Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find limit of $\frac{n(nx_n-\frac{1}{3})}{\ln n}$ knowing that $x_{n+1}=x_n-3x_n^2$ Let $x_n$ a sequence with $\frac{1}{3}>x_1>0$ and $x_{n+1}=x_n-3x_n^2$. Find the limit
$$\lim_{n\to \infty} \frac{n(nx_n-\frac{1}{3})}{\ln n}.$$
I proved that $x_n$ is convergent to $0$ and that $\lim\limits_{n\to \infty} nx_n= \frac{1... | Firstly,
$$\lim_{n\to \infty} \frac{n\left(nx_n-\frac{1}{3}\right)}{\ln n}=\lim_{n\to \infty} nx_n \cdot \lim_{n\to \infty} \frac{n-\frac{1}{3x_n}}{\ln n}=\frac{1}{3} \lim_{n\to \infty}\frac{n-\frac{1}{3x_n}}{\ln n}$$
Now, from Cesaro-Stolz:
$$
\begin{aligned}
\lim_{n\to \infty}\frac{n-\frac{1}{3x_n}}{\ln n} &= \lim_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3580201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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How do I solve this limit without l'Hopital? I tried the substitution $t=x-(\pi/3)$ but it doesn't help at all. I have also tried using $\sin(\pi/3)=\sqrt{3}/2$ but couldn't do anything useful then. I tried to factor the denominator and numerator, but it didn't help either. I want a solution without l'Hopital's rule.
$... | $$ \lim_{x \to \frac{\pi}{3}}\frac{\sin^2 x-\sin^2 \frac{\pi}{3}}{x^2-(\frac{\pi}{3})^2} =\\ \lim_{x \to \frac{\pi}{3}}\frac{(\sin x-\sin \frac{\pi}{3})(\sin x+\sin \frac{\pi}{3})}{(x-\frac{\pi}{3})(x+\frac{\pi}{3})}=\\
\lim_{x \to \frac{\pi}{3}}\frac{2\cos(\frac{x}{2}+\frac{\pi}{6})\sin(\frac{x}{2}-\frac{\pi}{6})(\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3580733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
$\sum _{n=0}^{\infty} \frac{1}{(n+1) (n+2)} \left(\frac{1}{\lfloor n \phi \rfloor +2}+\frac{1}{\lfloor n \phi ^{-1} \rfloor +2}\right)$ How to prove: $$\sum _{n=0}^{\infty} \frac{1}{(n+1) (n+2)} \left(\frac{1}{\lfloor n \phi \rfloor +2}+\frac{1}{\lfloor n \phi ^{-1} \rfloor +2}\right)=\frac{3}{4}$$
Here $\phi=\frac{1+\... | The term for $n=0$ gives us $\frac 1 2$.
For the others, use $\frac 1{(n+1)(n+2)} = \frac 1 {n+1} - \frac 1{n+2}$ break the sum into:
$$ \sum_{n=1}^\infty \frac 1{(n+1)(\lfloor n\phi\rfloor + 2)} - \sum_{n'=1}^\infty \frac 1 {(n'+2)(\lfloor n'\phi\rfloor + 2)} + \sum_{m'=1}^\infty \frac 1{(m'+1)(\lfloor m'\phi^{-1}\rfl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
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} |
$\int_0^\pi\left|\frac{\sin {nx}}{x}\right|dx\ge \frac{2}{\pi}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)$
Question: Show that $$\int_0^\pi\left|\frac{\sin {nx}}{x}\right|dx\ge \frac{2}{\pi}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right).$$
My approach: We know that $$1+\frac{1}{2}+\cdots+\frac{1}{n}=\int_0^1\left(1+... | Let $t_k=\frac{k\pi}{n}$ for $k=0,1,2,\ldots,n$. We have
$$\int_{t_{k-1}}^{t_k}\left|\frac{\sin nx}{x}\right|dx=\int_0^\pi\frac{\sin u}{(k-1)\pi+u}du,$$
where $u=nx-(k-1)\pi$. For $0\le u \le \pi$, $(k-1)\pi \le (k-1)\pi+u\le k\pi$, so
$$\int_{t_{k-1}}^{t_k}\left|\frac{\sin nx}{x}\right|dx\ge \int_0^\pi\frac{\sin u}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
LU factorization of a singular matrix I am trying to find the LU factorisation of the following matrix: $$A=\begin{pmatrix} 1 & 0 & 3 \\ 2 & 2 & 2 \\ 3 & 6 & -3 \end{pmatrix}.$$ Note that $A$ is singular.
I proceed with Gaussian elimination:
$$A=\begin{pmatrix} 1 & 0 & 3 \\ 2 & 2 & 2 \\ 3 & 6 & -3 \end{pmatrix}\implie... | $$A=\begin{pmatrix} 1 & 0 & 3 \\ 2 & 2 & 2 \\ 3 & 6 & -3 \end{pmatrix}=\begin{pmatrix}
a& 0 & 0\\
b& c &0 \\
d& e & f
\end{pmatrix}\times \begin{pmatrix}
1&0 &3 \\
0 & 2 &-4 \\
0 & 0 & 0
\end{pmatrix}$$
multiply it and equate to original matrix and find $a,b,c,d,e,f$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If the value of integral in the image below is π then what is the value of y? I could not simplify
$$
\int_0^1 \sqrt{-1 + \sqrt{\frac{1+y}{x} - y}}\ dx
$$
I tried integration it in an online integrator but trust me the result is seriously daunting to be back traced to $\pi$ as a value, thereby determining $y$. So I am... | Define
$$f \colon [-1,\infty) \to [0,\infty), \, f(y) = \int \limits_0^1 \sqrt{\sqrt{\frac{1+y}{x} - y}-1} \, \mathrm{d} x \, .$$
Obviously, $f(-1) = 0$. For $y > -1$ we have
\begin{align}
f(y) &= \int \limits_0^1 \sqrt{\sqrt{\frac{1+y}{x} - y}-1} \, \mathrm{d} x \stackrel{\frac{1+y}{x} - y = \frac{1}{t^2}}{=} (1+y) \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3591928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
how to show $B(n,m)=\frac{(n-1)!(m-1)!}{(n+m-1)!}$?
I have done the following transformation:
$(1-x)^{m-1}=u$ $\Rightarrow -(m-1)(1-x)^{m-2}dx=du$
$x^{n-1}dx=dv$ $\Rightarrow \frac{x^n}{n}=v$
Hence $$\int^1_0x^{n-1}(1-x)^{m-1}dx= (1-x)^{m-1}\frac{x^n}{n}\bigg |^1_0+ \int\frac{x^n}{n}(m-1)(1-x)^{m-2}dx=(1-1)^{m-1}\frac... | Beginning with the result you have shown, repeated application yields the following chain of equalities:
\begin{align*}
B(n,m) &= \frac{m-1}{n}\cdot B(n+1,m-1) \\
&= \frac{m-1}{n}\cdot \frac{m-2}{n+1}\cdot B(n+2,m-2) \\
&= \cdots \\
&= \frac{m-1}{n}\cdot \frac{m-2}{n+1} \cdot \cdots \cdot \frac{1}{n+m-2} \cdot B(n+m-1,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3592924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Complex number proof involving angles I need to show that (3 + i)^3 = 18 + 26i and use this to show that the angle AOC = 3AOB, where O, A, B, C are points in the plane given by O = (0, 0), A = (1, 0), B = (3, 1) and C = (18, 26).
This is what I have done:
Expand (3 + i)^3
(3 + i)^3 = (3 + i)(3 + i)(3 + i)
= 9 + 3... | Use $\sin 3\theta = 3\sin\theta - 4\sin^3 \theta$ (Mathworld [16])
$\sin\angle AOB=\frac1{\sqrt{10}}$ and $\sin\angle AOC=\frac{26}{10\sqrt{10}}$
so we have:
$RHS=\frac3{\sqrt{10}}-\frac4{10\sqrt{10}}=\frac{26}{10\sqrt{10}}=LHS$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3595535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Evaluate: $\sum_{k=0}^{n}(-4)^k\frac{{n+k \choose 2k}}{ak+b}$ From this Question:
How does one find the closed form for $(1)?$
$$\sum_{k=0}^{n}(-4)^k\frac{{n+k \choose 2k}}{ak+b}=F(a,b)\tag1$$
$$\sum_{k=0}^{n}(-4)^k\frac{{n+k \choose n-k}}{ak+b}=F(a,b)\tag2$$
This one is from the Question:
$F(0,1)=(-1)^n(2n+1)$
We g... | You can find a closed form as follow :
$$ F(a,b)=\left(\sum_{k=0}^n \int_0^1(-4t)^{ak+b-1}\binom{n+k}{2k}dt \right) $$
Finite sum
$$ F(a,b)=\left( \int_0^1\sum_{k=0}^n(-4t)^{ak+b-1}\binom{n+k}{n-k}dt \right) $$
Then using the coefficient extraction function $[X^i]$ for a given $i$.
\begin{align*}
\sum_{k=0}^n(-4t)^{ak+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3597406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Calculate the sum $\sum_{n=1}^{\infty}\Big((2n-1)\Big(\sum_{k=n}^{\infty}\frac{1}{k^2}\Big)-2\Big) $ Calculate the sum $$\sum_{n=1}^{\infty}\Big((2n-1)\Big(\sum_{k=n}^{\infty}\frac{1}{k^2}\Big)-2\Big) $$
The only thing I can think of is that $\sum_{n=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}.$ However, I can't go any fu... | For $ n\in\mathbb{N} $, define the sequence $ \left(u_{n,k}\right)_{k} $ as follows : $ u_{n,k}=\left\lbrace\begin{aligned}\frac{2n-1}{k^{2}\left(4k^{2}-1\right)},\ \ \ \textrm{If }k\geq n\\ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \textrm{If }k<n\end{aligned}\right. $
Since $ \sum\limits_{n=1}^{+\infty}{u_{n,k}}=\frac{1}{k^{2}\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3597766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that $\lim_{(x;y)\rightarrow (0;0)} \frac{xy(x-y)}{x^4+y^4} = 0$ using the squeeze theorem My attempt:
$$0 \le \frac{|xy(x-y)|}{|x^4+y^4|} = \frac{|x^2y-xy^2|}{|x^4+y^4|} \le \frac{|x^2y|+|xy^2|}{x^4+y^4} \le \frac{|x|^2|y|+|x||y|^2}{x^4+y^4} = \frac{|x|^2|y|+|x||y|^2}{(x^2+y^2)^2-(\sqrt{2}xy)^2} = ?$$
What do ... | This limit is not $0$ (I bet it doesnt even exist).
If we move towards $0$ on the line $2x=y$ we have
$$\frac{xy(x-y)}{x^4+y^4}=\frac{2y^3}{17y^4}=\frac{2}{17y}$$
and as $y\to0$ this is $\pm\infty$ (depending on whether $y>0$ or $y<0$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3598113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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limit of the sequence $a_n=n\left[\sin\left(\frac{1+n^3}{n^2}\right)-\sin n\right]$ I started by writting
$$
\begin{eqnarray*}
a_n &=& n \left[\sin\left(\frac{1+n^3}{n^2}\right)-\sin\, n\right]=\frac{\sin\left(\displaystyle\frac{1}{n^2}+n\right)-\sin\, n}{\displaystyle\frac{1}{n}} \\
&=&\frac{2\sin\left(\displaysty... | The basic formula is $sin(x+y)-sin(x-y)=2cos(x)sin(y)$. Here $x+y=n+\frac{1}{n^2}$ and $x-y=n$, resulting in $x=n+\frac{1}{2n^2}$ and $y=\frac{1}{2n^2}$. So $a_n=2ncos(n+\frac{1}{2n^2})sin(\frac{1}{2n^2})$ which is the expression in the book.
Your error was in getting expressions for $x$ and $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3599670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\left(A+B\right)^{2}\nmid A^{2n+1}+B^{2n+1}$ Let it be A and B two coprime positive integers. I know how to prove by induction that $A+B\mid A^{2n+1}+B^{2n+1}$, but I am having a bit trouble proving that $\left(A+B\right)^{2}\nmid A^{2n+1}+B^{2n+1}$.
For the case $n=1$ I have managed to prove it in the foll... | The statement is false. We show how to find counterexamples.
Since we know that $$A^{2n+1}+B^{2n+1}=(A+B)(A^{2n}-A^{2n-1}B+\dots-AB^{2n-1}+B^{2n}),$$ it is equivalent to finding $A,B$ such that $$A+B\mid A^{2n}-A^{2n-1}B+\dots-AB^{2n-1}+B^{2n},$$ which is the same as $$A^{2n}-A^{2n-1}B+\dots-AB^{2n-1}+B^{2n}\equiv 0\pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3606907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove $\frac{v}{\pi} \int_0^1 \frac{1}{x^2 \sqrt{1-x^2}} e^{\frac{-v^2}{2x^2}} dx =\frac{1}{\sqrt{2\pi}} e^{-\frac{v^2}{2}}$ If $v>0$ why
$$\frac{v}{\pi} \int_0^1 \frac{1}{x^2 \sqrt{1-x^2}} e^{\frac{-v^2}{2x^2}} dx
=\frac{1}{\sqrt{2\pi}} e^{-\frac{v^2}{2}}$$.
I saw it in
I tried to solve it by $t=\frac{1}{x^... | First substitute $y=\frac{v^2}{2x^2}$. Then $x^2=\frac{v^2}{2y}$ and $\mathrm dx=\color{green}{-\frac{v}{2\sqrt 2} y^{-\frac32}}\,\mathrm dy$. Thus $$\int_0^1 \frac{1}{\color{blue}{x^2} \color{orange}{\sqrt{1-x^2}}} e^{\frac{-v^2}{2x^2}} \,\mathrm dx=\color{green}-\frac{\color{blue}2\color{green}v}{\color{green}{2\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3607629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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how to compute $\lim_{n \to \infty}\sqrt{n}\int_{0}^{1}(1-x^2)^n$? I need to compute $$\lim_{n \to \infty}\sqrt{n}\int_{0}^{1}(1-x^2)^n dx.$$
I proved that
for $n\ge1$,
$$\int_{0}^{1}(1-x^2)^ndx={(2n)!!\over (2n+1)!!},$$
but I don't know how to continue from here.
I also need to calculate $\int_{0}^{1}(1-x^2)^ndx$ for ... | Redoing what has been done
many many many times before.
$\begin{array}\\
I_n
&=\int_0^1 (1-x^2)^n dx\\
I_0
&=\int_0^1 dx\\
&= 1\\
I_1
&=\int_0^1 (1-x^2) dx\\
&=1-\dfrac13\\
&=\dfrac23\\
I_n
&=\int_0^1 (1-x^2)^n dx\\
&=x(1-x^2)^n|_0^1+\int_0^1 2x^2n(1-x^2)^{n-1} dx\\
&\qquad\text{integrating by parts}\\
&\qquad f = (1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3609906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Laurent expansion of $f(z)=\frac{1}{1+z^2}+\frac{1}{3-z}$ in the region $|z|>3$. Laurent expansion of $f(z)=\frac{1}{1+z^2}+\frac{1}{3-z}$ in the region $|z|>3$.
According to the solution the expansion is
$$\sum_{n=0}^{\infty}(-1)^nz^{-2n}+\sum_{n=0}^{\infty}\frac{3^n}{z^{n+1}}.$$
But I don't understand why. My work is... | I think the given answer, the first summatoy BEGINS in $n=1$.
$$\frac{1}{z^2}\frac{1}{1-(-z^2)} = - \frac{-z^{-2}}{1-(-z^2)}= -\sum_{n=1}^\infty (-z^2)^n=\sum_{n=1}^\infty (-1)^{n+1} z^{2n}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3612168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Players and tournaments 16 players take part in a tennis tournament.
The order of the matches is chosen at random.
If there is a player better than the other one then the better one wins find the :
a)probability that all the four best place reach the semifinals
b)probability that sixth best reaches the semifinals
My... | I am assuming the way the tournament works is that there is a first round with $8$ games, then the $8$ winners of the first round play a second round of $4$ games, then the winners of the second round enter the semi-final. In order for the top $4$ players to all play in the semi-final, no two of them can play against ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3613700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $abcd \le 3+ ab+ac+ad+bc+bd+cd$ when $(a^2+1)(b^2+1)(c^2+1)(d^2+1)=16$ Mathematica proves the inequality (the code on demand) $$abcd \le 3+ ab+ac+ad+bc+bd+cd$$ under the condition $$(a^2+1)(b^2+1)(c^2+1)(d^2+1)=16$$
The equality is attained at
$$\begin{align}
a &=\frac{\frac{927}{512}-\frac{73 \left(1196032-\... | Since $$(x^2+y^2)(z^2+t^2)=(xz+yt)^2+(xt-yz)^2,$$ we obtain:$$16=(a^2+1)(b^2+1)(c^2+1)(d^2+1)=((ab-1)^2+(a+b)^2)((cd-1)^2+(c+d)^2)=$$
$$=((ab-1)(cd-1)-(a+b)(c+d))^2+((ab-1)(c+d)+(cd-1)(a+b))^2=$$
$$=(1+abcd-ab-ac-ad-bc-bd-cd)^2+$$
$$+(abc+abd+acd+bcd-a-b-c-d)^2\geq$$
$$\geq(abcd+1-ab-ac-ad-bc-bd-cd)^2,$$ which gives
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3615144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I find this matrix? a) Let $\mathbf{P}$ be the $2\times 2$ matrix that projects vectors onto $\mathbf{u} = \begin{pmatrix}2\\ -1 \end{pmatrix}$. That is,
$\mathbf{P} {v} = \operatorname{proj}_{u} ({v}) = \text{Projection of $\mathbf{v}$ onto $\mathbf{u}$}.$
Using the geometric meaning of the matrix and the pict... | You should set $(x,y)^T=(2,-1)^T$ as you did for your answer in part (a). This will not give you the values you seek, but will produce simultaneous equations for them. By repeating the same process, but for the other values obtained at the end of part (a), you will arrive at the values of $a,b,c,d$ and therefore $P$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3618121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Four roots of the polynomial $x^4+px^3+qx^2+rx+1$ Let $a, b, c, d$ be four roots of the polynomial $x^4+px^3+qx^2+rx+1$ prove that $(a^4+1)(b^4+1)(c^4+1)(d^4+1)=(p^2+r^2)^2+q^4-4pq^2r$
Please provide hint.
| $$\prod_{cyc}(a^4+1)=((a^2b^2-1)^2+(a^2+b^2)^2)((c^2d^2-1)^2+(c^2+d^2)^2)=$$
$$=((a^2b^2-1)(c^2+d^2)+(c^2d^2-1)(a^2+b^2))^2+$$
$$+((a^2b^2-1)(c^2d^2-1)-(a^2+b^2)(c^2+d^2))^2=$$
$$=\left(\sum_{cyc}a^2b^2c^2-\sum_{cyc}a^2\right)^2+\left(a^2b^2c^2d^2-\frac{1}{6}\sum_{sym}a^2b^2\right)^2.$$
Can you end it now?
| {
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"url": "https://math.stackexchange.com/questions/3619135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Number of ways to color in a $2\times5$ grid
How many ways are there to color in each unit square of a $2\times5$ grid red or blue so that no $2\times2$ square is allowed to be the same color?
So I thought about doing an inclusion-exclusion method to approach this problem, but my main issue with this approach was th... | Let $a_n$ denote the number of good colorings of a $2\times n$ grid. You want to compute $a_5$. The initial conditions are $a_0 = 1$ and $a_1 = 4$. To derive a recurrence relation for $a_n$, consider the four possibilities for the two cells in the leftmost column. You will find that $a_n = 3 a_{n-1} + 2 a_{n-2}$, y... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Langley's Adventitious Angles${}$ I've been running in circles and couldn't give a rigorous mathematical proof that the angle is x = 20°.
Any idea? This is my try:
I got the answer $x=20^\circ$ using a computer program: https://www.geogebra.org/classic/qt79hpec
| This is a variant of the original Langley's puzzle, which has a straightforward trigonometric solution. Apply the sine rule to the triangles ADE, ADB and BDE
$$\frac{\sin x}{\sin 10}\cdot \frac{\sin 20}{\sin (30+x)}\cdot \frac{\sin 80}{\sin 60}
=\frac{DA}{DE}\cdot \frac{DE}{DB}\cdot \frac{DB}{DA} = 1$$
which simplifies... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3620825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Showing that $\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}$ via Fourier Transform If $a,b>0$, how can I prove this using Fourier Series
$$\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}.$$
I tried to split the product and calculate the integral using Parceval's Theorem... | I know what is following is not what is asked but it's a simpler way.
$a\neq b$
\begin{align}J&=\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx\\
&=2\int_{0}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx\\
&=\frac{2b^2}{b^2-a^2}\int_0^\infty \frac{1}{x^2+b^2}\,dx-\frac{2a^2}{b^2-a^2}\int_0^\infty \frac{1}{x^2+a^2}\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3625913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
If $a,b,c$ are the sides of a triangle, then $\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$ is: If $a,b,c$ are the sides of a triangle, then $\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$ is:
$A)$ $\le3$ , $B)$ $\ge3$, $(C)$ $\ge2$, $(D)$ $\le2$
My attempt is as follows:-
$$\dfrac{1}{2}\left(\dfrac{a}{s-a}+... | Note that $f(x)=\frac x{s-x}$ is a convex function. Then, Jensen’s inequality leads to,
$$\begin{align}
& \dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}\\
=& \frac12 \left(\dfrac{a}{s-a}+\dfrac{b}{s-b}+\dfrac{c}{s-c}\right) \\
\ge &\frac12\cdot3\cdot\frac{ \frac{a+b+c}3}{s -\frac{a+b+c}3}=\frac32\cdot\frac{\frac{2s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3626167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Find complex roots of quartic function $(3z + 1)(4z + 1)(6z + 1)(12z + 1) = 2$ I found a math problem involving complex number
Find all complex number z such that
$$(3z + 1)(4z + 1)(6z + 1)(12z + 1) = 2$$
The complex number form is z = a + bi
If I multiply all the factor to make it to the standard form of a polynom... | Let $z=\frac{x}{12}.$
Thus, $$(3z + 1)(4z + 1)(6z + 1)(12z + 1)-2=$$
$$=\left(\frac{x}{4}+1\right)\left(\frac{x}{3}+1\right)\left(\frac{x}{2}+1\right)(x+1)-2=$$
$$=\frac{1}{24}((x+4)(x+3)(x+2)(x+1)-48)=$$
$$=\frac{1}{24}((x^2+5x+4)(x^2+5x+6)-48)=$$
$$=\frac{1}{24}(x^2+5x+12)(x^2+5x-2)=$$
$$=(12z^2+5z+1)(72z^2+30z-1).$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3626979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Show value of $\frac{\tan(x+A)}{\tan(x-A)}$ can not lie between $\tan^2(\pi /4 -A)$ and $\tan^2(\pi /4 +A)$ Let $$y=\frac{\tan(x+A)}{\tan(x-A)}.$$
Show that $y$ can't lie between $\tan^2(\pi /4 -A)$ and $\tan^2(\pi /4 +A)$.
I tried by contradiction. Assumed $$\tan^2\left(\frac{\pi}{4} -A\right)<y$$
and the counterpart'... | Use the short hands $t=\tan x$, $a=\tan A$ to rewrite $y=\frac{\tan(x+A)}{\tan(x-A)}$ as a quadratic equation in $t^2$ as you attempted,
$$(1+y)at^2+(1+a)(1-y)t+a(1+y)=0$$
Considering that the range over which $y$ can't lie corresponds to no real solutions for $t$, which requires that the discriminant is negative, i.e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3629741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$3^n$ does not divide $4^n+5$ for $n\geq 2$ Question as in the title : does anyone know how to prove that $3^n$ does not divide $4^n+5$ for $n\geq 2$ or find a counterexample ?
My thoughts : (1) I have checked that this is true for $n\leq 1000$.
(2) I asked a similar question recently, and it was successfully solved wi... | This follows from an effective abc conjecture.
If $4^n+5=3^nm$ then the quality of this $(a,b,c)$-triple is
\begin{align*}
q(4^n,5,3^nm)&=\frac{\log(3^nm)}{\log(\mathrm{rad}(4^n\cdot5\cdot3^nm))}\geq\frac{\log(4^n+5)}{\log(30m)}=\frac{\log(4^n+5)}{\log(30)+\log(4^n+5)-\log(3^n)}
\end{align*}
which is larger than 2 for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3631619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 0
} |
Prove that sum of matrices equals zero
The matrix $A$ has size $3 \times 3$ and we know that for any column
vector $v\in \mathbb{R}^{3}$ the vectors $Av$ and $v$ are orthogonal. Prove that
$A^{T} + A = 0$, where $A^{T}$ is the transposed matrix $A$.
So if
$$A = \begin{pmatrix}
a_{11} & a_{12} & a_{13}\\
... | It suffices to show $\langle (A^\intercal + A) v, v \rangle = 0$ for all $v$. This is true since
\begin{align} \langle (A^\intercal + A) v, v \rangle &= \langle A^\intercal v, v \rangle + \langle Av, v \rangle \\ &= \langle v, Av \rangle + \langle Av, v \rangle \\ &= 2 \cdot \langle Av, v \rangle \\ &= 2 \cdot 0 \\ &= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3633769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
If $\sec x + \csc x =p$ has four distinct solutions between $(0,2\pi)$, then which if the following is incorrect?
a) $p^2-8>0$
b) $p=\sqrt 2$
c) $p=-\sqrt 2$
d) $p=0$
My attempt
$$\frac{\sec x +\csc x}{2} \ge \sqrt {\sec x \csc x}$$
$$\frac{\sin x +\cos x}{\sin x \cos x }\ge 2\sqrt {\frac{1}{\sin x \cos x}}$$
$$\sin ... | Squaring an equation may introduce unwanted roots, so it is generally avoided. Rewrite the equation, instead, as
$$p(x) = \frac{\sin x +\cos x}{\frac12\sin 2x} =\frac{2\sqrt2 \cos(x-\frac\pi4)}{2\cos^2(x-\frac \pi4)-1}$$
So, $p(x)$ is a function of $\cos(x-\frac\pi4)$ and symmetric around $\frac\pi4$ and $\frac{5\pi}4$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3635138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
If $a+b+c=0$, then $a^3+b^3+c^3$ is ... $0$? $1$? $a^3b^3c^3$? $3abc$? Many mistakes in this post. ( See comments below). I let it as it is, as an example of what shouldn't be done.
If $a+b+c=0$, then $a^3+b^3+c^3 = \ldots $
A. $\;0\quad$ B. $\;1\quad$ C. $\;a^3b^3c^3\quad$ D. $\;3abc$
Source: 4/12/2020, Competitiv... | Normally $$a^3+b^3+c^3 = (a+b+c)^3-3(a+b)(a+c)(b+c)$$
If $a+b+c = 0$
Then $a^3+b^3+c^3 = -3(a+b)(a+c)(b+c)$
Setting $c = -a-b$
$$a^3+b^3+c^3 = -3ab(b+a)$$
$$a^3+b^3+c^3 = 3abc$$
Recheck you expansion of $(a+b)^3$ in the second line of your work, you've replace $3$ with $2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3636449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove by induction $u_{n+2}+u_n=4u_{n+1}$ for all $n\in \mathbb{N}$, where $u_n=(2+\sqrt{3})^n+(2-\sqrt{3})^n$. Prove by induction $u_{n+2}+u_n=4u_{n+1}$ for all $n\in \mathbb{N}$, where $u_n=(2+\sqrt{3})^n+(2-\sqrt{3})^n$.
For $n=1$,
My efforts:
LHS= $u_3+u_1=(2+\sqrt{3})^3+(2-\sqrt{3})^3+ (2+\sqrt{3})^1+(2-\sqrt{3... | To make it a bit easier to deal with, the relation you're asked to prove is true for all $n \in \mathbb{N}$ can be rewritten as
$$u_{n+2} = 4u_{n+1} - u_{n} \tag{1}\label{eq1A}$$
You're given that
$$u_n = (2 + \sqrt{3})^n + (2 - \sqrt{3})^n \tag{2}\label{eq2A}$$
for all $n \in \mathbb{N}$. Using \eqref{eq2A} to define ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3638022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve this optimization problem with equality constraints
Here's what I have tried:
I multiplied the second equation by $\sqrt[3]{abc}$ and get
$$\sqrt[3]{a^5c}+\sqrt[3]{b^5a}+\sqrt[3]{c^5b}=0$$
By manipulating the first expression, I obtain
$$
{1\over a^4b^4c^4}(a^5c+b^5a+c^5b)^2
$$
For simplicity, I attempt t... | Because $x+y+z=0$, $z=-x-y$, and the objective function $f(x,y,z)$ becomes
$$
f(x,y,z)\triangleq\left(x^3+y^3+z^3\over xyz\right)^2=\left[x^3+y^3-(x+y)^3\over xy(-x-y)\right]^2
$$
According to binomial theorem, we have $(x+y)^3=x^3+3x^2y+3xy^2+y^3$, so we can further simplify $f(x,y,z)$ into
$$
\begin{aligned}
f(x,y,z)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3638556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the supremum and infimum of a set of natural numbers
Let $A = \left\{ 2(-1)^{n+1} + (-1)^{ \frac{n(n+1)}{2} } \left( 2 +
\dfrac{3}{n} \right) : n \in \mathbb{N} \right\}.$ Our goal is find $\sup A$ and $\inf
A$.
Attempt
At first glance, it looks like a very formidable set, and I can't see any obvious way bu... | Let
$$F(n) = 2(-1)^{n+1} + (-1)^\frac{n(n+1)}{2}\left(2+\frac{3}{n}\right)$$
For $n \ge 3$ we have $| F(n)| \le 5$, so $-5 \le F(n) \le 5$ when $n \ge 3$.
Since $F(1) = -3$, $F(2)=-5.5$ and $F(3)=5$, we have $\sup A = 5$ and $\inf A = -5.5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3639208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Geometry Problem on $\triangle ABC$ and Angle Chasing
$\triangle ABC$ is an isosceles triangle with $AB=BC$ and $\angle ABD=60^{\circ}$, $\angle DBC=20^{\circ}$ and $\angle DCB=10^{\circ}$. Find $\angle BDA$.
My approach: Let $\angle BDA=x$. Let $AB=BC=p$. Applying sine law in $\triangle ADB$, $\dfrac{p}{\sin x}=\dfr... |
$\angle ABC=\angle ABD+\angle DBC=80^\circ$.
\begin{align*}
AB&=BC\\
\implies \angle CAB&=\angle BCA=(180^\circ-\angle ABC)/2=50^\circ.
\end{align*}
Erect an equilateral triangle $ACE$ on base $AC$.
Then $\triangle$s $ABE, CBE$ are congruent in opposite sense because $AB=CB$, $AE=CE$ and $BE$ is common. Thus $$\angle ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3642725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
Evaluating $ 4\sum^{30}_{n=1} n\;T(n)$, where $T(n) = \cos^2(30^\circ -n) - \cos(30^\circ -n)\cos(30^\circ +n) +\cos^2(30^\circ +n)$
For $n$ measured in degrees, let
$$T(n) = \cos^2(30^\circ -n) - \cos(30^\circ -n)\cos(30^\circ +n) +\cos^2(30^\circ +n)$$
Evaluate $$ 4\sum^{30}_{n=1} n \cdot T(n)$$
I have tried t... | Since
$$\cos^2(30^\circ-n)-\cos(30^\circ-n)\cos(30^\circ+n)+\cos^2(30^\circ+n)=\cos(30^\circ-n)[\cos(30^\circ-n)-\cos(30^\circ+n)] +\cos^2(30^\circ+n)=(\frac{\sqrt{3}}{2}\cos n +\frac{1}{2}\sin n )2\sin30°\sin n +(\frac{\sqrt{3}}{2}\cos n -\frac{1}{2}\sin n )^2=
\frac{3}{4}$$
$$ 4\sum^{30}_{n=1} n \cdot T(n)=4\frac{3}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3643516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the indicial equation for $r_1$ and $r_2$ The given expression is:
$$x^2y^{''}+xy^{'}+(x^2-\frac{1}{4})y=0$$
First we need to convert the given into its indicial equation:
$$x^2\sum^\infty_{n=0}(n+r)(n+r-1)C_nx^{n+r-2}+x\sum^\infty_{n=0}(n+r)C_nx^{n+r-1}+x^2\sum^\infty_{n=0}C_nx^{n+r}-\frac{1}{4}\sum^\infty_{n=0}C... | $$\sum^\infty_{n=0}\left[(n+r)(n+r-1)+(n+r)-\frac{1}{4}\right]C_nx^{n+r}+\sum^\infty_{n=0}C_nx^{n+r+2}=0$$
$$\sum^\infty_{n=0}\left[(n+r)^2-\frac{1}{4}\right]C_nx^{n+r}+\sum^\infty_{n=2}C_{n-2}x^{n+r}=0$$
Changing the indices:
$$ \color {blue} {C_0x^r(r^2-\frac 14)}+C_1x^{r+1}\left(r^2+2r+\frac 34 \right)+\sum^\infty_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3644966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Why does $\sum_{i,j}^\infty \frac{1}{i(i^2+j^2)}$ converge? According to Mathematica the series $\sum_{i,j=1}^\infty \frac{1}{i(i^2+j^2)}$ converges. Is there a simple argument to obtain the convergence? Maybe a dominating convergent series?
| By the integral test
$$\sum_{i=1}^\infty \frac{1}{i^3+ij^2}$$
is convergent and
$$\sum_{i=1}^\infty \frac{1}{i^3+ij^2} \leq \frac{1}{1+j^2}+ \int_{x=1}^\infty \frac{1}{x^3+xj^2} dx = \frac{1}{1+j^2}+ \int_{x=1}^\infty \frac{x}{x^4+x^2j^2} dx \\
=\frac{1}{1+j^2}+ \frac{1}{2}\int_{u=1}^\infty \frac{1}{u(u+j^2)} du \\ =\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3650867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solve $x^4y^{\prime\prime} = (y-xy^\prime)^3, y(1) = y^\prime(1) = 1$ $\lambda^4x^4\lambda^{n-2}y^{\prime\prime} = (\lambda^ny-\lambda x\lambda^{n-1}y^\prime)^3 \Rightarrow \lambda^{n+2}x^4y^{\prime\prime} = (\lambda^n(y-xy^\prime))^3$.
$\lambda^{n+2} = \lambda^{3n} \Rightarrow n+2 = 3n \Rightarrow n = 1$.
Let $x = e^... | Taking $y = u x$, the equation becomes
$$ u'' = - \dfrac{u' ((u')^2 x^2 + 2)}{x} $$
with $u(1) = 1$, $u'(1) = 0$.
If $u' = v$, this is
$$ v' = - \dfrac{v (v^2 x^2 + 2)}{x},\ v(1) = 0$$
Now we could go on to solve this differential equation in general, but for this specific initial condition it's obvious that the solut... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove $ \frac1{\sqrt{(z+a)(z+b)}}=\frac1\pi \int_a^b \frac1{\sqrt{(t-a)(b-t)}}\frac1{t+z}dt $ Reading papers online i found the following definition
$$ \frac1{\sqrt{(z+a)(z+b)}}=\frac1\pi \int_a^b \frac1{\sqrt{(t-a)(b-t)}}\frac1{t+z}dt $$
for $$ b>a \,\, and \,\,\,z \in \Bbb C |(-\infty,-a] $$
Question : How can this ... | Let $t = \frac{a+b}{2}+\frac{b-a}{2}\sin\theta$. Then
$$\pi I(z) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{d\theta}{\frac{a+b}{2}+\frac{b-a}{2}\sin\theta+z} = \int_0^\pi \frac{d\theta}{\frac{a+b}{2}+\frac{b-a}{2}\cos\theta+z}$$
Then denoting $\frac{a+b}{2}+z \equiv k_1$ and $\frac{b-a}{2} \equiv k_2$ we get that
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3654645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Calculating matrix determinant This is a very easy determinant to calculate, but I get two different results when I calculate it in two different ways.
\begin{equation}
A = \begin{bmatrix} 1 & 2 & 3 \\
0 & 1 & 2 \\
2 & 2 & 4
\end{bmatrix}
\end{equation}
When I used Laplace expansion right away I got:
\begin{equation}
\... | $r_1 \leftarrow 2r_1 -r_3$ is not allowed.
If you want to do row addition it is:
$r_1 \leftarrow r_1+\lambda r_2$ or $r_1 \leftarrow r_1+\lambda r_3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3656372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove and give geometric meaning of $|\sqrt{a^2+b^2} - \sqrt{a^2+c^2} | < |b-c| $
Carefully, prove that
$$ | \sqrt{a^2+b^2} - \sqrt{a^2+c^2} | \leq |b-c| $$
and give a geometric interpretation.
pf
We observe that $| \sqrt{a^2+b^2} - \sqrt{a^2+c^2} | = \dfrac{ |b^2 - c^2 | }{\sqrt{a^2+b^2} + \sqrt{b^2+c^2} } \leq \dfr... | Let $B(a,b)$ and $C(a,c)$ be the two vecrors. Then,
$$| |B|-|C||<| B-C| = |BC|$$
which translates To
$$| \sqrt{a^2+b^2} - \sqrt{a^2+c^2} | \leq |b-c| $$
Geometrically, the above inequality means the length difference of two triangle sides is less than the length of the third side.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3657100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Is there formula that solves quartic equation $ax^4+bx+c=0$ In general form, a quartic equation is $ax^4+bx^3+cx^2+dx+e=0$. I was thinking of the quartic equation of the form
$$ax^4+bx+c=0$$
which resembles a depressed cubic equation.
Does anyone know a formula that solves this type of depressed quartic equation, siml... | The quartic equation of the form
$$ax^4+bx+c=0$$
has the following formulas for its four roots
$$x= -s \pm \sqrt{ \frac b{4as}-s^2},\>\>\> s \pm \sqrt{ -\frac b{4as}-s^2} $$
where $s^2$ satisfies the cubic equation $s^6 -\frac c{4a} s^2-\left(\frac b{8a}\right)^2=0$. Take the example of
$$x^4+2x+\frac12=0$$
for which $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3658305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Homework Problem, Power Series Limit I am looking to find the solution for:
$$\lim_{x\rightarrow 0} {5^{tan^2(x) + 1} - 5 \over 1 - cos^2(x)}$$
A hint was provided:
transform: ${5^{tan^2(x) + 1} - 5 \over 1 - cos^2(x)}$ to $5y{5^{y-1} - 1 \over y-1}, y = {1\over cos^s(x)} \rightarrow 1$
The transformation is straigt fo... | I’m not sure you’re doing the differentiation right. We have
$$\lim_{y\to 1} \ 5y \ \frac{5^{y-1} -1}{y-1} $$
$$=5(1)\cdot \ \lim_{y\to 1} \frac{(5^{y-1} -1)’}{(y-1)’}$$
$$= 5\cdot\lim_{y\to 1} \frac{5^{y-1} \cdot \ln 5}{1}$$
$$=5\ln 5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3659907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Maclaurin series expansion for $\dfrac{1}{2t}\left(1+t-\dfrac{(1-t)^2}{2 \sqrt {t}}\log \left(\dfrac{1+\sqrt{t}}{1-\sqrt{t}}\right)\right)$
Show that the expansion of
$$\dfrac{1}{2t}\left(1+t-\dfrac{(1-t)^2}{2 \sqrt{t}} \log \left(\dfrac{1+\sqrt{t}}{1-\sqrt{t}}\right)\right)$$
as a Maclaurin series in power of $t$ to ... | We have $\frac12\log\frac{1+z}{1-z}=\sum_{n=0}^\infty\frac{z^{2n+1}}{2n+1}$, hence $\frac{1}{2\sqrt{t}}\log\frac{1+\sqrt{t}}{1-\sqrt{t}}=\sum_{n=0}^{\infty}\frac{t^n}{2n+1}$ and $$\frac{(1-t)^2}{2\sqrt{t}}\log\frac{1+\sqrt{t}}{1-\sqrt{t}}=(1-2t+t^2)\sum_{n=0}^{\infty}\frac{t^n}{2n+1}\\=1+\left(\frac13-2\right)t+\sum_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3662808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\prod_{k=1}^{\infty}\frac{{5^\frac {1}{2}}^k+{3^\frac{1}{2}}^k}{2}$ While doing questions on series and products, I tried hard to solve this infinite product, but no methods has worked so far.Well, this is from Johns Hopkins math tournament
Question:- Evaluate $\prod_{k=1}^{\infty}\frac{{5^\frac {1}{2}}^k+{3^... | Let $\alpha = 5, \beta = 3$. Recall the identity
$$x + y = \frac{x^2 - y^2}{x-y}$$
The product you have is a telescoping one.
$$\prod_{k=1}^N \frac{\alpha^{2^{-k}} + \beta^{2^{-k}}}{2}
= \frac{1}{2^N} \prod_{k=1}^N
\frac{\alpha^{2^{1-k}} - \beta^{2^{1-k}}}{\alpha^{2^{-k}} - \beta^{2^{-k}}}
= \frac{\alpha - \beta}{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3663757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Binomial theorem and multinomial coefficient I have a question to which I could not find an answer in the forum.
I am trying to solve a bracelet problem. Actually I already solved it but with some kind of a cheating (using wolframaplha to expand the Dihedral group formula). The problem that I try to solve is: how many... | For the first question, note that for example in $(x^2+y^2+z^2)^6$, each term will be of the form $kx^{2a}y^{2b}z^{2c}$ for some integers $k,a,b,c$. So it's not possible to have an exponent of $3$ or $5$. A similar consideration eliminates every term but the two you have mentioned.
For multinomial coefficients in the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3666805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Combinations of elements from multiple sets with constraints Given four sets of numbers:
$$A=\{1,2,3,4\}$$
$$B=\{1,2,3,4\}$$
$$C=\{1,2,3,5\}$$
$$D=\{1,2,6\}$$
I need to find number of all possible combinations of numbers from these sets in following format:
$$(a,b,c,d)$$
where $a\in A, b \in B, c \in C, d \in D$ and wh... | We calculate the wanted number with the help of rook polynomials. We consider a $(6\times4)$-rectangular board, where the columns represent the sets $A,B,C,D$ and the rows the elements of the sets. A valid quadruple $(a,b,c,d)\in A\times B\times C\times D$ can be represented by placing four non-attacking rooks on the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3666963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Evaluate logarithmic integral $\int_{-\infty}^{\infty}\frac{\ln{(x^2+a^2)}}{x^2+b^2}\,dx$ I need help to evaluate the integral with the residue theorem:
$$\int_{-\infty}^{\infty}\frac{\ln{(x^2+a^2)}}{x^2+b^2}\,dx$$
where a,b>0 real numbers.
I think I could consider the contour integral where C is the half circle in the... | Let $I(a)=2\int_{0}^{\infty}\frac{\ln{(x^2+a^2)}}{x^2+b^2}\,dx$ and evaluate
$$I’(a) =\int_{0}^{\infty}\frac{4adx}{(x^2+a^2)(x^2+b^2)}=
\frac{2\pi}{b(a+b)}$$
Then,
\begin{align}
\int_{-\infty}^{\infty}\frac{\ln{(x^2+a^2)}}{x^2+b^2}\,dx
& = I(a)=I(0)+ \int_{0}^{a}I’(t)dt\\
&=2 \int_{0}^{\infty} \frac{\ln{x^2}}{x^2+b^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3671716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Wirtinger's inequality variation
If $f \in C^1[0,1]$ with $f'(0) = f(1) = 0$, then$$\|f\|_2\leq\frac2\pi\|f'\|_2.$$
Elaboration:
Assume the Sturm-Liouville operator $A: D \longrightarrow L^2(0,1)$ where the domain is
$$
D = \{f \in C^1[0,1]: f'' \in L^2(0,1), f'(0) = f(1) =0\}
$$
and
$$
Af(x) = f''(x)-\lambda f(x), ... | $\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}{}}$Note that after changing the domain $D$ to$$
D_1 = \{f \in C^1([0, 1]) \mid f'' \in L^2([0, 1]),\ f'(1) = f(0) = 0\},
$$
the Sturm-Liouville theorem implies that $\{ψ_n(x) \mid n \in \mathbb{N}_+\}$ is also an orthonormal basis of $C^1([0, 1])$, where $ψ_n(x) = \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3677796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Using a solution to show no others are possible in positive odd integer If I have an equation say $$3(1+x+x^2)(1+y+y^2)(1+z+z^2)+1=4x^2y^2z^2 \quad (1)$$ and I know a non negative integer solution $x=4, y=64, z=262144$ then no odd positive integer solutions can possibly exist. I know that one way to show it for this ex... | The given equation is $3(1+x+x^2)(1+y+y^2)(1+z+z^2)=4x^2y^2z^2-1$.
Let $x \leq y \leq z$. Then $x=1$ has no solution as in this case the LHS is at least $9y^2z^2>4y^2z^2$.
Thus, let $x \geq 3$.
Also note that $3$ divides the LHS so $3$ does not divide $xyz$; hence $x \geq 5$.
Divide the original equation by $x^2y^2z^2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3677979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find min,max value of $P=2x-y$ Let $x,y$ such that $\sqrt{\left(x+2\right)^2+y^2}+\sqrt{\left(x-2\right)^2+y^2}=6$. Find minimize, maximize value of $$P=2x-y$$
We have :
$$y=2x-P$$ so from condition:
$$\sqrt{\left(x+2\right)^2+\left(2x-P\right)^2}+\sqrt{\left(x-2\right)^2+\left(2x-P\right)^2}=6$$
i solved if $x=\frac ... | The condition it's the equation of the ellipse with focuses $(2,0)$ and $(-2,0)$ and $a=3,$
which gives $b^2=5$ and an equation of the ellipse:
$$\frac{x^2}{9}+\frac{y^2}{5}=1.$$
Also, easy to get this equation by squaring twice.
Now, let $2x-y=k$.
Thus, $$y=2x-k$$ and the equation $$\frac{x^2}{9}+\frac{(2x-k)^2}{5}=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3678662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Number of straight lines can be drawn
How many different straight lines can be drawn using $9$ points on the triangle of the figure below?
My try: Considering points other than $A$, $B$ and $C$ we got $2\times3 + 2\times1 + 3\times1 = 11$. For those three points, number of lines are $1 \times1 + 2 \times 1 + 3\times... | We have $9$ points and a straight line is determined by two distinct points.
We can use the binomial coefficients.
In how many ways can we choose pairs of points?
$$\binom{9}{2}=\frac{9\cdot 8}{2}=36$$
But, $A,D,E, B$ are collinear as well as $A,F,G,H,C$ and $B,I,C$.
So, we have $\binom{4}{2}+\binom{5}{2}+\binom{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3681201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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What is the integral of $\sqrt{\tan x}$ I just need to know why is my method wrong:
Let $I=\int\sqrt{\tan x} dx$, let $\tan(x)=t^2$ then, \begin{align*}&\sec^2(x)dx=2tdt\\
\implies&(1+\tan^2(x))dx=2tdt\\
\implies &dx=2tdt/(1+t^2)\end{align*}
So \begin{align*}I &= \int t\cdot \frac{2t}{1+t^2} dt\\
&= \int\frac{2t^2}{1... | Let's flesh out @QuantumApple's hint. With $\tan x=t^2$,$$\begin{align}\int\sqrt{\tan x}dx&=\int\frac{2t^2dt}{1+t^4}\\&=\int\frac{1}{2}\left(\tfrac{1}{1-\sqrt{2}t+t^{2}}+\tfrac{1}{1+\sqrt{2}t+t^{2}}-\tfrac{1}{\sqrt{2}}\left(\tfrac{\sqrt{2}-2t}{1-\sqrt{2}t+t^{2}}+\tfrac{\sqrt{2}+2t}{1+\sqrt{2}t+t^{2}}\right)\right)dt\\&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3683735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Express $ \operatorname{gcd}\left(5^{m}+7^{m}, 5^{n}+7^{n}\right) $ in terms of $m$ and $n$
Let $m$ be a positive integer with $\operatorname{gcd}(m, n)=1 .$ Express
$
\operatorname{gcd}\left(5^{m}+7^{m}, 5^{n}+7^{n}\right)
$
in terms of $m$ and $n$
My work -
let $d=\operatorname{gcd}(5^m +7^m,5^n +7^n)$ then
$5... | For $k$ odd we have that
$$5^k+7^k\equiv 0\mod 3$$
and
$$5^k+7^k\equiv 5+7\equiv 4\mod 8$$
For $k$ even we have that
$$5^k+7^k\equiv 2\mod 3$$
and
$$5^k+7^k\equiv 1+1\equiv 2\mod 8$$
From here we have that if $m,n$ are both odd(i.e. if $m+n$ is even), then $\gcd(5^m+7^m,5^n+7^n)=12$.
Otherwise, if between $m$ and $n$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3687612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Find the point where L intersects the curve The curve $x^2 - 2y^2 = 1$ includes the point $(1, 0)$. Let $L$ be the line through $(1, 0)$ having slope $m$. Find the other point where $L$ intersects the curve.
The line can be written as $y=m(x-1)$. Substituting this value of y in the original curve gives us:
$x^2(1-2m^2... | Note the equation should actually be $$x^2(1-2m^2) +4m^2x \color{red} - 2m^2 -1=0$$ Performing long division by $(x-1)$ yields $$(x-1) \left( (1-2m^2)x +(2m^2+1) \right) =0 \\ \implies x=1, \frac{2m^2+1}{2m^2-1}$$ and the required point is $$\left ( \frac{2m^2+1}{2m^2-1} , \frac{2m}{2m^2-1} \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3691404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Can this function be defined in a way to make it continuous at $x=0$? We have $$f=\frac{x}{\vert x-1 \vert - \vert x +1 \vert}$$
If we want to "define" this function to be continuous at $x=0$, it's limit at $0$ must equal $f(0)$. So we should find this limit and assign it to be equal to $f(0)$, then the function is con... | There is an error at the end of your calculations. In the case $-1 < x < 0$ we have
$$
\left\vert 1 - \frac{1}{x} \right\vert-\left\vert 1+\frac{1}{x}\right\vert=
\left(1-\frac{1}{x}\right)+\left(1+\frac{1}{x}\right)=2
$$
because the argument of the first absolute value is positive, and the argument of the second absol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3700549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Let chord of contact be drawn from every point on the circle $x^2+y^2=100$ to the ellipse [CONT..]
Let chord of contact be drawn from every point on the circle $x^2+y^2=100$ to the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ such that all lines touch a standard ellipse. Find $e$ for the ellipse
Let the point $(h,k)$ lie ... | What you are looking for is the curve that the chord of contact envelops. You can find (one half of) the solution ellipse by letting $k=\sqrt{100-h^2}$ and substituting into the chord of contact $\frac{hx}{4}+\frac{ky}{9}-1=0.$ The curve the family of lines $\frac{hx}{4}+\frac{\sqrt{100-h^2}y}{9}-1=0$ envelops can be ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3700829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Closed form of $\int_0^\infty \arctan^2 \left (\frac{2x}{1 + x^2} \right ) \, dx$
Can a closed form solution for the following integral be found:
$$\int_0^\infty \arctan^2 \left (\frac{2x}{1 + x^2} \right ) \, dx\,?$$
I have tried all the standard tricks such as integration by parts, various substitutions, and parame... | Here is another solution with a generalization:
Let $r=\sinh\alpha$ and $s=\sinh\beta$. Then
$$\begin{aligned}
&\int_{0}^{\infty} \arctan\left(\frac{2rx}{1+x^2}\right)\arctan\left(\frac{2sx}{1+x^2}\right) \, \mathrm{d}x\\
&= \pi \left( \alpha \sinh\beta+\beta\sinh\alpha+(\cosh\alpha+\cosh\beta)\log\left(\frac{e^{\al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
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Find angles of triangle $DEC$ $ABC$ is an equaliterial triangle. $AC∥MN$ and $M$ and $N$ lies at $AB$ and $BC$ respectively. $D$ is centroid of $MBN$, $E$ is a midpoint of $AN$. Then, find the angles of the triangle $DEC$.
First of all I am sure that it can be solved by rotation. Unfortunately, I haven't managed to pro... | Another way to solve it is by analytical geometry. Considering the figure of the preceding answer, setting the vertices $B(0,0)$, $C(1,0)$, and $A(1/2,\sqrt{3}/2)$, and placing the point $N$ in $(k,0)$ (with $0 \leq k \leq 1$), we directy get that:
*
*the coordinates of $D$ are $\Big(k/2,k/(2 \sqrt{3})\Big)$;
*the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3709130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Why does simplifying $\arcsin(2x\sqrt{1-x^2})$ two ways give different results?
Find derivative of
$y=\arcsin(2x \sqrt{1-x^2}) $ in domain $\frac{-1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$
If you put $x=\sin\theta$ then
$$ y= \arcsin(2sin(\theta) \sqrt{1- sin^2 ( \theta)})$$
$$y= \arcsin( sin2\theta)$$
$$y= 2\theta$$
... | There is nothing wrong with your algebra. It’s just that you need to take extra care when dealing with the domain. Note that $\arcsin(\sin \theta=\theta$ only when $-\frac{\pi}{2} \le \theta\le \frac{\pi}{2}$.
Substituting $x=\sin\theta$ gives $$y= \arcsin(2\sin\theta|\cos\theta|)=\arcsin|\sin 2\theta| $$ Now, $$-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3709296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Let $a, b, c, d \in R^+$ such that $a + b + c + d = 1$. Prove that $\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} \geq \frac{1}{8}$ Let $a, b, c, d \in R^+$ such that $a + b + c + d = 1$. Prove that,
$$\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} \geq \frac{1}{8}$$
Well from their s... | No. Your proof is not correct.
After your first step we need to prove a wrong for $a\rightarrow0^+$ inequality.
I think it's better to use Holder: $$\sum_{cyc}\frac{a^3}{b+c}\geq\frac{(a+b+c+d)^3}{4\sum\limits_{cyc}(b+c)}=\frac{1}{8}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3709447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate: $\int \frac{t^5-t^3}{t^{10}-1}\, dt. $ None of my attempts (such as letting $x=t^5$ or splitting up the integrand into two parts etc.) succeeded. Basically I couldn't find a useful thing on manipulating the given integrand. Please give me some hints.. Thanks in advance.
| And I'll finish the thing. $I=\int\frac{t^5-t^3}{t^{10}-1}\,\mathrm{d}t$
Let $u=t^2,du=2t\,dt$, by J.G.'s answer
$$I=\frac12\int\frac{u\,\mathrm{d}u}{u^4+u^3+u^2+u+1}$$
Using $$u^4+u^3+u^2+u+1=\left(u^2+\frac{1+\sqrt{5}}{2}u+1\right)
\left(u^2+\frac{1-\sqrt{5}}{2}u+1\right)$$ both parenthesis have no real roots, so we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3709902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Find all functions $f : \mathbb R \to \mathbb R$ such that: $f\left(x^3\right)+f\left(y^3\right) = (x + y)f\left(x^2\right)+f\left(y^2\right)-f(xy)$ Here is the question:
Find all functions $f : \mathbb R \to \mathbb R$ such that:
$$f\left(x^3\right)+f\left(y^3\right) = (x + y)f\left(x^2\right)+f\left(y^2\right)-f(xy)... | Here's how to get $f(x^2) = f(y^2)$.
If $x+y-1 \neq 0$, we must have $f(x^2) = f(y^2)$. F any $x^2$ and $y^2$, pick a $z$ such that $x+z-1$ and $y+z-1$ are nonzero, which is always possible by taking $z \neq 1-x, 1-y$. Then $f(x^2) = f(z^2) = f(y^2)$. So $f(x^2) = f(y^2)$ for all reals $x, y$, and hence $f$ is constant... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3712136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute a matrix $P$ such that $P$ satisfies $3I+P+P^2=\left(\begin{smallmatrix}3&0&0\\3&6&0\\0&0&6\end{smallmatrix}\right)$ Question: Compute a matrix $P$ such that the matrix $P$ satisfies $3I + P + P^2 = \begin{pmatrix} 3 & 0 & 0\\ 3 & 6 & 0 \\ 0 & 0 & 6 \end{pmatrix}$
Attempt: First, I simplify the equation and ge... | This is a bit of a brute-force diagonalisation approach - I do feel as though there should be a simpler method, but this at least gives you an answer.
We can reduce this to solving an equation of the form $X^2=C$ by completing the square, as we would do in the real case. Note that (dropping the now-implicit identity m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3712724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Can any cyclic polynomial in $a, b, c$ be expressed in terms of $a^2b+b^2c+c^2a$, $a+b+c$, $ab+bc+ca$ and $abc$? Problem. Let $f(a,b,c)$ be a cyclic polynomial in $a, b, c$.
Can $f$ always be expressed as $g(a+b+c, ab+bc+ca, abc, a^2b+b^2c + c^2a)$ for some polynomial $g(p, q, r, Q)$?
Motivation: If we want to prove $... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.
Thus, $$\sum_{cyc}a^2b=\frac{1}{2}\sum_{cyc}2a^2b=\frac{1}{2}\sum_{cyc}(a^2b+a^2c-(a^2c-a^2b))=$$
$$=\frac{1}{2}(9uv^2-3w^3)+\frac{1}{2}(a-b)(a-c)(b-c).$$
Now, for $a\geq b\geq c$ we obtain:
$$\sum_{cyc}a^2b=\frac{1}{2}(9uv^2-3w^3)+\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate: $\lim_{n\to\infty} n^3\int_n^{2n} \frac x{1+x^5}\, dx. $ Is there a way to evaluate: $$\lim_{n\to\infty}n^3\int_n^{2n}\frac x{1+x^5}\, dx,$$ without evaluating the indefinite integral: $$\int \frac x{1+x^5}\, dx\;?$$ Please suggest. Thanks in advance.
| Under $x^{-3}\to x$,
$$ \frac{1}{n^3}\int_{n}^{2n}\frac{x}{1+x^5}dx=\frac{n^3}{3}\int_{\frac1{8n^3}}^{\frac1{n^3}}\frac{dx}{1+x^{\frac{5}{3}}}. $$
By the MVT for integrals, there is $c_n\in (\frac1{8n^3},\frac1{n^3})$ such that
$$ \frac{1}{n^3}\int_{n}^{2n}\frac{x}{1+x^5}dx=\frac1{n^3}\bigg(\frac1{8n^3}-\frac1{n^3}\big... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine all pairs of positive integers $(a,b)$ such that $a^2+b^2+ab$ is a perfect square. Consideration via mod $4$ shows that $a,b=0$ mod $4$ or one of between $a,b$ is $=1$ mod $4$ while the other is $=0 $ mod $4$.
Considering $(a+b)^2,a^2+an+b^2,(a+b-1)^2$ as we can deduce that $a,b>2$.
| Just another way to approach the problem.
Since $$a^2+b^2+ab=(a+b)^2-ab $$ is a squared integer, there exists an integer $n$ such that
$$ (a+b)^2-(a+b-n)^2=ab$$
Solving it for $a$, we have
$$a = \frac{n (2 b - n)}{b - 2 n} $$
with the immediate condition $b \neq 2 n$. However, we also have to set two other conditions.... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Permutations of numbers with difference 1 that are lesser than h How many are the permutations of $n$ numbers where the first and last number are $1$, the difference between the numbers is $1$ and the numbers are lesser than h?
For example, for $n = 5$ and $h = 6$ $(h > 4)$ the permutations are $9$:
$ 1-2-3-2-1 $
$ 1-2... | These "permutations" are related to Motzkin paths. Compare this picture with your example.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3716697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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proving some interesting properties of these matrices let X and Y be two matrices different from I, such that $XY=YX$ and $X^n-Y^n$ is invertible for some natural number n .If
$$X^n-Y^n = X^{n+1}-Y^{n+1} = X^{n+2}-Y^{n+2}$$,
then prove that $I-X,I-Y $ are singular and $X+Y=XY+I$
my approach:
I tried to pre multiply and... | From $X^n-Y^n=X^{n+1}-Y^{n+1}=X^{n+2}-Y^{n+2}$, we obtain
\begin{align}
X^n(I-X) &= Y^n(I-Y),\tag{1}\\
X^{n+1}(I-X) &= Y^{n+1}(I-Y).\tag{2}
\end{align}
Subtract both sides of $(1)$ by $X^n(I-Y)$, we obtain
\begin{align}
(Y-X)X^n &= (Y^n-X^n)(I-Y).\tag{3}
\end{align}
Substitute the LHS of (1) into the RHS of (2), we get... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3717102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Shuffling the digits of an integer so that the ratio between the resulting numbers is fixed. Suppose we have a 3-digit integer with all digits distinct. If we create its two shift numbers (resulting by rotation of the original digits), the ratio between the two subsequent numbers (if we order them in ascending or desce... | We want to find triples of integers $(x,y,z)$ such that
$$0\lt x\le 9, 0\le y\le 9,0\le z\le 9\tag1$$
$$x\not=y,y\not=z,z\not=x\tag2$$
$$\frac{100z+10x+y}{100y+10z+x}=\frac{100y+10z+x}{100x+10y+z}\tag3$$
From $(3)$, we get
$$ yz-x^2=10(xz-y^2)\implies yz\equiv x^2\pmod{10}\tag4$$
and
$$x=-5z+\sqrt{25z^2+10y^2+yz}\impli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3717744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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If $f(x)=2|2x-5|+3$, with $x\geq0$, find all $k$ such that $f(x) = kx + 2$ has exactly two roots
If $f(x) = 2|2x-5| + 3 $, $x\geq0$, find the values of $k$ such that the equation $f(x) = kx + 2$ has exactly two roots.
My attempt :-
$2(2x -5)+3 = kx + 2$ ,
$x \geq 2.5$
$x =\frac{9}{4-k}$
Then
$\frac{9}{4-k} \geq \frac... | The set of lines $y = kx + 2$ are the non-vertical lines passing through the point $(0,2)$, while the graph of $y = 2|2x - 5| + 3 = 4|x - {5 \over 2}| + 3$ consists of two rays containing $({5 \over 2}, 3)$, one going upwards with slope $4$ and the other going upwards with slope $-4$.
In order for a line containing $(0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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$\frac{dy}{dx} - {8} -{2}x^2+{4}y^2+y^2x^2 = 0.$ how should I procced from here having the equation $$\frac{dy}{dx} - {8} -{2}x^2+{4}y^2+y^2x^2 = 0.$$
I am getting to the following $$\frac{1}{2^{\frac{3}{2}}}\ln \left(y+\sqrt{2}\right)-\frac{1}{2^{\frac{3}{2}}}\ln \left(y-\sqrt{2}\right)=\frac{x^3}{3}+4x+c$$ from here ... | $$\frac{dy}{dx} = 8+2x^2-4y^2-y^2x^2 = (4+x^2)(2-y^2)$$
$$\frac{dy}{2-y^2}=(4+x^2)dx \iff 2^{-3/2}\ln \Big(\Big \vert\frac{y+\sqrt{2}}{y-\sqrt{2}}\Big \vert\Big) = 4x+\frac{1}{3}x^3+C$$
And this is where you got to. Now we can multiply by $2^{3/2}$ and take $\exp$ of both sides to get
$$\Big \vert \frac{y+\sqrt{2}}{y-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Base conversion problems Let $b$ be an integer greater than 2, and let $N_b = 1_b + 2_b + \cdots + 100_b$ (the sum contains all valid base $b$ numbers up to $100_b$). Compute the number of values of $b$ for which the sum of the squares of the base $b$ digits of $N_b$ is at most 512.
Since we are adding up to $100$ in b... | We have
$$
2N = b^2(b^2+1)=b^4+b^2 = 10100_b
$$
You need to show that
$$
10100_b =
\left\lbrace\begin{array}{c}
2 \cdot \left[ \tfrac{b}{2},0,\tfrac{b}{2},0 \right]_b
&\text{for even $b$}\\
2 \cdot \left[ \tfrac{b-1}{2},\tfrac{b+1}{2},0,0 \right]_b
&\text{for odd $b$}\\
\end{array}\right.
$$
where $[d_3,d_2,d_1,d_0]_b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3722225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Distribution of $\frac{2X_1 - X_2-X_3}{\sqrt{(X_1+X_2+X_3)^2 +\frac{3}{2} (X_2-X_3)^2}}$ when $X_1,X_2,X_3\sim N(0,\sigma^2)$
Given that $X_1, X_2, X_3 $ are independent random variables form $N(0, \sigma^2 )$, I have to indicate that the statistic given below has a $t$ distribution or not.
\begin{equation}
\frac{2X_1... | Consider the orthogonal transformation
$$\begin{pmatrix}Y_1 \\ Y_2 \\ Y_3\end{pmatrix}=\begin{pmatrix}\frac{2}{\sqrt 6} &-\frac1{\sqrt 6} & -\frac1{\sqrt 6} \\ \frac1{\sqrt 3} & \frac1{\sqrt 3} & \frac1{\sqrt 3}\\ 0 & \frac1{\sqrt 2} & -\frac1{\sqrt 2} \end{pmatrix}\begin{pmatrix}X_1 \\ X_2 \\ X_3\end{pmatrix}$$
So if... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3722577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Integrate $\int_0^2 \frac{\ln\left(1+x\right)}{x^2-x+1} \mathop{dx}$ Challenge problem $$\int_0^2 \frac{\ln\left(1+x\right)}{x^2-x+1} \mathop{dx}$$
First thought $u=1+x$, $$ \int_1^3 \frac{\ln{(u)}}{u^2-3u+3} \mathop{du}$$
Here complex analysis or what?
Tips please.
| Via the substitution $u=\sqrt{3}v$ we
get
\begin{equation*}
I=\int_{1}^{3}\dfrac{\ln(u)}{u^2-3u+3}\,\mathrm{d}u =I_1+I_2
\end{equation*}
where
\begin{equation*}
I_1=\dfrac{1}{\sqrt{3}}\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\dfrac{\ln(\sqrt{3})}{v^2-\sqrt{3}v+1}\,\mathrm{d}v =\left[\dfrac{\ln 3}{\sqrt{3}}\arctan\left(2v-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3730856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Arrange irrationals in ascending order:$ 2^{\sqrt{\frac{5}{3}}},3^{\sqrt{\frac{3}{5}}},5^{\sqrt{\frac{4}{15}}},29^{\frac{1}{\sqrt{15}}} $ How to arrange following irrational numbers in ascending order:
$$ 2^{\sqrt{\frac{5}{3}}}, 3^{\sqrt{\frac{3}{5}}}, 5^{\sqrt{\frac{4}{15}}}, 29^{\frac{1}{\sqrt{15}}} $$
My try:
Using ... | Equate the powers by taking LCM of denominators and compare the base-numbers as follows
$$ 2^{\sqrt{\dfrac{5}{3}}}, 3^{\sqrt{\dfrac{3}{5}}}, 5^{\sqrt{\dfrac{4}{15}}}, 29^{\dfrac{1}{\sqrt{15}}} $$
$$ 2^{\dfrac{5}{\sqrt{15}}}, 3^{\dfrac{3}{\sqrt{15}}}, 5^{\dfrac{2}{\sqrt{15}}}, 29^{\dfrac{1}{\sqrt{15}}} $$
$$ (2^5)^{\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Easier approach to $\int_0^{\infty} \frac{\mathrm{e}^{-x} \cosh(2x/5)}{1 + \mathrm{e}^{-2x}} \, \mathrm{d}x$? After playing with this integral for a little bit last night, I eventually resorted to complex analysis to solve it.
Can this be solved without complex analysis? It feels like there ought to be a way. If not,... | Starting from @Luis Sierra's answer
$$\begin{equation}
I=\frac{1}{2}\int\limits_{0}^{1} \frac{t^{\frac{2}{5}}}{1+t^{2}} \,dt +\frac{1}{2}\int\limits_{0}^{1} \frac{t^{-\frac{2}{5}}}{1+t^{2}}\,dt
\end{equation}$$
Using the quite standard
$$J_a=\int_0^1 \frac {t^a}{1+t^2}\, dt=\frac{1}{4} \left(\psi \left(\frac{a+3}{4}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Show that what is the graph of each one of these equations. Given the following equations:
$$1)\;\;x^{2}+2y^{2}+z^{2}-2x+4z-22=0$$
$$2)\;\;5x^{2}+6y^{2}+4z-4x=14$$
$$3)-x^{2}+y^{2}-z^{2}-2x+2z=0$$
$$4) x=z^2$$
Show that what is the graph of each one of these equations.
$$1)$$
$$x^{2}+2y^{2}+z^{2}-2x+4z-22=0$$
$$\frac{... |
Equation $1$. Ellipsoid.
The skirt is due to the real hack in the code to deal with complex values.
It shouldn't be there.
Equation $2$. Dome.
Equation $3$. Hyperboloid of one sheet. Note the y axis is vertical.
Octave:
figure 1;
tx = ty = [-5:0.1:5]';
[xx, yy] = meshgrid (tx, ty);
zplus = -2 + real(sqrt(27 - (xx-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3737706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What Is Bigger $100^{100}$or $\sqrt{99^{99} \cdot 101^{101}}$ Hello every what is bigger $100^{100}$or $\sqrt{99^{99} \cdot 101^{101}}$?
I tried to square up and I got $100^{200}$ or $99^{99} \cdot 101^{101}$ and I don't have an idea how to continue.
| It is “well-known” that $\log(u) \le u-1$ for all $u>0$, with strict inequality for $u \ne 1$ (see for example for each $x>1 , \frac{x-1}{x}\ < \ln x < x-1$).
It follows that for all real $x > 1$
$$
\log \left( \frac{x^{2x}}{(x-1)^{x-1} (x+1)^{x+1}}\right)
= (x-1) \log \left( \frac{x}{x-1}\right) + (x+1) \log \left( \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Three boxes with balls Box $U_1$ contains $1$ white ball and $2$ black balls. Box $U_2$ contains $2$ white balls and $2$ black balls. We extract without reinsertion two balls from every boxes. The four balls are put in a third box $U_3$ initially empty. We randomly extract a ball from $U_3$. Find the probability that t... | In the end, it all drills down to selecting one ball. With probability half, it is a ball from $U_1$ and with probability half, it is a ball from $U_2$.
The probability of choosing a white ball from each box is known, so the total probability is $\tfrac{1}{2}\cdot\tfrac{1}{3}+\tfrac{1}{2}\cdot\tfrac{1}{2}=\tfrac{5}{12}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $P(x) \bmod (x^2-x-2)$
Let $P(x)$ be a polynomial.
If $P(x) \bmod (x+1)=0$ and $P(x) \bmod (x-2)=6$, then evaluate $P(x) \bmod (x^2-x-2)$.
My attempt:
$$P(x)=k(x)(x-2)+6=k(x)(x+1)-3(k(x)-2)$$
$$\Longrightarrow (k(x))-2) \bmod (x+1)=0$$
$$\Longrightarrow \left(\dfrac{P(x)-6}{x-2}-2\right) \bmod (x+1)=0$$
$$\L... | Your solution is correct, but here is a different way to approach it. Though not stated, I'll assume we are working in $\mathbb{Q}[x]$. Notice that $(x^2 - x - 2) = (x-2)(x+1)$, i.e. it is the product of the moduli in your first two congruence equations. A quick check shows that indeed these moduli, $x+1$ and $x-2$, ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $(n + \frac{1}{2})\log(1+\frac{1}{n})$ is increasing in $n$? Is $(n + \frac{1}{2})\log(1+\frac{1}{n})$ increasing in $n$?
I attempted to differentiate $p_n=(n + \frac{1}{2})\log(1+\frac{1}{n})$ with respect to $n$.
$p_n^{'} = \log(1+\frac{1}{n}) + \frac{(n+\frac{1}{2})}{(1+\frac{1}{n})}(-\frac{1}{n^2}) = \... | As $n\to\infty$, $\log\left(1 + \dfrac1n\right)\to0$. Similarly, as $n\to\infty$, $\dfrac1n\to0$ and $\dfrac1{n+1}\to0$ and, therefore, $\dfrac12\left(\dfrac1n + \dfrac1{n + 1}\right)\to0$. So, $p_n^{'}\to0$ as $n\to\infty$ and, therefore, $p_n$ is not increasing in $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving $\int_{-1}^{1}8x^3-5x^2+4dx$ I've done the following so far:
$$\left.\int_{-1}^{1}\left(8x^3-5x^2+4\right)dx=\left(\frac{8}{4}x^4-\frac{5}{3}x^3+4x\vphantom{\int}\right)\right|_{-1}^{1}$$
$$=\left(\frac{8}{4}-\frac{5}{3}+4\right)=\frac{13}{3}$$
However, I double-checked on wolfram alpha and the solution is actu... | The value of last limit must be $4-(-4)=8$
$$\int_{-1}^1(8x^3-5x^2+4)dx=\left(4x^4-\frac53x^3+4x\right)_{-1}^1$$
$$=\left(4(1)^4-\frac53(1)^3+4(1)-4(-1)^4+\frac53(-1)^3-4(-1)\right) $$
$$=\frac{14}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3742430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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How prove this $\sum_{i=n+2}^{+\infty}\frac{1}{i^2}>\frac{2n+5}{2(n+2)^2}$ let $n$ be postive integer,show that
$$\sum_{i=n+2}^{+\infty}\dfrac{1}{i^2}>\dfrac{2n+5}{2(n+2)^2}\tag{1}$$
I know
$$\sum_{i=n+2}^{+\infty}\dfrac{1}{i^2}>\int_{n+2}^{+\infty}\dfrac{1}{x^2}dx=\dfrac{1}{n+2}$$
But $$\dfrac{1}{n+2}-\dfrac{2n+5}{2(n... | At least, for "large" values on $n$, we can show it since
$$\sum_{i=n+2}^{+\infty}\dfrac{1}{i^2}=\psi ^{(1)}(n+2)$$
So, using the asymptotics of the digamma function and Taylor expansions, we have
$$\sum_{i=n+2}^{+\infty}\frac{1}{i^2}-\frac{2n+5}{2(n+2)^2}=\frac{1}{6 n^3}+O\left(\frac{1}{n^4}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Derangement related problem-bijective functions $f: A \to A$ such that $f(x) \neq x$ and $f(1) \neq 2$ Let $R$ denote the set of all nonempty relations on the set $A$, where $A = \{1, 2, 3, 4, 5\}$. A function $f(x)$ is chosen at random from set $R$. The probability that $f(x)$ is bijective, $f(x)\ne x$ such that $x\in... | Let us find the number of derangements $\left\{ 1,2,3,4,5\right\} \to\left\{ 1,2,3,4,5\right\} $
that satisfy $f\left(1\right)=2$.
This equals the number of bijections $\left\{ 2,3,4,5\right\} \to\left\{ 1,3,4,5\right\} $
that satisfy $f\left(k\right)\neq k$ for $k\in\left\{ 3,4,5\right\} $.
Let $A_{k}$ denote the set ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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how can I integrate $\int \dfrac{2x+3}{x^3-9x}dx$? How can I integrate this $$\int \dfrac{2x+3}{x^3-9x}dx?$$
My work
$$\dfrac{2x+3}{x^3-9x}=\dfrac{2x+3}{x(x+3)(x-3)}$$
$$=\dfrac{A}{x-3}+\dfrac{B}{x}=\dfrac{C}{x+3}$$
after solving, I got $A=1/2$, $B=-1/3$, $C=-1/6$
partial fractions decompositions:
$$\dfrac{2x+3}{x^3-9x... | hint for substitution
Write the integral as
$$\int \frac{2dx}{x^2-9}+\int \frac{3xdx}{x^2(x^2-9)}$$
for the first, you can put $$x=3\cosh(t)$$
and for the second, $$x^2=u$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3744234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How to show that $J_{n+1} = \frac{3n-1}{3n} J_n$? Let $$J_n := \int_{0}^{\infty} \frac{1}{(x^3 + 1)^n} \, {\rm d} x$$
where $n > 2$ is integer. How to show that $J_{n+1} = \frac{3n-1}{3n} J_n$?
| Hint:
Let $y=\dfrac x{(x^3+1)^m}$
$$\dfrac{dy}{dx} =\dfrac1{(x^3+1)^m}+\dfrac{(-m)x(3x^2)}{(x^3+1)^{m+1}} =\cdots =\dfrac{3m}{(x^3+1)^{m+1}}-\dfrac{3m-1}{(x^3+1)^m}$$
Integrate both sides wrt $$\dfrac x{(x^3+1)^m}=3mI_{m+1}-(3m-1)I_m$$ where $$I_n=\int\dfrac{dx}{(x^3+1)^n}$$
Now $\dfrac x{(x^3+1)^m}\big|_0^\infty=0-0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Prove The following inequality $(ax+by)^2 \le ax^2+by^2$ for $a+b=1$ Prove The following inequality $(ax+by)^2 \le ax^2+by^2$ for $a+b=1, 0 \le a,b \le 1$
I tried expanding the equation and substituting $b=1-a$
\begin{equation}
(ax+by)^2=a^2x^2+2abxy+b^2y^2=a^2x^2+2axy-2a^2xy+b^2y^2
\end{equation}
The middle member $2a... | WLOG $a=\sin^2t,b=\cos^2t$
$$ax^2+by^2-(ax+by)^2=\sin^2t\cos^2t(x^2+y^2-2xy)=?$$
Alternatively, $$a(1-a)+b(1-b)=a+b-((a+b)^2-2ab)=\cdots=2ab$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3749567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Verifying the period of $f(x)=\sin(x)+\cos(x/2)$ It seems clear from the graph of $f(x)=\sin(x)+\cos(x/2)$ that the period $p$ of the function is equal to $4\pi$.
To verify that $4\pi$ is a period of $f(x)$, note that
\begin{align}
\sin(x + 4\pi) + \cos\left(\frac{x + 4\pi}{2}\right) & =\sin(x)\cos(4\pi)+\cos(x)\sin(4... | Let $$\sin(x + T) + \cos(\frac{x+T}{2}) = \sin(x) + \cos(\frac{x}{2})$$And $T\gt 0$. Then we have $$\sin(x+T) - \sin(x) = \cos(\frac{x}{2}) - \cos(\frac{x+T}{2}) \implies$$
$$2\sin(\frac{T}{2})\cos(\frac{2x+T}{2}) = -2\sin(\frac{2x + T}{4})\sin(\frac{-T}{4}) $$ So then $$\sin(\frac{T}{4}) = 0$$ Or $$2\cos(\frac{T}{4})\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3749717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Proof that any number is equal to $1$ Before I embark on this bizzare proof, I will quickly evaluate the following infinite square root; this will aid us in future calculations and working:
Consider $$x=\sqrt{2+\sqrt{{2}+\sqrt{{2}+\sqrt{{2}...}}}}$$
$$x^2-2=\sqrt{2+\sqrt{{2}+\sqrt{{2}+\sqrt{{2}...}}}}=x \implies x^2-x-... | Although they look similar at first glance, there is no reason for the two sequences
$$ \sqrt{2+2},\ \sqrt{2+\sqrt{2+2}},\ \sqrt{2+\sqrt{2+\sqrt{2+2}}}, \ldots $$
and
$$\sqrt{2+x^2+1/x^2},\ \sqrt{2+\sqrt{2+x^4+1/x^4}},\ \sqrt{2+\sqrt{2+\sqrt{2+x^8+1/x^8}}}, \ldots $$
to have the same limit unless $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Floor function equation $⌊x + 1/2⌋ + ⌊x⌋ = \frac12 x^6$ So in this floor equation $⌊x + 1/2⌋ + ⌊x⌋ = \frac12 x^6$, I've tried putting $x = n + e$, where $0 \le e < 1$, but I didn't get anything useful. What should be an approach in these situations?
| Suppose $x = n + e$ and consider two cases $e < 0.5$ and $e \geq 0.5$.
*
*$0\leq e < 0.5$
$\begin{alignat*}{2}
&2\cdot⌊n + e + 0.5⌋ + 2\cdot⌊n + e⌋ = (n + e)^6\\
&2\cdot n + 2\cdot n = (n + e)^6\\
&4\cdot n = (n + e)^6 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\\
&2\cdot \sqrt{n} = (n + e)^3\\
&(\sqrt{n} + 1)^2 - (n + 1) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to integrate $\int \sqrt{1-\dfrac{1}{25x^2}}\ dx$? How to integrate the following:
$$\int \sqrt{1-\dfrac{1}{25x^2}}\ dx$$
What I did is:
$$\int \sqrt{1-\dfrac{1}{25x^2}}\ dx=\int \dfrac{\sqrt{25x^2-1}}{5x}\ dx$$
I substituted $5x=\sec\theta$, $dx=\dfrac{1}{5}\sec\theta\tan\theta\ d\theta $
$$=\int \dfrac{\sqrt{\sec... | you can also do it with $t^2=25x^2-1$. To get
\begin{align*}
\int \frac{\sqrt{25x^2-1}}{5x} \, dx&=\frac{1}{5}\int \frac{t^2}{t^2+1} dt\\
&=\frac{1}{5}\left[\int \frac{t^2+1-1}{t^2+1} dt\right]\\
&=\frac{1}{5}\left[t-\int \frac{1}{t^2+1} dt\right]\\
&=\frac{t}{5}-\frac{\arctan t}{5}+c,
\end{align*}
where $t=\sqrt{25x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Evaluating the integral $\int_{f}^{g} \exp \left[-\frac{1}{d}\left(c+b x+a x^{2}\right)\right] d x$ I am trying to evaluate the following definite integral and find an analytic solution:
\begin{equation}
\int_{f}^{g} \exp \left[-\frac{1}{d}\left(c+b x+a x^{2}\right)\right] d x
\end{equation}
I have tried looking here, ... | assuming that ad>0
apply linearity
$$\int \ e^{- \frac{ax^2+bx+c}{d} }\, dx =e^{-c/d}\int \ e^{-\frac {ax^2+bx}{d}}dx$$
now solve
$$\int \ e^{-\frac {ax^2+bx}{d}}dx=\int \ e^{-\frac {ax^2}{d}-\frac{bx}{d}}dx$$
complete the square
$$\int\ e^{-\frac {b^2} {4ad}- (\frac{ x\sqrt{a} }{ \sqrt{d} } + \frac{b}{2 \sqrt{ad} } )... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3757891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $n$-th term formula of Fibonacci sequence using generating function I am trying to get the formula $F_n = \frac{\phi^n - \psi^n}{\phi - \psi}$ using generating functions. I managed to find that $G_F(x) = \frac{1}{1 - x - x^2}$ then I used partial fraction decomposition to find that $$G_F(x) = \frac{1}{\phi - \p... | Another way is to use exponential generating functions. Start with $F_{n + 2} = F_{n + 1} + F_n$, $F_0 = 0, F_1 = 1$. Define:
$\begin{align*}
\widehat{F}(z)
&= \sum_{n \ge 0} F_n \frac{z^n}{n!}
\end{align*}$
Now you see that:
$\begin{align*}
\frac{d}{d z} \widehat{F}(z)
&= \sum_{n \ge 0} F_{n + 1} \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim_{x\to 0} \frac{\sqrt{ax+b}-1}{x}=1$ My answer
Let $\sqrt{ax+b}=y$
Then
$$\lim_{y\to \sqrt b} \frac{(y-1)a}{y^2-b}$$
Let $b=1$
Then $$\lim _{y\to 1} \frac{a}{\frac{y^2-1}{y-1}}$$
$$=\frac a2 =1$$
$$a=2$$
The answer is correct, but this relies on assuming $b=1$, and that doesn’t seem appropriate. What is the c... | $$\frac{\sqrt{ax+b}-1}{x}=\frac{ax+b-1}{x(\sqrt{ax+b}+1)}.$$
Now, we see that we need $b=1$ (otherwise, the limit does not exist) and $\frac{a}{\sqrt{b}+1}=1,$
which gives also $a=2$.
If $b\neq1$ for $x\rightarrow0$ we obtain:
$$\frac{\sqrt{ax+b}-1}{x}=\frac{ax+b-1}{x(\sqrt{ax+b}+1)}=\frac{a}{\sqrt{ax+b}+1}+\frac{b-1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3762841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solving $\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{4-2x}$ Can we find the solutions for this equation?
$$\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{4-2x}, \quad x \in \mathbb{R}$$
I tried to amplify the second square root in the $LHS$ with the conjugate and then use AM-GM in order to find whe... | \begin{align*}
&\Rightarrow\sqrt{4-2x}\\
&=\sqrt{2+2(1-x)}&(x\in[0,2])\\
&=\sqrt{[1-\sqrt{2x-x^2}]+[1+\sqrt{2x-x^2}]+2\color{red}{(1-x)}}\\
&=\sqrt{[1-\sqrt{2x-x^2}]+[1+\sqrt{2x-x^2}]+2\color{red}{\sqrt{(1-x)^2}}}&(\color{red}{x\in[0,1]})\\
&=\sqrt{[1-\sqrt{2x-x^2}]+[1+\sqrt{2x-x^2}]+2\sqrt{1-(2x-x^2)}}\\
&=\sqrt{[1-\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Simplifying trigonometry function by substitution I want to simplify the following expression: $\frac{{\left( {1 + \sqrt 3 \tan {1^{\circ}}} \right)\left( {1 + \sqrt 3 \tan {2^{\circ}}} \right)\left( {\tan {1^{\circ}} + \tan {{59}^{\circ}}} \right)\left( {\tan {2^{\circ}} + \tan {{58}^{\circ}}} \right)}}{{\left( {1 + {... | Generalization:
$$\dfrac{(1+\tan y\tan z)(\tan z+\tan(y-z))}{1+\tan^2z} =\cdots =\dfrac{\cos(y-z)\sin(z+y-z)\cos^2z}{\cos y\cos z\cos z\cos(y-z)} =\tan y$$
$$\text{ if }\cos(y-z)\cos z\ne0$$
Here $y=60^\circ$
Then $z=1^\circ, 2^\circ$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $|a + b| = |a| + |b| \iff a\overline{b} \ge 0$ I'm reading a complex analysis book. In this book, the author establishes the following statement
If $a,b \in \mathbb{C}$, then $|a + b| = |a| + |b| \iff \left(a\overline{b}\in \mathbb{R}\right) \wedge \left(a\overline{b}\ge 0\right)$
This statement didn't s... | First off: Intuitively the statement reads “The distances to the origin of two complex points $a$ and $b$ add up to the distance to the origin of $a + b$ if and only if they lie on the same ray from origin.” This is because $\overline b$ is $b$ reflected on real line, which can be interpreted as “$b$, just with its ang... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3765616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Prove that in a triangle $\sum\limits_{cyc}\frac{w_bw_c}{w_a}\geq\frac{3}{4}\left(\sum\limits_{cyc}\frac{a^2w_a}{w_bw_c}\right)\geq\sqrt{3}s$ Let $ ABC$ is a triangle, $ w_a, w_b, w_c$ are bisectors of angles, $ h_a, h_b, h_c$ are altitudes respectively, $ r$ is radius of the incircle, prove that:$$ \frac {w_bw_c}{w_a}... | The right inequality.
In the standard notation we need to prove that:
$$\sum_{cyc}\frac{a^2\cdot\frac{2bc\cos\frac{\alpha}{2}}{b+c}}{\frac{2ac\cos\frac{\beta}{2}}{a+c}\cdot\frac{2ab\cos\frac{\gamma}{2}}{a+b}}\geq\frac{2(a+b+c)}{\sqrt3}$$ or
$$\sum_{cyc}\frac{a^2\cdot\frac{2bc\sqrt{\frac{1+\frac{b^2+c^2-a^2}{2bc}}{2}}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3766609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate the following limit: $\lim_{x\to 0}\frac{12^x-4^x}{9^x-3^x}$? How can I compute this limit
$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}\text{?}$$
My solution is here:
$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}=\dfrac{1-1}{1-1} = \dfrac{0}{0}$$
I used L'H$\hat{\mathrm{o}}$pital's rule:
\begin{align*}
\lim_{x\t... | If you add and subtract $1$ from numerator and denominator
$$
\lim_{x\to0}\frac{(12^x-1)-(4^x-1)}{(9^x-1)-(3^x-1)}
$$
then dividing each term by $x$
$$
\lim_{x\to0}\frac{\dfrac{12^x-1}{x}-\dfrac{4^x-1}{x}}{\dfrac{9^x-1}{x}-\dfrac{3^x-1}{x}}
$$
Now, all four limits have the form
$$
\lim_{x\to0}\frac{a^x-1}{x}=\log a
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3767178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
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.