Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove that $f(n)=(n!)^{\frac{1}{n}}-\frac{n+1}{2}$ is a monotone decreasing sequence Prove that the sequence $f(n)=(n!)^{\frac{1}{n}}-\frac{n+1}{2}$ is a monotone decreasing sequence for $n>2.$
We have to show that $f(n+1)<f(n)$ for all $n>2.$ We have
$$
f(n+1)-f(n)=((n+1)!)^{\frac{1}{n+1}}-\frac{n+2}{2}-\left((n!)^{\f... | We have
$$\begin{align}\frac{\left(\frac12+n!^{1/n}\right)^{n+1}}{(n+1)!}&=\frac{(n!^{1/n})^{n+1}+(n+1)\cdot\frac12(n!^{1/n})^{n}+{n+1\choose2}\cdot\frac14(n!^{1/n})^{n-1}+\ldots}{(n+1)!}
\\&>\frac{n!^{1/n}}{n+1}+\frac12+\frac n{8\cdot n!^{1/n}}\end{align}$$
Using $a x+\frac bx\ge 2\sqrt{ab}$ (arithmetic-geometric ine... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int\frac{d\theta}{1+x\sin^2(\theta)}$ We have to evaluate $$\int\frac{d\theta}{1+x\sin^2(\theta)}$$
Here is all my steps:
$$\tan\frac{\theta}{2}=\omega\Rightarrow \sin(\theta)=\frac{2\omega}{1+\omega^2}\Rightarrow 1+x\sin^2{(\theta)}=1+\frac{x 4\omega^2}{(\omega^2+1)^2}$$
Therefore: $$\int\frac{d\theta}{1+... | Hint: Let $$\sin \theta = \pm\frac{\omega}{\sqrt {1 +\omega^2}} \implies 1 +x\sin ^2 \theta = 1 + \frac{x\omega^2}{1 + \omega^2} = \frac{1 + (1 + x) \omega^2}{1 + \omega^2 }$$
where $\omega = \tan \theta$. Then
$$\int \frac{1 + \omega^2}{1 +(1 + x)\omega^2} d\theta = \int \frac{\sec^2 \theta}{1 + (1 +x)\tan^2 \theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculus differentiable gra Hey guys so stuck on a calculus question. So far all I know is that $d$ should equal $4$. I then got the derivative of the a b c d function to $3ax^2 + 2bx + c$ , subbed in $(0,4)$ to get $c$, which was also $4$. Just wondering if I'm on the right track and where I should go from here.
A ... | ok, so
$y = $
$x + 4, x ≤0$
$ax^3+ bx^2+ cx + d, 0 < x < 4$
$4 – x, x ≥4$
and y has to be differentiable everywhere (and therefor continuous, too)
so, the derivative at x = 0 and at x = 4 have to be the same when approached from both sides, as do the values of y.
Let $P(x) = ax^3 + bx^2 + cx + d$
and
$DP(x) = 3ax^2 +... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove the inequality $ (1+a+ab)(1+b+bc)(1+c+ca) \leq (1+a+a^2)(1+b+b^2)(1+c+c^2)?$ For $a,b,c>0$ prove the inequality
$$
(1+a+ab)(1+b+bc)(1+c+ca) \leq (1+a+a^2)(1+b+b^2)(1+c+c^2).
$$
I know that I should use the multiplicative rearrangement inequality but I am not sure how to choose the involved sequences c... | HINT: use that $$x^2+y^2+z^2\geq xy+xz+yz$$ for all real $x,y,z$ is hold.
And your inequality can be written as
$$(ab)^2+(ac)^2+(bc)^2-a^2bc-ab^2c-abc^2+a^2b+ac^2+b^2c-3abc+a^2+b^2+c^2-ab-ac-bc\geq 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1320267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine Positive Integer in a Series Let $a_{n} = 4 - 3n,$ $\forall$ Integers $n \geq 1.$
$$f(x, y) = x + \sum_{i=1}^{\infty}\left [\left(\frac{\prod_{j=1}^{i}a_{j}}{3^i\cdot i!} \right )x^{a_{i+1}}y^{i}\right]$$
$\forall$ Real $x$ and $y.$ How can I find a positive integer $k$ such that $f(19, k) = 20$.
| Consider the function $F(x, y) = \sqrt[3]{x^3 + y}.$ If we write the function in the form $(x^3 + y)^{1/3},$ we can use the generalized version of the binomial theorem to expand this. We get
$$(x^3 + y)^{1/3} = \binom{1/3}{0}x + \binom{1/3}{1}x^{-2}y + \binom{1/3}{2}x^{-5}y^2 \cdots$$
Also,
$$\begin{align*} \binom{1/3}... | {
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"timestamp": "2023-03-29T00:00:00",
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Integral using contour integration Here is the integral I want to evaluate:
$$\int_{0}^{2\pi} \frac{dx}{a+b \cos x }, \quad a>b >0$$
Apparently there are limitations as to what values $a, b$ are supposed to take but let us not concern about this. Since using the sub $u =\tan \frac{x}{2}$ (the Weiersstrass sub) results ... | The issue I see with your formula is that scaling $a$ and $b$ by $\lambda$ fails to scale the result by $\lambda^{-1}$. Anyway, I do not know how you computed the residue(s), but you did that wrong somehow.
Also, the substitution is $x=i^{-1}\ln z$, not $x=i\ln z$. Anyway, we get
$$\frac{2}{i}\oint\limits_{|z|=1}\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving that $\Delta x^{(n)} = n x^{(n-1)}$ Define $\Delta f(x) = f(x+1) - f(x)$ (the difference operator). Define $x^{(n)} = x(x-1) \dots (x-n+1)$ (the falling factorial function). There's a rather simple theorem which shows that $\Delta x^{(n)} = n x^{(n-1)}$:
$\Delta x^{(n)} = (x+1)^{(n)} - x^{(n)}$
$= (x+1)(x+1-1)(... | That first term is the product of $n$ factors of the form $x+1-k$, where $k$ starts at $0$ and increases by $1$ from one factor to the next. The highest value of $k$ must therefore by $n-1$, and the last factor is therefore $x+1-(n-1)=x-n+2$.
If you went all the way to $x-n=(x+1)-(n+1)$, you’d have $k$ running from $0$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How can we simplify the expression $P+\sqrt{P^2+\sqrt{P^4+\sqrt{ P^8+\cdots)}}}$? Is there a way to reduce the expression
$P+\sqrt{P^2+\sqrt{P^4+\sqrt{ P^8+\cdots)}}}$?
| So our target expression is:
$$ P + \sqrt{P^2 + \sqrt{P^4...}}$$
An indirect approach that is quite simple is to consider
$$ P \left( 1 + \sqrt{1 + \sqrt{1 ...}} \right)$$
It follows that of we distribute the P we will have
$$ P +P \sqrt{1 + \sqrt{1 ...}} $$
Which yields
$$ P + \sqrt{P^2 + P^2\sqrt{1 + ...}}$$
And aga... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Finding the maximum of an expression in three variables I am trying to find the max of $\frac{(a)(a-1)(9-a)}{2} + \frac{(b)(b-1)(9-b)}{2} + \frac{(c)(c-1)(9-c)}{2}$ subject to $a+b+c=9$ over nonnegative integers. I originally tried to see if something could work with Jensen's or calculus, and I have a conjecture that I... | $\dfrac{(a)(a-1)(9-a)}{2} + \dfrac{(b)(b-1)(9-b)}{2} + \dfrac{(c)(c-1)(9-c)}{2}$
$=\dfrac{-\sum a^3+10\sum a^2-9\sum a}2$
$\sum a=9$
$\sum a^2=(\sum a)^2-2\sum ab=9^2-2\sum ab$
$\sum a^3=(\sum a)\{(\sum a)^2-3\sum ab\}=9[9^2-3\sum ab]$
Finally $2[(\sum a)^2-3\sum ab]=\sum(a-b)^2\ge0$
$\implies6\sum ab\le2(\sum a)^2=2\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving a three variable equation I have three given values, suppose a=1.86, b=2.6 and c=4.2. Now I have to figure out x,y,z such that
*
*$x\gt 0,y\gt 0$ and $z\gt 0$
*$x+y+z=1$
*$a*x\gt 1, b*y\gt 1$ and $cz\gt 1$
I need a generalized solution steps for this to implement in programming.
Thanks.
| $\left\{ \begin{array}{l}
x > \frac{1}{a} \Rightarrow x = \frac{1}{a} + \varepsilon \\
y > \frac{1}{b} \Rightarrow y = \frac{1}{b} + \varepsilon \\
z > \frac{1}{c} \Rightarrow z = \frac{1}{c} + \varepsilon
\end{array} \right.$ where $0<\varepsilon$.
$\Rightarrow \frac{1}{a} + \varepsilon + \frac{1}{b} + \varepsilon ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Differential equation $y\cdot y'' - (y')^2 - 1 = 0$ I'm trying to solve this equation but at the end I'm stuck and can't reache the answer. I use the substitutions $y'=p$ and $y'' = p'\cdot p$:
$$y \cdot p'p - p^2 - 1 = 0 \implies y\cdot p \frac {dp}{dy} - (p^2 + 1) = 0 \implies \int \frac {p}{p^2 + 1}dp = \int \frac {... | You obtained :
$$y + \sqrt{y^2 - \frac{1}{C_1^2}} = \exp {\frac {x+C_2}{C_1}} $$
it remains to solve this equation for $y$
$$\sqrt{y^2 - \frac{1}{C_1^2}} = -y+\exp {\frac {x+C_2}{C_1}} $$
$$y^2 - \frac{1}{C_1^2} = \left( -y+\exp {\frac {x+C_2}{C_1}}\right)^2 $$
Then, simplify and express $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1332524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How exactly do we do Gauss elimination?
This is a matrix:
$$\begin{bmatrix}
1 & 1 & 1\\
1 & 2 & 3\\
1 & 3 & k
\end{bmatrix}\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=
\begin{bmatrix}
3\\
6\\
4+k
\end{bmatrix}$$
Find $k$ so that it has no unique solution. Solve the equations for this value of $k$.
I found $k = 5$ by ... | By Gauss-Jordan elimination I get the following:
$$\left(\begin{array}{ccc|c}
1 & 1 & 1 & 3 \\
1 & 2 & 3 & 6 \\
1 & 3 & k & 4+k
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 1 & 1 & 3 \\
0 & 1 & 2 & 3 \\
0 & 2 &k-1& 1+k
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 1 & 1 & 3 \\
0 & 1 & 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Sum of Square-Weights For positive reals $a,b,c$, prove that $$\frac{a^3+b^3+c^3}{3abc}+\sum_{\text{cyc}} \frac{a(b+c)}{b^2+c^2}\geq 4$$
I've heard of a lemma stating if a polynomial expression $f(a,b,c)$ satisfies both
$f(a,b,c)=f(b,a,c)$ and $f(a,a,c)=0$, then it is divisable by $(a-b)^2$. I tried splitting up the e... | $\sum_{\text{cyc}} \dfrac{a(b+c)}{b^2+c^2} -3=\sum_{\text{cyc}}(\dfrac{a(b+c)}{b^2+c^2} -1 )=\sum_{\text{cyc}}\dfrac{b(a-b)+c(a-c)}{b^2+c^2}=\sum_{\text{cyc}}(\dfrac{b(a-b)+}{b^2+c^2}+\dfrac{a(b-a)}{a^2+c^2})=\sum_{\text{cyc}}(\dfrac{(a-b)^2(ab-c^2)}{(b^2+c^2)(a^2+c^2)})=\sum \dfrac{(a-b)^2(ab)}{(b^2+c^2)(a^2+c^2)}-\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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The number of ways to write $10$ as the sum of five natural numbers not equal to $3$ How many answers are there for the equation
$$x_1+x_2+x_3+x_4+x_5=10$$
given that $x_1,x_2\dots x_5\in\Bbb{Z^{0+}}\setminus\{3\}$.
| The number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 = 10$$
is the number of ways four addition signs can be inserted into a row of ten ones, which is $\binom{10 + 4}{4} = \binom{14}{4}$. From these, we must exclude those solutions in which one or more of the numbers is equal to $3$. Since $3 \cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1335327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
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Taylor series expansion of function $f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}}$ Determine the Taylor series expansion of function $f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}}$.
$f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}} = \arcsin(x)\frac{1}{\sqrt{1-x^2}}$
It is known:
(1.) $\arcsin(x) = \sum_{n=0}^\infty\frac{(2n-1)!!x^{2n+1}}{2... | Do not multiply, tackle it directly by looking at the Taylor recipe...
Just a short answer for your check:
$$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=0}^{\infty}\frac{4^n (n!)^2}{(2n+1)!}x^{2n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $7\arctan^2\varphi+2\arctan^2\varphi^3-\arctan^2\varphi^5$ Let $\varphi=\frac{1+\sqrt5}2$ (the golden ratio).
How can I simplify the following expression?
$$7\arctan^2\varphi+2\arctan^2\varphi^3-\arctan^2\varphi^5$$
| The answer of @Vladimir Reshetnikov: is hard to improve on, so we'll try to answer some of the questions posed in the comments.
If $\phi=\frac{1+ \sqrt{5}}{2}$ is the positive root of $x^2 - x-1=0$ then
\begin{eqnarray}
\tan(2 \arctan(\phi^n)) = \frac{2\phi^n}{ 1- \phi^{2n}} = \begin{cases} -\frac{2}{L_{n}} \text{ if... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1338044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 3,
"answer_id": 0
} |
Irrational numbers, induction I have $\sqrt[3]{2}^{2^n}$.
Can I prove that this number is irrational by showing that $3$ does not divide $2^n$?
| If $\dfrac{p}{q} = 2^{\frac{2^n}{3}}, (p,q) = 1 \Rightarrow \dfrac{p^3}{q^3} = 2^{2^n}\Rightarrow p^3=2^{2^n}q^3\Rightarrow 2\mid p^3\Rightarrow 2\mid p$. Now if $p = 2^k$, then $p^3 = 2^{3k} \Rightarrow q^3 = 2^{2^n-3k}$. Observe $3\nmid 2^k \Rightarrow 2^n - 3k \geq 1 \Rightarrow 2\mid q^3 \Rightarrow 2\mid q \Righta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1339817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding value of an expression If $x^2-3x-1=0$ then find the value of $(x^6+1)/x^3$ I tried to solve the quadratic but it became too complicated any way of doing this without a calculator
| $$x-\frac1x=3.$$
Then
$$\left(x-\frac1x\right)^2+3=x^2+1+\frac1{x^2}=12.$$
Then
$$\left(x^2+1+\frac1{x^2}\right)\left(x-\frac1x\right)=x^3-\frac1{x^3}=36.$$
Finally,
$$\left(x^3+\frac1{x^3}\right)^2=\left(x^3-\frac1{x^3}\right)^2+4=1300.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Long polynomial expansion with 34 roots
This is a very tricky problem, I just need a few hints. I think the $(-x^{17})$ is also there for a specific trick. In the end if it is $ax^{17}$, I see that $a = 17 - 1 + 1 = 17$.
Also, another possible approach is:
$$(1 + x + \cdots + x^{17})^2 = x^{17}$$
$$1 + x + \cdots + x... | From the update to the question:
The value of $|z_k| < 1$, because $(1 + x + ... + x^{17})^2 > x^{17}$.
By the geometric series formula this changes to:
$$\left(\sum_{n=0}^{17} x^n \right)^2 = x^{17} \space \text{where} \space |x| < 1$$
$$ \left( \frac{1 - x^{18}}{1-x} \right)^2 = x^{17}$$
$$(1 - x^{18})^2 = (1-x)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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I need help with a Finite Series Problem:
Find the sum to $n$ terms of
\begin{eqnarray*}
\frac{1}{1\cdot 2\cdot 3} + \frac{3}{2\cdot 3\cdot 4} + \frac{5}{3\cdot 4\cdot 5} +
\frac{7}{4\cdot 5\cdot 6}+\cdots \\
\end{eqnarray*}
Answer:
The way I see it, the problem is asking me to find this series:
\begin{eqnarray... | There is more simple way (for me). You have
$$a_n=\frac{2n-1}{n(n+1)(n+2)}=-\frac52\frac{1}{n+2} + \frac{3}{n+1} - \frac{1}{2n};$$
hence
$$S_N = \sum_{n=1}^N a_n = -\frac52\left(H_{N+2}-1-\frac12\right) + 3(H_{N+1}-1) - \frac12 H_N,$$
where $H_N$ is $N$-th harmonic number. Simplify it:
$$S_N = -\frac52\left(H_N + \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341486",
"timestamp": "2023-03-29T00:00:00",
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What is the remainder when $6\times7^{32} + 7\times9^{45}$ is divided by $4$?
What is the remainder when $6\times7^{32} + 7\times9^{45}$ is divided by $4$ ?
$7 \equiv 3 \pmod 4$
$7^2 \equiv 9 \pmod 4\equiv 1 \pmod 4$
$(7^2)^{16} \equiv 1^{16} \pmod 4$
i.e $7^{32} \equiv 1 \pmod 4$
Similarly $9 \equiv 1 \pmod 4$ imp... | The relation "go modulo 4" is very nice. It respects addition and multiplication.
So
$$
6 \cdot 7^{32} + 7 \cdot 9^{45}
$$
is the same as
$$
6 \cdot 1 + 7 \cdot 1
$$
by what you have already computed!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Intersection of a cone and a plane In $\mathbb{R}^3$, given the cone $K$ and the plane $E_c$ with the equations $4x^2=y^2+z^2$ and $z=c(1-x)$.
How do I find out which different geometric objects I get for all $c\geq 0$ if I intersect both $K$ and $E_c$?
| Expanding on the above comment, if we substitute $z = c(1-x)$ in the equation of $K$ we get $4x^2 = y^2 + c^2(1-x)^2$, i.e.
$$
(4-c^2)x^2 - y^2 + 2c^2 x = c^2 \label{eq:1} \tag{1}
$$
Now all we have to do is distinguish a few cases.
If $c = 0$ then \eqref{eq:1} becomes
$$
0 = 4x^2 - y^2 = (2x + y)(2x - y)
$$
so $E_0 \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I evaluate this : $\int_{0}^{\infty} \ln \left( 1 + \frac{a^{2}}{x^{2}}\right)\ dx $ for $a > 0$? How do I evaluate this integral if I suppose that $a > 0$
$$\int_{0}^{\infty} \ln \left( 1 + \frac{a^{2}}{x^{2}}\right)\ \mathrm{d}x .$$
For $a=2$ I got $2\pi$ I think the result will be $a\pi$.
| According to an answer by rae306 on Art of Problem Solving:
Using integration by parts:
$$\int_0^\infty \ln\left(1+\frac{a^2}{x^2}\right)\,dx\,\,\begin{bmatrix}u=\ln\left(1+\frac{a^2}{x^2}\right)& du=\frac{\frac{-2a^2}{x}}{x^2+a^2}\,dx \\ dv=dx& v=x\end{bmatrix}\\=\underbrace{\left.x\cdot \ln\left(1+\frac{a^2}{x^2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
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If $f(x+y^3)=f(x)+[f(y)]^3$ and $f'(0)\ge0$, what can $f(10)$ be? A real valued function satisfies the condition: $f(x+y^3)=f(x)+[f(y)]^3$ for all real $x$, $y$.
If $f'(0)\ge0$ how to find $f(10)$ ?
| Here is my attempt:
Let $(x,y)=(0,t)$:$$
f(0+t^3) = f(t^3) = f(0) + [f(t)]^3
$$We can use this to find $f(0)$:$$
f(0^3)=f(0)+[f(0)]^3 \implies f(0) - f(0) = [f(0)]^3 \implies f(0) = 0
$$This means our function simplifies to:$$
f(t^3) = [f(t)]^3
$$Next let $(x,y)=(t,c)$ where $c$ is some arbitrary constant with respect ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integrating $\int \frac{x^2}{1+x^5} \, dx $ I just encountered the following integral
$$\int \frac{x^2}{1+x^5} \, dx $$
At first it appeared to be simple, but I don't know how to solve it. Please share any ideas.
| In this problem, the hard part is the algebra.
\begin{align}
x^5 + 1 & = (x+1)(x^4-x^3+x^2-x+1) \\[10pt]
& = (x+1)\left(x^2 - 2x\cos\frac\pi5 + 1\right)\left(x^2 - 2x\cos\frac{3\pi}5 + 1\right)
\end{align}
These two quadratic factors are irreducible, as can be seen by the fact that their discriminants are negative.
Nex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Trigonometry equation maximum Given the equation: $\cos x + \sqrt3 \sin x = a^2$ find the maximum value for $a$ for which the equation has solutions and for this case solve the equation, $a \in \mathbb{R}$.
I'm guessing I need to find the maximum for the function and for this I have to differentiate it and solve it wh... | More generally,
and with no originality:
If
$f(x)
=a \sin x + b \cos x
$,
let
$c = \sqrt{a^2+b^2}$.
Then
$f(x)
=c(\frac{a}{c} \sin x + \frac{b}{c} \cos x)
=c(A \sin x + B \cos x)
$.
Since
$A^2+B^2 = 1$,
there is an angle
$\theta$
such that
$\cos \theta = A$
and
$\sin \theta = B$.
Therefore
$f(x)
=c(\cos \theta \sin x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Differentiate the Function: $g(y)=ln\frac{(2y+1)^5}{\sqrt{y^2+1}}$ $g(y)=ln\frac{(2y+1)^5}{\sqrt{y^2+1}}$
$g(y)= ln(2y+1)^5-ln\sqrt{y^2-1}$
g'(y)=$\frac{5(2y+1)^4\cdot (2)}{(2y+1)^5}-\frac{2(y^2+1)(2y)}{(y^2+1)^2}$
At this point I can cancel the (2y+1) from the numerator and denominator of the first rational equation a... | Hint
$$g=\log\frac{(2y+1)^5}{\sqrt{y^2+1}}=5\log(2y+1)-\frac 12 \log(y^2+1)$$ $$g'=5 \frac 2{2y+1}-\frac 12 \frac {2y}{y^2+1}=\frac {10}{2y+1}-\frac {y}{y^2+1}=\frac{8 y^2-y+10}{(2y+1)(y^2+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do I show that:if$p$ is prime $>5$ then $p^4-20p^2+19$ is always divisible by $180$.? Is there someone who can show me How do i show that :If $p$ is a prime number greater than $5$ then : $$p^4-20p^2+19$$ is always divisible by $180$.
Note : i think should factor $p^4-20p^2+19=$ as:$p^4-20p^2+19 = (p^2-1)(p^2-1... | Let $p^4-20p^2+19 = (p^2-1)(p^2-19)$, and $180 = 6^2\cdot 5$. But $p\equiv \pm 1\pmod{6}$, so $p^2\equiv 1\pmod{6}$, thus $p^2-1 \equiv 1-1 = 0 \pmod{6}$ and $p^2-19 \equiv 1-19 = -18 \equiv 0 \pmod{6}$, therefore $6^2 \mid p^4-20p^2+19$. On the other hand $p\not \equiv 0\pmod{5}$, so $p^2\equiv \pm 1\pmod{5}$ and $p^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find the value of $x$ such that $\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$ Find the value of $x$, $$\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$$
Help guys please, I have tried and I got, $x=-2, x=1$, and I think it's wrong
| We can solve this problem completely, without looking at anything larger than a quartic polynomial.
Let $f(x)=\sqrt{4+\sqrt{4-x}}$. The equation we are trying to solve is $f(f(x))=x$.
Observe that $f(x)$ is strictly decreasing from $f(0)=\sqrt6\gt0$ to $f(4)=2\lt4$, which means the equation $f(x)=x$ has a unique sol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Summation Theorem how to get formula for exponent greater than 3 I'm studying in the summer for calculus 2 in the fall and I'm reading about summation. I'm given these formulas:
\begin{align*}
\sum_{i=1}^n 1 &= n, \\
\sum_{i=1}^n i &= \frac{n(n+1)}{2},\\
\sum_{i=1}^n i^2 &= \frac{n(n+1)(2n+1)}{6},\\
\sum_{i=1}^n i^3 &... | For any polynomial of $p$th powers $S_p(n) = \sum_{k=1}^n k^p$ $(p \in \Bbb N)$ there is a formula of higher degree $S_p(n)=\sum_{k=1}^{p+1} a_kn^k$
The question is how to find the unknown $a_k$ coefficients which are real numbers.
So use the following formulas to find them:
$a_k= \frac{1}{k!} \cdot \sum_{j=1}^{p+1-k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1353294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 9,
"answer_id": 5
} |
How can we solve a multi-variable recurrence relation in closed form, when the number of terms is also variable? Consider the formula
$f(x, y) = f(x, y-1) + 2 \sum\limits_{i=1}^{x-1} f(i, y-1) $
The factor '2' makes this not expressible cleanly as $f(x, y) = f(x, y-1) + f(x-1, y)$, which is solved here using generating... | Argh... better write this as:
$$
f(k + 1, n + 1) = f(k + 1, n) + 2 \sum_{1 \le i \le n} f(i, n)
$$
Define $F(x, y) = \sum_{k, n \ge 1} f(k, n) x^k y^n$, multiply the recurrence by $x^k y^n$, sum over $k \ge 1$ and $n \ge 1$. Recognize the resulting sums
$$
\frac{F(x, y) - \sum_{k \ge 1} f(k, 1) x^k - \sum_{n \ge 1} f(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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quadratic approximation of $e^{x^3}$ at $x=0$ we know that
$$P(x)=f(a)+f'(a)x+\frac{f''(a)}{2}x^2 $$
therefore
$f(0)=1$
if $x^3=u$, so $$f'(e^u)=e^{x^3}3x^2$$ and $$f''(e^u) = 4(e^u)'3x^2+e^u (3x^2)'=e^{x^3}3x^23x^2+e^{x^3}6x=e^{x^3}(9x^4+6x)$$
$$P(x)=1+e^{x^3}3x^3+\frac{e^{x^3}(9x^4+6x)}{2}x^2 $$
is it correct?
can yo... | We can just kick it and do: $$e^x = \sum_{n \geq 0}\frac{x^n}{n!} \implies e^{x^3} = \sum_{n \geq 0}\frac{x^{3n}}{n!},$$so that: $$e^{x^3} = 1 + x^3 + \frac{x^6}{2} + \frac{x^9}{6}+\cdots.$$We have no $x$ and $x^2$ terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1357847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that $\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{1\cdot 2\cdot 3\cdot \ldots \cdot n}\leq 2^{n}$ for all $n\in\mathbb{N}$.
Show that $$\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{1\cdot 2\cdot 3\cdot \ldots \cdot n}\leq 2^{n}\qquad (n\in \mathbb{N}).$$
I want to show the last step, that is, the inductiv... | For the step from $n=k$ to $n=k+1$, we multiply by $\frac{2k+1}{k+1}$, which is less than $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Proof of $\sum^{2N}_{n=1} \frac{(-1)^{n-1}}{n} = \sum^{N}_{n=1} \frac{1}{N+n}$ The title pretty much summarizes my question. I am trying to prove the following:
$$\displaystyle \forall N \in \mathbb{N}: \sum^{2N}_{n=1} \frac{(-1)^{n-1}}{n} = \sum^{N}_{n=1} \frac{1}{N+n}.$$
I tried proving this using induction. Startin... | It's easy to prove it using integration,
$$\sum_{k=1}^r\frac{1}{k+r}=\int_0^1\sum_{k=1}^r x^{k+r-1}\ dx=\int_0^1\frac{x^r-x^{2r}}{1-x}\ dx$$
$$=\int_0^1\frac{1-x^{2n}-(1-x^r)}{1-x}\ dx=H_{2r}-H_r\tag1$$
$$\overline{H}_{2r}=\sum_{k=1}^{2r}\frac{(-1)^{k-1}}{k}=\sum_{k=1}^{2r}(-1)^{k-1}\int_0^1 x^{k-1}\ dx=\int_0^1\sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
How many four digit numbers are perfect square whose first and last two digits are same? I tried it by assuming the number as $\sqrt{1100a+11b}$ and than tried to find figure out perfect square but I am unable to approach further.
| It's easy to solve the problem with brute force, but here is a handmade solution.
So, we are trying to find integers $a,b,x$ so that $1100a+11b=x^2$, $x>0$, $1\leq a\leq9$ and $0\leq b\leq9$.
(In fact, we must have $30<x<100$.)
Considering the equation modulo 11, we get $x^2\equiv0\pmod{11}$.
It is easy to check (there... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
About $(x^3 - 4)^2 - x^6 + 2x^5 = 2x^5 -8x^3 + 16$ Studying polynomials I got the follows:
$$
(x^3 - 4)^2 - x^6 + 2x^5 = 2x^5 -8x^3 + 16
$$
I can't understand from where we got this $-8x^3$.
I got to simplify this polynomial just to:
$$
2x^5 + 16
$$
Can someone help me understand from where we got the expression $-... | When expanding $(x^3-4)^2$, you get $x^6-4x^3-4x^3+16=x^6-8x^3+16.$ Then, the $x^6$ is subtracted off, and you are left with $2x^5-8x^3+16$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the matrix of a linear transformation with respect to given (non-standard) ordered bases I am working on the following question from a study guide for my exam on Friday.
Let $L: ℝ^3 → ℝ^3$ be a linear transformation, represented in standard basis by: $$ L(e_1) = \begin{pmatrix}-3 \\ -1 \\ -2 \end{pmatrix}, L... | You should apply $L$ to each column of $B$, then write them in linear combination of $B'$. The coefficients are the columns of the transformation. For example,
$$LB_1=\begin{pmatrix}-3&1&-5\\-1&4&-6\\-2&2&-5\end{pmatrix}\begin{pmatrix}0\\-1\\2\end{pmatrix}=\begin{pmatrix}-11\\-16\\-12\end{pmatrix}=-25\begin{pmatrix}1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1361917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find numbers $\overline{abcd}$ so that $\overline{abcd}+\overline{bcd}+\overline{cd}+d+1=\overline{dcba}$ Find the numbers $\overline{abcd}$, with digits not null that satisfy the equality
\begin{equation}\overline{abcd}+\overline{bcd}+\overline{cd}+d+1=\overline{dcba}\end{equation}
where
\begin{equation}\overline{abc... | I'm late to the game here, but I find it interesting that no one used the mod 3 identity for digit sums in base 10, which lets you go from the inside digits out, instead of the outside in.
\begin{align}
\overline{abcd} + \overline{bcd} + \overline{cd} + d + 1 &= \overline{dcba}&(\textrm{mod 3}) \\
(a + b + c + d) + (b ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the relation between $a,b $ and $c$ in quadratic equation.
If the roots of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0,\ \ \{a,b,c,x\}\in \mathbb{R}$ are equal, then $a,b,c$ are in
Options
$a.)\ AP\\
b.)\ GP\\
\color{green}{c.)\ HP}\\
d.)\ \text{cannot be determined}\\$
by using discriminant property
$[b(c-a)]^2-4a... | $$\begin{align}&b^2(c-a)^2=4ac(b-c)(a-b)\\&\Rightarrow b^2(c^2-2ac+a^2)=4ac(ab-b^2-ca+bc)\\&\Rightarrow b^2c^2-2ab^2c+a^2b^2=4a^2bc-4ab^2c-4a^2c^2+4abc^2\\&\Rightarrow b^2c^2+a^2b^2=4a^2bc-2ab^2c-4a^2c^2+4abc^2\end{align}$$Dividing the both sides by $a^2b^2c^2$, we have$$\begin{align}&\Rightarrow \frac{1}{a^2}+\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Polynomials: Linking equations via their roots The equation $ax^3+bx^2+cx+d=0$ has solutions $1,p$ and $q$. The equation $x^3+sx^2+tx+r=0$ has solutions $1, 1/p$ and $1/q$. Show that $r=a/d$ , and find an expression for $s$ in terms of $c$ and $d$ , simplifying your answer.
| By the fundamental theorem of algebra:
\begin{align}
ax^3+bx^2+cx+d&=a(x-1)(x-p)(x-q)\\&=a\left(x^3-(1+p+q)x^2+(p+q+pq)x-pq\right)\end{align}
and
\begin{align}
x^3+sx^2+tx+r&=(x-1)\left(x-\frac{1}{p}\right)\left(x-\frac{1}{q}\right)\\& = x^3-\left(1+\frac{1}{p}+\frac{1}{q}\right)x^2+\left(\frac{1}{p}+\frac{1}{q}+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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how to prove that the circle $(x-a)^2+(y-b)^2=a^2+b^2$ is passing through point $(0,0)$ How can one prove that the circle $(x-a)^2+(y-b)^2=a^2+b^2$ is passing through point $(0,0)$?
I know that if i put: $x=y=0$, i will get: $(0-a)^2+(0-b)=a^2+b^2=a^2+b^2$
But that's not a proof but checking.
Thanks.
| Notice, the equation $(x-a)^2+(y-b)^2=a^2+b^2$ represents a circle with a radius $\sqrt{a^2+b^2}$ & center at the point $(a, b)$.
Now, the distance of the given point $(0, 0)$ from the center $(a, b)$ of the circle $$=\sqrt{(a-0)^2+(b-0)^2}$$ $$=\sqrt{a^2+b^2}$$ $$=\text{radius of circle}$$
But the above result is tr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find the matrix $\mathbf{A}$ if $A\binom{7}{-1} = \binom{6}{2}.$ Find the $2\times2$ matrix $A$ where $A^2=A$ and
$$A\begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}.$$
I tried plugging in: $A= \begin{pmatrix}a&b\\c&d\end{pmatrix}$ but that became messy very quickly. I got the equations:
$7a... | Since
$A^2= A, \tag{1}$
we have
$A(I - A) = (I - A)A = A - A^2 = 0, \tag{2}$
whence, for any vector $x$,
$A(I - A)x = (I - A)Ax = 0. \tag{3}$
(3) indicates that vectors $Ax \ne 0$ in the image of $A$ are in the kernel of $I - A$, i.e., are eigenvectors of $A$ with eigenvalue $1$, and likewise that vectors $0 \ne (I - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
IMC 2011 Day 1 Problem 2 - Linear Algebra I have been reading the solutions of a past IMC paper (from 2011, Day 1)
and I did not understand the solution to Problem 2 completely.
Problem 2: Does there exist a real $3$x$3$ matrix $A$, such that $tr(A)=0$ and $A^2+A^t=I$?
Official solution:
"The Anwser is NO. Suppose that... | So first of all, you're right, they made a mistake, in the stated setting, we have for the eigenvalues of $A$ the possible set of
$$\left\{0,1,(\frac{-1+\sqrt 5}{2}),(\frac{-1-\sqrt 5}{2})\right \}
$$
which gives us, since for $\lambda$ being an eigenvalue to the eigenvector $v$ of $A$
$$
A(Av)=A\lambda v=\lambda Av=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Induction Proof using factorials Recall that for $n \in N$, $n! = 1 \cdot 2 \cdots n$.
Prove the following for each $n \in N$:
$$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \cdots + \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}$$
I understand how to do the proof, but in the inductive step I am facing some difficulty proving... | $$1-\frac{1}{(k+1)!}+\frac{k+1}{(k+2)!}$$
Note that to simplify this we need a common denominator. Let it be $(k+2)!$. Recall that $(k+1)! = (k+1)(k)(k-1)(k-2) \cdots$ So to get a $(k+2)!$ in the denominator of the fraction we must multiply the numerator and denominator by $k+2$ and get:
\begin{align*}
1-\frac{k+2}{(k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Baby Rudin claim: $1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}...$ converges This sequence is a rearrangement of the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}...$. Note that at this point in the text we do not have any theorem about the convergence ... | Here is a general result:
Let $s$ be an infinite sum over a sequence $(a_n)_{n\in\mathbb N}$, where $a_n$ are a permutation of $\frac{(-1)^{k+1}}{k}$, but where the sub sequences $(a_{n_k})_{k\in\mathbb N},\space a_{n_k}=-\frac{1}{2k}$ and $(a_{m_k})_{k\in\mathbb N},\space a_{m_k}=\frac{1}{2k-1}$ exist with $n_k<n_{k+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Differentiate the Function: $y=2x \log_{10}\sqrt{x}$ $y=2x\log_{10}\sqrt{x}$
Solve using: Product Rule $\left(f(x)\cdot g(x)\right)'= f(x)\cdot\frac{d}{dx}g(x)+g(x)\cdot \frac{d}{dx}f(x)$
and $\frac{d}{dx}(\log_ax)= \frac{1}{x\ \ln\ a}$
$(2x)\cdot [\log_{10}\sqrt{x}]'+(\log_{10}\sqrt{x})\cdot [2x]'$
$y'=2x\frac{1}{\sqr... | Firstly, note that $$\log_{10} \sqrt{x} = \dfrac{1}{2} \log_{10} x$$
Then we have that $$ \begin{aligned} f(x) = \log_{10} (x) & \iff 10^{f(x)} = x \\ & \implies f(x) \log 10 = \log x \\ & \implies \dfrac{\text{d}}{\text{d}x} f(x) \log 10 = \dfrac{\text{d}}{\text{d}x} \\ & \implies f'(x) \log 10 = \dfrac{1}{x} \\ & \im... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
Is there a way to show the following equality without induction I wanted to show the following equality without using induction:
$$
\sum_{k=2}^n \frac{1}{k(k-1)} = \frac{n-1}{n}
$$
Any hint on how to do it?
| Hint: Partial fraction decomposition yields $$\frac{1}{k(k-1)} = \frac{1}{k-1} - \frac{1}{k}$$ which makes your series a special little beast called a telescoping series, try writing out the first few terms and see if you can spot a pattern that allows cancelling.
In particular, you get $$\sum_{k=2}^n \frac{1}{k(k-1)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Diagonalize a symmetric matrix let $$A = \left(\begin{array}{cccc} 1&2&3\\2&3&4\\3&4&5 \end{array}\right)$$
I need to find an invertible matrix $P$ such that $P^tAP$ is a diagonal matrix and it's main diagonal may have only the terms from the set $\{ 1,-1,0 \}$
I'd be glad if you could explain to me how to solve this. ... | This trick is due to Hermite. It is especially useful when you have a symmetric matrix of integers. First, we write a certain function in three variables,
$$ f(x,y,z) = x^2 + 3 y^2 + 5 z^2 + 8 yz+ 6 zx +4xy, $$ because this is exactly the result of calculating $v^t A v,$ with
$$
v =
\left(
\begin{array}{c}
x \\
y \\
z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
find $\frac{dy}{dx}$ in terms of $x$ and $y$ of $x^2y^2=\frac{(y+1)}{(x+1)}$ - basic question
Find $\frac{dy}{dx}$ in terms of $x$ and $y$ of $x^2y^2=\frac{(y+1)}{(x+1)}$
Ok so using the product rule on the LHS and the quotient rule on the RHS I differentiated both sides of the equation and got:
$2xy(x\frac{dy}{dx}+y... | Notice, $$x^2y^2=\frac{y+1}{x+1}$$ $$x^2(x+1)y^2=y+1$$ $$(x^3+x^2)y^2=y+1$$ Now, differentiating both the sides w.r.t. $x$ by applying chain-rule as follows $$\frac{d}{dx}(x^3
+x^2)y^2=\frac{d}{dx}(y+1)$$ $$(3x^2+2x)y^2+2(x^3+x^2)y\frac{dy}{dx}=\frac{dy}{dx}$$ $$2(x^3+x^2)y\frac{dy}{dx}-\frac{dy}{dx}=-(3x^2+2x)y^2$$ $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$
Given $|x|<1 $ prove that $\\1+2x+3x^2+4x^3+5x^4+...=\frac{1}{(1-x)^2}$.
1st Proof: Let $s$ be defined as
$$
s=1+2x+3x^2+4x^3+5x^4+\cdots
$$
Then we have
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s... | As I'd suggested like Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $,
Using Generalized Binomial Expansion, $$(1+y)^n=1+ny+\frac{n(n-1)}{2!}y^2+\frac{n(n-1)(n-2)}{3!}y^3+\cdots$$ given the converge holds
Comparing with given ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 15,
"answer_id": 3
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Is this problem wrongly built? Or is there a solution which I don't know how to arrive at? I was solving a Cauchy-Schwarz's inequality based problem. Given that $x^2+y^2+z^2=1$ I am supposed to show that $x+y+z \le 6$. After struggling for a while I realised that I could solve this inequality had the condition been $x^... | We have $x^2\le1$, and thus $x\le1$. Similarly, we have $x+y+z\le3<6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$\int\dfrac{dx}{x^2(x^4+1)^{3/4}}$
Evaluate $$\large{\int\dfrac{dx}{x^2(x^4+1)^{3/4}}}$$
I thought of rewriting this as $$\large{\int\dfrac{dx}{x^5(1+\frac{1}{x^4})^{3/4}}}$$ and substituting $$u^4=\left(1+\frac{1}{x^4}\right)\Rightarrow u=\left(1+\frac{1}{x^4}\right)^{1/4}$$ and subsequently I got $$du=\dfrac{1}{4}\... | $\bf{My\; Solution::}$ Let $\displaystyle I = \int\frac{1}{x^2\left(x^4+1\right)^{\frac{3}{4}}}dx\;,$ Now Let
Let $\displaystyle x = \frac{1}{t}\;,$ Then $\displaystyle dx = -\frac{1}{t^2}dt\;,$ Then Integral Convert into
$\displaystyle = -\int \frac{t^3}{(1+t^4)^{\frac{3}{4}}}dt\;,$ Now Let $(1+t^4) = u\;, $ Then $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
given $2f(x) + f(1-x) = x^2$ find $f(-5)$ I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.
A function $f$ has property that $2f(x)+ f(1-x) = x^2$ for any number of x. What is the number for $f(-5)$?
(A) $\frac {15}4$ (B) $\frac... | We have, $$2f(x)-f(1-x)=x^2$$ $$\implies 2f(6)-f(1-6)=6^2$$ $$\implies 2f(6)-f(-5)=36\tag 1$$ $$ 2f(-5)-f(1-(-5))=(-5)^2$$ $$\implies 2f(-5)-f(6)=25$$ $$\implies 4f(-5)-2f(6)=50\tag 2$$ Now, adding (1) & (2), we get $$3f(-5)=36+50=86$$ $$\implies f(-5)=\frac{86}{3}$$ Your answer is correct. There may be some printing m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Number of divisors of the form $(4n+1)$
Find the number of divisors of $$2^2\cdot3^3\cdot5^3\cdot7^5$$ which are of the form $(4n+1)$
I know how to find the total number of divisors. But, to find the number of divisors of the form $(4n+1)$, I'm thinking of listing down the divisors and then finding, but that'd be v... | Number of divisors of $$N= 2^2\times3^3\times5^3\times7^5,$$which are of the form $4n+1$ exculuding
$$\begin{align}1 &=\{ \text{number of terms in product}\}\\
&=(1+3^2)(1+5+5^2+5^3)(1+7^2+7^4)+\{\text{number of terms in product}\}\\
&= ( 3+3^3)(7+7^3+7^5)(1+5+5^2+5^3)-1\\
&= 2\times4\times3+2\times3\times4-1\\
&= 47
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
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Prove that if $x \neq 0$, then if $ y = \frac{3x^2+2y}{x^2+2}$ then $y=3$ The problem is the following (Velleman's exercise 3.2.10):
Suppose that $x$ and $y$ are real numbers. Prove that if $x \neq 0$, then if $ y = \frac{3x^2+2y}{x^2+2}$ then $y=3$.
My approach so far: Suppose $x \neq 0$. Suppose $ y = \frac{3x^2+2... | Givens
*
*$x\neq 0$
*
*$y = \frac{3x^2+2y}{x^2+2}$
Goal
*
*$y=3$
Proof sketch:
*
*Suppose $x\neq 0$.
*Suppose $y = \frac{3x^2+2y}{x^2+2}$. This means that $y (x^2+2)=3x^2+2y \Rightarrow x^2y+2y=3x^2+2y $.
$\qquad \quad$ [Proof of $y=3$ goes here.]
$\qquad$ 3. Therefore $y=3$
*Thus, if $x\neq ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Proving that $1\cdot3+3\cdot5+5\cdot7+\cdots+(2n-1)(2n+1)={n(4n^2+6n-1) \over 3}$ by induction for $n\geq 1$
Prove using mathematical induction that
$$1\cdot3+3\cdot5+5\cdot7+\cdots+(2n-1)(2n+1)= {n(4n^2+6n-1) \over 3}.$$
Step 1: If we assume that the equation is true for a natural number, $n=k$, then we get
$$1\cdot... | From the formula for the sum of the square of the first $n$ natural numbers we have
\begin{align*}
\sum_{k=1}^n(2k-1)(2k+1)&=\sum_{k=1}^n(4k^2-1)\\
&=4\sum_{k=1}^nk^2-\sum_{k=1}^n1\\
&=\frac{4n(n+1)(2n+1)}{6}-n\\
&=\frac{n\left[4(n+1)(2n+1)-6\right]}{6}\\
&=\frac{n(4n^2+6n-1)}{3}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Inequality - Cauchy Schwarz Let $a, b, c, d > 0 \in \mathbb{R}$ such that $a^2 + b^2 + c^2 + d^2 = 4$. Show that:
$S = \frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \geq 4$
My approach:
I used the Cauchy-Schwarz inequality to show that $S \geq a + b + c + d$ but that is useless as $a + b + c + d \leq 4$... | By Holder's Inequality,
$$S^2\sum_{cyc} a^2b^2\ge (a^2+b^2+c^2+d^2)^3$$
So it remains to note that by AM-GM
$$\sum_{cyc}a^2b^2 = (a^2+c^2)(b^2+d^2) \le \frac{\left((a^2+c^2)+(b^2+d^2)\right)^2}4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solve the equation $(m^2-m-2)x=m^2+4m+3$ Here's how I solve it
I think that m is the variable (am I right?).
Then
$$m^2x-mx-2x-m^2-4m-3=0$$
$$m^2(x-1)-m(x+4)-(2x+3)=0$$
$$D=x^2+8x+16+4(x-1)(2x+3)$$
$$=x^2+8x+16+4(2x^2-2x+3x-3)$$
$$=9x^2+12x+4$$
$$=(3x+2)^2$$
$$m=\frac{x+4\pm (3x-2))}{2(x-1)}$$
$$m_1=\frac{4x+2}{2x-2... | $$(m^2-m-2)x=m^2+4m+3$$
$$(m-2)(m+1)x=(m+1)(m+3)$$
$$m=-1\,\,\,\,\,\text{or}\,\,\,\,\,(m-2)x=(m+3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Is the area of this pentagon $4-\sqrt 5$?
Consider a regular pentagon with vertices (in clockwise order) $A, B, C, D, E$, let $A'$ be the point of intersection of $BD$ and $CE$, let $B'$ be the point of intersection of $CE$ and $DA$, and so on. If $\triangle AC'D'$ has area 1, what is the area of the pentagon $A'B'C'D... | Let $\color{red}{C'D'=a}$ in isosceles $\triangle AC'D'$ then the angle of vertex A is given as $$\angle C'AD'=\frac{180^\circ}{5}=36^\circ\implies \angle C'AM=\frac{\angle C'AD'}{2}=18^\circ$$ Now, drop a perpendicular say $AM$ from vertex $A$ to the side $C'D'$ in $\triangle AC'D'$, we get $$\tan\angle C'AM=\frac{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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find x in $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$
Which one satisfies the equation $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$
(A)$27$ (B)$32$ (C)$45$ (D)$52$ (E)$63$
let $a = 6+\sqrt x , b=6-\sqrt x$
cube each side
\begin{align}
(\sqrt[3]a + \sqrt[3]b)^3 &= (\sqrt[3]3)^3 \\
(\sqrt[3... | \begin{align*} a + b + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= 3 \\
6+\sqrt{x} + 6-\sqrt{x} + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= 3 \\
12 + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= 3 \\
3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= -9 \end{align*}
Divide by $3\sqrt[3]{ab}$:
$$\sqrt[3]{b} + \sqrt[3]{a} = \frac{-9}{3\sqrt[3]{ab}}$$
Us... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
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How should I go about solving this definite integral? The integral is:
$$\int_{-1}^1\sqrt{4-x^2}dx$$
I'm having difficulty figuring out how to go about this. I attempted to use u-substitution, both by substituting $u$ for $\sqrt{4-x^2}$ entirely, and then just $x^2$, but I quickly realized that neither of those work. I... | Let $x = 2 \sin \theta$. Then $dx = 2\cos \theta \; d\theta$. When $x = -1, \theta = -\frac{\pi}{6}$ and when $x = 1, \theta = \frac{\pi}{6}$. Therefore
$$\int_{-1}^1 \sqrt{4-x^2} \; dx = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \sqrt{4 - 4 \sin^2 \theta} \; 2\cos \theta \; d\theta = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Solve for $x$ - Logarithm Equation $\ln x+\ln(x+1)=\ln 2$ My attempt:
$\ln x(x+1)=\ln 2$
$e^{\ln x(x+1)}=e^{\ln 2}$
$x(x+1)=2$
$x^2+x-2=0$
$(x-1)(x+2)=0$
therefore $x=1, -2$
| $$\ln { x+\ln { \left( x+1 \right) =\ln { 2 } } } \\ \ln { x\left( x+1 \right) =\ln { 2 } } \\ x\left( x+1 \right) =2\\ x^{ 2 }+x-2=0\\ \left( x-1 \right) \left( x+2 \right) =0\\ { x }_{ 1 }=-2,{ x }_{ 2 }=1\\ $$
x should be $x>0$ hence ${ x }_{ 2 }=1$ is root
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Show that $B$ is invertible if $B=A^2-2A+2I$ and $A^3=2I$ If $A$ is $40\times 40$ matrix such that $A^3=2I$ show that $B$ is invertible where $B=A^2-2A+2I$.
I tried to evaluate $B(A-I)$ , $B(A+I)$ , $B(A-2I)$ ... but I couldn't find anything.
| $B=A^2-2A+2I$. Now multiplying the sides respectively by $A$ and $A^2$, and using the assumption $A^3=2I$ yields:
$$AB=BA=-2A^2+2A+2I$$
$$A^2B=BA^2=2A^2+2A-4I$$
Therefore
$$AB+A^2B=4A-2I$$
$$AB+2B=-2A+6I$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 9,
"answer_id": 8
} |
This expression is always a perfect square How to show that for $x,y\in \Bbb R$, the expression $xy+\left(\frac{x-y}{2} \right)^2$ is always a perfect square?
For example $x=7, y=3$, $7\times 3+\left(\frac{7-3}{2} \right)^2=25=5^2$
| $$xy+\left(\frac{x-y}{2} \right)^2\\
=\frac{4xy+x^2+y^2-2xy}{4}\\
=\left(\frac{x+y}{2}\right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Let $D,E,F$ be (respectively) the projections of $O$ on $BC,CA,AB$. Prove that $\cot{\angle ADB} + \cot{\angle BEC} + \cot{\angle CFA} =0$
Let $O$ be an arbitrary point located inside the triangle $ABC$. Let $D, E, F$ be (respectively) the projections of $O$ on $BC, CA, AB$. Prove that $$\cot{\measuredangle ADB} + \co... | This is a nice question. I am assuming all the angles are directed. (Directed angles modulo $180^\circ$ are sufficient here, since $\cot$ is periodic with period $180^\circ$.)
We direct our three sidelines $BC$, $CA$ and $AB$. All distances along these lines will thus be understood to mean directed distances. Let $X$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Number Theory Problem involving fractional part of a number If
$x = ( 9 + 4 \sqrt {5} )^{48}$
where $x = [x] + f$, where $[x]$ is he integral part of $x$ , and $x$ is its fractional part
How do I go about finding the value of $x(1-f)$ ?
Thanks!
| Imagine expanding $A=(9+4\sqrt{5})^{48}+(9-4\sqrt{5})^{48}$ using the Binomial Theorem. The terms that involve odd powers of $\sqrt{5}$ cancel, so $A$ is an integer.
Note that $(9-4\sqrt{5})^{48}$ is positive and close to $0$. It follows that the integer part of $(9+4\sqrt{5})^{48}$ is $A-1$, so the fractional part $f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Expressing the integral in terms of the original variable In evaluating the integral:
$$ \int{dx\over(a^2-x^2)^{3/2}} $$ or $$ \int{dx\over(a^2-x^2)^{1/2}\ (a^2-x^2)}$$
Let $ x=a\sin\theta $ and $ dx=a\cos\theta\ d\theta $. Then
$$ \int{{a\cos\theta\ d\theta}\over{a\cos\theta\ (a^2-a^2\sin^2\theta)}} = {1\over a^2}\in... | Solution without trigonometric functions.
Given
$$
\int \frac{dx}{ \sqrt{ a^2 - x^2 }^3 }.
$$
Write this as
$$
\int \frac{dx}{ \sqrt{ a^2 - x^2 }^3 }
= \frac{1}{a^2} \int \frac{a^2 - x^2 + x^2}{ \sqrt{ a^2 - x^2 }^3 } dx
$$
or
$$
\int \frac{dx}{ \sqrt{ a^2 - x^2 }^3 }
= \frac{1}{a^2} \int \frac{dx}{ \sqrt{ a^2 - x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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An infinite sum of lengths. Let both the base and height of the following triangle have the length 6m. If the lines drawn inside the triangle bisect each right angle formed while proceeding. Then find the length of all these lines. The reason i could not do it is that i did not find any series representation of the len... | First length element is
$$
\frac{1}{2} \sqrt{2} a
$$
Second length element is
$$
\frac{1}{2} a
$$
Now you repeat this for a smaller triangle with
$$
\frac{1}{2} a
$$
Then
$$
\Bigg( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \Bigg)
\Big( 1 + \sqrt{2} \Big) a
$$
Result
$$
= \Big( 1 + \sqrt{2} \Big) a
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Find min & max of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$ Problem: Find the maximum and minimal value of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$.
Since $x^2 > x$ (edit $x^2 \geq x$) for all $x \in \mathbb{R}$, $f$ is bowl-ish with a minimal value in the bottom.
This is a critical point which means t... | $$2(x^2+y^2)-(x+y)^2=(x-y)^2\ge0\iff(x+y)^2\le2(x^2+y^2)=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
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A puzzling step in a solution for "Find $\sin(x)$ and $\cos(x)$, if $a\sin(x)+b\cos(x)=c$" A textbook problem:
Find $\sin(x)$ and $\cos(x)$, if $a\sin(x)+b\cos(x)=c$
The solution from the textbook:
Let's divide each term of this equation by $\sqrt{a^2+b^2}$:
$$\frac{a}{\sqrt{a^2+b^2}}\sin(x)+\frac{b}{\sqrt{a^2+b^2... | $$\sin^2(x-\phi)+\cos^2(x-\phi)=1$$
$$\sin(x-\phi)=\pm\sqrt{1-\cos^2(x-\phi)}$$
But $\cos(x-\phi)=\frac{c}{\sqrt{a^2+b^2}}$
Therefore $\sin(x-\phi)=\pm\sqrt{1-\cos^2(x-\phi)}=\pm\sqrt{1-(\frac{c}{\sqrt{a^2+b^2}})^2}=\pm\sqrt{1-\frac{c^2}{a^2+b^2}}$
This gives: $\pm\sqrt{\frac{a^2+b^2}{a^2+b^2}-\frac{c^2}{a^2+b^2}}=\pm... | {
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"url": "https://math.stackexchange.com/questions/1384945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Proving $(1 + \frac{1}{n})^n < n$ for natural numbers with $n \geq 3$. Prove with induction on $n$ that \begin{align*} \Bigl(1+ \frac{1}{n}\Bigr)^n < n \end{align*} for natural numbers $n \geq 3$.
Attempt at proof: Basic step. This can be verified easily.
Induction step. Suppose the assertion holds for $n >3$, then we... | HINT:
$$
\frac{(n+1)^2-1}{n+1}<\frac{(n+1)^2}{n+1}.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For positive integers $x, y, k$, prove that $4^x (4^y+1)=k(k+1)$ implies $x = y$ In the proof that I read, even $k$ implies $4^x=k$ and $4^y+1=k+1$. I am wondering why we don't need to factorize $4^y+1$ into $pq$, such that $p, q > 1$, and solve for $4^x p=k$, $q=k+1$.
| Note that $k^2 < k(k+1) = 4^{x+y}+4^x < (2^{x+y}+2^x)^2$, i.e. $k < 2^{x+y}+2^x$,
and that $(2^{x+y})^2 < 4^{x+y}+4^x = k(k+1) < (k+1)^2$, i.e. $2^{x+y}-1 < k$.
Since $2^x$ divides $4^x(4^y+1)$, either $2^x$ divides $k$, or $2^x$ divides $k+1$.
If $k$ is a multiple of $2^x$, then since $2^{x+y}-1 < k < 2^{x+y}+2^x$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How prove this inequality $\left|\int_{a}^{b}f(x)dx\right|\leq \frac{1}{8}(b-a)^{2}\int_{a}^{b}\left|f''(x)\right|dx$ Let $f(x)\in \mathcal C^2([a,b]),f(\frac{a+b}{2})=0$.
Show that$$\left|\int_{a}^{b}f(x)dx\right|\leq \frac{1}{8}(b-a)^{2}\int_{a}^{b}\left|f''(x)\right|dx.$$
I try to use Taylor's Theorem with Integral... | Changing the order of integration, we obtain
\begin{align}
\int_{\frac{a+b}{2}}^b \int_{\frac{a+b}{2}}^x (x-t)f''(t)\,dt\,dx
&= \int_{\frac{a+b}{2}}^b \int_t^b (x-t)\,dx\, f''(t)\,dt\\
&= \int_{\frac{a+b}{2}}^b \frac{(b-t)^2}{2}f''(t)\,dt.
\end{align}
Now we can take the absolute value to get
\begin{align}
\biggl\lvert... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Finding $\frac{ \mathrm{d}y}{ \mathrm{d}x}$ when given that $\sqrt{1-x^2} + \sqrt{1-y^2} ={a}(x-y)$ Finding $\frac{ \mathrm{d}y}{ \mathrm{d}x}$ when given that $\sqrt{1-x^2} + \sqrt{1-y^2} = a(x-y)$. $a$ is a constant.
I have the final answer, which is $$\frac{ \mathrm{d}y}{ \mathrm{d}x} = \sqrt{\frac{1-y^2}{1-x^2}}.$$... | Solve explicitly for $y$:
$y = \frac{a^2 x-2 a \sqrt{1-x^2}-x}{a^2+1}$.
Then take the derivative:
${d y \over d x} = \frac{a^2+\frac{2 a x}{\sqrt{1-x^2}}-1}{a^2+1}$
or
$-\frac{2 a \sqrt{1-x^2}}{a^2+1}-\frac{2 x}{a^2+1}+x$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of non-real roots of equation? What is the sum of all non-real, complex roots of this equation -
$$x^5 = 1024$$
Also, please provide explanation about how to find sum all of non real, complex roots of any $n$ degree polynomial. Is there any way to determine number of real and non-real roots of an equation?
Please ... | $\bf{My\; Solution::}$ Given $$x^5 = 1024 = 2^{10} = 4^5\Rightarrow x^5-4^5 = 0$$
So $$(x^5-4^5) = (x-5)\cdot (x^4+x^3\cdot 4+x^2\cdot 4^2+x\cdot 4^3+4^4) = 0$$
Above we use the formula $\displaystyle x^n-a^n = (x-a)\cdot (x^{n-1}+x^{n-2}\cdot a+x^{n-3}\cdot a^2+.......+a^{n-1})$
So We Get $x=4(\bf{Real \; Root})$ and ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int \frac{a}{x(x^2-a^2)^{1/2}} dx$ I was trying to integrate $\int \frac{a}{x(x^2-a^2)^{1/2}} dx$ and by applying on what i saw on the formula of inverse trigonometric functions, there is formula like $\frac{1}{a}Arcsec\frac{u}{a}$ = $\int \frac{du}{u\sqrt(u^2-a^) }$ so my answer is $\frac{1}{a}Arcsec\frac{x... | Notice, we have $$\int\frac{adx}{x\sqrt{x^2-a^2}}$$ Let $x=a\sec \theta \implies a\sec\theta\tan \theta d\theta=dt$
$$\int\frac{a^2\sec\theta\tan \theta d\theta}{a\sec\theta\sqrt{a^2 \sec^2\theta-a^2}}$$
$$=\int\frac{a^2\sec\theta\tan \theta d\theta}{a^2\sec\theta\tan \theta}$$ $$=\int d\theta=\theta+c_1$$ Now, setting... | {
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How to solve $\displaystyle\lim_{x\to 0}\tfrac{\sqrt{x+25}-5} {\sqrt{x+16}-4}$ \begin{eqnarray}
\\&\lim_{x\to 0}\frac{\sqrt{x+25}-5} {\sqrt{x+16}-4}
\end{eqnarray}
Undefined limit
\begin{eqnarray}
\frac{0} {0}
\end{eqnarray}
| Hint: By multiplying the numerator and denominator by $(\sqrt{x+25}+5)(\sqrt{x+16}+4)$, and then using the difference of squares identity $(a-b)(a+b) = a^2-b^2$, we have:
$\dfrac{\sqrt{x+25}-5}{\sqrt{x+16}-4}$ $= \dfrac{(\sqrt{x+25}-5)(\sqrt{x+25}+5)(\sqrt{x+16}+4)}{(\sqrt{x+16}-4)(\sqrt{x+16}+4)(\sqrt{x+25}+5)}$ $= \d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve trigonometric inequality $ \sin x \sin 2x - \cos x \cos 2x > \sin 6x $ Solve this trigonometric inequality:
$$ \sin x \sin 2x - \cos x \cos 2x > \sin 6x $$
My steps:
$$ \cos x \cos 2x - \sin x \sin 2x < - \sin 6x $$
$$ \cos 3x < \sin (-6x)$$
$$ \cos 3x < \cos (\frac{\pi}{2}+6x) $$
From this we get:
$$ 3x > \d... | $$ \sin (x) \sin (2x) - \cos (x) \cos (2x) > \sin (6x) \Longleftrightarrow$$
$$ -\cos(3x) > \sin (6x) $$
So we find $4$ solutions:
$$x=\frac{2\pi n +\pi}{3}$$
$$\frac{2\pi n -\pi}{3}<x<\frac{12\pi n -5\pi}{18}$$
$$\frac{4\pi n -\pi}{6}<x<\frac{12\pi n -\pi}{18}$$
$$\frac{4\pi n +\pi}{6}<x<\frac{2\pi n +\pi}{3}$$
With $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate the integral $PV\int_{-\infty}^{\infty} \frac{1}{(x^2+1)(x^2+2x+2)}dx$ Using the Cauchy Integral Principal to evaluate:
$$PV\int_{-\infty}^{\infty} \frac{1}{(x^2+1)(x^2+2x+2)}dx$$
I know this integral has a pole at $x=i$, $x = -1-i$ and $x = -1+i$. Can someone please show me how to use Cauchy Principal value ... | Let $C_R$ denote the closed semi-circle of radius $R$ centered at $z=0$ in the lower-half plane oriented in anti-clockwise fashion.
$$
\int_{C_R}{dz\over(z^2+1)(z^2+2z+2)}=\\\int_{\pi}^{2\pi}{iRe^{it}dt\over(R^2e^{2it}+1)(R^2e^{2it}+2Re^{it}+2)}+\int_{R}^{-R}{dx\over(x^2+1)(x^2+2x+1)}
$$
Now $|R^2e^{2it}+1|=R^2|e^{2it}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Can someone show me HOW to do this, I don't just want the answer $4n$ to the power of $3$ over $2 = 8$ to the power of negative $1$ over $3$
Written Differently for Clarity:
$$(4n)^\frac{3}{2} = (8)^{-\frac{1}{3}}$$
EDIT
Actually, the problem should be solving $4n^{\frac{3}{2}} = 8^{-\frac{1}{3}}$. Another user edit... | $${ \left( 4n \right) }^{ \frac { 3 }{ 2 } }={ \left( 8 \right) }^{ -\frac { 1 }{ 3 } }\\ \left( { \left( 4n \right) }^{ \frac { 3 }{ 2 } } \right) ^{ 2/3 }=\left( { \left( { 2 }^{ 3 } \right) }^{ -\frac { 1 }{ 3 } } \right) ^{ 2/3 }\\ 4n=2^{ -\frac { 2 }{ 3 } }\\ n=\frac { 2^{ -\frac { 2 }{ 3 } } }{ 4 } =\fr... | {
"language": "en",
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Number of Interesting Quadruples
Define an ordered quadruple of integers $(a, b, c, d)$ as interesting if $1 \le a<b<c<d \le 10$, and a+d>b+c. How many interesting ordered quadruples are there?
This is a bit of trouble here actually, I am to use $a + d \gt b + c$ as a constraint.
Without any restrictions (no $a + d ... | For each set $a,b,c,d$ with $a+d \neq b+c$, either $a+d \gt b+c$ or $(10-d)+(10-a) \gt (10-b)+(10-c)$. Exactly half the selections with $a+d \neq b+c$ will meet your restriction.
I haven't found a neat way to count the cases $a+d=b+c$. We can note that there are two ways for two numbers to sum to $5$, two ways to s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\frac{7}{5} \equiv 11 \pmod{12}$. Why is it $11$? I know this equation is true but I don't get the reason $\frac{7}{5}$ is congruent to $11$ here.
The quotient is supposed to be $1.4$ in non modular arithmetic and I don't get where $11$ has come from.
| $\frac{7}{5}$ is not an integer, the congruence relation $\bmod n$ is defined over the integers only.
On the other hand there is such thing as an inverse when working with modular arithmetic. Just like when working in the real numbers we have an inverse for $a$, we also have inverses in modular arithmetic. The inverse ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all integer values for $m$ such that $x_1,x_2\in\mathbb{Z}$ We have $(m+1)x^2-(2m+1)x-2m=0$ where $m\in\mathbb{R}\backslash\left\{-1\right\}$. We need to find all integer values for $m$ such that both roots of the equation are integers.
Here is all my steps:
*
*$x_{1,2}=\frac{(2m+1)\pm\sqrt{12m^2+12m+1}}{2(m+1)... | Let's express $x$:
$$
-m = \frac{x^2-x}{x^2-2x-2}.
$$
So,
$$
-m - 1 = \frac{x+2}{x^2-2x-2}.
$$
Denominator never equal to $0$ for integers $x$. If $|x+2|<|x^2-2x-2|$, $m$ cannot be integer (unless $x=-2$; we'll check this case later). Solve this inequality:
$$
|x+2|<|x^2-2x-2|\Leftrightarrow (x^2-2x-2)^2 - (x+2)^2 > 0 ... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Then the value of $ [f(2)] $ where [.] represents the greatest integer function is?
A differentiable function f is satisfying the relation $$f(x+y) = f(x) + f(y) + 2xy(x+y) - \dfrac{1}{3} $$ $ \forall $ $ x , y $ belongs to $\Re$ and $$lim_{h \to 0} \dfrac{3f(h)-1}{6h} = \dfrac{2}{3}. $$Then the value of $ [f(2)] $ ... | Using the formula $$\displaystyle f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}...............(1)$$
Now Given $$\displaystyle f(x+y) = f(x)+f(y)+2xy(x+y)-\frac{1}{3}$$
So Put $x=h\;$ We get $$\displaystyle f(x+h) = f(x)+f(h)+2xh(x+h)-\frac{1}{3}$$
So $$\displaystyle f'(x) = \lim_{h\rightarrow 0}\frac{f(x)+f(h)+2xh... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solution of System of linear equations I have three equations:
$$
\begin{cases}
4y + z = 2\\
2x + 6y - 2z = 3\\
4x + 8y - 5z = 4
\end{cases}
$$
Applying Gauss elimination I get:
$$ \left[
\begin{array}{ccc|c}
1&0&-\frac{7}{4} & 0\\
0 & 1 & \frac{1}{4} & \frac{1}{2} \\
0 & 0 & 0 & 0
\end{array... | You can arbitrarily choose $z$. Then, $x$ and $y$ can be computed from that:
$$
x - \frac{7}{4}z = 0 \;\Rightarrow\; x = \frac{7}{4} z\\
y + \frac{1}{4}z = \frac{1}{2} \;\Rightarrow\; y=\frac{1}{2} - \frac{1}{4}z
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The sum of the lengths of the hypotenuse and another side of a right angled triangle is given. The sum of the lengths of the hypotenuse and another side of a right angled triangle is given.The area of the triangle will be maximum if the angle between them is:
$(A)\frac{\pi}{6}\hspace{1cm}(B)\frac{\pi}{4}\hspace{1cm}(C)... | The area of the triangle can be expressed as
$$\frac 12ab=\frac 12a\sqrt{(k-a)^2-a^2}=\frac{a}{2}\sqrt{k^2-2ak}=\frac{\sqrt k}{2}\sqrt{a^2k-2a^3}$$
Here, let $f(a)=a^2k-2a^3$. Then, we have
$$f'(a)=2ak-6a^2=2a(k-3a).$$
So, since we know that the maximum of $f(a)$ is $f(k/3)$, it follows that the maximum of the area of ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Another integral $\int \frac{3 x^2+2 x+1}{ \left(x^3+x^2+x+2\right) \sqrt{1+\sqrt{x^3+x^2+x+2}}} \, dx$ Here is an indefinite integral that is similar to an integral I wanna propose for a contest. Apart from
using CAS, do you see any very easy way of calculating it?
$$\int \frac{1+2x +3 x^2}{\left(2+x+x^2+x^3\right) \... | HINT:
As $\dfrac{d(2+x+x^2+x^3)}{dx}=1+2x+3x^2,$
let $\sqrt{1+\sqrt{2+x+x^2+x^3}}=u\implies 2+x+x^2+x^3=(u^2-1)^2$
and $(1+2x+3x^2)dx=(u^2-1)2u\ du$
Now use Partial Fraction Decomposition,
$\dfrac1{(u^2-1)^2}=\dfrac A{u-1}+\dfrac B{(u-1)^2}+\dfrac C{u+1}+\dfrac D{(u+1)^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 0
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selection of balls of three colors with restrictions I have asked a similar question here and answers were very helpful. I tried doing similar questions and could solve them comfortably. However, I myself came up with a question like this and wondering how to solve this.
Suppose there are 20 black balls, 10 yellow bal... | Your approaches to both problems are correct.
Since balls of the same color are indistinguishable, the number of ways $15$ balls can be selected from $20$ black balls, $10$ yellow balls, and $5$ brown balls is the number of nonnegative integer solutions of the equation
$$x_1 + x_2 + x_3 = 15 \tag{1}$$
subject to the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Find $\lim_{n\to\infty}\left(2n\int_{0}^{1}\frac{x^n}{1+x^2}dx\right)^n$ Find this limits
$$\lim_{n\to\infty}\left(2n\int_{0}^{1}\dfrac{x^n}{1+x^2}dx\right)^n\tag{1}$$
since following links post this question have solve it.$$\lim_{n\to\infty}2n\int_{0}^{\frac{\pi}{4}}\tan^n{x}dx=\dfrac{1}{2}$$
Solving $\lim_{n\to\inft... | Integration by parts gives:
$$\int_{0}^{1}nx^{n-1}\frac{2x}{1+x^2}\,dx = \left.x^n\frac{2x}{1+x^2}\right|_{0}^{1}-2\int_{0}^{1}x^{n}\frac{(1-x^2)}{(1+x^2)^2}\,dx$$
so performing it twice:
$$\int_{0}^{1}nx^{n-1}\frac{2x}{1+x^2}\,dx = 1-2\left.\frac{x^{n+1}}{n+1}\frac{(1-x)^2}{(1+x^2)^2}\right|_{0}^{1}+2\int_{0}^{1}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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$\int_{0}^{\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}$ equals $\frac{\pi}{5050}$ For $a\geq2$,if the value of the definite integral $\int_{0}^{\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}$ equals $\frac{\pi}{5050}$.Find the value of $a$.
Substituting $x-\frac{1}{x}=t$ does not seem to work here,what is th... | The substitution $t=1/x$ works. Then, $dx=-\frac{1}{t^2}dt$ and
$$\begin{align}
I&=\int_0^{\infty}\frac{1}{a^2+\left(x-\frac1x\right)^2}\,dx\\\\
&=\int_0^\infty \frac{1}{a^2+\left(x-\frac1x\right)^2}\frac{1}{x^2}\,dx\\\\
&=\frac12\int_0^\infty \frac{1+\frac{1}{x^2}}{a^2+\left(x-\frac1x\right)^2}\,dx\\\\
&=\frac12\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Angle chase:In $\Delta ABC, AB=AC $ and $\angle BAC=20°.$ If $CD$ is the median from $C$ to side $AB$, find $\angle ADC$. In $\triangle ABC, AB=AC $ and $\angle BAC=20^\circ$ If $CD$ is the median from $C$ to side $AB$, find $\angle ADC$.
| Notice, we have $$\angle A+\angle B+\angle C=180^\circ$$ But $AB=AC\iff \angle B=\angle C$ Hence, we get $$\angle A+\angle C+\angle C=180^\circ$$ $$20^\circ+2\angle C=180^\circ$$ $$\angle C=\frac{180^\circ-20^\circ}{2}=80^\circ$$
Let the unknown angle $\angle ADC=x$
Applying sine rule in $\triangle ACD$ as follows $$\f... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $(1)\frac{1}{2}\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq\frac{5}{6}$ $(2)2e^{-1/4}<\int_{0}^{2}e^{x^2-x}dx<2e^2$ Prove that $(1)\frac{1}{2}\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq\frac{5}{6}$
$(2)2e^{-1/4}<\int_{0}^{2}e^{x^2-x}dx<2e^2$
I tried to prove it but my answer is not correct.
For first part,As $0\leq x\leq2\... | Both answers are fine, I would just point out than since $f(x)=\frac{\sqrt{x}}{2+x}$ is a concave function on $[0,4]$ and $e^{x^2-x}$ is a convex function on $[0,2]$, both inequalities can be improved through the Hermite-Hadamard inequality, i.e. by using the usual rectangle/trapezoid method for approximating an integr... | {
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"source": "stackexchange",
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Given $a, b,c,d$ non-negative and $a+b+c+d=4$. Prove that $a^3b+b^3c+c^3d+d^3a+23abcd \le 27.$
This is a problem on Mathlinks.ro and it have had no solution. So I hope it would have a nice answer and as simply as possible.
Given $a, b,c,d$ are non-negative numbers and $a+b+c+d=4$. Prove that $$a^3b+b^3c+c^3d+d^3a+23ab... | We can trade off the cyclic symmetry to reduce the number of variables to consider. The homogeneous form is:
$$4^4(a^3b+b^3c+c^3d+d^3a+23abcd) \le 27(a+b+c+d)^4$$
WLOG, let $a$ be the minimum among $a, b, c, d$; and so we have non-negatives $x, y, z$ s.t. $b = a+x, \, c = a+y, \, d = a+z$. Hence we get an inequality ... | {
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"answer_id": 0
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The set of real values of $x$ satisfying the equation $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]=5$ The set of real values of $x$ satisfying the equation $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]=5$,(where $[]$ denotes the greatest integer function) belongs to the interval $(a,\frac{b}{c}]$,where $a,b,... | So first of all, for $x=4$ you have $1$, for $x>4$ you have $0$. Next, your function has jumps whenever $x=3/n$ or $x=4/m$ for some integers $n,m$. Unless both of these happen at once (for instance, at $x=1$), these jumps are by one unit at a time. Can you find the closest four jumps to the left of $x=4$? (Hint: the fi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Simplifying a complication max operation I have dervied an inequality and have arrived to the following
$$\max\{1, \frac{b}{2}+1\} \leq \max\{a, \frac{b}{2}+ \frac{a}{2}\}$$
I am trying to simplify further and arrived to the following conclusions
$$a \geq 1$$
$$ a \geq \frac{b}{2}+1$$
$$ a\geq 2 $$
$$\frac{b}{2}+ \fr... | Starting with
$$\max\{2, \frac{b}{2}+1\}
\leq \max\{a, \frac{b}{2}+ \frac{a}{2}\}
$$
Let's look at the left side.
Case 1:
$2 > \frac{b}{2}+1
$.
Then
$b < 2$.
This then becomes
$2
\le \max\{a, \frac{b}{2}+ \frac{a}{2}\}
< \max\{a, \frac{1}{2}+ \frac{a}{2}\}
= \max\{a, \frac{a+1}{2}\}
$.
If
$a > \frac{a+1}{2}
$,
then
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
convergence proof without finding 'N' I tried to prove $\sqrt{n^2 +n}-n$ converges to $\frac{1}{2}$ I am not sure what I have proved is correct.
$$\left|\frac{n}{\sqrt{n^2+n} +n} - \frac{1}{2} \right| = \left| \frac{2n - 2(\sqrt{n^2+n}+n)}{2(\sqrt{n^2+n}+n)} \right|< \left|\frac{2n - 2\cdot 2n}{2\cdot 2 n} \right|<\lef... | Let $\epsilon>0$ be given. Then, we have
$$\begin{align}
\left|\frac{n}{\sqrt{n^2+n} +n} - \frac{1}{2} \right| &= \left| \frac{2n - \left(\sqrt{n^2+n}+n\right)}{2\left(\sqrt{n^2+n}+n\right)} \right|\\\\
&= \left| \frac{n -\sqrt{n^2+n}}{2\left(n+\sqrt{n^2+n}\right)} \right|\\\\
&= \left| \frac{n -\sqrt{n^2+n}}{2\left(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
computing an integral (fraction of real powers) I want compute the following integral depending on the real parameters $\alpha, \beta > 0$ and $C >0$
$$ \int_0^1 \frac{u^{2\beta}}{C+u^{2(\alpha+\beta)}} du$$
Thanks a lot for any clue !
| Consider
\begin{align}
I(\alpha, \beta) &= \int_{0}^{1} \frac{u^{2\beta}}{c+u^{2(\alpha + \beta)}} \, du \\
&= \frac{1}{c} \, \sum_{n=0}^{\infty} \frac{1}{c^{n}} \, \int_{0}^{1} u^{2\beta + 2 n (\alpha + \beta)} \, du \\
&= \frac{1}{2c} \, \sum_{n=0}^{\infty} \frac{1}{c^{n}} \, \frac{1}{ n \alpha + (n+1) \beta} \\
&= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the image of $|z+1|=2$ under $f(z) = \frac{1}{z}$ where $z \in \mathbb C$
Find the image of $|z+1|=2$ under $f(z) = \frac{1}{z}$ where $z \in \mathbb C$
My attempt:
Let $z = x + iy$
$\displaystyle |z+1|=2 \iff | (x + iy)+1|=2 \iff |(x+1) +iy|=2 \iff (x+1)^2 + y^2 = 4$
Let $w = u + iv$
Now let $\displaystyle w = ... | Proceeding with your result,
$$
\bigg(\frac{u}{u^2 + v^2} +1\bigg)^2 + \bigg(- \frac{v}{u^2 + v^2}\bigg)^2 = 4
$$ we get
$$
\bigg(\frac{u}{u^2 + v^2}\bigg)^2+2\frac{u}{u^2 + v^2}+1 + \bigg(\frac{v}{u^2 + v^2}\bigg)^2 = 4
$$ or
$$
\frac{u^2}{(u^2 + v^2)^2}+ \frac{v^2}{(u^2 + v^2)^2}+\frac{2u}{u^2 + v^2} = 3
\\$$$$
2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Trying to solve the trig equation $\sqrt{3+4\cos^2(x)}=\frac{\sin(x)}{\sqrt 3}+3\cos(x)$ The equation is
$$\sqrt{3+4\cos^2(x)}=\frac{\sin(x)}{\sqrt 3}+3\cos(x)$$
My solution goes like this
$$
\begin{cases}
3+4\cos^2(x)=\frac{\sin^2(x)}{3}+\frac{6}{\sqrt 3}\sin(x)\cos(x)+9\cos^2(x) \\
\frac{\sin(x)}{\sqrt 3}+3\cos(x) \g... | The solution of the quadratic equation $4t^2 -3\sqrt3 t -3 = 0$ is wrong. Here, $D = b^2 -4ac$ i.e. D=75.
Hence, $t = \frac{3√3 + or - √D}{2*4}$
$t = \frac{3√3 + 5√3}{8}$
And $t = \frac{3√3 -5√3}{8}$
Hence, $ t = √3, -\frac{\sqrt3}{4}$
Now you will get the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Integration of the square root of a quadratic I am in the tricky situation of trying to integrate the following.
$$\sqrt{4 a^2 (y-b)^2+c^4}$$
$a, b$ and $c$ are all known constants.
Can anybody provide insight as to how to do this?
I have tried to rearrange to fit the form:
$$\int (ax+b)^{\alpha}dx = \dfrac1a \cdot... | Before applying Inverse Trigonometric Substitution, I prefer to preprocess the expression as follows
$$\begin{array}{lll}
\int\sqrt{4a^2(y-b)^2+c^4}dy &=& \displaystyle\int\sqrt{\frac{4a^2c^4(y-b)^2}{c^4}+c^4}dy\\
&=& \displaystyle \int c^2\sqrt{\frac{4a^2(y-b)^2}{c^4}+1}dy\\
&=& \displaystyle \int c^2\sqrt{\frac{(2ay-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.