Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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AM-GM inequality: $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \geq a + b + c + d$
Let $a, b, c, d > 0$. Prove that $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \geq a + b + c + d$.
I'm supposed to prove this by AM-GM, but I can't see how. Any help would be appreciated.
| By AM-GM $\frac{a^2}{b}+b\geq2a$.
Thus, $\sum\limits_{cyc}\frac{a^2}{b}\geq\sum\limits_{cyc}(2a-b)=\sum\limits_{cyc}a$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove that:$\sum_{n=0}^{\infty}{2^{n+3}(n^2+n+\phi)\over (n+1)(2n+1)(2n+3){2n\choose n}}=\phi\pi^2+8\phi^2\pi-8\phi^3\sqrt{5}$ $\phi$ is the golden ratio
$$\sum_{n=0}^{\infty}{2^{n+3}(n^2+n+\phi)\over (n+1)(2n+1)(2n+3){2n\choose n}}=\phi\pi^2+8\phi^2\pi-8\phi^3\sqrt{5}$$
I try:
$$S=\sum_{n=0}^{\infty}{2^{n+3}\over {2n\choose n}}\left[{A\over n+1}+{B\over 2n+1}+{C\over 2n+3}\right]$$
$n^2+n+\phi=A(2n+1)(2n+3)+B(n+1)(2n+3)+C(n+1)(2n+1)$
$A=-\phi$, $B=\phi-0.25$ and $C=\phi+0.75$
$$S=\sum_{n=0}^{\infty}{2^{n+3}\over {2n\choose n}}\left[{-\phi\over n+1}+{\phi-0.25\over 2n+1}+{\phi+0.75\over 2n+3}\right]$$
If we know the closed form
$$\sum_{n=0}^{\infty}{2^n\over {2n\choose n}(an+b)}=F(a,b)$$
Then it is easy to prove the above series.
I manage to find
$$\sum_{n=0}^{\infty}{2^n\over {2n\choose n}(2n+1)}={\pi\over 2}$$
Any help? Thank you.
| Note that $$\frac{1}{\dbinom{2n}{n}}=\left(2n+1\right)B\left(n+1,n+1\right)=\left(2n+1\right)\int_{0}^{1}x^{n}\left(1-x\right)^{n}dx
$$ hence $$\sum_{n\geq0}\frac{2^{n+3}}{\dbinom{2n}{n}\left(n+1\right)}=\sum_{n\geq0}\frac{2^{n+3}}{n+1}\left(2n+1\right)\int_{0}^{1}x^{n}\left(1-x\right)^{n}dx
$$ $$=8\int_{0}^{1}\sum_{n\geq0}\frac{2n+1}{n+1}\left(2x\left(1-x\right)\right)^{n}=8\int_{0}^{1}\int_{0}^{1}\frac{1+2xy-2x^{2}y}{\left(1-2xy+2x^{2}y\right)^{2}}dxdy
$$ $$=16\int_{0}^{1}\frac{dx}{1-2x\left(1-x\right)}+4\int_{0}^{1}\frac{\log\left(1-2x\left(1-x\right)\right)}{x\left(1-x\right)}dx
$$ $$=\color{red}{8\pi-\pi^{2}}.$$ Note that the last integral can be calculated using the substitution $v=2x-1$. The other cases can be treated in the same way.
| {
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"url": "https://math.stackexchange.com/questions/2072228",
"timestamp": "2023-03-29T00:00:00",
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If $x = y$, $p$ = what? It is given that,
$$
\frac{\sqrt{2x+3y}+\sqrt{2x-3y}}{\sqrt{2x+3y}-\sqrt{2x-3y}}=p
$$
If $x=y$, $p$ =?
| We have: $\dfrac{\sqrt{2x+3y}+\sqrt{2x-3y}}{\sqrt{2x+3y}-\sqrt{2x-3y}}=p$
Let's replace $x$ with $y$:
$\Rightarrow p=\dfrac{\sqrt{2y+3y}+\sqrt{2y-3y}}{\sqrt{2y+3y}-\sqrt{2y-3y}}$
$\hspace{9 mm} =\dfrac{\sqrt{5y}+\sqrt{-y}}{\sqrt{5y}-\sqrt{-y}}$
$\hspace{9 mm} =\dfrac{\sqrt{5y}+\sqrt{y}\hspace{0.5 mm}i}{\sqrt{5y}-\sqrt{y}\hspace{0.5 mm}i}$
Then, let's multiply this fraction by the complex conjugate of its denominator:
$\hspace{9 mm} =\dfrac{\sqrt{5y}+\sqrt{y}\hspace{0.5 mm}i}{\sqrt{5y}-\sqrt{y}\hspace{0.5 mm}i}\cdot{\dfrac{\sqrt{5y}+\sqrt{y}\hspace{0.5 mm}i}{\sqrt{5y}+\sqrt{y}\hspace{0.5 mm}i}}$
$\hspace{9 mm} =\dfrac{\big(\sqrt{5y}+\sqrt{y}\hspace{0.5 mm}i\big)^{2}}{\big(\sqrt{5y}\big)^{2}-\big(\sqrt{y}\hspace{0.5 mm}i\big)6{2}}$
$\hspace{9 mm} =\dfrac{5y+2\cdot{\sqrt{5y}}\cdot{\sqrt{y}\hspace{0.5 mm}i}+y\cdot{i^{2}}}{5y-y\cdot{i^{2}}}$
$\hspace{9 mm} =\dfrac{5y+2{\sqrt{5y^{2}}\hspace{0.5 mm}i}-y}{5y+y}$
$\hspace{9 mm} =\dfrac{4y+2{\sqrt{5y^{2}}\hspace{0.5 mm}i}}{6y}$
$\hspace{9 mm} =\dfrac{2y+2\sqrt{y^{2}}\sqrt{5}\hspace{0.5 mm}i}{3y}$
$\hspace{9 mm} =\dfrac{2y+2y\sqrt{5}\hspace{0.5 mm}i}{3y}$
$\hspace{9 mm} =\dfrac{2+2\sqrt{5}\hspace{0.5 mm}i}{3}$; $y\neq{0}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2073892",
"timestamp": "2023-03-29T00:00:00",
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Prove or disprove the inequality $\tan x \geq x + x^3$ Given the inequality $\tan x \geq x + x^3$. Prove or disprove it.
$x \in (0, \pi/2)$.
Hints would be appreciated.
| The MacLaurin series of $\tan$ is given by
$$\tan(x) = x + \frac{x^3}{3} + \frac{2x^5}{15} + \cdots$$
Hence taking the difference $x^3 + x - \tan(x)$ becomes
$$\frac{2x^3}{3} - \frac{2x^5}{15} - \cdots$$
You could then suspect that for small $x$ (especially when $x^3 \gg x^5$), we will have that $x^3 +x > \tan(x)$.
For example $x = \pi/6$ gives you that:
*
*$x^3 + x = \pi/6 + \pi^3/216 \approx 0.66715$
*$\tan(x) = \tan (\pi/6) = 1/\sqrt{3} \approx 0.55735$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve equation in determinant Let $ a,b,c,m,n,p\in \mathbb{R}^{*} $, $ a+m+n=p+b+c $. Solve the equation:
$$
\begin{vmatrix}
x & a & b &c \\
a & x & b &c \\
m &n & x &p \\
m& n& p& x
\end{vmatrix}
=0
$$
I had used the Schur complement ($\det(M)=\det(A)\cdot (D-C\cdot A^{-1}\cdot B)$, for $ M= \begin{bmatrix}
A &B \\
C & D
\end{bmatrix}) $ but it didn't help me.
| Subtract the first line from the second:
$$\begin{vmatrix}
x & a & b &c \\
a & x & b &c \\
m &n & x &p \\
m& n& p& x
\end{vmatrix}=\begin{vmatrix}
x-a & a-x & 0 &0 \\
a & x & b &c \\
m &n & x &p \\
m & n& p& x
\end{vmatrix}=(x-a)\begin{vmatrix}
1 &-1 & 0 &0 \\
a & x & b &c \\
m &n & x &p \\
m & n& p& x
\end{vmatrix}=$$
$1.$ Multiply the first by $-a$ and add on the second
$2.$ Multiply the first by $-m$ and add on the third
$3.$ Multiply the first by $-m$ and add on the forth
$$(x-a)\begin{vmatrix}
1 &-1 & 0 &0 \\
0 & x+a & b &c \\
0 &m+n & x &p \\
0 &m+ n& p& x
\end{vmatrix}=$$
Forth line minus third:
$$(x-a)\begin{vmatrix}
1 &-1 & 0 &0 \\
0 & x+a & b &c \\
0 &m+n & x &p \\
0 &0& p-x& x-p
\end{vmatrix}=(x-a)(x-p)\begin{vmatrix}
1 &-1 & 0 &0 \\
0 & x+a & b &c \\
0 &m+n & x &p \\
0 &0& -1& 1
\end{vmatrix}=$$
Laplace theorem on the first column
$$(x-a)(x-p)\begin{vmatrix}
x+a & b &c \\
m+n & x &p \\
0& -1& 1
\end{vmatrix}=0$$
Add second and third column on the second
$$(x-a)(x-p)\begin{vmatrix}
x+a & b+c &c \\
m+n & x+p &p \\
0& 0& 1
\end{vmatrix}=0$$
Laplace on the third line
$$(x-a)(x-p)\begin{vmatrix}
x+a & b+c \\
m+n & x+p \\
\end{vmatrix}=0$$
Now $a+m+n=p+b+c=k$ then
$$(x-a)(x-p)\begin{vmatrix}
x+a & k-p \\
k-a & x+p \\
\end{vmatrix}=0$$
Add first line on the second
$$(x-a)(x-p)\begin{vmatrix}
x+a & k-p \\
x+k & x+k \\
\end{vmatrix}=(x-a)(x-p)(x+k)\begin{vmatrix}
x+a & k-p \\
1 & 1 \\
\end{vmatrix}=0$$
$$(x-a)(x-p)(x+k)(x+a+p-k)=0$$
$$(x-a)(x-p)(x+k)(x+a+p-k)=0$$
| {
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"url": "https://math.stackexchange.com/questions/2074189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Evaluating $\int_0^{1/2} \log(1-x) \log(1-2x) \ dx$. While messing around with Wolfram Alpha, I discovered that
$$\int_0^{\frac{1}2} \log(1-x) \log(1-2x) \ dx = 1 - \frac{\pi^2}{24} - \frac{\log(2)}2.$$
I've tried all sorts of standard tricks, but I cannot seem to prove it. Can someone prove this beautiful identity?
| With the substitution $u = 1-2x$, $x = (1+u)/2$, $dx = -du/2$, we get $$\begin{align*} \int_{x=0}^{1/2} \log (1-x) \log (1-2x) \, dx &= -\frac{1}{2} \int_{u=1}^0 \log \left(\frac{1+u}{2}\right) \log u \, du \\ &= \frac{1}{2} \int_{u=0}^1 \log u \,(\log (1+u) - \log 2) \, du \\
&= \frac{1}{2} \left( -\log 2 \int_{u=0}^1 \log u \, du + \int_{u=0}^1 \log u \, \log (1+u) \, du \right) \\ &= \frac{\log 2}{2} + \frac{1}{2} I,\end{align*}$$ where $$\begin{align*} I &= \int_{u=0}^1 \log u \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} u^k \, du \\ &= \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \int_{u=0}^1 u^k \log u \, du \\
&= \sum_{k=1}^\infty \frac{(-1)^k}{k(k+1)^2}, \end{align*}$$ with the evaluation of the last step being accomplished by a trivial integration by parts. Partial fraction decomposition gives $$\frac{1}{k(k+1)^2} = \frac{1}{k} - \frac{1}{k+1} - \frac{1}{(k+1)^2},$$ and using the fact that $$\sum_{k=1}^\infty \frac{(-1)^k}{k} = -\log 2,$$ and $$\sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6},$$ it is relatively straightforward to obtain the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075380",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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$\lfloor \sqrt{n-100} \rfloor$ dividing $n$
For how many positive integers $100 < n \le 10000$ does $\lfloor \sqrt{n-100} \rfloor$ divide $n$?
Let $n = 100+k^2+c$, where $k$ is a nonnegative integer and $c$ an integer such that $0 \leq c < 2k+1$. Then $k = \lfloor \sqrt{n-100} \rfloor$ and $1 \leq k \leq 99$ and we have $$\dfrac{n}{\lfloor \sqrt{n-100} \rfloor} = \dfrac{100}{k}+\dfrac{c}{k}+k.$$
How do we continue from here?
| First, fix $k = \lfloor \sqrt{n-100} \rfloor \in [1,98]$. Then, $100+k^2 \le n \le 100+k^2+2k$.
Exactly two of the integers $\{100+k^2+c : c \in \mathbb{N}, 1 \le c \le 2k\}$ will be divisible by $k$ (since these are $2k$ consecutive integers).
Also, $100+k^2$ will be divisible by $k$ iff $100$ is divisible by $k$, which happens exactly for $k = 1,2,4,5,10,20,25,50$ (a total of $8$ values of $k$).
Therefore, for $k = 1,2,4,5,10,20,25,50$, there are $3$ integers of the form $n = 100+k^2+c$ where $0 \le c \le 2k$ that are divisible by $k$, and for all other values of $k \in [1,98]$ (a total of $98-8 = 90$ values of $k$), there are $2$ such integers.
Finally, for $k = 99$, we must have $99^2+100 = 9901 \le n \le 10000$. The only value of $n$ in this range which is divisible by $k = 99$ is $9999$.
Thus, the total number of integers $100 < n \le 10000$ such that $\lfloor \sqrt{n-100} \rfloor$ divides $n$ is $8 \cdot 3 + 90 \cdot 2 + 1 = 205$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show $\forall a,b\in\mathbb{R}^*\quad 3+\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge 2 (\frac{a}{b}+\frac{b}{a})$ Let $a,b \in \mathbb{R}^*$ Show that :
$$3+\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge 2\left(\frac{a}{b}+\frac{b}{a}\right)$$
Let $a,b\in\mathbb{R}^*$
let $t=\dfrac{a}{b}+\dfrac{b}{a}$ , we've :
$$
\begin{aligned}
3+\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge 2\left(\frac{a}{b}+\frac{b}{a}\right) &\iff
3+\frac{a^2}{b^2}+\frac{b^2}{a^2}-2\left(\frac{a}{b}+\frac{b}{a}\right)\\
&\iff t^{2}-2t+1\geq 0\\
&\iff \left(t-1\right)^{2}\geq 0
\end{aligned}$$
since $\forall t\in \mathbb{R}\quad (t-1)^{2}\geq 0$ holds then also $$\forall a,b\in \mathbb{R}^*\quad 3+\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge 2\left(\frac{a}{b}+\frac{b}{a}\right) $$ holds
Am i right ? is there any other ways
| your inequality is equivalent to $${\frac { \left( {a}^{2}-ab+{b}^{2} \right) ^{2}}{{b}^{2}{a}^{2}}}\geq 0$$ and this is true for all $$a,b>0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $2(bc^2+ca^2+ab^2) = b^2c+c^2a+a^2b+3abc, $ then prove that triangle $ABC$ is equilateral In a triangle $ABC,$ if $2(bc^2+ca^2+ab^2) = b^2c+c^2a+a^2b+3abc, $ then prove that triangle
$ABC$ is equilateral
$\displaystyle \frac{2(bc^2+ca^2+ab^2)}{abc} = \frac{b^2c+c^2a+a^2b+3abc}{abc}$
$\displaystyle 2\left(\frac{c}{a}+\frac{a}{b}+\frac{b}{c}\right) = \left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)+3$
let $\displaystyle \frac{a}{b} = x,\frac{b}{c}=y,\frac{c}{a}=z$
so $\displaystyle 2(x+y+z) = \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3$
wan,t be able to proceed after that, could some help me
| Because
$$0=\sum\limits_{cyc}(2b^2a-a^2b-abc)=\sum\limits_{cyc}(a^3-abc-(b^3-2b^2a+a^2b))=$$
$$=(a+b+c)\sum\limits_{cyc}(a^2-ab)-\sum\limits_{cyc}b(a-b)^2=$$
$$=\sum\limits_{cyc}(a-b)^2\left(\frac{1}{2}(a+b+c)-b\right)=\frac{1}{2}\sum\limits_{cyc}(a-b)^2(a+c-b).$$
Id est, the given it's
$$(a-b)^2(a+c-b)+(b-c)^2(b+a-c)+(c-a)^2(c+b-a)=0,$$
which says that $a=b=c$.
Done!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Square root positive/negative What is the square root of $9$? Is it always $\pm 3$ or just positive $3$?
Trying to find the solution set of this equation : $x-3 = \sqrt{x+3}$
I want to understand the concept of square root to solve the problem.
Thanks
| Well, let us start with something we all know and love, the quadratic formula:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
And I have a question for you. Why is it that there is a $\pm$ in front of the square root? Isn't it already the case that we have things like $\sqrt9=\pm3$? And the answer is no. By definition, we have it that $\sqrt x\ge0$, and thus, $\sqrt9=3$ since $-3<0$.
Also notice that $\sqrt x$ is undefined for $x<0$, so by looking at the problem, we can immediately deduce that
$$x-3=\sqrt{x+3}\ge0\implies x\ge3$$
$$\sqrt{x+3}\implies x+3\ge0\implies x\ge-3$$
Now, upon squaring the original equality, we have
$$x^2-6x+9=(x-3)^2=(\sqrt{x+3})^2=x+3$$
$$x^2-7x+6=0$$
$$(x-6)(x-1)=0$$
$$\implies x=1,6$$
Checking this, only the second value is $\ge3$ and checking them in the square roots, we find that
$$-2=1-3\ne\sqrt{1+3}=2$$
$$3=6-3=\sqrt{6+3}=3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078896",
"timestamp": "2023-03-29T00:00:00",
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Recurrence relation with quadratic equation Situation: working on a homework problem for my discrete math class that I think is solved, but I am now wondering if my solution is right. This a is a recurrence relations problem with a quadratic equation to find the roots and then an extra term to deal with.
The question:
Solve homogenous problem:
$A_n = A_{n-1} + A_{n-2} + 2^n$
for $n \ge 2$.
My solution is:
Roots $= \dfrac{1 \pm \sqrt{5}}{2}$
So I have
$A_n = C_1\left( \dfrac{1 + \sqrt{5}}{2}\right)^n
+ C_2\left( \dfrac{1 - \sqrt{5}}{2}\right)^n
$ where $C_1$ and $C_2$ are constants.
For the $2^n$, I have set the following:
$A(2^n) - A(2^{n-1}) - A(2^{n-2}) = 2^n$,
$A(1 -1/2 - 1/4) = 1$
then $A((4-2-1)/4) = 1$ or $A=4$.
For a particular solution, $A_n = 4(2)^n$.
Putting this all together, I have:
$A_n = C_1\left( \dfrac{1 + \sqrt{5}}{2}\right)^n
+ C_2\left( \dfrac{1 - \sqrt{5}}{2}\right)^n
+ 4(2)^n$.
Am I on the right track? Any hints would be appreciated.
| Yeah! Seems quite good. Btw here's another nice way to eliminate the extra term and work out the recurrence:
Multiplying the giving recurrence by a factor of 2, seems like a hint to to me.
$$2A_n = 2A_{n-1} + 2A_{n-2} + 2^{n+1}$$
And also observe that
$$A_{n+1} = A_{n} + A_{n-1} + 2^{n+1}$$
From where we can easily eliminate the extra term and get $A_{n+1} - 3A_n + A_{n-1} + 2A_{n-2} = 0$ whose characteristic polynomial is $x^3-3x^2+x+2 = 0$ which in turn factorizes as $(x-2)(x-\varphi)(x+\varphi^{-1}) = 0$ where $\varphi = \frac{1+\sqrt{5}}{2}$ which is exactly what you got.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\ln(x+\sqrt{x^2 + 4}) - \ln2$ is odd I know that a function is odd when
$$f(-x) = -f(x)$$
Therefore I can say that if for a function $$-f(x) + f(x) = f(-x) + f(x) = 0$$
Then the function is odd!
I tried to use this trick to prove that $f(x) = \ln\left(x+\sqrt{x^2 + 4}\right) - \ln2$ is odd.
However, I would want to prove directly that $$f(-x) = -f(x)$$
In other words, I want to solve $$\ln\left(-x+\sqrt{(-x)^2 + 4}\right) - \ln2$$
and to come at the end to this:
$$-\ln\left(x+\sqrt{x^2 + 4}\right) + \ln2$$
This was my approach:
$$\ln\left(-x+\sqrt{(-x)^2 + 4}\right) - \ln2$$
$$\ln\left(-x+\sqrt{x^2 + 4}\right) - \ln2$$
$$\ln\left(\frac{-x+\sqrt{x^2 + 4}}{2}\right)$$
$$\ln\left(\left(\frac{2}{-x+\sqrt{x^2 + 4}}\right)^{-1}\right)$$
$$-\ln\left(\frac{2}{-x+\sqrt{x^2 + 4}}\right)$$
Here I got stuck. I want to get to $-\ln\left(x+\sqrt{x^2 + 4}\right) + \ln2$ but if I use $\ln\left(\frac ab\right) = \ln a - \ln b$ then I will get back to $f(-x)$ and not to $-f(x)$.
Any help?
| A faster approach. If we set $f(x)=\log\left(\frac{x+\sqrt{x^2+4}}{2}\right)$ and $x=2\sinh\theta$ we have
$$f(2\sinh\theta) = \log\left(\frac{2\sinh\theta+2\cosh\theta}{2}\right) = \log(e^\theta) = \theta $$
and since $\theta\mapsto 2\sinh\theta$ is a bijective odd function from $\mathbb{R}$ to $\mathbb{R}$,
$$ f(-x) = -f(x) $$
readily follows. That also implies $f(x) = \text{arcsinh}\left(\frac{x}{2}\right)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$x-\frac1x+y-\frac1y=4\,$ has no rational solutions? Do there exist rational numbers $x,y$ such that $x-\frac1x+y-\frac1y=4?$
I think no, because the equation reduces to $\frac{x^2-1}{x}+\frac{y^2-1}{y}=4\implies x^2y+x(y^2-4y-1)-y=0$ in two variables is irreducible in the field $\mathbb{Q}$(Eisenstein) Is my reasoning right? Thanks beforehand.
| WLOG, suppose that $x,y>0$ (otherwise $x'=-\dfrac{1}{x}$),then let $x= \tfrac ab, y= \tfrac cd$ with $a,b,c,d \in \mathbb{N}, \; \gcd (a,b)= \gcd (c,d)=1$. The equation is equivalent to $$\dfrac{a}{b}-\dfrac{b}{a}+\dfrac{c}{d}-\dfrac{d}{c}=4 \iff cd(a^2-b^2)+ab(c^2-d^2)=4abcd.$$
From here, since $ab|(a^2-b^2)cd+(c^2-d^2)ab$,we obtain $ab \mid cd(a^2-b^2)$ but since $\gcd (a,b)=1$ so $ab \mid cd$. Similarly, we obtain $cd \mid ab$. Thus, $ab=cd$. Since $\gcd (a,b)= \gcd (c,d)=1$ so there exists pairwise relatively prime numbers $m,n,p,q$ such that $a=mp,c=mq, b=nq,d=np$. Therefore, the equation becomes $$\left( p^2+q^2 \right) \left( m^2-n^2 \right)= 4mnpq.$$
If $2 \nmid mnpq$ then $8 \mid LHS$ but $RHS \equiv 4 \pmod{8}$, a contradiction.
If $2 \mid m$ (or $2 \mid n$) then $2 \nmid npq$ (or $2 \nmid mpq$) so $LHS \equiv 2 \pmod{4}$ and $4 \mid RHS$, a contradiction.
If $2 \mid pq$. WLOG, suppose that $2 \mid p$ then $2 \nmid qmn$. This follows that $2 \nmid p^2+q^2$. Also note that $\gcd (p^2+q^2,pq)= \gcd (m^2-n^2,mn)=1$ so $p^2+q^2=mn, m^2-n^2=4pq$. This follows $$\begin{aligned} 2 \left( \frac{m+n}{2} \right)^2 & = n^2+(p+q)^2, \\ 2x^2 &=z^2+w^2, \\ x^2 & = \left( \frac{z+w}{2} \right)^2+ \left( \frac{z-w}{2} \right)^2, \end{aligned}$$ with $x= \tfrac{m+n}{2}, z=n, w=p+q$. Since $x, \tfrac{z+w}{2}, \tfrac{z-w}{2}$ are pairwise relatively prime numbers and $2 \nmid n$ so there exist $k,l \in \mathbb{N}$ with $k>l, \gcd (k,l)=1$ such that $$x= k^2+l^2, \frac{z+w}{2}=2kl, \frac{z-w}{2}=k^2-l^2 \; \text{or} \; \frac{z+w}{2}=k^2-l^2, \frac{z-w}{2}=2kl.$$
We find $z=n= k^2-l^2+2kl$ and $\tfrac{m+n}{2}=x= k^2+l^2$ so $m= k^2+3l^2-2kl$. Hence, $$m^2-n^2=(k^2+3l^2-2kl)^2-(k^2-l^2+2kl)^2= 8l(l-k)(k^2+l^2) <0,$$ a contradiction since $4pq=m^2-n^2>0$.
Thus, there is no rational numbers $x,y$ such that $x- \frac 1x +y- \frac 1y=4$. $\blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 1
} |
Prove $\int_0^1 {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \, dx={1\over 2}$ using an alternative method Prove that
$$\int_0^1 {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \,dx={1\over 2}\tag1$$
My try:
$u=1+\sqrt{5}x$ then $du=\sqrt{5} \, dx$
$${1\over \sqrt 5}\int_1^{1+\sqrt{5}}(4u^{-3}-u^{-2}) \, du$$
$$\left. {1\over \sqrt{5}}(-2u^{-2}+u^{-1}) \right|_1^{1+\sqrt{5}}={1\over 2}$$
Prove $(1)$ using an alternative method other than substitution method.
| $$
{3-\sqrt{5}x\over (1+\sqrt 5 \, x)^3} = \frac A {1+\sqrt 5\,x} + \frac B {(1+\sqrt 5\,x)^2} + \frac C {(1+\sqrt 5\,x)^3}
$$
Find $A$, $B$, and $C$ and then integrate each term separately.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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The sum of digits in a 2-digit number
The sum of digits in a two digit number formed by the two digits from $1$ to $9$ is $8$. If $9$ is added to the number then both the digits become equal. Find the number.
My attempt:
Let the two digit number be $10x+y$ where, $x$ is a digit at tens place and $y$ is the digit at unit's place. According to question:
$$x+y=8$$
I could not figure out the other condition. Please help. Thanks in advance.
| Let the $2$-digit number be $\overline{ab} = 10a + b$.
If $9$ is added to the number then both the digits become equal.
This means that $a = b$, so the number plus $9$ is $10a + a = 11a$, which is always $0 \text{ mod } {11}$ (divisible by $11$).
Therefore $\overline{ab} + 9 \equiv 0 \text{ mod } {11} \Rightarrow \overline{ab} \equiv 2 \text{ mod } {11}$.
But $a + b \equiv 8 \text{ mod } {11}$. Thus $(10a+b)-(a+b) \equiv 9a \equiv -2a \equiv 2-8 \text{ mod } {11}$, so $a \equiv 3 \text{ mod } {11}$.
As $a+b = 8$, hence $\overline{ab} = \boxed{35}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 5
} |
Integration of inverse function We have to integrate the folowing
I tried and substitued x=atan$^2 \theta$
But stuck in that
| Hint: integrate by parts:
$$
\int \arcsin\sqrt{\frac{x}{x+a}}\ \mathrm dx=x\arcsin\sqrt{\frac{x}{x+a}}-\int \frac{1}{2} \sqrt{\frac{a x}{(a+x)^2}}\ \mathrm dx
$$
and let $\sqrt x=u$ to integrate the second integral:
$$
\frac{1}{2} \sqrt{\frac{a x}{(a+x)^2}}\ \mathrm dx=\frac{\sqrt{a} u^2}{a+u^2}\ \mathrm du
$$
The final solution is
\begin{equation} x \sin ^{-1}\left(\sqrt{\frac{x}{a+x}}\right)+\frac{(a+x) \sqrt{\frac{a x}{(a+x)^2}} \left(\sqrt{a} \tan ^{-1}\left(\frac{\sqrt{x}}{\sqrt{a}}\right)-\sqrt{x}\right)}{\sqrt{x}}\end{equation}
which, if $a$ and $x$ are positive, simplifies to
\begin{equation}(a+x) \tan ^{-1}\left(\sqrt{\frac{x}{a}}\right)-\sqrt{a x}\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $\int\frac{\mathrm d \, x}{x^2-1}= \operatorname{arctanh} ~x + C = \operatorname{arccoth} ~x + K$ correct? My professor gave us an integral but I think it's incorrect.
Is this correct?
$$
\DeclareMathOperator\arctanh{arctanh}
\DeclareMathOperator\arccoth{arccoth}
\int\frac{\mathrm d \, x}{x^2-1}= \arctanh ~x + C = \arccoth ~x + K$$
| It is useful to rewrite $\DeclareMathOperator\arctanh{arctanh}\DeclareMathOperator\arccoth{arccoth} \arccoth x$ as an inverse function. In this case, let $f(x)=\coth x=\frac{e^x+e^{-x}}{e^x-e^{-x}}=\frac{e^{2x}+1}{e^{2x}-1}$. To determine the inverse, just solve for $x$ in the equation $y=\frac{e^{2x}+1}{e^{2x}-1}$.
\begin{align*}
y&=\frac{e^{2x}+1}{e^{2x}-1}\\
e^{2x}y-y&=e^{2x}+1\\
e^{2x}(y-1)&=y+1\\
e^{2x}&=\frac{y+1}{y-1}\\
2x&=\ln\left(\frac{y+1}{y-1}\right)\\
x&=\frac{1}{2}\ln\left(\frac{y+1}{y-1}\right)\\
\arccoth x&=\frac{1}{2}\ln\left(\frac{x+1}{x-1}\right)
\end{align*}
You will get something similar for $\arctanh x$!
\begin{equation*}
\arctanh x=\frac{1}{2}\ln\left(\frac{x+1}{1-x}\right)
\end{equation*}
For the integral, use partial fractions. The rewrite is
\begin{equation*}
\frac{1}{x^2-1}=\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)
\end{equation*}
\begin{equation*}
\int\frac{1}{x^2-1}\mathrm{d}x=\frac{1}{2}\left(\ln\left|\frac{x-1}{x+1}\right|\right)+C
\end{equation*}
Your lecturer was close! for $|x|>1$, we have $\int\frac{1}{x^2-1}\mathrm{d}x=-\arccoth x+C$ and for $|x|<1$, we have $\int\frac{1}{x^2-1}\mathrm{d}x=-\arctanh x+C$.
I hope this helps (and is all correct)!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084893",
"timestamp": "2023-03-29T00:00:00",
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If, in a triangle, $\cos(A) + \cos(B) + 2\cos(C) = 2$ prove that the sides of the triangle are in AP By using the formula :
$$
\cos(A)+\cos(B)+\cos(C) = 1 + 4 \sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right)
$$
I've managed to simplify it to :
$$
2\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)=\sin\left(\frac{C}{2}\right)$$
But I have no idea how to proceed.
| From :
$$
\cos(A)+\cos(B)+2\cos(C)=2
\\
\implies \cos(A)+\cos(B)=2-2\cos(C)
\\
\implies \cos(A)+\cos(B)=2[1-\cos(C)]
\\
\\\implies \cos(A)+\cos(B)=4\sin^2\left(\frac{C}{2}\right)
$$
Using Prosthaphaeresis Formulas :
$$
\cos A+\cos B=2\cos\dfrac{A+B}2\cos\dfrac{A-B}2
$$
And substituting this formula in the first equation, we have :
$$
2\cos\dfrac{A+B}2\cos\dfrac{A-B}2=4\sin^2\left(\frac{C}{2}\right)
$$
Since
$A+B+C=\pi$
$$
\frac{A+B}{2}=\frac{\pi}{2}-\frac{C}{2}
$$
Taking cosines on both sides :
$$
\cos\left(\frac{A+B}{2}\right)=\cos\left(\frac{\pi}{2}-\frac{C}{2}\right)=\sin\left(\frac{C}{2}\right)
$$
Using this :
$$
2\cos\dfrac{A+B}2\cos\dfrac{A-B}2=4\cos^2\left(\frac{A+B}{2}\right)
\\
\implies \cos\dfrac{A-B}2=2 \cos\dfrac{A+B}2
$$
Multiplying both sides by $2\sin\dfrac{A+B}2$ :
$$
2\sin\dfrac{A+B}2\cos\dfrac{A-B}2=4\sin\dfrac{A+B}2\cos\dfrac{A+B}2
$$
Using Another Prosthaphaeresis Formula :
$$
\sin A+\sin B=2\sin\dfrac{A+B}2\cos\dfrac{A-B}2
$$
Applying the reverse in the obtained equation, we get :
$$
\sin A+\sin B=2\sin\dfrac{A+B}2\cos\dfrac{A-B}2=4\sin\dfrac{A+B}2\cos\dfrac{A+B}2
$$
Using the sine double angle formula $2\sin(\alpha)\cos(\alpha)=\sin(2\alpha)$ :
$$
\sin A+\sin B=2\sin(A+B)=2\sin C
\\
\implies \sin A + \sin B = 2\sin C
$$
Using the sine rule :
$$
a + b = 2c
$$
Hence, the sides of the triangle are in A.P.
| {
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"url": "https://math.stackexchange.com/questions/2084979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help in understanding a simple proof. Let $Ax + By + C = 0$ be a general equation of a line and $x\cos \alpha + y\sin \alpha - p = 0$ be the normal form of the equation.
Then,
$${-p\over C } = { \cos \alpha\over A} = { \sin\alpha\over B}\tag{1}$$
$${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} \tag{2}$$
$${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} = {\sqrt{\sin^2 \alpha + \cos^2 \alpha}\over \sqrt{A^2 + B^2}} = {1\over \sqrt{A^2 + B^2}} \tag{3}$$
$$\therefore \bbox[ #FFFDD0, 10px, Border:2px solid #DC143C]{p = {C\over \sqrt{A^2 + B^2}}, \cos \alpha = {-A\over \sqrt{A^2 + B^2}},\sin\alpha = {-B\over \sqrt{A^2 + B^2}}} $$
I did not get the $(3)$ part. Where does $\displaystyle{\sqrt{\sin^2 \alpha + \cos^2 \alpha}\over \sqrt{A^2 + B^2}}$ come from ?
| Let's modify ${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} $
to ${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} =K$
And ${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} = {\sqrt{\sin^2 \alpha + \cos^2 \alpha}\over \sqrt{A^2 + B^2}} = {1\over \sqrt{A^2 + B^2}} \tag{3}$
to $K = {p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} = {K\sqrt{A^2 +B^2}\over \sqrt{A^2 + B^2}}={\sqrt{(-AK)^2 +(-BK)^2}\over \sqrt{A^2 + B^2}}={\sqrt{\sin^2 \alpha + \cos^2 \alpha}\over \sqrt{A^2 + B^2}} = {1\over \sqrt{A^2 + B^2}} \tag{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Floor function and inequality I'm trying to work through a problem and I need to find a nice way to show that
$$\lfloor x^2\rfloor +\lfloor2x\rfloor\leq\lfloor x^4\rfloor$$
for $x\in[\sqrt[4]{3},\infty)$.
I know one possible way is to break down that interval into all the places where the three floor values increase by $1$. However, I was wondering if there is a way to do this in 1-2 lines.
| Hint: by definition of the greatest integer function:
$$
\begin{cases}
\begin{align}
\lfloor x^2\rfloor +\lfloor2x\rfloor & \leq x^2 + 2 x \\
\lfloor x^4\rfloor & \gt x^4 - 1
\end{align}
\end{cases}
$$
Therefore a sufficient condition for the inequality to hold is:
$$
x^2+2x \le x^4 - 1 \;\; \iff \;\; x^4-x^2-2x-1 = x^4 - (x+1)^2 = (x^2-x-1)(x^2+x+1) \ge 0
$$
The latter holds for $x \ge \frac{1+\sqrt{5}}{2}$ which leaves the interval $\big[\sqrt[4]{3}, \frac{1+\sqrt{5}}{2}\big)$ to be checked by hand.
[ EDIT ] To answer a posted comment, below is the breakdown on the interval $\big[\sqrt[4]{3}, \frac{1+\sqrt{5}}{2}\big)\,$.
$$
\begin{array}{c|lcr}
x \quad & \quad \lfloor x^2\rfloor +\lfloor2x\rfloor\leq\lfloor x^4\rfloor \\
\hline
\big[\sqrt[4]{3}, \,\sqrt{2}\big) \quad & \quad \quad 1\;+\;2\;=\;3 \\
\big[\sqrt{2}, \,\sqrt[4]{5}\big) \quad & \quad \quad 2\;+\;2\;=\;4 \\
\big[\sqrt[4]{5}, \,1.5\big) \quad & \quad \quad 2\;+\;2\;\lt\;5 \\
\big[1.5, \,\sqrt[4]{6}\big) \quad & \quad \quad 2\;+\;3\;=\;5 \\
\big[\sqrt[4]{6}, \frac{1+\sqrt{5}}{2}\big) \quad & \quad \quad 2\;+\;3\;\lt\;6 \\
\end{array}
$$
It follows that the inclusive inequality holds for $x \ge \sqrt[4]{3}\,$, and is a strict inequality for $x \ge \sqrt[4]{6}$.
[ EDIT #2 ] Added the missing break at $\sqrt[4]{5}$ in the table above.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Zeros of a Fourth Degree Polynomial Given that $\left(\sqrt3+\sqrt5\right)$ is one zero of a fourth-degree polynomial with integer coefficients and leading coefficient 1, how can the constant term of this polynomial be found?
I know that $\left(\sqrt3-\sqrt5\right)$ must also be a root because it is the conjugate. How can I determine the other two roots (and ultimately, the constant term) beyond what I have right now?
$(x^2-2\sqrt3x-2)(x-r_1)(x-r_2)$
Thanks!
| We know that $\sqrt 3+\sqrt 5$ is a root of the polynomial $\sqrt 3+\sqrt 5=x$, so we begin from there and eliminate radicals to obtain a polynomial over $\Bbb Z$:
\begin{align}
\sqrt 3+\sqrt 5&=x\\
\sqrt 3&=x-\sqrt 5\\
3&=x^2-2\sqrt 5 x+5\\
2\sqrt 5x&=x^2+2\\
20x^2&=x^4+4x^2+4\\
0&=x^4-16x^2+4.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How is $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}} = \sqrt{2}-1$? I wondered, why this:
$$\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$$
is equal to $\sqrt{2}-1$.
Can anyone explain me, why this is equal? :/
| Since I've expanded on my initial "joke comment", I might as well make it a full joke answer :) By that I mean, nobody in their right mind would take this approach to actually verify that the two quantities are equal: instead, what follows is a good, but limited, way to produce expressions with radicals that look different, but are really not.
The trigonometric function $\tan$ (tangent) has a wide variety of half-angle formulas. I would like to use the following two:
\begin{align*}
\tan \frac\theta 2 &= \frac{1 - \sin \theta}{\cos \theta} \tag{1}\\[10pt]
\tan \frac\theta 2 &= \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \tag{2}
\end{align*}
Evaluating the first at $\theta = \frac{\pi}{4}$, we have that
\begin{align*}
\tan \frac{\pi}{8} = \tan \frac{\pi/4}{2} &= \frac{1 - \sin(\pi/4)}{\cos(\pi/4)} \\[7pt]
&= \frac{1 - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} \\[7pt]
&= \left(1 - \frac{1}{\sqrt{2}}\right) \cdot \frac{\sqrt{2}}{1} \\[5pt]
&= \sqrt{2} - 1
\end{align*}
Now, using the second identity with $\theta = \pi/4$, we have
\begin{align*}
\tan \frac{\pi}{8} = \tan \frac{\pi/4}{2}
&= \sqrt{\frac{1 - \cos(\pi/4)}{1 + \cos(\pi/4)}} \\[7pt]
&=\sqrt{\frac{1 - \frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}}} \\[7pt]
&=\sqrt{\frac{1 - \frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}} \cdot \left(\frac{\sqrt 2}{\sqrt 2}\right)} \\[7pt]
&= \sqrt{\frac{\sqrt{2} - 1}{\sqrt{2} + 1}}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if $n+1$ is divisible by $8$, then the sum of $n$'s factors is also divisible by $8$
Let $n$ be a positive integer. Prove that if $n+1$ is divisible by $8$, then the sum of $n$'s factors is also divisible by $8$.
If $n+1$ is divisible by $8$, then $n \equiv 7 \pmod{8}$, but I didn't see how to separate it into cases here.
| Let $n + 1 = 8m$.
We show first that if $d$ is a divisor of $n = 8m - 1$, then $d^2 - 1$ is divisible by $8$. Now $d$ must be odd (because $n$ is), so $d - 1$ and $d + 1$ are consecutive even numbers, so one must be a multiple of $4$, and there product $d^2 - 1$ must be a multiple of $8$.
Now $8m - 1$ cannot be a square (because squares are congruent to $0$ or $1 \bmod 8$), so its divisors come in pairs $d, \frac{8m - 1}{d}$. But $d + \frac{8m - 1}{d} = \frac{(d^2-1)+8m}{d}$ is divisible by $8$ (because $d^2 - 1$ and $8m$ are and no factor of $8$ can divide $d$). Hence the sum of all the divisors of $n - 1$ is divisible by $8$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Computing $\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}$
What is $\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}$ ?
Here are a few remarks:
*
*Since $x\mapsto \frac{2^x}{x}$ is increasing when $x\geq 2$, one might be tempted to use the integral test. This fails: when doing so, one gets $a_n\leq \sum_{k=1}^n \frac{2^k}{k}\leq b_n$ where $a_n\sim \frac{2^n}{\ln (2)n}$ and $b_n\sim \frac{2^{n+1}}{\ln (2)n}$.
Unfortunately $b_n$ is too big and this estimate doesn't yield the limit.
*Here's my solution: since it's easy to sum $2^k$ and the difference $\frac{1}{k}-\frac{1}{k+1}$ is small, it's natural to try summation by parts: $$\begin{align} \sum_{k=1}^n \frac{2^k}{k}
&=\frac{S_n}{n+1}-1+\sum_{k=1}^n S_k \left(\frac{1}{k}-\frac{1}{k+1} \right)\quad \text{where} \; S_n=\sum_{k=0}^n 2^k\\
&= \frac{2^{n+1}}{n+1} + \sum_{k=1}^n \frac{2^{k+1}}{k(k+1)} - \underbrace{1 - \sum_{k=1}^n\left(\frac{1}{k(k+1)}\right) - \frac{1}{n+1}}_{\text{bounded}}\\
\end{align}$$
Intuition suggests $\displaystyle \sum_{k=1}^n \frac{2^{k+1}}{k(k+1)}=o\left(\frac{2^n}n \right)$ but it's not immediate to prove. I had to resort to another summation by parts! Indeed
$$\begin{align}\small\sum_{k=1}^n \frac{2^{k+1}}{k(k+1)}&= \small 2\left[ \frac{2^{n+1}}{n(n+1)} + 2\sum_{k=1}^n \left(\frac{2^{k+1}}{k(k+1)(k+2)}\right)-\frac 12 -2\sum_{k=1}^n \left(\frac{1}{k(k+1)(k+2)}\right) - \frac{1}{n(n+1)}\right]\\
&\small\leq \frac{2^{n+2}}{n(n+1)}+\frac{2^{n+2}}{n(n+1)(n+2)}\cdot n \\
&\small= o\left(\frac{2^n}n \right)
\end{align}$$
Hence $$\sum_{k=1}^n \frac{2^k}{k} = \frac{2^{n+1}}{n+1} + o\left(\frac{2^n}n \right)$$ and $$\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k} = 2$$
This solution is quite tedious and computational... That's why I'm looking for a shorter or smarter solution that avoids summation by parts (integration by parts is easy to perform on functions, it just gets quite heavy with series).
| I encountered this limit when computing
$$
\sum_{k=0}^n\frac1{\binom{n}{k}}=\frac{n+1}{2^n}\sum_{k=0}^n\frac{2^k}{k+1}
$$
in equation $(8)$ of this answer, and for which an asymptotic expansion is given in equation $(6)$ of this answer:
$$
\begin{align}
\frac{n}{2^n}\sum_{k=1}^n\frac{2^k}k
&=\sum_{k=1}^n\frac1{\binom{n-1}{k-1}}\\
&\sim2+\frac2{n-1}+\frac4{(n-1)(n-2)}+\frac{12}{(n-1)(n-2)(n-3)}\\
&+\frac{48}{(n-1)(n-2)(n-3)(n-4)}+O\left(\frac1{n^5}\right)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $\tan 7°30' = \sqrt {6} - \sqrt {3} + \sqrt {2} - 2$ Prove that:
$$\tan 7°30' = \sqrt {6} - \sqrt {3} + \sqrt {2} - 2$$
My Work:
I guess that I have to use the formula :
$$\tan A = \frac {2 \tan(\frac {A}{2})}{1-\tan^2 (\frac {A}{2})}$$
But, I am not being able to use it. Please help me.
| You may prove that identity in a very geometric flavour, i.e. by bisecting twice a $30^\circ$ angle, through the Pythagorean theorem and the bisector theorem. It is best to keep the expressions of the involved lengths as simple as possible during the process. So, let we consider a triangle $BAE$ with $BA=\sqrt{3}, AE=2, EB=1$. Let $AD$ be the bisector of $\widehat{BAE}$ and $AC$ be the bisector of $\widehat{BAD}$.
By the bisector theorem, we have:
$$ BD = \frac{\sqrt{3}}{2+\sqrt{3}} = 2\sqrt{3}-3. \tag{1}$$
By the Pythagorean theorem, it follows that:
$$ AD^2 = 3+(2\sqrt{3}-3)^2 = 24-12\sqrt{3} = 6(\sqrt{3}-1)^2 \tag{2} $$
hence $AD=3\sqrt{2}-\sqrt{6}$. By the bisector theorem again,
$$ BC = BD\cdot\frac{AB}{AB+AD},\quad \frac{BC}{BA} = \frac{BD}{BA+AD} = \frac{2\sqrt{3}-3}{\sqrt{3}+3\sqrt{2}-\sqrt{6}}\tag{3} $$
so:
$$ \tan(7^\circ 30') = \frac{2-\sqrt{3}}{1+\sqrt{6}-\sqrt{2}}\tag{4}$$
and the claim turns out to be equivalent to:
$$ (1+\sqrt{6}-\sqrt{2})(\sqrt{6}-\sqrt{3}+\sqrt{2}-2)=(2-\sqrt{3})\tag{5} $$
that is tedious but straightforward to check.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove the following inequality $x^2+y^2+1>x\sqrt{y^2+1}+y\sqrt{x^2+1}$ Can anybody help me prove this inequality ?
| It is not difficult to show that $x-\sqrt{\strut x^2+1}<0$ for all $x\in\mathbb{R}$. Then $$\left(x-\sqrt{\strut x^2+1}\right)\left(y-\sqrt{y^2+1}\right)>0.$$ We get $$xy+\sqrt{\strut x^2+1}\sqrt{y^2+1}>x\sqrt{y^2+1}+y\sqrt{\strut x^2+1}\tag{1}$$ From the other hand, since $(a-b)^2\geqslant 0$, then $$ab\leqslant\frac{a^2+b^2}{2}$$ for all $a,b\in\mathbb{R}$. Hence $$xy+\sqrt{\strut x^2+1}\sqrt{y^2+1}\leqslant\frac{x^2+y^2}{2}+\frac{x^2+1+y^2+1}{2}=x^2+y^2+1.$$ Joining this with (1) we get the desired inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove that $(x+ {\sqrt{1+x^2}} ) ( y+ {\sqrt {1+y^2}}) = 1$ if $(x+ {\sqrt{1+y^2}} ) ( y+ {\sqrt {1+x^2}}) = 1$ Let $x,y$ be real numbers such that :
$(x+ {\sqrt{1+y^2}} ) ( y+ {\sqrt {1+x^2}}) = 1$.
Prove that :
$(x+ {\sqrt{1+x^2}} ) ( y+ {\sqrt {1+y^2}}) = 1$.
I tried taking $x=y$. It simplifies everything a lot. But I'm not able to progress when both $x$ and $y$ are in the same equation.
| HINT:
Write $$x = \frac{1}{2}(s-\frac{1}{s})\\
y = \frac{1}{2}(t-\frac{1}{t})$$
with $s$, $t>0$. Then
$$( x+ \sqrt{1+y^2})(y + \sqrt{1+x^2}) -1 = (s t -1) \frac { s t (s+t)^2 + (s-t)^2}{ 4 s^2 t^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $ \sin \alpha = \frac 45 $ and $ \cos \beta = \frac{5}{13} $, prove that $ \cos \frac{\alpha-\beta}{2} = \frac{8}{\sqrt {65}} $ I can solve it easily if I assume that $ 0 < \alpha, \beta < \frac{\pi}{2}$
But there is no mention of the quadrants in which $ \alpha $ and $ \beta $ lie in.
Is the question wrong ?
| Hint:
$\sin \alpha>0$ then $0<\alpha<\pi$ and $\cos \beta>0$ then $-\frac{\pi}2<\beta<\frac{\pi}2$ and $-\frac{\pi}2<-\beta<\frac{\pi}2$
Thus $$-\frac{\pi}2<\alpha-\beta<\frac{3\pi}2$$
and
$$-\frac{\pi}4<\frac{\alpha-\beta}2<\frac{3\pi}4$$
1) $\sin^2\alpha+\cos^2\alpha=1$, then $\cos\alpha=\pm\sqrt{1-\sin^2\alpha}$ and $\sin \beta=\pm\sqrt{1-\cos^2\beta}$
2) $\cos (\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$
3)$|\cos \frac{\alpha-\beta}2|=\sqrt{\frac{1+\cos(\alpha-\beta)}2}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Need to prove divisibility of $7^p-5^p-2$ by $3$ by using Fermat's little Theorem Let $p > 2$ be a prime number. Prove that $$7^p-5^p-2$$ can be divided by $6p$.
I have already proved that it can be divided by $2$ and $p$, but how can I prove that it can be divided by $3$ ?
| Fermat's little theorem asserts that if $p$ is prime, then for every integer $n$, we have :
$$n^p-n\equiv 0\quad (mod\,p)$$
Since $5$ and $7$ are primes :
$$7^p-5^p-2=(7^p-7)-(5^p-5)\equiv0\quad(mod\,p)$$
But $p$ is also odd, so that the relation : $7^p-5^p-2\equiv1-(-1)^p-2\quad (mod\,6)$ takes the form :
$$7^p-5^p-2\equiv0\quad(mod\,6)$$
At this point, we have proven that if $p$ is an odd prime, then $p\mid(7^p-5^p-2)$ and $6\mid(7^p-5^p-2)$
If $p>3$, then $\gcd(p,6)=1$ and we conclude that $6p\mid(7^p-5^p-2)$.
And if $p=3$, then a direct calculation shows that $$7^3-5^3-2\equiv(7\times(-5))-5\times(7)-2=72\equiv0\quad(mod\,18)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$\int \frac{dx}{x\sqrt{x^4-16}}$ While I know that the answer is $-\frac{1}{8}\arctan \frac{4}{\sqrt{x^4-16}},$ I couldn't reach it. When I solved it I reached a different answer, if someone can point at what point I made a mistake and what is the way to solve it I will be thankful.
I will start with taking $u=x^2$, so $du =2x \;dx,$ and $dx = du/2x.$
So I will multiple the integral by $2$ making it $2\int \frac{du}{\sqrt{u^2-4^2}}$ after that I will use the $\cosh^{-1}x$ to make it $2\cosh^{-1}(x^2/4) + c$
| $\int \frac{1}{x\sqrt{x^4 - 16}} dx$
Multiply and dividide by $x^3$
$\int \frac{x^3}{x^4 \sqrt{x^4 - 16}} dx$
Put $x^4 - 16 = u^2$
$4x^3 dx = 2u du$
$x^3 dx = \frac{1}{2} u du$
On putting in integral,
= $\frac12 \int \frac{u}{(u^2 + 16) \sqrt{u^2}} du$
= $\frac12 \int \frac{u}{u(u^2 + 16)} du$
= $\frac12 \int \frac{1}{(u^2 + 16)} du$
= $\frac12 \int \frac{1}{(u)^2 + (4)^2} du$
= $\frac12 \tan^{-1} \frac{u}{4} + c$
Then replace value of u.
As you want to know how to do according to your method.
We have $x^2 = u$
$2x dx = du$
$dx = \frac{1}{2x} du$
Then from integral we have,
$\int \frac{1}{2x} \cdot \frac{1}{x \sqrt{u^2 - 16}} du$
= $ \frac{1}{2} \int \frac{1}{x^2 \sqrt{u^2 - 16}} du$
= $ \frac{1}{2} \int \frac{1}{u \sqrt{u^2 - 16}} du$
Then solve it.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all three-digit numbers $\overline{abc}$ such that 6003 digit number $\overline{abcabcabc.....abc}$ is divisible by 91? Find all three-digit numbers $\overline{abc}$ such that $6003$ digit number $\overline{abcabcabc......abc}$ is divisible by 91?Here $\overline{abc}$ occurs $2001$ times.I know the divisibility rule for 91 which states to subtract $9$ times the last digit from the rest and, for large numbers,to form the alternating sum of blocks of three numbers from right to left. However, I am not able to see how I could apply this rule to determine the numbers $\overline{abc}$. How can I solve this?
| The given number can be written as follows,
$abc(1+10^3+10^6+\cdots+10^{6000})$
Now, $91|1001=1+10^3$ . The sum $S=1+10^3+10^6+\cdots+10^{6000}$ has $2001$ terms, therefore, $91$ and $(1+10^3)+10^6(1+10^3)+\cdots+10^{1999}(1+10^3)+10^{6000}$ are relatively prime $\implies$ $abc$ is a multiple of 91.
Therefore, the required numbers are $91\times n$ , where $n={2,3,4,5,6,7,8,9}$ i.e., the required numbers are :
$182,273,364,455,546,637,728,819$ and $910$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is it true that $\,(1+\sin n)^{1/n}\to 1$? In order to show that $\lim_{n\to\infty}(1+\sin n)^{1/n}=1$, we need to show that $\,1+\sin n\,$ cannot not become too small. It suffices for example to show that
$$
1+\sin n\ge \frac{c}{n^k},
$$
for some $c,k>0$. This could be a consequence of showing that there exist $d,m>0$, such that
$$
\Big|\,\pi-\frac{p}{q}\,\Big|\ge \frac{d}{q^{m}}
$$
for every rational $p/q$ (cf. irrationality measure).
It would be very interesting if we could produce a more elementary proof.
| Consider $f(x)=(1+\sin x)^{\frac{1}{x}}$ . The binomial expansion is valid for $|\sin x|<1$.
Case I
$x\neq\frac{\pi}{2}+2\pi k$ and $x\neq\frac{3\pi}{2}+2\pi k$, where $k\in\mathbb{Z}$. Hence we may expand binomially
$f(x)=1+\frac{\sin x}{x}+\frac{\frac{1}{x}(\frac{1}{x}-1)}{2!}\sin^2x+\dots+\frac{\frac{1}{x}(\frac{1}{x}-1)\dots(\frac{1}{x}-k+1)}{k!}\sin^kx+\dots$
EDIT (after comments made by Andréas)
$f(x)=1+\sum_{k=-x}^\infty\frac{(\frac{1}{x})!}{(x+k)!(\frac{1}{x}-x-k)!}\sin^{x+k}x$
Consider the coefficient of $\sin^{x+k}x$.
$g(x)=\frac{(\frac{1}{x})!}{(x+k)!(\frac{1}{x}-x-k)!}=\frac{\frac{1}{x}(\frac{1}{x}-1)\dots(\frac{1}{x}-x-k+2)(\frac{1}{x}-x-k+1)}{(x+k)!}$
Now taking the limit as $x\to\infty$,
\begin{align*}
\lim_{x\to\infty}g(x) & =\lim_{x\to\infty}\frac{1}{x}\lim_{x\to\infty}\frac{(-1)(-2)\dots(-x-k+2)(-x-k+1)}{(x+k)!}\\
& =\lim_{x\to\infty}\frac{(-1)^{x+k-1}}{x}\lim_{x\to\infty}\frac{(x+k-1)!}{(x+k)!}\\
& =\lim_{x\to\infty}\frac{(-1)^{x+k-1}}{x(x+k)}\\
& =0
\end{align*}
Every term in the sum tends to zero, leaving $\lim_{x\to\infty}f(x)=1$.
Case II
$x=\frac{\pi}{2}+2\pi k$
Now $f(x)=2^{\frac{1}{x}}$ and clearly $\lim_{x\to\infty}f(x)=1$ again.
Case III
$x=\frac{3\pi}{2}+2\pi k$
Now $f(x)=0^{\frac{1}{x}}$ and therefore $\lim_{x\to\infty}f(x)=0$.
We may conclude that
$\lim_{x\to\infty}f(x)=
\begin{cases}
0& x=\frac{3\pi}{2}+2\pi k\\
1&\text{otherwise}
\end{cases}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Evaluating an expression given values of symmetric polynomials Evaluate $\dfrac x{yz} + \dfrac y {xz} + \dfrac z y$
Given, $z+y+x=4, \qquad xyz=-60, \qquad xy+xz+yz=-17$
How do we do this? I found a common denominator, and substituted it for $-60$, but I am unaware of how to proceed.
Someone already asked the question but there is no useful answer. These are the possible answers:
A. $4/17$
B. $−5/6$
C. $17/60$
D. $−33/60$
E. $33/60$
| \begin{align}
& \frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy} = \frac{x^2 + y^2 + z^2}{xyz}
\end{align}
$x^2 + y^2 + z^2 = (x+y+z)^2 - 2(xy + xz +yz)$
so: \begin{align} \frac{x^2 + y^2 + z^2}{xyz} = \frac{(x+y+z)^2 - 2(xy + xz +yz)}{xyz} = \frac{16 - 2(-17)}{-60} = \frac{50}{-60} = -\frac{5}{6}
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/2103370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the sum of this infinite series: $\sum_{n=1}^{ \infty } \frac1n \cdot \frac{H_{n+2}}{n+2}$ How do I find this particular sum?
$$\sum_{n=1}^{ \infty } \frac1n \cdot \frac{H_{n+2}}{n+2}$$
where $H_n = \sum_{k=1}^{n}\frac1k$.
This was given to me by a friend and I have absolutely no idea how to proceed as I have never done these questions before. If possible, please give a way out without using polylogarithmic functions or other non-elementary functions.
| Actually the calculation for this sum is very simple and what we need is the sum of telescopic series. In fact
\begin{align}
\sum_{n=1}^\infty\frac{H_{n+2}}{n(n+2)}&=\frac12\sum_{n=1}^\infty H_{n+2}\left(\frac{1}{n}-\frac1{n+2}\right)\\\\
&=\frac{1}{2}\sum_{n=1}^\infty\left(\frac{1}{n}\left(H_{n}+\frac{1}{n+1}+\frac{1}{n+2}\right)-\frac{H_{n+2}}{n+2}\right)\\\\
&=\frac{1}{2}\sum_{n=1}^\infty\left(\frac{H_n}{n}+\frac{1}{n(n+1)}+\frac{1}{n(n+2)}-\frac{H_{n+2}}{n+2}\right)\\\\
&=\frac{1}{2}\sum_{n=1}^\infty\left(\frac{H_n}{n}-\frac{H_{n+2}}{n+2}\right)+\frac12\sum_{n=1}^\infty\left(\frac{1}{n(n+1)}+\frac{1}{n(n+2)}\right)\\\\
&=\frac12\left(H_1+\frac{H_2}2\right)+\frac78\\\\
&=\frac{7}{4}.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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If $-1 < a < 1$ then $\sqrt[4]{1-a^2} + \sqrt[4]{1-a} + \sqrt[4]{1+a} < 3$?
Prove that if $-1<a<1$, then :
$\sqrt[4]{1-a^2} + \sqrt[4]{1-a} + \sqrt[4]{1+a} < 3 $
I do not know how to even approach this problem. Those fourth roots confuse me a lot.
Any help would be appreciated.
Also I would like to know how the left hand side of the inequality behaves if there are no given bounds for the value of $a$.
Thanks in advance :) .
| Let $f:[-1,1] \to \mathbb{R}$, $f(x)=\sqrt[4]{1-x}+\sqrt[4]{1+x}+\sqrt[4]{1-x^2}$.
$f$ is obvously continous and differentable on $(-1,1)$
Then:
$$f'(x)=\frac{\sqrt[4]{(1-x)^3}-\sqrt[4]{(1+x)^3}-2x}{4\sqrt[4]{(1-x^2)^3}}$$
Of course $4\sqrt[4]{(1-x^2)^3} \geq 0$ for all $x\in[-1,1]$.
*
*for $x>0$ :
$\sqrt[4]{(1-x)^3}<\sqrt[4]{(1+x)^3}$ and $-2x<0$, so $f'(x)<0$ and $f$ is decreasing
*for $x<0$ :
$\sqrt[4]{(1-x)^3}>\sqrt[4]{(1+x)^3}$ and $-2x>0$, so $f'(x)>0$ and $f$ is increasing
*for $x=0$ : $f'(x)=0$ and $f(x)=3$
Now we can deduce, that $$\forall{x\in [1,1]}: f(x)\leq 3$$
Bounds for $a$ are required, because the number under fourth root must be nonnegative
| {
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"timestamp": "2023-03-29T00:00:00",
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Factorizing $z^2-z\frac{a^2 +b^2}{ab}+1 = (z-a/b)(z-b/a)$ The intermediate factorization calculation $z^2-z\frac{a^2 +b^2}{ab}+1 = (z-a/b)(z-b/a)$ was coined straightforward by Riley's Mathematical Methods for Physics and Engeneering (eq.24.66).
I've tried completing the square,
$$
z^2-z\frac{a^2 +b^2}{ab}+1 = 0 \\
\bigg( z-\frac{a^2 +b^2}{2ab} \bigg)^2 +1 -\frac{(a^2+b^2)^2}{(2ab)^2} = 0 \\
\bigg( z-\frac{a^2 +b^2}{2ab} \bigg)^2 = \frac{(a^2+b^2)^2}{(2ab)^2}-1 \\
z= \frac{a^2 +b^2}{2ab} \pm \sqrt{\frac{(a^2+b^2)^2}{(2ab)^2}-1} \ , \\
$$
which doesn't nearly suggest the simple end result.
I'm sure it will roll out after reorganizing terms, but I'm interested in an alternative approach which more naturally suggests the end result of $(z-a/b)(z-b/a)$, reflecting the word-use of straightforward in the text.
Context.
Using contour integration to evaluate $I = \int_{0}^{2\pi} \frac{\cos(2\theta)}{a^2+b^2-2ab\cos(\theta)}d\theta$
| it must be $$z_{1,2}=\frac{a^2+b^2}{2ab}\pm\sqrt{\left(\frac{a^2+b^2}{2ab}\right)^2-1}$$
and we can factorize to $$-\frac{(a-bz)(az-b)}{ab}$$ and $$ab\ne 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral related to Pythagoras theorem ${2\over \pi}\int_{0}^{\infty}{(ab)^3\over (a^2+b^2x^2)(b^2+a^2x^2)}\mathrm dx=h^2$. Integral related to Pythagoras theorem
Triangle ABC is a right angle triangle, where Angle $ABC=90^o$.
$h$ is perpendicular to the hypotenuse AC and meet at angle ABC.
Where $a$ and $b$ are two small sides
How can I Show that h can be represented in term of this integral $(1)$
$${2\over \pi}\int_{0}^{\infty}{(ab)^3\over (a^2+b^2x^2)(b^2+a^2x^2)}\mathrm dx=h^2\tag1$$.
Any hints on this can be relate to Pythagoras theorem
Basic formulas : $AC^2=AB^2+BC^2$ and area, $A={bh\over 2}$
| Hint: Observe if you have a right triangle with legs $a, b$ and $h$ is as specified, then we see that
\begin{align}
\text{Area} = \frac{a\cdot b}{2} = \frac{h\cdot \sqrt{a^2+b^2}}{2} \ \ \Rightarrow \ \ h^2 = \frac{a^2b^2}{a^2+b^2}.
\end{align}
One can show that the integral
\begin{align}
\frac{2}{\pi}\int^\infty_0 \frac{(ab)^3dx}{(a^2+b^2x^2)(b^2+a^2x^2)} = \frac{(ab)^3}{ab(a^2+b^2)}= \frac{a^2b^2}{a^2+b^2} = h^2.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of $\int\frac{dx}{x+ \sqrt{x^2-x+1}}$ Evaluate :
$$\frac{dx}{x+ \sqrt{x^2-x+1}}$$
After dividing and multiply with $x-\sqrt{x^2-x+1}$, I get
$x+\ln |x-1|-\int \frac{\sqrt{x^2-x+1}}{x-1}dx$.
Is $\int \frac{\sqrt{x^2-x+1}}{x-1}dx$ is integrable in terms of elementary functions?
| We have $$I =\int \frac {\sqrt {x^2-x+1}}{x-1} dx =\frac {1}{2 }\int \frac {\sqrt {(2x-1)^2+3}}{x-1} dx $$ Substituting $u =2x-1$, we get, $$I =\frac {1}{2} \int \frac {\sqrt{u^2+3}}{u-1} du $$ Now perform hyperbolic substitution $u =\sqrt {3} \operatorname {sinh}(v)$ and simplifying gives us $$I =\frac {3}{2} \int \frac {(\operatorname {cosh }(v))^2}{\sqrt {3}\operatorname{sinh}(v) -1} dv $$ which can be solved by Weierstrass substitution. Hope it helps.
| {
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A 9 dice are rolled at the same Time?what is the possible number of combinations, so that we have ( exactly )one 1 or at (least) a 4 or (both) A 9 dies are rolled at the same Time !
what is the possible number of combinations, so that we have ( exactly )one 1 or at (least) a 4 or (both)
My solution ;
Let C = exactly one 1
B = at least a 4
$Let A = C \vee B \vee ( C \wedge B) $
$ \neg A = \neg C \wedge \neg B \wedge (\neg B \vee \neg C ) $
$ \neg A = \neg C \wedge \neg B \wedge \neg C \vee \neg C \wedge \neg B \wedge \neg B$
$ \neg A = \neg C \wedge \neg B $
$ \neg C$ = no 1 $\vee$ at least two 1's
$ C_{1} $ = no 1 ;
$ C_{2} $ = at least two 1's
$ \neg C = C_{1} \vee C_{2} $
$ \neg B$ = no 4
$ \neg A = (C_{1} \wedge \neg B ) \vee (C_{2} \wedge \neg B ) $
$ |C_{1} \wedge \neg B| $ = no 1 and no 4 = $ \binom{9+4-1}{9} $
$ |C_{2} \wedge \neg B| $ = at least two 1's and no 4 = $ \binom{7+4-1}{
7}+\binom{6+4-1}{6}+\binom{5+4-1}{5}+\binom{4+4-1}{4}+\binom{3+4-1}{3}+\binom{2+4-1}{1}+\binom{1+4-1}{1}+1 $
$ C_{1} \wedge \neg B $ and $ C_{2} \wedge \neg B $ are disjoint set
$|\neg A| $ = $ |C_{2} \wedge \neg B| +|C_{1} \wedge \neg B| $
$ |A| = \binom{9+6-1}{9}- |\neg A|$
Is my answer correct ?
| Rolling 9 identical dice, in general, corresponds to placing 9 identical balls in 6 boxes.
To take care of the conditions placed, use stars and bars for the following $2$ cases:
*
*Zero $1's$, at least one $4$:
Consider as if only boxes $2-6$ left, with one $4$ preplaced $\to\binom{8+5-1}{8}$
*Exactly one $1$:
Preplace one $1$, again consider as if only boxes $2-6$ left $\to \binom{8+5-1}{8}$
Add up.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\int^{\frac{\pi}{2}}_0 \frac{\sin x}{\sin x+\cos x+1}dx$ What results?
$$\int^{\frac{\pi}{2}}_0 \frac{\sin x}{\sin x+\cos x+1}dx$$
my try : $u= \tan \frac{x}{2 } $
but : What is the short way?
| \begin{eqnarray}
\sin x+\cos x+1 &=& 1+\sin x+\cos x \\
&=& (\sin\frac{x}{2}+\cos\frac{x}{2})^2+(\cos^2\frac{x}{2}-\sin^2\frac{x}{2}) \\
&=& (\sin\frac{x}{2}+\cos\frac{x}{2})2\cos\frac{x}{2}
\end{eqnarray}
We have by substitution $x=\dfrac{\pi}{2}-u$
\begin{eqnarray}
\int^{\frac{\pi}{2}}_0 \frac{\sin u}{\sin u+\cos u+1}du
&=&
\int^{\frac{\pi}{2}}_0 \frac{\cos x}{\sin x+\cos x+1}dx\\
&=&
\int^{\frac{\pi}{2}}_0 \frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}{(\sin\frac{x}{2}+\cos\frac{x}{2})2\cos\frac{x}{2}}dx\\
&=&
\frac12\int^{\frac{\pi}{2}}_0 \frac{(\cos\frac{x}{2}-\sin\frac{x}{2})}{\cos\frac{x}{2}}dx\\
&=&
\frac12\int^{\frac{\pi}{2}}_0 1-\tan\frac{x}{2}dx\\
&=&
\frac{\pi}{4}+\ln\cos\frac{x}{2}\Big|_0^\frac{\pi}{2}\\
&=&
\color{blue}{\frac{\pi}{4}-\frac12\ln2}
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How do we evaluate $\int\frac{dx}{\sin^{4}{x}+1}$? How do we evaluate $$\displaystyle\int{\dfrac{dx}{\sin^{4}{x}+1}}?$$
don't know how to solve this, please help.
| $\displaystyle \int \dfrac{dx}{\sin^4 x+1}$
Multiplying top and bottom with $\csc^4 x$ gives
$\displaystyle \int \dfrac{\csc^4 x}{1+\csc^4 x}\,dx$
$=\displaystyle \int \dfrac{\csc^2 x(1+\cot^2 x)}{1+(1+\cot^2 x)^2}\,dx$
Let $u=\cot x\implies du=-\csc^2 x\,dx$
$=\displaystyle \int \dfrac{-(1+u^2)}{1+(1+u^2)^2}\,du$
$=\displaystyle -\int \dfrac{1+u^2}{(1+u^2)^2-i^2}\,du$
$=\displaystyle -\int \dfrac{1+u^2}{(u^2+1+i)(u^2+1-i)}\,du$
$=\displaystyle -\dfrac{1}{2}\int \dfrac{(u^2+1+i)+(u^2+1-i)}{(u^2+1+i)(u^2+1-i)}\,du$
$=\displaystyle -\dfrac{1}{2}\int \left[\dfrac{1}{u^2+1-i}+\dfrac{1}{u^2+1+i}\right]\,du$
$=\displaystyle -\dfrac{1}{2}\int \left[\dfrac{1}{u^2+(\sqrt{1-i})^2}+\dfrac{1}{u^2+(\sqrt{1+i})^2}\right]\,du$
$=-\dfrac{1}{2}\left[\dfrac{1}{\sqrt{1-i}}\arctan\left(\dfrac{u}{\sqrt{1-i}}\right)+\dfrac{1}{\sqrt{1+i}}\arctan\left(\dfrac{u}{\sqrt{1+i}}\right)\right]+C$
$=-\dfrac{1}{2}\left[\dfrac{1}{\sqrt{1-i}}\arctan\left(\dfrac{\cot x}{\sqrt{1-i}}\right)+\dfrac{1}{\sqrt{1+i}}\arctan\left(\dfrac{\cot x}{\sqrt{1+i}}\right)\right]+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2109901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
series: $\frac{1}{2\cdot 3\cdot 4}+\frac{1}{4\cdot 5\cdot 6}+\frac{1}{6\cdot 7\cdot 8}+\cdots$
We have the infinite series:$$\frac{1}{2\cdot 3\cdot 4}+\frac{1}{4\cdot 5\cdot 6}+\frac{1}{6\cdot 7\cdot 8}+\cdots$$
This is not my series: $\frac{1}{1\cdot 2\cdot 3}+\frac{1}{2\cdot 3\cdot 4}+\frac{1}{3\cdot 4\cdot 5}+\frac{1}{4\cdot 5\cdot 6}+\cdots$ so I cannot use $\sum_{k=1}^n \frac{1}{k(k+1)(k+2)}$
My attempt:
I know that this type of series solved by making it telescoping series but here, I am unable find general term of the series. Thank you.
| Let
$$ f(x)=\sum_{n=1}^\infty\frac1{n(n+1)(n+2)}x^{n+2}. $$
Then
$$ f'(x)=\sum_{n=1}^\infty\frac1{n(n+1)}x^{n+1},f''(x)=\sum_{n=1}^\infty\frac1{n}x^{n},f'''(x)=\sum_{n=1}^\infty x^{n-1}=\frac1{1-x}. $$
So
$$ f''(x)=-\ln(1-x)$$
and
$$ f(1)=-\int_0^1\int_0^x\ln(1-t)dtdx=\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Volume of a piece of ellipsoid I want to compute the volume of $A:=\{(x,y,z)\in\mathbb{R}^3:\frac{x^2}{4}+\frac{y^2}{9}+\frac{z^2}{25}\leq 1; x\geq\frac{y^2}{9}+\frac{z^2}{25}\}$ (which I think is a piece of ellipsoid) so I set up the following integral: $\iiint_A 1 dxdydz=\int_{x=0}^{x=2}\int_{-3\sqrt{x}}^{3\sqrt{x}}\int_{-\frac{5}{3}\sqrt{9x^2-y^2}}^{\frac{5}{3}\sqrt{9x^2-y^2}}1 dzdydz=30\pi$ which should be equal to the required volume.
Is this correct? (i.e. is the integral I computed equal to the volume of $A$)
Best regards,
lorenzo.
| The intersection between the ellipsoid and the paraboloid is given by
$$
\frac{x^2}{4}+x=1\quad\Rightarrow\quad x=2(\sqrt{2}-1)>0
$$
So the projection of this intersection in the plane $x=0$ is the ellipse
$$
\frac{y^2}{9}+\frac{z^2}{25}=2(\sqrt{2}-1)
$$
Let $D$ be the region bounded by this ellipse:
$$
D=\{(y,z)|\frac{y^2}{9}+\frac{z^2}{25}\le2(\sqrt{2}-1) \}
$$
Your solid is thus defined by
$$
E=\{(x,y,z)|(y,z)\in D,\; \frac{y^2}{9}+\frac{z^2}{25} \le x \le 2\sqrt{1-\frac{y^2}{9}-\frac{z^2}{25}} \}
$$
To integrate this, consider the non distorted version of your problem: suppose the ellipsoid is a sphere with radius $1$ and that the paraboloid has equation $x=y^2+z^2$. In this case, $D$ is a disc with radius $R=\frac{\sqrt{5}-1}{2}$, and the volume equals:
$$
V_0=\int_0^{2\pi}\int_0^R\int_{r^2}^{1-r^2}rdr d\theta = \frac{5\pi(3-\sqrt{5})}{12}
$$
You can adapt this last equation with the coefficients of the initial equations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2111832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $\lim_{x\to 0} \frac{\sin(2x)-\sin(Ax)}{x+x^3} = A^2$ How do I solve for $A$? $$\lim_{x\to 0} \frac{\sin(2x)-\sin(Ax)}{x+x^3} = A^2$$
Since the denominator evaluates to $0$, I tried doing $$\lim_{x\to 0} [\sin(2x)-\sin(Ax)]=A^2 \cdot \lim_{x\to0}[x+x^3]$$ but it would go into $0=0$.
If I checked from the graph, then I believe that $A=1$.
Any advice on how to do this? I haven't learned l'Hopital's rule yet so I would rather avoid using it for now.
Thanks!
| $$\begin{align}\lim_{x\to 0} \frac{\sin(2x)-\sin(Ax)}{x+x^3}&=\lim_{x\to 0} \left[\frac{\sin(2x)-\sin(Ax)}{x+x^3}\cdot\frac{\frac{1}{x}}{\frac{1}{x}}\right]\\&=\lim_{x\to 0} \frac{2\frac{\sin(2x)}{2x}-A\frac{\sin(Ax)}{Ax}}{\frac{x+x^3}{x}}\\
&=\lim_{x\to 0} \frac{2\frac{\sin(2x)}{2x}-A\frac{\sin(Ax)}{Ax}}{1+x^2}\\
&=\frac{2(1)-A(1)}{1+0}\\
&=2-A.\end{align}$$
Since we want to have
$$\lim_{x\to 0} \frac{\sin(2x)-\sin(Ax)}{x+x^3}=A^2,$$ we get
$$2-A=A^2,$$
that is, $$A^2+A-2=0\iff (A+2)(A-1)=0.$$
Thus, $A=-2$ or $A=1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Question on proof of $1+2+\dots+n=\frac{n(n+1)}{2}$ by induction. I saw some video where it needs to prove $1+2+\dots+n=\frac{n(n+1)}{2}$ inductively. So it has to be true if $k=1$ and $k+1$ are true.
So, for $k=1$:
$$1=\frac{1(1+1)}{2}=\frac{1(2)}{2}=\frac{2}{2}=1$$
it is valid.
For $k+1$ here is the proof he does:
$$
\begin{align}
1+2+\dots+k+k+1&=\frac{(k+1)(k+2)}{2} &(1)\\
\frac{k(k+1)}{2}+k+1&=\frac{(k+1)(k+2)}{2} &(2)\\
&=\frac{k^2+2k+k+2}{2}&(3)\\
&=\frac{k^2+3k+2}{2}&(4)\\
&=\frac{(k+1)(k+2)}{2}&\text{factoring (4)}
\end{align}$$
Therefore this formula is valid for $k+1$.
But is this true? I think not. He is just undoing what he have just done. To prove it I think I need to do this:
$$
\begin{align}
1+2+\dots+k+k+1&=\frac{(k+1)(k+2)}{2}\\
\frac{k(k+1)}{2}+k+1&=\frac{(k+1)(k+2)}{2}\\
\frac{k(k+1)+2(k+1)}{2}&=\frac{(k+1)(k+2)}{2}\\
\frac{(k+1)(k+2)}{2}&=\frac{(k+1)(k+2)}{2}\\
(k+1)(k+2)&=(k+1)(k+2)\\
k+1&=k+1\\
k&=k\\
0&=0\text{ or }1=1
\end{align}$$
Therefore, $k+1$ is valid in this formula.
Is this right or am I making a mistake somewhere?
| The second way is correct but I would be more carefull and use like this:
$$1+2+...+k+(k+1)=\frac{k(k+1)}{2}+k+1=\frac{k^2+3k+2}{2}=\frac{(k+1)(k+2)}{2}$$
P.s: Not use the equality in the first place. The problem using the equality is that you need guarantee equivalence in every step what is not so easy in many problems.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Solve for x (quadratic equation) How do you solve the following equation:
\begin{align*}
\sqrt{2017 + \sqrt{2017 - x}} &= x
\end{align*}
I tried squaring it twice, but then I am left with quadratic equation that I can not solve.
| Let $2017-x=y^2$, where $y\geq0$.
Hence, $x+y>0$ and $2017+y=x^2$, which gives $y+x=x^2-y^2$ or $x-y=1$ and the rest is smooth.
We have $y=x-1$, $2017+x-1=x^2$ or $$x^2-x-2016=0$$ and since $x>0$,
we get the answer $\{\frac{1+\sqrt{8065}}{2}\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many solutions does a trigonometric function have $0\le x \le 2\pi$? How do I algebraically determine how many solutions a trigonometric solution have $0\leq x\leq 2\pi$? So far I have been drawing graphs for each question and counted the solutions but I want a way to do this algebraically. Many people tell me to use the double angle identities but I haven't learned it yet. I know that the period of $\sin$ and $\cos$ are $2\pi$ and $\tan$ is $\pi$. One of the questions on my test was like..
Determine how many solutions do the following equations have for $0\leq x\leq 2\pi$.
a) $\sin(3x)=-1/4 $
b)$(\tan(2x))^2=1$
Thanks in advance.
| If $0\le x \le 2\pi$ then $0 \le 3x \le 6 \pi$
$\sin (\theta) = -1/4$ will have two solutions in $0 \le \theta \le 2\pi$[$*$] so it will have two solutions in $2\pi \le \theta \le 4\pi$ and two solution is $4\pi \le \theta \le 6\pi$. So there are six solutions for $0 \le \theta = 3x \le 6\pi$ or in other words $0 \le x \le 2\pi$.
$\tan (\theta)^2 = 1 \implies \tan (\theta) = \pm 1$.
$\tan(\theta) = 1 \implies \sin \theta = \cos \theta$ has 2 solution in $0\le \theta \le 2\pi$. (At $\theta = \pi/4$ and $\theta = \pi + \pi/4$).
$\tan(\theta) = -1 \implies \sin \theta = -\cos \theta$ has 2 solutions as well. (At $\pi - \pi/4$ and $2\pi - \pi/4$).
So $\tan (\theta)^2 = 1$ has 4 solutions in $0 \le \theta \le 2\pi$.
If $\theta = 2x$ and $0 \le x \le 2\pi$ then $0 \le 2x = \theta \le 4\pi$ and so there are $8$ solutions for $0 \le 2x = \theta \le 4\pi$.
[$*$] Note: we don't have to actually solve the equation. We know there will be one solution in each the third and fourth quadrant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find $\tan(-\frac{5\pi}{16})$ with half-angle formulas? How to find $\tan(-\frac{5\pi}{16})$ with half-angle formulas?
I tried the $\pm \sqrt{\frac{1-\cos{A}}{1+\cos{A}}}$ and $\frac{\sin{A}}{1+\cos{A}}$ but got stuck once there were square roots on top and bottom like $\frac{\sqrt{...}}{1-\sqrt{...}}.$
Using the cosine over cosine in square root I got up to
$$=-\sqrt{ \frac{ 1+\sqrt{\frac{1+\cos(5\pi/3)}{2}}}{1-\sqrt{\frac{1+\cos(5\pi/3)}{2}}} }$$
| Apart from a couple of threes that might be fours ? You are fine upto here
\begin{eqnarray*}
\tan(-\frac{5\pi}{16})=-\sqrt{ \frac{ 1+\sqrt{\frac{1+\cos(5\pi/4)}{2}}}{1-\sqrt{\frac{1+\cos(5\pi/4)}{2}}}}.
\end{eqnarray*}
Now $\cos(5\pi/4)=\frac{ 1}{\sqrt{2}}$ & after a little bit algebra ...
\begin{eqnarray*}
\tan(-\frac{5\pi}{16})=-\sqrt{ \frac{ 8^{1/4}+\sqrt{\sqrt{2}-1}}{ 8^{1/4}-\sqrt{\sqrt{2}-1}}}.
\end{eqnarray*}
My Casio fx-83MS gives both sides equal to $-1.4966 \cdots $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Solve the equation $ (100 a+10b+c)^2 =(a+b+c)^5.$ Find a three-digits number $\overline{abc}$ such that $\overline{abc}^2=(a+b+c)^5.$
It is easy to see that
$$
(a+b+c)^5 \leq 999^2 \implies a+b+c< \sqrt[5]{999^2}\leq 15
$$
and
$$
(100 a+10b+c)^2<15^5 \implies 100 a+10b+c>\sqrt{15^5} \leq 871.
$$
Also
$$
(100 a+10b+c)^2 =(a+b+c) \mod 2
$$
implies $a+b=0 \mod 2$.
Similarly
$$
(100 a+10b+c)^2 = c^2 = (a+b+c) \mod 5.
$$
But it not enough to find the solution $243$.
No more ideas.
| You should be able to quickly see that $(a+b+c)$ must be a square. Since $999<1024$, we know that $(a+b+c)<16$ so we have ${a+b+c}\in\{1,4,9\}$, and thus $\sqrt{(a+b+c)^5}\in\{1,32,243\}$, the last of which produces a valid solution. Since you form $\overline{abc}$ as a number I'll assume $a$ cannot be zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2121275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Algebraic manipulation: $\sum_{r=1}^n \frac{n-r-1}{n-2} \frac{n-r}{n-1} \frac{1}{n} = \frac{1}{3}$ If you're up for a tedious algebraic manipulation, I'm stuck.
The following is true: $\sum_{r=1}^n \frac{n-r-1}{n-2} \frac{n-r}{n-1} \frac{1}{n} = \frac{1}{3}$.
However, I can't for the life of me figure out the how to show it.
| The sum can rewrite in this way:
$$\frac{1}{(n-2)(n-1)n}\sum_{r=1}^n(n^2-nr-n-nr+r^2+r)=$$
$$=\frac{1}{(n-2)(n-1)n}\left(n^3-n(\frac{n(n+1)}{2})-n^2-n(\frac{n(n+1)}{2})+(\frac{n(n+1)(2n+1)}{6})+(\frac{n(n+1)}{2})\right)=$$
$$=\frac{1}{6(n-2)(n-1)n}(6n^3-3n^3-3n^2-6n^2-3n^3-3n^2+2n^3+3n^2+n+3n^2+3n)=$$
$$=\frac{1}{6(n-2)(n-1)n}(2n^3-6n^2+4n)=\frac{1}{6(n-2)(n-1)}(2n^2-6n+4)=$$
$$=\frac{1}{3(n-2)(n-1)}(n^2-3n+2)=\frac{(n-2)(n-1)}{3(n-2)(n-1)}=\frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find the derivative of $f=\arcsin\left(\frac{2x}{1+x^2}\right)$ I'm trying to find the derivative of $f=\arcsin\left(\frac{2x}{1+x^2}\right)$. I think I'm mistaken and perhaps using the chain rule incorrectly.
Let $g(x) = \frac{2x}{1+x^2}$ and let $h(x) = \arcsin x$
According to the chain rule -
$$f'(x) = \frac{1}{\sqrt{1-\frac{2x}{1+x^2}}}⋅((2⋅(1+x^2 )-2x⋅2x)/(1+x^2 )^2 ) = \cdots \frac{-2(x^4-1)}{x-1}$$
Is this a correct usage of the chain rule?
| \begin{align}
\frac d {dx} \arcsin\left( \frac{2x}{1+x^2} \right) & = \frac 1 {\sqrt{1-\left( \frac{2x}{1+x^2} \right)^2}} \cdot \frac d {dx} \frac{2x}{1+x^2} & & \text{by the chain rule} \\[10pt]
& = \frac 1 {\sqrt{1-\left( \frac{2x}{1+x^2} \right)^2}} \cdot \frac{2(1-x^2)}{(1+x^2)^2} & & \text{by the quotient rule} \\[10pt]
& = \frac 1 {\sqrt{\frac{(1+x^2)^2 - (2x)^2}{(1+x^2)^2} }} \cdot \frac{2(1-x^2)}{(1+x^2)^2} \\[10pt]
& = \frac{1+x^2}{\sqrt{(1+x^2)^2 - (2x)^2}} \cdot \frac{2(1-x^2)}{(1+x^2)^2} \\[10pt]
\end{align}
Now notice that $$ (1+x^2)^2 - (2x)^2 = (1+2x^2+x^4) - 4x^2 = 1-2x^2 + x^4 = (1-x^2)^2 $$
so we get
\begin{align}
\sqrt{(1+x^2)^2 - (2x)^2} = |1-x^2|.
\end{align}
If $-1\le x\le1$ then this is $1-x^2$ and so by cancellations from the numerator and denominator the derivative simplifies to
$$
\frac 2 {1+x^2} \qquad \left( = 2 \frac d {dx} \arctan x \right).
$$
However, since arcsine is not differentiable at $\pm1,$ the chain rule does not tell us that the composite function is differentiable at points where $2x/(1+x^2)\in\{\pm1\}.$ Those two points must therefore be examined separately.
If $x>1$ or $x<-1$ then $|1-x^2| = -(1-x^2)$ and the answer we get will be multiplied by $-1$.
Notice that $2x/(1+x^2)=\pm1$ when $x=\pm1$ and has absolute extreme values at those points. Thus $2x/(1+x^2)$ is never outside the domain of the arcsine function.
So the derivative is $\dfrac{-2}{1+x^2}$ when $x>1$ or $x<-1$.
When $x=\pm1$ then $2x/(1+x^2)=\pm1$ and one can find the two one-sided derivatives at those points. Lo and behold, they are not equal, so the function is not differentiable at those two points.
$$
\frac d {dx} \arcsin\left( \frac{2x}{1+x^2}\right) = \begin{cases} \dfrac 2 {1+x^2} & \text{if } -1<x<1, \\[8pt] \dfrac{-2}{1+x^2} & \text{if } x>1 \text{ or } x<-1. \end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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convergence of $\int_{\mathbb{R}^3}\frac{dx\,dy\,dz}{1+x^4+y^4+z^4}$ I need to determine whether the integral
$$\int_{\mathbb{R}^3}\frac{dx\,dy\,dz}{1+x^4+y^4+z^4}$$
converges or diverges.
I have no good idea for a substitution, but I tend to believe that the way to solve it is to find some upper bound of the function that converges, or vice versa:
a lower bound that diverges.
Any ideas?
| $$\iiint_{\mathbb{R}^3}\frac{d\mu}{1+(x^2+y^2+z^2)^2} = \int_{0}^{+\infty}\frac{4\pi\rho^2}{1+\rho^4}\,d\rho = \pi^2\sqrt{2}$$
and in a similar way
$$\iiint_{\mathbb{R}^3}\frac{d\mu}{1+\frac{1}{3}(x^2+y^2+z^2)^2} = \int_{0}^{+\infty}\frac{4\pi\rho^2}{1+\frac{1}{3}\rho^4}\,d\rho = \pi^2 3^{3/4}\sqrt{2} $$
hence the given integral is convergent since $x^4+y^4+z^4\geq \frac{1}{3}\left(x^2+y^2+z^2\right)^2$.
The exact value of the integral is given by
$$ \iiint_{\mathbb{R}^3}\int_{0}^{+\infty}\exp\left(-t(1+x^4+y^4+z^4)\right)\,dt\,d\mu = \int_{0}^{+\infty}\left(\frac{2\,\Gamma\left(\frac{5}{4}\right)}{t^{1/4}}\right)^3 e^{-t}\,dt$$
that is:
$$ 8\,\Gamma\left(\frac{1}{4}\right)\,\Gamma\left(\frac{5}{4}\right)^3 = \color{red}{\frac{1}{8}\,\Gamma\left(\frac{1}{4}\right)^4}.$$
The second approach also shows that given $\alpha,\beta,\gamma>0$,
$$ \iiint_{(0,+\infty)}\frac{dx\,dy\,dz}{1+x^{\alpha}+y^{\beta}+z^{\gamma}}$$
is convergent iff $\color{red}{\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}<1}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find the minimal-polynomial of $\begin{pmatrix} 0&1&0&1\\1&0&1&0\\0&0&0&0\\0&0&0&0\end{pmatrix}$
$E:=\begin{pmatrix} 1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix} A:= \begin{pmatrix} 0&1&0&1\\1&0&1&0\\0&0&0&0\\0&0&0&0\end{pmatrix} A^2:= \begin{pmatrix} 1&0&1&0\\0&1&0&1\\0&0&0&0\\0&0&0&0\end{pmatrix}$
My attempt:
The idea is to find a linear combination $lin(E+A+A^2+...+A^n) = A^{n+1}$
I tried it by taking the first column of each matrix and look for a solution:
(1) E|A
$x\begin{pmatrix} 1\\0\\0\\0\end{pmatrix} = \begin{pmatrix} 0\\1\\0\\0\end{pmatrix}$
This can obviously not work
(2) $lin(E+A) = A^2$
$x\begin{pmatrix} 1\\0\\0\\0\end{pmatrix} +y \begin{pmatrix} 0\\1\\0\\0\end{pmatrix} = \begin{pmatrix} 1\\0\\0\\0\end{pmatrix}$
Which works for $x=1, y=0.$
Now the minimalpolynomial should be $0\cdot E+ 1\cdot A + A^2 \rightarrow 0+x+x^2$, but it seems to be wrong in further calculations.
| The quick way to deduce the minimal polynomial (or Jordan normal form or anything else) is to note that $A = B \otimes C$ where
$$
B = \pmatrix{1&1\\0&0}, \quad C = \pmatrix{0&1\\1&0}
$$
and $\otimes$ denotes the Kronecker product.
Since $B$ has minimal polynomial $x^2 - x$ and $C$ has minimal polynomial $x^2 - 1$, the minimal polynomial of $A$ is the least common multiple $x^3 - x$.
Alternatively, my approach would be to calculate the characteristic polynomial, then try whichever factors make sense. In this case, we find the characteristic polynomial to be
$$
x^2(x-1)(x+1)
$$
the minimal polynomial must divide this. However, the zeros of the minimal polynomial must include all eigenvalues. So, the only possibilities are that the minimal polynomial is the same as the characteristic polynomial, or that it's the factor $x(x-1)(x+1)$.
Yet another method:
It seems clear to me that $E,A,$ and $A^2$ are linearly independent. We must therefore use another power of $A$. Upon calculation, it's clear that $A^3 = A$, which is to say that $A^3 - A = 0$. Since we've excluded the possibility of degree $2$ minimal polynomials, the answer must be $x^3 - x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proof convergence of series using limit of summand I have the following series: $\sum^\infty_{n=0}(\frac{n+(-1)^n}{2n+10})^{\frac{n}{3}}$ and I have to check wether it converges or not. Which convergence test should I take? I think the limit of the summand is zero so the series should converge but how do I prove that?
| $$
\begin{align}
\,& 0\le\left[\frac{n+\color{red}{(-1)^n}}{2n+10}\right]^{\frac{n}{3}}\le\left[\frac{n+\color{red}{1}}{2n+10}\right]^{\frac{n}{3}} \space\implies \\[4mm]
\,& 0\le\sum_{n=0}^{\infty}\left[\frac{n+(-1)^n}{2n+10}\right]^{\frac{n}{3}}\le\sum_{n=0}^{\infty}\left[\frac{n+1}{2n+10}\right]^{\frac{n}{3}} \space\rightarrow{\small\text{converges by ratio test or root test}} \\[6mm]
\,& \small\lim_{n\rightarrow\infty}\left|\frac{a_{\small n+1}}{a_{\small n}}\right|=\lim_{n\rightarrow\infty}\frac{\left[\frac{(n+1)+1}{2(n+1)+10}\right]^{\frac{n+1}{3}}}{\left[\frac{n+1}{2n+10}\right]^{\frac{n}{3}}}=\lim_{n\rightarrow\infty}\sqrt[3]{\frac{1+2/n}{2+12/n}}\,\left[\frac{(n+2)(2n+10)}{(n+1)(2n+12)}\right]^{\frac{n}{3}} =\color{red}{\frac{1}{\sqrt[3]{2}}}\,\lt1
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2130751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding the area between a line and a curve
The two equations are $x+1$ and $4x-x^2-1$.
The answer is $\frac{1}{6}$, but I've done it 4 different times and gotten -$\frac{15}{2}$ each time.
My working:
*
*$x+1$ = $4x-x^2-1$
*$x^2-3x+2 = 0$
*$(x-1)(x-2)$ means $x=1$ or $x=2$
*$\int_1^2$ $3x-x^2$
*$[\frac{3x^2}{2}-\frac{x^3}{3}]_1^2$
*$\frac{3(1)^2}{2}-\frac{(1)^3}{3}$ = $\frac{3}{2}-\frac{1}{3}$
*$\frac{3(2)^2}{2}-\frac{(2)^3}{3}$ = $\frac{12}{2}-\frac{8}{3}$
*($\frac{3}{2}-\frac{1}{3}$)-($\frac{12}{2}-\frac{8}{3}$) = $-\frac{9}{2}-\frac{9}{3}$
*-$\frac{15}{2}$
| So the area is $$\begin{align}A&=\int_{1}^2\big[(4x-x^2-1)-(x+1)\big]dx\\
&=\int_{1}^2(-x^2+3x-2)dx\\
&=\Big[\frac{-x^3}{3}+\frac{3x^2}{2}-2x\Big]_{1}^{2}\\
&=\left(-\frac{8}{3}+6-4\right)-\left(-\frac{1}{3}+\frac{3}{2}-2\right)\\
&=-\frac{8}{3}+2+\frac{1}{3}-\frac{3}{2}+2\\
&=-\frac{7}{3}+4-\frac{3}{2}\\
&=-\frac{7}{3}+\frac{5}{2}\\
&=\frac{1}{6}\\
\end{align}$$
Edit: I got down voted. Well, there is nothing I can do. My only point is to show to the OP that his integral $$\int_{1}^2(3x-x^2)dx$$ is wrong. Maybe he thought that $1$ cancels out. So, I showed how he should have the correct computations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2131864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Calculating area of sphere with constraint on zenith Let $\mathbb S_R$ be the sphere of radius $R$ centered about the origin. Consider $$A_R= \left\{ (x,y,z)\in \mathbb S_R \mid x^2+y^2+(z-R)^2\leq 1 \right\}.$$
I want to calculate the area of this region of the sphere with radius $R$ about the origin. I already calculated the area of the region of $\mathbb S_R$ of vectors with zenith $\leq \alpha$ is given by $2\pi R^2(1-\cos \alpha)$.
For $A_R$, here's my idea. I want to find the zenith $\alpha$ which satisfies $R\sin \alpha=\sin \alpha+R$, since this is the zenith of points of both $\mathbb S_R$ and the unit sphere translated up $R$ units at points which are of the same height. Then, I just want to integrate the zenith $0\leq \phi\leq \alpha$.
Solving gives $\phi=\arcsin \frac R{R-1}$ and things get a little messy.
On the other hand, another student posted a solution which makes sense. He just writes $\alpha$ should satisfy $2R^2-2R^2\cos\alpha=1$ and then obtains the area of $A_R$ is $\pi$, independently of the radius.
Why is my method wrong?
Picture
| I think @David K has addressed your point. Here is how I would proceed:
Parametrize the surface as follows:
\begin{cases}
x=x\\
y=y\\
z=\sqrt{R^2-x^2-y^2}
\end{cases}
with $(x,y)\in D$, where $D$ is projection of the intersection of both spheres in the $xy$-plane. When both spheres intersect, we have
\begin{cases}
x^2+y^2+z^2 = R^2 \\
x^2+y^2+(z-R)^2 = 1
\end{cases}
which leads to $x^2+y^2=1-\frac{1}{4R^2}$. In other words, $D$ is a disc with radius $\sqrt{1-\frac{1}{4R^2}}=\frac{\sqrt{4R^2-1}}{2R}$ (assuming $R\ge \frac{1}{{2}}$). This parametrization only works when the radius of the bottom sphere is smaller than (or equal to) the radius of $D$, i.e. when $R\le \frac{\sqrt{4R^2-1}}{2R}$, or $R\ge 1/\sqrt{2}$.
We need the norm of the normal vector to $S$, i.e., the norm of $$\pmatrix{1\\0\\\frac{-x}{\sqrt{R^2-x^2-y^2}}}\times\pmatrix{0\\1\\\frac{-y}{\sqrt{R^2-x^2-y^2}}}=\pmatrix{\frac{x}{\sqrt{R^2-x^2-y^2}} \\ \frac{y}{\sqrt{R^2-x^2-y^2}} \\ 1} $$
which is
$$
\sqrt{1+\frac{x^2+y^2}{R^2-x^2-y^2}}=\frac{R}{\sqrt{R^2-x^2-y^2}}
$$
It follows that the wanted area equals
$$
\iint_D \frac{R}{\sqrt{R^2-x^2-y^2}}\; dA = \int_0^{2\pi}\int_0^{\frac{\sqrt{4R^2-1}}{2R}} \frac{R}{\sqrt{R^2-r^2}}\; r dr d\theta = 2\pi R^2-\pi |2R^2-1|\\
$$
Remember that we assumed that $R\ge 1/\sqrt{2}$, so this expression equals $\pi$.
If $0<R\le\frac{1}{2}$ the intersection is at most a point, and it is the entire surface of the sphere of radius $R$ that we want, i.e., $4\pi R^2$. And for the other values of $R$ ($1/2\le R\le 1/\sqrt{2}$), we are better off in spherical coordinates and the same reasoning also leads us to $\pi$: you should get
$$
\int_0^{2\pi} \int_0^{\cos^{-1}(1-\frac{1}{2R^2})} R^2\sin\phi\; d\phi d\theta = \pi\quad (1)
$$
So in summary the area equals
\begin{cases}
4\pi R^2 \quad &\mbox{if }0<R\le \frac{1}{2}\\
\pi \quad &\mbox{if } R\ge 1/2
\end{cases}
Note: spherical coordinates are much more interesting here. Equation $(1)$ is simpler to find, to solve, and it is valid for all $R\ge 1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2134064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find limit of $\lim_{n\to\infty} (\sqrt[3]{n^2+5}-\sqrt[3]{n^2+3}$)? I am stuck on this limit.
$$\lim_{n\to\infty} \sqrt[3]{n^2+5}-\sqrt[3]{n^2+3}$$
I couldn't find the limit using the basic properties of limits, since that just yields: $$\infty-\infty$$ which is undefined. Could I get any hints for finding this limit?
| $\lim_{n\to\infty} \sqrt[3]{n^2+5}-\sqrt[3]{n^2+3}
$
Since
$a^3-b^3
=(a-b)(a^2+ab+b^2)
$,
$a-b
=\dfrac{a^3-b^3}{a^2+ab+b^2}
$.
Therefore
$a^{1/3}-b^{1/3}
=\dfrac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}
$.
Since,
for $x > 0$,
$1 < (1+x)^{1/3}
< 1+x/3
$
(cube both sides),
we have,
if $u > 0$,
$n^{2/3}
< (n^2+u)^{1/3}
= n^{2/3}(1+u/n^2)^{1/3}
< n^{2/3}(1+u/(3n^2))
= n^{2/3}+u/(3n^{4/3}))
$.
Therefore,
if $a = n^2+u$
and $b = n^2+v$,
$\begin{array}\\
3n^{4/3}
&< a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\\
&< (n^{2/3}+u/(3n^{4/3})))^2+(n^{2/3}+u/(3n^{4/3})))(n^{2/3}+v/(3n^{4/3})))+(n^{2/3}+v/(3n^{4/3})))^2\\
&= n^{4/3}( (1+u/(3n^2))^2+(1+u/(3n^2))(1+v/(3n^2))+(1+v/(3n^2))^2)\\
&= n^{4/3}(3+(2u+u+v+2v)/(3n^2)+(u^2+uv+v^2)/(3n^2)^2\\
&= 3n^{4/3}(1+(u+v)/(3n^2)+(u^2+uv+v^2)/(3n^4))\\
\end{array}
$
Therefore
$$a^{1/3}-b^{1/3}
=\dfrac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}
< \dfrac{u-v}{3n^{4/3}}
$$
and
$$a^{1/3}-b^{1/3}
> \dfrac{u-v}{3n^{4/3}(1+(u+v)/(3n^2)+(u^2+uv+v^2)/(3n^4))}.
$$
Immediately we have that
$\lim_{n \to \infty} a^{1/3}-b^{1/3}
=0
$.
More precisely,
we also have
$$\dfrac{u-v}{3}
> n^{4/3}(a^{1/3}-b^{1/3})
>\dfrac{u-v}{3}\dfrac1{1+(u+v)/(3n^2)+(u^2+uv+v^2)/(3n^4)}
$$
so that
$$\lim_{n \to \infty} n^{4/3}(a^{1/3}-b^{1/3})
=(u-v)/3
=2/3
$$
since
$u=5, v=3$.
Note:
We can get explicit bounds
by using
$\dfrac1{1+x}
\gt 1-x
$
for $0 < x < 1$
and choosing
$n$ large enough
compared with
$u$ and $v$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2137187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
How do you project vectors? So, I am relatively new to the concept of vectors. I'm not to sure on how to "project a vector", as shown below.
For $\textbf{v} = \begin{pmatrix} 7 \\ 4 \end{pmatrix}$ and $\textbf{w} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$, compute $\text{proj}_{\textbf{w}} \textbf{v}$.
And... For $\textbf{v} = \begin{pmatrix} -10 \\ 6 \end{pmatrix}$ and $\textbf{w} = \begin{pmatrix} 15 \\ -9 \end{pmatrix}$, compute $\text{proj}_{\textbf{w}} \textbf{v}$.
These questions are virtually the same, but I don't know how to calculate the projections, thanks!
| Recall what is the application of the dot product between vectors in physics:
$$
\mathbf{v} \cdot \mathbf{w} = \text{scalar} = \left| {proj_\mathbf{w} \mathbf{v}} \right| \cdot \left| \mathbf{w} \right|
$$
then
$$
\left| {proj_\mathbf{w} \mathbf{v}} \right| = \frac{{\mathbf{v} \cdot \mathbf{w}}}
{{\left| \mathbf{w} \right|}}
$$
and
$$
proj_\mathbf{w} \mathbf{v} = \left| {proj_\mathbf{w} \mathbf{v}} \right|\frac{\mathbf{w}}
{{\left| \mathbf{w} \right|}} = \left( {\frac{{\mathbf{v} \cdot \mathbf{w}}}
{{\left| \mathbf{w} \right|^2 }}} \right)\mathbf{w} = \left( {\frac{{\mathbf{v} \cdot \mathbf{w}}}
{{\mathbf{w} \cdot \mathbf{w}}}} \right)\mathbf{w}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2140959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Inequality trouble: $(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2) \geq (ab+bc+ca)^3$ The following inequality is exercise 1.8 from this book.
For any real $a,b,c$, prove the following $$(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2) \geq (ab+bc+ca)^3.$$
I've managed to prove this via brut-force and Muirhead's inequality (Very unsatisfying). However I'm having difficulty understanding the solution in the back of the book. I'm also interested if anyone else knows of any nice proofs of this inequality?
In particular, the solution at the back says the following (word for word):
If we can show that $\frac{27}{64}(a+b)^2(b+c)^2(c+a)^2\geq (ab+bc+ca)^2$, then the conclusion follows. Denote $S_1=a+b+c$, $S_2=ab+bc+ca$ and $S_3=abc$. We need to show that $27(S_1S_2-S_3)^2\geq 64S_2^3.$
It then goes onto prove the last inequality via cases. However I'm struggling to see how it is sufficient to prove either of those two inequalities.
We can rewrite $a^2+ab+b^2 = (\frac{a+b}{2})^2+\frac{3}{4}(a^2+b^2)$ and so by AM-GM it is sufficient to prove $$\sqrt{\frac{27}{64}(a+b)^2(b+c)^2(c+a)^2}\sqrt{(a^2+b^2)(b^2+c^2)(c^2+a^2)} \geq (ab+bc+ca)^3 .$$ However $$\sqrt{(a^2+b^2)(b^2+c^2)(c^2+a^2)} \geq (ab+bc+ca)^2 $$ is clearly wrong and so this does not justify the first claim in the solution. This does give me an idea of where the $\frac{27}{64}$ term comes from however. Other than this I'm a bit stuck and any help would be appreciated.
Edit: I realised straight after posting that $a^2+ab+b^2 \neq (\frac{a+b}{2})^2+\frac{3}{4}(a^2+b^2)$. So in fact my work is nonsense and can be ignored.
| It is a typo. In fact $27(S_1S_2-S_3)^2\geq 64S_2^3$ (which is correct) expands to
$$
\frac{27}{64}(a+b)^2(b+c)^2(c+a)^2\geq (ab+bc+ca)^3
$$
with exponent $3$ instead of $2$ on the right-hand side.
Then the wanted inequality follows because
$$
a^2 + ab + b^2 = \frac 34 (a+b)^2 + \frac 14 (a-b)^2 \ge \frac 34 (a+b)^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2142139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
plugging in terms of a sequence I am having some trouble understanding how to plug terms into the following sequence. I was given the following,
$$a_1=1, a_{n+1}=\frac{a_n}{\sqrt{n}}$$ for all n greater or equal to 1.
Plugging in terms we get the following,
$$a_1=1$$
$$a_2=a_1=1$$
$$a_3=\frac{a_2}{\sqrt{2}}=\frac{1}{\sqrt{2}}$$
$$a_4=\frac{a_3}{\sqrt{3}}=\frac{1}{\sqrt{6}}$$
How are they getting these terms $a_2,a_3,a_4$?
| We have that for $n \in \mathbb Z^+$, $$a_{n+1} = \frac {a _n}{\sqrt {n}} \tag {1}$$
Thus, using the relation $(1)$ successively, we get, $$a_2 = a_{1+1} = \frac {a_1}{\sqrt {1}} = \frac {1}{1} =1 $$ $$a_3 = a_{2+1} = \frac {a_2}{\sqrt {2}} = \frac {1}{\sqrt {2}} $$ $$a_4 = a_{3+1 } = \frac {a_3}{\sqrt {3}} = \frac {1}{\sqrt {2}\times \sqrt {3}} = \frac {1}{\sqrt {6}} $$
Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Integer solutions of $(b^2+1)(c^2+1)=a^2+1$ As the title says, I'm interested in integer solutions of the equation $(b^2+1)(c^2+1)=a^2+1$. Is it possible to parametrize the solutions as in the case of Pythagorean triples? If yes, then how would one proceed to find a parametrization in this case?
As suggested in the comments, the Pythagorean quadruples could be a good idea: we find that
$c = \frac{p^2-n^2-m^2}{2mp},\ b = \frac{p^2-n^2-m^2}{2np},\ a = \frac{(p^2+n^2+m^2)(p^2-n^2-m^2)}{4mnp^2}$ where $m,n,p$ are coprime and all the fraction are integers... I don't know if we can eliminate one of the $m,n,p$ using this...
Another take, which shows that there are infinitely many solutions: for fixed $b$ we can write $a^2-(b^2+1)c^2=b^2$ which is a Pell equation. This has a solution $a=b,c=0$ so it has infinitely many... Parametrizations, algorithms are available...
| Partial answer :
Let $n\in\mathbb{N}$ and :
$$(a,b,c)=(n,n+1,n^2+n+1)$$
We have :
$$(a^2+1)(b^2+1)=(n^2+1)(n^2+2n+2)=n^4+2n^3+3n^2+2n+2=(n^2+n+1)^2+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Compute the following without the calculator
$$4\left(5+3\sqrt2\over 2\right)^4-16\left(5+3\sqrt2\over 2\right)^3-17\left(5+3\sqrt2\over 2\right)^2+27\left(5+3\sqrt2\over 2\right)-3$$
Please solve the following equation without using calculator.
Substituting $\left(5+3\sqrt2\over 2\right)$ to x must be the first step, but then I don't know how to factor it.
| Let $f(x)=4x^4-16x^3-17x^2+27x-3$. Then $g(x)=f(x)-4=(4x^2 - 20x + 7)(x^2 + x - 1)$.
Since $a=(5+3\sqrt{2})/2$ satisfies $4a^2-20a+7=0$ we know that $g(a)=0$ and hence
$f(a)=4$. So the answer is $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to use powers on matrices In the questions compute $\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}^6$ and $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{99}$, how would you solve these?
| Since first part is answered by Doug M , for the second bit we can approach by the method of induction.
We consider this matrix ,$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{n}$
Lets check for n=2,
$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}.\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$.
Lets check for n=3,
$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}.\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}$.
I think we got a pattern!
So, our hypthesis is $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{n} = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$ .To prove our hypothesis we use first principle of mathematical induction.
Let us assume that this form is true for $n = k$ that is multiplying $k$ times which gives us $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{k} =\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$
Now if we prove it for $n=k+1$ then it's true for all $n \geq 1$
So, consider $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{k+1} = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{k}.\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$
Now $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{k} = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$ from our assumption, so $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{k+1} = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}.\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & (k+1) \\ 0 & 1 \end{pmatrix}$ from our first case.
Hence this holds for any $n \geq 1$,so as a particular case of yours,for $n=99$,this case also holds that is $\begin{pmatrix} 1 & 1 \\ 0 & 1
\end{pmatrix}^{99} = \begin{pmatrix} 1 & 99 \\ 0 & 1 \end{pmatrix}$.
Hope this helps!
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute $5!25! \mod 31$ For an exercise, I was asked to compute $5!25! \mod 31$. I noticed that $5! = 120 \equiv -4 \equiv 27 \mod 31$. Therefore we have that
$$5!25! \equiv 27 \cdot 25! \mod 31.$$
Because of the congruence of Wilson, I also know that $30! \equiv -1 \mod 31$.
We have that $30! \equiv 30 \cdot 29 \cdot 28 \cdot 26 \cdot 27 \cdot 25! \equiv -1 \mod 31$, so I computed
$$(30 \cdot 29 \cdot 28 \cdot 26) \equiv (-1 \cdot (-2) \cdot (-3) \cdot (-5)) \equiv 30 \equiv -1 \mod 31.$$ Hence we find that $-1 \cdot (27 \cdot 25!) \equiv -1 \mod 31$ and therefore $27 \cdot 25! \equiv 1 \mod 31$. This proves that $5!25! \equiv 1 \mod 31$.
Is this correct?
| $$5!25!\equiv\left(15!\right)^2\equiv1^2=1$$
| {
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If $x,y,z>0$, prove that $\frac{x+y+z}{3\sqrt{3}} \geq \frac{yz+zx+xy}{\sum \limits_{cyc} \sqrt{x^2+xy+y^2}}$
If $x,y,z > 0$, prove that $\dfrac{x+y+z}{3\sqrt{3}} \geq \dfrac{yz+zx+xy}{ \sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}}$
with equality if and only if $x=y=z$.
SOURCE :CRUX (Page Number 20 ; Question Number 805)
I tried various approaches like C-S, Holder and Schur, but failed. The problem is very cleverly devised because of which it seems invulnerable to the common tricks.
Any help will be gratefully acknowledged.
Thanks in Advance. :-)
| Let $x^2+y^2+z^2=k(xy+xz+yz)$.
Hence, $k\geq1$ and we need to prove that
$$\sum_{cyc}\sqrt{x^2+xy+y^2}\geq\frac{3\sqrt3(xy+xz+yz)}{x+y+z}$$ or
$$\sum_{cyc}(2x^2+xy)+2\sum_{cyc}\sqrt{(x^2+xy+y^2)(x^2+xz+z^2)}\geq\frac{27(xy+xz+yz)^2}{(x+y+z)^2}$$
Now, by C-S $$\sqrt{(x^2+xy+y^2)(x^2+xz+z^2)}\geq x^2+\frac{x(y+z)}{2}+yz.$$
Thus, it remains to prove that
$$\sum_{cyc}\left(2x^2+xy+2\left(x^2+\frac{x(y+z)}{2}+yz\right)\right)\geq\frac{27(xy+xz+yz)^2}{(x+y+z)^2}$$ or
$$\sum_{cyc}(4x^2+5xy)(x+y+z)^2\geq27(xy+xz+yz)^2$$ or
$$(4k+5)(k+2)\geq27,$$
which is obvious.
Done!
| {
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how to prove this question of eigen-values and eigen-vectors? If
$$
A=\begin{bmatrix}
\sin\theta & \csc\theta & 1 \\
\sec\theta & \cos\theta & 1 \\
\tan\theta & \cot\theta & 1 \\
\end{bmatrix}
$$
then prove that there does not exist a real value of $\theta$ for which characteristics roots of $A$ are $-1,1,3$
i tried to solve as follows,
sum of eigen value $$=\sin\theta +\cos\theta + 1=-1+1+3=3$$
$$\sin\theta +\cos\theta = 2$$
but what to do next.
| \begin{eqnarray}
a\sin\theta+b\cos\theta&=&\sqrt{a^2+b^2}\left[\frac{a}{\sqrt{a^2+b^2}}\sin\theta+\frac{b}{\sqrt{a^2+b^2}}\cos\theta\right]\\
&=&\sqrt{a^2+b^2}\left[\sin\phi\sin\theta+\cos\phi\cos\theta\right]\\
&=&\sqrt{a^2+b^2}\cos(\theta-\phi)
\end{eqnarray}
Where $$\phi=\arcsin\frac{a}{\sqrt{a^2+b^2}}$$
So the result is a sinusoidal with a phase shift and an amplitude of $\sqrt{a^2+b^2}$
When $a=b=1$ what will be the amplitude?
| {
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What is nth number on this series? I have series of
1+5+12+22+.....
where d=3n+1 where n=0,1,2...
but i can not find the nth number on this series ? can anyone explain how to get nth number on this series?
| We have $$T_1 =1$$ $$T_2 =5 =1+4$$ $$T_3 = 12 = 1+4+(4+3) $$ $$T_4 =22 =1+4+ (4+3) + (4+3+3) $$ $$T_5 =35 = 1+4+(4+3)+(4+3+3)+(4+3+3+3) $$ $$\vdots $$
We can thus write $$T_n = 1 + 4 +(4+3) +(4+2\times 3) + \cdots +(4+(n-2)\times 3) $$ $$= 1 + (n-1)\times 4 + (1+2+3+\cdots + (n-2))\times 3$$ $$=1+(n-1)\times 4 + (\sum_{i=1}^{n-2} i)\times 3$$ $$=1+(4n-4) + 3(\frac {(n-2)(n-1)}{2})$$ $$ = (4n-3) + \frac {3n^2-9n+6}{2}$$ $$\boxed {T_n = \frac {3n^2-n}{2}} $$
Hope it helps.
| {
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Calculating limit of recursive sequence I am preparing for a test and wanted to ask you
$a_0 = 1; a_{n+1} = \sqrt{a_n} + \frac{15}{4} $
I already showed its strictly monotonically increasing. Now im trying to calculate the limit.
$$\lim a_{n+1} = \lim a_n \Leftrightarrow a = \sqrt{a} + \frac{15}{4} \Leftrightarrow a = (a- \frac{15}{4})^2 \Leftrightarrow 0 = a^2 - \frac{17a}{2} + \frac{225}{16}$$
$$\Longrightarrow a_1 = 2.25 , a_2 = 6.25$$
So you basically take the first limit $a_1 = 2.25$ . Is that correct? Is there better way of calculating the limit?
Thank you
| The missing part is $a_n$ has upper bound.
Using induction, we'll show $a_n \le \frac {25} 4, \forall n \ge 1$.
For $n=1$ it's obvious. Suppose it's true for $n$. Then $a_{n+1}=\sqrt{a_n} + \frac{15}{4} \le \frac 5 2 + \frac {15} 4 = \frac {25} 4$
| {
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How to establish the identity of the infinite sum How to prove the following identity?
$$\sum_{n=-\infty}^{\infty}\frac{1}{(z+n)^2 +a^2} = \frac{\pi}{a}\cdot\frac{\sinh 2\pi a}{\cosh 2\pi a - \cos 2\pi z}$$
| The trick here is to use $$\frac{1}{(z+w)^2+a^2} \times \pi\cot(\pi
w)$$ as shown at the following MSE
link. The sum term
is also quadratric in $n$ so the estimates of the integrals presented
there apply to the present case as well.
We get for the residues at $w = -z \pm ia$ the closed form
$$\left.\frac{1}{2(z+w)} \pi\cot(\pi w)\right|_{w=-z\pm ia}.$$
Recall that
$$\cot(v) = i \frac{\exp(iv)+\exp(-iv)}{\exp(iv)-\exp(-iv)}.$$
Introducing $x=\exp(\pi i z)$ and $y=\exp(\pi a)$
we get for the two residues
$$\frac{\pi}{2a} \frac{1/x/y+xy}{1/x/y-xy}
- \frac{\pi}{2a} \frac{y/x+x/y}{y/x-x/y}
= \frac{\pi}{2a}
\left(\frac{1+x^2y^2}{1-x^2y^2} - \frac{y^2+x^2}{y^2-x^2}\right)
\\ = \frac{\pi}{2a}
\frac{y^2+x^2y^4-x^2-x^4y^2-y^2+x^2y^4-x^2+x^4y^2}
{(1-x^2y^2)(y^2-x^2)}
\\ = \frac{\pi}{2a}
\frac{2x^2y^4-2x^2}{(1-x^2y^2)(y^2-x^2)}
= \frac{\pi}{a}
\frac{y^2-1/y^2}{(y^2-x^2-x^2y^4+x^4y^2)/y^2/x^2}
\\ = \frac{\pi}{a}
\frac{y^2-1/y^2}{1/x^2 - 1/y^2 - y^2 + x^2}.$$
Flip the sign to get
$$\bbox[5px,border:2px solid #00A000]{
\frac{\pi}{a} \frac{\sinh(2\pi a)}{\cosh(2\pi a)-\cos(2\pi z)}.}$$
Observe that when $z = q \mp ia$ with $q$ an integer we have
$$\cos(2\pi z) = \cos(2\pi q \mp 2\pi i a) = \cosh(2\pi a)$$
and the formula becomes singular. This is correct however since in
this case the sum term $$\frac{1}{(z+n)^2+a^2}$$ is singular as well
namely when $n = -q$ and the sum is undefined.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the Taylor series for $f(x) =\ln(1+x)$ Find the Taylor series for $f(x) = \ln (1+x)$ centered at $x = 0$ using the formula for Taylor Series.
|
$$\ln(1+x)=x-\dfrac {x^2}2+\dfrac {x^3}3-\dfrac {x^4}4+\&\text{c}=\sum\limits_{r=0}^{\infty}\dfrac {(-1)^{r}x^{r+1}}{r+1}$$
The Taylor Series for $f(x)$ at the point $x=a$ is
$$f(x)=f(a)+f^1(a)(x-a)+\dfrac {f^2(a)}{2!}(x-a)^2+\dfrac {f^3(a)}{3!}(x-a)^3+\cdots+\dfrac {f^r(a)}{r!}(x-a)^r$$
Since we want it at $x=0$, $a=0$ and simply plug them into the formula to get the expansion!
Note that another way is to start with the exponential theorem$$a^x=1+cx+\dfrac {c^2x^2}{2!}+\dfrac {c^3x^3}{3!}+\&\text{c}$$
Where $$c=(a-1)-\dfrac 12(a-1)^2+\dfrac 13(a-1)^3-\&\text c$$
And by setting $x=1,\ x=c$, we obtain$$a=e^c\implies c=\ln a$$
And therefore,$$\begin{align*} & \ln a=(a-1)-\dfrac 12(a-1)^2+\dfrac 13(a-1)^3-\&\text{c}\\ & \implies \ln(x+1)=x-\dfrac {x^2}2+\dfrac {x^3}3-\dfrac {x^4}4+\&\text c\end{align*}$$
| {
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Evaluating $\prod_{r=1}^{\infty} \frac{\sin \frac {a}{2^r}}{\tan^2 \frac {a}{2^r} \tan \frac {a}{2^{r-1}}+\tan \frac {a}{2^{r}}}$
The question is to evaluate $$\prod_{r=1}^{\infty} \frac{\sin \frac {a}{2^r}}{\tan^2 \frac {a}{2^r} \tan \frac {a}{2^{r-1}}+\tan \frac {a}{2^{r}}}$$
I could rewrite the denominator as $$\tan\frac {a}{2^r}\left(\tan \frac {a}{2^r}\tan \frac{a}{2^{r-1}}+1\right)$$which is same as $$\tan \frac{a}{2^r}-\tan\frac {a}{2^{r-1}}$$And so the product transforms to $$\prod_{r=1}^{\infty} \frac{\sin \frac {a}{2^r}}{\tan \frac{a}{2^r}-\tan\frac {a}{2^{r-1}}}$$ I have no idea on how to proceed after this. Any help is appreciated. Thanks.
| The product can be rewritten as
$$\prod_{r=1}^{\infty} \frac{\cos{\left ( 2^{-r} a \right )}}{1+\tan{\left ( 2^{-r} a \right )} \tan{\left ( 2^{-r+1} a \right )}} $$
Use the tangent half-angle formula to simplify the denominator:
$$\begin{align}1+\tan{\left ( 2^{-r} a \right )} \tan{\left ( 2^{-r+1} a \right )} &= 1+\sqrt{\frac{1-\cos{\left ( 2^{-r} a \right )}}{1+\cos{\left ( 2^{-r} a \right )}}} \frac{\sqrt{1-\cos^2{\left ( 2^{-r} a \right )}}}{\cos{\left ( 2^{-r} a \right )}}\\ &= \frac1{\cos{\left ( 2^{-r} a \right )}} \end{align}$$
The product is then equal to
$$\left [ \prod_{r=1}^{\infty} \cos{\left ( 2^{-r} a \right )} \right ]^2 $$
Note that
$$ \prod_{r=1}^{\infty} \cos{\left ( 2^{-r} a \right )} = \frac{\sin{a}}{a}$$
Therefore, the product is equal to
$$\frac{\sin^2{a}}{a^2} $$
ADDENDUM
The product in the square brackets may be evaluated with ease using the sine double angle forumla. The product of the cosines is equal to
$$\lim_{n \to \infty} \frac{\sin{a}}{2 \sin{(a/2)}} \frac{\sin{(a/2)}}{2 \sin{(a/2)}} \cdots \frac{\sin{(a/2^{n-1})}}{2 \sin{(a/2^n)}} = \lim_{n \to \infty} \frac{\sin{a}}{2^n \sin{(a/2^n)}}$$
The result follows.
| {
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$\lim_{ x \to 0 }\left( \frac{\sin 3x}{x^3}+\frac{a}{x^2}+b \right)=0$ if :
$$\lim_{ x \to 0 }\left( \frac{\sin 3x}{x^3}+\frac{a}{x^2}+b \right)=0$$
then $a+b=?$
Without the use of the L'Hôspital's Rule
My Try :
$$\lim_{ x \to 0 }\left( \frac{ax+bx^3+\sin 3x}{x^3} \right)=0$$
$$\lim_{ x \to 0 }\left( \frac{x(a+bx^2)+\sin 3x}{x^3} \right)=0$$
$$\lim_{ x \to 0 }x(a+bx^2)+\sin 3x=0 $$
now ?
| Why not to use Taylor series $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^6\right)$$ $$\sin(3x)=3 x-\frac{9 x^3}{2}+\frac{81 x^5}{40}+O\left(x^6\right)$$
$$\frac{\sin 3x}{x^3}+\frac{a}{x^2}+b =\frac{a+3}{x^2}+\left(b-\frac{9}{2}\right)+\frac{81 x^2}{40}+O\left(x^3\right)$$ Then $???$
| {
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Proof that expression is positive if conditions is met Is there a way to show that this expression is always positive as long as $b>0$ and $r>x$ ? Assume $r>0$ and $x>0$
\begin{equation}
b-1+\sqrt{(1+b)^2-\frac{4rb}{x}}
\end{equation}
It's simple with $b>1$ but I can't figure it out if it is true for lower values.
| The statement is obviously not true if $(1+b)^2 - \frac {4br}x < 0$.
Case 1: $(1+b)^2 - \frac {4br}x < 0$ or in other words
$(1+b)^2 < \frac {4br}x$ or
$\frac {(1+b)^2*|x|}{4|b|} < r$. (Assuming $b \ne 0$. If $b = 0$ then $(1+b)^2 - \frac {4br}x = (1+b)^2 = 1$.)
The statement is false and yields non-real number.
Case 2:
$(1+b)^2 - \frac {4br}x = 0$ or $r =\frac {(1+b)^2*|x|}{4|b|}$
In this case $b-1 + \sqrt{(1+b)^2 - \frac {4br}x} = b-1 > 0 \iff b > 1$
Case 3:$r <\frac {(1+b)^2*|x|}{4|b|}$
$b-1 + \sqrt{(1+b)^2 - \frac {4br}x} > 0 \iff$
$\sqrt{(1+b)^2 - \frac {4br}x} > 1-b$.
Case 3a: This will be true for $b \ge 1$.
If $b < 1$ then this will be true if
$(1+b)^2 - \frac {4br}x > (1-b)^2 \iff$
$1 + b^2 + 2b - \frac {4br}x > 1 + b^2 -2b \iff$
$4b > \frac {4br}x \iff \frac b{|b|} > \frac rx$
Case 3b: $1> b > 0$ then true if $|x| > r$
Case 3c: $-1< b < 0$ then this is true if $0 > -|x| >r$
So given $b > 0$ and $r > x > 0$ we must have $b \ge 1$ and $x< r \le\frac {(1+b)^2x}{4b}$ or in other words $1 < \frac rx \le \frac{(1+b)^2}{4b}$.
But if $b = 1$ we have $1 < \frac rx \le 1$ so $b > 1$. But we must also have the condition $x< r \le\frac {(1+b)^2x}{4b}$.
| {
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Does removing all numbers in the harmonic series with a units digit of 9 affect the series? The harmonic sequence is $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}\ldots$ diverges. There is a simple reason why:
$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}\ldots$ is obviously greater than $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8}+\frac{1}{8}+\frac{1}{8}\ldots$ , which is infinte.
But, if you remove all numbers that include a "9", the series converges. The reason for this is that as the amount of digits in the number increases, it will be harder and harder to avoid the digit "9", and it will slowly become impossibly hard.
I was wondering that if you remove every number with a units digit of 9, will the series converge or diverge?
| The series $\sum\limits_{n=1}^\infty\frac{1}{2n}$ diverges because $\sum\limits_{n=1}^N\frac{1}{2n}=\frac12\sum\limits_{n=1}^N\frac{1}{n}$. You get rid of all odd units digits this way, still diverging.
The series $\sum\limits_{n=1}^\infty\dfrac{1}{10^k n}$ for any fixed $k\in\mathbb N$ gives you $0$s in the first $k$ places of each number, and still diverges.
You could also take away all terms with leading digit $9$ and still get a divergent series. In fact, only taking terms with leading digit $1$ yields
$$1+\frac1{10}+\cdots+\frac{1}{19}+\frac{1}{100}+\cdots+\frac{1}{199}+\cdots\geq 1+10\cdot\frac{1}{20}+100\cdot\frac1{200}+\cdots.$$
Combining this with the previous trick of multiplying by $10^k$, you could have only leading digits $1$, with an arbitrary but fixed number of ending $0$s, and still diverge.
| {
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Tangent at a given point I have a function $\frac{x^2+2}{x-1}$. I want to find a tangent at a given point of $x=1+\sqrt{3}$.
At first, I found a value of the function at given point. This is what I got: $2+\frac{6}{\sqrt{3}}$
Then, I found a derivative of this point and I think it's $0$. So, for me the tangent is $y=(2+\frac{6}{\sqrt{3}})$ but, the proper result is $-2\sqrt{3}-2$. What am I doing wrong?
| The derivative of a function $f$ at a point $x_0$ is the slope of the tangent line to $f$ at $x_0.$
So, let $f(x)=\frac{x^2+2}{x-1}$ and $x_0=1+\sqrt{3}.$
$$\Rightarrow f'(x)=\frac{(x-1)\cdot 2x - (x^2+2)\cdot 1}{(x-1)^2}=\frac{x^2-2x-2}{(x-1)^2}.$$
and $$f'(x_0)=f'(1+\sqrt{3})=\frac{(1+\sqrt{3})^2-2(1+\sqrt{3})-2}{(1+\sqrt{3}-1)^2}=0.$$
Now, to get the equation of the line we have $$y-f(x_0)=f'(x_0)(x-x_0)$$ but $f'(x_0)=0$ so we have $$y=f(x_0).$$
$$\Rightarrow y=f(x_0)=f(1+\sqrt{3})=\frac{(1+\sqrt{3})^2+2}{(1+\sqrt{3})-1}=\frac{6+2\sqrt{3}}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{6\sqrt{3}+6}{3}=2\sqrt{3}+2.$$
Therefore, $$y=2\sqrt{3}+2.$$
| {
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Solving: $3^m-2=n^2$
Solve $3^m-2=n^2$ for positive pairs of integers $(n,m)$.
My try:
$3^m-6=n^2-4\implies 3(3^{m-1}-2)=(n+2)(n-2)\implies 3=n+2\ or\ 3=n-2$ so we got $n=\pm 5 , \pm 1$. Used $\pm$ as we have $n^2$.
Putting $n=\pm 5 , \pm 1$ gives $m=3,1$ respectively. We kick out negative ones as problem ask for positive ones only.
But how do I prove that $n+2=3$ not something like multiples of $3$ and similar for $n-2$. More precisely something like the following:
$n+2=3d$, $n-2=k$ and $dk=3^{m-1}-2$.
Please help.
| This is too long for a comment. My approach is not "elementary" since I use factorization in quadratic rings, more exactly, I'll work in $\Bbb{Z}[\sqrt{-2}]$ which is known to be a Euclidean domain and hence a UFD (unique factorization domain). Moreover, in this ring every element $x+y\sqrt{-2}$ has norm $N(x+y\sqrt{-2})=(x+y\sqrt{-2})(x-y\sqrt{-2})=x^2+2y^2$.
If $m$ is even, then we would have a difference of two squares equal to $2$, but it's easy to see that in that case there are no solutions.
Then $m$ is odd. If $m=1$ we have $3-2=n^2$ which give us $n=1$. So from now on we assume that $m\ge 3$. We rewrite the equation in the following form $$3^m=n^2+2=(n+\sqrt{-2})(n-\sqrt{-2}).$$
We claim that $n+\sqrt{-2}$ and $n-\sqrt{-2}$ are coprime. Indeed, if $d=\gcd(n+\sqrt{-2},n-\sqrt{-2})$, then $d\mid (n+\sqrt{-2})-(n-\sqrt{-2})=2\sqrt{-2}$, so $N(d)\mid N(2\sqrt{-2})=8$. On the other hand, $d\mid n+\sqrt{-2}$, which implies that $N(d)\mid N(n+\sqrt{-2})=n^2+2=3^m$ and therefore we deduce that $N(d)$ is an odd positive divisor of $8$, hence it must be $N(d)=1$ and this implies that $d$ is an unit in $\Bbb{Z}[\sqrt{-2}]$.
Since $n+\sqrt{-2}$ and $n-\sqrt{-2}$ are coprime and its product is a $m$-th power, by the fundamental theorem of the arithmetic in $\Bbb{Z}[\sqrt{-2}]$ we can conclude that there are units $u, v\in \Bbb{Z}[\sqrt{-2}]$ and elements $a_1\pm b_1\sqrt{-2}\in \Bbb{Z}[\sqrt{-2}]$ such that $$n+\sqrt{-2}=u(a_1+b_1\sqrt{-2})^m,$$ $$n-\sqrt{-2}=v(a_1-b_1\sqrt{-2})^m.$$
But, it's known that the units of $\Bbb{Z}[\sqrt{-2}]$ are $1$ and $-1$, both being $m$-th powers (indeed, $1^m=1$ and $(-1)^m=-1$), thus we can assume that both $n+\sqrt{-2}$ and $n-\sqrt{-2}$ are pure $m$-th powers, i.e., $$n+\sqrt{-2}=(a+b\sqrt{-2})^m,$$ $$n-\sqrt{-2}=(a-b\sqrt{-2})^m.$$
Now, we have $(a+b\sqrt{-2})^m(a-b\sqrt{-2})^m=(n+\sqrt{-2})(n-\sqrt{-2})=n^2+2=3^m$, then $$((a+b\sqrt{-2})(a-b\sqrt{-2}))^m=(a^2+2b^2)^m=3^m.$$
Therefore we deduce that $a^2+2b^2=3$, which has the solutions $a=\pm 1$ and $b=\pm 1$. Let's suppose initially that both $a$ and $b$ are positive, i.e., $a=b=1$. This lead us to $$n+\sqrt{-2}=(1+\sqrt{-2})^m.$$
Comparing the imaginary parts of both expressions in last equation gives us $$1=\sum_{k=0}^{\frac{m-1}{2}}(-2)^k \binom{m}{2k+1}.$$
I'm stuck here because the RHS alternates positive with negative values and I don't know how that expression can be bounded. Besides we can't try to use a $p$-adic argument similar to the solution of the equation $x^m-1=y^2$ because we can't remove $1$ from the LHS.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2170137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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The number of 3 digit numbers of the form xyz such that $x
The number of 3 digit numbers of the form $xyz$ such that $x<y$ and $z\leq y$ is N such that n is a 3 digit number of the form $abc$ then find $a+c-b$
My approach:
Case 1:
$x<y$
$z<y$
$x,y,z\neq 0$
Choosing 3 digits out of 9 numbers: $\binom 93$, greatest of them will always be $y$, hence ways of arranging is (exchanging x and z) $2\cdot\binom 93=54$.
Case 2:
$x,y$ can't be $0$. Taking $z=0$, number of ways of selecting the other 2 digits are $\binom 92=36$. There is only one way to arrange them.
case 3:
$y=z$,
_99, values for $x=8$
_88, values for $x=7$
_77, values for $x=6$
.
.
.
_11, values for $x=0$
Total possibilities=$8+7+6+\cdots+1=36$
Hence total permutations=$54+36+36=126$
This gives $a+c-b=5$ which is not the right answer. Why is this logic wrong?
| A trial and error method looks better here.
Note that, here $$1\le x \le 8$$ $$\max(x,z) \le y\le 9$$ $$0\le z\le 9$$
Now, for $x=1$, and for all $z$, $y$ can take $8+8+8+7+6+5+4+3+2+1=52$ values.
Again, for $x=2$, and for all $z$, $y$ can take $7+7+7+7+6+5+4+3+2+1=49$ values.
So, it can be checked that, in general for $x=r$ when $1\le r \le 8$, and for all $z$, $y$ can take $(r+2)\cdot (9-r)+(8-r)+\ldots + 1= (r+2)\cdot (9-r)+\sum_{i=1}^{(8-r)} i$ values.
The value of N $$=\sum_{r=1}^8 (r+2)\cdot (9-r)+(8-r)+\ldots + 1$$ $$=\sum_{r=1}^8 (r+2)\cdot (9-r)+\sum_{r=1}^8 \sum_{i=1}^{(8-r)} i$$
$$=\sum_{r=1}^8 (7r+18-r^2)+\sum_{r=1}^8 \frac{(8-r)(8-r+1)}{2}$$
Hope this helps you.
| {
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"url": "https://math.stackexchange.com/questions/2171143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\min\int_0^\pi {f^2(x)dx}$, assume $\int_0^\pi{f(x)\sin xdx} = \int_0^\pi{f(x)\cos xdx} = 1.$ $f(x)$ is continuous on $[0,\pi]$ and $\int_0^\pi{f(x)\sin xdx} = \int_0^\pi{f(x)\cos xdx} = 1.$
Find $\min\int_0^\pi {f^2(x)dx}.$
I try to solve this problem by this:
$$\begin{array}{l}
{\left( {\int\limits_0^\pi {f(x)\sin xdx} } \right)^2} \le \left( {\int\limits_0^\pi {{f^2}(x){{\sin }^2}xdx} } \right)\left( {\int\limits_0^\pi {dx} } \right) \le \pi \int\limits_0^\pi {{f^2}(x){{\sin }^2}xdx} \\
{\left( {\int\limits_0^\pi {f(x)\cos xdx} } \right)^2} \le \pi \int\limits_0^\pi {{f^2}(x){{\cos }^2}xdx} \\
\Rightarrow \pi \int\limits_0^\pi {{f^2}(x)\left( {{{\sin }^2}x + {{\cos }^2}x} \right)dx} \ge 1 + 1 = 2
\end{array}$$
The thing is I can't find $f(x)$ to let the equation happens. Any help? Thank you in advance.
| Let us consider the Fourier series of $f$
\begin{align}
f(x) = \frac{1}{2}a_0+\sum^\infty_{n=1} a_n \cos nx+ \sum^\infty_{n=1} b_n \sin nx
\end{align}
then that means
\begin{align}
\int^\pi_0 f^2(x)\ dx= \frac{1}{4}a_0^2+\frac{\pi}{2}\sum^\infty_{n=1}(a_n^2+b_n^2).
\end{align}
Since
\begin{align}
\int^\pi_0 f(x) \cos x\ dx = 1
\end{align}
then this means $a_1= 2/\pi$ and likewise $b_1 = 2/\pi$. It follows
\begin{align}
\int^\pi_0 f^2(x)\ dx \geq \frac{4}{\pi}= \frac{\pi}{2}(a_1^2+b_1^2).
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/2171237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Reducibility of $x^4+1$ in prime fields In Dummit and Foote's "Abstract Algebra" there is a proof for the reducibility of $x^4+1$ in which it's stated:
For odd primes p, $p^2\equiv1\pmod 8$. Hence $x^8-1\mid x^{p^2-1}-1 $.
Why does $x^8-1\mid x^{p^2-1}-1$ follow?
| As a side answer, not to the original question, you can get this result fairly explicitly.
If $-1=a^2$, for some $a$, then $x^4+1=(x^2+a)(x^2-a)$.
If $2=a^2$ for some $a$, then $x^4+1=(x^2+ax+1)(x^2-ax+1)$.
If $-2=a^2$ for some $a$, then $x^4+1=(x^2+ax-1)(x^2-ax-1)$.
But if $-1$ and $2$ are not squares, modulo $p$, then $-2$ is a a square, modulo $p$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2173029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Intersection of two lines - Have I got it right? I have been asked to solve where these two lines intersect, it has been a very long time since I have done this, my numbers seem too big. Can someone check these for me?
2x + y = 6
x + 2y = 12
(find x) (plug into first equation)
x + 2y = 12 2(-2y+12)+y=6
x = -2y + 12 -4y+24+y=6
-3y=6-24
-3y=18
3y=18
y=18-3
y=15
(put y back in)
x=-2(15)+12
x=-30+12
x=-18
Intersection at (-18,15)
| By elimination:
$$\begin{cases}2x + y = 6,\\
x + 2y = 12.\end{cases}$$
Subtract twice the first equation from the second and get
$$-3x=0.$$
By substitution:
From the first equation, $y=6-2x$, and plugging in the second, $x+12-4x=12$, giving $$-3x=0.$$
By Cramer:
$$x=\frac{\begin{vmatrix}6&1\\12&2\end{vmatrix}}{\begin{vmatrix}2&1\\1&2\end{vmatrix}}=0.$$
In all three cases, $y=6$ by the first equation.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Show that $\frac 1x+\frac 1y =(\frac 27)^a$ does not produce integer solutions for $a > 3$. Show that $\frac 1x+\frac 1y =(\frac 27)^a$ does not produce integer solutions for $a > 3$.
I have shown it is not possible for $a=4$, but not for any a greater than $4$.
| Rewrite the equation into
$$7^a\cdot(x+y)=2^a\cdot xy$$
From this, we conclude that there is some positive integer $k$ with
\begin{align}
xy&=k\cdot 7^a\tag{1}\label{eq1}\\
x+y&=k\cdot 2^a\tag{2}\label{eq2}
\end{align}
Now, from equation $\eqref{eq1}$, we conclude that if $p\neq 7$ is prime and $p|k$, then $p|x$ or $p|y$.
On the other hand, equation $\eqref{eq2}$ implies that if $p\neq2$ is prime and $p$ divides both $x$ and $k$, then $p|y$, and similarly exchanging the roles of $x$ and $y$.
Together, this means that any prime $p\neq 2,7$ in the factorization of $k$ also factors both $x$ and $y$. Say $p$ is one such prime and let $e_x$ be the exponent of $p$ in the factorization of $x$, and similarly for $e_y$. From equation $\eqref{eq1}$, $p$ has an exponent of $e_x+e_y$ in the factorization of $k$, but from equation $2$ that exponent is $\max\{e_x,e_y\}$.
It follows that no prime $p\neq 2,7$ is in the factorization of $k$, so $k,x$ and $y$ are all of the form $2^s7^t$. We may then rewrite this system. In the obvious notation, equation $\eqref{eq1}$ yields
\begin{align}
s_x+s_y&=s_k\\
t_x+t_y&=t_k+a
\end{align}
Equation $\eqref{eq2}$ then yields
$$2^{s_x}7^{t_x}+2^{s_y}7^{t_y}=2^{s_k+a}7^{t_k}$$
We may assume without loss of generality that $s_x\geq s_y$. Then:
$$2^{s_x-s_y}7^{t_x}+7^{t_y}=2^{s_k-s_y+a}7^{t_ k}=2^{s_x+a}7^{t_ k}$$
so if $s_x>s_y$, the RHS is odd while the LHS is odd. It follows that $s_x=s_y$ and the equation may be simplified to
$$7^{t_x}+7^{t_y}=2^{s_x+a}7^{t_k}$$
Now, we may assume without loss of generality that $t_x\geq t_y$. Then
$$7^{t_x-t_y}+1=2^{s_x+a}7^{t_k-t_y}=2^{s_x+a}7^{t_x-a}$$
We must then have $t_x=a$, for otherwise the RHS is either fractional, or a multiple of $7$ (while the LHS is not). This then implies $t_y=t_k$. Hence
$$7^{a-t_k}+1=2^{s_x+a}$$
If either $a$ were even or $s_x$ were positive and even, we could factor $2^{s_x+a}-1$ as
$$\Big({\left(2^{s_x}\right)}^{a/2}-1\Big)\Big({\left(2^{s_x}\right)}^{a/2}+1\Big)$$
Since these factors differ by $2$, at most one can be a multiple a $7$, and this would imply that the smallest is $1$, so that $\frac{as_x}2=1$. Since $a\geq 4$, this is impossible, and $a$ must be odd and $s_x$ must be either odd or $0$.
If $s_x$ were odd, then $a+s_x$ is even, and we may repeat the factorization argument:
$$\Big(2^{(s_x+a)/2}-1\Big)\Big(2^{(s_x+a)/2}+1\Big)$$
Once again $\frac{s_x+a}2=1$ cannot be true for $a\geq 4$, so $s_x$ must be $0$. We then have
$$7^{a-t_k}=2^a-1,$$
where $a\geq 5$ is odd. By looking into the equaiton above modulo $4$, we find that $t_k$ is even. Then, with $a=2\beta+1$ and $t_k=2\sigma$ we may rewrite the equation as
$$7\cdot {\left(7^2\right)}^{\beta-\sigma}=2^{2\beta+1}-1$$
Finally, looking at the equation above modulo $16$, we find that the LHS is congruent to $7$ while the RHS is congruent to $-1\equiv 15$. This completes the proof that when $a\geq 4$, no solution exists. $\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2174985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What does this series converge to, if anything? $$\arctan{1} + \arctan{\frac{1}{2}} + \arctan{\frac{1}{3}} + \arctan{\frac{1}{4}} ...= ?$$
The infinite series for arctan is
$$\arctan{x} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} ...$$
So I want to sum up the $\arctan{1\over n}$ where $n$ starts at $1$ and goes to infinity.
I originally thought the resulting series can be written this way:
$$(1 + \frac{1}{2} + \frac{1}{3} ...) - \frac{(1 + \frac{1}{2} + \frac{1}{3} ...)^3}{3} + \frac{(1 + \frac{1}{2} + \frac{1}{3} ...)^5}{5} - \frac{(1 + \frac{1}{2} + \frac{1}{3} ...)^7}{7} ...$$
But that is wrong. The right way to "insert" the series is:
$$1 + \frac{1}{2} + \frac{1}{3} ... - \frac{1}{3} - \frac{(\frac{1}{2})^3}{3} - \frac{(\frac{1}{3})^3}{3} ... + \frac{1}{5} + \frac{(\frac{1}{2})^5}{5} + \frac{(\frac{1}{3})^5}{5} ... ...$$
So it looks like a bunch of harmonic serieses manipulated. Harmonic series diverges, but I remember that doesn't necessarily mean a similar series diverges. I remember from Calculus 2 that, for example, $\lim_{n\to\infty}$ of $\sin(n)\over n$ converges to zero even though $\sin(n)$ does not converge.
So how do I figure out what this series converges to, if anything?
| Based on this (On the arctangent inequality.) answer, we have:
$$\frac{\arctan x}{x} \geq 1/2$$
for $x \in (0,1]$.
So letting $x = \frac{1}{n}$, we have $\arctan \frac{1}{n} \geq \frac{1}{2n}$ for each $n\geq 1$, so by the comparison test, your series diverges
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Factor $(x+y)^4+x^4+y^4$ Title says it all, I just want to know how to factor $(x+y)^4+x^4+y^4$. I only know that it's possible to factor, but got no idea how to do it. If it were a single-variable polynomial I could try to find rational roots or something, but I'm lost with this one.
| $$(x+y)^4+x^4+y^4=(x^2+y^2+2xy)^2+(x^2+y^2)^2-2(xy)^2=\\
[(x^2+y^2+2xy)^2-(xy)^2]+[(x^2+y^2)^2-(xy)^2]=\\
(x^2+y^2+xy)(x^2+y^2+3xy)+(x^2+y^2+xy)(x^2+y^2-xy)=\\
(x^2+y^2+xy)(2x^2+2y^2+2xy)=2(x^2+y^2+xy)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2176542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that $F(xy) =F(x) + F(y)$ when $F(x)$ is not $\ln(x)$ So we have this function for $x > 0$.
$$\int_{1}^{x} \frac{1}{t}\text{d}t$$
Show that $F(xy) = F(x) + F(y)$ without assuming $F(x) = \ln(x)$.
I came so far as this point, but I can't crack the last step.
$$\int_{1}^{x} \frac{1}{t}\text{d}t + \int_{x}^{xy} \frac{1}{t}\text{d}t$$
which is:
$$F(x) + \int_{x}^{xy} \frac{1}{t}\text{d}t$$
However, I am having difficulties proving that:
$$\int_{x}^{xy} \frac{1}{t}\text{d}t = F(y)$$
Help would be very very appreciated!
| Spoiler below... Don't forget to change both upper and bottom integral bounds when using an integral sustitution
This is the full answer...
$$ F(xy):=\int_{1}^{xy} \frac{1}{t}dt $$
$$ let \,\,\,\begin{matrix} t = x \cdot u & x \cdot u = xy \\ dt = x \cdot du & x \cdot u = 1\end{matrix}
\,\,\,OR\,\,\,
\begin{matrix} u = y \\ u = \frac{1}{x} \end{matrix} $$
$$ then \,\,\,\,F(xy)=\int_{1}^{xy} \frac{1}{t}dt = \int_{\frac{1}{x}}^{y} \frac{1}{x \cdot u} (x \cdot du) = \int_{\frac{1}{x}}^{y} \frac{1}{u} du$$
$$ splitting \int_{\frac{1}{x}}^{y} \frac{1}{u} du = \int_{\frac{1}{x}}^{1} \frac{1}{u} du + \int_{1}^{y} \frac{1}{u} du$$
creating one last u-substitition
$$ let \,\,\,\begin{matrix} v = x \cdot u & x \cdot 1 = x = v \\ \frac{dv}{x} = du & x\cdot (1/x) = 1 = v \end{matrix} \,\,\, which\,\, becomes
$$
$$ F(xy)= \int_{1}^{x} \frac{1}{v/x} (dv/x) + \int_{1}^{y} \frac{1}{u} du = \int_{1}^{x} \frac{1}{v} dv + \int_{1}^{y} \frac{1}{u} du = F(x) + F(y)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Decomposition into partial fractions of an inverse of a generic polynomial with three distinct roots. Let $d \ge 2$ be an integer. Let $\left\{ m_j \right\}_{j=1}^d$ be strictly positive integers and $\left\{ b_j \right\}_{j=1}^d$ be parameters. Define the following quantity:
\begin{equation}
{\mathfrak F}_d(x) := \frac{1}{\prod\limits_{j=1}^d (x+b_j)^{m_j}}
\end{equation}
Below we decompose the quantity above into partial fractions for $d=3$ using differentiation with respect to the $b$-parameters. We have:
\begin{eqnarray}
&&{\mathfrak F}_d(x) = \\
&&\sum\limits_{\begin{array}{r}1 \le l_1 \le m_1 \\ l_1 \le l \le m_1 \end{array}} \binom{m_1+m_2-1-l}{m_2-1} \binom{l+m_3-1-l_1}{m_3-1} \frac{(-1)^{m_2+m_3}}{(x+b_1)^{l_1} (b_1-b_3)^{l+m_3-l_1} (b_1-b_2)^{m_1+m_2-l}} + \\
&&\sum\limits_{\begin{array}{r}1 \le l_1 \le m_2 \\ l_1 \le l \le m_2 \end{array}} \binom{m_1+m_2-1-l}{m_1-1} \binom{l+m_3-1-l_1}{m_3-1} \frac{(-1)^{m_1+m_3}}{(x+b_2)^{l_1} (b_2-b_3)^{l+m_3-l_1} (b_2-b_1)^{m_1+m_2-l}} + \\
&&\sum\limits_{\begin{array}{r}1 \le l_1 \le m_3 \\ l_1 \le l \le m_3 \end{array}} \binom{m_2+m_3-1-l}{m_2-1} \binom{l+m_1-1-l_1}{m_1-1} \frac{(-1)^{m_1+m_2}}{(x+b_3)^{l_1} (b_3-b_1)^{l+m_1-l_1} (b_3-b_2)^{m_2+m_3-l}}
\end{eqnarray}
Now the question is to provide the result for arbitrary $d$.
| The result is given below:
\begin{eqnarray}
&&{\mathfrak F}_d(x) =
\sum\limits_{k=1}^d
\sum\limits_{\begin{array}{rrr} 1&\le l_{d-2}& \le m_k \\l_{d-2}& \le l_{d-3} &\le m_k\\&\vdots&\\l_1&\le l_0&\le m_k\end{array}}
\prod\limits_{j=-1}^{d-3} \binom{l_j + m_{f_{k,j}}-1-l_{j+1}}{m_{f_{k,j}}-1} \frac{1}{\left(b_k-b_{f_{k,j}}\right)^{l_j + m_{f_{k,j}}-l_{j+1}}} \cdot \frac{(-1)^{M-m_k}}{\left(x+b_k\right)^{l_{d-2}}}
\end{eqnarray}
subject to $l_{-1}= m_k$.
Here $f_{k,j} := (k-2-j) 1_{j\le k-3} + (j+3) 1_{j>k-3}$ and $M= \sum\limits_{j=1}^d m_j$.
As a sanity check we analyze special cases.
Firstly let us take $m_1=\cdots=m_d=1$ then all the $l$-indices are equal to one and we immediately get:
\begin{equation}
rhs = \sum\limits_{k=1}^d \prod\limits_{j=-1}^{d-3} \frac{1}{(b_k-b_{f_{k,j}})^1} \cdot \frac{(-1)^{d-1}}{(x+b_k)^1}
\end{equation}
as it should be.
Now let us take $m_1=\cdots=m_d=2$. In this cases there are two cases.
(A) $l_{d-2}=2$ then $(l_{d-3},\cdots,l_0,l_{-1})=(2,\cdots,2)$ or (B) $l_{d-2}=1$ then $(l_{d-3},\cdots,l_0,l_{-1})=(1,\cdots,1,2,\cdots,2)$ where the number of one's can be zero but the number of two's has to be strictly positive. This yields:
\begin{eqnarray}
rhs&=& \sum\limits_{k=1}^d \prod\limits_{j=-1}^{d-3} \frac{1}{(b_k-b_{f_{k,j}})^2} \cdot \frac{(-1)^{2(d-1)}}{(x+b_k)^2} +\\
&&\sum\limits_{k=1}^d \prod\limits_{j=-1}^{d-3} \frac{1}{(b_k-b_{f_{k,j}})^2} \cdot \left\{\sum\limits_{j=-1}^{d-3} \frac{2}{b_k-b_{f_{k,j}})^1}\right\} \frac{(-1)^{2(d-1)}}{(x+b_k)^1}
\end{eqnarray}
as it should be.
Finally we take the case $m_1=\cdots=m_d=3$. Then there are three cases. (A) $l_{d-2}=3$ then $(l_{d-3},\cdots,l_0,l_{-1})=(3,\cdots,3)$ or (B) $l_{d-2}=2$ then $(l_{d-3},\cdots,l_0,l_{-1})=(2,\cdots,2,3,\cdots,3)$ or (C) $l_{d-2}=1$ then $(l_{d-3},\cdots,l_0,l_{-1})=(1,\cdots,1,2,\cdots,2,3,\cdots,3)$. In the last two cases the number of three's has to be strictly positive. This immediately gives:
\begin{eqnarray}
rhs&=& \sum\limits_{k=1}^d \prod\limits_{j=-1}^{d-3} \frac{1}{(b_k-b_{f_{k,j}})^3} \cdot \frac{(-1)^{3(d-1)}}{(x+b_k)^3} +\\
&&\sum\limits_{k=1}^d \prod\limits_{j=-1}^{d-3} \frac{1}{(b_k-b_{f_{k,j}})^3} \cdot \left\{\sum\limits_{j=-1}^{d-3} \frac{3}{b_k-b_{f_{k,j}})^1}\right\} \frac{(-1)^{3(d-1)}}{(x+b_k)^2}+\\
&&\sum\limits_{k=1}^d \prod\limits_{j=-1}^{d-3} \frac{1}{(b_k-b_{f_{k,j}})^3} \cdot \left\{
\sum\limits_{-1\le j_1<j_2\le d-3}
\frac{3^2}{\prod\limits_{\xi=1}^2 (b_k-b_{f_{k,j_\xi}})^1}+
\sum\limits_{j=-1}^{d-3} \frac{6}{(b_k-b_{f_{k,j}})^2}
\right\} \frac{(-1)^{3(d-1)}}{(x+b_k)^1}
\end{eqnarray}
again as it should be.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2177956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $x,y>0$ and $x^2+y^3\ge x^3+y^4$, prove that $x^3+y^3 \le 2$. As in the title. If $x,y>0$ and $x^2+y^3\ge x^3+y^4$, prove that $$x^3+y^3 \le 2.$$
This seems to be a very tricky one. I tried applying various inequalities like AM-GM, unfortunately, none of techniques I'm familiar with seem to work here.
I'd greatly appreciate any hints.
| We can try to maximize $f(x,y)=x^3+y^3$ subject to
\begin{align}
&x>0\\
&y>0\\
&x^2+y^3-x^3-y^4\geq 0
\end{align}
Since $\nabla f(x,y)=(3x^2,3y^2)$, $f$ always increases away from the origin in the first quadrant. Hence, a maximum must occur on $x^2+y^3-x^3-y^4=0$.
At this point we can use Lagrange multipliers. A maximum must satisfy
\begin{align}
&x>0\\
&y>0\\
&x^2+y^3-x^3-y^4=0\\
&(3x^2,3y^2)=\lambda\cdot(2x-3x^2,3y^2-4y^3)
\end{align}
for some real $\lambda$. Then
$$(1+\lambda)3x-2\lambda=0\\3(1-\lambda)+4\lambda y=0.$$
Notice we may not have $\lambda=0$, nor $\lambda=\pm1$. Hence
$$x=\frac{2\lambda}{3(1+\lambda)}\,\,\,\,\,\,y=\frac{-3(1-\lambda)}{4\lambda}$$
Since $x$ must be positive, we find that $\lambda >0$ or $\lambda <-1$. Since $y$ must be positive, we find that $\lambda >1$ or $\lambda <0$. Hence, the only options are $\lambda>1$ or $\lambda <-1$.
Substituting $x$ and $y$ into $x^2+y^3-x^3-y^4=0$, we get a rational function in $\lambda$ whose numerator must equal $0$. It has a single root in $(-\infty,-1)\cup(1,\infty)$, at $\lambda=-3$.
With this value, we get $x=y=1$, so the maximum is attained at $(1,1)$ with precisely the value $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2178764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find all pairs of prime numbers $p$ and $q$ such that $\,p^2-p-1=q^3.$ I'm preparing a mathematical olympiad and our group is stuck in this problem. Here is all we've done:
First, let's rewrite the equation
$$p^2-p-1=q^3\Rightarrow p(p-1)=(q+1)(q^2-q+1)$$
It's obvious that $q$ must be less than $p$. Then, $p|(q^2-q+1)$ and then equation is of the form
$$p-1=k(q+1)$$
For some integer $k$. Putting that onto the first equation
$$\left(\frac{p-1}{k}-1\right)^3=q^3=p^2-p-1$$
Which is the same as
$$p^2-(2+3k+k^3)p+(3k^2+3k+1)=0$$
Since $p$ is prime $p|3k^2+3k+1$, a possible solution is $p=3k^2+3k+1$. Substituting
\begin{align*}
(3k^2+3k+1)^2-(2+3k+k^3)(3k^2+3k+1)+3k^2+3k+1=-k^2(k-3)(3k^2+3k+1)=0
\end{align*}
$k$ cannot be 0, so it's unique possible integer value is $3$ and $p=37$. Solving, $q=11$ which is a valid solution but we don't know if there are more or how to prove there aren't more, please help us continue.
| Since we want $$p^2-(2+3k+k^3)p+(3k^2+3k+1)=0$$ to have integer solutions its Discriminant has to be a perfect square
i.e. $k^2(k^4+6k^2+4k-3)$ has to be a perfect square thus
$k^4+6k^2+4k-3$ has to be a perfect square
But $(k^2+3)^2 \le k^4+6k^2+4k-3<(k^2+4)^2$ thus we have
$4k-3=9 \Rightarrow k=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2180688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Use extended Euclidean algorithm to find $j,k$ such that $52j+15k=3$ Is this questions suggesting that $\gcd(52,15)=3$ i.e. $52j+15k=3$?
if it is then why am I getting $1$ when I am computing the $\gcd$.
$$
\gcd(52,15)
= \gcd(15,7)
= \gcd(7,1)
= \gcd(1,0)
= 1
$$
Am I going in the right direction or not?
| First, we find $\gcd(\color{#c00}{52}, \color{#0a0}{15})$ using the Euclidean Algorithm.
$$
\begin{align*}
\color{#c00}{52}&=3 * \color{#0a0}{15} + \color{#00c}{7}\\
\color{#0a0}{15}&=2 * \color{#00c}{7} + 1\\
\color{#00c}{7}&=1*\color{#00c}{7}+0
\end{align*}
$$
So, $\gcd(\color{#c00}{52}, \color{#0a0}{15})=1$. You correctly calculated the $\gcd$ to be one. Now we rewrite the previous equations in terms of the remainders so we can substitute them back in:
$$
\begin{align*}
\color{#00c}{7}&=\color{#c00}{52}-3*\color{#0a0}{15}\\
1&=\color{#0a0}{15}-2*\color{#00c}{7}
\end{align*}
$$
which gives
$$
\begin{align*}
1&=\color{#0a0}{15}-2*(\color{#c00}{52}-3*\color{#0a0}{15})\\
1&=\color{#0a0}{15}-2*\color{#c00}{52}+6*\color{#0a0}{15}\\
1&=\color{#00c}{7}*\color{#0a0}{15}-2*\color{#c00}{52}.
\end{align*}
$$
To finish, we multiply the whole equation by $3$:
$$
\begin{align*}
3(1)&=3(\color{#00c}{7}*\color{#0a0}{15}-2*\color{#c00}{52})\\
3&=3*\color{#00c}{7}*\color{#0a0}{15}-3*2*\color{#c00}{52}\\
3&=21*\color{#0a0}{15}-6*\color{#c00}{52}.
\end{align*}
$$
Thus, we have that $j=-6$ and $k=21$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving the given identity For every continuous periodic function $F(\theta)$ with period $2 \pi$ and for $0 < \rho < 1,$ prove that
$$\lim_{\rho \rightarrow 1}\frac{1}{4\pi \rho} \int_{-\pi}^{\pi} F(\theta ) \left( \frac{1-\rho^2}{1-2 \rho \cos n\theta + \rho^2} - \frac{1-\rho^2}{1+2 \rho \cos n\theta + \rho^2} \right)~d\theta=\frac{1}{2n}\sum_{k=1}^{2n}(-1)^k~F \left(\frac{k \pi }{n} \right).$$
Any suggestions of proving this is much appreciated.
| We can prove the identity by elementary methods.
Since $F(\theta )$ and $\cos n\theta $ are periodic
\begin{align}
&\int_{-\pi}^{\pi} F(\theta ) \left( \frac{1-\rho^2}{1-2 \rho \cos n\theta + \rho^2} - \frac{1-\rho^2}{1+2 \rho \cos n\theta + \rho^2} \right)~d\theta\\
=&\int_{0}^{2\pi} F(\theta ) \left( \frac{1-\rho^2}{1-2 \rho \cos n\theta + \rho^2} - \frac{1-\rho^2}{1+2 \rho \cos n\theta + \rho^2} \right)~d\theta
\end{align}
holds.
Making the change of variable $\varphi =n\theta $, we have
\begin{align}
I(\rho )&:=\frac{1}{2\pi }\int_{0}^{2\pi} F(\theta )\frac{1-\rho^2}{1-2 \rho \cos n\theta + \rho^2}d\theta \\
&= \frac{1}{2n\pi }\int_{0}^{2n\pi} F\left(\frac{\varphi }{n}\right)\frac{1-\rho^2}{1-2 \rho \cos \varphi + \rho^2}d\varphi \\
&=\frac{1}{2n\pi}\sum_{k=1}^n \int_{2(k-1)\pi}^{2k\pi} F\left(\frac{\varphi }{n}\right)\frac{1-\rho^2}{1-2 \rho \cos \varphi + \rho^2}d\varphi \\
&=\frac{1}{2n\pi}\sum_{k=1}^n \int_{0}^{2\pi} F\left(\frac{\varphi }{n}+\frac{2(k-1)\pi}{n}\right)\frac{1-\rho^2}{1-2 \rho \cos \varphi + \rho^2}d\varphi \\
&=\frac{1}{2\pi}\int_{0}^{2\pi} G(\varphi )\frac{1-\rho^2}{1-2 \rho \cos \varphi + \rho^2}d\varphi ,
\end{align}
where $G(\varphi )=\frac{1}{n}\sum_{k=1}^n F\left(\frac{\varphi }{n}+\frac{2(k-1)\pi}{n}\right).$
Since $G(\varphi )$ is continuous and periodic with period $2\pi$, letting $\rho $ to $1$ we have$$
\lim_{\rho \to 1} I(\rho )=G(0)=\frac{1}{n}\sum_{k=1}^n F\left(\frac{2(k-1)\pi}{n}\right).
$$
Similarly we can prove \begin{align}
\frac{1}{2\pi }\int_{0}^{2\pi} F(\theta )\frac{1-\rho^2}{1+2 \rho \cos n\theta + \rho^2}d\theta
&=\frac{1}{2\pi}\int_{0}^{2\pi} G(\varphi )\frac{1-\rho^2}{1+2 \rho \cos \varphi + \rho^2}d\varphi\\
&=\frac{1}{2\pi}\int_{0}^{2\pi} G(\varphi )\frac{1-\rho^2}{1-2 \rho \cos (\varphi-\pi) + \rho^2}d\varphi
\\
&\to G(\pi)=\frac{1}{n}\sum_{k=1}^n F\left(\frac{(2k-1)\pi}{n}\right).
\end{align}
Thus we have \begin{align}
&\lim_{\rho \to 1}\frac{1}{4\pi \rho} \int_{-\pi}^{\pi} F(\theta ) \left( \frac{1-\rho^2}{1-2 \rho \cos n\theta + \rho^2} - \frac{1-\rho^2}{1+2 \rho \cos n\theta + \rho^2} \right)~d\theta\\
&=\frac{1}{2n}\left(\sum_{k=1}^{n} F\left(\frac{2(k-1)\pi}{n}\right)-\sum_{k=1}^{n} F\left(\frac{(2k-1)\pi}{n}\right)\right)\\
&=\frac{1}{2n}\sum_{k=1}^{2n}(-1)^k~F \left(\frac{k \pi }{n} \right).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2183718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
To find the integral surface of given differential equation. Find the integral surface of the differential equation $(x-y)p+(y-x-z)q=z$ passing through the circle C: $z=1, x^2+y^2=1$
Clearly the Lagrange's auxillary equations are
$\frac{dx}{P} = \frac{dy}{Q}. =\frac{dz}{R}$
Where P=$(x-y)$ ,Q=$y-x-z$ & R=$z$
on comparing the given P.D.E with
the general quasilinear equation
P(x,y,z) p+Q(x,y,z)q=R(x,y,z)
I obtained two solutions
$x+y+z=c_1$
&
$\frac{x-y+z}{z^2}=c_2$
Then I substituted $x=s$ where S is a parameter.
So $y=\sqrt{1-s^2}$ & $z=1$.
I need to find a relation in $c_1$ &$c_2$
Then substitute for $c_1$ &$c_2$ to find the final integral surface passing through given circle.
How can I proceed now?
| Your calculus is correct. The characteristic equations are :
$$\begin{cases}
x+y+z=c_1\\
\frac{x-y+z}{z^2}=c_2
\end{cases}$$
The general solution of the PDE, expressed on the form of implicit equation, is :
$$F(X,Y)=0 \quad \begin{cases}
X=x+y+z\\
Y=\frac{x-y+z}{z^2}
\end{cases}$$
where $F(X,Y)$ is any differentiable equation of two variables.This function has to be determined according to the conditions :
First condition : $z=1
\begin{cases}
X=x+y+1\\
Y=x-y+1
\end{cases}
\quad\to\quad
\begin{cases}
x=\frac{X+Y}{2}-1\\
y=\frac{X-Y}{2}
\end{cases}$
Second condition : $x^2+y^2=1 \quad\to\quad \left(\frac{X+Y}{2}-1\right)^2+\left(\frac{X-Y}{2}\right)^2=1$
After simplification : $X(X-2)+Y(Y-2)=0$ which determines the function
$$F(X,Y)=X(X-2)+Y(Y-2)$$
Thus, with
$\begin{cases}
X=x+y+z\\
Y=\frac{x-y+z}{z^2}
\end{cases}$
the particular solution according to the specified condition is :
$$(x+y+z)(x+y+z-2)+\frac{x-y+z}{z^2}\left(\frac{x-y+z}{z^2}-2\right)=0$$
This is the equation of the surface passing through the specified circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2184361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Counting the Number of Real Roots of $y^{3}-3y+1$ Here's my question:
How many real roots does the cubic equation $y^3-3y +1$ have?
I graphed the function and it crossed the x-axis $3$ times. But my professor doesn't want a graphical explanation. So in that case, I was looking at the Fundamental Theorem of Algebra and states that a polynomial of degree n can have at most n distinct real roots. so therefore, there must be 3 real roots?
EDIT
It seems that there are numerous ways to approach this problem after all. And we can expand this to other types of polynomials as well, not just cubics.
| Set $x=2\cos\varphi$; then the equation becomes
$$
4\cos^3\varphi-3\cos\varphi=-\frac{1}{2}
$$
that is,
$$
\cos3\varphi=\cos\frac{2\pi}{3}
$$
Thus we get
$$
3\varphi=\frac{2\pi}{3}
\qquad\text{or}\qquad
3\varphi=\frac{2\pi}{3}+2\pi
\qquad\text{or}\qquad
3\varphi=\frac{2\pi}{3}+4\pi
$$
and therefore
$$
x=2\cos\frac{2\pi}{9}
\qquad\text{or}\qquad
x=2\cos\frac{8\pi}{9}
\qquad\text{or}\qquad
x=2\cos\frac{14\pi}{9}
$$
Note that choosing $3\varphi=-\frac{2\pi}{3}$ wouldn't give different solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2185090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 15,
"answer_id": 8
} |
Calculating Euler Number limit Please, so far I did
$$\lim_{x\to +\infty}\left(\frac{x^2-x+1}{x+2}\right)^{\frac{1}{x-1}},$$ but I can write
$$\frac{x^2-x+1}{x+2}=1+\frac{x^2-2x-1}{x+2}=1+\frac{1}{\frac{x+2}{x^2-2x-1}}.$$
But
$$\lim_{x\to +\infty}\frac{x+2}{x^2-2x-1}=0,$$ so I can not use
$$e =\lim_{N\to \infty}(1+\frac{1}{N})^N$$
| Let $\displaystyle y=\left(\frac{x^2-x+1}{x+2}\right)^{\frac{1}{x-1}}$ then
$$\ln y= \frac{1}{x-1}\ln(x^2-x+1)-\frac{1}{x-1}\ln(x+2)$$
$$\lim_{y\rightarrow \infty} \ln y=\lim_{x\rightarrow \infty}\frac{2x-1}{x^2-x+1}-\lim_{x\rightarrow \infty}\frac{1}{x+2}=0,\ L'Hopital$$
Then $\lim_{y\rightarrow \infty}y=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Trigonometric inequality with weird angle I have this inequality $\log_{\tan x}\sqrt{\sin^2x-\frac{5}{12}}<1$ that I would like to solve.
So $\tan x>0$ and $ \tan x\ne1$. First I try when $\tan x>1$.
Then I have $\sqrt{\sin^2x-\frac{5}{12}}<\tan x$
I have to solve the system
*
*$\sin^2x-\frac{5}{12}\ge0$ (I can't solve this one ($\frac{5}{12}$ troubles me I souldn't use a calculator or a table to solve this))
*$\tan x>0$
*$\sin^2x-\frac{5}{12}<\tan^2x$ and this one
| $1)$ If $\tan x>1\to \frac{\pi}{4}<x<\frac{\pi}{2}$ or $\frac{5\pi}{4}<x<\frac{3\pi}{2}$ $(1)$ then:
$$\sqrt{\sin^2x-\frac{5}{12}}<\tan x\to \sin^2x-\frac{5}{12}<\tan^2x\\
\sin^2x-\tan^2x<\frac{5}{12}\to -\frac{\sin^4x}{\cos^2 x}<\frac{5}{12}\\
-12\sin^4x<5(1-\sin^2x)\to12\sin^4x-5\sin^2x+5>0$$
which is true for any $x\in \Bbb R$ $(2)$.
We also must have
$$\sin^2x\ge\frac{5}{12}\to\sin x\le-\frac{\sqrt{15}}{6} \text{ or } \sin x\ge\frac{\sqrt{15}}{6}\quad (3)$$
So, considering the intersection of $(1), (2)$ and $(3)$ we get the solution
$$\frac{\pi}{4}<x<\frac{\pi}{2} \text{ or } \frac{5\pi}{4}<x<\frac{3\pi}{2}$$
Can you do the same idea for $0<\tan x <1$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Maximize $P=\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+y^{2}}}+\frac{1}{\sqrt{1+z^{2}}}$ For $x,y,z$ are positive real numbers that satisfy $xy+yz+xz=1$. Maximize $$P=\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+y^{2}}}+\frac{1}{\sqrt{1+z^{2}}}.$$
I think if we let $x=\tan A;y=\tan B;z=\tan C$, then
$$P\Leftrightarrow \displaystyle \text {sin}A+\text {sin}B+ \text{sin}C\leq \frac{3}{2}. \hspace{2cm}(1)$$
But I can't prove $(1)$.
| Like How to show that the triangle is equilateral triangle?,
$\cot A\cot B+\cot B\cot C+\cot C\cot A=1$ with $A+B+C=\pi$
Now use this
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
integrate $\int\frac{x\cdot dx}{(x^3+1)^2}$ What methods are there to integrate: $$\int\frac{x\cdot dx}{(x^3+1)^2}$$
I know about partial fractions:
$$\int\frac{x\cdot dx}{(x^3+1)^2} $$
$$= \int\frac{x\cdot dx}{((x+1)(x^2-x+1))^2} $$
$$= \int \left(\frac{A}{x+1}+\frac{Bx+C}{(x+1)^2} + \frac{Dx+E}{x^2-x+1} + \frac{Fx^3+Gx^2+H+I}{(x^2-x+1)^2}\right)dx$$ and after this solving is easy, i was trying to do the same many times, but i can't find coefficients because mistakes or something other.
I want to know about another methods to solve it.
| $$
\begin{align}
\int\frac{x\,\mathrm{d}x}{\left(x^3+1\right)^2}
&=-\frac13\int\frac1x\,\mathrm{d}\frac1{x^3+1}\tag{1}\\
&=-\frac1{3x\left(x^3+1\right)}-\frac13\int\frac{\mathrm{d}x}{x^2\left(x^3+1\right)}\tag{2}\\
&=-\frac1{3x\left(x^3+1\right)}-\frac13\int\left(\frac1{x^2}-\frac{x}{x^3+1}\right)\,\mathrm{d}x\tag{3}\\
&=-\frac1{3x\left(x^3+1\right)}+\frac1{3x}+\frac19\int\left(\frac{x+1}{x^2-x+1}-\frac1{x+1}\right)\,\mathrm{d}x\tag{4}\\
&=-\frac1{3x\left(x^3+1\right)}+\frac1{3x}-\frac19\log(x+1)+\frac19\int\frac{x-\frac12+\frac32}{\left(x-\frac12\right)^2+\frac34}\,\mathrm{d}x\tag{5}\\
&=\small\frac{x^2}{3\left(x^3+1\right)}-\frac19\log(x+1)+\frac1{18}\log\left(x^2-x+1\right)+\frac1{3\sqrt3}\tan^{-1}\left(\frac{2x-1}{\sqrt3}\right)+C\tag{6}\\
&=\frac{x^2}{3\left(x^3+1\right)}+\frac1{18}\log\left(\frac{x^3+1}{(x+1)^3}\right)+\frac1{3\sqrt3}\tan^{-1}\left(\frac{2x-1}{\sqrt3}\right)+C\tag{7}
\end{align}
$$
Explanation:
$(1)$: prepare to integrate by parts
$(2)$: integrate by parts
$(3)$: partial fractions
$(4)$: partial fractions
$(5)$: integrate
$(6)$: integrate
$(7)$: simplify
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Find all solutions in positive integers to $a^b -b^a = 3$
Find all solutions in positive integers to $a^b -b^a = 3$
It appears that the only solution is $a=4, b=1$.
Modular arithmetic is not of much help (apart from deriving that $a,b$ must have opposite parity).
We know that $a^x$ dominates the polynomial $x^a$ but this does not seem to yield the result.
Tried hard to estimate a bound for $a^b - b^a$ but with no success.
Any help is appreciated. This is a problem from a regional mathematics Olympiad.
Added on 26 March, 2017:
I was trying to find a solution that involves no Calculus (or a minimum of Calculus) since this is a problem from a Junior Olympiad in which the students do not have a knowledge of Calculus. The following argument seems to work. Please point out flaws/mistakes, if any.
Lemma 1 For any $n \geq 4$, $$n^{n+1} - (n+1)^n > (n-1)^n - n^{n-1}$$
Proof
\begin{align*}
&\qquad n^{n+1} - (n+1)^n > (n-1)^n - n^{n-1} \\
&\Leftrightarrow n^{n+1} + n^{n-1} > (n-1)^n + (n+1)^n \\
&\Leftrightarrow n+\frac{1}{n} > \left(1-\frac{1}{n}\right)^n + \left(1+\frac{1}{n}\right)^n
\end{align*}
Since $\left(1+\frac{1}{n}\right)^n < 3$ and $\left(1-\frac{1}{n}\right)^n < 1$, we have
$$\left(1+\frac{1}{n}\right)^n+\left(1-\frac{1}{n}\right)^n < 4 < n +\frac{1}{n}$$
Lemma 2 For any $k$ and $n \geq 4$,
$$n^{n+k} -(n+k)^n > n^{n+1} - (n+1)^n$$
Proof
\begin{align*}
&\qquad n^{n+k} -(n+k)^n > n^{n+1} - (n+1)^n \\
&\Leftrightarrow n^{n+k} -n^{n+1} > (n+k)^n - (n+1)^n \\
&\Leftrightarrow n^k - n > \left(1+\frac{k}{n}\right)^n - \left(1+\frac{1}{n}\right)^n
\end{align*}
Since $\left(1+\frac{k}{n}\right)^n < 3^k$ and $\left(1-\frac{1}{n}\right)^n < 1$, we have
$$\left(1+\frac{k}{n}\right)^n+\left(1-\frac{1}{n}\right)^n < 3^k +1 < n^k - n $$
as $n > 4$.
Now suppose that $a^b - b^a = 3$. Let $a \geq 4$. If $b = a+k > a$, then we have
\begin{align*}
a^{a+k} - (a+k)^a &> a^{a+1} - (a+1)^a \\
&> (a-1)^a - a^{a-1} \\
&> \cdots \\
&> 3^4 - 4^3 = 17
\end{align*}
Thus there are no solutions with $b > a \geq 4$.
If $a > b \geq 4$, then
$$a^b - b^a = -(b^a - a^b) \leq -17 $$
from what we have seen above. Thus there are no solutions if $a > b \geq 4$. Thus all solutions can be only in the range $1 \leq a,b \leq 4$. Clearly, in this range only $a=4, b=1$ satisfies the given equation.
| Consider $$f_a(x)=x\ln(a)-a\ln(x)$$ with $a,x\ge 3$
We have $$f_a'(x)=\ln(a)-\frac{a}{x}>0$$ for $x>\frac{a}{\ln(a)}$
In particular, $f_a(x)$ is strictly increasing for $x\ge a$ and we have $$f_a(a+1)=(a+1)\ln(a)-a\ln(a+1)$$
The function $g(x)=(x+1)\ln(x)-x\ln(x+1)$ has derivate $g'(x)=\ln(x)-\ln(x+1)+\frac{x+1}{x}-\frac{x}{x+1}=\ln(\frac{x}{x+1})+\frac{2x+1}{x^2+x}>\ln(\frac{x}{x+1})+\frac{1}{x}=\frac{1}{x}-\ln(1+\frac{1}{x})>0$ for $x\ge 3$
So, we have $f_a(x)\ge 0.235566$ for $3\le a\le x$ hence $$x\ln(a)\ge a\ln(x)+0.235566$$ implying $$a^x>1.26x^a$$
Since $x^a\ge 27$, we can conclude $$a^x>x^a+7$$
This rules out $3\le a<x$
The case $3\le x<a$ can be ruled out analogue by changing the roles of $a$ and $x$. $a=x$ can be rules out anyway.
Now, we show $2^a>a^2+3$ for $a\ge 5$ by induction
$a=5$ : $2^5=32>28=5^2+3$
$$2^{a+1}=2\cdot 2^a>2(a^2+3)=2a^2+6>a^2+2a+4=(a+1)^2+3$$
for $a^2>2a-2$ , or $a^2-2a+2=(a-1)^2+1>0$ , which is true for every $a$
The case $x=1$ and the remaining cases for $x=2$ can be verified easily.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2195113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Prove that all odd derivate of $\tan(x)$ at $x=0$ is at least $1$. This was the original excercise:
Prove that $$\frac{\sin{x}+\tan{x}}{2} \geq x$$ where $x \in (0,\frac{\pi}{2})$
This is how I did it:
Knowing that
\begin{align}\sin{0}&=0\\
\cos{0}&=1\\
\frac{\mathrm{d} \sin{x}}{\mathrm{d} x}&=\cos{x}\\
\frac{\mathrm{d} \cos{x}}{\mathrm{d} x}&=- \sin{x}\end{align}
and the Taylor series of $f(x)$ is: $$\sum_{n=0}^{\infty} \frac{x^n\frac{\mathrm{d}^n f(0)}{\mathrm{d} x^n}}{n!}$$
we can easily see that the Taylor series of $\sin(x)$ is:
$$\sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$$
We know what the tan(x) function is odd, so the n-th derivate will be $0$ for all even $n$. For odd numbers, it will be $1,2,16,272,\cdots$ So if we sum the $2$ Taylor series, we will get this:
$$\sin{x}+\tan{x}=2x+\frac{x^3}{3!} (-1+1)+\frac{x^5}{5!}(16+1)+\frac{x^7}{7!}(272-1)+...$$
Where we can see that all part will be positive (or $0$) on the given interval. Back to the original inequality:
$$2x+\frac{x^3}{3!} (-1+1)+\frac{x^5}{5!}(16+1)+\frac{x^7}{7!}(272-1)+...\geq 2x$$
$$\frac{x^3}{3!} (-1+1)+\frac{x^5}{5!}(16+1)+\frac{x^7}{7!}(272-1)+... \geq 0$$
Which is true in the given interval.
But how could I prove that all odd derivate of $\tan(x)$ at $x=0$ is at least $1$?
| An alternative proof for the desired inequality without using
Taylor series would be to notice that
$$
f(x) = \frac 12 (\sin x + \tan x)
$$
satisfies $f(0) = 0$ and
$$
f'(x) = \frac 12 \left(\cos x + \frac{1}{\cos^2 x} \right) \ge
\sqrt{\cos x \cdot \frac{1}{\cos^2 x}} = \frac{1}{\sqrt{\cos x}}
\ge 1
$$
for $0 \le x < \frac \pi 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Factorisation of quartic equation. I'm given a logarithm problem which is $$\log_{2} (x^3+1)-2\log_{2}x=\log_{2}(x^2-x+1)-2$$
I'm stuck in the step of $x^4-5x^3+x^2-4=0$
By many times of trial and error, I got $(x^2-4x-4)(x^2-x+1)=0$
My question, is there any standard way to factorise the equation without trials and errors? Thanks in advance.
| Hint: the quartic can be avoided altogether by noting that $x^3+1=(x+1)(x^2-x+1)\,$. Given that $x^2-x+1 \gt 0$ on $\mathbb{R}$ it follows that $x^3+1$ and $x+1$ have the same sign, which must be positive for the $\log_2$ to be defined. Then the equation simplifies to:
$$
\require{cancel}
\log_{2} (x+1)+\cancel{\log_{2} (x^2-x+1)}-2\log_{2}x=\cancel{\log_{2}(x^2-x+1)}-2 \\
\iff\quad \log_2{}\frac{x+1}{x^2}=-2 \quad\iff\quad \frac{x+1}{x^2}=\frac{1}{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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When, where and **how often** do you find polynomials of higher degrees than two in mathematical, pure/applied, research? A formula for solving a polynomial of degree three, see this link; $ax^3+bx^2+cx+d=0$, is
$$\begin{align}
x\quad&=\quad
\sqrt[3]{
\left(
\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}
\right)
+
\sqrt{
\left(
\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}
\right) ^2
+
\left(
\frac{c}{3a} - \frac{b^2}{9a^2}
\right) ^3
}
}\\
&+\quad
\sqrt[3]{
\left(
\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}
\right)
-
\sqrt{
\left(
\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}
\right) ^2
+
\left(
\frac{c}{3a} - \frac{b^2}{9a^2}
\right) ^3
}
}
\;-\;\frac{b}{3a}
\end{align}$$
Unlike quadratic, cubic, and quartic polynomials, the general quintic
cannot be solved algebraically in terms of a finite number of
additions, subtractions, multiplications, divisions, and root
extractions, as rigorously demonstrated by Abel (Abel's impossibility
theorem) and Galois. However, certain classes of quintic equations can
be solved [...] Source: http://mathworld.wolfram.com/QuinticEquation.html
At levels of $5^{\text{th}}$ degree polynomials, things are starting to look really serious in my eyes. My question is:
If it is possible to not answer subjectively: When, where and how often do you find polynomials of higher degrees than two in mathematical, pure/applied, research?
| Cubic polynomials are ubiquitous in computer-aided design and computer graphics.
They also are the basis for computer fonts.
Finite element analysis is based on polynomial functions. Isogeometric analysis uses NURBS.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2197171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 13,
"answer_id": 10
} |
Inequality involving the tangent function How can one show that
$$x \tan x > \frac{4x - \pi}{\pi - 2x},$$
for $x \in \left(0,\frac{\pi}{2}\right)$.
Clearly, for $x \in \left(0,\frac{\pi}{4}\right)$ as $4x - \pi < 0$ and $\pi - 2x > 0$ the term $\frac{4x - \pi}{\pi - 2x}$ is negative while $x \tan x$ is always positive, so one need only show the inequality is true for $x \in \left(\frac{\pi}{4},\frac{\pi}{2}\right)$.
| For $0<x\leq\frac{\pi}{4}$ our inequality is obviously true.
Let $\frac{\pi}{4}<x<\frac{\pi}{2}$ and $t=\tan{x}$.
Hence, $t>1$ and we need to prove that
$$(\pi-2x)x\tan{x}>4x-\pi$$ or
$$2x^2\tan{x}+(4-\pi\tan{x})x-\pi>0$$ or
$$x<\frac{-4+\pi\tan{x}+\sqrt{16+\pi^2\tan^2x}}{4\tan{x}}$$
or $f(x)>0$, where
$$f(x)=\frac{-4+\pi\tan{x}+\sqrt{16+\pi^2\tan^2x}}{4\tan{x}}-x.$$
We see that
$$f'(x)=\frac{\cos^2x\sqrt{16+\pi^2\tan^2x}-4}{\sin^2x\sqrt{16+\pi^2\tan^2x}}.$$
We'll prove that $f'(x)<0$ for all $\frac{\pi}{4}<x<\frac{\pi}{2}$.
Indeed, it's enough to prove that
$$\cos^2x\sqrt{16+\pi^2\tan^2x}<4$$ or
$$\sqrt{16+\pi^2t^2}<4(1+t^2)$$ or
$$16t^4+(32-\pi^2)t^2>0,$$
which is obvious.
Id est, $$f(x)>\lim\limits_{x\rightarrow\frac{\pi}{2}^+}f(x)=\frac{\pi}{4}+\frac{\pi}{4}-\frac{\pi}{2}=0$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2199134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.