:
$$572\equiv 2\pmod {10} \\
572^2 \equiv 2^2 \equiv 4\pmod{10} \\
572^3 \equiv 2^3 \equiv 8\pmod{10} \\
572^4 \equiv 2^4 \equiv 6\pmod{10} \\
572^5 \equiv 2^5 \equiv 2\pmod{10} \\
572^6 \equiv 2^6 \equiv 4\pmod{10} \\
(...)$$
I can see that this goes 2,4,8,6 and then repeats. I remember that the gist of the exercise is to find the remainder based on this repetition. How do I do that? I know that $572^{42} \equiv 2^{42}\equiv ? \pmod {10}$. How do I simplify that 42 and answer this using that repetition?
| Another approach is:
The reminder is repeating between:
$$R=\{2,4,8,6\}$$
Solve the following, since $|R|=4$
$$42\equiv x\pmod{4}$$
Have $x=2$, which means the reminder is the second term in $R$ that is $4$
Therefore $$572^{42}\equiv4\pmod{10}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3428727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all triples of non-negative real numbers $(a,b,c)$ Find all triples of non-negative real numbers $(a,b,c)$ such that:
$a^2+ab=c$
$b^2+bc=a$
$c^2+ca=b$.
This question was problem number of 3 in the RMO(India) Olympiad in 2019 held on $10^{th}$ November. link .
My attempt-
Assume $a \geq b\geq c$,
$\therefore a^2 \geq b^2$ and $ab\geq bc$.
Adding these two, $a^2+ab\geq b^2+bc$
implies $c \geq a$.
Which can be possible only if $a=c$, which also implies $a=b=c$.
Substituting in the base equations,
$a^2+a^2=a$
$\therefore (a,b,c)=(0,0,0)$ or $(0.5,0.5,0.5)$
I want to know if my method is correct because it seems a lot different than the one provided in the solutions.
| Eliminating $b,c$ from our system we get for $a$ the following equation:
$$2 a^8+a^7+a^6+3 a^5+6 a^4+2 a^3-a^2-a=0$$
and this is $$a (2 a-1) \left(a^6+a^5+a^4+2 a^3+4 a^2+3 a+1\right)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3431214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Simplifying the result of integration of $\int\frac{e^x + e^{3x}}{1-e^{2x}+e^{4x}}\mathop{dx}$
Evaluate:
$$
\int\frac{e^x + e^{3x}}{1-e^{2x}+e^{4x}}\mathop{dx}
$$
I'm trying to simplify my answer so that it matches the keys section, no success so far. The integral itself is pretty simple. Factor $e^x$ in the denominator and then make an obvious substitution:
$$
I = \int \frac{e^x(1 + e^{2x})}{1-e^{2x}+e^{4x}}\mathop{dx}\\
t = e^x\, , dt = e^x\mathop{dx}\\
$$
Thus:
$$
\begin{align}
I&=\int \frac{1+t^2}{1-t^2 + t^4}\mathop{dt} \\
&={1\over 2}\int\left(\frac{1}{t^2 + \sqrt3t + 1} + \frac{1}{t^2-\sqrt3t+1}\right)\mathop{dt} \\
&={1\over 2}\int\left(\frac{1}{\left(t+{\sqrt3\over 2}\right)^2+{1\over 4}} + \frac{1}{\left(t-{\sqrt3\over 2}\right)^2+{1\over 4}}\right)\mathop {dt}
\end{align}
$$
Which after some further substitutions yields:
$$
\boxed{I = \arctan(2e^x+\sqrt3) + \arctan(2e^x-\sqrt3)}\tag1
$$
However, the answer section suggests that:
$$
I = \arctan(2\sinh x)\tag2
$$
Which matches my answer up to a constant , $-{\pi \over 2}$ in this case.
Even though the answer is correct, I would still like to see how I could arrive from $(1)$ to $(2)$, I've given it several tries without any luck. I would appreciate it if someone could show me why:
$$
\arctan(2\sinh x) = \arctan(2e^x+\sqrt3) + \arctan(2e^x-\sqrt3) - {\pi\over 2}
$$
Thank you!
As pointed out in the comments by @mickep, there is a way to directly arrive at the desired result. I would like to elaborate on it here. Instead of factoring $e^x$ one could factor $e^{2x}$ which would give:
$$
\begin{align}
I &= \int \frac{e^{2x}(e^{-x}+e^{x})}{e^{2x}(e^{-2x} - 1 + e^{2x})}\mathop{dx}\\
&= \int \frac{e^{-x}+e^{x}}{e^{-2x} - 1 + e^{2x}}\mathop{dx} \\
&= \int \frac{2(e^{-x}+e^{x})}{2(e^{-2x} - 1 + e^{2x})}\mathop{dx}\\
&= \int \frac{2\cosh x}{e^{-2x} - 1 + e^{2x}}\mathop{dx} \\
&= \int \frac{2\cosh x}{4{e^{-2x} - 2e^{x}e^{-x} + e^{2x}\over 4} + 1}\mathop{dx}\\
&= \int \frac{2\cosh x}{(2\sinh x)^2 + 1}\mathop{dx}
\end{align}
$$
Now using a substitution $t = 2\sinh x$, one may obtain:
$$
I = \int \frac{\mathop{dt}}{t^2 + 1} = \arctan(t) = \arctan(2\sinh x) + C
$$
By this approach, we have arrived at the desired result.
| Let's start off from
$$\sinh(x) = \frac{e^x-e^{-x}}{2} = -\frac{e^{-x}-e^x}{2} = -\frac{\frac{1}{e^x}-e^x}{2} = -\frac{1-e^{2x}}{2e^x} = -\frac{4-4e^{2x}}{8e^{2x}} = \\ = -\frac{1-(4e^{2x}-3)}{8e^x} = -\frac{1-(2e^x+\sqrt{3})(2e^x-\sqrt{3})}{8e^x}.$$
For conveniency, I'll rewrite this as:
$$2\sinh(x)=-\frac{1-(2e^x+\sqrt{3})(2e^x-\sqrt{3})}{4e^x}.\tag{1}$$
Now since
$$\cot(\arctan(x)) = \frac{1}{\tan(\arctan(x))} = \frac{1}{x}$$
We can just take the $\cot(\arctan())$ of the reciprocal of the right side of $(1)$:
$$2\sinh(x) = -\cot\left(\arctan\left(\frac{4e^x}{1-(2e^x+\sqrt{3})(2e^x-\sqrt{3})}\right)\right) = \\ = -\cot\left(\arctan\left(\frac{2e^x+\sqrt{3}+2e^x-\sqrt{3}}{1-(2e^x+\sqrt{3})(2e^x-\sqrt{3})}\right)\right)\tag{2}$$
Now we'll show a formula for $\arctan(x) + \arctan(y)$:
$$\tan(\alpha+\beta) = \frac{\tan(\alpha)+\tan(\beta)}{1+\tan(\alpha)\tan(\beta)}$$
$$\alpha+\beta = \arctan\left(\frac{\tan(\alpha)+\tan(\beta)}{1+\tan(\alpha)\tan(\beta)}\right)$$
Now with the substitution $\alpha = \arctan(x)$ and $\beta = \arctan(y) \leftrightarrow x = \tan(\alpha), y = \tan(\beta)$:
$$\arctan(x) + \arctan(y) = \arctan\left(\frac{x+y}{1-xy}\right) \tag{3}$$
Now we'll use $(3)$ with $x=2e^x+\sqrt{3}, y=2e^x-\sqrt{3}$ on $(2)$:
$$2\sinh(x) = -\cot(\arctan(2e^x+\sqrt{3})+\arctan(2e^x-\sqrt{3}))\tag{4}$$
Now finally, we'll use the equality that
$$\tan\left(x-\frac{\pi}{2}\right) = -\cot(x)\tag{5}$$
Putting $(5)$ into $(4)$ we get:
$$2\sinh(x) = \tan\left(\arctan(2e^x+\sqrt{3})+\arctan(2e^x-\sqrt{3})-\frac{\pi}{2}\right)$$
Or:
$$\arctan(2\sinh(x)) = \arctan(2e^x+\sqrt{3})+\arctan(2e^x-\sqrt{3})-\frac{\pi}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3431568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Proof $a_{n}$ by induction This topic is a continuation of Prove that $a_{n} = a_{n-1} + a_{n-2}$.
I need to prove that $a_{n} = (\frac{5+3\sqrt{5}}{10})(\frac{1+\sqrt{5}}{2})^{n} + (\frac{5-3\sqrt{5}}{10})(\frac{1-\sqrt{5}}{2})^{n}$ for all $n \geq 1$.
This is how I tried to prove this.
Basic step: $a_{1} = (\frac{5+3\sqrt{5}}{10})(\frac{1+\sqrt{5}}{2})^{1} + (\frac{5-3\sqrt{5}}{10})(\frac{1-\sqrt{5}}{2})^{1}$ which is equal to $2$ and is correct.
Inductive step: Let $k \geq 1$ and let the statement apply to each $k$ where $1 \geq j \geq k$. From the previous topic we know that $a_{k} = a_{k-1} + a_{k-2}$ and so that $a_{k+1} = a_{k} + a_{k-1}$. Because of the hypothesis we know that the statement applies for $a_{k}$ and $a_{k-1}$. As $a_{k+1}$ exists of the sum of $a_{k}$ and $a_{k-1}$, we can conclude that the statement also applies for $a_{k+1}$.
Is this proof by induction valid?
| Rather than solve the problem through the standard methods for more general linear recurrences, I'll address your questions about induction:
*
*Your choice of induction hypothesis is right. We already know that $a_{n+2}=a_{n+1}+a_n$ for all $n\geq 1$ from your previous question. Here you want to assume that $a_{n+1}$ and $a_n$ can be written explicitly as
$$a_{n+1}=\left(\frac{5+3\sqrt{5}}{10}\right)\left(\frac{1+\sqrt{5}}{2}\right)^{n+1} + \left(\frac{5-3\sqrt{5}}{10}\right)\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}$$
and
$$a_n=\left(\frac{5+3\sqrt{5}}{10}\right)\left(\frac{1+\sqrt{5}}{2}\right)^{n} + \left(\frac{5-3\sqrt{5}}{10}\right)\left(\frac{1-\sqrt{5}}{2}\right)^{n}$$
respectively. Using that and the fact that $a_{n+2}=a_{n+1}+a_n$ you want to show that $a_{n+2}$ can be written in the same way, only with $n+2$ in the corresponding exponents. This is a matter of simple algebraic manipulation: write out the sum, factor out $\lambda_\pm=\frac{1\pm\sqrt 5}{2}$ and note that these numbers are roots of $x^2=x+1$ as @lhf pointed out.
*That brings us to the following problem: when $n=1$, we have $a_3=a_2+a_1$. This means that $a_1$ is not the only base case: you need to prove $a_2$ as a base case as well. More generally: if your inductive step $P(n)$ depends on $P(n-1),P(n-2),...,P(n-k)$, then you need to prove $k$ base cases. See the Wikipedia page on strong induction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find $I=\int_0^1\frac{\arctan^2x}{1+x}\left(\frac{\ln x}{1-x}+\ln(1+x)\right)dx$ $$I=\int_0^1\frac{\arctan^2x}{1+x}\left(\frac{\ln x}{1-x}+\ln(1+x)\right)dx=-\frac{\pi^4}{512}+\frac{3\pi^2}{128}\ln^22+\frac{\pi}{8}G\ln2-\frac{21}{64}\zeta(3)\ln2$$
This integral was proposed to me by a friend, but without the solution.
I tried the integration by parts, but too complex.
I tried to find a closed form but without result. $$I=\int_0^1\frac{\arctan^2x}{1±x}\ln{x}dx$$
| Continuing on @AliShather's argument, let's focus on the integral
\begin{align*}
I_1&=\int_0^1\frac{\arctan^2 x\ln x}{1-x^2}dx\\
&\overset{\rm IBP}=-\int_0^1\frac{\arctan^2 x\mathop{\rm artanh} x}{x}+2\underbrace{\frac{\arctan x\mathop{\rm artanh} x \ln x}{1+x^2}}_{x\mapsto \frac{1-x}{1+x}}dx\\
&=-\int_0^1\frac{\arctan^2 x\mathop{\rm artanh} x}{x}-2{\frac{\left(\arctan x-\frac{\pi}{4}\right)\mathop{\rm artanh} x \ln x}{1+x^2}}dx\\
&=-\int_0^1\frac{\arctan^2 x\mathop{\rm artanh} x}{x}dx-\frac{\pi}{4}\int_0^1\frac{\mathop{\rm artanh} x\ln x}{1+x^2}dx\\
&=-A-\frac{\pi}{4}B
\end{align*}
where $\mathop{\rm artanh}x=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$.
By converting into multiple integrals then integrating back,
\begin{align*}
A&=\int_{[0,1]^4}\frac{x^2}{\left(1-s^2x^2\right)\left(1+t^2x^2\right)\left(1+u^2x^2\right)}dV\\[3px]
&=\mbox{*Partial fractions*}\\[3px]
&=-\int_0^1\frac{\mathop{\rm artanh}x}{x}\left(2\arctan x-\arctan\frac{1}{x}\right)\arctan\frac{1}{x}\,dx\\
&=-\frac{\pi^4}{64}+\pi\underbrace{\int_0^1\frac{\arctan x \mathop{\rm artanh}x}{x}dx}_{J}
\end{align*}
where we used the identity $\arctan x+\arctan\frac{1}{x}=\frac{\pi}{2}$.
By $x\mapsto\frac{1-x}{1+x}$,
$$J=\int_0^1\frac{\left(\arctan x-\frac\pi 4\right)\ln x}{1-x^2}dx=\frac{\pi^3}{32}+\int_0^1\frac{\arctan x\ln x}{1-x^2}dx$$
Then by using IBP directly to J,
\begin{align*}
J&=-\int_0^1\frac{\arctan x\ln x}{1-x^2}+\frac{\mathop{\rm artanh}x \ln x}{1+x^2}dx\\&=\frac{\pi^3}{64}-\frac{1}{2}\underbrace{\int_0^1\frac{\mathop{\rm artanh}x \ln x}{1+x^2}dx}_B
\end{align*}
By here,
$$I_1=\frac\pi 4 B=\frac{3\pi^4}{256}-\frac{G\pi\ln 2}{4}-\frac \pi 2\mathfrak{I}\mathop{\rm Li_3}\frac{1+i}{2}+\frac{\pi^2\ln^2 2}{64}$$
Finally using the closed form of $I_2$ by @AliShather and $I=I_1+I_2$ we should get the desired identity, where of course the closed form of B was not necessary and can be replaced by the partial steps used in the evaluation of B.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving the inequality $\sum_{\text{cyc}} \frac{a}{a^2+b^3+c^3}\le\frac1{5abc}$ Let $a,b,c> 0$ be three real numbers such that $a+b+c=1$. I want to prove that
$$\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+a^3+c^3}+\frac{c}{c^2+a^3+b^3}\le\frac{1}{5abc}.$$
My attempt: Using AM-GM on each denominator gives (here, LHS denotes the left-hand side)
$$LHS\le\frac{1}{3a^{\frac12}bc}+\frac{1}{3b^{\frac12}ac}+\frac{1}{3c^{\frac12}ab}.$$
However, I think that my attempt doesn't work because the last expression can get larger than $\frac{1}{5abc}$. In fact, if we multiply with $3abc$ then the original inequality is:
$$\sqrt a+\sqrt b+\sqrt c\le\frac35$$
However, as $a,b,c<1$, $$\sqrt a +\sqrt b + \sqrt c> a+b+c=1>\frac35.$$ So my upper bound is always larger than the given one.
| Also, by AM-GM and C-S we obtain:
$$\sum_{cyc}\frac{a}{a^2+b^3+c^3}=\sum_{cyc}\frac{a}{a^3+b^3+c^3+a^2b+a^2c}\leq\sum_{cyc}\frac{a}{3abc+a^2b+a^2c}=$$
$$=\sum_{cyc}\frac{1}{3bc+ab+ac}\leq\frac{1}{(3+1+1)^2}\sum_{cyc}\left(\frac{3^2}{3bc}+\frac{1^2}{ab}+\frac{1^2}{ac}\right)=\frac{1}{5}\sum_{cyc}\frac{1}{ab}=\frac{1}{5abc}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3442048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $3^{2n-1} + 2^{n+1}$ is divisible by $7$ for all values of $n$ I have tried to prove this through mathematical induction but I can't seem to prove that the proposition works for $k+1$.
| $3^{2n-1} \begin{matrix} n& 1 & 2 & 3 & 4 & 5 & \cdots \\ mod 7 &3 & 6 & 5 & 3 & 6 & \cdots \end{matrix}$
$2^{n+1} \begin{matrix} n& 1 & 2 & 3 & 4 & 5 & \cdots \\ mod 7 &4 & 1 & 2 & 4 & 1 & \cdots \end{matrix}$
Since $3+4 = 7$, $6+1=7$, $5+2 = 7$ we can see that $3^{2n-1}+2^{n+1} \equiv 0 \pmod{7}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3443306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability of taking out coins by order
The question goes as following:
There is a jar with 65 coins. 30 are green, 15 are yellow and 20 are red.
When taking out 20 coins from the jar, one by one, what is the probability that the first three coins and last three coins selected will be green?
My logic was that if I only look at the first 3 and last 3 selected, there are $30 \cdot 29\cdot 28\cdot 27\cdot 26\cdot 25$ options for them to be green, and a total of $65\cdot 64\cdot 63\cdot 62\cdot 61\cdot 60$ options to choose the first and last 3.
So the probability should be $ (30 \cdot 29\cdot 28\cdot 27\cdot 26\cdot 25)\div (65\cdot 64\cdot 63\cdot 62\cdot 61\cdot 60) $
Would appreciate any comments regarding if this answer makes sense or not.
| Let's see for $7$ balls:
$$P(\color{green}{G_1G_2G_3}(\color{green}G \text{ or } G')_4\color{green}{G_5G_6G_7})=\\
P(\color{green}{G_1G_2G_3G_4G_5G_6G_7})+P(\color{green}{G_1G_2G_3}G'_4\color{green}{G_5G_6G_7})=\\
\color{green}{\frac{30}{65}\cdot \frac{29}{64}\cdot \frac{28}{63}}\cdot \color{green}{\frac{27}{62}\cdot \frac{26}{61}\cdot \frac{25}{60}\cdot \frac{24}{59}}+
\color{green}{\frac{30}{65}\cdot \frac{29}{64}\cdot \frac{28}{63}}\cdot \frac{35}{62}\cdot \color{green}{\frac{27}{61}\cdot \frac{26}{60}\cdot \frac{25}{59}}=\\
\color{green}{\frac{30}{65}\cdot \frac{29}{64}\cdot \frac{28}{63}}\cdot \color{green}{\frac{27}{62}\cdot \frac{26}{61}\cdot \frac{25}{60}}\cdot \left(\frac{\color{green}{24}+35}{59}\right)=\\
\color{green}{\frac{30}{65}\cdot \frac{29}{64}\cdot \frac{28}{63}}\cdot \color{green}{\frac{27}{62}\cdot \frac{26}{61}\cdot \frac{25}{60}}$$
Similarly:
$$P(\color{green}{G_1G_2G_3}(\color{green}G \text{ or } G')_4...(\color{green}G \text{ or }G')_{17}\color{green}{G_{18}G_{19}G_{20}})=\\
\color{green}{\frac{30}{65}\cdot \frac{29}{64}\cdot \frac{28}{63}}\cdot \color{green}{\frac{27}{62}\cdot \frac{26}{61}\cdot \frac{25}{60}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3447903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Obtain the maximum and minimium of the function $f$ I have to calculate the max and min of the function
\begin{equation}
f(x_1,x_2 \ldots ,x_n)=x_1^3 + x_2^3+\cdots + x_n^3
\end{equation}
subject to
\begin{equation}
x_1 + x_2+\cdots + x_n =0
\end{equation}
\begin{equation}
x_1^2 + x_2^2+\cdots + x_n^2 =1
\end{equation}
I have applied the Lagrange theorem but i don't conclude anything. Could somebody helps me?
| The Lagrange multipliers tell us that there exist $\lambda$ and $\mu$ such that for all $i=1,\ldots,n$
\begin{equation}
3 x_i^2 = 2 \lambda x_i + \mu
\end{equation}
by summing these equalities one gets $3 = n\mu$. We still don't know $\lambda$ but as all the $x_i$ satisfy the same 2nd degree equation, it means that the $x_i$ can take only two values.
Multiplying by $x_i$ and summing, we get $3 f = 2 \lambda$ at the given point, so the equation can be rewritten
\begin{equation}
x_i^2 = f x_i + \frac{1}{n}
\end{equation}
Suppose that $q$ of the $x_i$ have value $\frac{f}{2} + \sqrt{\frac{f^2}{4} + \frac{1}{n}}$ and that $p=n-q$ of them have the value $\frac{f}{2} - \sqrt{\frac{f^2}{4} + \frac{1}{n}}$. The sum is
\begin{equation}
0 = \sum_i x_i = n \frac{f}{2} + (q-p) \sqrt{\frac{f^2}{4} + \frac{1}{n}}
\end{equation}
It follows that $q-p \neq 0$ otherwise $f=0$ and it is obviously not the maximum (if $n>2$). Furthermore, $\sqrt{\frac{f^2}{4} + \frac{1}{n}} = \frac{q+p}{p-q} \frac{f}{2}$ which implies that $q$ of the $x_i$ have the value $f \frac{p}{p-q}$ and $p$ of the $x_i$ have the value $-f\frac{q}{p-q}$. It also implies that $f$ and $p-q$ have the same sign. Summing the $x_i^2$ yields
\begin{equation}
1 = \sum x_i^2 = f^2 \left(\frac{q p^2}{(p-q)^2} + \frac{p q^2}{(p-q)^2}\right) = f^2\frac{npq}{(p-q)^2}
\end{equation}
Hence one has $f = \frac{p-q}{\sqrt{n p q}}$ and $q$ of the $x_i$ have the value $\frac{p }{\sqrt{npq}}$
and $p$ of them have the value $\frac{-q}{\sqrt{n p q}}$. Such a point satisfies all the constraints and the equation above with the Lagrange multipliers.
It remains to select, among all the integer choices of $p+q=n$ the value that maximizes (or minimizes, which is the opposite), the quantity
\begin{equation}
f = \frac{p-q}{\sqrt{npq}}
\end{equation}
The maximum is reached for $p=n-1$ and $q=1$ because the function $\frac{1-2x}{\sqrt{x(1-x)}}$ decreases in $[0,1]$. This gives a maximum of
\begin{equation}
f = \frac{n-2}{\sqrt{n(n-1)}}
\end{equation}
| {
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"timestamp": "2023-03-29T00:00:00",
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Help evaluating a multivariable limit I tried approaching from y=0, from x=0, from x=y, I also tried using polar coordinates, I always get 0/0. What do I need to do to?
$$\lim_{(x,y)\to(1,1)} \dfrac{(x-y)(x+y-2)}{\sqrt{(x-1)^2+(y-1)^2}}$$
Mulivariable limit problem
| Set $x=1+r\cos\theta$, $y=1+\sin\theta$. You obtain
\begin{align}
\frac{(x-y)(x+y-2)}{\sqrt{(x-1)^2+(y-1)^2}}&=\frac{(r\cos\theta-r\sin\theta)(r\cos\theta+r\sin\theta)}{\sqrt{r^2}}\\&=r(\cos^2\theta-\sin^2\theta)=r\cos2\theta,
\end{align}
which tends to $0$ as $r$ tends to $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Computation of a 2D limit I came across this limit and I used Mathematica to solve it, (it shows that it is equal to $0$).
Any thoughts on how to prove this ? The limit is:
$$\lim \limits_{(x,y)\to (4,0)} \frac{y^2(x^2+y^2)}{(x-4)^2+y^2}$$
Thanks.
| Let $u=x-4\to 0$ then
$$\lim \limits_{(x,y)\to (4,0)} \frac{y^2(x^2+y^2)}{(x-4)^2+y^2}=\lim \limits_{(u,y)\to (0,0)} \frac{y^2((u+4)^2+y^2)}{u^2+y^2}$$
and for
*
*$y=0$
$$\frac{y^2((u+4)^2+y^2)}{u^2+y^2}=0$$
but for
*
*$u=y$
$$\frac{u^2((u+4)^2+u^2)}{2u^2}=\frac{2u^2+8u+16}{2}=u^2+4u+8\to 8$$
therefore the given limit doesn't exist.
| {
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Prove that $\sum_\text{cyc} \frac{ab}{ab+b^2+ca}\le 1$ Hello ladies and gentlemen, here I have another inequality that I am struggling with:
Let $a,b,c>0$ Then
$$\sum_\text{cyc} \frac{ab}{ab+b^2+ca}\le 1.$$
I try to show $$\frac{ab}{ab+b^2+ca}\le\frac{a}{a+b+c}$$ but this is doesn't work because $$\frac{ab}{ab+b^2+ca}=\frac{a}{a+b+\frac{ca}b}$$ which is in general not less than $\frac{a}{a+b+c}$.
So what can I do? Do I have to use some Hölder?
| Multiply both sides with $\prod_{\text{cyc}} ab+b^2+ac$, subtract $ab\cdot(bc+c^2+ab)\cdot(ac+a^2+bc)$ from both sides and you get by expanding that the original inequality is equivalent to:
\begin{equation}\tag 1\label 1\sum_{\text{cyc}} a^4bc+\sum_{\text{cyc}} a^3b^3\geq 2\sum_{\text{cyc}} a^3b^2c.\end{equation}
By AM-GM, we have \begin{align}2a^3b^3 + a^3c^3&\geq 3a^3b^2c, \\ 2a^3c^3 + b^3c^3&\geq 3a^2bc^3, \\ 2b^3c^3 + a^3b^3&\geq 3ab^3c^2.\end{align}
By summing these inequalities we get $$\sum_{\text{cyc}} a^3b^3\geq \sum_{\text{cyc}} a^3b^2c.$$ Similarly, we have $$2a^4bc+ab^4c\geq3a^3b^2c,\\\vdots$$
which gives $$\sum_{\text{cyc}} a^4bc\geq \sum_{\text{cyc}} a^3b^2c.$$ Hence, \eqref{1} is proven and hence also the original inequality.
| {
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} |
Logical statements: $A+B+C=A+B+C+AB+BC+AC$
Let $A,B, C$ be logical statements.
Then:
$A+B+C=A+B+C+AB+BC+AC$
Prove it without a table.
My attempt:
$$A=A\cdot(1+B+C)$$
$$B=B\cdot(1+A+C)$$
$$C=C\cdot(1+A+B)$$
$$\implies A+B+C= A+AB+AC+B+AB+BC+C+BC+AC$$
$$\implies A+B+C=A+B+C+AB+AB+BC+BC+AC+AC$$
$$\implies A+B+C=A+B+C+AB+BC+AC$$
Is this correct?
| Yes. it is completely correct.
You could have shortened it a bit by writing
$$A+B+C+AB+BC+AC = A(1+B+C) + B(1+C) + C $$ $$
\stackrel{1+x=1,1x=x}{=}A+B+C$$
| {
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Convergence of $\int_0^{\infty}\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $
Find out if the following integral diverges or converges:
$$
\int_0^\infty \frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx
$$
First I split the integral as $\displaystyle \int_0^1 \frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx + \int_1^\infty \frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $.
*
*For $\int_0^{1}\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $ , I prove $\ln(1+x^2)< x^2$, using that I can prove $\int_0^1\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $ converges, but it take too long so is there shorter way to do this problem?
*For $\int_1^{\infty}\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $ , I have no idea how to do this problem.
| Just for the fun of it.
The convergence issues having been clearly explained and one answer for the value showing that a CAS is able to find a (nasty) antiderivative, the result is in fact
$$\int_0^{\infty}\frac{\log (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx=\frac{1}{3} \log \left(\frac{1}{4} \left(-1+i+\sqrt{-1-2 i}\right)^{-\sqrt{4-2 i}}\,\,
\left(-1-i+\sqrt{-1+2 i}\right)^{-\sqrt{4+2 i}}\right)$$
| {
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Prove that the minimum values of $x^2+y^2+z^2$ is $27$ with given condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$. Question: Prove that the minimum values of $x^2+y^2+z^2$ is $27$, where $x,y,z$ are positive real variables satisfying the condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$.
From AM$\ge$ GM, we have $\left( \dfrac{x^2+y^2+z^2}{3}\right)^3\ge (xyz)^2=(xy+yz+zx)^2$. Is it possible to show thae result from this relation?
| You can still finish it off if you notice that $( xy+yz+zx )^2 = (xyz)^2 = xy\cdot yz \cdot zx \le \dfrac{(xy+yz+zx)^3}{27} \implies xy+yz+zx \ge 27 \implies \dfrac{x^2+y^2+z^2}{3} \ge \sqrt[3]{27^2} = 9 \implies x^2+y^2+z^2 \ge 9\cdot 3 = 27 $.
| {
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Show that following determinant is divisible by $\lambda^2$ and find the other factor. Show that $\begin{vmatrix}
a^2+\lambda &ab &ac \\
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}=0$ is divisible by $\lambda^2$ and find the other factor.
My attempt is as follows:-
$$R_1\rightarrow R_1+R_2+R_3$$
$$\begin{vmatrix}
a(a+b+c)+\lambda &b(a+b+c)+\lambda &c(a+b+c)+\lambda \\
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}=0$$
$$C_1\rightarrow C_1-\dfrac{a}{b}C_2$$
$$C_2\rightarrow C_2-\dfrac{b}{c}C_3$$
$$\begin{vmatrix}
\lambda-\dfrac{a\lambda}{b}&\lambda-\dfrac{b\lambda}{c} &c(a+b+c)+\lambda \\
-\lambda & \lambda & bc \\
0 & -\lambda & c^2+\lambda
\end{vmatrix}=0$$
Taking $\lambda^2$ common
$$\lambda^2\begin{vmatrix}
1-\dfrac{a}{b}&1-\dfrac{b}{c} &c(a+b+c)+\lambda \\
-1 & 1 & bc \\
0 & -1 & c^2+\lambda
\end{vmatrix}=0
$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
b-a&c-b &c(a+b+c)+\lambda \\
-b & c & bc \\
0 & -c & c^2+\lambda
\end{vmatrix}=0
$$
$$R_1\rightarrow R_1-R_3$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
b-a&2c-b &ca+bc \\
-b & c & bc \\
0 & -c & c^2+\lambda
\end{vmatrix}=0$$
$$R_1\rightarrow R_1-R_2$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
2b-a&c-b &ca \\
-b & c & bc \\
0 & -c & c^2+\lambda
\end{vmatrix}=0$$
Now expanding it
$$\dfrac{\lambda^2}{bc}\left(c(2b^2c-abc+abc)+(c^2+\lambda)(2bc-ac+bc-b^2)\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(2b^2c^2+(c^2+\lambda)(3bc-ac-b^2)\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(2b^2c^2+3bc^3-ac^3-b^2c^2+3bc\lambda-\lambda ac-\lambda b^2\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(b^2c^2+3bc^3-ac^3+3bc\lambda-\lambda ac-\lambda b^2\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(c^2(b^2+3bc-ac\right)+\lambda(3bc-ac-b^2)=0$$
So another factor seems to be $\dfrac{1}{bc}\left(c^2(b^2+3bc-ac)+\lambda\left(3bc-ac-b^2\right)\right)$
But actual answer is $a^2+b^2+c^2+\lambda$.
I tried to find my mistake, but everything seems correct. What am I missing here? Please help me in this.
| Alternatively:
$$\begin{vmatrix}
a^2+\lambda &ab &ac \\
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}=
\begin{vmatrix}
a^2 &ab &ac \\
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}+
\begin{vmatrix}
\lambda&ab&ac \\
0 & b^2+\lambda & bc \\
0 & bc & c^2+\lambda
\end{vmatrix}=\\
a^2\begin{vmatrix}
1 &b &c \\
b & b^2+\lambda & bc \\
c & bc & c^2+\lambda
\end{vmatrix}+
\lambda^2(b^2+c^2+\lambda)=\\
a^2\left(\begin{vmatrix}
1 &b &c \\
b & b^2 & bc \\
c & bc & c^2+\lambda
\end{vmatrix}+
\begin{vmatrix}
1 &0 &c \\
b &\lambda & bc \\
c & 0 & c^2+\lambda
\end{vmatrix}\right)+\lambda^2(b^2+c^2+\lambda)=\\
a^2\left(b^2\begin{vmatrix}
1 &1 &c \\
1 & 1& c \\
c & c & c^2+\lambda
\end{vmatrix}+\lambda^2\right)+\lambda^2(b^2+c^2+\lambda)=\\
a^2\lambda^2+\lambda^2(b^2+c^2+\lambda)=\\
\lambda^2(a^2+b^2+c^2+\lambda).$$
| {
"language": "en",
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Infinite series from Euler I am reading Euler's Introduction to analysis of the infinite" and is stumped at this section:
How does he derive the part that I circled in red?
I thought:
$\dfrac{a+bz+cz^2+...}{z(1-\alpha\cdot z-\beta\cdot z^2...)}=A+Bz+Cz^2+Dz^3+Ez^4...$
is gonna be:
$a+bz+cz^2+...=z(1-\alpha\cdot z-\beta\cdot z^2...)(A+Bz+Cz^2+Dz^3+Ez^4...)$
$=(z-\alpha\cdot z^2-\beta\cdot z^3...)(A+Bz+Cz^2+Dz^3+Ez^4)=(Az-\alpha\cdot Az^2-\beta\cdot Az^3...)+(Bz^2-\alpha\cdot B\cdot z^3-\beta\cdot B\cdot z^4...)$
So how does he have a fraction there, I don't understand.
Could you help me, please?
| You are happy with geometrically expanding ?
\begin{eqnarray*}
\frac{1}{1- \alpha z - \beta z^2 -\cdots} &=& 1+(\alpha z + \beta z^2 -\cdots)+( \alpha z + \beta z^2 +\cdots)^2+ \cdots \\
&=& 1+pz+qz^2+ \cdots.
\end{eqnarray*}
Now let multiply in the numerator
\begin{eqnarray*}
\frac{a+bz+cz^2+\cdots }{1- \alpha z - \beta z^2 -\cdots} &=& (a+bz+cz^2+\cdots)( 1+pz+qz^2+ \cdots) \\
&=& A+Bz+Cz^2 + \cdots. \\
\end{eqnarray*}
Now if we divide by $z$, we get
\begin{eqnarray*}
\frac{a+bz+cz^2+\cdots }{z(1- \alpha z - \beta z^2 -\cdots)} &=&&=& \frac{A}{z}+B+Cz + \cdots. \\
\end{eqnarray*}
As required ?
| {
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Find the closed form for this series I found this interesting series from the , it is from an old math books. It is as followed:
$\dfrac{1}{2}-\dfrac{x^2}{6}+\dfrac{x^4}{12}-\dfrac{x^6}{20}+\dfrac{x^8}{30}-...$
I notice that one can rewrite this series as followed:
$\dfrac{1}{2}-\dfrac{x^2}{2\cdot 3}+\dfrac{x^4}{3\cdot 4}-\dfrac{x^6}{4\cdot 5}+\dfrac{x^8}{5\cdot 6}-...$
So the general formula for this series is
$$\sum_{n=0}^{\infty} \dfrac{(-1)^{n}x^{2n}}{(n+1)(n+2)}$$
Is there a closed form for this series?
| $$\sum_{n=0}^{\infty} \dfrac{(-1)^{n}x^{2n}}{(n+1)(n+2)}=\sum_{n=0}^\infty(-1)^nx^{2n}\int_0^1\int_0^1 y^nz^{n+1}dydz$$
$$=\int_0^1\int_0^1z\sum_{n=0}^\infty(-x^2yz)^ndydz$$
$$=\int_0^1\int_0^1\frac{z}{1+x^2yz}dydz$$
$$=\int_0^1z\left(\int_0^1\frac{dy}{1+x^2yz}\right)dz$$
$$=\int_0^1z\left(\frac{\ln(1+x^2z}{x^2z}\right)dz$$
$$=\frac1{x^2}\int_0^1\ln(1+x^2z)dz$$
$$=\frac1{x^2}\cdot\frac{(1+x^2)\ln(1+x^2)-x^2}{x^2}$$
| {
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"source": "stackexchange",
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Number of empty subway cars I have task:
There are $9$ passengers and they get into empty $5$-car subway train. What is the probability of the fact, that exactly two cars will be empty?
I know that the power of all possibilities is $5^9$. Then I thought to pick two empty cars that is ${5\choose 2}$ possibilities, then pick three people that get to the rest of the cars, that is $9\cdot 8\cdot 7$ and the rest of the people have $3^6$ possibilities. I checked it on my calculator and it seems that the power of this event is bigger than the power of $\Omega$. What is wrong?
| The three cars must have at least $1$ passenger each. So the cases can be easily calculated:
$$\{1,1,7\} \Rightarrow {9\choose 1}{8\choose 1}3=216\\
\{1,2,6\} \Rightarrow {9\choose 1}{8\choose 2}6=1512\\
\{1,3,5\} \Rightarrow {9\choose 1}{8\choose 3}6=3024\\
\{1,4,4\} \Rightarrow {9\choose 1}{8\choose 4}3=1890\\
\{2,2,5\} \Rightarrow {9\choose 2}{7\choose 2}3=2268\\
\{2,3,4\} \Rightarrow {9\choose 2}{7\choose 3}6=7560\\
\{3,3,3\} \Rightarrow {9\choose 3}{6\choose 3}1=1680\\
$$
The sum is $18150$.
The three cars with passengers can be any of the five: ${5\choose 3}=10$.
Hence, there are $18150\cdot 10=181500$ ways to arrange $9$ passengers in $5$ cars so that exactly $2$ will remain empty. The probability is:
$$P=\frac{181500}{5^9}\approx 0.09.$$
| {
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Solve degree-four polynomial equation $\left(x^2-3x-5\right)^2-3\cdot \left(x^2-3x-5\right)-5=x$ How to solve
$$\left(x^2-3x-5\right)^2-3\cdot \left(x^2-3x-5\right)-5=x?$$
I think there is a trick, because, if $X=x^2-3x-5$, I get
$$ X^2-3X-5=x.$$
I don't know how to continue.
I've also tried to factorize but it does not work. Any idea?
| Factorize as follows,
$$(x^2-3x-5)^2-3(x^2-3x-5)^2-4=x+1$$
$$(x^2-3x-9)(x^2-3x-4)=x+1$$
$$(x^2-3x-9)(x-4)(x+1)-(x+1)=0$$
$$(x+1)(x^3-7x^2+3x+35)=0$$
$$(x+1)[(x^3-5x^2)-(2x^2-3x-35)]=0$$
$$(x+1)[x^2(x-5)-(2x+7)(x-5)]=0$$
$$(x+1)(x-5)(x^2-2x-7)=0$$
$$(x+1)(x-5)(x-1+2\sqrt2)(x-1-2\sqrt2)=0$$
Thus, the roots are $x=-1, 5, 1\pm 2\sqrt2$.
| {
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Let $A$ be a real $2×2$ matrix such that $A^6=I$. The total number of possibilities for the characteristic polynomial of $A$ is: Let $A$ be a real $2×2$ matrix such that $A^6 = I$ (where $I$ denote the identity $2×2$ matrix). The total number
of possibilities for the characteristic polynomial of $A$ is:
Annihilating polynomial is $x^6-1=(x^3-1)(x^3+1)=(x-1)(x^2+x+1)(x+1)(x^2-x+1).$ Characteristic polynomial divides any annihilating polynomial of degree greater than or equal to the degree of the characteristic polynomial.
So, Possible CHaracteristic polynomials are $(x-1)(x+1),(x^2+x+1),(x^2-x+1)$.We can not take any 2-degree polynomial factor from the annihilating polynomial. Since complex roots occur in pairs. But the answer given is $5$. Can you help me where is my mistake?
| To find the possibilities for the characteristic polynomial of an $n\times n$ matrix $A$, it helps to know some facts about the minimal polynomial of $A$. Let $c(x)$ be the characteristic polynomial of $A$, and let $m(x)$ be the minimal polynomial. Here are some facts about $m(x)$:
*
*The degree of $m(x)$ is at least 1, and the lead coefficient of $m(x)$ is 1.
*For any polynomial $p(x)$, if $p(x)$ annihilates $A$, then $m(x)$ divides $p(x)$. Hence, if a polynomial $p(x)$ has smaller degree than $m(x)$, then $p(x)$ won't annihilate $A$. This is why $m(x)$ is called the minimal polynomial.
*The Cayley-Hamilton theorem says that $c(x)$ annihilates $A$. So $m(x)$ divides $c(x)$. Hence if $A$ is $n\times n$, then the degree of $m(x)$ is at most $n$.
*$m(x)$ and $c(x)$ have the same roots. The roots of $m(x)$ are the eigenvalues of $A$. Note that even though $m(x)$ and $c(x)$ have the same roots, $m(x)$ and $c(x)$ may be different since the multiplicities of the roots may be different.
Now let's let $A$ be $2\times2$, and suppose $A^6=I$, so that $x^6-1$ annihilates $A$.
Then $m(x)$ is of degree at most 2, and $m(x)$ divides $x^6-1$. Hence $m(x)$ is either $x-1$, $x+1$, $(x-1)(x+1)$, $x^2-x+1$, or $x^2+x+1$.
$m(x)$ divides $c(x)$, and $m(x)$ and $c(x)$ have the same roots, so $c(x)$ must respectively be either $(x-1)^2$, $(x+1)^2$, $(x-1)(x+1)$, $x^2-x+1$, or $x^2+x+1$.
| {
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Why is $\frac{1}{\sqrt 5}\left[\left(\frac{1+\sqrt 5}{2}\right)^n-\left(\frac{1-\sqrt 5}{2}\right)^n\right]$ an integer?
I am looking for a proof on why
$$\frac{1}{\sqrt 5}\left[\left(\frac{1+\sqrt 5}{2}\right)^n-\left(\frac{1-\sqrt 5}{2}\right)^n\right]$$
an integer.
I have seen many proofs on this, but they all refer to a properties of Fibonacci numbers, which shouldn't be necessary.
I am trying to see why it is true using purely elementary results such as the binomial formula. Clearly this reduces to
$$\frac{1}{2^n\sqrt 5}\sum_{k=0}^n {n\choose k }\left(1-(-1)^k\right) 5^{k/2}$$
I am looking for a "divisibility" argument to see why this is an integer.
| One of the easiest proofs goes by induction.
*
*Set $a= \frac{1+\sqrt{5}}{2}$ and $b= \frac{1-\sqrt{5}}{2}$
Then you have
$$ab = -1 \mbox{ and } a+b = 1$$
Induction start $n= 1$: $\frac{1}{\sqrt 5}\left(a-b\right) = 1$ (Note, that for $n=0$ it is trivially true.)
Induction hypothesis: $\frac{1}{\sqrt 5}\left(a^k-b^k\right)$ is integer for $0\leq k \leq n$.
Induction step $n\to n+1$ ($n \geq 1$):
\begin{eqnarray} \frac{1}{\sqrt 5}\left(a^{n+1}-b^{n+1}\right)
& = & \frac{1}{\sqrt 5}\left((a+b-b)a^{n}-(b+a-a)b^{n}\right) \\
& = & \frac{1}{\sqrt 5}\left((a+b)(a^{n}-b^{n})-ba^n + ab^n \right) \\
& = & \underbrace{\frac{1}{\sqrt 5}\left((a+b)(a^{n}-b^{n})\right)}_{integer} + \frac{1}{\sqrt 5}\left(-\underbrace{ba}_{=-1}a^{n-1} + \underbrace{ab}_{=-1}b^{n-1} \right) \\
& = & \underbrace{\frac{1}{\sqrt 5}\left((a+b)(a^{n}-b^{n})\right)}_{integer} + \underbrace{\frac{1}{\sqrt 5}\left(a^{n-1} - b^{n-1} \right)}_{integer} \\
\end{eqnarray}
| {
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Does exist $a,b,c \in \Bbb N$ such that $(a+b)(b+c)(c+a)=340$? Does exist $a,b,c \in \Bbb N$ such that $(a+b)(b+c)(c+a)=340$?
$340=2\cdot2\cdot5\cdot17$
I just noticed that $(a+b)+(b+c)+(c+a)=2(a+b+c)$ can it be useful to prove first equation?
Also I tried to construct numbers $a,b,c$ such that $a+b$ would be even, $b+c$ odd, $c+a$ odd and I couldn't get $340$ so I made a prediction that it is impossible to get it. But how should I prove it?
| We may assume that $a \le b \le c$. Then
$ 2a \le a+b \le 2b $
$ 2b \le b+c \le 2c $
$ 2a \le c+a \le 2c $
and so
$ 8a^3 \le 8a^2b \le 340 \le 8 bc^2 \le 8 c^3$. Therefore, $a \le 3$ and so $a \in \{0,1,2,3\}$. Consider all four cases.
*
*$a=0$. Then $b(b+c)c=340$ and so $2b^3 \le 340$. Therefore, $b \le 5$ and so $b \in \{1,4,5\}$ because $b$ divides $340$. But then $c \not \in \mathbb N$.
*$a=1$. Then $b \le 4$ and so $b \in \{2,4,5\}$ because $b+1$ divides $340$. Again, no $c$ works.
etc...
| {
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Inequality Proof $(x^2+1)(y^2+1)(z^2+1)\leq...$ I want to show, that for positive numbers $x,y,z$ with $xy,yz,zx\geq1$, $(x^2+1)(y^2+1)(z^2+1)\leq\left(\left(\frac{x+y+z}{3}\right)^2+1\right)^3$.
Using the AM-QM inequality for $(x^2+1),(y^2+1)$ and $(z^2+1)$, I already got $(x^2+1)(y^2+1)(z^2+1)\leq\left(\frac{x^2+y^2+z^2}{3}+1\right)^3$.By equivalent transformations of $\left(\frac{x^2+y^2+z^2}{3}+1\right)^3\leq\left(\left(\frac{x+y+z}{3}\right)^2+1\right)^3$ I get $0\leq\frac{2}{3}\left(xy+yz+zx-x^2-y^2-z^2\right)$, which cannot be true. Where did I miss something? Is the AM-QM inequality too weak?
Thank you!
| Hint: First show that, if $ab\geq 1$,
$$(a^2+1)(b^2+1)\leq \left(\left(\frac{a+b}{2}\right)^2+1\right)^2$$
(hint for this: expand out and use that $(a-b)^2\geq 0$ and $ab\geq 1$). Then apply this to get that
$$(a^2+1)(b^2+1)(c^2+1)(d^2+1)\leq \left(\left(\frac{a+b+c+d}{4}\right)^2+1\right)^4$$
if $ab,bc,cd,da,ac,bd\geq 1$. Now, set $d=\frac{a+b+c}{3}$.
| {
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On the biconditional $I(n^2) = 2 - \frac{5}{3q} \iff (k = 1 \land q = 5)$, where $q^k n^2$ is an odd perfect number MOTIVATION
Let $N$ be an odd perfect number given in the so-called Eulerian form
$$N = q^k n^2,$$
i.e., $q$ is the special / Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
In what follows, I let
$$I(x)=\frac{\sigma(x)}{x}$$
denote the abundancy index of the positive integer $x$. ($\sigma(x)$ is the sum of divisors of $x$.)
CLAIM
If $q^k n^2$ is an odd perfect number given in Eulerian form, then
$$I(n^2) = 2 - \frac{5}{3q} \iff (k = 1 \land q = 5).$$
PROOF OF CLAIM
Let $q^k n^2$ be an odd perfect number given in Eulerian form.
Suppose that
$$I(n^2) = 2 - \frac{5}{3q}.$$
Since $q^k n^2$ is perfect, then we have
$$I(q^k n^2) = I(q^k)I(n^2) = 2$$
where we have used the fact that $I$ is multiplicative.
Hence,
$$I(n^2) = I(q^k)I(n^2) - \frac{5}{3q} \implies \frac{5}{3q} = I(n^2)\bigg(I(q^k) - 1\bigg) \geq I(n^2)\bigg(1+\frac{1}{q}-1\bigg).$$
This implies that
$$I(n^2) \leq \frac{5}{3}.$$
Assume to the contrary that
$$I(n^2) = 2 - \frac{5}{3q} < \frac{5}{3}.$$
Then we have
$$\frac{6q - 5}{3q} < \frac{5}{3} \implies 18q - 15 < 15q \implies 3q < 15 \implies q < 5,$$
contradicting $q \geq 5$.
Added to the Proof of Claim (Dec 15 2019) Hence,
$$I(n^2) = 2 - \frac{5}{3q} \implies I(n^2) = \frac{5}{3} \implies (k=1 \land q=5)$$
while the proof of the direction
$$(k=1 \land q=5) \implies I(n^2) = \frac{5}{3} \implies I(n^2) = 2 - \frac{5}{3q}$$
is trivial.
QUESTION
It can be proved (page 17) that
$$I(n^2) \leq 2 - \frac{5}{3q}$$
holds in general for an odd perfect number $q^k n^2$ given in Eulerian form.
My question is: Can a biconditional similar to the one proved here be derived for the case
$$I(n^2) < 2 - \frac{5}{3q}?$$
| Let $q^k n^2$ be an odd perfect number given in Eulerian form.
Since the equation
$$I(n^2) = 2 - \frac{5}{3q}$$
is true if and only if the conjunction
$$k=1 \land q=5$$
holds, then $I(n^2) = 2 - {5/(3q)}$ is true if and only if
$$I(n^2) = 2 - \frac{5}{3\cdot{5}} = 2 - \frac{1}{3} = \frac{5}{3}.$$
It follows that
$$I(n^2) < 2 - \frac{5}{3q} \iff I(n^2) \neq \frac{5}{3}.$$
Suppose that $I(n^2) > 5/3$. We get
$$2 - \frac{5}{3q} > I(n^2) > \frac{5}{3} \implies q > 5.$$
(Another way to get the same conclusion is to use the inequality
$$I(n^2) \leq \frac{2q}{q+1},$$
which holds generally.)
Now, assume that $I(n^2) < 5/3$. We obtain
$$\frac{2(q-1)}{q} < I(n^2) < \frac{5}{3} \implies q < 6 \implies q = 5.$$
Thus,
$$I(5^k) = I(q^k) = \frac{2}{I(n^2)} > \frac{6}{5} \implies k > 1.$$
We summarize our results below:
If $I(n^2) > 5/3$, then $q > 5$.
If $I(n^2) < 5/3$, then $q = 5$ and $k > 1$.
Since Cohen and Sorli (2012, page 4) have proved that $q=5$ and $k=5$ is untenable, then we have
$$I(n^2) < \frac{5}{3} \implies \bigg(q=5 \land k>1\bigg) \implies \bigg(q=5 \land k \geq 9\bigg) \implies I(n^2) < \frac{2}{I(5^9)} = \frac{1953125}{1220703} \approx 1.60000016384,$$
whence there is no contradiction, since it is known (unconditionally) that $I(n^2) > 8/5$ holds.
| {
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"url": "https://math.stackexchange.com/questions/3477159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve $x^{\log(x)-1 } = \frac{1}{\sqrt[4]{10}}$ Question:
$x^{\log(x)-1 } = \frac{1}{\sqrt[4]{10}}$
My attempts to solve:
$x^{\log(x)-1 } = \frac{1}{\sqrt[4]{10}}$
$x^{\log(x)-\log(10) } = 10^\frac{-1}{4}$
$x^{\log(\frac{x}{10}) } = 10^\frac{-1}{4}$
$-\frac{1}{4}\log{_x 10} = \log(\frac{x}{10})$
$-\frac{1}{4}\log{_x 10} = \frac{\log{_x}x}{\log{_x}10}$
$-\frac{1}{4}\log{_x 10} = \frac{1}{\log{_x}10}$
$-\frac{1}{4}\log{_x 10^2} = 1$
$-\frac{1}{2}\log{_x 10} = 1$
$\log{_x 10} = -2$
$10 = x^{-2}$
$10 = \frac{1}{x^2}$
$x^2 = \frac{1}{10}$
$x = {\sqrt{\frac{1}{10}}}$
But this is still wrong. the correct answer is $x = {\sqrt{10}}$. Can someone please point out which part i have done wrong? Thanks.
| There is a mistake later as well, but here's the first, from the fourth to fifth line:
$$
\boxed{
\log \left(\frac x{10}\right) \neq \frac{\log_x x}{\log_x {10}}}
$$
Because the left hand side is $\log_{10} x - 1 = \frac{1}{\log_{x}10} - 1$ while the RHS is $\frac 1{\log_x 10}$ , and these two are clearly not equal, in fact they differ by one.
Thanks to this a second error, namely $(\log_x 10)^2 \neq \log_x (10^2)$ becomes inconsequential(this is the transition from sixth to seventh line).
Correcting that to $\log\left(\frac{x}{10}\right) = \log_{10} x - 1$ gives $$
\frac{-1}{4} \log_{x} 10 = \log_{10} x - 1
$$
now use the fact that $\log_{10}x \log_{x}10 = 1$ to see that if $y = \log_{x} 10$ then $-y = \frac 4y - 4$, etc. , it is a quadratic equation which you can solve easily to get the desired $y,x$ values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3479328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Finding $\int \sqrt{3-2x-x^2}dx$ $\int \sqrt{3-2x-x^2}\,dx$
First I did:
$$\begin{align}\int \sqrt{3-2x-x^2}\,dx &= \int \sqrt{-((x+1)^2-4)}\,dx \\
&=\int \sqrt{4-(x+1)^2}\,dx \\
\end{align}$$
Then I set $(x+1)=2\sin(t)$, $dx = 2\cos(t)\,dt$
$$\begin{align}
\int \sqrt{4-(2\sin(t))^2}2\cos(t)\,dt &= \int\sqrt{4-4\sin^2(t)}\,\,2\cos(t)\,dt \\
&= \int4\sqrt{1-\sin^2(t)}\cos(t)\,dt \\
&=4\int \sqrt{1-\sin^2(t)}\cos(t)\,dt \\
&=4\int\cos^2(t)\,dt \\
&=4\int\frac{1+\cos(2t)}\,2\,dt \\
&=2 \int 1+\cos(2t) \,dt \\
&=2t + \sin(2t) + C
\end{align}$$
$x+1 = 2\sin(t) \Leftrightarrow \frac{x+1}{2} = \sin(t) \Leftrightarrow t = \arcsin(\frac{x+1}{2})$
Replacing,
$$2\arcsin\left(\frac{x+1}{2}\right) + \sin\left(2\arcsin\left(\frac{x+1}{2}\right)\right) + C$$
But my book's solution is $$\frac{x+1}{2}\sqrt{3-2x-x^2}+2\arcsin\left(\frac{x+1}{2}\right)$$
So... what went wrong?
| Hint:
$$\sin\left(2\arcsin\left(\dfrac u2\right)\right) = u\sqrt{1 - \dfrac{u^2}4}$$
In your case, $u = x + 1$. This may be helpful.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3480370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
$\int \frac{dx}{a^2-x^2}=\frac{1}{2a}\ln\left|\frac{x+a}{x-a}\right|+C$
$\int\frac{dx}{a^2-x^2}=\frac{1}{2a}\ln\left|\frac{x+a}{x-a}\right|+C$
how to derive this formula?
$\int \frac{dx}{a^2-x^2}=\int\frac{dx}{(a-x)(a+x)}\boxed{=}\\\frac{1}{(a-x)(a+x)}=\frac{A}{a-x}+\frac{B}{a+x}\\1=A(x+a)+B(a-x)\\
x=a:1=A(a+a)\Rightarrow A=\frac{1}{2a}\\x=-a:1=B(a-(-a))=1=2aB\Rightarrow B=\frac{1}{2a}\\\frac{1}{(a+x)(a-x)}=\frac{\frac{1}{2a}}{a-x}+\frac{\frac{1}{2a}}{a+x}=\frac{1}{2a}\left(\frac{1}{a-x}+\frac{1}{a+x}\right)\\\int\frac{dx}{(a-x)(a+x)}=\int\frac{1}{2a}\left(\frac{1}{a-x}+\frac{1}{a+x}\right)dx=\frac{1}{2a}\left(\ln|a-x|\boxed{+}\ln|a+x|\right)=????$
where is mistake?
| Beginning from your partial fractions:
\begin{align} \int{\frac{dx}{a^2-x^2}} &= \frac{1}{2a} \int{\frac{1}{a-x} + \frac{1}{a+x} \; dx} \\ &= \frac{1}{2a} \left(-\ln|a-x|+\ln|a+x|\right) + C \\ &= \frac{1}{2a} \; \ln\left|\frac{x+a}{x-a}\right| + C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3480583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\frac{a}{7a+b}+\frac{b}{7b+c}+\frac{c}{7c+a}\le\frac38$ Suppose that $a,b,c>0$. How to prove $$\frac{a}{7a+b}+\frac{b}{7b+c}+\frac{c}{7c+a}\le\frac38$$
?
My first idea: By AM-GM, $$7a+b\geq \sqrt{7ab}$$ so $$\sum_{cyc} \frac{a}{7a+b}\le\sum_{cyc}\sqrt{\frac{a}{7b}}$$ but I am not sure if we can continue from here.
Also I try Cauchy-Schwarz: $$\sum_{cyc} \frac{a}{7a+b}\le\sqrt{a^2+b^2+c^2}\sqrt{\sum_{cyc} \frac{1}{(7a+b)^2}}.$$
Now what?
| Can be much detail, please
$$\frac{3}{7}-\frac{1}{7}\sum_{cyc}\frac{b^2}{7ab+b^2}\leq\frac{3}{7}-\frac{1}{7}\cdot\frac{(a+b+c)^2}{\sum\limits_{cyc}(7ab+b^2)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Find $\lim_{x\to0}\frac{1-\cos x\cos2x\cos3x}{x^2}$
Find $$\lim_{x\to0}\dfrac{1-\cos x\cos2x\cos3x}{x^2}$$
My attempt is as follows:-
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(2\cos x\cos2x\right)\cos3x}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(\cos3x+\cos x\right)\cos3x}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(\cos^23x+\cos x\cos3x\right)}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(\dfrac{1+\cos6x}{2}+\dfrac{1}{2}\left(\cos4x+\cos2x\right)\right)}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{4}\left(1+\cos6x+\cos4x+\cos2x\right)}{x^2}$$
Applying L'Hospital as we have $\dfrac{0}{0}$ form
$$\lim_{x\to0}\dfrac{-\dfrac{1}{4}\left(-6\sin6x-4\sin4x-2\sin2x\right)}{2x}$$
Again applying L'Hospital as we have $\dfrac{0}{0}$ form
$$\lim_{x\to0}\dfrac{-\dfrac{1}{4}\left(-36\cos6x-16\cos4x-4\cos2x\right)}{2}=\dfrac{36+16+4}{8}=\dfrac{56}{8}=7$$
But actual answer is $3$, what am I messing up here?
| Consider that
$$
1-\cos x\cos2x=1-\cos x+\cos x-\cos x\cos2x=(1-\cos x)+\cos x(1-\cos 2x)
$$
Going a further step on,
$$
1-\cos x\cos2x\cos3x=(1-\cos x)+\cos x(1-\cos2x)+\cos2x(1-\cos3x)
$$
and, by induction,
$$
1-\prod_{k=1}^n\cos(kx)=\sum_{k=1}^{n}\cos((k-1)x)(1-\cos(kx))
$$
Now
$$
\lim_{x\to0}\frac{1-\cos(kx)}{x^2}=\lim_{x\to0}k^2\,\frac{1-\cos(kx)}{(kx)^2}=\frac{k^2}{2}
$$
Therefore
$$
\lim_{x\to0}\frac{1-\prod_{k=1}^n\cos(kx)}{x^2}=\sum_{k=1}^n\frac{k^2}{2}=\frac{n(n+1)(2n+1)}{12}
$$
For $n=3$ we get
$$
\frac{3\cdot4\cdot7}{12}=7
$$
Your method in the now deleted answer is good as well, but you computed wrongly the derivative: if $f(x)=\cos x\cos2x\cos3x$, then
$$
\frac{f'(x)}{f(x)}=\frac{-\sin x}{\cos x}+\frac{-2\sin2x}{\cos2x}+\frac{-3\sin3x}{\cos3x}
$$
so
$$
f'(x)=-\cos x\cos2x\cos3x(\tan x+2\tan2x+3\tan3x)
$$
and so applying l'Hôpital yields
$$
\lim_{x\to0}\frac{\cos x\cos2x\cos3x(\tan x+2\tan2x+3\tan3x)}{2x}
$$
and you can evaluate this as
$$
\frac{1}{2}(1+2\cdot2+3\cdot3)=\frac{14}{2}=7
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding solutions if existed in this number theory - prime numbers problem. I was working on a problem involving prime number and I reached a point where I could not solve an equation, and the equation is.
Given that $a,b$ and $c$ are positive integers which satisfies that both $a$ and $b $ are less than $c$.
And satisfy the following equation
$$ac=b^2 +1$$
Find solutions if theres any, if there exists infintly many solutions find $a$ and $ab$ as functions in tearms of $c$.
(The original problem was if given a number $s$, find the biggest prime divisor for ($s^2 +1$). And after some conclusions I came up with the above equation and didnt know how to find a solution, so if the first problem cannot be answered I hope that the second one can be).
| Changing your letters to become $$ y^2 - z x = -1. $$
The automorphisms (with positive determinant) of the quadratic form $y^2 - zx$ are known, and parametrized by the modular group. So, given $ps-qr = 1,$ so
$$
\det
\left(
\begin{array}{cc}
p & q \\
r & s \\
\end{array}
\right) = 1
$$
we construct
$$
W =
\left(
\begin{array}{ccc}
p^2 & 2pq & q^2 \\
pr & ps+qr& qs \\
r^2 & 2rs & s^2 \\
\end{array}
\right)
$$
The Hessian matrix of the quadratic form is
$$
H =
\left(
\begin{array}{ccc}
0 & 0 & -1 \\
0 & 2 & 0 \\
-1 & 0 & 0 \\
\end{array}
\right)
$$
and we have arranged $$ W^T H W = H. $$
This means that, if
$$
\left(
\begin{array}{c}
x \\
y \\
z \\
\end{array}
\right)
$$
solves $y^2 - zx = -1,$ so does every
$$
\left(
\begin{array}{ccc}
p^2 x + 2pq y + q^2 z \\
pr x + (ps+qr) y + qs z \\
r^2 x +2rs y + s^2 z \\
\end{array}
\right)
$$
For example, if we take $x=1, y=0, z=1$ to begin,
$$
\left(
\begin{array}{ccc}
p^2 + q^2 \\
pr + qs \\
r^2 + s^2 \\
\end{array}
\right)
$$
is a solution vector whenever $ps-qr = 1$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$, find the value of $p^3+q^3+r^3$. If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$, find the value of $p^3+q^3+r^3$.
Here's what I have got,
By Vieta's rule
$p+q+r=1\text{. ...........}(1)$
$pq+qr+pr=1\text{. ...........}(2)$
$pqr=2\text{. ...........}(3)$
Squaring $(1)$,
$p^2+q^2+r^2+2(pq+qr+pr)=1\text{. ...........}(4)$
From $(2)$,
$p^2+q^2+r^2=-1\text{. ...........}(5)$
Putting the roots and adding these equations,
$p^3-p^2+p-2=0$
$q^3-q^2+q-2=0$
$r^3-r^2+r-2=0$
We get,
$(p^3+q^3+r^3)-(p^2+q^2+r^2)+(p+q+r)-6=0$
Putting the values,
$(p^3+q^3+r^3)-(-1)+1-6=0$
$(p^3+q^3+r^3)=4$
Am I doing something wrong in my solution?
Because the answer given is -5.
Any help would be appreciated.
| You can use the strict $p^3+q^3+r^3-3pqr=(p+q+r)(p^2+q^2+r^2-pq-qr-rp)$. It's easy to compute $p^2+q^2+r^2$ and use Viete's rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Finding $n$ if $|\sum_{r=0}^{3n-1}\beta^{2^r}|=4\sqrt{2}$ where $\beta=\exp(i2\pi/7)$
Find $n$ if $\left|\displaystyle\sum_{r=0}^{3n-1}\beta^{2^r}\right|=4\sqrt{2}$ where $\beta=\exp(i2\pi/7)$.
My Attempt
$$\begin{aligned}\left|\sum_{r=0}^{3n-1}\beta^{2^r}\right|&=\left|\sum_{r=0}^{n}(\beta+\beta^2+\beta^4)\right|\\ &=n|\beta|\left|1+\beta+\beta^3\right|\\&=n\sqrt{\left(1+\cos\left(2\pi/7\right)+\cos\left(6\pi/7\right)\right)^2+\left(\sin\left(2\pi/7\right)+\sin(6\pi/7)\right)}\\&=n\sqrt{1+4\sin^2(3\pi/14)+4\sin(\pi/14)\sin(3\pi/14)}\end{aligned}$$
I'm not sure how to proceed on the simplification of the term inside the square roots. Any hints are appreciated.
| Let $z=\beta+\beta^2+\beta^4$, then $\bar{z}=\beta^3+\beta^5+\beta^6$.
Notice that $\beta^7=1$ and $\beta+\beta^2+\beta^3+\beta^4+\beta^5+\beta^6=-1$.
So $\left|z\right|^2=z\bar{z}=(\beta+\beta^2+\beta^4)(\beta^3+\beta^5+\beta^6)$
$=\beta^4+\beta^6+\beta^7+\beta^5+\beta^7+\beta^8+\beta^7+\beta^9+\beta^{10}$
$=3+\beta+\beta^2+\beta^3+\beta^4+\beta^5+\beta^6=2$.
Thus $\left|\beta+\beta^2+\beta^4\right|=\sqrt{2}$, and we have $n=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find the remainder when $f(x) = x^{2016}+2x^{2015}-3x+4$ is divided by $g(x)=x^2+3x+2$ Find the remainder when $f(x) = x^{2016}+2x^{2015}-3x+4$ is divided by $g(x)=x^2+3x+2$
I try to factor them, but I failed. I know that it's impossible to divide it
| First note $x^2+3x+2=(x+1)(x+2)$.
Modulo $x+2$, $x\equiv-2$, so $f(x) = x^{2016}+2x^{2015}-3x+4\equiv2^{2016}-2^{2016}+6+4=\color{blue}{10}.$
Modulo $x+1$, $x\equiv-1$, so $f(x) = x^{2016}+2x^{2015}-3x+4\equiv1-2+3+4=\color{blue}{6}.$
$\color{red}1(x+2)+\color{red}{-1}(x+1)=1$,
so by the Chinese remainder theorem $f(x)\equiv \color{blue}{10}(x+1)\color{red}{(-1)}+\color{blue}6(x+2)\color{red}{(1)}\bmod x^2+3x+2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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I am stuck on problem involving absolute values and inequalities The problem states to solve for $x$, and then to write the answer with absolute value notation.
The problem is: $(x-2)/(x-4) > (x+2)/(x)$
The correct answer is abs$(x-2) > 2$
(abs stands for absolute value).
Can someone explain to me how to arrive at this answer? Thank you.
| First, the given inequality is invalid when $x = 0$ or $x = 4$, since it contains undefined expressions in either of these cases.
Suppose $x \neq 0$ and $x \neq 4$. Then we may multiply both sides of the inequality by $x(x-4)$ to clear denominators. We should be careful: for some values of $x$, in particular $0 < x < 4$, we are multiplying both sides of the inequality by a negative number, so the sense of the inequality reverses; for other values of $x$, we are multiplying by a positive number, so the sense does not reverse. In particular, we obtain \begin{align*}
(x(x-2) > (x+2)(x-4) &\text{ and } (x < 0 \text{ or } 4 < x)) \text{ or }\\
(x(x-2) < (x+2)(x-4) &\text{ and } 0<x<4) \text{.}
\end{align*}
This is written in the form "(case 1) or (case 2)". In the first case, we expand and then subtract terms to get
$$ x^2 - 2x > x^2 - 2 x - 8 \text{ , so}$$
$$ 8 > 0 \text{,} $$
which is always true. Doing the same to the second case, we obtain
$$ 8 < 0 \text{,} $$
which is never true. So we have simplified to
\begin{align*}
(\text{True} &\text{ and } (x < 0 \text{ or } 4 < x)) \text{ or }\\
(\text{False} &\text{ and } 0<x<4) \text{.}
\end{align*}
"True or (anything)" is always true and "False and (anything)" is always false, so actually we have
$$ x < 0 \text{ or } 4 < x \text{.} $$
(The next steps are very clear if you plot the current solution set on a number line.) The midpoint of $0$ and $4$ is $2$, so we subtract $2$ from both equations to reveal
$$ x-2 < -2 \text{ or } 2 < x-2 \text{,} $$
or what is the same thing,
$$ |x-2| > 2 \text{.} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 2
} |
Prove that $\sum_{k=0}^{\lfloor (n-1)/2 \rfloor} (-1)^k {n+1 \choose k} {2n-2k-1 \choose n} =\frac{ n(n+1)}2 $ $$\sum_{k=0}^{\Big\lfloor \frac{(n-1)}{2} \Big\rfloor} (-1)^k {n+1 \choose k} {2n-2k-1 \choose n} = \frac{n(n+1)}{2} $$
So I feel like $(-1)^k$ is almost designed for the inclusion-exclusion principle. And the left-hand side looks like some sort of pairing, so I am interested in some combinatorics proof like below-linked question. But using a generating function is always helpful.
[EDIT] now I am probably equally, if not more interested in a generating function solution now that I see below answer that completely makes sense to me, but with some issues in signs..
Evaluation of a sum of $(-1)^{k} {n \choose k} {2n-2k \choose n+1}$
| We seek to show that
$$\sum_{k=0}^{\lfloor (n-1)/2 \rfloor}
(-1)^k {n+1\choose k} {2n-2k-1\choose n}
= \frac{1}{2} n (n+1).$$
The LHS is
$$\sum_{k=0}^{\lfloor (n-1)/2 \rfloor}
(-1)^k {n+1\choose k} {2n-2k-1\choose n-1-2k}
\\ = [z^{n-1}] (1+z)^{2n-1}
\sum_{k=0}^{\lfloor (n-1)/2 \rfloor}
(-1)^k {n+1\choose k}
z^{2k} (1+z)^{-2k}.$$
Now the coefficient extractor $[z^{n-1}]$ combined with the $z^{2k}$
term enforces the range, making for a zero contribution when $2k\gt
n-1$ and we may continue with
$$[z^{n-1}] (1+z)^{2n-1}
\sum_{k\ge 0} (-1)^k {n+1\choose k}
z^{2k} (1+z)^{-2k}
\\ = [z^{n-1}] (1+z)^{2n-1}
\left(1-\frac{z^2}{(1+z)^2}\right)^{n+1}
\\ = [z^{n-1}] \frac{1}{(1+z)^3} (1+2z)^{n+1}.$$
This is
$$\sum_{q=0}^{n-1} (-1)^q {q+2\choose q}
{n+1\choose n-1-q} 2^{n-1-q}.$$
Observe that
$${q+2\choose q} {n+1\choose n-1-q}
= \frac{(n+1)!}{q!\times 2! \times (n-1-q)!}
= {n+1\choose 2} {n-1\choose q}. $$
This yields for the sum
$${n+1\choose 2} \sum_{q=0}^{n-1} (-1)^q {n-1\choose q}
2^{n-1-q}
\\ = {n+1\choose 2} (-1+2)^{n-1}
= {n+1\choose 2}.$$
We have the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3497919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
If $z = \frac{(\sqrt{3} + i)^n}{(\sqrt{3}-i)^m}$, find the relation between $m$ and $n$ such that $z$ is a real number. I am given the following number $z$:
$$z = \dfrac{(\sqrt{3} + i)^n}{(\sqrt{3} - i)^m}$$
with $n, m \in \mathbb{N}$. I have to find a relation between the natural numbers $n$ and $m$ such that the number $z$ is real. I know that for a complex number to be real, its imaginary part must equal $0$, but I can't isolate the imaginary part. This is as far as I got:
$$\sqrt{3} + i =
2 \bigg (\frac{\sqrt{3}}{2} + i\frac{1}{2} \bigg) =
2 \bigg( \cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6} \bigg ) $$
$$\sqrt{3} - 1 =
2 \bigg ( \dfrac{\sqrt{3}}{2} - i \dfrac{1}{2} \bigg) =
2 \bigg ( \cos \dfrac{\pi}{6} - i \sin \dfrac{\pi}{6} \bigg ) =
2 \bigg ( \cos \dfrac{11\pi}{6} + i \sin \dfrac{11\pi}{6} \bigg )$$
So I got the numerator and the denominator in a form that I can use DeMoivre's formula on. So, next I'd have:
$$z = \dfrac{\bigg [2 \bigg ( \cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6} \bigg ) \bigg ]^n}
{\bigg [2 \bigg( \cos \dfrac{11 \pi}{6} + i \sin \dfrac{11 \pi}{6} \bigg ) \bigg ]^m }$$
$$z = 2^{n - m} \cdot \dfrac{\cos \dfrac{n \pi}{6} + i \sin \dfrac{n \pi}{6}}
{\cos \dfrac{11 m \pi}{6} + i \sin \dfrac{11 m \pi}{6}}$$
But this is where I got stuck. I still can't isolate the imaginary part of $z$ in order to equal it to $0$.
| $\sqrt {3} - i$ is the conjugate to $\sqrt {3} + i$
$\frac {(\sqrt 3 + i)^n}{(\sqrt 3 - i)^m} = \frac {(\sqrt 3 + i)^{n+m}}{(\sqrt 3 - i)^m(\sqrt 3 + i)^m}$
This will rationalize the denominator
$\frac {(\sqrt 3 + i)^n}{(\sqrt 3 - i)^m} = \frac {(\sqrt 3 + i)^{n+m}}{4^{m}}$
We must find where $(\sqrt 3 + i)^{n+m}$ is real
$\sqrt 3 + i = 2(\cos \frac {\pi}{6} + i\sin \frac {\pi}{6})\\
(\sqrt 3 + i)^{n+m} = 2^{n+m}(\cos \frac {\pi}{6} + i\sin \frac {\pi}{6})^{n+m}\\
(\sqrt 3 + i)^{n+m} = 2^{n+m}(\cos \frac {(n+m)\pi}{6} + i\sin \frac {(n+m)\pi}{6})$
if $(\sqrt 3 + i)^{n+m}$ is real $\sin \frac {(n+m)\pi}{6} = 0$
$\frac {(n+m)\pi}{6} = k\pi\\
n+m = 6k$
$6$ divides $n+m$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3498784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Calculate the limit of function I was trying to find it for some time, but couldn't, so please help me.
$ \lim_{x\to\infty} ( \sqrt[100]{(x + 3*1)(x + 3*2)...(x +3*100)} - x)$
| \begin{gather*}
\lim _{x\leadsto \infty }(\sqrt[100]{}( x+3\times 1)( x+3\times 2) ...( x+3\times 100) \ -\ x\\
Taking\ x\ common\ \ from\ all\ brackets\ ,\ \\
\lim _{x\leadsto \infty }( x\sqrt[100]{}( 1+3\times 1/x)( 1+3\times 2/x) ...( 1+3\times 100/x) \ -\ x\\
Apply\ the\ binomial\ series\ expansion\ because\ \frac{1}{x} \leadsto 0\\
\lim _{x\leadsto \infty }( x( 1+\frac{1}{100}\left(\frac{3\times 1}{x} +\frac{3\times 2}{x} +....\frac{3\times 100}{x}\right) \ -\ x\\
\lim _{x\leadsto \infty }\frac{1}{100}\left(\frac{3\times 1}{1} +\frac{3\times 2}{1} +....\frac{3\times 100}{1}\right) \ \ =\ \frac{3}{200}( 100\ \times 101) \ =\frac{303}{2} \ \\
\\
\\
\\
\end{gather*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3499311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\lim_{x \to 1}\frac{\sin(x^2-x)}{x^2-1}$ and $\lim_{x \to 1}\frac{\ln(x^2+x-1)}{\ln(x^3+x-1)}$ I found these limits and I was unable to solve them due to the occurring indeterminations
$$\lim_{x \to 1}\frac{\sin(x^2-x)}{x^2-1}$$
$$\lim_{x \to 1}\frac{\ln(x^2+x-1)}{\ln(x^3+x-1)}$$
Can someone help me, please?
| Since $\lim_{t \to 0}\frac{\log_{a}\left(t+1\right)}{t}=\log_{a}\left(e\right)$,
we have:
$$\lim_{x \to 1}\frac{\ln(x^2+x-1)}{\ln(x^3+x-1)}$$$$=\lim_{x \to 1}\frac{\ln\left(1+\left(\color{blue}{x^{2}+x-2}\right)\right)}{\color{blue}{x^{2}+x-2}}\cdot\frac{x^{2}+x-2}{x^{3}+x-2}\cdot\frac{\color{red}{x^{3}+x-2}}{\ln\left(1+\left(\color{red}{{x^{3}+x-2}}\right)\right)}=3/4$$
note that: $\frac{x^{2}+x-2}{x^{3}+x-2}=\frac{\left(x+2\right)\left(x-1\right)}{\left(x^{2}+x+1\right)\left(x-1\right)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
A Diophantine equation: solve $(3x^2+y^2-4y-17)^3-(2x^2+2y^2-4y-6)^3=(x^2-y^2-11)^3$ (without using Fermat's last theorem) Solve this Diophantine equation: $(3x^2+y^2-4y-17)^3-(2x^2+2y^2-4y-6)^3=(x^2-y^2-11)^3$
My attempt (use Fermat's last theorem)
$$(3x^2+y^2-4y-17)^3-(2x^2+2y^2-4y-6)^3=(x^2-y^2-11)^3$$
$$\Leftrightarrow (3x^2+y^2-4y-17)^3=(x^2-y^2-11)^3+(2x^2+2y^2-4y-6)^3$$
Use Fermat's last theorem, we get:
$$
\left\{
\begin{array}{c}
x^2-y^2-11=0 \\
3x^2+y^2-4y-17=2x^2+2y^2-4y-6
\end{array}
\right.$$
or
$$
\left\{
\begin{array}{c}
2x^2+2y^2-4y-6=0\\
3x^2+y^2-4y-17=x^2-y^2-11
\end{array}
\right.$$
Then we continue...
My question is, is there another way to solve that other than Fermat's last theorem? I see that using Fermat's last theorem is like crack a nut by a sledgehammer.
| Let
$$3x^2 + y^2 - 4y -17 = a\text{ ,}$$
$$-(2x^2 +2y^2 +4y -6)= b$$
and observe
$$x^2-y^2 -11= a+b$$
Your problem may be reexpressed as $$a^3 + b^3=(a+b)^3$$ implying
$$3ab(a+b)=0.$$ This gives your proposed solution set, plus the case where $a=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3502327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How many solutions does the expression $3^x(3^x+1)+2=|3^x-1|+|3^x+2|$ have? I can solve this by opening every single modulus, doing alternate combinations of $\pm$ and checking whether every $3^x$ is $>0$ or not. But that consumes a lot of time. Is there is a shorter way to solve such questions.
The answer is 1
| Simplify the equation by substituting $a=3^x$ and using the fact that $a>0$.
$a(a+1)+2=|a-1|+a+2$ since $a+2>0$
There are only two cases to be checked:
Case 1: $a^2 + a+2=a-1+a+2$
$\Longleftrightarrow a^2 -a+1=0$
Here $D=b^2-4ac=1-4=-3<0$
Therefore no real solutions.
Case 2: $a^2 + a+2=-a+1+a+2$
$\Longleftrightarrow a^2+a-1=0$
$\Longleftrightarrow a= \frac{-1+ \sqrt 5}{2} $ since $a>0$ we neglect $a= \frac{-1- \sqrt 5}{2} $
Therefore, there's only one root.
Further, if you want to find $x$,
$3^x=\frac{-1+ \sqrt 5}{2} $
Take $\log$ on both sides,
$x= \log_3\frac{\sqrt 5 -1}{2}$
This is the only solution to the expression given.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3503998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Calculating limit approaches inifnity having infinite terms On my sample calculus mid-term exam, there is a weird question that asks us to calculate the limit that has an infinite term:
$\displaystyle\lim_{n \to \infty} {\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}}$
I am not sure what kind of technique should be used here.
Giving that this is mid-term of the first-year university calculus exam, we have only learned L'hopital's Rule and other basic techniques. But I don't see any technique that can fit to solve this problem
Thanks!
| Note that
$$\displaystyle\lim_{n \to \infty} {\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}}=\displaystyle\lim_{n \to \infty} \frac{1}{n}{\sum_{k=1}^n\frac{1}{1+\frac{k}{n}}}$$
This is a Riemann Sum for $f(x)=\frac{1}{1+x}$ over the interval $[0,1]$, with the satndard partition $x_i=\frac{k}{n}$ for $1 \leq k \leq n$ and the right hand points of the interval as intermediate points.
Therefore
$$\displaystyle\lim_{n \to \infty} {\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}}=\int_0^1 \frac{1}{1+x}dx =\ln(1+x)|_0^1=\ln(2)$$
P.S. One can reach the came conclusion by using the well known identity
$$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}+...-\frac{1}{2n}$$ and the standard definition of the Euler -Mascheroni constant, but this approach is typically beyond calculus.
Just for fun, here is how you get the limit with the E-M constant
$$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}+...-\frac{1}{2n} \\
=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n}-2 (\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2n})\\
=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n}- (\frac{1}{1}+\frac{1}{2}+...+\frac{1}{n})\\
=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n}- \ln(2n)- (\frac{1}{1}+\frac{1}{2}+...+\frac{1}{n} -\ln(n))+\ln(2)
$$
This converges to $\gamma-\gamma+\ln(2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3504583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Can we deduce the characteristic polynomial for this matrix? Given a square $n \times n$ matrix $A$ that satisfies $$\sum\limits_{k=0}^n a_k A^k = 0$$ for some coefficients $a_0, a_1, \dots, a_n,$ can we deduce that its characteristic polynomial is $\sum\limits_{k=0}^n a_k x^k$?
| The answer is no.
You are given a degree $n$ polynomial $p(x)$ that an $n\times n$ matrix $A$ satisfies. This is not enough information to find the characteristic polynomial $c_A(x)$, although you will be able to narrow it down to finitely many possibilities.
Let's look at an example to see why. Suppose your friend picks the matrix $A$. Suppose he doesn't tell you $A$, but he does tell you $p(x)$ and asks you to guess $c_A(x)$. Suppose your friend picked
$$A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$
and told you that $p(x)=x^3-6x^2+11x-6$.
You could then reason that $p(x)=(x-1)(x-2)(x-3)$. Which would mean that the minimal polynomial $m_A(x)$ must be one of the following: $(x-1)$, $(x-2)$, $(x-3)$, $(x-1)(x-2)$, $(x-1)(x-3)$, $(x-2)(x-3)$, $(x-1)(x-2)(x-3)$.
Hence the characteristic polynomial $c_A(x)$ must be one of the following: $(x-1)^3$, $(x-2)^3$, $(x-3)^3$, $(x-1)^2(x-2)$, $(x-1)(x-2)^2$, $(x-1)^2(x-3)$, $(x-1)(x-3)^2$, $(x-2)^2(x-3)$, $(x-2)(x-3)^2$, $(x-1)(x-2)(x-3)$.
Since your friend picked
$$A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$
the characteristic polynomial is $c_A(x)=(x-1)^2(x-2)$, but you can't prove that, because for all you know he may have picked
$$B=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$$
which also satisfies $p(x)=x^3-6x^2+11x-6=(x-1)(x-2)(x-3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3506928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 3
} |
Solving $3|2-x| + |2x-4| = 8$ I'm not sure how to solve $3|2-x| + |2x-4| = 8$. I don't exactly need an answer, the method is what I'm interested in. I've done equations with one absolute value element. >_<
| Well, it THIS case, $|2x-4| = 2\cdot |x-2|= 2\cdot |2-x|$ so $3|2-x| + |2x-4| = 3|2-x|+2|2-x| = 5|2-x| =8$ so
$|2-x| =|x-2|= \frac 85$ and so $x-2 =\pm \frac 85$ and $x = \frac 85 + 2=\frac {18}5=3\frac 35$ or $x = 2-\frac 85 = \frac 25$.
But in general do cases.
Do cases:
$2-x \ge 0$ if $x \le 2$ and $2-x < 0$ if $x > 2$.
And $2x-4 \ge 0$ if $x \ge 2$ and $2x-4 < 0$ if $x < 2$.
So do cases:
Case 1: $x < 2$ and $2-x >0$ so $|2-x| = 2-x$ and $2x-4< 0$ so $|2x-4| = 4-2x$ and we have
$3|2-x| + |4-2x| = 8$
$3(2-x) + (4-2x) = 8$
$10 - 5x = 8$
$5x =2$ and $x =\frac 25$.
Case 2: $x = 2$ and $2-x=0$ and $2x-4=0$ and so we have
$3|2-x| + |4-2x| = 8$
$3*0 + 0 = 8$
$0 = 8$ which is impossible.
Case 3: $x > 2$ and $2-x < 0$ so $|2-x| = x-2$ and $2x -4 > 0$ so $|2x-4| = 2x -4$
So
$3|2-x| + |2x-4| = 8$
$3(x-2) + (2x-4) = 8$
$5x - 10 = 8$
$x = \frac {18}5 = 3\frac 35$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find summation definition of a certain function Consider a tuple $T$ to be an object (list of numbers) similar to a row vector of $n$ elements which are real numbers: $$T=(x_{1}, x_{2},\ldots, x_{n}); n\in \Bbb N, x_{i}\in \Bbb R$$
A function $F$ can be defined from the tuples $T$ to $\Bbb R$ which takes the alternating sum of all possible products of $2$ elements. For example, for $n=3$:
$$F(T) = x_{1}x_{2}-x_{2}x_{3}+x_{3}x_{1}$$
and for $n=4$ $$F(T)=x_{1}x_{2}-x_{2}x_{3}+x_{3}x_{4}-x_{1}x_{3}+x_{2}x_{4}-x_{1}x_{4}$$
It is pretty obvious that the number of terms is $n \choose 2$ and if $n \choose 2$ is even, there are as much positive terms as there are negatives.
I came up with a faulty general definition that doesn't seem to get the signs properly:$$\sum^{n-1}_{k=1}\sum^{n-k}_{j=1}\left[(-1)^{j+1}*x_{k}*x_{k+j}\right]$$
This doesn't work because the signs on some of the elements are flipped. You are supposed to place the signs in that manner: you first take all products of neighboring elements and multiply by $-1$ each time, then the elements distance $1$ apart, then distance $2$ apart and so on. I am asking to correct me.
| We build multiplication tables of $x_j\ast x_k$ for small $n$ and check for a pattern of the signs. Since the pattern differs slightly for $n$ even and $n$ odd we consider $n=6$ and $n=7$.
Case $n=6$:
We have in this case
\begin{align*}
&x_1x_2-x_1x_3+x_1x_4-x_1x_5+x_1x_6\\
&\qquad-x_2x_3+x_2x_4+x_2x_5-x_2x_6\\
&\qquad\qquad\quad\ +x_3x_4-x_3x_5-x_3x_6\\
&\qquad\qquad\qquad\qquad\ \ -x_4x_5+x_4x_6\\
&\qquad\qquad\qquad\qquad\qquad\qquad+x_5x_6
\end{align*}
The tables below give the signs of the terms $x_j\ast x_k$ and the values $j+k \mod 4$ which show a correlation to the plus and minus signs.
$$
\begin{array}{c|cccccc|cccccc}
\ast&x_2&x_3&x_4&x_5&x_6&\qquad\qquad\qquad\ast&x_2&x_3&x_4&x_5&x_6\\
\hline
x_1&+&-&-&+&+&\qquad\qquad\qquad x_1&3&0&1&2&3&\\
x_2&&-&+&+&-&\qquad\qquad\qquad x_2&&1&2&3&0&\\
x_3&&&+&-&-&\qquad\qquad\qquad x_3&&&3&0&1&\\
x_4&&&&-&+&\qquad\qquad\qquad x_4&&&&1&2&\\
x_5&&&&&+&\qquad\qquad\qquad x_5&&&&&3&
\end{array}
$$
From the tables above we obtain for even $n$:
\begin{align*}
\sum_{{1\leq j<k\leq n}\atop{{j+k \equiv 2\ \mathrm{mod}\,(4)}\atop{j+k \equiv 3\ \mathrm{mod}\,(4)}}}x_jx_k
-\sum_{{1\leq j<k\leq n}\atop{{j+k \equiv 0\ \mathrm{mod}\,(4)}\atop{j+k \equiv 1\ \mathrm{mod}\,(4)}}}x_jx_k
=\color{blue}{\sum_{1\leq j<k\leq n}(-1)^{\left\lfloor\frac{1}{2}\left(j+k+2-4\left\lfloor\frac{j+k+2}{4}\right\rfloor\right)\right\rfloor}x_jx_k}
\end{align*}
Case $n=7$:
We have in this case
\begin{align*}
&x_1x_2+x_1x_3-x_1x_4-x_1x_5+x_1x_6+x_1x_7\\
&\qquad-x_2x_3-x_2x_4+x_2x_5+x_2x_6-x_2x_7\\
&\qquad\qquad\quad\ +x_3x_4+x_3x_5-x_3x_6-x_3x_7\\
&\qquad\qquad\qquad\qquad\ \ -x_4x_5-x_4x_6+x_4x_7\\
&\qquad\qquad\qquad\qquad\qquad\qquad+x_5x_6+x_5x_7\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\ -x_6x_7\\
\end{align*}
The tables below give the signs of the terms $x_j\ast x_k$ and the values $j+k \mod 4$ which show a correlation to the plus and minus signs.
$$
\begin{array}{c|ccccccc|ccccccc}
\ast&x_2&x_3&x_4&x_5&x_6&x_7&\qquad\qquad\qquad\ast&x_2&x_3&x_4&x_5&x_6&x_7\\
\hline
x_1&+&+&-&-&+&+&\qquad\qquad\qquad x_1&3&0&1&2&3&0&\\
x_2&&-&-&+&+&-&\qquad\qquad\qquad x_2&&1&2&3&0&1&\\
x_3&&&+&+&-&-&\qquad\qquad\qquad x_3&&&3&0&1&2&\\
x_4&&&&-&-&+&\qquad\qquad\qquad x_4&&&&1&2&3&\\
x_5&&&&&+&+&\qquad\qquad\qquad x_5&&&&&3&0&\\
x_6&&&&&&-&\qquad\qquad\qquad x_6&&&&&&1&
\end{array}
$$
From the tables above we obtain for odd $n$:
\begin{align*}
\sum_{{1\leq j<k\leq n}\atop{{j+k \equiv 0\ \mathrm{mod}\,(4)}\atop{j+k \equiv 3\ \mathrm{mod}\,(4)}}}x_jx_k
-\sum_{{1\leq j<k\leq n}\atop{{j+k \equiv 1\ \mathrm{mod}\,(4)}\atop{j+k \equiv 2\ \mathrm{mod}\,(4)}}}x_jx_k
=\color{blue}{\sum_{1\leq j<k\leq n}(-1)^{\left\lfloor\frac{1}{2}\left(j+k+1-4\left\lfloor\frac{j+k+1}{4}\right\rfloor\right)\right\rfloor}x_jx_k}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3512161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve inverse trigonometric equation $\frac{\pi}{6}=\tan^{-1} \frac{11}{x} -\tan^{-1} \frac{1}{x}$ How do I go about solving for $x$ when I have:
$\frac{\pi}{6}=\tan^{-1} \left( \frac{11}{x} \right)-\tan^{-1}\left( \frac{1}{x} \right)$.
| Make a right triangle ABC with C=90, AC=x, BC=11. Try to represent $\sin B$ in two ways: the straightforward way:
$$
\sin B=\frac{x}{\sqrt{121+x^2}},
$$
and - let $D$ be on $BC$ with $CD=1$, apply law of sine to triangle ADB and get
$$
\frac{\sqrt{1+x^2}}{\sin B}=\frac{10}{\sin 30}=20,
$$
thus get
$$
\frac{x}{\sqrt{121+x^2}}=\frac{\sqrt{1+x^2}}{20},
$$
square this we get $400x^2=(121+x^2)((1+x^2)$. You can let $y=x^2$ and that is a 2nd order equation you can solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3513047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
A contestant participates in a game show where three important prizes are offered. A contestant participates in a game show where three important prizes are offered. His chances of winning the three prizes are $\frac{1}{6}$, $\frac{1}{3}$ and $\frac{1}{2}$, respectively.
*
*How many prizes can the contestant expect to win?
*Let $X$ denote the number of prizes won by the contestant. Find the
probability distribution function of $X$.
For 1. Using the Poisson model I obtained that $$P(0)=\frac{10}{36}, P(1)=\frac{17}{36}, P(2)=\frac{8}{36}, P(3)=\frac{1}{36}$$ and I think that the correct answer is 1 because it has the highest probability.
Is it correct?
For 2. Is it alright if I write directly $$
\begin{pmatrix}
0 & 1 & 2 & 3 \\
\frac{10}{36} & \frac{17}{36} & \frac{8}{36} & \frac{1}{36}\\
\end{pmatrix}
$$
| $X$ does not have a Poisson distribution because it only takes finitely many values. Observe that $X=X_1+X_2+X_3$ where $X_1\sim\mathrm{Ber}\left(\frac16\right)$, $X_2\sim\mathrm{Ber}\left(\frac13\right)$, and $X_3\sim\mathrm{Ber}\left(\frac12\right)$ are independent. Therefore
\begin{align}
\mathbb P(X=0) &= \mathbb P(X_1=0,X_2=0,X_3=0)\\ &= \mathbb P(X_1=0)\mathbb P(X_2=0)\mathbb P(X_3=0)\\
&=\left(1-\frac16\right)\left(1-\frac13\right)\left(1-\frac12\right)\\
&= \frac5{18},
\end{align}
\begin{align}
\mathbb P(X=1) &= \mathbb P\left(\bigcup_{i=1}^3 \{X_i=1\}\cap\bigcap_{j\in\{1,2,3\}\setminus\{i\}}\{X_j=0\} \right)\\
&= \sum_{i=1}^3 \mathbb P(X_i=1)\prod_{j\in\{1,2,3\}\setminus i\}}\mathbb P(X_j=0)\\
&= \mathbb P(X_1=1)\mathbb P(X_2=0)\mathbb P(X_3=0)+\mathbb P(X_1=0)\mathbb P(X_2=1)\mathbb P(X_3=0)+\mathbb P(X_1=0)\mathbb P(X_2=0)\mathbb P(X_3=1)\\
&= \frac16\left(1-\frac13\right)\left(1-\frac12\right) +\left(1-\frac16\right)\frac13\left(1-\frac12\right) + \left(1-\frac16\right)\left(1-\frac13\right)\frac12 \\&= \frac{17}{36},
\end{align}
\begin{align}
\mathbb P(X=2) &= \mathbb P(X_1+X_2+X_3=2)\\
&= \mathbb P(X_1=1)\mathbb P(X_2=1)\mathbb P(X_3=0)+\mathbb P(X_1=1)\mathbb P(X_2=0)\mathbb P(X_3=1)+\mathbb P(X_1=0)\mathbb P(X_2=1)\mathbb P(X_3=1)\\
&= \frac16\cdot\frac13\left(1-\frac12\right)+ \frac16\left(1-\frac13\right)\frac12 + \left(1-\frac16\right)\frac13\cdot\frac12\\
&= \frac29,
\end{align}
and
\begin{align}
\mathbb P(X=3) &= \mathbb P(X_1+X_2+X_3=3)\\ &= \mathbb P(X_1=1)\mathbb P(X_2=1)\mathbb P(X_3=1)\\
&= \frac16\cdot\frac13\cdot\frac12 = \frac1{36}.
\end{align}
The expected value of $X$ is
\begin{align}
\mathbb E[X] &= \sum_{i=0}^3 i\cdot\mathbb P(X=i)\\
&= 0\cdot\frac5{18} + 1\cdot\frac{17}{36} + 2\cdot\frac29 + 3\cdot\frac1{36}\\
&= 1.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Flux through the Surface Find the flux through the surface $\iint_S F\cdot NdS$ where N is the normal vector to S.
i) $F=3z\hat i-4\hat j+y\hat k$
$~~~S:z=1-x-y$ (first octant)
ii) $F=x\hat i+y\hat j-2z\hat k$
$~~~~S:\sqrt{a^2-x^2-y^2}$
I have evaluated $N$ vector as : $-\hat i-\hat j\over\sqrt2$ for i) and -$\frac{x}{\sqrt{a^2-x^2-y^2}}\hat i-\frac{y}{\sqrt{a^2-x^2-y^2}}\hat j$ for ii).
I can calculate $F\cdot N$ through this but I am unable to convert $dS$ into $dxdy$ using projections and find the limits. Can somebody give me a very easy explanation on how to convert using these two examples?
| Take the example: $F=x\hat i+y\hat j-2z\hat k$; $~S: z=\sqrt{a^2-x^2-y^2}$. Rewrite the surface as $f(x,y,z)=x^2+y^2+z^2 = a^2$ and calculate its unit normal vector
$$N=\frac{(f_x, f_y, f_z)}{\sqrt{f_x^2+ f_y^2+ f_z^2}}=\frac1a(x,y,z)$$
Then,
$$F\cdot N = (x,y,-2z)\cdot \frac1a(x,y,z)=\frac1a (x^2+y^2-2z^2)=a-\frac3a z^2=a(1-3\cos^2\theta)$$
where the spherical coordinate $z=a\cos\theta$ is used in the last step. The corresponding surface element on the sphere $S:\>\> x^2+y^2+z^2=a^2$ is
$$dS = a^2\sin\theta \>d\theta d\phi$$
Thus, the surface integral over the half-sphere for $z>0$ is
$$S =\int_{z>0}F\cdot N \>dS = a^3\int_0^{\pi/2}\int_0^{2\pi} (1-3\cos^2\theta)\sin\theta \>d\theta d\phi=0
$$
Edit: In Cartesian coordinates, we have $F\cdot N = \frac{3(x^2+y^2)-2a^2}{a}$ and $dS = \frac a{\sqrt{a^2-x^2-y^2}}dxdy$. The surface integral instead is
$$S =\int_{z>0}F\cdot N \>dS = \int_{z>0} \frac{3(x^2+y^2)-2a^2}{\sqrt{a^2-x^2-y^2}}dxdy=0
$$
For the example: $F=3z\hat i-4\hat j+y\hat k$; $~S:z=1-x-y$ in the first octant. Calculate the unit normal vector and $F\cdot N$,
$$N=\frac{(f_x, f_y, f_z)}{\sqrt{f_x^2+ f_y^2+ f_z^2}}=\frac1{\sqrt3}(1,1,1)$$
$$F\cdot N = (3z,-4,y)\cdot \frac1{\sqrt3}(1,1,1)=\frac1{\sqrt3} (3z-4+y)
=-\frac1{\sqrt3} (1+3x+2y)$$
Then, use the standard surface element formula
$$dS = \sqrt{1+z_x^2+z_y^2}\>dxdy= \sqrt3 \>dxdy$$
As a result, the integral over the first-octant surface $~S:z=1-x-y$
$$S =\int_{x,y>0}F\cdot N \>dS = -\int_0^1\int_0^{1-x} (1+3x+2y)dydx=-\frac43
$$
(Note: the flux integrals over axis planes in both examples are relatively straightforward if required.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3515937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Derivative of $\dfrac{\sqrt{3-x^2}}{3+x}$ I am trying to find the derivative of this function
$f(x)=\dfrac{\sqrt{3-x^2}}{3+x}$
$f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}}\frac{d}{dx}(3-x)(3+x)-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-2x(3+x)}{2\sqrt{3-x^2}}-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-x(3+x)}{\sqrt{3-x^2}}-\sqrt{3-x^2}}{(3+x)^2}$
$\dfrac{-x(3+x)}{\sqrt{3-x^2}(3+x)^2}-\dfrac{\sqrt{3-x^2}}{(3+x)^2}$
$\dfrac{-x}{\sqrt{3-x^2}(3+x)}-\dfrac{\sqrt{3-x^2}}{(3+x)^2}$
At this point, I want to transform this derivative into the form of $\dfrac{3(x+1)}{(3+x)^2\sqrt{3-x^2}}$
How do I do this? This form is given by Wolfram:
https://www.wolframalpha.com/input/?i=derivative+%283-x%5E2%29%5E%281%2F2%29%2F%283%2Bx%29
| your first line is not entirely clear
\begin{eqnarray*}
f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}} \color{red}{ \left( \frac{d}{dx}(3-x^{\color{blue}2}) \right)}(3+x)-\sqrt{3-x^2}}{(3+x)^2}
\end{eqnarray*}
But everything is right after this and you just need to do a common denominator at the end & you will get the Wolfie answer ... look again, there is a minus sign!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3525571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
} |
Minimum value of $\frac{ax^2+by^2}{\sqrt{a^2x^2+b^2y^2}}$ If $$x^2+y^2=1$$
Prove that Minimum value of $$f(x,y)=\frac{ax^2+by^2}{\sqrt{a^2x^2+b^2y^2}}$$ is
$$\frac{2\sqrt{ab}}{a+b}$$
My try:
I used basic Trigonometry:
Let $x=\cos t$ and $y=\sin t$
Then we get $$f(x,y)=g(t)=\frac{a\cos^2 t+b\sin^2 t}{\sqrt{a^2\cos^2 t+b^2\sin^2 t}}$$
Now let $$p=\cos(2t)$$
then we get a single variable function as:
$$h(p)=\frac{1}{\sqrt{2}}\frac{(a+b)+p(a-b)}{\sqrt{a^2+b^2+p(a^2-b^2)}}$$
where $p \in [-1, 1]$
Now we can find critical point and find minimum.
Is there a better approach, i tried lagrange multipliers but very tedious
| Let us show
$$
\Big(\ a^2x^2+b^2y^2\ \Big)(x^2+y^2)\cdot\frac{4ab}{(a+b)^2}\le \Big(\ ax^2+by^2\ \Big)^2\ .
$$
It is useful to substitute $u=x^2$, $v=y^2$, so let us show:
$$
4ab\Big(\ a^2u+b^2v\ \Big)(u+v)\le \Big(\ au+bv\ \Big)^2\cdot (a+b)^2 \ .
$$
It turns out that after moving all terms to the R.H.S. we can factor and restate equivalently:
$$
0\le (au-bv)^2(a-b)^2\ ,
$$
which is true. To see this, compare with $(S+T)^2(a+b)^2-(S-T)^2(a-b)^2=(\dots)^2-(\dots)^2=4(Sa+Tb)(Ta+Sb)$, where $S=au$, $T=bv$.
Note: The inequality of Cauchy-Schwarz gives:
$$
\Big(\ (ax)^2+(by)^2\ \Big)(x^2+y^2)\ge \Big(\ ax^2+by^2\ \Big)^2\ .
$$
so the maximum of the given expression is one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How does this fraction $\frac{1}{y} = \frac{1}{a}$ become $y = a$? I was learning SET. To solve a problem for Y at some point I stopped right here $\frac{1}{y} = \frac{1}{a}$, and I couldn't go further. Then I opened the solution book and there I found that everything I did was right up-to this $\frac{1}{y} = \frac{1}{a}$ line. And I also found that, the solution book solved the problem just adding this line $y = a$ after my line ($\frac{1}{y} = \frac{1}{a}$), but there is no any explicit explanation. Can anybody expert in mathematics please tell me how this fraction ($\frac{1}{y} = \frac{1}{a}$) has been converted to $y = a$? Thanks in advance guys!
| There’s plenty of ways. You can take reciprocals on both sides, since $r=s$ implies $\frac{1}{r} = \frac{1}{s}$ when they are not zero. So you get
$$\frac{1}{\quad\frac{1}{y}\quad} = \frac{1}{\quad\frac{1}{a}\quad}$$
and this is the same as $y=a$.
Or you can multiply through by $ay$ and then cancel. You get
$$\begin{align*}
\frac{1}{y} &= \frac{1}{a}\\
\frac{ay}{y} &= \frac{ay}{a}\\
\frac{a}{1} &= \frac{y}{1}\\
a &= y
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3529082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Torsion group of $y^2 = x^3 \pm nx$ Let $n\in \mathbb{Z}$ be not divisible by $15$ and suppose $n$ and $-n$ are both not perfect squares. Prove that at least one of the elliptic curves $Y^2 = X^3 + nX$ and $Y^2 = X^3 - nX$ has torsion group given by $\{{\bf o},(0,0)\}$ (where ${\bf o}$ is the point at infinity).
Here is my progress:
The discriminant is $\Delta = \pm 4n^3$ so we may use reduction with any odd prime $p \nmid n$. Since $n$ is not divisible by $15$, at least one of $3\nmid n$ and $5\nmid n$ holds.
*
*Suppose $3\nmid n$ - then we can use $p=3$. Since $x^3 \equiv x \pmod 3$, for the reduced curves we can equivalently consider $Y^2 = (n+1)X$ and $Y^2 = (1-n)X$. If $n\equiv 1 \pmod 3$, then the former has only ${\bf {\underline{o}}}, (0,0), (2,\pm 1)$ as points, while the ones on the latter are ${\bf {\underline{o}}}, (0,0), (1,0), (2,0)$. The points are exchanged when $n\equiv 2 \pmod 3$.
*Suppose $5\nmid n$ - then we can use $p=5$. For $n\equiv 1,2,3,4 \pmod 5$ a direct verification shows that $Y^2 = X^3 + nX$ has $4,2,6,8$ points respectively; hence $Y^2 = X^3 - nX$ has $8,6,2,4$ points, respectively.
Hence currently I have that at least one of the curves has $1, 2$ or $4$ torsion points. It is also easy to check that $(0,0)$ is always a point of order $2$. So how to rule out the case of $4$ torsion points?
Any help appreciated!
| Since $n$ and $-n$ are not perfect squares, the point $(0, 0)$ is the only point of order $2$ on both curves.
Now if $P = (x, y)$ is a rational point of order $4$ on the curve $Y^2 = X^3 - nX$, then by the doubling formula, we have $$X(2P) = \frac{x^4 + 2nx^2 + n^2}{4y^2}.$$ However, $2P$ must be the point $(0, 0)$, hence $x^4 + 2nx^2 + n^2 = (x^2 + n)^2$ must be zero, which means $-n = x^2$ is the square of a rational number, and therefore a perfect square. This contradicts the assumption.
Changing $n$ to $-n$ shows that the curve $Y^2 = X^3 + nX$ has no point of order $4$ either.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3530422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove that $x + \frac{2x^3}{3} + \cdots + \frac{2\cdot 4 \cdot \cdots 2nx^{2n+1}}{3\cdot 5 \cdot (2n+1)}+\cdots = \frac{\arcsin(x)}{\sqrt{1-x^2}}$
Prove that $x + \dfrac{2x^3}{3} + \cdots + \dfrac{2\cdot 4 \cdot \cdots 2nx^{2n+1}}{3\cdot 5 \cdot (2n+1)}+\cdots = \dfrac{\arcsin(x)}{\sqrt{1-x^2}}.$
Let $f(x) = x + \dfrac{2}3x^3 +\dfrac{8}{15}x^5 + \cdots + \dfrac{1}2n!\cdot\dfrac{n!}{(2n+1)!}(2x)^{2n+1}+\cdots.$
Then $xf(x) = x^2 + \dfrac{2}3 x^4 + \dfrac{8}{15}x^6 + \cdots + \dfrac{1}{4}n!\cdot \dfrac{n!}{(2n+1)!}(2x)^{2n+2}+\cdots.$
Also, $f'(x) = 1 + 2x^2 + \dfrac{8}3 x^4 + \cdots + n!\cdot \dfrac{n!}{(2n)!}(2x)^{2n}+\cdots$
and $(1-x^2)f'(x) =1+x^2 + \dfrac{2}3 x^4 + \cdots + [n!\cdot \dfrac{n!}{(2n)!}2^{2n}-(n-1)!\cdot \dfrac{(n-1)!}{(2n-2)!}2^{2n-2}]x^{2n}+\cdots\\
=1+x^2 + \dfrac{2}3 x^4 + \cdots + \dfrac{1}{4}\cdot(n-1)!\cdot \dfrac{(n-1)!}{(2n-1)!}\cdot(2x)^{2n}+ \cdots.$
Hence the derivative of $\sqrt{1-x^2}f(x)$ is $\sqrt{1-x^2}f'(x)-\dfrac{xf(x)}{\sqrt{1-x^2}} = \dfrac{1}{\sqrt{1-x^2}}((1-x^2)f'(x) - xf(x))=\dfrac{1}{\sqrt{1-x^2}}.$ Integrating, we see that $\sqrt{1-x^2}f(x) =\arcsin(x)+C.$ Plugging in the constant $C=0$ and dividing both sides by $\sqrt{1-x^2}$ gives the desired result.
Are there other approaches to solving this problem?
| Like this
$$ \sum_{n=0}^{\infty} \frac{x^{2n+1}(2n)!!}{(2n+1)!!}=\sum_{n=0}^{\infty} \int_{0}^{\frac{\pi}{2}}x^{2n+1}\sin^{2n+1} t dt=\int_{0}^{\frac{\pi}{2}} \frac{x \sin t}{1-x^{2}\sin^{2} t} dt=-\int_{0}^{\frac{\pi}{2}}\frac{1}{1-x^{2}+x^{2}\cos^{2} t} d(x\cos t)=\frac{1}{\sqrt{1-x^{2}}} \arctan{\frac{x}{\sqrt{1-x^2}}}=\frac{\arcsin x}{\sqrt{1-x^2}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3534871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Find whether $\sum_{n=1}^\infty \frac{2^{3n-3}3^{n+2}}{5^{2n-2}} $ is geometric and convergent and if so, state the value of the sum $$\sum_{n=1}^\infty \frac{2^{3n-3}3^{n+2}}{5^{2n-2}} $$
First I tried to find the ratio by doing $\frac{a_{n+1}}{a_n}$. Many calculations were made, and the result was $24/25$, so it is geometric. This is between -1 and 1, and therefore the series is convergent.
Then I did $$\sum_{d=1}^\infty \frac{a_1}{1-24/25} = (...) = 675$$
Is this correct?
| HINT
You have
$$
\frac{2^{3n-3}3^{n+2}}{5^{2n-2}}
= \frac{2^{-3} 3^2}{5^{-2}} \frac{2^{3n} 3^n}{5^{2n}}
= \frac{25 \cdot 9}{8} \frac{8^n 3^n}{25^n}
= \frac{25 \cdot 9}{8} \left(\frac{24}{25}\right)^n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Calculate $\int_{0}^{3}\sqrt{4-x^2}\,dx$ using a Riemann sum What is the Riemann intergral of $\sqrt{4-x^2}$ from $0$ to $3$?
I tried to write down the Riemann sum $\sum_{i=1}^n\sqrt{4-\frac{(3i)^2}{n^2}}\frac{3}{n}$. Then I tried to take limit as $n \to \infty$, but I don't know how to evaluate it. I don't know what to do with the square root.
| Building on (Unfortunately, it does not fit in the comment section) from Integrand and Dark Malthrop). Hence posting as a full version.
We can start with the series expansion,
\begin{eqnarray*}
\sqrt{a^{2}-x^{2}} &=& \sum_{m=0}^{\infty}{(-1)^{m} a^{1-2m} \binom{\frac{1}{2}}{m} x^{2 m}}
\end{eqnarray*}
\begin{eqnarray*}
\int_{}^{}{\sqrt{a^{2}-x^{2}} dx}&=&\int { \sum_{m=0}^{\infty}{(-1)^{m} a^{1-2m} \binom{\frac{1}{2}}{m} x^{2 m}} dx} \\
&=& \sum_{m=0}^{\infty}{\int {(-1)^{m} a^{1-2m} \binom{\frac{1}{2}}{m} x^{2 m}} dx} \\
&=& \sum_{m=0}^{\infty}{(-1)^{m} a^{1-2m} \binom{\frac{1}{2}}{m} \frac{x^{2 m+1}}{2m+1} } \\
&=& \frac{1}{2} \left( a x \sqrt{1-\frac{x^{2}}{a^{2}} } + a^{2} \sin^{-1}\left(\frac{x}{a}\right) \right)
\end{eqnarray*}
With $a=2$, this get to,
\begin{eqnarray*}
\int{\sqrt{4-x^{2}} dx} &=& \left( x \sqrt{1-\frac{x^{2}}{4 } } + 2 \sin^{-1} \frac{x}{2} \right)
\end{eqnarray*}
I don't see a reason why we cannot extend the limit beyond 2, aside from the fact that, it goes outside $\mathbb{R}$. As mentioned by many people here, when the upper limit is $2$, it becomes to $ \left( 2 \sqrt{1-\frac{2^{2}}{4 } } + 2 \sin^{-1} \frac{2}{2} \right) =\pi$. If the limit is indeed as posited in the original formulation, it becomes,
\begin{eqnarray*}
\int_{0}^{3}{\sqrt{4-x^{2}} dx} &=& \left( 3 \sqrt{1-\frac{3^{2}}{4 } } + 2 \sin^{-1} \frac{3}{2} \right) \\
&=& \jmath \frac{3 \sqrt{5}}{2} + 2\sin^{-1} \frac{3}{2}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Limit of $\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$ $$\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$$
I tried to used $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ but it did not worked out so I tried to use the squeeze theorem.
$$0=\sqrt[3]{x^3}-\sqrt{x^2}\leq \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}\leq\sqrt[3]{8x^3}-\sqrt{4x^2}=2-2=0$$
But on the right hand I have inrcrased $\sqrt{x^2-2x}$ rather then deceased
Another attempt:
$$\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}=\lim_{x\to \infty} \sqrt[3]{x^3(1+\frac{2}{x^2})}-\sqrt{x^2(1-\frac{2}{x})}=\\=\lim_{x\to \infty} x\sqrt[3]{(1+\frac{2}{x^2})}-x\sqrt{(1-\frac{2}{x})}=\lim_{x\to \infty} x[\sqrt[3]{(1+\frac{2}{x^2})}-\sqrt{(1-\frac{2}{x})}]$$
| $\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}
=\lim_{x\to \infty} \frac{(\sqrt[3]{x^3+2x}^6-\sqrt{x^2-2x}^6)}{(\sqrt[3]{x^3+2x}^5+\sqrt[3]{x^3+2x}^4\sqrt{x^2-2x}+\sqrt[3]{x^3+2x}^3\sqrt{x^2-2x}^2+\sqrt[3]{x^3+2x}^2\sqrt{x^2-2x}^3+\sqrt[3]{x^3+2x}\sqrt{x^2-2x}^4+\sqrt{x^2-2x}^5)}=
=\lim_{x\to \infty} \frac{((x^3+2x)^2-(x^2-2x)^3)}{(\sqrt[3]{x^3+2x}^5+\sqrt[3]{x^3+2x}^4\sqrt{x^2-2x}+\sqrt[3]{x^3+2x}^3\sqrt{x^2-2x}^2+\sqrt[3]{x^3+2x}^2\sqrt{x^2-2x}^3+\sqrt[3]{x^3+2x}\sqrt{x^2-2x}^4+\sqrt{x^2-2x}^5)}=
=\lim_{x\to \infty} \frac{x^6+4x^4+4x^2-x^6+6x^5-12x^4+8x^3}{(\sqrt[3]{x^3+2x}^5+\sqrt[3]{x^3+2x}^4\sqrt{x^2-2x}+\sqrt[3]{x^3+2x}^3\sqrt{x^2-2x}^2+\sqrt[3]{x^3+2x}^2\sqrt{x^2-2x}^3+\sqrt[3]{x^3+2x}\sqrt{x^2-2x}^4+\sqrt{x^2-2x}^5)}=\frac{6}{6}=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
$a_n=\sqrt{2-a_{n-1}},a_1=\sqrt{2}$ Given
$$a_n=\sqrt{2-a_{n-1}},a_1=\sqrt{2}$$
I calculated $a_1$ to $a_5$
$$\sqrt{2},
\sqrt{2-\sqrt{2}},
\sqrt{2-\sqrt{2-\sqrt{2}}}, \\
\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}, \\
\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}}$$
which made me think of $\sin/\cos$. So I divided each by $2$, calculated $\arcsin(x)$ and got $$\left\{\frac{\pi }{4},\frac{\pi }{8},\frac{3 \pi }{16},\frac{5 \pi }{32},\frac{11 \pi }{64}\right\}$$
I found that this could be a formula
$$\frac{\pi}{3} \cdot (-1)^n \cdot\frac{(-2)^n-1}{2^{n+1}}$$
So $$\sin \left(\frac{\pi}{3} \cdot (-1)^n \cdot\frac{(-2)^n-1}{2^{n+1}}\right)$$ is $a_n/2$, but this way seems not natural. Is there a more natural way?
| If the question is about finding the limit, let's consider $a_{n+1}=f(a_n)$, where $f(x)=\sqrt{2-x}$. Then we have
If $0\leq x \leq \sqrt{2}$ then $0\leq f(x)\leq\sqrt{2}$
Indeed $0\leq x\leq\sqrt{2}
\Rightarrow 0\geq-x \geq -\sqrt{2}
\Rightarrow 2\geq 2-x \geq 2-\sqrt{2}>0
\Rightarrow \sqrt{2}\geq \sqrt{2-x}=f(x)\geq 0$.
Now, let's use Banach fixed-point theorem, and MVT, given $f'(x)=-\frac{1}{2\sqrt{2-x}}$, for $\forall x,y \in[0,\sqrt{2}], x<y$, there $\exists\varepsilon\in (x,y)$ s.t.
$$|f(x)-f(y)|=|f'(\varepsilon)|\cdot |x-y|=
\frac{1}{2\sqrt{2- \varepsilon}}\cdot|x-y|<
\frac{1}{2\sqrt{2-\sqrt{2}}}\cdot|x-y|$$
since $\varepsilon\in[0,\sqrt{2}]$ as well. It's not to difficult to check that $0<\frac{1}{2\sqrt{2-\sqrt{2}}}<1$.
So, the limit exists and you can legitimately use $L=\sqrt{2-L}$ to find it, considering that $L\in[0,\sqrt{2}]$ of course, since all $(a_n)_{n>0} \subset[0,\sqrt{2}]$.
Remark: More interesting results here.
The trick that is typically applied for the $+$ with $\cos$ may not easily apply for $\arcsin$ and $\sin$ since:
$$\sin{\frac{\pi}{8}}=\frac{1}{2}\sqrt{2-\sqrt{2}}$$
$$\sin{\frac{\pi}{16}}=\frac{1}{2}\sqrt{2-\sqrt{2\color{red}{+}\sqrt{2}}}$$
I'd rather try induction, given that
$$\sin{\frac{\color{red}{1}\cdot\pi}{4}}=\frac{\sqrt{2}}{2}$$
$$\sin{\frac{\color{red}{1}\cdot\pi}{8}}=\frac{1}{2}\sqrt{2-\sqrt{2}}$$
$$\sin{\frac{\color{red}{3}\cdot\pi}{16}}=\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{2}}}$$
$$\sin{\frac{\color{red}{5}\cdot\pi}{32}}=\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}$$
$$\sin{\frac{\color{red}{11}\cdot\pi}{64}}=\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}}$$
$$\sin{\frac{\color{red}{21}\cdot\pi}{128}}=\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}}}$$
where $\{1,1,3,5,11,21\}$ is the begining of the Jacobsthal sequence, according to OEIS.
And of course:
$$\sin^2\left(\frac{\pi}{4}-\frac{x}{2}\right)=\frac{1}{2}\left(1-\sin{x}\right)\tag{1}$$
Jacobsthal sequence is $J_{n+1}=2^n-J_n$ and, assuming induction hypothesis, we have
$$\sqrt{\frac{1}{2}\left(1-\sin\left(\color{red}{J_n}\frac{\pi}{2^{n+1}}\right)\right)}=\\
\sqrt{\frac{1}{2}\left(1-\frac{1}{2}\sqrt{2-\sqrt{2-...-\sqrt{2-\sqrt{2}}}}\right)}=\\
\sqrt{\frac{1}{4}\left(2-\sqrt{2-\sqrt{2-...-\sqrt{2-\sqrt{2}}}}\right)}=\\
\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{2-...-\sqrt{2-\sqrt{2}}}}}\overset{(1)}{=}\\
\sin\left(\frac{\pi}{4}-J_n\frac{\pi}{2^{n+2}}\right)=
\sin\left(\left(2^n-J_n\right)\frac{\pi}{2^{n+2}}\right)=
\sin\left(\color{red}{J_{n+1}}\frac{\pi}{2^{n+2}}\right)$$
The positive sign of the square root above, while appling $(1)$, is justified by $\frac{\pi}{4}>\frac{\pi}{4}-J_n\frac{\pi}{2^{n+2}}>0$, where the $\sin$ function is positive. As a result:
$$2\sin\left(J_{n}\frac{\pi}{2^{n+1}}\right)=\underbrace{\sqrt{2-\sqrt{2-\sqrt{2-...-\sqrt{2-\sqrt{2}}}}}}_{n\text{ times}} \tag{2}$$
Jacobsthal sequence also has a closed form of
$$ J_n = \frac{2^n - (-1)^n}{3}$$
which can be solved using, for example, characteristic polynomials (more than half of the work is done here), leading to
$$2\sin\left(\frac{2^n-(-1)^n}{2^{n+1}}\cdot\frac{\pi}{3}\right)=\underbrace{\sqrt{2-\sqrt{2-\sqrt{2-...-\sqrt{2-\sqrt{2}}}}}}_{n\text{ times}} \tag{3}$$
Remark: It is worth noting this question is not duplicating this family of questions.
| {
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"url": "https://math.stackexchange.com/questions/3538786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is $x^3 + x + 1$ for $x=\frac{\sqrt 5 - 1}{2}$
What is $x^3 + x + 1$ for $x=\frac{\sqrt 5 - 1}{2}$
Of course we can subtitute and expand it
${(\frac{\sqrt 5 -1}{2})}^3 + \frac{\sqrt 5 -1}{2} + 1$
Is there a better way to do it?
| Yes. $\dfrac{\sqrt{5}-1}{2}$ is a root of $x^2+x-1=0$. So $x^2=1-x$.
Multiply by $x$ to get
$$x^3=x-x^2=x-1+x=2x-1$$
Therefore
$$x^3+x+1=3x=\frac{3}{2}(\sqrt{5}-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Calculating Inverse modulo 101 A lot of problems are calculating $n\mod m$ with $n<m$. But I have the following problem: Find the inverse of $71 \mod 101$. Now using Euclid, I get the $\gcd(71,101)=1$ and
$$
1=3-(1\cdot(8-2\cdot(11-1\cdot(30-2\cdot(71-2\cdot(101-71)))))
$$
I know the solution is $37$ but I am not able to get there...
The above I get with:
$$
101=1\cdot71+30, 71=2\cdot30+11, 30=2\cdot11+8, 11=1\cdot8+3, 8=2\cdot3+2, 3=1\cdot2 +1
$$
How do I get the $37$?
| $1=3-1\times2=3-1\times(8-2\times3)=3\times3-1\times8=3\times(11-1\times8)-1\times8$
$=3\times11-4\times8=3\times11-4\times(30-2\times11)=11\times11-4\times30$
$=11\times(71-2\times30)-4\times30=11\times71-26\times30=11\times71-26\times(101-71)$
$=\color{red}{37}\times71-26\times101$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3541509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluating Diverging Integral? I'm looking at:
$$\int_{-\infty}^\infty \sin(2cx)\frac{1-2x^2-\cos(2x)}{2x^3}dx$$
How come this converges for $c \in \mathbb{R}$? And how does mathematica compute it as:
$$\frac{\pi \left(-2 \left(c^2+1\right) \left| c\right| +(c-1) c \left| c-1\right| +c (c+1)
\left| c+1\right| \right)}{2 c}$$
| First, the convergence analysis:
*
*Observe that for $x$ small we have $1 - 2x^2 - \cos(2x) = O(x^4)$, and so the integral is not singular near the origin.
*When $|x|$ is large, the integral $\displaystyle \int\sin(2cx) \frac{1 - \cos(2x)}{2x^3} dx$ is absolutely integrable.
*The remaining term is $\displaystyle \int \frac{\sin(2cx)}{x} dx$ which rescales to the well-known sinc integral.
To evaluate the integral $\displaystyle \int \sin(2cx) \frac{1 - \cos(2x)}{2x^3} dx$ observe that since the function is even we can write it as
$$ 2 \lim_{a \to 0^+} \lim_{b \to \infty} \int_a^b \frac{\sin(2cx)}{2x^3} - \frac{\sin(2cx + 2x) + \sin(2cx - 2x)}{4x^3} dx $$
Integrating by parts we have that
$$ \int \frac{\sin(2Ax)}{2x^3} dx = - \frac{\sin(2Ax)}{x^2} + 2A \int \frac{\cos(2Ax)}{x^2} dx \\= - \frac{\sin(2Ax)}{x^2} - 2A \frac{\cos(2Ax)}{x} + (2A)^2 \int \frac{\sin(2Ax)}{x} ~dx $$
Doing some algebra we are essentially down to looking at
$$ \lim_{a \to 0^+} \frac{\sin(2ca)}{a^2} + 2c \frac{\cos(2ca)}{a} - \frac{\sin(2(c+1)a)}{2a^2} - (c+1) \frac{\cos(2(c+1)a)}{a} - \frac{\sin(2(c-1)a)}{2a^2} - (c-1) \frac{\cos(2(c-1)a)}{a} $$
plus a bunch of sinc integrals. This limit can be evaluated using L'Hospital's rule to yield zero.
This means that the original integral can be simplified as the sum of a bunch of sinc integrals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3542216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to calculate $\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}$ $$\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}$$
someone please help i’m not sure how to compute this. i’ve tried to do it this way:$$\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})} = \frac {1-1/7^{4x} }{1+ 1/21^{4x}} $$
$$\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}
= \dfrac {\lim\limits_{ x \to \infty } 1-1/7^{4x} }{ \lim\limits_{ x \to \infty } 1+ 1/21^{4x} }$$
but that’s defiantly wrong
| $$\begin{align}
\frac13 \frac{7^{2x} + 7^{-2x}}{7^{2x} - 7^{-2x}}
&= \frac13 \frac{7^{2x}(7^{2x} + 7^{-2x})}{7^{2x}(7^{2x} - 7^{-2x})} \\
&= \frac13 \frac{7^{4x} + 1}{7^{4x} - 1} \\
&= \frac13 \frac{(7^{4x} -1) + 2}{7^{4x} - 1} \\
&= \frac13 \Big(1 + \frac{2}{7^{4x} - 1}\Big) \\
&= \frac13 + \frac{2}{3(7^{4x} - 1)}
\end{align}$$
Now, what happens if $x$ goes to infinity?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3544652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Evaluate $\lim_{n\to\infty} \sin^2 (\pi\sqrt{n^{2014} + n^{2012} + 1})$ I tried to use $\sin^2(\pi x) = \sin^2 (\pi x - \pi)$
So,
$$\lim_{n\to\infty}\sin^2 (\pi\sqrt{n^{2014} + n^{2012} + 1}) \\
=\lim_{n\to\infty}\sin^2 (\pi(\sqrt{n^{2014} + n^{2012} + 1} - 1)) \\
\lim_{n\to\infty}\sin^2 (\frac{\pi (n^{2014} + n^{2012})}{\sqrt{n^{2014} + n^{2012} + 1}})$$
And I got trouble on here, Thanks for any hint.
| NOTE: this is not a complete answer but a small collection of some partial results.
First remark
$$
\begin{gathered}
\pi \sqrt {n^2 + n + 1} = \pi n\left( {1 + \frac{{\text{1}}}
{n} + \frac{1}
{{n^2 }}} \right)^{1/2} = \hfill \\
\hfill \\
= \pi n\left( {1 + \frac{1}
{{2n}} + o\left( {\frac{1}
{n}} \right)} \right) = \pi n + \frac{\pi }
{2} + o\left( 1 \right) \hfill \\
\end{gathered}
$$
this means that
$$
\begin{gathered}
\sin ^2 \left( {\pi \sqrt {n^2 + n + 1} } \right) = \sin ^2 \left( {\pi n + \frac{\pi }
{2} + o\left( 1 \right)} \right) = \hfill \\
\hfill \\
= \sin ^2 \left( {\frac{\pi }
{2} + o\left( 1 \right)} \right) \hfill \\
\end{gathered}
$$
and therefore
$$
\mathop {\lim }\limits_{n \to + \infty } \left[ {\sin ^2 \left( {\pi \sqrt {n^2 + n + 1} } \right)} \right] = \mathop {\lim }\limits_{n \to + \infty } \left[ {\sin ^2 \left( {\frac{\pi }
{2} + o\left( 1 \right)} \right)} \right] = 1
$$
This prove that, in general, it is not enough to consider only the main term of the polynomial under the square root.
Second remark
Let us consider a more general case
$$
\begin{gathered}
\pi \sqrt {n^{2k} + n^k + 1} = \pi n^k \left( {1 + \frac{{\text{1}}}
{{n^k }} + \frac{1}
{{n^{2k} }}} \right)^{1/2} = \hfill \\
\hfill \\
= \pi n^k \left( {1 + \frac{1}
{{2n^k }} + o\left( {\frac{1}
{{n^k }}} \right)} \right) = \pi n^k + \frac{\pi }
{2} + o\left( 1 \right) \hfill \\
\end{gathered}
$$
and as before we have that
$$
\mathop {\lim }\limits_{n \to + \infty } \left[ {\sin ^2 \left( {\pi \sqrt {n^{2k} + n^k + 1} } \right)} \right] = \mathop {\lim }\limits_{n \to + \infty } \left[ {\sin ^2 \left( {\frac{\pi }
{2} + o\left( 1 \right)} \right)} \right] = 1
$$
On the other side, if we consider $n^{2k}+n^h+1$ with $0 \leq h<k$ we have that
$$
\begin{gathered}
\pi \sqrt {n^{2k} + n^h + 1} = \pi n^k \left( {1 + \frac{{\text{1}}}
{{n^{2k - h} }} + \frac{1}
{{n^{2k} }}} \right)^{1/2} = \hfill \\
\hfill \\
= \pi n^k \left( {1 + \frac{{\text{1}}}
{{2n^{2k - h} }} + o\left( {\frac{{\text{1}}}
{{n^{2k - h} }}} \right)} \right) = \pi n^k + \frac{\pi }
{{2n^{k - h} }} + o\left( {\frac{1}
{{n^{k - h} }}} \right) \hfill \\
\end{gathered}
$$
thus
$$
\mathop {\lim }\limits_{n \to + \infty } \left[ {\sin ^2 \left( {\pi \sqrt {n^{2k} + n^k + 1} } \right)} \right] = \mathop {\lim }\limits_{n \to + \infty } \left[ {\sin ^2 \left( {\frac{\pi }
{{2n^{k - h} }} + o\left( {\frac{1}
{{n^{k - h} }}} \right)} \right)} \right] = 0
$$
Third remark:
My conjecture about the case $k<h<2k$ is motivated by numerical trials like that below which is relative to the case $n^{20}+n^{18}+1$
I have some ideas but not yet a proof. Of course, if my conjecture is true the proposed limit does not exists.
Fourth remark:
Let be $h=k+1$ with $k \geq 3$. Then
$$
\begin{gathered}
\pi \sqrt {n^{2k} + n^{k + 1} + 1} = \pi n^k \left( {1 + \frac{{\text{1}}}
{{n^{k - 1} }} + \frac{1}
{{n^{2k} }}} \right)^{1/2} = \hfill \\
\hfill \\
= \pi n^k \left( {1 + \frac{{\text{1}}}
{{2n^{k - 1} }} - \frac{1}
{{8n^{2k - 2} }} + o\left( {\frac{1}
{{n^{2k - 2} }}} \right)} \right) = \hfill \\
\hfill \\
= \pi n^k + \frac{{\pi n}}
{2} - \frac{1}
{{8n^{k - 2} }} + O\left( {\frac{1}
{{n^{k - 2} }}} \right) \hfill \\
\hfill \\
\end{gathered}
$$
Therefore
$$
\begin{gathered}
\sin ^2 \left( {\pi \sqrt {n^{2k} + n^{k + 1} + 1} } \right) = \sin ^2 \left( {\pi \sqrt {n^{2k} + n^{k + 1} + 1} } \right) \hfill \\
= \sin ^2 \left( {\pi n^k + \frac{{\pi n}}
{2} - \frac{1}
{{8n^{k - 2} }} + O\left( {\frac{1}
{{n^{k - 2} }}} \right)} \right) = \hfill \\
\hfill \\
= \sin ^2 \left( {\frac{{\pi n}}
{2} - \frac{1}
{{8n^{k - 2} }} + O\left( {\frac{1}
{{n^{k - 2} }}} \right)} \right) \hfill \\
\hfill \\
\end{gathered}
$$
so that the limit does not exists.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Whats maximum and minimum of $4x + y^2$ without Lagrange
Whats maximum and minimum of $4x + y^2$ without Lagrange
constraints = $2x^2 + y^2 =4$
Substitute $y^2= 4-2x^2$ into $4x+y^2$ we get $4x+4-2x^2$ manipulate it into $-2(x-1)^2 + 6$ the maximum is 6. Or using differentiation we get $x=1$ and only produces maximum value. What about the minimun? Also find the maximum in different way?
| $f(x)=4x+4-2x^2=$
$-2(x-1)^2+6$;
1)$2x^2+y^2=4$;
An ellipse: $x^2/(√2)^2+y^2/2^2=1$.
Semi minor axis : $√2$ ;
$y^2=0: x_{1,2}=\pm √2$;
Minimum:
$x=-√2$: $f_{min}(-√2)=-2(-√2-1)^2 +6=-2(1+√2)^2 +6$.
See also comments.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3546579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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A question involving inequality $x=cy+bz\ ,\ y=az+cx$ and $z=bx+ay$ has a non zero solution and atleast one of a,b,c is a proper fraction(a rational number whose absolute value is less than 1) , prove that $a^3+b^3+c^3<3$ and $abc>-1$.
For non zero solution implies , the determinant of following marix be zero because if AX=B (A$^{-1}$ should not exist for infinite solution).And there is no possibility of no solution because we already know (0,0,0) is a solution.
$$
\begin{bmatrix}
-1 & c & b \\
c & -1 & a\\
b & a & -1 \\
\end{bmatrix}
$$
Giving us $a^2+b^2+c^2+2abc=1$. After this I am totally stuck. Somehow we have to use proper fraction condition .Thanks for the help
| Let $|c|<1.$
Thus, $$1=a^2+b^2+c^2+2abc=(a+bc)^2+b^2+c^2-b^2c^2\geq b^2+c^2-b^2c^2,$$
which gives
$$(1-b^2)(1-c^2)\geq0$$ and $$b^2\leq1.$$
Similarly, $$a^2\leq1.$$
Thus, $$-1<abc<1$$ and $$a^3+b^3+c^3<3.$$
| {
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"url": "https://math.stackexchange.com/questions/3547569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\frac{a}{a+2b+c}+\frac{b}{b+2c+a}+\frac{c}{c+2a+b}\geq\frac{3}{4}$ I found the following exercise:
Prove that
$$\frac{a}{a+2b+c}+\frac{b}{b+2c+a}+\frac{c}{c+2a+b}\geq\frac{3}{4}$$
for any positive $a$, $b$, $c$.
I tried substituting the denominators but it led me nowhere. I don really have other ideas.
| Using Cauchy-Schwarz:
$$ \left(\sum_{cyc}\frac{a^2}{a^2+2ab+ca}\right)\left[\sum_{cyc}(a^2+2ab+ca)\right] \geq \left(\sum_{cyc}\frac{a}{\sqrt{a^2+2ab+ca}}\cdot \sqrt{a^2+2ab+ca}\right)^2$$
$$=(a+b+c)^2$$
Since $\displaystyle\sum_{cyc}(a^2+2ab+ca) =a^2+b^2+c^2+3(ab+bc+ca)$, we get:
$$\sum_{cyc}\frac{a^2}{a^2+2ab+ca}\geq \frac{(a+b+c)^2}{a^2+b^2+c^2+3(ab+bc+ca)}$$
Therefore
$$\sum_{cyc} \frac{a}{a+2b+c}=\sum_{cyc}\frac{a^2}{a^2+2ab+ca}\geq \frac{(a+b+c)^2}{a^2+b^2+c^2+3(ab+bc+ca)}$$
It remains to prove:
$$\frac{(a+b+c)^2}{a^2+b^2+c^2+3(ab+bc+ca)} \geq \frac{3}{4}$$
or after expanding
$$a^2+b^2+c^2\geq ab+bc+ca$$
which is
$$\frac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]\geq 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3548517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove $\int_a^b \frac{1}{x}\sqrt{-(x-a)(x-b)}dx = (\frac{a+b}{2}-\sqrt{ab})\pi$ Does anyone know how to solve $$\int_a^b \frac{1}{x}\sqrt{-(x-a)(x-b)}dx$$
After trying Wolfram Alpha, I conclude that it possibly equals $(\frac{a+b}{2}-\sqrt{ab})\pi = \frac{(\sqrt{a}-\sqrt{b})^2}{2}\pi$.
The book I am reading says one can use residue theorem to quickly obtain the result, but the details are not written and I cannot figure out the solution.
P.S. One can prove from this integral that arithmetic mean $\geq$ geometric mean.
| Noticing that $(x-a)(b-x)=\left(\frac{b-a}{2}\right)^2-\left(x-\frac{a+b}{2}\right)^2$, we let $x-\frac{a+b}{2}=\frac{b-a}{2} \sin \theta$ and transforms the integral into $$
\begin{aligned}I&=\left(\frac{b-a}{2}\right)^2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos ^2 \theta}{\frac{b-a}{2} \sin \theta+\frac{a+b}{2}} d \theta \\ &=h^2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos ^2 \theta}{h \sin \theta+k} d \theta\end{aligned}
$$
where $h=\frac{b-a}{2}$ and $k=\frac{a+b}{2}$.
$$
\begin{aligned}
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos ^2 \theta}{h \sin \theta+k} d \theta =& \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1-\sin ^2 \theta}{h \sin \theta+k} d \theta \\
=& \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(-\frac{\sin \theta}{h}+\frac{k}{h^2}+\frac{1-\frac{k^2}{h^2}}{h \sin \theta+k}\right) d \theta\\
=& {\left[\frac{\cos \theta}{h}+\frac{k}{h^2} \theta\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}+\frac{h^2-k^2}{h^2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{d \theta}{h \sin \theta+k} } \quad (*)\\=& \frac{k \pi}{h^2}-\frac{k^2-h^2}{h^2} \cdot\frac{\pi}{\sqrt{k^2-h^2}}\\=&\frac{\pi}{h^2}\left(k-\sqrt{k^2-h^2}\right)
\end{aligned}
$$
We now conclude that
$$\boxed{ \int_a^b \frac{1}{x} \sqrt{(x-a)(b-x)} d x=\left(\frac{a+b}{2}-\sqrt{a b}\right) \pi}$$
Note:
Please refer to the post for (*).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3549527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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How to prove that $\tan \left( \frac{\pi}{2} - \theta \right) = \cot \theta$ I am asked to simplify the following expression:
$$\tan \left( \frac{\pi}{2} - \theta \right)$$
The book gives me the answer $\cot \theta$ but, when I try to derive that formula using
$$\tan \left( A \pm B \right) = \frac{\tan A \pm \tan B}{1 \mp \tan A \cdot \tan B}$$
I get $\tan \frac{\pi}{2}$ as one of the terms (which is undefined).
$$\tan \left( A \pm B \right) = \frac{\tan A \pm \tan B}{1 \mp \tan A \cdot \tan B} = \frac{\tan \frac{\pi}{2} - \tan \theta}{1 + 0 \cdot \tan \theta}$$
How do I proceed?
Thank you.
| $\tan\left(\frac{\pi}{2}-\theta\right)=\frac{\sin\left(\frac{\pi}{2}-\theta\right)}{\cos\left(\frac{\pi}{2}-\theta\right)}=\frac{\sin\frac{\pi}{2} \cos\theta-\cos\frac{\pi}{2} \sin\theta}{\\cos\frac{\pi}{2}\cos\theta+\sin\frac{\pi}{2}\sin\theta}=\frac{\cos\theta}{\sin\theta}=\cot\theta.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3550289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $\left\lfloor \sum_{k=1}^n\sqrt[2k+1]{\frac{2k+1}{2k-1}}\right\rfloor$
If $n$ is a positive integer, find:
$$\left\lfloor \sum_{k=1}^n\sqrt[2k+1]{\frac{2k+1}{2k-1}}\right\rfloor$$
I computed a few values for small $n$ and it seems it is always $n$, so I think I should prove:
$$n\le \sum_{k=1}^n\sqrt[2k+1]{\frac{2k+1}{2k-1}} < n+1$$
The left part is easy, because each fraction is greater than $1$:
$$\sum_{k=1}^n\sqrt[2k+1]{\frac{2k+1}{2k-1}} > \sum_{k=1}^n 1 = n$$
but for the right I could not prove.
| We can also use AM-GM:
$$\sqrt[2k+1]{\frac{2k+1}{2k-1}}=\sqrt[2k+1]{1+\frac{2}{2k-1}} < \frac{1+\frac{2}{2k-1}+1+\ldots+1}{2k+1}$$
$$=\frac{2k+1+\frac{2}{2k-1}}{2k+1}=1+\frac{2}{(2k-1)(2k+1)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is this algebra allowed Given
$$\frac{\frac{a}{a+b}}{\frac{c}{c+d}} = 1.8 \text{ (1)}$$
and
$$\frac{a}{a+b}-\frac{c}{c+d} = \frac{2}{15} \text{ (2)}$$
and
$a = 30$
and
$a+b+c+d=220 \text{ (3)}$
Solve the remaining
This is what I did
According to $(1)$ $$\frac{a}{a+b} = 1.8 \frac{c}{c+d}$$
$$\Leftrightarrow \frac{5}{9} \cdot \frac{a}{a+b} = \frac{c}{c+d}$$
then using $(2)$
$$\frac{a}{a+b} - \frac{c}{c+d} = \frac{2}{15}$$
$$\frac{30}{30+b} - \frac{5 \times 30}{9(30+b)} = \frac{2}{15} \implies b = 70$$
using $(3)$
$(30+70) + (c+d) = 220 \implies (a+c) = 120$
going back to $(2)$
$$\frac{a}{a+b} - \frac{c}{c+d} = \frac{2}{15}$$
$$\frac{3}{10} - \frac{c}{120} = \frac{2}{15} \implies c = 20, d = 100$$
Therefore : $a = 30, b = 70, c = 20, d = 100$
Confused if some of these algebra isn't allowed like rearranging then using it in another equation like:
$$\frac{c}{c+d} = \frac{5}{9} \times \frac{a}{a+b}$$
in the initial step.
| Your answer is perfectly fine. Giving fractional algebraic values to another variable fraction is as good as substituting non-fractional variables such as $p=q$. You can always do this unless the denominator equals $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to simplify the trigonometric functions sine and cosine in a derivative? I am learning calculus and came across the following problem:
Find $\frac{dx}{d\theta}$ when $x=\sin^2\theta \cos^3\theta$
I solved this using the product rule.
$$u= \sin^2\theta \qquad v=\cos^3\theta$$
$$u'= \sin2\theta \qquad v'=-3\cos^2\theta \sin\theta$$
$$\begin{align}
\frac{dx}{d\theta}&= \sin^2\theta(-3\cos^2\theta \sin\theta) +\cos^3\theta(\sin2\theta) \\
&= -3\cos^2\theta \sin^3\theta + \cos^3\theta \sin2\theta
\end{align}$$
The problem is that this does not match any of the options the exercise provides, so how can I simplify it further? Are there common methods for simplification?
| With
$u = \sin^2\theta, \; v = \cos^3 \theta, \tag 1$
the chain rule yields
$u'(\theta) = \dfrac{d\sin^2 \theta}{d\theta} = \dfrac{d\sin^2 \theta}{d\sin \theta}\dfrac{d\sin \theta}{d\theta} = 2(\sin \theta)(\cos\theta), \tag 2$
and
$v'(\theta) = \dfrac{d\cos^3\theta}{d\cos \theta} \dfrac{d\cos \theta}{d\theta} = 3(\cos^2 \theta)(-\sin \theta) = -3(\cos^2 \theta)(\sin \theta); \tag 3$
then
$\dfrac{d((\sin^2 \theta )(\cos^3 \theta)) }{d\theta} = \dfrac{d(uv)}{d\theta} = u'(\theta)v(\theta) + u(\theta)v'(\theta)$
$= (2(\sin \theta)(\cos \theta))(\cos^3 \theta) + (\sin^2 \theta)((-3\cos^2\theta)(\sin \theta))$
$= 2(\sin \theta)(\cos^4 \theta) -3(\sin^3\theta)(\cos^2 \theta)$
$= (\sin \theta)(\cos^2 \theta)(2\cos^2 \theta - 3\sin^2 \theta), \tag 4$
which agrees with the expression given by Blue in his/her comment to the question itself.
| {
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"url": "https://math.stackexchange.com/questions/3556475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all the relative extrema of the function $f(x)=2x-24x^{1/3}$ Find all the relative extrema of the function $f(x)=2x-24x^{1/3}$
Solution:
Step 1: Find the values of $x$ where $f'(x)=0$ and $f'(x)$ DNE.
$f'(x)=2-8x^{\frac{-2}{3}}=2-\frac{8}{x^{\frac{2}{3}}}$
We can see that $f'(x)$ DNE when $x=0$. Let's figure out what values of $x$ give $f'(x)=0$
$f'(x)=2-8x^{\frac{-2}{3}}=2-\frac{8}{x^{\frac{2}{3}}}=0$
$\rightarrow x^{\frac{2}{3}}=4$
$\rightarrow $x^2=64$
$\rightarrow x= \pm 8$
Thus the critical points are $x=0,x=8,x=-8$.
Step 2: Make a number line and plot the sign of $f'(x)$ in each section to figure out where $f(x)$ is increasing and decreasing
$f'(-10) \cong 2.77 >0$
$f'(-1) \cong -6 <0$
$f'(1) \cong -6 <0$
$f'(9) \cong .151 >0$
So $f(x)$ is increasing as $x$ increasing towards -8, and then $f(x)$ decreases as $x$ approaches zero. Thus $f(-8)$ corresponds to a relative maxmimum.
Also, $f(x)$ is decreasing as $x$ increasing towards 8, and then $f(x)$ is increasing as $x$ approaches $\infty$. Thus $f(8)$ corresponds to a relative minimum.
Step 3: Find the corresponding $y$ values
$f(-8) = 32$
$f(8) = -32$
Thus we have a relative maximum at $(-8,32)$ and a relative minimum at $(8,-32)$
| Let $t=x^{1/3}$ so $x=t^3$.
Then we want extrema of $2t^3-24t$.
These occur when $6t^2-24=0$; i.e., $t=\pm2$.
Then $x=\pm8$, and $f(x)=\mp32$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3558930",
"timestamp": "2023-03-29T00:00:00",
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Finding series from its sum, then finding its reciprocal's sum If $$\sum_{r=1}^nt_r=\frac{n(n+1)(n+2)}{12}$$, then value of $$\sum_{r=1}^n\frac{1}{t_r}$$ is
Now, how can we find the series (i.e. formulae of $n^{th}$ term) from the sum?
The answer is $\frac{4n}{n+1}$
| Observe that
$$
t_n=\frac{n(n+1)(n+2)}{12}-\frac{(n-1)n(n+1)}{12}=\frac{n(n+1)}{4}
$$
and
$$
\frac{1}{t_n}=\frac{4}{n(n+1)}=\frac{4}{n}-\frac{4}{n+1},
$$
and hence
$$
\sum_{k=1}^n\frac{1}{t_k}=\left(\frac{4}{1}-\frac{4}{2}\right)+\left(\frac{4}{2}-\frac{4}{3}\right)+\cdots+
\left(\frac{4}{n-1}-\frac{4}{n}\right)+\left(\frac{4}{n}-\frac{4}{n+1}\right)\\ =\frac{4}{1}-\frac{4}{n+1}=\frac{4n}{n+1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3559514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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proving the inequality $n \cdot a^{n-1} \cdot(b-a) < b^n - a^n < n \cdot b^{n-1} \cdot (b-a)$ let $b>a>0$ and $n>1$ I need to prove that:
$n \cdot a^{n-1} \cdot(b-a) < b^n - a^n < n \cdot b^{n-1} \cdot (b-a)$
I though of proving it using induction, or to build a function on where n is the variable but failed both times.
| There are two ways to prove this.
Method 1:
Using the fact that $\dfrac{b^n-a^n}{b-a}=b^{n-1}+a^{n-2}b+\cdots+a^{n-1}$, then
$$b^{n-1}+a^{n-2}b+\cdots+a^{n-1}>a^{n-1}+a^{n-1}+\cdots+a^{n-1}=na^{n-1}\\b^{n-1}+a^{n-2}b+\cdots+a^{n-1}<b^{n-1}+b^{n-1}+\cdots+b^{n-1}=nb^{n-1} \\na^{n-1}<\dfrac{b^n-a^n}{b-a}<nb^{n-1} \\ na^{n-1}(b-a)<b^n-a^n<nb^{n-1}(b-a)$$
Method 2:
Let $f(x)=x^n$, then $f'(x)=nx^{n-1}$.
By Mean Value Theorem, we know that $f'(a)<\dfrac{f(b)-f(a)}{b-a}<f'(b)$.
$$na^{n-1}<\dfrac{b^n-a^n}{b-a}<nb^{n-1} \\ na^{n-1}(b-a)<b^n-a^n<nb^{n-1}(b-a)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3559796",
"timestamp": "2023-03-29T00:00:00",
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Find convolution of $f=x+2$ and $g=\frac{\sin\frac{7x}{2}}{\sin\frac{x}{2}}$.
I have to find the convolution of the functions $f$ and $g$ on the interval $(-\pi, \pi)$.
$$
f(x) = x+2;\ \ \ g(x) =\frac{\sin\frac{7x}{2}}{\sin\frac{x}{2}}
$$
Here is my attempt:
$$
\begin{aligned}
&(f*g)(x)=\int\limits_{-\pi}^{\pi}f(t)g(x-t)dt=\int\limits_{-\pi}^{\pi}(2+t)\cdot\frac{\sin\frac{7(x-t)}{2}}{\sin\frac{x-t}{2}}dt=I_1+I_2\ \text{where}\\
&I_1=2\int\limits_{-\pi}^{\pi}\frac{\sin\frac{7(x-t)}{2}}{\sin\frac{x-t}{2}}dt=4\cdot\int\limits_{\frac{-\pi-x}{2}}^{\frac{\pi-x}{2}}\frac{\sin 7z}{\sin z}dz\\
&I_2=\int\limits_{-\pi}^{\pi}t\cdot\frac{\sin\frac{7(x-t)}{2}}{\sin\frac{x-t}{2}}dt
\end{aligned}
$$
The problem is to evaluate $I_1$ and especially $I_2$. Perhaps there is an easier way of cracking this problem?
| You can first evaluate Fourier series for $f(x)$ and then use its relation to the Dirichlet kernel:$$S_n(x)=\int_{-\pi}^{\pi}f(x)D_n(x-t)dt$$ where $D_n(t)=\frac{1}{2\pi}\frac{\sin((n+\frac{1}{2})t)}{\sin(\frac{t}{2})}$ and $S_n(x)$ is n-th partial sum of Fourier series.
In your case, $g(t)=\frac{\sin\frac{7t}{2}}{\sin\frac{t}{2}}$ equals $2\pi \cdot D_3(t)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3560140",
"timestamp": "2023-03-29T00:00:00",
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Compute $\lim_{n\to\infty}(1+a)(1+a^2)(1+a^4)...(1+a^{2^n})$? Compute
$$\lim_{n\to\infty}(1+a)(1+a^2)(1+a^4)...(1+a^{2^n})$$
Where $|a|<1$.
| $$(1-a)f=(1-a)(1+a)(1+a^2)(1+a^4)(1+a^8)...(1+a^{2^n})=(1-a^{2^{n+1}})$$
Use $(x-y)(x+y)=x^2-y^2$ sucsessively to get
$$f_n=\frac{1-a^{2^{n
+1}}}{1-a}.$$
As $|a|<1$
$$\lim_{n \rightarrow \infty} f_n= \frac{1}{1-a}.$$
| {
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Find where $f(x)=\frac{x^2+4}{x^2-4}$ is concave upwards and concave downwards Find where $f(x)=\frac{x^2+4}{x^2-4}$ is concave upwards and concave downwards
Solution: We are going to take the second derivative, find the critical points, and then test each region. Lets get to the second derivative by doing the quotient rule twice:
$f'(x) = \frac{(x^2-4)\frac{d}{dx}(x^2+4)-(x^2+4)\frac{d}{dx}(x^2-4)}{(x^2-4)^2}$
$=\frac{(x^2-4)(2x)-(x^2+4)(2x)}{(x^2-4)^2}$
$=\frac{-16x}{(x^2-4)^2}$
Okay, theres the first derivative. Let's do it again!!
$f''(x)=\frac{(x^2-4)^2 \frac{d}{dx}(-16x) - (-16x)\frac{d}{dx}(x^2-4)^2}{(x^2-4)^4}$
Note that we have to the chain rule to evluate $\frac{d}{dx}(x^2-4)^2$.
$\rightarrow f''(x)=\frac{(x^2-4)^2 (-16) - (-16x)(2(x^2-4)\frac{d}{dx}(x^2-4))}{(x^2-4)^4}$
$\rightarrow =\frac{(x^2-4)^2 (-16) - (-16x)(2(x^2-4)(2x)}{(x^2-4)^4}$
$\rightarrow =\frac{(x^2-4)^2 (-16) + 64x^2(x^2-4)}{(x^2-4)^4}$
Canceling off one of the $(x^2-4)$ from each term in the numerator and the denominator...
$\rightarrow =\frac{(x^2-4) (-16) + 64x^2}{(x^2-4)^3}$
$\rightarrow =\frac{-16x^2+64+64x^2}{(x^2-4)^3}$
$\rightarrow =\frac{64+48x^2}{(x^2-4)^3}$
Few.. okay the hard part is over. Now we need to find the points to chop our number line up from, so we have to solve $f''(x)=0$ and find where $f''(x)$ DNE.
Let's solve where $f''(x)=0$
$f''(x) =\frac{64+48x^2}{(x^2-4)^3}=0$
$\rightarrow 64+48x^2 = 0$
$\rightarrow x= \pm \sqrt{-64/48}$
So $f''(x)=0$ only when $x$ is imaginary... so those values are irrelevant.
On the other hand, $f''(x)$ DNE when $x = \pm 2$, because the denominator will be zero. So these are the values we split our number line up by.
Since $f''(-3)=\frac{496}{125}>0$, we have that $f(x)$ is concave up on $(-\infty,-2)$.
Since $f''(0)=-1<0$, we have that $f(x)$ is concave down on $(-2,2)$.
Since $f''(3)=\frac{496}{125}>0$, we have that $f(x)$ is concave up on $(2,\infty)$.
| It's good to keep in mind that the critical points of $f''(x)$ are places where either $f''(x)=0$ or $f''(x)$ is undefined. Hence concavity can change at places where $f''(x)=0$ and at places where $f''(x)$ is undefined.
We were given that
$$f(x)=\frac{x^2+4}{x^2-4},$$
and you correctly calculated that
$$f''(x)=\frac{48x^2+64}{(x^2-4)^3}.$$
Note that there is no $x$ with $f''(x)=0$, but $f''(x)$ is undefined at $-2$ and $2$. So it is enough to check the concavity in the intervals $(-\infty,-2)$, $(-2,2)$, and $(2,\infty)$.
Since $f''(-3)=\frac{496}{125}>0$, we have that $f(x)$ is concave up on $(-\infty,-2)$.
Since $f''(0)=-1<0$, we have that $f(x)$ is concave down on $(-2,2)$.
Since $f''(3)=\frac{496}{125}>0$, we have that $f(x)$ is concave up on $(2,\infty)$.
| {
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Can a sum arithmetic square ever equal to power of two? Does there exist any $2^t,\ t\in\mathbb{Z}_+$ which can be express as Sum of arithmetic square number.
Or
Can it be shown that
$$\begin{split}2^t&\ne \sum_{q=0}^u (n+qd)^2=n^2+(n+d)^2+(n+2d)^2+\cdots+(n+ud)^2\end{split}$$
Where $n,u,d\in\mathbb{Z}_+$
Example: let $n=5,u=3,d=2$ so $5^2+7^2+9^2+11^2=276\ne 2^t$
$2^8= 1^2+5^2+7^2+9^2+10^2$ but $1,5,7,9,10 $ are not in arithmetic progression.
My incomplete attempt
Let $n,u,d\in\mathbb{Z}_+$
$$\begin{split}\sum_{q=0}^u (n+qd)^2 &=n^2+(n+d)^2+(n+2d)^2+\cdots+(n+ud)^2\\ &=n^2(u+1)+\frac{(u+1)u}{2}(2nd+d^2)+\frac{(u+1)u(u-1)}{3}d^2 \end{split}$$
Let
$$\begin{split}2^t &=\sum_{q=0}^u (n+qd)^2 \\ \implies 3\cdot 2^{t+1}&=6n^2(u+1)+3(u+1)u(2nd-d^2)+(u+1)u(u-1)2d^2 \\ &= (u+1)(6n^2+3u(2nd+d^2)+u(u-1)2d^2)\\ &(in\ case,\ u+1= 3) \\
\implies 2^t&= 3n^2+3(2nd+d^2)+2d^2\\ &= n^2+(n+1)^2+(n+2d)^2 \end{split}$$
Now we need to simplify for case, $6n^2+3u(2nd+d^2)+u(u-1)2d^2=3\cdot2^x$ and $u+1=2^y$ where $x+y=t+1$ but I'm stuck here. Thank you.
Related post:
Can a sum of consecutive $n$th powers ever equal a power of two?
Recent post
Can a Sum of distinct squares ever equal power of two?
| Let's focus on the case $n=d=1$.
$$\sum_{q=0}^u (n+qd)^2=\sum_{q=0}^u (1+q)^2=(u+1)+u(u+1)+\frac{u(u+1)(2u+1)}6$$
$$=(u+1)\frac{6+6u+u(2u+1)}6$$
Therefore, we have two cases: $\,u=2^k-1\,$ or $\,u=3\cdot 2^k-1$, for a certain $\,k$.
First case ($\,u=2^k-1\,$):
$$\sum_{q=0}^u (1+q)^2=2^{k-1}\frac{6+6(2^k-1)+(2^k-1)(2(2^k-1)+1)}3=2^{k-1}\frac{6\cdot2^k+(2^k-1)(2^{k+1}-1)}3=2^{k-1}\frac{(3+2^{k+1})2^k+1}3$$
And so, necessarily,
$$(3+2^{k+1})2^k+1=3\cdot2^t$$
for a certain $\,t$. But this is impossible, because the two sides have different parity.
Second case ($\,u=3\cdot 2^k-1\,$):
$$\sum_{q=0}^u (1+q)^2=2^{k-1}(6+6(3\cdot 2^k-1)+(3\cdot 2^k-1)(2(3\cdot 2^k-1)+1))=2^{k-1}(9\cdot 2^k(1+2^{k+1})+1)$$
So, in this case, necessarily
$$9\cdot 2^k(1+2^{k+1})+1=2^t$$
for a certain $\,t$. But, for the same reason as above, this is impossible.
I think that this argument can be extended to your general case.
| {
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} |
Finding the MacLaurin series of $\frac{x+3}{2-x}$ I did:
$$\frac{x+3}{2-x} = \frac{x}{2-x}+\frac{3}{2-x} = x(\frac{1}{2-x})+3(\frac{1}{2-x}) = \\
= \frac{x}{2}(\frac{1}{1-\frac{x}{2}})+\frac{3}{2}(\frac{1}{1-\frac{x}{2}}) = \\
= \frac{x+3}{2}\sum(\frac{x}{2})^n = ...?$$
The answer my professor got was $\frac{(3 + x)}{(2 - x)} = \frac{3}{2} + \sum_{n=1}^∞ \frac{x^n5}{ 2^{n+1}}$
Unfortunately I forgot to write down how he did it. How do I get that solution and is mine necessarily incorrect/incomplete?
| \begin{align}
\frac{x+3}2\sum_{n=0}^\infty\left(\frac{x}2\right)^n
&=\frac{x}2\sum_{n=0}^\infty\frac{x^n}{2^n}+\frac32\sum_{n=0}^\infty\frac{x^n}{2^n}\\
&=\sum_{n=0}^\infty\frac{x^{n+1}}{2^{n+1}}+\sum_{n=0}^\infty\frac{3x^n}{2^{n+1}}\\
&=\overbrace{\sum_{n=1}^\infty\frac{x^n}{2^n}}^{n+1\,\mapsto\, n}+\overbrace{\left(\frac32+\sum_{n=1}^\infty\frac{3x^n}{2^{n+1}}\right)}^{\text{remove constant term }(n=0)}\\
&=\frac32+\sum_{n=1}^\infty\left(\frac{x^n}{2^n}+\frac{3x^n}{2^{n+1}}\right)\\
&=\frac32+\sum_{n=1}^\infty\frac{5x^n}{2^{n+1}}\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3570936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Given positives $a, b, c$ such that $a + b + c = 3$, prove that $\sum_{cyc}\frac{1}{a^2 + 4b^2 + c^2} \le \frac{1}{2}$.
Given positives $a, b, c$ such that $a + b + c = 3$, prove that $$\frac{1}{c^2 + 4a^2 + b^2} + \frac{1}{a^2 + 4b^2 + c^2} + \frac{1}{b^2 + 4c^2 + a^2} \le \frac{1}{2}$$
We have that $$a^2 + 4b^2 + c^2 = a^2 + (a + b + c + 1)b^2 + c^2 = (b^2 + a)a + (b + 1)b^2 + (b^2 + c)c$$
$$\implies \sum_{cyc}\frac{1}{a^2 + 4b^2 + c^2} = \frac{1}{(a + b + c)^2} \cdot \sum_{cyc}\frac{(a + b + c)^2}{(b^2 + a)a + (b + 1)b^2 + (b^2 + c)c}$$
$$ \le \frac{1}{(a + b + c)^2} \cdot \sum_{cyc}\left(\frac{a}{b^2 + a} + \frac{1}{b + 1} + \frac{c}{b^2 + c}\right)$$
Furthermore, $\dfrac{a}{c^2 + a} + \dfrac{c}{a^2 + c} = 1 - \dfrac{(c + a - 2)ca}{(c^2 + a)(a^2 + c)}$, $$\sum_{cyc}\frac{1}{a^2 + 4b^2 + c^2} \le \frac{1}{(a + b + c)^2} \cdot \sum_{cyc}\left[\frac{1}{b + 1} - \frac{(c + a - 2)ca}{(c^2 + a)(a^2 + c)}\right] + \frac{1}{3}$$
Then I don't know what to do next.
| Another way.
We'll prove that our inequality is true for any real $a$, $b$ and $c$ such that $a+b+c=3.$
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v^2$ can be negative, and $abc=w^3$.
Thus, by my first proof we need to prove that:
$$\sum_{sym}(a^6-4a^5b+13a^4b^2-2a^4bc-6a^3b^3-12a^3b^2c+10a^2b^2c^2)\geq0$$ or
$$27w^6+A(u,v^2)w^3+B(u,v^2)\geq0,$$ where $A$ and $B$ are polynomials of $u$ and $v^2$.
We'll prove that even the following inequality is true.
$$\sum_{sym}(a^6-4a^5b+13a^4b^2-2a^4bc-6a^3b^3-12a^3b^2c+10a^2b^2c^2)\geq$$
$$\geq\frac{1}{3}\left(\sum_{cyc}(a^3-a^2b-a^2c+abc)\right)^2.$$
Since $$\sum_{cyc}(a^3-a^2b-a^2c+abc)=27u^3-27uv^2+3w^3-9uv^2+3w^3+3w^3$$ and $$27-\frac{1}{3}\cdot9^2=0,$$ we see that the last inequality is a linear inequality of $w^3$.
Thus, it's enough to prove the last inequality for extreme value of $w^3$,
which happens for equality case of two variables.
Since the last inequality is homogeneous, it's enough to assume $b=c=1$, which gives
$$2(a-1)^4(a^2+5)\geq\frac{1}{3}(a-1)^4(a+2)^2$$ or
$$(a-1)^4(5a^2-4a+26)\geq0,$$ which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $x^4+12x-5$ has roots $x_1,x_2,x_3,x_4$ find polynomial with roots $x_1+x_2,x_1+x_3,x_1+x_4,x_2+x_3,x_2+x_4,x_3+x_4$ I have the polynomial $x^4+12x-5$ with the roots $x_1,x_2,x_3,x_4$ and I want to find the polynomial whose roots are $x_1+x_2,x_1+x_3,x_1+x_4,x_2+x_3,x_2+x_4,x_3+x_4$.
I found the roots $x_1=-1+\sqrt{2},x_2=-1-\sqrt{2},x_3=1-2i,x_4=1+2i$. And after long computations the polynomial is $x^6+20x^2-144$. Are there clever way to find it?
| Let $s_1,p_1$ the sum and product of any two roots and $s_2,p_2$ the sum and product of the other two roots. From Vieta's:
$$
\begin{cases}
s_1+s_2=0 \\
s_1s_2+p_1+p_2=0\\
p_1s_2+p_2s_1=-12\\
p_1p_2=5
\end{cases}
$$
Substitute $s_2=-s_1$
$$
\begin{cases}
p_1+p_2=s_1^2\\
-p_1+p_2=\frac{12}{s_1}\\
p_1p_2=5
\end{cases}
$$
or
$$
\begin{cases}
p_1=\frac{1}{2}\left(s_1^2+\frac{12}{s_1}\right)\\
p_2 =\frac{1}{2}\left(s_1^2-\frac{12}{s_1}\right)\\
p_1p_2=5
\end{cases}
$$
Replace in the last equation:
$$\frac{1}{4}\left(s_1^2+\frac{12}{s_1}\right)\left(s_1^2-\frac{12}{s_1}\right)=5$$
or equivalently $s_1^6+20s_1-144=0$. Since $s_1$ can be the sum of any two roots, that means each of $x_1+x_2,x_1+x_3,x_1+x_4,x_2+x_3,x_2+x_4,x_3+x_4$ is a root of $X^6+20X-144$ and there are not other roots. Of course, any other polynomial $a(X^6+20X-144),\ a\in\mathbb{R}$ satisfies the requirements as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3574530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Prove $\cos x +\cos y = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$ Prove that $$\cos(x) + \cos(y) = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$$ holds true for any $x, y \in \mathbb{R}$.
Even though I managed to prove its brother $\sin(x) + \sin(y)$, I haven't been able to tackle this one.
Important identities needed for the proof:
$$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$$
$$\cos(x-y)=\cos(x)\cos(y)+\sin(x)\sin(y)$$
Let's go:
$$2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2}) = 2\cos(\frac{x}{2}+ \frac{y}{2})\cos(\frac{x}{2} - \frac{y}{2}) = $$
$$ = 2(\cos(\frac{x}{2})\cos(\frac{y}{2}) - \sin(\frac{x}{2})\sin(\frac{y}{2}))(\cos(\frac{x}{2})\cos(\frac{y}{2}) + \sin(\frac{x}{2})\sin(\frac{y}{2})) = $$
$$ = 2(\cos^2(\frac{x}{2})\cos^2(\frac{y}{2}) - \sin^2(\frac{x}{2})\sin^2(\frac{y}{2})) = $$
$$ = 2\cos^2(\frac{x}{2})\cos^2(\frac{y}{2}) - 2\sin^2(\frac{x}{2})\sin^2(\frac{y}{2})$$
Now I tried, I believe, almost every possible replacement by deriving from $\cos^2(x) + \sin^2(x) = 1$ and sadly nothing worked.
| Add these equations:
$\cos x=\cos\dfrac{x+y}2\cos\dfrac{x-y}2-\sin\dfrac{x+y}2\sin\dfrac{x-y}2$
$\cos y=\cos\dfrac{x+y}2\cos\dfrac{x-y}2+\sin\dfrac{x+y}2\sin\dfrac{x-y}2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3576912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to find the first four terms of the maclaurin series of $f(x) = e^{e^x}$ I tried substituting the maclaurin series for $e^x$ into the equation, but that doesn’t seem to give the right answer.
I’m not sure what do besides just taking the derivatives.
| We know
$$
e^x = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots
$$
so then
$$\begin{align}
\exp(e^x) &= \exp\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots\right)
\\ &= e^1 e^x e^{x^2/2} e^{x^3/6}\dots
\\
&= e \left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots\right)
\left(1+\frac{x^2}{2}+\frac{x^4}{8}+\dots\right)
\left(1+\frac{x^3}{6}+\dots\right)\cdots
\\ &=
e\left(1+x+x^2+\frac{5}{6} x^3 + \dots\right)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3577668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Use epsilon-delta definition of limit to establish the following: $\displaystyle\lim_{x\to 1}\frac{1}{2+\sqrt{x}}=\frac{1}{3}$ I understand that my solution here is probably not the most efficient (My professor's solution is "cleaner") but it is how my mind attacked the problem. I have been losing lots of points for minor details that I've been unable to see. Does the following proof hold? Am I making any major (or minor) errors?
\begin{align*}
\left| \frac{1}{2+\sqrt{x}}-\frac{1}{3}\right|&\leq\left|\frac{1}{2+\sqrt{x}}\right|+\left|\frac{1}{3}\right|<\epsilon~~~\mbox{(by triangle inequality)}\\
&\implies\left|\frac{1}{2+\sqrt{x}}\right|+\frac{1}{3}<\epsilon\\
&\implies\left|\frac{1}{2+\sqrt{x}}\right| < \epsilon-\frac{1}{3}\\
&\implies \frac{1}{2} < \epsilon-\frac{1}{3}~~~~\mbox{(Because, }\sqrt{x}~\mbox{only a real number when } x\geq 0.)\\
&\implies 1<2(\epsilon-\frac{1}{3})\\
&\implies \left|x-1\right|<2\epsilon-\frac{2}{3}=\delta~~~~\mbox{(Because, choosing }x~s.t.~0<x<2\implies~-1<x-1<1)\\
\end{align*}
$\therefore \left|x-1\right|<\delta\implies\left| \frac{1}{2+\sqrt{x}}-\frac{1}{3}\right|<\epsilon$ and $\displaystyle\lim_{x\to 1}\frac{1}{2+\sqrt{x}}=\frac{1}{3}$
| I think the following salvages the initial strategy while still holding true to the mathematical principals of epsilon-delta definition of limits. Feedback welcome. Please let me know if I have made another error. I posted this as an answer so that the original question could remain for reference.
\begin{align*}
\left| \frac{1}{2+\sqrt{x}}-\frac{1}{3}\right|&\leq\left|\frac{1}{2+\sqrt{x}}\right|+\left|\frac{1}{3}\right|<\epsilon~~~\mbox{(by triangle inequality)}\\
&\implies\left|\frac{1}{2+\sqrt{x}}\right| < \epsilon~~~~(|a|+|b|<\epsilon\implies|a|<\epsilon)\\
&\implies \frac{1}{2} < \epsilon~~~~\mbox{(Because, }\sqrt{x}~\mbox{only a real number when } x\geq 0.~\mbox{Hence, } \delta\leq 1)\\
&\implies 1<2\epsilon\\
&\implies |x-1|\leq\delta\implies-1\leq x-1\leq 1\implies 0\leq x\leq 2\\
&~~~~~~~~~~~~~~~~~~~~(\delta\mbox{-neighborhood of }x~\mbox{is }1~\mbox{or less.})\\
&\implies \left|x-1\right|<2\epsilon=\delta~~~~\mbox{(Because, }|x-1|\leq 1)\\
\end{align*}
$\therefore~\forall~\delta=\inf(1,2\epsilon),~\left|x-1\right|<\delta\implies\left| \frac{1}{2+\sqrt{x}}-\frac{1}{3}\right|<\epsilon$ and $\displaystyle\lim_{x\to 1}\frac{1}{2+\sqrt{x}}=\frac{1}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3578018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Confused about positive and negative signs: Find the value of $\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$. Without tables or a calculator, find the value of $\displaystyle\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$.
I do not understand how the positive/negative signs are obtained as shown in the book; is there a formula for expanding these kind of things (what kind of expression is it, by the way?)?
This is my solution:
$\displaystyle\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$
$= \displaystyle\frac{[(\sqrt5+2)^3+(\sqrt5-2)^3][(\sqrt5+2)^3-(\sqrt5-2)^3]}{8\sqrt5}$
$=\displaystyle\frac{(\sqrt5+2+\sqrt5-2)[(\sqrt5+2)^2\color{red}{+}(\sqrt5+2)(\sqrt5-2)+(\sqrt5-2)^2](\sqrt5+2-\sqrt5+2)[(\sqrt5+2)^2\color{red}{-}(\sqrt5+2)(\sqrt5-2)+(\sqrt5-2)^2]}{8\sqrt5}$
$=\displaystyle\frac{[2\sqrt5(5+4\sqrt5+4+\color{red}{5-4}+5-4\sqrt5+4][4(5+4\sqrt5+4\color{red}{-(5-4)}+(5-4\sqrt5+4)]}{8\sqrt5}$
$=\displaystyle\frac{2584\sqrt5}{8\sqrt5}$
$=323$
Because of the multiplication, I still got the same answer as given in the book. However, is the book or I correct in terms of the positive/negative signs(in red)?
| It might be interesting that you may avoid tedious calculations with roots if you use recurrence relations:
*
*Set $t_1 =2+\sqrt 5$ and $t_2 = 2-\sqrt 5$.
So, the searched for value is
$$\frac{t_1^6-t_2^6}{8\sqrt{5}}$$
This is $a_6$ in the recurrence relation
$$a_{n+2} - (t_1+t_2)a_{n+1} + t_1t_2 = a_{n+2} - 4a_{n+1} -a_n $$
with
$$a_0 = 0 \text{ and } a_1 = \frac{t_1-t_2}{8\sqrt{5}}=\frac 14$$
Now, calculating recursively you get
$$a_6 = 4\left(76+\frac 14\right)+18 = 323$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3578191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
$\left| \frac{1}{z}+ \sum_{n=1}^\infty \frac{2z}{z^2-n^2}\right| \le C + C \sum_{n=1}^\infty \frac{y}{y^2+n^2}$ for $y>1$ and $|x|\le 1/2$. Suppose $|x| \le \frac{1}{2}$, and $y>1$ for $z=x+iy$.
Then we have
$$\frac{1}{z} + \sum_{n=1}^\infty \frac{2z}{z^2-n^2} = \frac{1}{x+iy} + \sum_{n=1}^\infty \frac{2(x+iy)}{x^2-y^2-n^2+2ixy};$$ From this I would like to get the bound
$$\Bigg| \frac{1}{z}+ \sum_{n=1}^\infty \frac{2z}{z^2-n^2}\Bigg| \le C + C \sum_{n=1}^\infty \frac{y}{y^2+n^2}.$$
It is easy to deal with the first term since $|x+iy| \ge |y| > 1$.
However, I am having difficulty with the denominator of the second term.
We have $|x^2-y^2-n^2+2ixy| \ge |x^2-y^2-n^2|$. How could I get rid of $x^2$ to get a lower bound of a multiple of $y^2+n^2$?
| The idea is that under these restrictions, $z^2 \approx -y^2$ lies near the negative real axis, so that $|z^2 -n^2| \approx y^2 + n^2$.
Now the precise estimate: You have $x^2 \le \frac 14 \le \frac{y^2}{4}$, so that
$$
|z^2-n^2| \ge |\operatorname{Re}(z^2-n^2)| = y^2+n^2-x^2 \ge \frac 34 (y^2+n^2)
$$
and
$$
|2z| = 2 \sqrt{x^2+y^2} \le 2y \sqrt{\frac 5 4}
$$
It follows that
$$
\frac{2|z|}{|z^2-n^2|} \le \frac{4\sqrt{5}}{3} \frac{y}{y^2-n^2}
$$
and therefore
$$
\Bigg| \frac{1}{z}+ \sum_{n=1}^\infty \frac{2z}{z^2-n^2}\Bigg| \le C + C \sum_{n=1}^\infty \frac{y}{y^2+n^2}
$$
with $C = \frac{4\sqrt{5}}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3578505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show $\sum_{k=0}^{n}2^{n-k}\binom{a+k}{k}\frac{a-k}{a+k}=\binom{a+n}{n}$ How it can be shown that:
$$\sum_{k=0}^{n}2^{n-k}\binom{a+k}{k}\frac{a-k}{a+k}=\binom{a+n}{n}$$
Where $a \ne 0$
My try:
$$\sum_{k=0}^{n}2^{n-k}\binom{a+k}{k}\frac{a-k}{a+k}=\sum_{k=0}^{n}2^{n-k}\binom{a+k}{a}\frac{a-k}{a+k}$$
$$=\frac{1}{a}\sum_{k=0}^{n}2^{n-k}\binom{a+k-1}{a-1}\left(a-k \right)$$
$$=\sum_{k=0}^{n}2^{n-k}\binom{a+k-1}{a-1}-\frac{1}{a}\color{red}{\sum_{k=0}^{n}2^{n-k}\binom{a+k-1}{a-1}k }$$
On the other hand:
$$\color{red}{\sum_{k=0}^{n}2^{n-k}\binom{a+k-1}{a-1}k}=\sum_{k=0}^{n}\left(a+k-1\right)2^{n-k}\binom{a+k-2}{k-1} $$
But computing these expressions takes much time and I think there should be a better way, but I cannot find that way.
Please if it's possible, then do the proof using elementary ways.
| Here is an algebric proof :
Let $ a $ be a real, let's define the function $ f_{a} $ as follows : $$ \left(\forall x\in\left]-\frac{1}{2},\frac{1}{2}\right[\right),\ f_{a}\left(x\right)=\displaystyle\frac{1-2x}{\left(1-x\right)^{a+1}} $$
know : \begin{aligned} \left(\forall x\in\left]-\frac{1}{2},\frac{1}{2}\right[\right),\ f_{a}\left(x\right)&=\displaystyle\frac{2}{\left(1-x\right)^{a}}-\displaystyle\frac{1}{\left(1-x\right)^{a+1}}\\ &=2\displaystyle\sum_{n=0}^{+\infty}{\left(-1\right)^{n}\displaystyle\binom{-a}{n}x^{n}}-\displaystyle\sum_{n=0}^{+\infty}{\left(-1\right)^{n}\displaystyle\binom{-a-1}{n}x^{n}} \end{aligned}
Observe that forall $ \left(x,n\right)\in\mathbb{R}\times\mathbb{N} $ :
$ \ \ \ \ \ \ \ \ \ \ \left(-1\right)^{n}\binom{-x}{n}=\frac{\left(-1\right)^{n}}{n!}\prod\limits_{k=0}^{n-1}{\left(-x-k\right)}=\frac{1}{n!}\prod\limits_{k=0}^{n-1}{\left(x+k\right)}=\frac{1}{n!}\prod\limits_{k=0}^{n-1}{\left(x+n-1-k\right)}=\binom{x+n-1}{n} $
Thus, \begin{aligned}\left(\forall x\in\left]-\frac{1}{2},\frac{1}{2}\right[\right),\ f_{a}\left(x\right)=\displaystyle\sum_{n=0}^{+\infty}{\left(2\displaystyle\binom{a+n-1}{n}+\displaystyle\binom{a+n}{n}\right)x^{n}}\end{aligned}
Since $ \binom{a+n-1}{n}=\frac{a}{a+n}\binom{a+n}{n} $, we get, \begin{aligned}\left(\forall x\in\left]-\frac{1}{2},\frac{1}{2}\right[\right),\ f_{a}\left(x\right)=\displaystyle\sum_{n=0}^{+\infty}{\displaystyle\frac{a-n}{a+n}\displaystyle\binom{a+n}{n}x^{n}}\end{aligned}
Observe $ \left(\forall x\in\left]-\frac{1}{2},\frac{1}{2}\right[\right),\ \frac{f_{a}\left(x\right)}{1-2x}=\frac{1}{\left(1-x\right)^{a+1}} $, meaning that forall $ x \in\left]-\frac{1}{2},\frac{1}{2}\right[ $, \begin{aligned} \left(\displaystyle\sum_{n=0}^{+\infty}{\displaystyle\frac{a-n}{a+n}\displaystyle\binom{a+n}{n}x^{n}}\right)\left(\displaystyle\sum_{n=0}^{+\infty}{2^{n}x^{n}}\right)&=\displaystyle\sum_{n=0}^{+\infty}{\left(-1\right)^{n}\displaystyle\binom{-a-1}{n}x^{n}} \\ \iff \displaystyle\sum_{n=0}^{+\infty}{\left(\displaystyle\sum_{k=0}^{n}{2^{n-k}\displaystyle\binom{a+k}{k}\displaystyle\frac{a-k}{a+k}}\right)x^{n}}&=\displaystyle\sum_{n=0}^{+\infty}{\displaystyle\binom{a+n}{n}x^{n}}\end{aligned}
Which leads to the following : $$ \left(\forall n\in\mathbb{N}\right),\ \displaystyle\sum_{k=0}^{n}{2^{n-k}\displaystyle\binom{a+k}{k}\displaystyle\frac{a-k}{a+k}}=\displaystyle\binom{a+n}{n} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3579670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Interpreting an olympiad inequality as a convex function We need to prove the following inequality:
$$\frac{y^3xz}{x^3(xy+z^2)}+\frac{z^3xy}{y^3(x^2+yz)}+\frac{x^3yz}{z^3(xz+y^2)}\geq \frac{3}{2}$$
This equation is convex in each of the variables $x,y,z$. Moreover, its minimum seems to be at the origin (the equation is homogeneous, and there are not displacements of the vertex).
Can I then say that the minima is achieved when $x=y=z$? Clearly, when $x=y=z$, the value of the inequality is $\frac{3}{2}$
| By C-S and AM-GM we obtain: $$\sum_{cyc}\frac{x^3yz}{z^3(y^2+xz)}=\sum_{cyc}\frac{x^4y^2}{z^2xy(y^2+xz)}\geq\frac{\left(\sum\limits_{cyc}x^2y\right)^2}{\sum\limits_{cyc}z^2xy(y^2+xz)}=$$
$$=\frac{\left(\sum\limits_{cyc}x^2y\right)^2}{xyz\sum\limits_{cyc}(y^2z+z^2x)}=\frac{\left(\sum\limits_{cyc}x^2y\right)^2}{2xyz\sum\limits_{cyc}x^2y}=\frac{1}{2}\sum_{cyc}\frac{x}{z}\geq\frac{3}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate: $S=\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}$ Evaluate of this sum:
$$S=\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}$$
Expand out the sum:
$$S=\prod_{k=1}^{1}\frac{2k}{k+2}+\prod_{k=1}^{2}\frac{2k}{k+3}+\prod_{k=1}^{3}\frac{2k}{k+4}+\cdots$$
$$S=\frac{2}{3}+\frac{2}{4}\cdot\frac{4}{5}+\frac{2}{5}\cdot\frac{4}{6}\cdot\frac{6}{7}+
\frac{2}{6}\cdot\frac{4}{7}\cdot\frac{6}{8}\cdot\frac{8}{9}+\cdots+\frac{2^nn!}{(2n)!\div (n+1)!}$$
I don't know what to do next...
| Correcting $(2n)!$ to $(2n+1)!$ in OP's nth term, we have
$$S=\sum_{n=1}^{\infty} \frac{2^n n! (n+1)!}{(2n+1)!} =\sum_{n=1}^{\infty}\frac{2^n n ~\Gamma(n) \Gamma(n+2)}{\Gamma(2n+2)}=\sum_{n=1}^{\infty}2^n n~B(n,n+2)= \sum_{n=1}^{\infty} 2n\int_{0}^{\pi/2} 2^n \sin^{2n-1} x ~\cos^{2n+3} xdx.$$
$$S=\int_{0}^{\pi/2} \sqrt{2}\cos^4 x \sum_{n=1}^{\infty} n~(\sqrt{2} \sin x \cos x)^{2n-1}$$
Using the infinite GP result that $\sum_{1}^{\infty} n ~z^{2n-1} =\frac{z}{(1-z^2)^2}$ We get
$$S=\int_{0}^{\pi/2} \sqrt{2} \cos^4 x \frac{\sqrt{2} \sin x \cos x}{(1-2\sin^2 x\cos^2 x)^2} dx=
\int_{0}^{\pi/2} \frac{2 \sin x \cos^5x}{(1-2\sin^2 x \cos^2 x)^2} dx$$
Next using $\int_{0}^{a} f(x) dx= \int_{0}^{a} f(a-x) dx,$
we get $$S=\int_{0}^{\pi/2} \frac{2 \sin^5 x \cos x}{(1-2\sin^2 x \cos^2 x)^2} dx$$ Adding the last two integrals, we get
So $$2S=\int_{0}^{\pi/2} \frac{2\sin x \cos x(\sin^4 x+ \cos^4 x)}{(1-2\sin^2 x \cos^2 x)^2}=\int_{0}^{\pi/2} \frac{2\sin x \cos x}{(1-2\sin^2 x \cos^2 x)}dx=\int_{0}^{\pi/2} \frac{4\sin 2x dx}{1+\cos^2 2x}$$
$$\implies 2S=8\int_{0}^{\pi/4}\frac{\sin 2x dx}{1+\cos^2 2x}=-4\int_{1}^{0}\frac{dt}{1+t^2}=\pi \implies S=\frac{\pi}{2}$$
Lastly we have used $\cos 2x=t.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3585115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\lim\limits_{n\to \infty}\frac{1^p+3^p+\dots+(2n-1)^p}{n^{p+1}}$
Evaluate
$\lim\limits_{n\to \infty}\frac{1^p+3^p+\dots+(2n-1)^p}{n^{p+1}}$ using Stolz-Cesaro theorem .
Now this is my attempt :$\lim\limits_{n\to \infty}\frac{1^p+3^p+\dots+(2n-1)^p+(2n)^p-1^p-3^p-\dots-(2n-1)^p}{(n+1)^{p+1}-n^{p+1}}$=$\lim\limits_{n\to \infty}\frac{(2n)^p}{(n+1)^{p+1}-n^{p+1}}$ (and now i was thinking to use the binomial theorem )
$\lim\limits_{n\to \infty}\frac{(2n)^p}{n^{p+1}+{p+1 \choose p}n^p+\dots+1-n^{p+1}}$,which will eventually lead to the answer : $\frac{2}{p+1}$ . Is this correct ?
| No, not correct.
\begin{align*}
&\quad \ \ \frac {(2n)^p} {(n+1)^{p+1} - n^{p+1}} \\
&=\frac 1 n \cdot \frac {2^p} {(1 + 1/n)^{p+1} - 1} \\
&\sim \frac 1 n \cdot \frac {2^p} {(p+1)/n} \\
&= \frac {2^p} {p+1}. \tag {$n \to \infty$}
\end{align*}
$p$ is a real positive number, so generally the binomial theorem fails. Instead, a frequently used equivalent infinitesimal would work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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