Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Express $x$ in terms of $a$ and $b$: $\sin^{-1} {\frac{2a}{1+a^2}} + \sin^{-1}{\frac{2b}{1+b^2}} = 2\tan^{-1}x$
Find the value of $x$ from the following equation in terms of $a$ and $b$
$$\sin^{-1} {2a\over{1+a^2}} + \sin^{-1}{2b\over{1+b^2}} = 2\tan^{-1}x$$
I tried to expand the LHS using the formula $$\sin^{-1}c... | Let $\arctan a=u,a=\tan u$
If $-\dfrac\pi2\le2u\le\dfrac\pi2\iff -1\le a\le1$ $$P=\arcsin\dfrac{2a}{1+a^2}=2\arctan a$$
If $2u>\dfrac\pi2,P=\pi-2\arctan a$
If $2u<-\dfrac\pi2,P=-\pi-2\arctan a$
Now use my answer in showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$
| {
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"timestamp": "2023-03-29T00:00:00",
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Range of $y = \frac{x^2-2x+5}{x^2+2x+5}$? How do I approach this problem? My book gives answer as $[\frac{3-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2}]$. I tried forming an equation in $y$ and putting discriminant greater than or equal to zero but it didn't work. Would someone please help me?
I get $x^2 (y-1) + 2x (y+1) + (5y-5... | Hint:
The derivative of $\dfrac{x^2-2x+5}{x^2+2x+5}$ is
$\dfrac{4 (x^2 - 5)}{x^2+2x+5}$ and so the critical points are $\pm \sqrt 5$.
Consider also $\displaystyle\lim_{x\to\pm\infty}\dfrac{x^2-2x+5}{x^2+2x+5}=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim_{x\rightarrow0}{\frac{(x-\arctan(x))\ln(1+2\sin(x))}{(1+\cos{x})(e^x-1-x)^2}}$ using Taylor I want to evaluate the following limit:
$$\lim_{x\rightarrow0}{\frac{(x-\arctan(x))\ln(1+2\sin(x))}{(1+\cos{x})(e^x-1-x)^2}}$$
For example, we have $x-\arctan{x}$. They are both $0$. This seems to be the so called... | For the direct questions: One must, needs to, determine powers of the expansion beyond constant factors to determine the behavior of the expansions; Not every function has the same powers in an expansion. The result then becomes "at least get a few powers of $x$ of each function".
This becomes more familiar by the exa... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find this integral $I=\int\frac{x}{(1-3x)^{3/2}(x+1)^{3/2}}dx$ Find the integral
$$I=\int\dfrac{x}{(1-3x)^{3/2}(x+1)^{3/2}}dx$$
My try: $$(1-3x)(x+1)=-3x^2-2x+1=-3(x+1/3)^2+\dfrac{4}{3}$$
Thus
$$I=\int\dfrac{x}{(\frac{4}{3}-3(x+\frac{1}{3})^2)^{3/2}}dx$$
| Substitute $u=\sqrt{1-3x}$ thus $\mathrm{d}x=-\dfrac{2\sqrt{1-3x}}{3}\,\mathrm{d}u$
$$I={\displaystyle\int}\dfrac{2\left(u^2-1\right)}{\sqrt{3}u^2\left(4-u^2\right)^\frac{3}{2}}\,\mathrm{d}u$$
$$I=\dfrac{2}{\sqrt{3}}{\displaystyle\int}\left(\dfrac{1}{\left(4-u^2\right)^\frac{3}{2}}-\dfrac{1}{u^2\left(4-u^2\right)^\frac... | {
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Loney: If $\alpha$, $\beta$, $\gamma$ are the roots of $x^3 + px^2 + qx + p = 0$, then $\tan^{-1}\alpha + \tan^{-1}\beta + \tan^{-1}\gamma = n\pi$
If $\alpha, \beta, \gamma$ are the roots of the equation $$x^3 + px^2 + qx + p = 0,$$
prove that $$\tan^{-1}\left(\alpha\right) + \tan^{-1}\left(\beta\right) + \tan^{-1}\... | By Vieta’s formula we have
$$-abc=p$$
$$-(a+b+c)=p$$
$$ab+ac+bc=q$$
By using the formula
$$\tan^{-1}u+\tan^{-1}v=\tan^{-1}\frac{u+v}{1-uv}\mod{\pi}\qquad{uv\ne 1}$$, we have
$$
\begin{align}
\tan^{-1}a+ \tan^{-1} b+ \tan^{-1} c
&= \tan^{-1} \frac{a+b}{1-ab}+ \tan^{-1} c \\
&=\tan^{-1}\frac
{\frac{a+b}{1-ab}+c}
{1-\frac... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Linear transformations defined by $T(v) = Av$. Find all of possible $v$ I'm stuck on a problem. The problem is this:
The linear transformation $T : \Bbb{R}^4 \to \Bbb{R}^2$ is defined by $T(v) = Av$, where
$$A = \begin{bmatrix} 2 & -1 & 0 & 1 \\ 1 & 2 & 1 & -3 \end{bmatrix}$$
Find all vectors $v$ such that: $$T(v) = \b... | Your way to solve systems of equations is absolutely correct. But you solved the wrong system. What you did is solving the system of equations $Ax=0$. But you need to solve $Ax=\begin{bmatrix} 1 \\ 2 \end{bmatrix}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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maximum value of expression $(\sqrt{-3+4x-x^2}+4)^2+(x-5)^2$ maximum value of $\bigg(\sqrt{-3+4x-x^2}+4\bigg)^2+\bigg(x-5\bigg)^2\;\forall\;x\in[1\;,3]$
what i try
$\displaystyle -3+4x-x^2+16+8\sqrt{-3+4x-x^2}+x^2+25-10x$
$\displaystyle -6x+38+8\sqrt{-3+4x-x^2}$
using derivative it is very lengthy
help me how to solve,... | Given:
$\bigg(\sqrt{-3+4x-x^2}+4\bigg)^2+\bigg(x-5\bigg)^2\;\forall\;x\in[1\;,3]$
Denote $x=t+2$, then:
$$f(t)=\bigg(\sqrt{1-t^2}+4\bigg)^2+\bigg(t-3\bigg)^2\;\forall\;t\in[-1\;,1]\\
f'(t)=2\bigg(\sqrt{1-t^2}+4\bigg)\cdot \frac{-t}{\sqrt{1-t^2}}+2(t-3)=0 \Rightarrow\\
-2t-\frac{8t}{\sqrt{1-t^2}}+2t-6=0 \Rightarrow \... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Which pairs of positive integers (,) satisfy $^2−2^=153$? My attempt: Rearrange to $x^2=2^n + 153$ and with $2^n\geq 2\ $ it follows $x^2 \geq 155\ $.
The next square number is 169, so $x = 13$ and $n = 4$. A first solution. Since $2^n$ is even and 153 is odd, $x^2$ will be odd. So any candidate solution will have an e... | $n$ must be even.
$$ 153 = 3^2 \cdot 17 $$
If $$ x^2 - 2 y^2 $$
is divisible by $3,$ then both $x,y$ are divisible by $3.$ Since this $y$ would be a power of $2$ this is impossible.
So $n$ is even, $n=2k,$ and we actually have $$ x^2 - (2^k)^2 = 153 \; , $$
$$ (x+ 2^k) (x-2^k) = 153 $$
Umm. $$ (x+ 2^k) - (x-2^k) ... | {
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Prove that $(x−2y+z)^2 \geq 4xz−8y$ Let $x,y,z$ be nonnegative real numbers such that $x+z\leq2$
Prove that, and determine when equality holds.
$(x−2y+z)^2 \geq 4xz−8y$
Please correct me if my methods are incorrect or would lead nowhere.
I tried expanding the LHS of the inequality getting
$x^2+4y^2+z^2-4xy-4yz+2xz \... | consider,
$(x-2y+z)^2-4xz+8y=x^2+4y^2+z^2-4xy-4yz+2xz-4xz+8y$
$=x^2+4y^2+z^2-4xy-4yz-2xz+8y=x^2+4y^2+z^2-4y(x+z)-2xz+8y$
$=(x-z)^2+4y^2+4y(2-(x+z))\geq 0$
because $y$ is non negative and $x+z\leq 2$.
Hence that inequality holds.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How would I find the common ratio to determine the sum of the given geometric serie: How would I find the common ratio to compute the sum of the given geometric serie: $$\sum_{n=1}^\infty= \frac{(8^n+2^n)}{9^n}$$
| First note the evaluation of the relevant finite summation:
$$\sum _{n=1}^{N}{c}^{n}={\frac {{c}^{N+1}}{c-1}}-{\frac {c}{c-1}} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(0)$$
And then observe that your sum is the sum of two such summations,the key diff... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Finding Multivariable limits using polar coordinates How do I find the limit of this multivariable function as it goes to zero using polar coordinates?
$$ \frac{\sin (x^2 + y^2)}{(x^2 + y^2)^2}
$$
| Use
\begin{align}
x &= r \cos \theta \\
y &= r \sin \theta
\end{align}
So $x^2 + y^2 = r^2$ hence
\begin{equation}
\frac{\sin (x^2 + y^2)}{(x^2 + y^2)^2}
=
\frac{\sin r^2}{r^4}
\end{equation}
Using L'Hopital twice, we get
\begin{equation}
\frac{\sin r^2}{r^4}
\sim
\frac{2\cos\left(r^2\right)-4r^2\sin\left(r^2\rig... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Basis of image of operator I am trying to solve the following question:
I have found the Matrix $M(T)$ as:
\begin{bmatrix}
0&-2&2&0\\
-2&0&0&2\\
2&0&0&-2\\
0&2&-2&0\\
\end{bmatrix}
For $b$, I solve $MX=0$ and get the answer.
Can anyone please help me with the $c$ part?
| The "image of T" is the set of all vectors v such that Tu= v for some vector u. In particular, $\begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix}$ is in the image of T if and only if there exist $\begin{bmatrix} w \\ x \\ y \\ z \end{bmatrix}$ such that $\begin{bmatrix}0 & -2 & 2 & 0 \\ -2 & 0 & 0 & 2 \\ 2 & 0 & 0 & -2 \... | {
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"timestamp": "2023-03-29T00:00:00",
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Residue of $z_0=1$ for $f(z)=\frac{z^3+5}{z(z-1)^3}$
Consider $$f(z)=\frac{z^3+5}{z(z-1)^3}.$$ I am trying to find the residue of the pole of order $3$, $\ z_0=1$.
I know from calculations that $$\text{Res}(f,1)=\frac{1}{2}\lim_{z\to 1}\frac{\partial^2}{\partial z^2}\left(\frac{z^3+5}{z}\right)=6.$$
I wish to express... | First of all, note that$$\frac{z^3+5}{z(z-1)^3}=-\frac5z+\frac6{z-1}-\frac3{(z-1)^2}+\frac6{(z-1)^3}.$$So, what can we do with $-\frac5z$? We have\begin{align}-\frac5z&=-\frac5{1+(z-1)}\\&=-5\left(1-(z-1)+(z-1)^2-(z-1)^3+\cdots\right)\end{align}As you can see, this doesn't matter for compution of the residue, which is ... | {
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"timestamp": "2023-03-29T00:00:00",
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If $a^2+b^2 \gt a+b$ and $a,b \gt 0$ Prove that $a^3+b^3 \gt a^2+b^2$ I'm not too sure about this, I have been working on for some time and I reached a solution (not really too sure about)
Question: If $a^2+b^2 \gt a+b$ and $a,b \gt 0$ Prove that $a^3+b^3 \gt a^2+b^2$
My solution: Let $a \geq b$
From $a^2+b^2 \gt a+b$ ... | Also, we can make the following.
Since by the condition $1>\frac{a+b}{a^2+b^2},$ by C-S we obtain:
$$a^3+b^3>\frac{(a^3+b^3)(a+b)}{a^2+b^2}\geq\frac{(a^2+b^2)^2}{a^2+b^2}=a^2+b^2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding the result of the polynomial The question is that: Given that $x^2 -5x -1991 = 0$, what is the solution of $\frac{(x-2)^4 + (x-1)^2 - 1}{(x-1)(x-2)}$
I've tried to factorize the second polynomial like this:
$\frac{(x-2)^4 + (x-1)^2 - 1}{(x-1)(x-2)}
=\frac{(x-2)^4+x(x-2)}{(x-1)(x-2)}
=\frac{(x-2)((x-2)^3+x)}{(x... | Hint:
$\dfrac{(x-2)^4 + (x-1)^2 - 1}{(x-1)(x-2)}$ simplifies to $x^2 - 5 x + 8$
| {
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Proving $a\sqrt{7}>\frac{1}{c}$, where $a$ is an integer and $c = \lceil\frac{a}{\sqrt{7}}\rceil-\frac{a}{\sqrt{7}}$
Let $a \in \Bbb{N}$, and $c=\lceil \frac{a}{\sqrt{7}}\rceil-\frac{a}{\sqrt{7}}$. Prove that $$a\sqrt{7}>\frac{1}{c}$$
So the very original problem sounds like this:
It is given that $a,b \in \Bbb{N... | If $$0 <\sqrt{7} - \frac{a}{b} \leq \frac{1}{ab}$$ then (rearranging and squaring) $$\frac{a^2}{b^2} < 7 \leq \frac{(a^2 + 1)^2}{(ab)^2}$$ or, rearranging again, $$a^2 < 7b^2 \leq a^2 + 2 + \frac{1}{a^2}.$$
If $a \neq 1$, then $a^2 < 7b^2 < a^2 + 3$, which is impossible by Ross Millikan's remark on quadratic residues ... | {
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Q. Reducible polynomials in $\mathbb{Z}[x]$ Show that $x^3+ax^2+bx+1$ $\in \mathbb Z[x]$ is reducible on $\mathbb{Z}$ if and only if $a=b$ or $a+b=-2$.
If it is reducible, then it has root in $\mathbb Z$. Be $u$ the root, so i can write it as:
$x^3+ax^2+bx+1=(x-u)(x^2+cx+d)$
That developing the right expression, I ge... | If $x^3 + ax^2 + bx + 1$ is reducible in $\Bbb Z[x]$, then we have
$x^3 + ax^2 + bx + 1 = (x^2 + cx + d)(x + e)$
$= x^3 + ex^2 + cx^2 + ecx + dx + ed = x^3 + (c + e)x^2 + (ec + d)x + ed; \tag 1$
we know that $x^3 + ax^2 + bx + 1$ must factor in this form since (i.) a cubic, if reducible, may always be written as the pr... | {
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Geometric series and polynomials
i) Find a generating function expression of a sequence with terms
$$d_n=\sum_{p=0}^n p^3$$
using operations on the geometric series $\sum_{n\geq 0} x^n$
ii) Derive a polynomial (in $n$) expression for $d_n$.
for i) I got $x(1+4x+x^2)/(1-x)^5$
but I'm confused what to do for ii),... | Yes, the generating function is correct: just apply the operator $x\frac{d}{dx}$ to $\sum_{n\geq 0} x^n=1/(1-x)$ three times ($3$ is the exponent of $k$) and then divide by $(1-x)$ (for $\sum_{k=0}^n$):
$$f(x)=\frac{x(1+4x+x^2)}{(1-x)^5}$$
Now, in order to find a polynomial formula for $\sum_{k=0}^n k^3$ we have to ext... | {
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Antiderivative of $x\sqrt{1+x^2}$ I am attempting this problem given to me, but the answer key does not explain the answer. The question asks me to find the antiderivative of $x\sqrt{1+x^2}$.
My attempt:
$\frac{d}{dx}f(g(x)) = g'(x)f'(g(x))$
$g'(x)$ must be $x$, therefore $g(x)=\frac{1}{2}x^2$
$f'(x)$ must equal $\sqr... | Alternatively:
$$\int{x\sqrt{1+x^2} \,\mathrm{d}x} = \frac12\int{\sqrt{1+x^2} \,\mathrm{d}x^2}=\\
\frac12\int \sqrt{1+t} \, \ \mathrm dt=\frac12\cdot \frac23\cdot (1+t)^{3/2}+C=\frac13(1+x^2)^{3/2}+C.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Integrate $\int \frac {dx}{\sqrt {(x-a)(x-b)}}$ where $b>a$ Integrate: $\displaystyle\int \dfrac {dx}{\sqrt { (x-a)(x-b)}}$ where $b>a$
My Attempt:
$$\int \dfrac {dx}{\sqrt {(x-a)(x-b)}}$$
Put $x-a=t^2$
$$dx=2t\,dt$$
Now,
\begin{align}
&=\int \dfrac {2t\,dt}{\sqrt {t^2(a+t^2-b)}}\\
&=\int \dfrac {2\,dt}{\sqrt {a-b+t^2}... | Proceeding further from where you left
$$
\int \dfrac {2\,}{\sqrt {a-b+t^2}}\ dt
$$
$$
\int \dfrac {2}{\sqrt {(t)^2 + (\sqrt{a-b}) ^2}}\ dt
$$
Using
$$
\int \frac{1}{\sqrt{x^2 + a ^ 2}} = \log({x+\sqrt{x^2+a^2}})
$$
It becomes
$$
2 \cdot {\log (t+\sqrt{t^2 + (\sqrt{a-b}) ^2})}
$$
Then substitute back the value of t .... | {
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Can someone help me finish this: evaluate $S_n = \frac{x}{1-x^2}+\frac{x^2}{1-x^4}+ ... + \frac{x^{2^{n-1}}}{1-x^{2^{n}}}$ I am asked to find the closed form solution for the below.
$$S_n = \frac{x}{1-x^2}+\frac{x^2}{1-x^4}+ ... + \frac{x^{2^{n-1}}}{1-x^{2^{n}}}$$
Just writing out the $S_1, S_2, S_3$, I have managed to... | $$\begin{align} \frac{x}{1-x}-S_n &= \\
&= \frac{x}{1-x}-\frac{x}{1-x^2}-\frac{x^2}{1-x^4}- \ldots - \frac{x^{2^{n-1}}}{1-x^{2^{n}}} \\
&=\frac{x^2}{1-x^2}-\frac{x^2}{1-x^4}- \ldots - \frac{x^{2^{n-1}}}{1-x^{2^{n}}} \\
&=\ldots \\
&=\frac{x^{2^{n-1}}}{1-x^{2^{n-1}}}- \frac{x^{2^{n-1}}}{1-x^{2^{n}}} \\
&=\frac{x^{2^{n}}... | {
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How do I complete the square of $y= -4x^2-2x-4$? $y = -4x^2 - 2x - 4$
I just can't figure this out, do I divide the second number and the third number by $4$ then by $2$ and then add the product to the second one and subtract it from the third one?
| $-4x^2 -2x -4 = -(4x^2 + 2x + 4)$
The $4x^2 +2x + 4$ must come from some $(2x+b)^2$, to get the right square, and this has linear term $4b$, which should equal $2x$ so $b=\frac{1}{2}$.
Now $(2x + \frac12)^2 = 4x^2 + 2x + \frac14$, so we need an extra $3\frac34$ to get $4$, like we need. So in all
$$-4x^2 -2x -4 = -\le... | {
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Prove that the quantity is an integer I want to prove that $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6} \in \mathbb{Z}, \forall n \geq 1$.
I have thought to use induction.
Base Case: For $n=1$, $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}=\frac{1}{3}-\frac{1}{2}+\frac{1}{6}=0 \in \mathbb{Z}$.
Induction hypothesis: We suppose t... | If you want to use induction you want to show that
$$\frac{(k+1)^3}{3}-\frac{(k+1)^2}{2}+\frac{k+1}{6}$$ is an integer. You can expand all the terms and use the known fact about the expression for $k$ to get there.
Another approach is to note that $6$ is a common denominator and say you want to prove that the numera... | {
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} |
Let $a
Let $a<b$ and $a,b\in\Bbb R$. Then there is $c\in\Bbb R\setminus\Bbb Q$ such that $a<c<b$.
My attempt:
*
*$a+b$ is irrational
Let $c:=\dfrac{a+b}{2}$
*$a+b$ is rational
Let $x:=\dfrac{a+b}{\sqrt 2}$. Then $x$ is irrational.
*
*$a<x<b$
Let $c:=x$
*
*$b\le x$
Then $x-b<x-a$. Take $x'\in (x-b,x-a)$ ... | If $a, b$ both are rationals then, $\frac{b+a}{2}$ is rational. Then, lets pick $c= \frac{b+a}{2}+\frac{b-a}{2\sqrt{2}}$
If $a$ is rational, and $b$ is irrational, we take $c=\frac{b+a}{2}$
If $a, b$ are irrational:
$\frac{a+b}{2}$ is irrational then pick it as $c$.
Else we pick, $c=\frac{b+a}{2}+ (b-\frac{b+a}{2})/2$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2952803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Simple inequality with products of finite geometric series Given numbers $p$ and $q$ with $1\le p < q$, and integers $n$ and $m$ with $1\le n < m$, does this simple inequality hold:
$$\left(1+\frac{1}{p}+\frac{1}{p^2}+\frac{1}{p^3}+\dots+\frac{1}{p^n}\right)\left(1+\frac{1}{q}+\frac{1}{q^2}+\frac{1}{q^3}+\dots+\frac{1}... | Yes, it holds.
*
*If $1=p\lt q$, then the inequality is equivalent to$$(n+1)\cdot\frac{1-\frac{1}{q^{m+1}}}{1-\frac 1q}\lt (m+1)\cdot\frac{1-\frac{1}{q^{n+1}}}{1-\frac 1q},$$i.e.$$\frac{1-\frac{1}{q^{m+1}}}{1+m}< \frac{1-\frac{1}{q^{n+1}}}{1+n}\tag1$$Let $f(x):=\frac{1-\frac{1}{q^{x+1}}}{1+x}$ for $x\ge 1$. Then, $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2956935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the domain of the following function The given function is:
$f(x)=\sqrt{\log_{|x|-1}(x^2 + 4x +4)}$
My approach:
The argument $x^2+4x+4>0$ for all $x\neq-2$
Also, the base $|x|-1$ should be greater than 0 and not equal to 1.
$\therefore |x|-1>0$
$\implies |x|>1$
$\implies x>1$ or $x<-1$
And $|x|-1\neq1$
$\implies ... | HINT
Recall that
*
*for $a>1 \quad \log_a x \ge0 \iff x\ge 1$
*for $0<a<1 \quad \log_a x \ge 0 \iff 0<x\le 1$
therefore in order to have $\log_{|x|-1}(x^2 + 4x +4) \ge 0$ we need to consider two cases
*
*$|x|-1>1 \implies x^2 + 4x +4\ge 1$
*$0<|x|-1<1 \implies 0<x^2 + 4x +4\le 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2959308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Probability that second ball is magenta
attempt
Notice that in any urn, we have $r-1+n-r = n-1$ balls. We have to pick 2 balls, so the sample space size is ${n-1 \choose 2 }$. Now if we want the second ball to be mangenta, then we must have either $MM$ or $RM$. So the probability of this is
$$ P = \frac{ MM + RM }{{n-... | In the first urn there are $(r - 1)$ red balls i.e. $0$ red balls and $(n-r)$ i.e. $(n-1)$ magenta balls. There are two ways the second ball drawn will be magenta $RM$ and $MM$. So, the number of ways to select that (the first term in the square brackets shows for the number of favorable ways of $RM$ and the second one... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2959555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove $\frac{2^{4n}-(-1)^n}{17} \in \mathbb{N}$ by induction Here is my attempted proof:
$\forall n \in \mathbb{N}$, let $S_n$ be the statement: $\frac{2^{4n}-(-1)^n}{17} \in \mathbb{N}$
Base case: $S_1$: $\frac{2^{4(1)}-(-1)^1}{17} = \frac{16+1}{17} = 1 \in \mathbb{N}$
Inductive step: $\forall n \geq 1$, $S_n$ holds
$... | Your proof is fine.
But note:
$(a - b)(\sum\limits_{k=0}^{n-1} a^kb^{n-1 -k}) =$
$a(\sum\limits_{k=0}^{n-1} a^kb^{n-1 -k}) - (\sum\limits_{k=0}^{n-1} a^kb^{n-1 -k})=$
$(\sum\limits_{k=0}^{n-1}a^{k+1}b^{n-1 - k}) -(\sum\limits_{k=0}^{n-1}a^kb^{n-k}) = $
$(\sum\limits_{j=1}^n a^jb^{n-j})-(\sum\limits_{j=0}^{n-1}a^jb^{n-j... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2964749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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If $\frac{2x + 6} {(x + 2)^2}- \frac{2} {x + 2} = \frac{a} {(x + 2)^2}$, then what is $a$? $$\frac{2x + 6} {(x + 2)^2}- \frac{2} {x + 2}$$
The expression above is equivalent to $$\frac{a} {(x + 2)^2}$$
where $a$ is a positive constant and $x \neq -2$.
What's the value of $a$?
| Through basic substitution one finds:
$$a =\frac{2 \left(x^4+5 x^3+7 x^2+3 x-4\right)}{x}$$
not a positive constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2966822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Mean of two numbers by infinite sequences Consider two numbers $a$ and $b$, and the following sequence alternating between even and odd positions:
$$
a+2b+3a+4b+5a+6b\ldots,
$$
If we ''normalize''
$$
\frac{a+2b+3a+4b+\ldots}{1+2+3+4+\ldots},
$$
it turns out this ratio approaches the mean value of $a$ and $b$: $(a+b)/2$... | To give an evaluation, to make more rigourous, in the first case we have
$$1^n+2^n+3^n+4^n+\ldots+k^n\sim \frac{k^{n+1}}{n+1}$$
$$a+2^n b+3^n a+4^n b+\ldots+k^nb\sim a\left(\frac{k^{n+1}}{n+1}-2^n\frac{k^{n+1}}{(n+1)2^{n+1}}\right)+2^nb\frac{k^{n+1}}{(n+1)2^{n+1}}=$$$$=\frac12(a+b)\frac{k^{n+1}}{n+1}$$
and therefore
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2968055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Surface area of sphere and cone Given the sphere $x^2 + y^2 + z^2 = 1$ and the cone $z = \alpha \sqrt{x^2 + y^2}$, $\alpha > 0$, $S_1$ is the portion of the sphere inside the cone while $S_2$ is the portion of the cone inside the sphere. I need to find the value of $\alpha$ such that both surface areas are the same (it... | convert to spherical
$x = \rho\cos\theta\sin \phi\\
y = \rho\sin\theta\sin \phi\\
z = \rho\cos \phi$
$z= \alpha \sqrt{x^2 + y^2}$ becomes
$\cos \phi = \alpha \sin \phi\\
\tan\phi = \frac {1}{\alpha}\\
\phi = \arctan \frac 1{\alpha}$
$\int_0^{2\pi}\int_0^{\arctan{\frac 1\alpha}} \sin\phi \ d\phi\ d\theta$
Is the area of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2968365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $\lim_{n\rightarrow-\infty}\frac{3x^2+x}{2x^2+1}=\frac{3}{2}$ Using Definition Definition of Limit of Function As n $\rightarrow\infty$: Let $g:\mathbb{R}\rightarrow\mathbb{R}$ be a function and let $B\in\mathbb{R}$. If for all $\epsilon>0$, there exists $N>0$ such that $x<-N\Rightarrow |g(x)-B|<\epsilon$, we w... | Let $o<\epsilon <1/2$. Then $x <-\frac 1 {\epsilon} $ implies $|g(x)-2|=\frac {|x-3/2|} {2x^{2}+1} \leq \frac {|x|+3/2} {2x^{2}}=\frac {|x|} {2x^{2}}+\frac 3 {4x^{2}}=\frac {1} {2|x|}+\frac 3 {4x^{2}}< {\epsilon /2}+\frac {3\epsilon ^{2}} 4<\epsilon$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2969810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$ I have this problem which says that for any positive integer $n$, $n \neq 0$ the following inequality is true: $$\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}... | With $AM-GM$
\begin{align}
1.(1+\frac{1}{2^3})(1+\frac{1}{3^3})\cdots(1+\frac{1}{n^3})
&\leq\left(\dfrac1n(n+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\cdots+\frac{1}{n^3})\right)^n \\
&\leq\left(\dfrac1n(n+\sum_{n=1}^\infty\frac{1}{n^3}-1)\right)^n \\
&\leq\left(\dfrac1n(n+\zeta(3)-1)\right)^n \\
&\leq\left(1+\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2970739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 3
} |
Solve differential equation $f''''(x)=f'''(x)f''(x)f'(x)f(x)$ I met this DE recently, and I am utterly befuddled at how to solve it
$$f''''(x)=f'''(x)f''(x)f'(x)f(x)$$
I tried this:
$$\frac{f''''(x)}{f'''(x)}=f''(x)f'(x)f(x)$$
$$\ln|f'''(x)|=c_1+\int f(x)f'(x)f''(x)dx$$
I do not know how to solve the right side, though... | Hint:
Let $u=\dfrac{df}{dx}$ ,
Then $\dfrac{d^2f}{dx^2}=\dfrac{du}{dx}=\dfrac{du}{df}\dfrac{df}{dx}=u\dfrac{du}{df}$
$\dfrac{d^3f}{dx^3}=\dfrac{d}{dx}\left(u\dfrac{du}{df}\right)=\dfrac{d}{df}\left(u\dfrac{du}{df}\right)\dfrac{df}{dx}=\left(u\dfrac{d^2u}{df^2}+\left(\dfrac{du}{df}\right)^2\right)u=u^2\dfrac{d^2u}{df^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
sum of multiplicative inverses for $p=3k+1$ I want to prove that for any prime $p=3k+1$
$$\sum_{i=1}^ki^{-1}\equiv\sum_{i=1}^k\left(i+\frac{p-1}{2}\right)^{-1}\;(\text{mod}\;p)$$
but I can't seem to get anywhere.
Can anyone tell me how I can approach this kind of problem?
| Let $p=2h+1$. By Fermat little theorem, we have
\begin{align*} S:=\sum_{i=1}^k i^{-1}- \sum_{i=1}^k (i+h)^{-1}\equiv \sum_{i=1}^k i^{p-2}- \sum_{i=1}^k (i+h)^{p-2} \pmod p
\end{align*}
since $1\le i, i+h <p$. Then,
\begin{align*} S\equiv& -\sum_{i=1}^k\sum_{j= 0} ^{p-3}\binom{p-2}{j} i^j h^{p-2-j} \pmod p\\
\equiv& -\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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If $2 \mid F_n$, then $4 \mid F_{n+1}^2-F_{n-1}^2$, where $F_n$ is $n$-th Fibonacci number
I want to show that
*
*If $2 \mid F_n$, then $4 \mid F_{n+1}^2-F_{n-1}^2$
*If $3 \mid F_n$, then $9 \mid F_{n+1}^3-F_{n-1}^3$
where $F_n$ is the $n$-th Fibonacci number.
I have tried the following so far:
Since $F_... | Modulo two the Fibonacci sequence cyclically repeats the pattern of length three $0,1,1,0,1,1,0,1,1,\ldots$. So if $F_n$ is even then both $F_{n-1}$ and $F_{n+1}$ are odd. Therefore both $F_{n-1}^2$ and $F_{n+1}^2$ are congruent to $1\pmod 4$ (mod eight actually!). Therefore their difference is a multiple of eight.
Mod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2974491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Prove that $\sqrt[3]{5} + \sqrt{2}$ is irrational I tried with both squaring and cubing the statement, it got messy, here's my latest attempt:
Assume for the sake of contradiction: $\sqrt[3]{5} + \sqrt{2}$ is rational
$\sqrt[3]{5} + \sqrt{2}$ = $\frac{a}{b}$ $a,b$ are odd integers $> 0$ and $ b\neq 0$
${(\sqrt[3]{5}... | You can't do divisibility in irational and rational numbers. When you are operating with divisibility you have to have an integers. It is a relation defined on integer numbers.
Suppose it is rational, then exist rational number $q$ such that $$\sqrt[3]{5} + \sqrt{2}= q$$ so $$ 5 = (q-\sqrt{2})^3 = q^3-3q^2\sqrt{2}+6q-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2975313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
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Calculating coefficient I have a generating function,
$$ \frac{(1-x^7)^6}{(1-x)^6} $$
and I want to calculate the coefficient of $x^{26}$
Solution for this is,
$$ {26+5 \choose 5} - 6{19+5 \choose 5} + 15{12+5 \choose 5} - 20{5+5 \choose 5} $$
Is there formula for this? If there is, what is called?
If there is no formu... | It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series.
We obtain
\begin{align*}
\color{blue}{[x^{26}]}&\color{blue}{\frac{\left(1-x^7\right)^6}{\left(1-x\right)^6}}\\
&=[x^{26}]\left(1-x^7\right)^6\sum_{j=0}^{\infty}\binom{-6}{j}(-x)^j\tag{1}\\
&=[x^{26}]\left(1-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2979109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integral $\int_0^1 \frac{(1-x^4)^{3/4}}{(1+x^4)^2}$, may involve beta function Evaluate the integral: $$\int_0^1 \frac{(1-x^4)^{3/4}}{(1+x^4)^2} dx$$
I have used substitutions like $x^4 = u$, or $x^{-4} = u$. After many hours, I came up with $\frac{2}{1+x^4} = u$ which reduces the integral to :
$$\int_1^2 \frac{2^{3/4}... | Alternative approaches? The integral is a hypergeometric functions, and I would immediately try to get it because then I could always use some identities to simplify if needed.
Let's use the substitution $x^4=v$, which the OP rejected.
$$\int_0^1 (1-x^4)^{3/4}(1+x^4)^{-2} dx=\frac{1}{4} \int_0^1 v^{-3/4} (1-v)^{3/4}(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Calculate $\sum_{n=1}^{\infty} \arctan\bigl(\frac{2\sqrt2}{n^2+1}\bigr) $ $$ \lim_{n \to\infty} \sum_{k=1}^{n} \arctan\frac{2\sqrt2}{k^2+1}= \lim_{n \to\infty} \sum_{k=1}^{n} \arctan\frac{(\sqrt{k^2+2}+\sqrt2)-\sqrt{k^2+2}-\sqrt2)}{(\sqrt{k^2+2}+\sqrt2)(\sqrt{k^2+2}-\sqrt2)+1}= $$
$$\lim_{n \to\infty} \sum_{k=1}^{n} \... | Result
$$\pi -\frac{1}{2} \arctan \left(2 \sqrt{2}\right)\simeq 2.52611... $$
Derivation
More complicated than the elegant telescoping approach, but that was the way I found the result.
Observing that
$$\text{arctan}(t) = t \int_{0}^1 \frac{1}{1+t^2 x^2}\,dx$$
and interchanging sum and integral the sum can be written a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Proof verification of $\{x_n\} = \left(1 + {1\over 2n}\right)^n$ is an increasing sequence.
Let $n\in \mathbb N$ and:
$$
x_n = \left(1 + {1\over 2n}\right)^n
$$
Show that $\{x_n\}$ is an increasing sequence.
$\Box$ Consider ratio test of two consequent terms $x_n$ and $x_{n+1}$:
$$
\frac{x_{n+1}}{x_n} = \frac{\le... | $b_n=(1+1/n)^n$ is increasing. Your sequence is $x_n=\sqrt{b_{2n}}$, so it is increasing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2990359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Inverse of tridiagonal Toeplitz matrix Consider the following tridiagonal Toeplitz matrix. Let $n$ be even.
$${A_{n \times n}} = \left[ {\begin{array}{*{20}{c}}
{0}&{1}&{}&{}&{}\\
{1}&{0}&{1}&{}&{}\\
{}&{1}&{\ddots}&{\ddots}&{}\\
{}&{}&{\ddots}&{\ddots}&{1}\\
{}&{}&{}&{1}&{0}
\end{array}} \right]$$
What is the inverse... | The most direct way to establish the conjecture is to perform the multiplication $E = A^{-1} A$ with the conjectured $A^{-1}$ and show that the result is the unit matrix. Since $A$ is bi-diagonal, each entry of $E$ is the sum of at most two terms, so there is little confusion.
$$E_{ik} = \sum_{j=1}^n A^{-1}_{ij} A_{j ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2991218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to show $\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n}<2$ using induction I´d like to show that $$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n}+\frac{n+1}{2^{n+1}}<2$$ using the fact that $$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n}<2$$
I guess the answer use transitivity of natural numbe... | In order to use induction in the way you tried, you may show a stronger inequality which implies the original one, such as
$$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\dots+\frac{n}{2^n}\leq 2-\frac{n+a}{2^n}$$
where $a$ is a non-negative constant to be determined.
The basic step: for $n=1$ is $\frac{1}{2}\leq 2-\frac{1+a}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2992278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Evaluating $\frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C}$ for a triangle with sides $2$, $3$, $4$
What is
$$\frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C}$$
for a triangle with sides $2$, $3$, and $4$?
One can use Heron's formula to get $\sin A$, etc, and use $\cos A = (b^2+c^2-a^2)/(2bc... | Using an application of the Inscribed Angle Theorem, we get
$$
\begin{align}
2R\sin(A)=a\tag{1a}\\
2R\sin(B)=b\tag{1b}\\
2R\sin(C)=c\tag{1c}
\end{align}
$$
where $R$ is the radius of the circumcircle.
Furthermore, with $s=\frac{a+b+c}2$,
$$
\begin{align}
\text{Area}
&=\sqrt{s(s-a)(s-b)(s-c)}\tag2\\[3pt]
&=\frac12bc\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2992867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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System of equations $a(1 - b^2) = b(1 -c^2) = c(1 -d^2) = d(1 - a^2)$ Given a positive real number $t$, find the number of real solutions $a, b, c, d$ of the system
$$a(1 - b^2) = b(1 -c^2) = c(1 -d^2) = d(1 - a^2) = t$$
I have a solution
Let $f(x)=\frac t{1-x^2}$ and we get $f(f(f(f(a))))=a$ and so we are looking for... | I set your polynomial to $0$ and graphed it in Desmos with $t$ along the x-axis (and $x$ along the y-axis... oops!) For each value of $t$, I drew a vertical line and counted the number of intersections:
\begin{align*}
12 \quad &\text{when}\; 0<t<\frac{\sqrt{5}-1}{2}\\
11 \quad &\text{when}\; t=\frac{\sqrt{5}-1}{2}\\
12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2996282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Prove eigenvalues of a symmetric matrix are in a certain interval I am given a matrix $A=\begin{bmatrix}1&2&0\\0&1&2\\0&0&1\end{bmatrix}$. I am asked to compute $A^tA=\begin{bmatrix}1&2&0\\2&5&2\\0&2&5\end{bmatrix}$
and then to prove that the eigenvalues of $A^tA$ are all in $]0,8[$.
I have no idea how to prove this. ... | If $A^TAx = \lambda x$, then $x^TA^TAx = \lambda x^Tx$, so $\lambda ||x||^2 = ||Ax||^2$, where $||y|| = \sqrt{\sum y_i^2}$ is the norm of the vector $y$. It follows that $\lambda > 0$, since $A$ is invertible, so $||Ax|| \neq 0$ for $x \neq 0$.
To see that $\lambda < 8$, we need to show that for all $x$, we have $||Ax|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2998090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve this Semi-Linear PDE (Partial Differential Equation) with the Characteristic Method I need to solve this linear PDE:
$3u_x - 4u_y = y^2$
The initial condition provided is:
$ u (0,y)= sin(y)$
I need to use the Characteristic Method. I learned the method from this video.
I have reached an answer. However, I am not ... | $$u(x,y) = \frac{(y+\frac{4}{3}x)^3}{12}+\sin(y+\frac{4}{3}x) - \frac{y^3}{12}\quad\text{is correct}$$
Expanding leads to :
$$u(x,y)=\sin(y+\frac{4}{3}x)+\frac{y^2x}{3}+\frac{4yx^2}{9}+\frac{16x^3}{81}$$
So, there is no mistake in your calculus. There is a sign mistake in the handwritten page, which at end gives $\sin(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2998680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solving $8x^3 - 6x + 1$ using Cardano's method Solve for the first root of $8x^3 - 6x + 1 = 0$
After solving I get $\sqrt[3]{\frac{-1 + \sqrt{3}i}{16}} - \sqrt[3]{\frac{1 + \sqrt{3}i}{16}}$, which is not a solution to the cubic equation:
Here's how I come up with:
Using Cardano's method:
let $x = y - \frac{b}{3a}$
Then... | With free computer algebra system Maxima 5.42.1
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3000859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to solve the limit $\lim_{k \to \infty} \frac{(2k)!}{2^{2k} (k!)^2}$. How to solve this limit??
$$\lim_{k \to \infty} \frac{(2k)!}{2^{2k} (k!)^2}$$
It's a limit, not a series
| \begin{align}
\lim_{k\to\infty} \frac{(2k)!}{2^{2k}\cdot(k!)^2}
&=\lim_{k\to\infty} \frac{(2k)!}{2^k \cdot 2^k \cdot k! \cdot k!} \\
&=\lim_{k\to\infty} \frac{(2k)!}{(2^k \cdot k!)^2} \\
&=\lim_{k\to\infty} \frac{(2k)(2k-1)\cdots(2)(1)}{(2k)^2 (2k-2)^2 \cdots (4)^2 (2)^2} \\
&=\lim_{k\to\infty} \frac{(2k-1)(2k-3)\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3008771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What is $\frac{1}{|{x}|}-\frac{x^2}{|x|^3}$? What's the result of:
$$\frac{1}{|{x}|}-\frac{x^2}{|x|^3}$$
Is it
$$\frac{1}{|{x}|}-\frac{x^2}{|x|x^2}=\frac{1}{|{x}|}-\frac{1}{|x|}=0$$
or
$$\frac{1}{|{x}|}-\frac{x^2}{|x|^2x}=\frac{1}{|{x}|}-\frac{1}{x}\frac{x^2}{|x|^2}=\frac{1}{|{x}|}-\frac{1}{x}=\left\{\begin{matrix}\fr... | Yes the first one is correct, indeed we have $|x|^3=|x||x^2|=|x|x^2$ then
$$\frac{1}{|{x}|}-\frac{x^2}{|x|^3}=\frac{1}{|{x}|}-\frac{x^2}{|x|x^2}=\frac{1}{|{x}|}-\frac{1}{|{x}|}=0$$
For the same reason the second one is wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3010837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Find the range of $xy$ under the conditions: $x^2-xy+y^2=9$ and $|x^2-y^2|<9$ Assume $x,y \in \mathbb{R}^+$, and satisfied the following expression:
$$x^2-xy+y^2=9$$
$$\left|x^2-y^2\right| < 9$$
find the range of $xy$
My approach: $x^2-xy+y^2=9$ $\Rightarrow$ $xy+9=x^2+y^2 \geq 2xy$ $\Rightarrow$ $xy \leq 9$
But I do... | For the lower bound you can use the follwing facts:
*
*$x^2-xy+y^2 = (x-y)^2 +xy \Rightarrow 9-xy = (x-y)^2$
*$x^2-xy+y^2 = (x+y)^2 -3xy \Rightarrow 9+3xy = (x+y)^2$
*$\left|x^2-y^2\right| < 9 \Leftrightarrow (x+y)^2(x-y)^2 < 81$
Plugging 1. and 2. into 3. you get:
$$(x+y)^2(x-y)^2 < 81 \Leftrightarrow (9-xy)(9+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3013749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the Maclaurin series of $e^{\sin x}$ by comparing coefficients I believe I have found a nice way to find the Maclaurin series of $e^{\sin x}$. Please check if there are any mistakes with my working. Is this method well known?
| (Working finds up to $x^5$ and assumes knowledge that the Maclaurin series of $\cos x\approx1-\frac{x^2}{2}+\frac{x^4}{24}$.)
Let $f(x)= e^{\sin x}$, therefore $f'(x)=\cos x*e^{\sin x}=\cos x*f(x)$ by chain rule.
Assume that $f(x)$ has a Maclaurin series and let that series $f(x)=a+bx+cx^2+dx^3+ex^4+fx^5$, therefore $f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3015698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Conditional Probability of a Uniform Random Subset.
Question:
Let X = $({1,2,3,..,10})$ Let $Y$ be a uniformly random subset of $X$. Define the events:
A = “$Y$ contains at least $4$ elements",
B = “all elements of $Y$ are even".
What is $Pr(A|B)$?
Answer: 0.1875
Attempt:
I know that P(A $\bigcap$ B) / P(B) is what I... | $5$ elements of $X$ are even.
To have at least $4$ elements and all elements are even, you have either $4$ elements ($5$ ways, choose one to leave out) or $5$ elements ($1$ way).
$$\frac{1+5}{2^{5}}=\frac{6}{32}$$
Remark:
*
*We are choosing uniformly over all subsets. $P(B)= \frac{2^5}{2^{10}} \ne \frac{5}{10}$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3019627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove $a_n = \sqrt{n+1}-\sqrt{n}$ is monotone $n \ge 0$.
Prove $a_n = \sqrt{n+1}-\sqrt{n}$ is monotone $n \ge 0$.
To be monotone it must be either increasing or decreasing, so:
$a_n \ge a_{n-1}$ or $a_n \le a_{n-1}$
$\sqrt{n+1}-\sqrt{n} \ge? \sqrt{n}-\sqrt{n-1} $
$\sqrt{n+1}+\sqrt{n-1} \ge? \sqrt{n} + \sqrt{n}$
I k... | Note that, for $n\geq 1$,
$$ (\sqrt{n+1}+\sqrt{n-1})^2 = n+1+2\sqrt{n^2-1}+n-1 = 2n+2\sqrt{n^2-1}\leq 2n +2n=4n, $$
so that
$$ \sqrt{n+1}+\sqrt{n-1} \leq 2\sqrt{n} \quad \iff\quad \sqrt{n+1}- \sqrt n \leq \sqrt n - \sqrt{n-1}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3021788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Minimum value of the given function
Minimum value of $$\sqrt{2x^2+2x+1} +\sqrt{2x^2-10x+13}$$ is $\sqrt{\alpha}$ then $\alpha$ is________ .
Attempt
Wrote the equation as sum of distances from point A(-1,2) and point B(2,5) as
$$\sqrt{(x+1)^2 +(x+2-2)^2} +\sqrt{(x-2)^2 + (x+2-5)^2}$$
Hence the point lies on the line ... | If we write this as $$\sqrt{x^2+(x+1)^2}+\sqrt{(x-2)^2+(x-3)^2}$$ then we are searching for a point $T$ on line $y=x$ for which $TA+TB$ takes minumim where $A(0,-1)$ and $B(3,2)$.
Now this is well know problem froma ancient greek. Reflect $A$ acros this line and get $A'(-1,0)$. By triangle inequality we can see that $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3022822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
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Why is $\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x = \lim_{x\to -\infty}(-x)+x=\lim_{x\to -\infty}0 = 0$ not correct? $\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x = \lim_{x\to -\infty}(-x)+x=\lim_{x\to -\infty}0 = 0$
Apparently, the 2nd step is illegal here. Probably because for $x=-\infty$ I'd get $(+\infty-\infty)$ which is not p... | I will flesh out much of what would be considered a series of valid steps you may take in order to get something similar to the first step performed in the initial limit calculation. I will note that the second step that was taken is valid because we perform all of the algebra in the limit expression first, then we tak... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 6
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Proving $(\sec^2x+\tan^2x)(\csc^2x+\cot^2x)=1+2\sec^2x\csc^2x$ and $\frac{\cos x}{1-\tan x}+\frac{\sin x}{1-\cot x} = \sin x + \cos x $
Prove the following identities:
$$(\sec^2 x + \tan^2x)(\csc^2 x + \cot^2x) = 1+ 2 \sec^2x \csc^2 x
\tag i$$
$$\frac{\cos x}{1-\tan x} + \frac{\sin x}{1-\cot x} = \sin x + \cos x
\... | (i) Let $C=\cos(2x), S=\sin(2x)=2\sin(x)\cos(x)$
$S^2*RHS= S^2(1+\frac{2}{(S/2)^2})= S^2+8=9-C^2$
$\begin{align}S^2*LHS
&=(2+2\sin^2(x))(2+2\cos^2(x)) \cr
&=(2+(1-C))(2+(1+C))\cr
&=9-C^2 =S^2*RHS
\end{align}$
QED
(ii) Let $t=\tan(x/2),\text{then }\sin(x)=\frac{2t}{1+t^2}\text{ , }\cos(x)=\frac{1-t^2}{1+t^2}$
$1-\tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3027602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Volume between cone and sphere of radius $\sqrt2$ with surface integral Consider the cone $z^2=x^2+y^2$ between $z=0$ and $z=1$. Find the volume of the region above this cone and inside the sphere of radius $\sqrt2$ centered at the origin that encloses the cone.
The straightforward approach to this problem would have ... | We can also calculate by cap area and its solid angle.
Denoting radius by $ R={\sqrt 2}$
Cap Area $A= 2 \pi R. R (1-1/\sqrt 2 ) $
Solid Angle of Cap = $\dfrac{A}{ R^2} = 2 \pi (1-1/\sqrt 2 )$
Remaining Solid Angle $V_1 = 4 \pi -\dfrac{A}{ R^2} = 2 \pi (1+1/\sqrt 2 )$
Remaining Volume $V= \dfrac{R^3}{3}. V_1 = \dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3028125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Factor $10^n -1$ There is an identity or quick way to factor problems of the form $c^x -1$ e.g. $10^n-1$, what is it?
I mean that it can be represented in some way as multiple factors,e.g. $(x-a)(y-b)$, there is a way to expand the difference easily.
| Since it’s subtraction, you could use the difference of two squares I suppose.
$$a^2-b^2 = (a+b)(a-b) \implies a^c-b^c = \big(a^{\frac{c}{2}}+b^{\frac{c}{2}}\big)\big(a^{\frac{c}{2}}-b^{\frac{c}{2}}\big)$$
In this case, you have $b = 1$, so the factorization is
$$\big(c^{\frac{x}{2}}+1\big)\big(c^{\frac{x}{2}}-1\big) \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3029574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Simplify third degree polynomial equations. Given an equation:
$6x^3 - 13x^2 + 8x + 3 = 0$
Broken down to get one form
$(x + {3\over2})$
How can you divide the prior equation to know it will simplify to
$(x + {3\over2})(6x^2 + 4x + 2) = 0$
| Following @MartinArgerami's discovery that there was probably a typo.
if we have a degree of three polynomial we can use the Rational Zero Theorem as such:
Ration Zero Theorem: For $f(x) = a_nx^n + a_{n-1}x^{n-1}+...+a_2x^2 + a_1x + a_0$, every rational zero of $f(x)$, $\frac{p}{q}$ has $p$ as a factor of $a_0$ and $q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many non-negative solutions for $x_{1} + x_{2} + x_{3} + x_{4} = 40$ where $2 \leq x_{1} \leq 8, x_{2} \leq 4, x_{3} \geq 4, x_{4} \leq 5$? My solution:
We have:
$x_{1} + x_{2} + x_{3} + x_{4} = 40$ where $2 \leq x_{1} \leq 8, x_{2} \leq 4, x_{3} \geq 4, x_{4} \leq 5$
$\Leftrightarrow x_{2} + x_{3} + x_{4} = 40 - ... | Math answer
Note that the given constraints for $x_1, x_2$ and $x_4$ and $\sum\limits_{i = 1}^4 x_i = 40$ allows us to define
$$
\begin{aligned}
x_3 &= 40 - x_1 - x_2 - x_4 \\
&\ge 40 - 8 - 4 - 5 \\
&= 23.
\end{aligned}$$
This renders the constraint $x_3 \ge 4$ redundant. As a result, the required answer is $(8-2+1) \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3032553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Question about constructing fields
Find all the monic irreducible polynomials in $F_5[x]$ of degree two (aside
from $x^2-2$ and $x^2-3$, there are eight of them) Adjoining a root u of
these polynomials to $F_5$, construct eight fields $F_5(u)$ of $25$ elements.
Prove that each of these fields is isomorphic to $F... | For example, let's see that there is a root of $x^2 + x + 1$ in $F_5[\sqrt{2}]$.
If $\alpha^2 = 2$ and $\beta = x \alpha + y$ we have
$$\beta^2 + \beta + 1 = (2 x^2 + y^2 + y + 1) + (2 x y + x)\alpha = 0$$ if $x=y=2$. So we get an isomorphism from $F_5[x]/(x^2+x+1)$ into $F_5[\sqrt{2}]$. Since the cardinalities are e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3034349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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In how many ways 11 items can be distributed among 3 peoples such that, sum of items received by any two is more than the third
In how many ways $11$ items can be distributed among $3$ peoples such that, number of items received by any two person is more than the number of items received by the third person.
I have t... | We have $x_1+x_2>x_3$ or, $x_3<\frac{x_1+x_2+x_3}{2}$. And it holds for $ x_1, x_2$ also. Hence, $x_i$ can be $5$ maximum(as they are integers) for $i\in \{1,2,3\}$. So, we need to find the number of solution of the equation $$x_1+x_2+x_3=11,~~~ 1\le x_1,x_2,x_3\le 5$$
Which is equivalent to finding coeff. of $x^{11}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Let $x \in \mathbb{N}$ and let $p$ be a prime divisor of $x^4+x^3+x^2+x+1$, prove that $p = 5$ or $p \equiv 1 \mod{5}$
Let $x \in \mathbb{N}$ and let $p$ be a prime divisor of $x^4+x^3+x^2+x+1$, prove that $p = 5$ or $p \equiv 1 \mod{5}$
Note that $(x^4+x^3+x^2+x+1)(x-1) = x^5 - 1$.
I tried reducing the equation $\mo... | If $r= ord _p(x)$ then by Fermat $r\mid p-1$ and by proposition $r\mid 5$.
If $r=5$ then $p-1 =5k$ so $p\equiv_5 1$
If $r=1$ then $x=1$ so $p\mid 5$ and thus $p=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Divisor of $x^2+x+1$ can be square number? $$1^2+1+1=3$$
$$2^2+2+1=7$$
$$8^2+8+1=73$$
$$10^2+10+1=111=3\cdot37$$
There is no divisor which is square number.
Is it just coincidence? Or can be proved?
*I'm not english user, so my grammer might be wrong
| No, for $x=18$ we get $x^2+x+1=343=7^3$.
Here are the first few counterexamples:
$$
\begin{array}{rrl}
x & x^2+x+1 & \text{factorization}\\
18 & 343 & 7^3 \\
22 & 507 & 3 \cdot 13^2 \\
30 & 931 & 7^2 \cdot 19 \\
67 & 4557 & 3 \cdot 7^2 \cdot 31 \\
68 & 4693 & 13 \cdot 19^2 \\
79 & 6321 & 3 \cdot 7^2 \cdot 43 \\
116 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Fallacious moving of powers resulting with a correct trigonometric series identity.
Prove that
$$
\\ \sum_{r=0}^n \left( \frac{ (-1)^r {n \choose r} } {n+r+1}
( \sin^{2(n+r+1)}x + \cos^{2(n+r+1)}x )\right) = \sum_{r=0}^n \frac{ (-1)^r {n \choose r} } {n+r+1}
$$
for all values of $x$.
I came across this joke whic... | Let's prove $$
\\ \sum_{r=0}^n \left( \frac{ (-1)^r {n \choose r} } {n+r+1}
( \sin^{2(n+r+1)}x + \cos^{2(n+r+1)}x )\right) = \sum_{r=0}^n \frac{ (-1)^r {n \choose r} } {n+r+1} \tag1
$$
for all values of $x$.
By differentiating both sides with respect to $x$, on the right hand side one gets $0$, on the left hand side o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Integral of $(1-x^2)^{1/4}$ I came across this integral while originally solving the integral of $((1-\sqrt{x})/(1+\sqrt{x}))^{1/2}$ which led me to two integrals, one, $\sqrt{x}/(1-x^2)^{1/2}$ which on substituting $(1-x^2)^{1/2}$ as t gives $2(1-t^2)^{1/4}$, which is in the title as well and the other a simple $1/(1-... | Note that:
$$
\int \frac{\sqrt x}{\sqrt{1-x^2}}\,dx = \int (1-t^2)^{1/4}\frac{-2t}{2t\sqrt{1-t^2}}\,dt
$$
But in fact you can work directly with what you originally have:
$$ I= \int\sqrt{\frac{1-\sqrt x}{1+\sqrt x}}\,dx =\int \sqrt{\frac{1-u}{1+u}}2u\,du = \int 4(z^2-1)\sqrt{2-z^2} \,dz$$
using the substitutions $x=u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Does the constant $C$ in this solution to a differential equation equal infinity? The problem is $y' = -\frac{1}{t^2} - \frac{1}{t}y + y^2;\ y_p = \frac{1}{t}$. My solution is
$$\begin{align}
y = \frac{1}{t} + B &\implies y' = -\frac{1}{t^2} + B' \\
&\implies -\frac{1}{t^2} - \frac{1}{t}y + y^2 = -\frac{1}{t^2} + B' ... | This is a very common situation. A general solution for a nonlinear $n$-th order differential equation is a solution depending on $n$ independent parameters, but it does not necessarily give you all the solutions. As in this case, some may be obtained as limits where some parameters go to $\infty$. There are also ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof of this formula for $\sqrt{e\pi/2}$ and similar formulas. \begin{align}
\sqrt{\frac{e\pi}{2}}=1+\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot5}+\frac{1}{1\cdot3\cdot5\cdot7}+\dots+\cfrac1{1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\ddots}}}}
\end{align}
as seen here.
Is there other series that relate $\pi$ and $e$?
Also, ... | About 2 years ago I discovered a lot of pretty nice series that relate $\pi$ and $e$, for instance :
$$\sum_{n=1}^{\infty}\frac{n^2}{16n^4-1}=\frac{\pi}{32}\cdot\frac{e^{\pi}+1}{e^{\pi}-1}$$
$$\sum_{n=1}^{\infty}\frac{n^2}{4n^4+1}=\frac{\pi}{8}\cdot\frac{e^{\pi}-1}{e^{\pi}+1}$$
$$\sum_{n=1}^{\infty}\frac{n^2}{(4n^4+1)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Repeatedly dividing $360$ by $2$ preserves that the sum of the digits (including decimals) is $9$ Can someone give me a clue on how am I going to prove that this pattern is true or not as I deal with repeated division by 2? I tried dividing 360 by 2 repeatedly, eventually the digits of the result, when added repeatedly... | In general if an integer is divisible by 9 then its digital root is $9$. Any multiple of an integer divisible by $9$ will also be divisible by $9$ and have digital root $9$.
What's slightly surprising here is that this property is also preserved by division by $2$. Note that it doesn't work for, say, division by $3$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 6,
"answer_id": 3
} |
Algebraic Closed Form for $\sum_{n=1}^{k}\left( n- 3 \lfloor \frac{n-1}{3} \rfloor\right)$ Let's look at the following sequence:
$a_n=\left\{1,2,3,1,2,3,1,2,3,1,2,3,...\right\}$
I'm trying to calculate:
$$\sum_{n=1}^{k} a_n$$
Attempts:
I have a Closed Form for this sequence.
$$a_n=n- 3 \bigg\lfloor \frac{n-1}{3} ... | If $n \equiv 0 \pmod{3}$, i.e. say $n=3s$ (where $s \geq 1$), then $a_n=3s-3\lfloor s-\frac{1}{3}\rfloor=3s-3(s-1)=3$.
If $n \equiv 1 \pmod{3}$, i.e. say $n=3s+1$ (where $s \geq 0$), then $a_n=3s+1-3\lfloor s\rfloor=1$.
If $n \equiv 2 \pmod{3}$, i.e. say $n=3s+2$ (where $s \geq 0$), then $a_n=3s+2-3\lfloor s+\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Find the directrix of the parabola with equation $y=-0.5x^2+2x+2$
Find the directrix of the parabola with equation
$$y=-0.5x^2+2x+2$$
I did this:
$$a=-0.5, b = 2, c = 2$$
Formula for the directrix is:
$$y=-1/(4a)$$
$$y=-1/(4\cdot(-0.5))=3.5$$
This is not right:
What went wrong? What is the proper way to do it?
| Complete the square $$\begin{aligned} y&=-0.5x^2+2x+2\\ &=-0.5(x^2-4x)+2\\ &=-0.5(x^2-4x+4-4)+2\\ &=-0.5[(x-2)^2-4]+2\\&=-0.5(x-2)^2+4\end{aligned}$$
or equivalently $$y-4=-0.5(x-2)^2.$$
The directrix is given by $$y-4=-\frac{1}{4a}\quad \text{with}\; a=-0.5$$
or $$y=4+\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3066042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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$\frac{7x+1}2, \frac{7x+2}3, \frac{7x+3}4, \ldots ,\frac{7x+2016}{2017}$ are reduced fractions for integers $x\in(0,301)$.
BdMO 2017 junior catagory Question 7. $$\dfrac{7x+1}2, \dfrac{7x+2}3, \dfrac{7x+3}4, \ldots ,\dfrac{7x+2016}{2017}$$
Here $x$ is a positive integer and $x < 301$. For some values of $x$ it is po... | As noted by lab bhattacharjee, we need to see if $7x-1$ is co-prime to all $k=2,3,\ldots,2017$. If $0<x<301$, then $6\le 7x-1\le 7\times 300-1=2099$. Clearly, an integer in the interval $[6,2017]$ is not co-prime to all of the integers in $[2,2017]$ (in other words, $t$ is not co-prime to $t$ for $t\ge 2$). So we ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3066236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Prove that $3^{2n} +7$ is divisible by 8
Prove by induction that $3^{2n} +7$ is divisible by 8 for $n \in \Bbb N$
So I think I have completed this proof but it doesn't seem very thorough to me - is my proof valid?
If $n=1$ then $3^{2n} +7 = 16 = 2(8)$ so true when $n=1$
Assume true for $n=k$ so $$ 8\vert 3^{2k} +7$$
... | We can prove it like this :
$$ 9^n + 7 = 9^n + \sqrt[n]{7}^n $$
And by using factorization :
$$ 9^n + \sqrt[n]{7}^n = (9+7)(9^{n-1} -9^{n-2}\cdot7 + ... + 7^{n-1})$$
$$ 9^n + \sqrt[n]{7}^n = (16)(9^{n-1} -9^{n-2}\cdot7 + ... + 7^{n-1})$$
$$ \frac{9^n + 7}{16} = 9^{n-1} -9^{n-2}\cdot7 + ... + 7^{n-1}$$
We have $16$ divi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3067826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Power of matrix using diagonalization First one
$$\begin{pmatrix}
2& 3\\5
& 1
\end{pmatrix}^{20}$$
Second one
$$A=\begin{pmatrix}
4&0& 0\\0
& 3&0\\2 &0&2
\end{pmatrix}^{20}$$
$$P=\begin{pmatrix}
1&0& 0\\0
& 1&0\\1 &0&1
\end{pmatrix}$$
$$P^{-1}=\begin{pmatrix}
1&0& 0\\0
& 1&0\\-1 &0&1
\end{pmatrix}$$
I have some difficu... | $A$ has distinct eigenvalues, it must be diagonalizable.
\begin{align}\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}\begin{bmatrix}4 & 0 & 0 \\ 0 & 3 & 0 \\ 2 & 0 & 2 \end{bmatrix} \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} &= \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3068767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find shortest distance from the parabola $y=x^2-9$ to the origin.
Find shortest distance from the parabola $y=x^2-9$ to the origin.
First, I find minima of $\sqrt{x^2+(x^2-9)^2}$, so use derivative and ...
Is have an easier way?
| Here is an easier way: the shortest distance $r$ is taken at the minimum of
$r^2 = y^2 + x^2 = y^2 + y + 9$ This gets minimal at $y = -0.5$ (take the derivative or write $r^2 = (y+ 0.5)^2 + 8.75 \ge 8.75 $), so $r = \sqrt{y^2 + y + 9} \ge \sqrt{8.75} \simeq 2.9580$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3070891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Differentiate $f ( x ) = \frac { \ln \left( x ^ { 2 } \cos ( x ) \right) } { \sqrt { 1 - x ^ { 2 } } }$ could I get a clue as I dont even know where to begin , I thought about the quotient rule but thats making another quotient rule necessary and I have a cos x within a ln , I really dont even know where to begin for m... | We have $${d\over dx}\ln {x^2\cos x}={1\over x^2\cos x}\cdot {(2x\cos x-x^2\sin x)}\\{d\over dx}\sqrt{1-x^2}=-{x\over \sqrt{1-x^2}}$$therefore by defining $g(x)=\ln x^2\cos x$ and $h(x)=\sqrt{1-x^2}$ and using $\left({g\over h}\right)'={g'h-gh'\over h^2}$ we finally obtain$$f'(x){=\left({g\over h}\right)'=\left({g'h-gh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3072985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to apply CRT to a congruence system with moduli not coprime? $x=1 \pmod 8$
$x=5 \pmod{12}$
8 and 12 are not coprime, I could break it to:
$x=1 \pmod 2$
$x=1 \pmod 4$
and
$x=5 \pmod 3$
$x=5 \pmod 4$
But what are the next steps to solve it? By the way, $x$ should be $17$ not sure how to get that number ...
Thanks in... | Here is a way:
$$\begin{cases}
x\equiv 1\pmod 8\\x\equiv 5\pmod{12}
\end{cases}\iff \begin{cases}
x -1\equiv 0\pmod 8\\x -1\equiv 4\pmod{12}\end{cases}\iff
\begin{cases}
\frac{x -1}4\equiv 0\pmod 2\\\frac{x -1}4\equiv 1\pmod{3}
\end{cases}$$
Now set $y=\frac{x-1}4$. As $3-2=1$, the solutions of the last system of cong... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3075823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
evaluate this elliptic hyperbloid volume? How can I calculate the volume of this region in cylindrical coordinates?
$D=\{2x^2+y^2=z^2+4,|z| \le 2\}$
I think I got this wrong :
$$\operatorname{Volume} = 2\int_{0}^{2\pi}\int_{0}^{2} \int_{0}^{\sqrt{2r^2\cos^2\theta+r^2\sin^2\theta-4}}rdzdrd\theta$$
The problem is that th... | It's a single sheet hyperboloid and its central axis of symmetry is the $z$-axis.
Easiest way to find its volume is to consider that at every $z$ we have an ellipse, and integrate over the areas of those ellipses.
For a constant $z$ we have the ellipse:
$$\frac{x^2}{\frac{z^2+4}{2}}+\frac{y^2}{z^2+4}=1$$
It has semi ax... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3076175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determining the area of a right triangle, perimeter given, hypotenuse value given in terms of one of the legs. The problem states:
Right Triangle- perimeter of $84$, and the hypotenuse is $2$ greater than the other leg. Find the area of this triangle.
I have tried different methods of solving this problem using Pyt... | $\\ \textbf{Finding triples, given perimeter using Euclid's formula}$ where $P=perimeter$
$$P=(m^2-n^2 )+2mn+(m^2+n^2 )=2m^2+2mn\implies n=\frac{P-2m^2}{2m}\quad where \quad \biggl\lceil\frac{\sqrt{P}}{2}\biggr\rceil\le m \le \biggl\lfloor\sqrt{\frac{P}{2}}\biggr\rfloor$$
Here, the lower limit ensures that $m>n$ and t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3076504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Proving $\binom{n + m}{r} = \sum_{i = 0}^{r} \binom{n}{i}\binom{m}{r - i}$ To prove $$\binom{n + m}{r} = \sum_{i = 0}^{r} \binom{n}{i}\binom{m}{r - i},$$
I demonstrated that the equality is true for any $n,$ for $m = 0, 1,$ and for any $r < n + m$ simply by fixing $n$ and $r$ and inserting $0,1$ for $m.$ Then, I procee... | For a short reply, your induction proof has tiny problem, in that $r$ can take value $n+k$ for the $m = k+1$ induction step. So when you use induction hypothesis to get ${{n+k}\choose{r}} = \sum_{i=1}^r {n\choose i}{k\choose {r-i}}$, you can actually use it only for $r<n+k$. This is not big problem though as the $r = n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find range of $x$ if $\log_5\left(6+\frac{2}{x}\right)+\log_{1/5}\left(1+\frac{x}{10}\right)\leq1$
If $\log_5\left(6+\dfrac{2}{x}\right)+\log_{1/5}\left(1+\dfrac{x}{10}\right)\leq1$, then $x$ lies in _______
My Attempt
$$
\log_5\bigg(6+\dfrac{2}{x}\bigg)+\log_{1/5}\bigg(1+\dfrac{x}{10}\bigg)=\log_5\bigg(6+\dfrac{2}{x... | By your work we need to solve
$$\frac{x^2-2x-4}{x(x+10)}\geq0$$ and the domain gives $x>0$ or $-10<x<-\frac{1}{3}.$
The first by the interval's method gives
$$1-\sqrt5\leq x<0$$ or $$x\geq1+\sqrt5$$ or $$x<-10,$$ which with our domain gives the answer:
$$\left[1-\sqrt5,-\frac{1}{3}\right)\cup[1+\sqrt5,+\infty).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3084524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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The number of prime pairs of $x^2-2y^2=1$ How to find the number of pairs of positive integers $(x,y)$ where $x$ and $y$ are prime numbers and $x^2-2y^2=1$?
I am not getting any clue here.
| Since $2y^2$ is even, $x$ must be odd so that $x^2$ can be odd as well. This much should be obvious.
From there it's not too much of a leap to find that $x^2 \equiv 1 \pmod 4$. If $y$ is odd as well, then $y^2 \equiv 1 \pmod 4$, too, but $2y^2 \equiv 2 \pmod 4$, which means that $x^2 - 2y^2 \equiv 3 \pmod 4$, but clear... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3084983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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For $G$ the centroid in $\triangle ABC$, if $AB+GC=AC+GB$, then $\triangle ABC$ is isosceles. (Likewise, for the incenter.)
Let $G$ be the centroid of $\triangle ABC$. Prove that if $$AB+GC=AC+GB$$ then the triangle is isosceles!
Of course, the equality is true, when we have isosceles triangle, but the other way is n... | In the standard notation we obtain:
$$c+\frac{1}{3}\sqrt{2a^2+2b^2-c^2}=b+\frac{1}{3}\sqrt{2a^2+2c^2-b^2}$$ or
$$3(b-c)=\frac{3(b^2-c^2)}{\sqrt{2a^2+2b^2-c^2}+\sqrt{2a^2+2c^2-b^2}},$$ which gives $b=c$ or
$$\sqrt{2a^2+2b^2-c^2}+\sqrt{2a^2+2c^2-b^2}=b+c$$ or
$$\sqrt{(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)}=bc-2a^2,$$ which is im... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3086013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Determine the sequence generated by a generating function $A(z)=2z-1+\frac{1}{2z-2z^2}$
I have no clue how to solve this, I tried looking at other examples but I am just stuck, could anyone be so kind and explain how to solve this step by step?
| If you want to see the generating function:
$\begin{array}\\
A(z)
&=2z-1+\frac{1}{2z-2z^2}\\
&=2z-1+\frac1{2z}\frac{1}{1-z}\\
&=2z-1+\frac1{2z}\sum_{n=0}^{\infty} z^n\\
&=2z-1+\frac1{2z}+\sum_{n=1}^{\infty} \frac{z^{n-1}}{2}\\
&=2z-1+\frac1{2z}+\sum_{n=0}^{\infty} \frac{z^{n}}{2}\\
&=2z-1+\frac1{2z}+\frac12+\frac{z}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3088889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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$\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5}=3x$ Problem:
solve equation
$$\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5}=3x$$
I don't look for easy solution (square booth side and things like that...) I look for some tricks for "easy" solution because:
I would like to use substitution, but we have $3x$ and $-3x$, but I can't see... | Let $2x^2-3x+5=t^2$. Then:
$$\sqrt{t^2+6x}+t=3x \Rightarrow t^2+6x=t^2-6xt+9x^2 \Rightarrow \\
2t=3x-2 \Rightarrow 4t^2=9x^2-12x+4 \Rightarrow \\
4(2x^2-3x+5)=9x^2-12x+4 \Rightarrow \\
x^2=16 \Rightarrow x=4.$$
Note: The square roots of the LHS of the original equation exist for all $x\in \mathbb R$, however, the LHS i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Is $1111111111111111111111111111111111111111111111111111111$ ($55$ $1$'s) a composite number? This is an exercise from a sequence and series book that I am solving.
I tried manipulating the number to make it easier to work with:
$$111...1 = \frac{1}9(999...) = \frac{1}9(10^{55} - 1)$$
as the number of $1$'s is $55$.
Th... | More explicitly,
$$\begin{align*}
\frac{10^{55} - 1}{9}
&= \frac{(10^5)^{11} - 1}{9} \\
&= \frac{(10^5 - 1)(10^{50} + 10^{45} + 10^{40} + \cdots + 10 + 1)}{9} \\
&= \frac{(10 - 1)(10^4 + 10^3 + 10^2 + 10^1 + 1)(10^{50} + 10^{45} + \cdots + 1)}{9} \\
&= (10^4 + 10^3 + \cdots + 1)(10^{50} + 10^{45} + \cdots + 1).
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3091832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 3
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Calculate conditional probability; throwing the cube The symmetrical cube was threw $30$ times.
Calculate probability situation, when in first $20$ throws came out $4$ times number $3$, if in $30$ throws number $3$ came out $7$ times.
| So in this case we are dealing with conditional probability, I think.
Let's calculate $\mathbf{B}$
$7=7+0+0 \to $ $3$ times, because 7 could be anywhere
$7=6+1+0\to$ $6$ times
$7=5+1+1\to$ $3$ times
$7=5+2+0\to$ $6$ times
$7=4+3+0\to$ $6$ times
$7=4+2+1\to$ $6$ times
$7=3+2+2\to$ $3$ times
$7=3+1+3\to$ $3$ times
So
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3092539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Tough Irrational Equation highschool Have been trying to solve this irrational equation for a day but as it seems, i'm not going anywere with it. Can somebody offer me a tip ? Thanks!
*Tried a "t" substitution for x squared but it still yields a 4th degree polynomial equation instead of an 8th, which I think can be sol... | We have
$$\sqrt{\frac{x^2+2x\sqrt{1-x^2}+1-x^2}{2}}+2x^2-1=0$$ or
$$\frac{|x+\sqrt{1-x^2}|}{\sqrt2}+2x^2-1=0.$$
Now, $$x=\sqrt{1-x^2}$$ gives $x=\frac{1}{\sqrt2}$ which is not a root of our equation.
Thus, our equation is equivalent to
$$\frac{|2x^2-1|}{\sqrt2|x-\sqrt{1-x^2}|}+2x^2-1=0$$ or since $2x^2-1\leq0,$
$$(2x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3094528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integral with respect to $x +$ constant Is this a valid expression:
$$\int xd(x+5)$$
I am trying to calculate the value using a u-sub, of $u = x + 5$. So then $du = d(x+5)$ and so the result is:
$$\int (u - 5)du= \frac{u^2}{2} - 5u + C = \frac{(x+5)^2}{2} - 5(x+5) +C = \frac{x^2}{2} -12.5 + C $$
Or is it correct to jus... | Of course it's a valid expression. So is this. Note that [1] and [2] are the same expression. :-)
$$\begin{align}
\int xe^{-x^2}dx &= \int \frac{xe^{-x^2}dx}{1}\\
&= \int \frac{xe^{-x^2}dx}{1}\cdot\frac{\frac{d(-x^2)}{dx}}{\frac{d(-x^2)}{dx}}\\
&= \int \frac{xe^{-x^2}dx\cdot\frac{d(-x^2)}{dx}}{-2x}\\
&= \int \frac{xe^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving $2\sin\theta\cos\theta + \sin\theta = 0$
The question is to solve the following question in the range $-\pi \le \theta \le \pi$
$$2\sin\theta\cos\theta + \sin\theta = 0$$
I missed the obvious sin factorisation so proceeded as below. I see the correct solutions should be $\pm2/3\pi$ and the values when $\sin\... | I'll start by graphing this function x-axis is $\theta / \pi $ which shows the function is zero at 5 points.
\begin{align}
2 \cdot \sin(\theta)\cos(\theta) + \sin(\theta) & = 0 \\
sin(\theta) \cdot (2\cdot\cos(\theta)+1) & = 0
\end{align}
So either $\sin(\theta) = 0$ or $2\cdot\cos(\theta)+1 = 0 \Rightarrow \cos(\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3096324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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If $S=1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}\cdots$, then what is $\lfloor S \rfloor$?
If
$$S=1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}\cdots$$
then $$\lfloor S \rfloor = \text{?}$$
What I tried:
I know that
$$S=1+\frac{1}{2^4}+\frac{1}{3^4}+\cdots=\zeta(4)=\frac{\pi^4}{90}\approx 1.1$$
then $\lfloor ... | Hint: Note that
$$
\begin{align}
\frac1{(k-1)^3}-\frac1{k^3}
&=\frac{3k^2-3k+1}{k^3(k-1)^3}\\
&\gt\frac{3k^2-3k}{k^3(k-1)^3}\\
&=\frac3{k^2(k-1)^2}\\
&\gt\frac3{k^4}
\end{align}
$$
Therefore,
$$
\begin{align}
\sum_{k=n}^\infty\frac1{k^4}
&\lt\frac13\sum_{k=n}^\infty\left(\frac1{(k-1)^3}-\frac1{k^3}\right)\\
&=\frac1{3(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3097084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Differentiation uner the integral sign - help me find my mistake This is my integral:
$$I(a)=\int_0^\infty\frac {\ln(a^2+x^2)}{(b^2+x^2)}dx.$$
Taking the first derivative with respect to a:
$$I'(a)=\int_0^\infty \frac {2adx} {(a^2+x^2)(b^2+x^2)}.$$
This is how I did the partial fraction decomposition:
$\frac {2a} {(a^2... | Let’s redo the work$$\frac 1{(a^2+x^2)(b^2+x^2)}=\frac {Ax+B}{a^2+x^2}+\frac {Cx+D}{b^2+x^2}$$Multiplying both sides by the common denominator$$1=(Ax+B)(b^2+x^2)+(Cx+D)(a^2+x^2)$$To find values for $A$, $B$, $C$, and $D$, first set $x^2=-a^2$. Thus$$\begin{align*}1 & =(Ax+B)(b^2-a^2)\\ & =Ax(b^2-a^2)+B(b^2-a^2)\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3101376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluate the integral $\int_1^\infty \left(\frac{1+x}{1+x^2}\right)^2\,dx$ I have to prove that this integral is finite so that the series $\sum_1^\infty \left(\frac{1+n}{1+n^2}\right)^2$ converges , but I am not able to integrate the function . Please help
| $$
\begin{align}
\int\left(\frac{1+x}{1+x^2}\right)^2\,dx
&=\int\frac{1+2x+x^2}{\left(1+x^2\right)^2}\,dx\\
&=\int\left(\frac{1+x^2}{\left(1+x^2\right)^2}+\frac{2x}{\left(1+x^2\right)^2}\right)\,dx\\
&=\int\frac{1}{1+x^2}\,dx+\int\frac{1}{\left(1+x^2\right)^2}\frac{d}{dx}\left(1+x^2\right)\,dx\\
&=\arctan{x}+\int\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3102084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Use the definition of limit to show that $\lim_{x \to -1} \frac{x+5}{2x+3}=4$
Use the definition of limit to show that
$\lim_{x \to -1} \frac{x+5}{2x+3}=4$
We have $|\frac{x+ 5}{2x+3} -4|= |\frac{x+5-8x-12}{2x+3}|=|\frac{-7x-7}{2x+3}|=\frac{7}{|2x+3|}|x+1|$
To get a bound on the coefficient of $|x+1|$, we restrict $x... | You way of thinking is on the right track. First, we need to "guess" $\delta$ so we will make a draft of the calculation first.
This is just the "draft" :
\begin{align*}
\big|\frac{x+5}{2x+4}-4\big| &< \varepsilon \\
7\big|\frac{x+1}{2x+3}|&< \varepsilon \\
|x+1| &< \frac{\varepsilon}{7}|2x+3|
\end{align*}
So now, for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3103516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the limit of $\frac {x^2+x} {x^2-x-2}$ where $x \to -1$? I need to find the limit of $\frac {x^2+x} {x^2-x-2}$ where $x \to -1$. Right now I am getting $\frac{0}{0}$ if I don't factor first, or $\frac{2}{0}$ if I do.
Here are my factoring steps:
$\frac {x^2+x} {x^2-x-2}$
$=\frac{x(x+1)}{(x-2)(x+1)}$
replace $x$ wi... | $$
\begin{align}
x^2-x-2
&=x^2-2x\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2-2\\
&=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}-\frac{2\cdot4}{4}\\
&=\left(x-\frac{1}{2}\right)^2-\frac{9}{4}\\
&=\left(x-\frac{1}{2}\right)^2-\left(\frac{3}{2}\right)^2\\
&=\left(x-\frac{1}{2}-\frac{3}{2}\right)\left(x-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3104145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
show this inequality $a_{1}+a_{2}+\cdots+a_{n}\ge\sqrt{3}(a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n}a_{1})$
let $a_{1},a_{2},\cdots,a_{n}\ge 0,n\ge 3$,and such $$a^2_{1}+a^2_{2}+\cdots+a^2_{n}=1$$
show that
$$a_{1}+a_{2}+\cdots+a_{n}\ge\sqrt{3}(a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n}a_{1})$$
I can prove when $n=3$, it need to... | For $n=3$ you have a proof.
We'll prove that for all $n\geq4$ the following stronger inequality is true:
$$\sum_{k=1}^na_k\geq2\sum_{k=1}^na_ka_{k+1},$$ where $a_{n+1}=a_1$.
Indeed, we need to prove that
$$\sum_{k=1}^na_k^2\left(\sum_{k=1}^na_k\right)^2\geq4\left(\sum_{k=1}^na_ka_{k+1}\right)^2,$$ which is true because... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3108726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Better proof for $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$ Prove this $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$
My attempt
Proof - by using [axiomdistributive] and [axiommulcommutative]:
$$\begin{split}
&(x+y)(x^2 - xy + y^2)\\
&= (x+y)x^2 - (x+y)xy + (x+y)y^2\\
&= (x^3+x^2y) - (x^2y+xy^2) + (xy^2+y^3)\\
&= x^3 + x... | Maybe not easier, but given $$\frac{1-z^n}{1-z}=\sum_{i=0}^{n-1}z^i\\1-z^n=(1-z)\sum_{i=0}^{n-1}z^i\\$$ let $z=\frac{y}{x}$, then $$1-\left(\frac{y}{x}\right)^n=\left(1-\frac{y}{x}\right)\sum_{i=0}^{n-1}y^ix^{-i}$$ multiplying by $x^n$ yields $$x^n-y^n=(x-y)\sum_{i=0}^{n-1}y^ix^{n-1-i}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3110478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
What is the smallest number of $45^\circ$–$60^\circ$–$75^\circ$ triangles in non-trivial substitution tiling? Let base = $45^\circ$–$60^\circ$–$75^\circ$ triangle.
Over at What is the smallest number of bases that a square can be divided into? it was determined that 23 base were needed to make a $45^\circ$–$45^\circ$–$... | Hmm... This doesn't exactly fit your criterion in that it's not unique to the $45^∘–60^∘–75^∘$ triangle, but the central triangle here has no edges parallel to the edges of the original triangle.
The vertices are:
$$
\begin{array}{ccc}
\{0,0\} \\
\{1,0\} \\
\left\{\frac{1}{2} \left(3-\sqrt{3}\right),\frac{1}{2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3111813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Show that $f_A$ is an inner product Let $A$ be a $2 \times 2$ matrix with real entries. For $X, Y$ in $R^{2 \times 1}$ let
$f_A(X, Y) = Y^tAX$.
Show that $f_A$ is an inner product on $R^{2 \times 1}$ if and only if $A = A^t$, $A_{11} > 0$, $A_{22} > 0$,
and $det A > 0$.
I was able to solve the first part, assuming the ... | You're basically there with (conjugate) symmetry: note that $X^\top A Y = f(Y, X)$. It's also a scalar (well, a $1 \times 1$ matrix), so the transpose is superfluous.
As for the positive-definiteness, this is where we need to start making use of the individual entries of $A$. Suppose $X = \begin{bmatrix} x_1 \\ x_2 \en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3112211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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.