Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How do I solve $\int\frac{\sqrt{x}}{\sqrt{x}-3}\,dx?$ $$
\int \frac{\sqrt{x}}{\sqrt{x}-3}dx
$$
What is the most dead simple way to do this?
My professor showed us a trick for problems like this which I was able to use for the following simple example:
$$
\int \frac{1}{1+\sqrt{2x}}dx
$$
Substituting:
$u-1=\sqrt{x}$
be... | if $t=\sqrt{x}-3$, $x=\left(t+3\right)^2$ then $\dfrac{dt}{dx}=\frac{1}{2\sqrt{x}}\implies dt=\frac{dx}{2\sqrt{x}}\implies2\sqrt{x}dt=dx$
so$$\frac{\sqrt{x}}{\sqrt{x}-3}dx=\frac{\sqrt{x}}{\sqrt{x}-3}d2\sqrt{x}dt=2\frac{x}{t}dt=2\frac{\left(t+3\right)^2}{t}dt=2\left[\frac{t^2+6t}{t}+\frac{9}{t}\right]dt=2\left[t+6+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Prove for all $x \in \mathbb{R}$ at least one of $\sqrt{3}-x$ and $\sqrt{3}+x$ is irrational I have attempted to do some work on this, but I'm not sure if I'm heading in the correct direction. If I am heading in the correct direction, then I'm not quite sure where to take it from here.
Roughly what I have so far is:
W... | The sum of two rational numbers is rational, but
$$(\sqrt 3 -x)+(\sqrt 3+x)=2\sqrt 3$$
which is irrational. Therefore they cannot both be rational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If we know range of a function , how can construct range of other function?
If we know $\frac{-1}{2}\leq \frac{x}{x^2+1}\leq \frac{1}{2}$ how can obtain the range of the function below?
$$y=\frac{4 x}{9x^2+25}$$
The problem is what's the range with respect to $\frac{-1}{2}\leq \frac{x}{x^2+1}\leq \frac{1}{2}$
| $$y=\frac{4 x}{9x^2+25}\\y=\frac{4 x}{25(\frac{9}{25}x^2+1)}=\\y=\frac4{25}\frac{ x}{((\frac{3}{5}x)^2+1)}$$ now take $u=\frac{3x}{5}$
$$y=\frac4{25}\frac{ x}{((\frac{3}{5}x)^2+1)}=\\\frac4{25}\frac{ \frac{5}{3}u}{((u)^2+1)}=\\\frac{4}{15}\frac{u}{u^2+1}\\$$you know $\frac{-1}{2}\leq \frac{u}{u^2+1}\leq \frac{1}{2}$ so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find p is the prime number which $\frac{p+1}{2}$ and $\frac{p^2+1}{2}$ both are square number. Find p is the prime number which $\dfrac{p+1}{2}$ and $\dfrac{p^2+1}{2}$ both are square number.
I do not know how to use "p is prime" assumption given. I just know
$p=7$ is satisfied.
If $\dfrac{p+1}{2}=X^2$ and $\dfrac{p^2+... | \begin{align}
& p=2{x}^{2}-1 \\
& {{p}^{2}}=2{{y}^{2}}-1 \\
& \Rightarrow {{p}^{2}}-p=2{{y}^{2}}-2{{x}^{2}} \\
& \Rightarrow p(p-1)=2(y-x)(y+x)
\end{align}
Wlog, assume $x,y$ are positive and it is easy to see $p>2$.
Then, from $(4)$, we have $y> x$ and:
a) $p|y-x$ and $2(y+x)|p-1$
we get $p\leq y-x$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Basis for Kernel and Image of a linear map I have to find the basis for the kernel and image of the following linear map.
$ \phi: R^3 → R^2, ϕ \begin{pmatrix} \begin{pmatrix} x \\y \\z \end{pmatrix}\end{pmatrix}= \begin{pmatrix} x -y \\z \end{pmatrix} $
For the range, I think we can express any arbitrary linear transfo... | Everything looks good. A slightly more elementary (but less insightful) approach that works quite well for calculations that are a little harder to "guess" as you say is to form the matrix for the linear transformation by checking the basis:
$f(1,0,0)=(1,0), \, f(0,1,0)=(-1,0) \, , f(0,0,1)=(0,1)$
So, we get the matrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find all n for which $2^n + 3^n$ is divisible by $7$? I am trying to do this with mods. I know that:
$2^{3k+1} \equiv 1 \pmod 7$, $2^{3k+2} \equiv 4 \pmod 7$, $2^{3k} \equiv 6 \pmod 7$ and $3^{3k+1} \equiv 3 \pmod 7$, $3^{3k+2} \equiv 2 \pmod 7$, $3^{3k} \equiv 6 \pmod 7$, so I thought that the answer would be when $... | We need $$2^n+3^n\equiv0\pmod7\iff3^n\equiv-2^n\iff-1\equiv(3\cdot4)^n\equiv(-2)^n$$ using $2\cdot4\equiv1\pmod7\iff2^{-1}\equiv4$
$\implies(-2)^n\equiv-1\equiv(-2)^3\iff(-2)^{n-3}\equiv1$
$\implies n-3\equiv0\pmod6$ as $(-2)^3\equiv-1,(-2)^6\equiv1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Why isn't there a vertical asymptote at $x = -1$ for $f(x) = \frac{(x^3-1)(\ln(x+2))}{2x^3+3x^2-2x-3}$ Why isn't there a vertical asymptote at $x = -1$ for $f(x) = \frac{(x^3-1)(\ln(x+2))}{2x^3+3x^2-2x-3}$
I understand that there is a removable discontinuity at x = 1 since $x^3 - 1 = (x-1)(x^2+x+1)$ but I don't know ho... | hint
The denominator is
$$(2x+3)(x-1)(x+1) $$
and we use
$$\lim_{x\to-1}\frac {\ln (x+2)}{x+1}=$$
$$\lim_{X\to 0}\frac {\ln (X+1)}{X}=1$$
we find that
$$\lim_{x\to-1}f (x)=\frac {-2}{-2}=1$$
Thus, there is No vertical asymptote at $x=-1$ since the limit is finite.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove fixed point iteration of $g(x) = \frac{1}{2}x + \frac{A}{2x}$ converges to $\sqrt{A}$ If we define, for any positive number $A$:
\begin{align*}
g(x) &= \frac{1}{2}x + \frac{A}{2x} \\
g'(x) &= \frac{1}{2} - \frac{A}{2x^2} \\
\end{align*}
And we define the normal fixed point iteration sequence $\{ p_n \}_{n=0}^... | I am not sure that this could be an answer.
$$g(x) = \frac{1}{2}x + \frac{A}{2x}=x -\frac{x^2-A}{2x}$$
$$x_{n+1} = x_n -\frac{x_n^2-A}{2x_n}$$ is the Newton formula for solving $x^2-A=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2455260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What is $\lim _{ x\rightarrow 0 } \frac { f(x) -x }{ x^2 }$? Given that
$$f(x)=8x-f(3x)-\sin^2(2x),$$
find
$$\lim _{ x\rightarrow 0 } \frac { f\left( x \right) -x }{ x^2 }$$
| Note that
$$ f(x)=f(0)+f'(0)x+\frac12f''(0)x^2+O(x^3). $$
Then from $f(x)=8x-f(3x)-\sin^2(2x)$, one has
\begin{eqnarray}
&&\bigg[f(0)+f'(0)x+\frac12f''(0)x^2+O(x^3)\bigg]+\bigg[f(0)+3f'(0)x+\frac{9}2f''(0)x^2+O(x^3)\bigg]\\
&=&8x-\sin^2(2x)
\end{eqnarray}
which gives
$$ 2f(0)+4f'(0)x+5f''(0)x^2+O(x^3)=8x-\sin^2(2x). $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2455880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Then find the value of ... Let $a, \: b, \: c$ be three variables which can take any value (Real or Complex). Given that $ab + bc + ca = \frac{1}{2}$; $a + b + c = 2$; $abc = 4$. Then find the value of $$\frac{1}{ab + c - 1} + \frac{1}{bc + a - 1} + \frac{1}{ac + b - 1}$$
I always get stuck in such problems. Please giv... | Another way is to find a cubic $p(x)$ with roots $ab+c, bc+a, ca+b$ (which is easier than the original problem) and then use simple transformations. Here using elementary symmetric polynomials and Vieta, we get $p(x) = x^3-\frac52x^2- \frac52x-\frac{65}4$.
Now get the polynomial $p_1(x)$ which has roots $ab+c-1, bc+a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2456096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Binomial theorem relating proof There is this identity
$$1 -\frac{1}{2}\binom{n}{1}+\frac{1}{3} \binom{n}{2}- \frac{1}{4}\binom{n}{3}+....+(-1)^n \frac{1}{n+1}\binom{n}{n}$$
And we are supposed to prove it using these two identities
$$k\binom{n}{k} = n\binom{n-1}{k-1}$$
and
$$\binom{n}{0} + \binom{n}{1} + \binom{n}{... | \begin{eqnarray}
&&1 - \frac{1}{2}\binom{n}{1}+\frac{1}{3} \binom{n}{2}- \frac{1}{4}\binom{n}{3}+....+(-1)^n \frac{1}{n+1}\binom{n}{n}\\
&=&\sum_{k=0}^n(-1)^{k}\frac{1}{k+1}\binom{n}{k}\\
&=&\sum_{k=0}^n(-1)^{k}\frac{1}{n+1}\binom{n+1}{k+1}\\
&=&\frac{1}{n+1}\sum_{k=1}^{n+1}(-1)^{k}\binom{n+1}{k}\\
&=&-\frac{1}{n+1}\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2456255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $f(1)=1$, then is it true that $f(n)=n$ for all $n \in \mathbb{N}\cup\{0\}$. Let $f:\mathbb{N}\cup\{0\}\to\mathbb{N}\cup\{0\}$ be a function which satisfies $f(x^2+y^2)=f(x)^2+f(y)^2$ for all $x,y \in\mathbb{N}\cup\{0\}$. It' easy to see that $f(0)=0$ and $f(1)=0$ or $f(1)=1$. Suppose let's assume that $f(1)=1$. The... | Yes.
Suppose you already know that $f(n)=n$ for all $n < N$. You want to find $a,b,c$, all smaller than $N$, such that $N^2+a^2=b^2+c^2$. If you find such a triplet, you immediately conclude that $f(N)=N$, and you have the necessary inductive step.
For odd $N>6$, use
$N^2+\left(\frac{N-5}{2}\right)^2 = (N-2)^2+\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2456751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
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How to prove that $x^y - y^x = x + y$ has only one solution in positive integers? Prompted by this question, I tried to show that $(2,5)$ is the only solution in positive integers of $x^y - y^x = x+y$ (which would show, a fortiori, that it's the only solution in primes). It's convenient to rewrite the equation as $f(x,... |
If $(x,y) = (2,6)$ then $f(x,y) = 20$, and as $y$ increases above $6$, $f(x,y)$ increases.
To prove this rigorously, let us prove by induction that $$2^{X-1}\gt X+1\tag1$$ for $X\ge 6$.
The base case : $2^{6-1}=32\gt 7=6+1$.
Supposing that $(1)$ holds for some $X$ gives
$$2^{(X+1)-1}=2\cdot 2^{X-1}\gt 2(X+1)=X+X+2\gt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2457087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
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Limit of $\sqrt{x^2+3x}+x$ when $x\to-\infty$ Limit of $ \lim_{x\to -\infty}(\sqrt{x^2+3x}+x)$, I know that the final answer is $-3/2$, my question is about Wolfram Alpha step by step solution:
$$x+\sqrt{x^2+3x}=\frac{(x+\sqrt{x^2+3x})(x-\sqrt{x^2+3x})}{x-\sqrt{x^2+3x}}$$
$$=-\frac{3x}{x-\sqrt{x^2+3x}}$$
$$\lim_{x\to-\... | Yes, Wolfram is free to do that because for all finite $x$,
$$\frac{f(x)}{g(x)}=\frac{xf(x)}{xg(x)}$$ by "unsimplification", so that the limits are the same.
You can even write a "modified L'Hospital rule" theorem saying
$$\lim\frac fg=\lim\frac{f'}{g'}=\lim\frac{f+xf'}{g+xg'},$$ if that has any use.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2457183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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How do I find details of a parabola from its general two-degree equation? This general equation in two-degree represents a parabola: $$(ax + by)^2 + 2gx + 2fy + c = 0$$
How do I find the following from this equation:
*
*Vertex
*Focus
*Axis
*Length of Latus Rectum
*Co-ordinates of end points of Latus Rectum
*Equ... | $y'=bx-ay$ is parallel to the directrix and $x'=ax+by$ parallel to the axis. This we can see from comparing to the form I mentioned in the comments. So let these be the new coordinates. The inverse is $x=\frac{ax'+by'}{a^2+b^2}$, $\frac{bx'-ay'}{a^2+b^2}$, transforming your equation into $$x'^2+2g\frac{ax'+by'}{a^2+b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2457587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $2\log_bx + 2\log_b(1-x) = 4$ I need to solve $$2\log_bx + 2\log_b(1-x) = 4.$$
I have found two ways to solve the problem. The first (and easiest) way is to divide through by $2$:
$$\log_bx + \log_b(1-x) = 2.$$ Then, combine the left side: $$\log_b[x(1-x)] = 2,$$ and convert to the equivalent exponential form, $$... | You can see from the equation itself that are are, in general, two solutions: If $x$ is a solution, then so is $1-x$, since $1-(1-x)=x$.
As for the two spurious solutions $x=(1\pm\sqrt{1+4b^2})/2$ in your second approach, note that for these two, either $x$ or $1-x$ is negative. If the equation were written in the s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2458109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Proof that $(abc)^2 = \frac{(c^6-a^6-b^6)}{3}$ We are given the length of the sides of a right triangle T, where $a \leq b \leq c$. We are asked to prove if $T$ is a right triangle. then $(abc)^2 = (c^6-a^6-b^6)/3$.
What I tried:
I tried substituting with the Pythagorean theorem $a^2 + b^2 = c^2$.
Where I am stuck:
I a... | Hint : Cube the equation $\color{blue}{a^2+b^2=c^2}$
\begin{eqnarray*}
a^6+b^6+\color{red}{3} a^2 b^2 (\color{blue}{a^2+b^2}) =c^6.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2459375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How would one prove that for all positive real $a$ and $b$, $\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}$? I think that the best way to prove this would be to prove by contradiction. Am I right?
If so, can I just assume that $\sqrt{a+b} = \sqrt{a} + \sqrt{b}$ and say $a = 1$ and $b = 1$ and show that $\sqrt{1+1} \neq \sqrt{1} ... | If
$\sqrt{a + b} = \sqrt a + \sqrt b, \tag 1$
then, squaring,
$a + b = a + 2\sqrt a \sqrt b + b, \tag 2$
whence
$2\sqrt a \sqrt b = 0, \tag 3$
which forces
$a = 0 \vee b = 0. \tag 4$
Contradiction!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2459873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 0
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Finding the value of the definite integral $\int_0^2{x\int_x^2{\frac{dy}{\sqrt{1+y^3}}}}dx$ If $$f(x) = \int_x^2{\frac{dy}{\sqrt{1+y^3}}}$$
then find the value of $$\int_0^2{xf(x)}dx$$
I have no idea how to solve this question. Please help.
| Using integration by parts
$$\int_0^2xf(x)dx=\frac{1}{2}\int_{x=0}^{x=2}f(x)d(x^2)$$
$$=\frac{1}{2}x^2f(x)\bigg|_0^2-\frac{1}{2}\int_0^2x^2f'(x)dx$$
$$=2f(2)+\frac{1}{2}\int_0^2x^2\frac{1}{\sqrt{1+x^3}}dx$$
$$=0+\frac{1}{3}\sqrt{1+x^3}\bigg|_0^2$$
$$=\frac{1}{3}(3-1)$$
$$=\frac{2}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2460221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding integral solutions of $x+y=x^2-xy+y^2$
Find integral solutions of $$x+y=x^2-xy+y^2$$
I simplified the equation down to
$$(x+y)^2 = x^3 + y^3$$
And hence found out solutions $(0,1), (1,0), (1,2), (2,1), (2,2)$ but I dont think my approach is correct . Is further simplification required? Is there any other met... | $x+y=x^2-xy+y^2\to y^2-(x+1) y+x^2-x=0$
$y=\dfrac{1}{2} \left(1+x\pm\sqrt{-3 x^2+6 x+1}\right)$
discriminant must be positive $\Delta=-3 x^2+6 x+1\geq 0\to \dfrac{1}{3} \left(3-2 \sqrt{3}\right)\leq x\leq \dfrac{1}{3} \left(3+2 \sqrt{3}\right)$
which for integer $x$ means, $0\leq x \leq 2$
For $x=0$ we get $y=0;\;1$ s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2460474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proving $\frac{x+1}{y+1}$ is equal to or less than $\frac{x-1}{y-1}$ I know that $\frac{x+1}{y+1}$ is equal to (when $x = y$) or less than $\frac{x-1}{y-1}$. Suppose we cross multiply, then
*
*$xy - x+ y -1$ is less than or equal to $xy + x - y -1$
So,
*$y- x$ is less than or equal to $x- y$.
But when $y$ is greate... | You have
$$
f(x,y)=\frac{x+1}{y+1}-\frac{x-1}{y-1}=
\frac{(xy+y-x-1)-(xy+x-y-1)}{(y^2-1)}=
\frac{2(y-x)}{(y^2-1)}
$$
Then
$$
f(x,y)\begin{cases}
>0 & \text{if $y>x$ and $|y|>1$} \\[4px]
>0 & \text{if $y<x$ and $|y|<1$} \\[4px]
<0 & \text{if $y>x$ and $|y|<1$} \\[4px]
<0 & \text{if $y<x$ and $|y|>1$}
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2462867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Combinatorics - Hexadecimals, how many possible numbers.... with ascending order?
How many different numbers with 4 figures can you make out of the sixteen hexadecimal ’figures’ $\{0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F\}$
$16^4$ possible solutions
(i) How many numbers are ’real’ 4-figure numbers, meaning the first figure... |
How many $4$-character strings can be formed using the sixteen hexadecimal digits?
Your answer $16^4$ is correct?
How many $4$-digit numbers can be formed using the sixteen hexadecimal digits?
Your answer $16^4 - 16^3$ is correct.
How many $4$-digit hexadecimal numbers have distinct digits?
We have $15$ choices f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2463404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Why does the discriminant in the Quadratic Formula reveal the number of real solutions? Why does the discriminant in the quadratic formula reveal the number of real solutions to a quadratic equation?
That is, we have one real solution if
$$b^2 -4ac = 0,$$
we have two real solutions if
$$b^2 -4ac > 0,$$
and we have no ... | Because for $a\neq0$ we obtain:
$$ax^2+bx+c=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}\right).$$
Now, we see that if $b^2-4ac<0$ then $\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}>0$,
which says that the equation $ax^2+bx+c=0$ has no solutions.
For $b^2-4a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2464409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "59",
"answer_count": 10,
"answer_id": 0
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Find $\int\sqrt{t^{3/2}+1}\,dt$
Integrate $\int\sqrt{t^{3/2}+1}\,dt$
I tried all possible substituion but its not worked.
Wolfram alpha also gives something different answer
https://www.wolframalpha.com/input/?i=integrate+(%E2%88%9A(t%E2%88%9At%2B1))
Can someone give me just a hint please?
| When $|t|\leq1$ ,
$\int\sqrt{t^\frac{3}{2}+1}~dt$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^\frac{3n}{2}}{4^n(n!)^2(1-2n)}dt$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^\frac{3n+2}{2}}{4^n(n!)^2(1-2n)\dfrac{3n+2}{2}}+C$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^\frac{3n+2}{2}}{2^{2n-1}(n!)^2(1-2n)(3n+2)}... | {
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Quadrilateral in Square
$S$ is a unit square. Four points are taken randomly, one on each side of $S$. A quadrilateral is drawn. Let the sides of this quadrilateral be $a,b,c,d$. Prove that $2\leq{}a^2+b^2+c^2+d^2\leq{}4$.
My Efforts:
Let
$\begin{align}m^2+t^2&=a^2&\mathfrak{a}\\n^2+o^2&=b^2&\mathfrak{b}\\p^2+q^2&=... | For the lower bound, note by CS inequality, $$(a^2+b^2+c^2+d^2)\cdot8=(m^2+n^2+\cdots)(1+1+\cdots)\geqslant(m+n+\cdots)^2=4^2$$
| {
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Prove identity in a triangle I want to show that if $ABC$ is a triangle then
$$\sin^2(A/2)+ \sin^2(B/2) + \sin^2(C/2) =1-2\sin(A/2) \sin(B/2) \sin(C/2)$$
Well I eventually got it after much algebra, but I am looking for a shorter solution, or maybe even a geometric one?
| Let $\alpha=\pi-2A$, $\beta=\pi-2B$ and $\gamma=\pi-2C$.
Thus, $\alpha+\beta+\gamma=\pi$ and we need to prove that
$$\cos^2\alpha+\cos^2\beta+\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma=1,$$
which is obvious for acute-angled triangle $ABC$ (it's just law of cosines for new triangle).
In the general case we obtain:
$$\c... | {
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Solve Recurrence relation -:$T(n)=T(n-1)+6n^{2}+2n$
Solve Recurrence relation -:$T(n)=T(n-1)+6n^{2}+2n$
Base case $T(1)=8$
$T(n)=T(n-1)+6n^{2}+2n$
$T(n-1)=T(n-2)+6(n-1)^{2}+2(n-1)............(1)$
$T(n-2)=T(n-3)+6(n-2)^{2}+2(n-2)...........(2)$
Substituting $(1)\,\, \text{and} \,\,(2) \text{in our question},$
$T(n)=T... | I think it should be $$T(n)=T(1)+\sum_{k=2}^n(6k^2+2k)=T(1)+\sum_{k=1}^n(6k^2+2k)-8=$$
$$=T(1)+6\cdot\frac{n(n+1)(2n+1)}{6}+2\cdot\frac{n(n+1)}{2}-8=$$
$$=6\cdot\frac{n(n+1)(2n+1)}{6}+2\cdot\frac{n(n+1)}{2}$$
| {
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If $a+b+c=0$ prove that $ (a^2+b^2+c^2) \times (a^2+b^2+c^2) = 2(a^4+b^4+c^4)$
If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove
$$(a^2+b^2+c^2) \times (a^2+b^2+c^2) = 2(a^4+b^4+c^4)$$
What is a good way to do this?
This question came from answering this slightly harder question. Those answers were somewhat hard to unde... | Let $e_1 = a+b+c, e_2 = ab+bc+ac, e_3 = abc$.
Since $A = (a^2+b^2+c^2)^2-2(a^4+b^4+c^4)$ is symmetric and homogeneous of degree $4$, it is a linear combination of $e_1^4, e_1^2e_2, e_1e_3, e_2^2$, and when $e_1=0$ the only nonzero one is possibly $e_2^2$ :
There is a coefficient $k$ such that if $e_1=0$ then $A = ke_2... | {
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On the way finding limit.
Find The Limit $$\lim_{x\to0}\left(\frac{\tan(x)}{x}\right)^{\displaystyle\frac{1}{x^{2}}}$$
My Approach
We know that $$\lim_{x\to0}\frac{\tan(x)}{x}
= 1$$
So $$\lim_{x\to0}
\left[ 1+\left(\frac{\tan(x)}{x}-1\right)\right]
^{\displaystyle\left(\frac{1}{\left(\frac{\tan(x)}{x}-1\right)}\right... | $$\lim_{x\to0}\left(\frac{\tan(x)}{x}\right)^{\frac{1}{x^{2}}}=e^{\lim_{x\to0}\ln\left(\left(\frac{\tan(x)}{x}\right)^{\frac{1}{x^{2}}}\right)}=e^{\lim_{x\to0}\frac{\ln\left(\frac{\tan(x)}{x}\right)}{x^2}}=e^{\frac{0}{0}}\\\lim_{x\to0}\frac{\ln\left(\frac{\tan(x)}{x}\right)}{x^2}=\lim_{x\to0}-\dfrac{\tan\left(x\right)-... | {
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Inductive proof for recursive formula So, I have a recursion in which $$a_0 = 5$$ $$a_1 = 1$$ $$a_{n+2} = a_{n+1} + 2a_n$$
I should then prove by induction that the formula $a_n = 2^{n+1} + 3(-1)^n$ works for every number.
Anyway, I generally know how to use induction as proof but doesn't really have a clue on how to ... | Start with
$a_{n+2} = a_{n+1} + 2a_n$
and substitute
the induction assumption
$a_n = 2^{n+1} + 3(-1)^n
$
for $n$ and $n+1$.
This gives
$\begin{array}\\
a_{n+2}
&= a_{n+1} + 2a_n\\
&= (2^{n+2} + 3(-1)^{n+1}) + 2(2^{n+1} + 3(-1)^n)\\
&= 2^{n+2} + 3(-1)^{n+1} + 2^{n+2} + 6(-1)^n\\
&= 2^{n+3} + 3(-1)^{n+1} - 6(-1)^{n+1}\... | {
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Prove that $\int_0^\infty \frac{x\,\log x}{(1+x^2)^2} = 0$ I have been asked to prove that $$\int_0^\infty \frac{x\,\log x}{(1+x^2)^2}\,dx = 0$$
I tried proceeding as follows:
\begin{align}
\int_0^\infty \frac{x\,\log x}{(1+x^2)^2}\,dx &= \int_0^1 \frac{x\,\log x}{(1+x^2)^2}\,dx + \int_1^\infty \frac{x\,\log x}{(1+x^2)... | $I$ converges because
$$\dfrac{x\ln x}{(1+x^2)^2}\le \dfrac{x^2}{(1+x^2)^2}\le\dfrac{1}{1+x^2} ~~~x>1$$
and $$ \lim_{x\to 0}\dfrac{x\ln x}{(1+x^2)^2} = 0$$
that is $$x\mapsto \dfrac{x\ln x}{(1+x^2)^2}~~~\text{is continuous on [0,1]}$$
Now Set $x=\frac{1}{u}$ that is $dx = -\frac{1}{u^2}$
$$I=\int_0^\infty\dfrac{x\ln ... | {
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Find number of positive integer solutions Find number of positive integer solutions $(x,y,z)$ for the following equation:
$19x + 11y + 8z = 240$
I divided the equation by $8$ and then tried to equate remainders.
It yields that $3(x + y) = 8k$ for some constant $k$ or $x + y$ is a multiple of 8.
Can't choose which co... | I had a very similar method, which takes advantage of $19-11=8$ and also the factor $8$ generally. Rewrite as $$11(x+y)+8(x+z)=240=8\times 30$$
$x+y$ must be divisible by $8$ and the only possibilities are $x+y=8, 16$. In the first of these cases $x+y=8$ has seven solutions. In the second we find $x+z=8 (=30-2\times 11... | {
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Solve identity: $\frac {\sin x}{1-\cos x} = \frac {1+\cos x}{\sin x}$ $$\frac {\sin x}{1-\cos x} = \frac {1+\cos x}{\sin x}$$
The only way I can see of doing this is by cross multiplying but isn't that not allowed when trying to prove something?
| It's $$\sin^2x=1-\cos^2x$$ or
$$\sin^2x+\cos^2x=1.$$
Also, we have:
$$\frac {\sin x}{1-\cos x}=\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\sin^2\frac{x}{2}}=\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}}=\frac{2\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}=\frac{1+\cos{x}}{\sin{x}}$$
| {
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Check solution of recurrence relation $a_0 = 3$, $a_1 = 7$, $a_n = 3a_{n-1} - 2a_{n-2}$ for $n \geq 2$ $a_0 = 3, a_1 = 7, a_n = 3a_{n-1} - 2a_{n-2}$ for $n \geq 2$
I know I've got the wrong answer because my outputs don't match. So if someone could show me where I'm going wrong I would be super grateful.
Let $ y = \sum... | Alternatively:
$$a_n = 3a_{n-1} - 2a_{n-2} \Rightarrow a_n-2a_{n-1}=a_{n-1}-2a_{n-2}.$$
Change: $b_n=a_n-2a_{n-1}, b_0=1$ to get:
$$b_n=b_{n-1} \Rightarrow b_n=1.$$
Hence:
$$a_n-2a_{n-1}=1, a_0=3 \Rightarrow a_n=2^{n+2}-1.$$
| {
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Let $r$ be a root of the polynomial $p(x) = (\sqrt{5} - 2\sqrt{3})x^3 + \sqrt{3}x - \sqrt{5} + 1$. Find another polynomial $q(x)$ with integer coefficients such that $q(r) = 0$.
I have no clue how to do this question. Can't use rational root theorem and I see no feasible way to get the roots of $p(x)$. Any help would ... | Start by collecting all the like-terms with a particular radical and moving them to the left-hand-side; I chose $\sqrt{3}$ first. Move all terms without $\sqrt{3}$ to the right-hand-side. Then, factor out the chosen radical. This is shown in $(1)$. Next, square both sides, as shown in $(2)$. Since all terms with $\sqrt... | {
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What is the sum of $E=\frac{1}{3}+\frac{2}{3^{2}}+\frac{3}{3^{3}}+\frac{4}{3^{4}}+...$ What is the right way to assess this problem?
$$E=\frac{1}{3}+\frac{2}{3^{2}}+\frac{3}{3^{3}}+\frac{4}{3^{4}}+...$$
To find the value of the sum I tried to use the fact that it could be something convergent like a geometric series. H... | \begin{eqnarray}
E&=&\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+\ldots=\sum_{k=1}^\infty\frac{k}{3^k}=\frac{1}{3}\sum_{k=1}^\infty k\cdot\left(\frac{1}{3}\right)^{k-1}\\
&=&\frac{1}{3}\dfrac{d}{dx}\left(\sum_{k=0}^\infty x^k\right)\Big|_{x=\frac{1}{3}}=\frac{1}{3}\frac{d}{dx}\left(\frac{1}{1-x}\right)\Big|_{... | {
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Evaluate $\oint_C \dfrac{dz}{z^3 - 1}$, where $C$ is in the circle $|z + 1| = \dfrac{3}{2}$: Discrepancy Between Solutions. I have the following complex analysis contour integration problem:
Evaluate $\oint_C \dfrac{dz}{z^3 - 1}$, where $C$ is in the circle $|z + 1| = \dfrac{3}{2}$.
There is a discrepancy between my ... | First of all, what you say the instructor has makes no sense and the arithmetic in it is not correct. What is missing is a sum (with the $2\pi i$ factored out). Note that your residue at the primitive cube root of unity $\omega = -\dfrac12+i\dfrac{\sqrt3}2$ is $\dfrac 1{3\omega^2} = \dfrac13 \omega$. Adding this to its... | {
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Find $\alpha + \beta$ given that: $\alpha^3-6\alpha^2+13\alpha=1$ and $\beta^3-6\beta^2+13\beta=19$ , ( $\alpha , \beta \in \Bbb R)$
Find $\alpha + \beta$ given that: $\alpha^3-6\alpha^2+13\alpha=1$ and $\beta^3-6\beta^2+13\beta=19$ , ( $\alpha , \beta \in \Bbb R)$
I have a solution involving a variable change ($\alp... | No change of variables:
Adding the two equations and using $$\alpha^2 + \beta^2 = (\alpha+\beta)^2-2\alpha\beta\\
\alpha^3+\beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$$gives the equation
$$(\alpha + \beta)^3 - 6(\alpha + \beta)^2 + (13-3\alpha\beta)(\alpha + \beta) + 12\alpha\beta - 20$$
meaning that $\... | {
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Prove by induction that $1^1\cdot 2^2\cdot \dots\cdot n^n\leq \left(\frac{2n+1}{3}\right)^{\frac{n(n+1)}{2}}.$ How can I prove by induction that for every natural number $n$,
$$1^1\cdot 2^2\cdot \dots\cdot n^n\leq \left(\frac{2n+1}{3}\right)^{\frac{n(n+1)}{2}}.$$
| Hint. Instead of induction, I warmly recommend the use of the AGM inequality. Note that
$$\left(1^1\cdot 2^2\cdot \dots\cdot n^n\right)^{\frac{2}{n(n+1)}}$$
is the geometric mean of the numbers
$$1,2,2,3,3,3,\dots,\underbrace{n,\dots,n}_{\text{$n$ times}}.$$
What is their arithmetic mean?
$$\frac{1+2+2+3+3+3+\dots+\ove... | {
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Let $\alpha $ is a root of $x^2+3x+5=0$ then express Let $\alpha $ is a root of $x^2+3x+5=0$ then express $\alpha^4 + 5\alpha^3 + 7\alpha^2 + 8\alpha -9$ as a linear expression of $\alpha $.
My Attempts:
since $\alpha $ is a root of $x^2+3x+5=0$
We have,
$$\alpha^2 + 3\alpha +5=0$$.
..
| Because $$x^4+5x^3+7x^2+8x-9=$$
$$=x^2(x^2+3x+5)+2x(x^2+3x+5)-4(x^2+3x+5)+10x+11.$$
Thus, since $\alpha^2+3\alpha+5=0$, we got that our expression it's $10\alpha+11$.
Also, since a degree of the remainder should be less that $2$ we must get a linear expression of $\alpha$.
| {
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Rationalize denominator: $\frac{1}{\sqrt[3]{5}+\sqrt[3]{1}+\sqrt[3]{6}}$ $$\frac{1}{\sqrt[3]{5}+\sqrt[3]{1}+\sqrt[3]{6}}$$
So this is what I thought: the square root of 1 is obviously one, so I have $1^3 +(\sqrt[3]{5} + \sqrt[3]{6})$. In my head I see that this is the first part for the sum of cubes formula. I multipli... | Since $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
$$\frac{1}{\sqrt[3]{5}+\sqrt[3]{1}+\sqrt[3]{6}}=\frac{\sqrt[3]{5^2}+1+\sqrt[3]{36} - \sqrt[3]{5}-\sqrt[3]{6}-\sqrt[3]{30}}{12-3\sqrt[3]{30}}$$
and since $a^3-b^3=(a-b)(a^2+ab+b^2)$$$\frac{1}{\sqrt[3]{30}-4}=\frac{\sqrt[3]{30^2}+4\sqrt[3]{30}+16}{30-64}$$
you just... | {
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Calculus inequality involving sine and cosine community! I saw the following inequality in a calculus assignment, which I thought was harder to prove than I expected: For every $x\in\mathbb{R},$$$(\sin x + a \cos x)(\sin x + b \cos x)\leq 1+(\frac{a+b}{2})^2.$$
By means of a graphing device, I noticed that when $a=b$, ... |
For any trig0nometric expression of the form $$A\sin x+B\cos x=\sqrt{A^2+B^2}\left(\dfrac{A}{\sqrt{A^2+B^2}}\sin x+\dfrac{B}{\sqrt{A^2+B^2}}\cos x\right)$$
can be written in the form $\sqrt{A^2+B^2}\sin(x+\phi)$ with $\tan\phi=B/A$ and hence $\vert A\sin x+B\cos x|\le\sqrt{A^2+B^2}$ for any $A,B\in\Bbb{R}.$
In fa... | {
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"source": "stackexchange",
"question_score": "2",
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Distribute n integers into m (non distinct) bins without creating consecutive triples I have $[n] := \{0, 1, \cdots, n-1\}$ integers and $m$ bins (we don't distinguish bins). How many possible ways to distribute $m$ integers from $[n]$ into $m$ bins are there such that the distributions have no consecutive triple?
In o... | Let $A(n,m)$ be the number of ways to choose $m$ integers from $[n]$ without a consecutive triple. Clearly $A(n,m) = \binom{n}{m}$ when $m < 3$, and $A(n,m) = 0$ when $m > \left\lceil\frac23 n\right\rceil$.
For larger $m$ it might be easier to invert the question. How many subsets $S$ of size $m$ are there which do hav... | {
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Justify the recurrence relation $a_{n+2}=2a_{n+1}+a_n$ and find $a_n$ Let $a_n$ be the number of ways to color the squares of a $1$ x $n$ chessboard using the colors red, white, and blue, so that no red square is adjacent to a white square. Justify the relation $a_{n+2}=2a_{n+1}+a_n$ (for certain $n$), and then find $a... | Generating Function Approach
$
\begin{array}{cl}
\text{function}&\text{meaning}\\
x&\text{blue}\\
\frac{x^2}{1-x}&\text{blue followed by one or more red}\\
\frac{x^2}{1-x}&\text{blue followed by one or more white}\\
\frac1x&\text{remove initial blue}
\end{array}
$
The generating function is
$$
\begin{align}
\frac1x\sum... | {
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Find $\lim_{x\rightarrow0}\frac{3^x-5^x}{4^x-10^x}$ Find $$\lim_{x\rightarrow0}\frac{3^x-5^x}{4^x-10^x}$$
My work so far:
$$\lim_{x\rightarrow0}\frac{3^x-5^x}{4^x-10^x}=\frac{\ln3-\ln5}{\ln4-\ln10}$$
Is correct?
Add:
I used $a^x\sim 1+x\ln a$ for $x\rightarrow 0$
| L'Hospital's rule for $\frac{0}{0}$ limits works well in this case. An alternate is to use
$$a^{x} = e^{x \, \ln(a)} = 1 + \ln(a) \, x + \frac{\ln^{2}(a)}{2!} \, x^{2} + \mathcal{O}(x^{3})$$
which yields
\begin{align}
\frac{a^{x} - b^{x}}{c^{x} - d^{x}} &= \frac{(\ln(a) - \ln(b)) \, x + \frac{1}{2} \, (\ln^{2}(a) - \l... | {
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Inverse Z-Transform of $(1+2/z)^{-3}$ Working through the K.A Stroud Advanced Engineering Mathematics Textbook on my own, and have really got stuck on this question with no one to help out.
The question and my working can be found here
Basically, brought it down into 3 fractions, which I could transform up until the la... | $$\frac{z}{(z+2)^3}=\frac{z^{-2}}{(1+\frac{2}{z} )^3}=z^{-2}\sum_{k=0}^\infty \binom{k+2}{k}\left(-\frac{2}{z}\right)^k=
z^{-2}\sum_{k=0}^\infty \frac{1}{2}(k+1)(k+2)\left(-\frac{2}{z}\right)^k$$
$$\frac{z}{(z+2)^3}=\sum_{k=0}^\infty (-1)^k 2^{k-1}(k+1)(k+2)\frac{1}{z^{k+2}}$$
$k=n-2$
$$\frac{z}{(z+2)^3}=\sum_{n=-2}^\i... | {
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"answer_id": 0
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How to calculate $\lim_{n \to \infty} \sum_{k=1}^n\binom nk k!k\frac{1}{n^k}$? How to calculate this limit :
$$\lim_{n\rightarrow +\infty}u_n$$
with :
$$u_n=\sum_{k=1}^n\binom nk k!k\frac{1}{n^k}$$
We can write this :
$$u_n=\sum_{k=1}^n\frac{n}{n}\times\frac{n-1}{n}\times\cdots\times\frac{n-k+1}{n}\times k$$
But I can'... | Hint: The number $u_n$ has a nice telescoping property. Let's for example consider the case $n=5$.
\begin{align*}
\color{blue}{u_5}&=1+2\cdot\frac{4}{5}+3\cdot\frac{4}{5}\cdot\frac{3}{5}+4\cdot\frac{4}{5}\cdot\frac{3}{5}\cdot\frac{2}{5}
+\color{blue}{5}\cdot\frac{4}{5}\cdot\frac{3}{5}\cdot\frac{2}{5}\cdot\color{blue}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2499476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Why is $\sqrt x \times \sqrt y = \sqrt {xy}$ Sorry I do not know latex. Why is $\sqrt x \times \sqrt y = \sqrt {xy}$? It also applies for division, but why not addition and subtraction? i.e., why is $\sqrt x + \sqrt y$ not equal to $\sqrt {x+y}$?
Thank you for editing and answering.
| remember what $^2$ is:
$$u^2=u\cdot u\\\text{so: }\left(\sqrt{a}\cdot\sqrt{b}\right)^2=\sqrt{a}\cdot\sqrt{b}\cdot\sqrt{a}\cdot\sqrt{b}=\sqrt{a}\cdot\sqrt{a}\cdot\sqrt{b}\cdot\sqrt{b}=a\cdot b=\sqrt{ab}^2\implies\sqrt x \cdot\sqrt y = \sqrt {xy}\\\text{and you can do: }\left(\sqrt{a}+\sqrt{b}\right)^2=\left(\sqrt{a}+\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2501058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $(1+a)^7.(1+b)^7.(1+c)^7>7^7.a^4 b^4 c^4$ If a,b,c are positive prove that $(1+a)^7.(1+b)^7.(1+c)^7>7^7.a^4 b^4 c^4$
My approach
I tried (1+a)(1+b)(1+c) =1+[a+b+c+ab+bc+ac+abc]
Find the AM of Square Bracket, them AM $\ge$ GM
But if I add 1 then AM>GM. Please help me with my approach
| You almost did it.
(1+a)(1+b)(1+c) =1+a+b+c+ab+bc+ac+abc (Now subtract 1 and multiply and divide by 7)
= $\frac{7a+7b+7c+7ab+7bc+7ac+7abc}{7}\ge \sqrt[7]{(7)(7a)(7b)(7c)(7ab)(7bc)(7ac)(7abc)}$
add 1 on LHS
=>$1+\frac{7a+7b+7c+7ab+7bc+7ac+7abc}{7} > \sqrt[7]{(7)(7a)(7b)(7c)(7ab)(7bc)(7ac)(7abc)}$
Taking power 7 on bo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2501208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proof for an identity (from Ramanujan written) I saw an identity by Ramanujan
$$\forall n \in \mathbb{N} ,n>1 :\lfloor \sqrt n+\sqrt {n+2}+\sqrt{n+4} \rfloor=\lfloor \sqrt {9n
+17}\rfloor$$ I tried to prove it by limit definition . I post my trial below . If possible check my prove (right , wrong) ?
Then Is there more... | if $$|\sqrt n+\sqrt {n+2}+\sqrt{n+4} - \sqrt {9n
+17}|<\epsilon \\\Rightarrow \lfloor \sqrt n+\sqrt {n+2}+\sqrt{n+4} \rfloor=\lfloor \sqrt {9n
+17}\rfloor $$
so ,I will show $|\sqrt n+\sqrt {n+2}+\sqrt{n+4} - \sqrt {9n
+17}|<\epsilon$
$$\quad{|\sqrt n+\sqrt {n+2}+\sqrt{n+4} - \sqrt {9n
+17}|=\\|\sqrt n+\sqrt {n+2}+\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2501870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
let $f(\frac{x}{3})+f(\frac{2}{x})=\frac{4}{x^2}+\frac{x^2}{9}-2$ then find $f(x)$ let $f(\frac{x}{3})+f(\frac{2}{x})=\frac{4}{x^2}+\frac{x^2}{9}-2$
then find $f(x)$
My Try :
$$f(\frac{x}{3})+f(\frac{2}{x})=(\frac{2}{x})^2-1+(\frac{x}{3})^2-1$$
So we have :
$$f(x)=x^2-1$$
it is right ?Is there another answer?
| A plausible $f(x)$ is in fact $f(x)=x^2-1$. A proof of that is as follows.
Suppose $f(x)=x^2+h$ For $x=\sqrt6$ one has $\dfrac 2x=\dfrac x3$ so
$$2f(\frac{\sqrt6}{3})=2(\frac{\sqrt6}{3})^2-2=2((\frac{\sqrt6}{3})^2+h)\Rightarrow h=-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Probability that length of Randomly chosen chord of a circle Find Probability that length of Randomly chosen chord of a circle lies between $\frac{2}{3}$ and $\frac{5}{6}$ of its diameter.
My try: I assumed unit circle with center origin. Let two randomly chosen distinct points be $A(\cos \alpha, \sin \alpha)$ and $... | Let $r$ be the constant radius of the circle, let $L$ be the continuous random variable that represents the chord’s length, and let $\Theta$ be the continuous random variable that represents the counterclockwise angle measured from the first point chosen to the second point chosen.
As mentioned by others, per symmetry ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2503686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve $P(z)=0$, over complex field and factorise $P(z)=0$ over real field $P(z)=3z^4+10z^3+6z^2+10z+3$
The roots are $z=-3, -1/3, i,-i$ but I couldn't find $i,-i$ as the root.
Also the factorised version is meant to be $(z+3)(3z+1)(z^2+1)$ but I got something fabulous like $z^2(z+\frac{1}{z})(3z+\frac{3}{z}-10)$
I feel... | I like to write complete solution:
\begin{align}
P(z)
&= 3z^4+10z^3+6z^2+10z+3 \\
&= 3z^4+3 + 10z^3+10z+6z^2 \\
&= 3z^2\left(z^2+\dfrac{1}{z^2}\right)+10z^2\left(z+\dfrac{1}{z}\right)+6z^2 \\
&= z^2\Big[3\left(z^2+\dfrac{1}{z^2}\right)+10\left(z+\dfrac{1}{z}\right)+6\Big] \\
&= z^2\Big[3\left(z+\dfrac{1}{z}\right)^2+10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2505312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Inequality proof $\frac{2 \cdot a \cdot b }{a+b} < \frac {a+b}{2}$ This is very simple, but impossible to Google as you can understand.
I need to prove this inequality:
$$\frac{2 \cdot a \cdot b }{a+b} < \frac {a+b}{2}$$
where $ a < b$ and $a > 0$ and $b > 0$
I have tried to rework it to:
$$4 \cdot a \cdot b < {(a+b)}^... | We know that $(a-b)^2 > 0$ or $a^2 + b^2 -2ab > 0$
Adding $2ab$ term on both sides of the inequality we get
$a^2 + b^2 +2ab - 2ab > 2ab$
$a^2 + b^2 + 2ab > 4ab$
$(a+b)^2 > 4ab$
Equivalently
$\frac{(a+b)}{2} > \frac{2(ab)}{a+b}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2506815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove directly from the definition that $({1\over2}+\frac{1}{2^2}+...+\frac{1}{2^n})_n $ is Cauchy Prove directly from the definition that $({1\over2}+\frac{1}{2^2}+...+\frac{1}{2^n})_n$ is cauchy
I know from the definition of Cauchy that |$x_n$-$x_m$|< ϵ but how do you do this with |$\frac{1}{2^n}- \frac{1}{2^m}$|
wha... | To prove that the series satisfies the Cauchy condition you have to estimate, for $n,p\in\mathbb{N}$, the difference
$$
\sum_{j=n}^{n+p} \frac{1}{2^j} = \frac{1}{2^{n-1}}\left(1 - \frac{1}{2^{p+1}}\right) < \frac{1}{2^{n-1}}\,.
$$
Given $\varepsilon > 0$, it is enough to choose $N > 1 - \log_2\varepsilon$ so that
$$
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2515137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Divisibility theorem based proof for any square mod 4 being either 0 or 1 Kindly help me in understanding the below proof for given statement:
If $n$ is a square, then leaves a remainder $0$ or $1$ when divided by 4.
Proof: The divisibility theorem states that for two integers $a,b$ with $b>0$, then there is a unique p... | I think you missed the fact that the last $r$ is squared in the expression of $n$.
If $r=2$, then you have $$\begin{align}n=&16q^2+8qr+r^2\\=&16q^2+8q\cdot 2 + 2^2 \\=& 16q^2+16q+4\\=&4(4q^2+4q+1).\end{align}$$
For $r=3$, you missed a $4$ when writing the original solution, so you have
$$\begin{align}n=&16q^2+8qr+r^2\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2522766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
find all positive integers n with $\sigma(n)=12$ How do you find all positive integers n with $\sigma(n)=12$?
$\sigma(n)$ is the sum of the divisors of $n$.
The book gives the answer:
Each factor in the formula for $\sigma(n)$ must divide $12$. The only way to get factors, other than $1$, of $12$ for sums of this type... | Let $n = \prod p_i^{a_i}$ then
Then the factors are ... any combination of primes and powers. But the important thing to realize is: Suppose there is a prime factor $p$ and $n$ has $p^a$ power dividing it. And suppose that $m$ is factor that is not divisible by $p$. Then $m, pm, p^2m..... p^am$ are also factors. S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2523187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
General solution Find the general solution of $\sec4\theta-\sec2\theta =2$
My approach:
Converted them to cosines and then further simplified. I got $\theta = \frac{n\pi}{5}+\frac{\pi}{10}$ or $\theta = -n\pi+\frac{\pi}{2}$But my book stated the answer as $\theta = \frac{2n\pi}{5}\pm\frac{\pi}{10}$ or $\theta = 2n\pi\p... | You got the same answer, just in a different form
$$ \pm \frac{\pi}{2} + 2n\pi = -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \ \cdots = \frac{\pi}{2}+n\pi $$
$$ \pm \frac{\pi}{10} + \frac{2n\pi}{5} = -\frac{\pi}{10}, \frac{\pi}{10}, \frac{3\pi}{10}, \frac{5\pi}{10}, \ \cdots = \frac{\pi}{10} + \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2524318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Laplace Transform to solve system of differential equations I have the system:
\begin{align}
2\frac{dx}{dt} + \frac{dy}{dt} - 2x &= 1 \\
\frac{dx}{dt} + \frac{dy}{dt} - 7x-7y &=2
\end{align}
$y(0)=0, x(0)=0$
Which I am attempting to solve using Laplace Transform's. I change the system to
$2sx+sy-2x=1$
$sx-sy-7x-7y=2$
S... | Using the standard Laplace transform then the equations become
\begin{align}
2(s-1) \, \overline{x} + s \overline{y} &= \frac{1}{s} \\
(s-7) \overline{x} + (s-7) \overline{y} &= \frac{2}{s}.
\end{align}
Solving for $\overline{x}$ and $\overline{y}$ then
\begin{align}
\overline{x} &= - \frac{s+7}{s (s-2) (s-7)} \\
\ove... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2524812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
An identity involving the Gamma function I am trying to prove the following identity. Let $k\in\mathbb{N}$, $k\geq 2$. Then
$$
\sum_{i=1}^{k-1}\binom{k}{i}\prod_{l=2}^{i}(4l-6)\prod_{l=2}^{k-i}(4l-6)=\prod_{l=2}^{k}(4l-6).
$$
There is numerical evidence that the equation holds, also I have checked the identity for smal... | We start from
$$\sum_{q=1}^{n-1} {n\choose q}
\prod_{l=2}^q (4l-6) \prod_{l=2}^{n-q} (4l-6)
= \prod_{l=2}^n (4l-6).$$
This is
$$\sum_{q=1}^{n-1} {n\choose q}
2^{q-1} \prod_{l=2}^q (2l-3) 2^{n-q-1} \prod_{l=2}^{n-q} (2l-3)
= 2^{n-1} \prod_{l=2}^n (2l-3)$$
or
$$\sum_{q=1}^{n-1} {n\choose q}
\prod_{l=2}^q (2l-3) \prod_{l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2525372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding coefficient of $ x^{2017}$ in expansion of $(x +1+\frac{1}{x})(x^3 + 1 + \frac{1}{x^3})...(x^{2187} + 1 + \frac{1}{x^{2187}})$ I notice that $3^7=2187$ and this implies there are 8 terms in product.
The presence of $x^3$ and its powers gives a slight possible hint that $\omega$ might do some trick. But I don't ... | A nice example of telescoping products. It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series.
We obtain
\begin{align*}
\color{blue}{[x^{2017}]\prod_{j=0}^7\left(\frac{1}{x^{3^j}}+1+x^{3^j}\right)}
&=[x^{2017}]\prod_{j=0}^7\frac{1+x^{3^j}+\left(x^{3^j}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2525752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
recurrence relation, all terms of the sequence positive Let $a_1=a$, $a_2=\frac{1}{a}-a$, $a_{n+1}=\frac{n}{a_n}-a_n-a_{n-1}$ for $n=2,3,4,...$.
Find all $a$ such that $(a_n)$ is a sequence of positive reals.
My attempt was to look at $a_3=\frac{3a^2-1}{a-a^3}$, $a_4=\frac{8a^3-4a}{3a^4-4a^2+1}$ and a few more, $a_1>... | I post it as a curiosity. Considering the difference equation
$$
\frac{n}{x_n^2} = 1 + \frac{x_{n+1}+x_{n-1}}{x_n}
$$
and considering
$$
\Delta x_n = x_n-x_{n-1}
$$
we have
$$
\frac{n}{x_n^2} = 3+\frac{\Delta x_{n+1}-\Delta x_n}{x_n}
$$
which is a difference approximation for the DE
$$
\frac{t}{x^2(t)} = 3 + \frac{\dd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2525989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 4,
"answer_id": 0
} |
Solving system of linear equations with rationals Problem:$$(x+11):(x+6)=(y+12):(y+7)\land(y+1):(x-1)=y:x$$
When I tried using $\frac{a1}{b1}=\frac{a2}{b2}=\frac{k1a1+k2a2}{k1b1+k2b2}$ for $\frac{x+11}{x+6}=\frac{y+12}{y+7}$ where k1 = 1 and k2 = -1, this is what I got:$$\frac{x+11-y-12}{x+6-y-7}=\frac{x-y-1}{x-y-1}=1$... | $
\left\{
\begin{array}{l}
\dfrac{x+11}{x+6}=\dfrac{y+12}{y+7} \\
\dfrac{y+1}{x-1}=\dfrac{y}{x} \\
\end{array}
\right.
$
$
\left\{
\begin{array}{l}
(x+11)(y+7)=(x+6)(y+12) \\
x(y+1)=y(x-1) \\
\end{array}
\right.
$
$
\left\{
\begin{array}{l}
xy+7x+11y+77=xy+12x+6y+72 \\
xy+x=xy-y\\
\end{array}
\right.
$
$
\left\{
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2526489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Help thorougly proving a limit: $a>b>0; (a^n-b^n)/(a^n+b^n)$ as $n\to \infty$ Clearly, using basic calculus, I can find that the limit is 1. This is what I had so far:
$(1)$ $\frac {a^n-b^n}{a^n+b^n} = (2)$ $\frac {a^n}{a^n+b^n} - \frac {b^n}{a^n+b^n}
= (3)$ $\frac {1}{1+ \frac{b^n}{a^n}}-\frac {1}{1+ \frac{a^n}{b^n}}$... | If you mean that $n\rightarrow+\infty$ then
$$\frac{a^n-b^n}{a^n+b^n}=\frac{1-\left(\frac{b}{a}\right)^n}{1+\left(\frac{b}{a}\right)^n}\rightarrow\frac{1-0}{1+0}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2531193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int \frac{x+1}{x^2+2x}dx$ in two different ways I am trying to evaluate
$$\int \frac{x+1}{x^2+2x}dx$$
Sounds like $$\int \frac{x+1}{x^2+2x}dx=\int \frac{(x+1)}{(x+1)^2-1}dx\\u=x+1 \implies\int\frac{u}{u^2-1}du\\t=u^2 \implies \frac{1}{2}\int\frac{1}{t-1}dt $$ Which is also known as $$\frac{1}{2}\ln|t-1|=\fr... | These are not really differents ways – only variants. You can indeed use a partial fractions decomposition:
$$\frac{x+1}{x^2+2x}=\frac{x+1}{x(x+2)}=\frac Ax+\frac B{x+2},$$
determine $A$ and $B$, then the integrals of each term.
Or you can go faster and observe that
$$\int\frac{x+1}{x^2+2x}\mathop{}\!\mathrm d x=\int\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2537366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Trigonometric equation result differs from given i've got an equation:
$$\sin^6(x) + \cos^6(x) = 0.25$$
and i'm trying to solve it using the sum of cubes formula, like this:
$$ (\sin^2(x))^3 + (\cos^2(x))^3 = 0.25 $$
$$ (\sin^2(x) + \cos^2(x))^2 (\sin^4(x) - \sin^2(x)\cos^2(x) + \cos^4(x)) = 0.25 $$
$$ 1 - 3\sin^2(x)\c... | Observe that $$\dfrac{\pi n}2\pm\dfrac\pi4=\dfrac{\pi(2n\pm1)}4$$
Now set a few values of $n$ to check overlap which is evident as $2n+1=2(n+1)-1$
Actually, $\cos y=0\implies y=(2m+1)\dfrac\pi2$ where $m$ is any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2539488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $2^{3n+1} + 5$ is a multiple of 7 for all n ≥ 0. As the title states I need to prove that $(2^{3n+1}+5)$ is a multiple of 7 for all $n \geq 0$.
I can do this using induction but I also want to prove it using modular arithmetic. So here's what I've got so far.
Start with Fermat's little theorem:
\begin{al... | When you add $1$ to the exponent on $2$ you multiply by $2$, not by $5$ or $-2$. So $2^{3n+1}\equiv 1×2\bmod 7$ and the rest is easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2549706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove $5^n + 2 \cdot 3^n - 3$ is divisible by 8 $\forall n\in \mathbb{N}$ (using induction)
Prove $5^n + 2 \cdot 3^n - 3$ is divisible by 8 $\forall n\in \mathbb{N}$
Base case $n = 1\to 5 + 6 - 3 = 8 \to 8 \mid 8 $
Assume that for some $n \in \mathbb{N}\to 8 \mid 5^n + 2 \cdot 3^n - 3$
Showing $8 \mid 5^{n+1} + 2 ... | It is $4\cdot 3^n-12=4(3^n-3)$, since $3^n-3$ is even $2\mid 3^n-3$. Hence $8\mid 4\cdot 3^n-12$.
To make it clear: $3^n-3$ is even, since for $n\geq 1$ it is $3^n$ odd, and
"odd-odd=even" Since $(2k+1)-(2l+1)=2k-2l=2(k-l)$ which is even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2554949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
If $z=\tan(\frac{x}{2})$, show that $\sin(x)=\frac{2z}{1+z^2}$ So as the title states I've got the following problem:
If $z=\tan(\frac{x}{2})$, show that $\sin(x)=\frac{2z}{1+z^2}$
So I'd guess that this probably involves the formula for half-angles, but that is is a dead-end.
Any suggestions?
Thank you in advance!
| $$\frac{2\tan\left(\frac{x}{2} \right)}{1+\tan^2\left(\frac{x}{2} \right)}=$$
We know that $\tan^2(x)+1=\sec^2(x)$, so:
$$\frac{2\tan\left(\frac{x}{2} \right)}{\sec^2\left(\frac{x}{2} \right)}=$$
$$\frac{2\tan\left(\frac{x}{2} \right)}{\frac{1}{\cos^2\left(\frac{x}{2} \right)}}=$$
$$2\tan\left(\frac{x}{2} \right)\cos^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that the sequence $a_1=1$, $a_2=2$, $a_{n+2} = (a_{n+1}+a_n)/2$ converges by showing it is Cauchy
Show that the sequence $a_1=1$, $a_2=2$, $a_{n+2} = (a_{n+1}+a_n)/2$ converges by showing it is Cauchy.
My work :
Need to show that for every $\epsilon \gt 0$ there exist $N$ such that $n,m\ge N \implies | a_n - a_... | Note that for $n\geq 0$,
$$a_{n+2}-a_{n+1} =\frac{ (a_{n+1}+a_n)}{2}-a_{n+1}=\frac{a_n-a_{n+1}}{2}.$$
Hence
$$|a_{n+2}-a_{n+1}|= \frac{1}{2}|a_{n+1}-a_n|= \frac{1}{2^2}|a_{n}-a_{n-1}|= \frac{1}{2^n}|a_{2}-a_{1}|=\frac{1}{2^n}.$$
Now if $n>m\geq 1$ then, by the triangle inequality,
$$|a_{n}-a_{m}|\leq |a_{n}-a_{n-1}|+\d... | {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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What is the probability that this quadratic equation has real roots? (I've seen the other questions in this site similar to my problem, but they didn't help much). So, the problem is:
The numbers $B$ and $C$ are chosen at random between $−1$ and $1$,
independently of each other. What is the probability that the
qu... | I am not sure if the answer given in the book is exactly correct or not. Because I am finding a different answer while I tried to compute the probability.
What I attempted:- The numbers $B$ and $C$ are chosen randomly from $[-1,1]$. So, we can model them as uniform random variables taking values between $-1$ and $1... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Time and Work. How much work does C do per hour?
A, B and C need a certain unique time to do a certain work. C needs 1
hours less than A to complete the work. Working together, they require
30 minutes to complete 50% of the job. The work also gets completed if
A and B start working together and A leaves after 1 ... | Let the total work be 1, or 100%.
Let the efficiency of A,B and C be a,b and c(percentage of work be done in an hour)
(1) $\frac{1}{c}$ + 1 = $\frac{1}{a}$
(2) $(a+b+c)\cdot\frac{1}{2}$ = $\frac{1}{2}$
(3) $(a+b)\cdot1 + b\cdot3 = 1$
from (3) you get
a + 4b = 1
b = $\frac{1-a}{4}$
put it into (2)
a + $\frac{1-a}{... | {
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"url": "https://math.stackexchange.com/questions/2566277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Deriving the asymptotic estimate (9.62) in Concrete Mathematics I was reading Chapter 9: Asymptotics in Graham, Knuth, Patashnik: Concrete Mathematics, and I got stuck while deriving the following asymptotic estimate on page 466:
$$
\begin{equation}
g_n = \frac{e^{\pi^2/6}}{n^2} + O(\log n / n^3) \, ,
\quad \text{f... | We have $\enspace\displaystyle n g_n = \frac{1}{n}\sum\limits_{k\ge 0}g_k - \frac{1}{n}\sum\limits_{k\ge n}g_k + \frac{1}{n}\sum\limits_{0<k<n}\frac{k}{n-k}g_k $
with $\enspace\displaystyle g_n = \frac{G(1)}{n^2} + O\left(\frac{\ln n}{n}\right)^3 \enspace$ and $\enspace\displaystyle G(1)=e^{\zeta(2)}\,$ .
First part: ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Probability Problem with $10$ players being put on two teams Could somebody please check my answer to this problem?
Thanks
Bob
Problem:
$10$ kids are randomly grouped into an A team with five kids and a B team with five kids. Each
grouping is equally likely. There are three kids in the group, Alex and his two best frie... | Visualize $5$ slots each for team A and team B, viz $\;\;\boxed A\boxed A\boxed A\boxed A \boxed A\quad \boxed B\boxed B\boxed B\boxed B\boxed B$.
Alex can be on any team, occupying one slot.
P(Jose and Carl are both on the other team ) $= \dfrac59\cdot\dfrac48 = \dfrac5{18}$
Hence P(at least one of them is on Alex's... | {
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"url": "https://math.stackexchange.com/questions/2567048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Finding smallest possible value of expression with x and y I'm not supposed to use calculus here. I'm trying to find the smallest possible value of the expression $x^2+4xy+5y^2-4x-6y+7$ for real numbers $x$ and $y$.
Here's my attempt:
$x^2+4xy+5y^2-4x-6y+7=[(x-2)^2+3)+5((y-3/5)^2-9/25)]+4xy$
$3\le (x-2)^2+3$ for all re... | Let $u=x^2+4xy+5y^2-4x-6y+7$
$\iff x^2+4x(y-1)+5y^2-6y+7-u=0$
which is a quadratic equation in $x$
As $x$ is real, the discriminant must be $\ge0$
i.e., $$16(y-1)^2-4(5y^2-6y+7-u)\ge0$$
$$\iff u\ge y^2+2y+3=(y+1)^2+2\ge2$$
The equality occurs if $y+1=0$ and $x=-\dfrac{4(y-1)}2=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2569688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Slope of The Tangent to $r=7\sin \theta$ , $\theta = \frac{\pi}{6}$ Find slope of tangent to $r=7\sin \theta$ , $\theta = \frac{\pi}{6}$
using $$\frac{dy}{dx} = \frac{\frac{dr}{dθ}\sin \theta + r \cos \theta}{\frac{dr}{dθ}\cos \theta-r \sin \theta}$$
I got
$$\frac {14\cos \theta \sin \theta}{7 \cos^2 \theta-7\sin \thet... | Your derivation seems to be not correct.
Indeed:
$$x=r \cos \theta$$
$$y=r \sin \theta$$
$$r=7 \sin \theta \implies dr=7 \cos \theta d\theta$$
Thus:
$$dx=\cos \theta dr - r \sin \theta d\theta=7\cos^2 \theta-7\sin^2 \theta$$
$$dy=\sin \theta dr +r \cos \theta d\theta=7\cos \theta \sin \theta+7\cos \theta \sin \theta$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2571026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Is there a solution possible for the following integral? I want to solve the following integral
$$\int_0^{\infty}e^{-a(1+x^{m})^{\frac{2}{m}}}x\,{\rm d}x$$
where $a$ is real number but is not equal to $0$ and $m>2$. Any help or upper/lower bounds on this integral will be very helpful. Thank you.
| Using CAS like Mathematica and MellinTransform can be solve it for m even integers and $a>0$.
$$\int_0^{\infty } \exp \left(-a \left(1+x^m\right)^{2/m}\right) \, dx=\mathcal{M}_a\left[\int_0^{\infty } \exp \left(-a
\left(1+x^m\right)^{2/m}\right) \, dx\right](s)=\int_0^{\infty } \mathcal{M}_a\left[\exp \left(-a \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2571213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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system of equations with 3 variables in denominator I'm having problems with this system of equations:
$$\frac{3}{x+y}+\frac{2}{x-z}=\frac{3}{2}$$
$$\frac{1}{x+y}-\frac{10}{y-z}=\frac{7}{3}$$
$$\frac{3}{x-z}+\frac{5}{y-z}=-\frac{1}{4}$$
I've tried substitution method but that took me nowhere, I'm not sure I can solve ... |
$$\frac{3}{x+y}+\frac{2}{x-z}=\frac{3}{2}$$
$$\frac{1}{x+y}-\frac{10}{y-z}=\frac{7}{3}$$
$$\frac{3}{x-z}+\frac{5}{\color{red}{x-z}}=-\frac{1}{4}$$
I assume that the part in red should be $y-z$; if not, please clarify.
Hint. Letting:
$$a=\frac{1}{x+y} \;,\; b=\frac{1}{x-z} \;,\;c=\frac{1}{y-z}$$
turn the system in... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Why is function domain of fractions inside radicals not defined for lower values than those found by searching for domain of denominator in fraction? Consider function $y = \sqrt{\frac{1-2x}{2x+3}}$. To find the domain of this function we first find the domain of denominator in fraction:
$2x+3 \neq 0$
$2x \neq -3$
$x \... | If $x<-\frac{3}{2}$, then $2x+3<0$ and $1-2x>0$, so $\frac{1-2x}{2x+3}<0$, which means you can't take the square root and get a real answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2572000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A Junior Olympiad like system of linear equations Given a system of linear equations
$$\begin{align}\frac{x}{3}+\frac{y}{5}+\frac{z}{9}+\frac{w}{17} &=1 \\
\frac{x}{4}+\frac{y}{6}+\frac{z}{10}+\frac{w}{18} &=\frac{1}{2} \\
\frac{x}{5}+\frac{y}{7}+\frac{z}{11}+\frac{w}{19} &=\frac{1}{3} \\
\frac{x}{6}+\frac{y}{8}+\frac{... | Let $G(s) = \frac{x}{s+2} + \frac{y}{s+4} + \frac{z}{s+8} + \frac{w}{s+16} - \frac 1s$, and $F(s) = s(s+2)(s+4)(s+8)(s+16)G(s)$.
Clearly $F(s)$ is a degree 4 polynomial with roots being 1, 2, 3, and 4. Also, $F(0) = - 2\times4\times8\times16 = -2^{10}$. Hence
$$
F(s) = -\frac{2^{10}}{24}(s-1)(s-2)(s-3)(s-4) = -\frac{2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Does the convergence of $\sum (a_k)^2$ imply $\sum (a_k)^3$ convergence? Does the convergence of $\sum (a_k)^2$ imply $\sum (a_k)^3$ convergence? I feel like it definitely should but can't find a solid way to prove it ....
| It's true for real sequences (as shown in other answers), but false for complex sequences.
For example, if $\omega = e^{2\pi i/3}$ and
$$
a_{3k+m} = \frac{\omega^m}{(k+1)^{1/3}}
\qquad\text{for}\qquad
k \ge 0
,\qquad
m \in \{ 0,1,2 \}
,
$$
that is, if the sequence $(a_j)_0^\infty$ is
$$
\frac{1}{1^{1/3}},
\frac{\omega}... | {
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"url": "https://math.stackexchange.com/questions/2574186",
"timestamp": "2023-03-29T00:00:00",
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Muhammad Ali pulls tiles from bag $3.$ A bag contains three tiles marked $A$, $L$, and $I$. Muhammad wants to pick the letters $ALI$, in that sucession. Randomly, he pulls one til from the bag. If the letter $A$ is drawn, he keeps it. If the letter pulled is other than $A$, he puts it back into the back. He does the sa... | What is the probability that it takes n draws, imagine the sequence of draws $x_1, x_2, \cdots, x_n$.
For this this to be a valid draw we have some $x_i=A$ and $x_{n-1}=L$ such that $1 \le i <n-1$. And since it took $n$ draws then $x_n=I$
For a particular $i,j$ the probability is $p_n(i,j)$
$$p_n(i,j)=\left(\frac{2}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2575305",
"timestamp": "2023-03-29T00:00:00",
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Factorization of a 4th degree polynomial
Factorization of :
$$x^4-x^3+3x^2+3x+5$$
$$x^4-x^3+3x^2+3x+5=(x^4-x^3+2)+3(x^2+x+1)$$
what do i do ? please help me
| A possible error in the question, it might explain things. If we change 5 to 54, we get
$$ x^4 - x^3 + 3x^2 + 3x + 54 = ( x^2 -4 x + 9) ( x^2 + 3 x + 6) $$
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
First you look for rational roots. There are none. Then we nee... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Equation of the sphere that passes through 4 points Write he equation of the sphere that passes through points
$$a(-5,4,1),b(3,4,-5),c(0,0,4),d(0,0,0)$$
I tried to use four points to draw a geometric shape and then calculate the center of this shape on the basis of the circle that passing on four points. But I did not... | The equation of the sphere with a centre in $(x_0,y_0,z_0) $ and radius $r $ is: $(x-x_0)^2+(y-y^0)^2+(z-z_0)^2=r^2$. For us, the unknowns are $x_0,y_0,z_0,r $.
Plug the given four points in:
$$\begin {align}(-5-x_0)^2+(4-y_0)^2+(1-z_0)^2=r^2 \\ (3-x_0)^2+(4-y_0)^2+(-5-z_0)^2=r^2 \\ (-x_0)^2+(-y_0)^2+(4-z_0)^2=r^2 \\ ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given $3\cos x - 4 \sin x = 2$, find $3 \sin x + 4 \cos x$ without first solving for $x$ If $$3\cos{x}-4\sin{x}=2$$
find $$3\sin{x} +4\cos{x} $$
I have solved the equation for $x$, then calculated the required value, but I think there is a direct solution without solving the equation.
| $$a \cos(x) - b \sin(x) = \sqrt{a^2+b^2} \cos(x+y)$$ where $\cos(y) = a/\sqrt{a^2+b^2}$ and $\sin(y) = b/\sqrt{a^2 + b^2}$. Then $$a \sin(x) + b \cos(x) = \sqrt{a^2 + b^2} \sin(x+y)$$
In your case, with $a=3$ and $b=4$, $\sqrt{a^2+b^2}=5$, $\cos(x+y) = 2/5$ so $\sin(x+y) = \pm \sqrt{1-(2/5)^2} = \pm \sqrt{21}/5$, so ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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A double integral into polar coordinates I have the double integral
$$\int^{10}_0 \int^0_{-\sqrt{10y-y^2}} \sqrt{x^2+y^2} \,dx\,dy$$
And I am asked to evaluate this by changing to polar coordinates.
| Complete the square:
$$
10y-y^2 = 25 - (5-y)^2,
$$
so the graph of $x = -\sqrt{10y-y^2} = - \sqrt{5^2 - (5-y)^2}$ is the left half of the circle $x^2 + (5-y)^2 = 5^2.$
Exercises with polar coordinates will have shown you that $$\tag 1 r = 5\sin\theta$$ is that circle. If you multiply both sides of $(1)$ by $r,$ you get... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Sum of the infinite series The series is $$\frac{5}{1\cdot2}\cdot\frac{1}{3}+\frac{7}{2\cdot3}\cdot\frac{1}{3^2}+\frac{9}{3\cdot4}\cdot\frac{1}{3^3}+\frac{11}{4\cdot5}\cdot\frac{1}{3^4}+\cdots$$
This is my attempt:
$$T_n=\frac{2n+3}{n(n+1)}\cdot\frac{1}{3^n}$$
Assuming $$\frac{2n+3}{n(n+1)}=\frac{A}{n}+\frac{B}{n+1}$$
... | $\sum_{i=1}^{n} T_i =S_n - R_n$ , where
$S_n := \sum_{i=1}^{n} \dfrac{1}{i}\dfrac{1}{3^{i-1}};$
$R_n :=\sum_{i=1}^{n} \dfrac{1}{i+1} \dfrac{1}{3^i}.$
Change the dummy index in $R_n:$
$k= i +1$, then
$R_n=\sum_{k=2}^{n+1} \dfrac{1}{k}\dfrac{1}{3^{k-1}} =$
$(1 + \sum_{k=2}^{n}\dfrac{1}{k}\dfrac{1}{3^{k-1}}) -1 +
\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2590246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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What is the residue for this function How to find the residue of $\dfrac{1}{\tan(z)}$? I have calculated that it has a pole of order 2 . But I'm having trouble when using the residue theorem and I end up with a residue of $2$. However, wolfram alpha calculator says that the residue=$0$.
| Since, $ 1 + \cos z = 2\cos^2 \frac{z}{2} $, the function has second-order poles at $\frac{z}{2} = (n + \frac12)\pi$ or $z = (2n+1)\pi$
Let $w = z - (2n+1)\pi$, we can compute the residues
$$ \begin{align}
\operatorname*{Res}_{z=(2n+1)\pi} f(z) &= -\frac12 \lim_{z\to (2n+1)\pi} \frac{d}{dz}\left(\big(z -
(2n+1)\pi\b... | {
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"source": "stackexchange",
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lim$_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{3}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{9}}+...+\frac{1}{\sqrt{3n}+\sqrt{3n+3}}\right)$
Question
Edit:
My Approach:
$\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{3}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{9}}+...+\frac{1}{\sqrt{3n}+\sqrt{3n+3}... | More generally, by Stolz-Cesaro Theorem, if $a\geq 0$ then
$$\lim_{n\to\infty}\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{3k}+\sqrt{3k+a}}=
\lim_{n\to\infty}\left(\frac{1}{\sqrt{n}-\sqrt{n-1}}\cdot\frac{1}{\sqrt{3n}+\sqrt{3n+a}}\right)\\
=
\lim_{n\to\infty}\frac{\sqrt{n}+\sqrt{n-1}}{\sqrt{3n}+\sqrt{3n+a}}=\frac{1}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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a four variable inequality Let $a,b,c,d$ be non-negative real numbers satisfying $a+b+c+d=3$. Show
\begin{equation*}
\frac{a}{1+2b^3}+\frac{b}{1+2c^3}+\frac{c}{1+2d^3}+\frac{d}{1+2a^3} \geqslant \frac{a^2+b^2+c^2+d^2}{3}.
\end{equation*}
I got this inequality from an IMO preparation club. I no longer do contest math ... | By AM-GM $$\sum_{cyc}\frac{a}{1+2b^3}=3+\sum_{cyc}\left(\frac{a}{1+2b^3}-a\right)=3-\sum_{cyc}\frac{2ab^3}{1+2b^3}\geq$$
$$\geq3-\sum_{cyc}\frac{2ab^3}{3b^2}=\frac{(a+b+c+d)^2-2\sum\limits_{cyc}ab}{3}=$$
$$=\frac{a^2+b^2+c^2+d^2+ac+bd}{3}\geq\frac{a^2+b^2+c^2+d^2}{3}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$5$ divides $2ax+b$ for some integer $a$,given $5$ does not divide $x$ $x$ and $b$ are given.Let,$5$ does not divide $x$.
Then $5$ divides $2ax+b$ for some integer $a$.
How do i prove this? I have not gone through a course in number theory,so a general proof would be appreciated.
| The slightly stronger statement holds true that $\,2ax+b\,$ is a multiple of $\,5\,$ for an $\,a \in \{0, \pm 1, \pm 2\}\,$.
If $\,b\,$ is a multiple of $\,5\,$ then $\,2 \cdot 0 \cdot x+b=b\,$ is a multiple of $\,5\,$, so $\,a=0\,$ works.
Otherwise, note first that if $\,u\,$ is not a multiple of $\,5\,$ then $\,u^4 =... | {
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How to prove that this triangle is equilateral? Question: If $\cos A +\cos B +\cos C=\frac{3}{2}$, prove that the triangle is equilateral.
My attempt: According to cosine rule, $\cos A=\frac{b^2+c^2-a^2}{2bc}$, $\cos B=\frac{c^2+a^2-b^2}{2ca}$ and $\cos C=\frac{a^2+b^2-c^2}{2ab}$.
$\cos A +\cos B +\cos C=\frac{3}{2}$
$... | Or $$\sum_{cyc}(a^3-a^2b-a^2c+abc)=0.$$
Now, let $a\geq b\geq c$.
Thus,
$$0=\sum_{cyc}(a^3-a^2b-a^2c+abc)=\sum_{cyc}(a^3-a^2b-ab^2+abc)=\sum_{cyc}a(a-b)(a-c)\geq$$
$$\geq a(a-b)(a-c)+b(b-a)(b-c)=(a-b)^2(a+b-c)\geq0,$$
which says that the equality occurs for $a=b=c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2602051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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sum $\displaystyle \sum _{n=2}^{\infty}\frac{(-1)^n}{n^2+n-2}$ I have the following series:
$$\sum _{n=2}^{\infty}\frac{(-1)^n}{n^2+n-2}$$
I am not able to do the telescoping process in the above series. I converted it into the following partial fraction:
$$\sum _{n=2}^{\infty}\frac{(-1)^n}{(n+2)(n-1)}$$
But nothing se... | Writing the $n$th term of your sum in a form that telescopes can be avoid altogether by converting the problem to a double integral as follows.
Noting that
$$\frac{1}{n - 1} = \int_0^1 x^{n - 2} \, dx \qquad \text{and} \qquad \frac{1}{n + 2} = \int_0^1 y^{n + 1} \, dy,$$
the sum can be rewritten as
\begin{align*}
\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2603199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all local maxima and minima of $x^2+y^2$ subject to the constraint $x^2+2y=6$. Does $x^2+y^2$ have a global max/min on the same constraint? This is my method for the local max/min. Does this answer sound sensible? (Not sure how to go about checking for global max/min though...)
Method
$G(x,y)=x^2+2y-6$. Rewrite in... | From $x^2+2y=6 $ we get $ x^2=6-2y$
Substitute in $x^2+y^2$ to get $$x^2+y^2=y^2-2y+6$$ Note that $y^2-2y+6=(y-1)^2+5$.
Thus the minimum is attained when at $y=1$ at two points where and $x=2$ or $x=-2$
There is no global for $(y-1)^2+5$, hence there is no global maximum for $ x^2+y^2$ subject to $x^2+2y=6 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2603889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Calculating an inverse Laplace transform I have a pain for a problem... need to find the inverse Laplace transform of:
$$\frac{s^2+s+1}{(s+1)(s+2)^2(s^2+4s+9)}$$
Now, I get that we have to expand the partial fraction, I managed to get to this:
$$\frac{A}{(s+1)}+\frac{B}{(s+2)}+\frac{C}{(s+2)^2}+\frac{Ds+E}{(s^2+4s+9)}$... | Hint. Note that
$$s^2+4s+9=(s+2)^2+(\sqrt{5})^2$$
and take a look at "exponentially decaying
sine/cosine wave" in this Table.
P.S. Check your partial fraction decomposition. It should be
$$\frac{1/6}{(s+1)}-\frac{3/5}{(s+2)^2}-\frac{s/6-1/10}{(s^2+4s+9)}
=\frac{1/6}{(s+1)}-\frac{3/5}{(s+2)^2}-\frac{(s+2)/6-13/30}{(s+2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2604769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Integral of a Polynomial in Square Root
I need to solve the indefinite integral:
$$\int \frac{\sqrt{x^4+x^{-4}+2}}{x^3} dx,$$
but I can not find any technique which could solve it.
Can you help me please?
| It is :
$$(x^2+x^{-2})^2=x^4+x^{-4}+2$$
which means that your initial integral can be substituted to :
$$\int \frac{\sqrt{x^4+x^{-4}+2}}{x^3} dx = \int\frac{\sqrt{(x^2+x^{-2})^2}}{x^3}dx = \int \frac{x^2 + x^{-2}}{x^3}dx$$
since $x^2 + x^{-2} \geq 0 \space \forall x \in \mathbb R$.
So :
$$\int \frac{x^2 + x^{-2}}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2609487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find a number $M$, such that $|x^3-4x^2+x+1| < M$ for all $1The question states that I cannot use calculus, so finding max/min of the function is not an accepted solution.
My attempt at solving the problem:
$|(x^3-4x^2) + (x+1)|$ $\le$ $|x^3-4x^2| + |x+1|$
$= |x^2||x-4| + |x+1|$
$ = x^2|x-4| + x+1$ (because 1 < x < 3... | For all $1<x<3$ we have $$x^3-4x^2+x+1=x^3-4x^2+3x+1-2x=-x(3-x)(x-1)-2x+1<0.$$
Thus, we need to find the maximal value of $f(x)=-x^3+4x^2-x-1$ on $(1,3).$
We have $$f'(x)=-3x^2+8x-1,$$ which gives $$x_{max}=\frac{4+\sqrt{13}}{3}$$ and all $$M>f\left(\frac{4+\sqrt{13}}{3}\right)=\frac{13(5+2\sqrt{13})}{27}$$ is valid.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2610859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
An alternative way to find the sum of this series? $\displaystyle \frac{4}{20}$+$\displaystyle \frac{4.7}{20.30}$+$\displaystyle \frac{4.7.10}{20.30.40}$+...
Now I have tried to solve this in a usual way, first find the nth term $t_n$.
$t_n$= $\displaystyle \frac{1}{10}$($\displaystyle \frac{1+3}{2}$) ... | Through Euler's Beta function and the reflection formula for the $\Gamma$ function:
$$\sum_{n\geq 1}\frac{\prod_{k=1}^{n}(3k+1)}{10^n(n+1)!}=\sum_{n\geq 1}\frac{3^n\Gamma\left(n+\frac{4}{3}\right)}{10^n \Gamma(n+2)\Gamma\left(\frac{4}{3}\right)}=\frac{3\sqrt{3}}{2\pi}\sum_{n\geq 1}\left(\tfrac{3}{10}\right)^n B\left(\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2611305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Show that if $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=1$ then $xyz=0$ For any real numbers $x,y,z$ such that $x+y+z = x^2+y^2+z^2 = x^3+y^3+z^3 =1\\$ show that $x \cdot y \cdot z=0$.
I think that because $x \cdot y \cdot z = 0$ at least one of the three number should be equal to zero, but I'm stuck into relating the other things... | For $x:0\neq|x|<1$ we have $x^3<x^2$. Therefore if we have 3 numbers $x_1^2+x_2^2+x_3^2=1$ and $x_1x_2x_3\neq0$ we have that $x_i\neq0$ $\forall i$. Therefore for every $i$ we have the strict inequallity $x_i^3<x_i^2$ therefore $x_1^3+x_2^3+x_3^3<1$. Therefore one of the $x_i$ is $0$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 3
} |
Three touching circles inscribed in a rectangle 'Three equal circles, with radius r, are inscribed in a rectangle in a way that all three of them touch each other only once. Find the area of rectangle.' The thing i can't understand is that the arrangement of circles seem limitless to me, do they need to be touching t... | A little thought will reveal that, if the three-circle grouping is to be tangent to the sides of a rectangle, then (at least) one of those circles must be tangent to two adjacent sides of the rectangle. This gives rise to the following diagram:
Let $\bigcirc O$, of radius $r$, be tangent to two sides of the rectangle.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.