Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
What is the probability that the sum of 6 four-sided dice is less than or equal to 14? The solution given is shown below.
My question is how did they count the numerator like that?
What is the explanation for it please?
$$\begin{align}
\frac{C_6^{14}-C_1^6\times C_6^{10}+C_2^6 \times C_6^6}{4^6}&=\frac{3003-6\times 210... | The following solution uses a generating function. Readers not familiar with generating functions can find several resources in the answers to this question: How can I learn about generating functions?
There are $4^6$ possible outcomes when rolling six four-sided dice, all of which we assume are equally likely. We wan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4627424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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How can I integrate $\int\sqrt{\frac{x^2+bx+c}{x^2+ex+f}}dx?$ How can I integrate
$$\int\sqrt{\dfrac{x^2+bx+c}{x^2+ex+f}}\,dx?$$ I was thinking a substitution $$t=\frac{x^2+bx+c}{x^2+ex+f},$$ which inverts as follows:
$$(x^2+ex+f)t=x^2+bx+c$$
$$(t-1)x^2+(et-b)x+ft-c=0$$
$$x=\dfrac{b-et+\sqrt{(b-et)^2-4(t-1)(ft-c)}}{2(t... | Going off of Travis Willse's answer, I notice that
$$\int \frac{x^2+ax+b}{\sqrt{x^4+\alpha x^2+\beta x+\gamma}}dx$$ can be addressed as follows:
Since $$\frac{1}{\sqrt{x^4+\alpha x^2+\beta x+\gamma}}=\sum_{p,q,r\geqslant 0 }\binom{-\frac{1}{2}}{p,q,r,-\frac{1}{2}-p-q-r}\alpha^{q}\beta^r\gamma^{-\frac{1}{2}-p-q-r}x^{4p+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4629947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find $a+b +c$, if $\sin{x}+\sin^2{x}=1$ and $\cos^{12}{x}+a\cos^{10}{x}+b\cos^8{x}+c\cos^6{x}=1$ There is my problem :
Find $a+b +c$,
if $$\sin{x}+\sin^2{x}=1$$ and $$\cos^{12}{x}+a\cos^{10}{x}+b\cos^8{x}+c\cos^6{x}=1$$
I'm sorry, I can't solve this problem but I really want to know the solution.
I know that $\cos^2{x}... | With what you already know, $$\sin^6x+a\sin^5x+b\sin^4x+c\sin^3x-1=0\qquad \sin^2x+\sin x-1=0$$ Now let $X=\sin (x)$. You would like the polynomial $X^2+X-1$ to divide $X^6+aX^5+bX^4+cX^3-1$. How to make that happen?
There would need to be a 4th degree polynomial $v=X^4+v_3X^3+v_2X^2+v_1X+1$ as the quotient.
$$\left(X^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4631715",
"timestamp": "2023-03-29T00:00:00",
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Number of distinct arrangement of $(a,b,c,d,e)$
If $a<b<c<d<e $ be
positive integer such that
$a+b+c+d+e=20$.
Then number of distinct
arrangement of $(a,b,c,d,e)$ is
Here the largest value of $e$ is $10$
like $a\ b\ c\ d\ e$
as $ \ \ 1\ 2\ 3\ 4\ 10$
And least value is $6$
like $ a\ b\ c\ d\ e$
as $\ \ 2\ 3\... | This is the same as the number of solutions in positive integers of:
$$5a+4w+3x+2y+z=20 \tag{1}, \\ \text{ where }w=b-a, x=c-b, y=d-c, z=e-d.$$
Equation $(1)$ tells us immediately that $a \in \{1, 2 \}$ and if $a=2$ there is only one possible solution.
Thus, we want to add $1$ to the number of possible solutions in pos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4634663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Verify my proof: "for all odd integers $a$ and $b$, $b^2-a^2 \neq 4$" I'm learning math without a math professor. I need some feedback from community regarding my proof.
The book is : "Discrete Mathematics with Applications" by Susanna S. Epp, 5th edition.
Exercise 16 from page 226. Please verify my proof.
Prove the f... | I'm merely critiquing your presentation.
Prove the following statement: "for all odd integers $a$ and $b$, $b^2-a^2 \neq 4$"
Proof. Let's prove by contradiction. Let say there are two integers $a$ and $b$ and $b^2-a^2 = 4$. Because $a$ and $b$ are odd integers, then $a=2n+1$ and $b=2k+1$, where $n \in Z$ and $k \in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4640389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How many methods are there to evaluate $\int_0^{\infty} \frac{1}{\left(x+\frac{1}{x}\right)^{2n}}$? Background
As I had found the integral
$$I=\int_0^{\infty} \frac{1}{\left(x+\frac{1}{x}\right)^2} d x =\frac{\pi}{4}, $$
by using $x\mapsto \frac{1}{x}$ yields
$\displaystyle I=\int_0^{\infty} \frac{\frac{1}{x^2}}{\left(... | Letting $x=\tan \theta$ transforms the integral
$$
\begin{aligned}
I_n & = \int_0^{\infty} \frac{x^{2 n}}{\left(1+x^2\right)^{2 n}} d x \\&=\int_0^{\frac{\pi}{2}} \frac{\tan ^{2 n} \theta}{\sec ^{4 n} \theta} \sec ^2 \theta d \theta \\
& =\int_0^{\frac{\pi}{2}} \sin ^{2 n} \theta \cos ^{2n-2} \theta d \theta \\
& =\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4647346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Proving $\sin\frac{\pi}{13}+\sin\frac{3\pi}{13}+\sin\frac{4\pi}{13}=\frac12\sqrt{\frac{13+3\sqrt{13}}2}$
Prove that $$\sin\left(\frac{\pi}{13}\right)+\sin\left(\frac{3\pi}{13}\right)+\sin\left(\frac{4\pi}{13}\right)=\frac{1}{2}\sqrt{\frac{13+3\sqrt{13}}{2}}$$
My Attempt
Let $$x = \frac{1}{2}\sqrt{\frac{13+3\sqrt{13}}... | Let
$$
s=\sin\left(\frac{\pi}{13}\right)+\sin\left(\frac{3\pi}{13}\right)+\sin\left(\frac{4\pi}{13}\right)\tag1
$$
If $\alpha=e^{\pi i/13}$, then
$$
\alpha^{13}+1=0\tag2
$$
and
$$
2is=\underbrace{\alpha+\overbrace{\quad\alpha^{12}\quad}^{-\alpha^{-1}}}_{2i\sin\left(\frac{\pi}{13}\right)}+\underbrace{\alpha^3+\overbrace... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4647658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Do we have a simpler method for computing $\int_{-\infty}^{\infty} \frac{\ln \left(x^2+ax+b\right)}{1+x^2} d x$, where $b> \frac{a^2}{4} $? Background
After finding the exact value of the integral in my post, I start to investigate a similar integral
$$I(a):
=\int_{-\infty}^{\infty} \frac{\ln \left(x^2+ax+b\right)}{1+... | Consider, instead
\begin{align}
&\int_{-\infty}^{\infty} \frac{\ln (x^2+2x\sqrt b \sin \theta+b)}{x^2+1} \ d x\\
=& \int_{-\infty}^{\infty}\bigg(\ln (x^2+b)+ \int_0^{\theta}\frac{2x\sqrt b \cos t}{x^2+2x\sqrt b \sin t+b} dt\bigg) \frac{dx}{x^2+1}\\
=& \ 2\pi\ln(1+\sqrt b) -\int_0^{\theta}\frac{2\pi \sqrt b \sin t}{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4648645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Evaluate $\int_{0}^{\pi/2}x\sin^a (x) dx$, $a>0$
I want to evaluate $$\int_{0}^{\pi/2}x\sin^a (x)\, dx$$ where $a>0$ is a real number.
I tried: $$I(a)= \int_{0}^{\pi/2}x\sin^a(x)\,dx = \int_{0}^{1}\frac{\arcsin x}{\sqrt{1-x^2}}x^a\,dx$$
$$ I(a)=\sum_{m\geq 1}\frac{4^m}{2m\left(2m+a\right)\binom{2m}{m}}$$
$$I(a)=\frac... | An elementary (enough) deduction of the (already mentioned) equality $$\boxed{I(a)=\frac{\pi^{3/2}}{4}\frac{\Gamma\left(\frac{a+1}2\right)}{\Gamma\left(\frac{a+2}2\right)}-\sum_{n=1}^\infty\frac{\prod_{k=1}^{n-1}(a+2k)^2}{\prod_{k=1}^{2n}(a+k)}.\quad(\Re a>-1)}$$
We note that $$(a+1)\big(I(a)-I(a+2)\big)=\int_0^{\pi/2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4648938",
"timestamp": "2023-03-29T00:00:00",
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If $x^2-kx+1=0$, find the value of $x^3+\frac{1}{x^3}$. I have the following question:
If $x^2-kx+1=0$, find the value of $x^3+\frac{1}{x^3}$.
Using the first equation, I rearrange to get $x^2=kx-1$. Then, I multiply both sides by x to get get $x^3=(kx-1)^{1.5}$. I can’t think of any other way than to substitute $(kx... | I don't see any way to continue from what you've done, since you're dealing with fractional powers of a polynomial, which are generally hard to work with. Instead, since $x \neq 0$, we can divide both sides by $x$ below to get
$$x^2 + 1 = kx \;\; \to \; \; x+\frac{1}{x}=k$$
Cubing both sides then gives
$$\begin{equatio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4652280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof that $n^3+2n$ is divisible by $3$ I'm trying to freshen up for school in another month, and I'm struggling with the simplest of proofs!
Problem:
For any natural number $n , n^3 + 2n$ is divisible by $3.$
This makes sense
Proof:
Basis Step: If $n = 0,$ then $n^3 + 2n = 0^3 +$
$2 \times 0 = 0.$ So it is divisi... | $$n^3+2n=n(n^2+2)$$
If $n$ is divisible by $3$, then obviously, so is $n^3+2n$ because you can factor out $n$.
If $n$ is not divisible by $3$, it is sufficient to show that $n^2+2$ is divisible by 3. Now, if $n$ is not divisible by $3$, $n=3k+1$ or $n=3k+2$ for some integer $k$. Plug that into $n^2+2$ and you'll get $9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 13,
"answer_id": 1
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Area of a quadrilateral
The perpendicular bisector of the line
joining $A(0,1)$ and $C(-4,7)$ intersects
the $x$-axis at $B$ and the $y$-axis at $D$.
Find the area of the quadrilateral.
Thank you in advance!
| The perpendicular bisector will pass through the mid-point: $(-2, 4)$ of the line passing through the points $A(0, 1)$ & $C(-4, 7)$ & have a slope: $\frac{2}{3}$ normal to the line $AC$.
The equation of perpendicular bisector is given as $$y-4=\frac{2}{3}(x-(-2))$$ $$\implies y=\frac{2x+4+12}{3}=\frac{2x+16}{3}$$ By... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Proving $ { 2n \choose n } = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} $ How to prove this binomial identity :
$$ { 2n \choose n } = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} $$
The left hand side arises while solving a standard binomial problem the right hand side is given in the solution , I checked usi... | Your original identity, ${ 2n \choose n } = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!}$, can be rewritten (by multiplying both sides by $(n!)^2$) as $(2n)!=2^n\cdot 1\cdot 3\cdot 5\cdots (2n-1)\cdot n!$. Now, $2^n\cdot 1\cdot 3\cdot 5\cdots (2n-1)\cdot n!=$ $1\cdot 3\cdot 5\cdots (2n-1)\cdot 2\cdot 4\cdot 6\cdots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/7778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it w... | Define the following series for $ x > 0 $
$$\frac{\sin x}{x} = 1 - \frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots\quad.$$
Now substitute $ x = \sqrt{y}\ $ to arrive at
$$\frac{\sin \sqrt{y}\ }{\sqrt{y}\ } = 1 - \frac{y}{3!}+\frac{y^2}{5!}-\frac{y^3}{7!}+\cdots\quad.$$
if we find the roots of $\frac{\sin \sqrt{y}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/8337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "814",
"answer_count": 48,
"answer_id": 33
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Funny identities Here is a funny exercise
$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$
(If you prove it don't publish it here please).
Do you have similar examples?
| $$
\dfrac{1}{2}=\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\cdots}}}}}}
$$
and more generally we have
$$
\dfrac{1}{n+1}=\frac{\dfrac{1}{n(n+1)}}{\dfrac{1}{n(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/8814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "281",
"answer_count": 63,
"answer_id": 56
} |
Find a closed form for $\sum_{k=1}^{x-1} a^{1/k}$ Please find a closed form for partial sum of a function
$$f(x)=a^{1/x}$$
I want it to be expressed in terms of bounded number of elementary functions and/or well known special functions.
No computer algebra systems I have tried so far could find a satisfactory solution.... | Just for fun here are some bounds for $\sum_{k=1}^n a^{1/k}$ for $ a \ge 1.$
We have
$$a^{1/k} = 1 + \frac{\log a}{1! k} + \frac{(\log a)^2}{2! k^2}
+ \frac{(\log a)^3}{3! k^3} + \cdots$$
and so
$$\sum_{k=1}^n a^{1/k} = \zeta_n(0) + (\log a)\zeta_n(1) + \frac{(\log a)^2}{2!}\zeta_n(2) +
\frac{(\log a)^3}{3!}\zeta_n(3) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/9417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Proving an identity involving terms in arithmetic progression. If $a_1,\ldots,a_n$ are in arithmetic progression and $a_i\gt 0$ for all $i$, then how to prove the following two identities:
$ (1)\large \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \cdots + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}... | For the first one, use induction (or) note that $\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}} = \frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d}$, where $d$ is the common difference between the successive terms. Now use the telescopic summation to cancel out the terms in the numerator and massage it to get the final expression on the right ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Find the value of $\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ}) $ How to find the value of
$$\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ})$$
manually ?
| If $\cos3\theta=\frac12,$
$$\tan3\theta\cot\theta-4\cos\theta$$
$$=\frac{\sin3\theta\cos\theta}{\cos3\theta\sin\theta}-4\cos\theta$$
$$=\frac{\sin3\theta\cos\theta-4\cos3\theta\sin\theta\cos\theta}{\cos3\theta\sin\theta}$$
$$=\frac{\sin3\theta\cos\theta-\sin2\theta}{\frac12\sin\theta}\text { as }\sin2x=2\sin x\cos x\te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/10661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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Proving trignometrical identities: If $\sin(A+B) + \sin(B+C) + \cos(C-A) = \frac{3}{2}$ show that, $$1.\sin A + \cos B + \sin C = 0$$ $$2. \cos A + \sin B +\cos C = 0$$
| The problem seems to be missing some assumptions, as noted by Americo.
For instance, If $B = 0$ and $A=C$ are acute angles, such that $\sin A = 1/4$ the we have that
$\sin(A+B) + \sin(B+C) + \cos(A-C) = 3/2$, but none of
$\sin A + \cos B + \sin C$ or $\cos A + \sin B + \cos C$ are $0$.
In any case, this looks like a p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/10702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Nice expression for minimum of three variables? As we saw here, the minimum of two quantities can be written using elementary functions and the absolute value function.
$\min(a,b)=\frac{a+b}{2} - \frac{|a-b|}{2}$
There's even a nice intuitive explanation to go along with this: If we go to the point half way between two... | First, define
$$
\Delta=|a-b|+|b-c|+|c-a|\newcommand{\Mu}{\mathrm{M}}\tag{1}
$$
It is somewhat intuitive that
$$
\frac{\Delta}{2}=\max(a,b,c)-\min(a,b,c)\tag{2}
$$
For example, if $a\ge b\ge c$ then $|a-b|+|b-c|+|c-a|=2a-2c$.
Next, define
$$
\Sigma=a\left(1-\frac{|b-c|}{\Delta}\right)+b\left(1-\frac{|c-a|}{\Delta}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/13253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "66",
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"answer_id": 3
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Does closed-form expression exist for integral $\int \frac{dx}{1-x^n}$ Playing with integrals on the form $$\int \frac{1}{1-x^n}\,dx$$ I noticed that for odd values of n > 5, it doesn't appear to be possible to express the integral as a closed-form expression. Is this true? How can I prove it?
| All of those integrals have closed form expressions, for both even and odd $n$.
The polynomial $x^n-1$ can be factored over the complex numbers as
$$x^n - 1 = (x-1)(x-\zeta)(x-\zeta^2)\cdots(x-\zeta^{n-1}),$$
and when $n$ is odd, the only real solution is $x-1$ and the complex factors pair up in conjugate pairs to giv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/19100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Find matrices $A$ and $B$ given $AB$ and $BA$ Given that:
$$AB= \left[ {\matrix{
3 & 1 \cr
2 & 1 \cr
} } \right]$$
and
$$BA= \left[ {\matrix{
5 & 3 \cr
-2 & -1 \cr
} } \right]$$
find $A$ and $B$.
| First note that both the products are of full rank and hence $A$ and $B$ are also of full rank. So we would expect one set of solutions with one degree of freedom since if $A$ and $B$ satisfy our equations, so will $kA$ and $\frac{1}{k}B$.
Let $$A = \left[ {\matrix{
a_1 & a_2 \cr
a_3 & a_4 \cr
} } \right],B... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How can I prove that $n^7 - n$ is divisible by $42$ for any integer $n$? I can see that this works for any integer $n$, but I can't figure out why this works, or why the number $42$ has this property.
| As $42=2.3.7$. Therefore,we need to check that $n^7-n$ is divisible by $2,3$ and 7.
For divisibility by $2$, by fermat's little theorem, $n^2=n\pmod 2 \implies {(n^2)}^3.n=n^4\pmod 2=n^2\pmod 2=n\pmod 2 \implies n^7-n=0\pmod2$.
For divisibility by 3, $n^3=n\pmod 3\implies n^7=n^3\pmod 3=n\pmod 3 \implies n^7-n=0\pmo... | {
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"url": "https://math.stackexchange.com/questions/22121",
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Prove that $x^{2} \equiv 1 \pmod{2^k}$ has exactly four incongruent solutions
Prove that $x^{2} \equiv 1 \pmod{2^k}$ has exactly four incongruent solutions.
My attempt:
We have,
$x^2 - 1 = (x - 1) \times (x + 1)$, then
$(x - 1)(x + 1) \equiv 0 \pmod{2^k}$
which implies,
$2^k|(x - 1)$ or $2^k|(x + 1) \implies x \equ... | Existence is easy:
The solutions are $\{1,2^{k-1}-1,2^{k-1}+1,2^k-1\}$.
Squaring them gives $\{1, 2^{2k-2}-2^k+1, 2^{2k-2}+2^k+1, 2^{2k}-2^{k+1}+1\}$.
Reducing $\pmod {2^k}$ gives 1 in each case (given that $2k-2 > k+1$, which forces $k \ge 3$).
For uniqueness you could use the fact that the units group of $\mathbb{Z}/... | {
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If $p - a \equiv -a \pmod{p}$ then what is $\frac{p-1}{2} \equiv ? \pmod{p}$? If $p - a \equiv -a \pmod{p}$ then what is $\frac{p-1}{2} \equiv ? \pmod{p}$? Where $p$ is an odd prime.
I read in the book, they claimed:
$$p - 1 \equiv -1 \pmod{p}$$
$$p - 2 \equiv -2 \pmod{p}$$
$$p - 3 \equiv -3 \pmod{p}$$
$$ ... $$
$$\fra... | The sequence as written doesn't make sense, since the last term does not follow the pattern of the previous ones. However, you can argue that
$$p - \left(\frac{p-1}{2}\right) \equiv -\frac{p-1}{2}\pmod{p}.$$
This does follow the same pattern as the rest of the terms.
And since
$$p - \frac{p-1}{2} = \frac{2p-p+1}{2} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/25913",
"timestamp": "2023-03-29T00:00:00",
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Constants of integration in integration by parts After finishing a first calculus course, I know how to integrate by parts, for example, $\int x \ln x dx$, letting $u = \ln x$, $dv = x dx$: $$\int x \ln x dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2x} dx.$$
However, what I could not figure out is why we assume from $dv... | Take your example, $$\int x\ln x\,dx.$$
Note $x\gt 0$ must be assumed (so the integrand makes sense).
If we let $u = \ln x$ and $dv= x\,dx$, then we can take $v$ to be any function with $dv = x\,dx$. So the "generic" $v$ will be, as you note, $v = \frac{1}{2}x^2 + C$. What happens then if we use this "generic" $v$?
\b... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
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Three-variable system of simultaneous equations
$x + y + z = 4$
$x^2 + y^2 + z^2 = 4$
$x^3 + y^3 + z^3 = 4$
Any ideas on how to solve for $(x,y,z)$ satisfying the three simultaneous equations, provided there can be both real and complex solutions?
| For a fixed number of variables and a fixed power $n$ the sum of powers $$x^n + y^n + z^n + ... + w^n$$ is a symmetric polynomial.
It is expressible in terms of elementary symmetric polynomials. The elementary symmetric polynomials for three variables are
*
*$e_1 = x + y + z$
*$e_2 = x y + x z + y z$
*$e_3 = x y z... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find all irreducible monic polynomials in $\mathbb{Z}/(2)[x]$ with degree equal or less than 5
Find all irreducible monic polynomials in $\mathbb{Z}/(2)[x]$ with degree equal or less than $5$.
This is what I tried:
It's evident that $x,x+1$ are irreducible. Then, use these to find all reducible polynomials of degree ... | Extrapolated Comments converted to answer:
First, we note that there are $2^n$ polynomials in $\mathbb{Z}_2[x]$ of degree $n$.
A polynomial $p(x)$ of degree $2$ or $3$ is irreducible if and only if it does not have linear factors. Therefore, it suffices to show that $p(0) = p(1) = 1$. This quickly tells us that $x^2 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Another magic re-write question, picture! It's the part before and after "Thus".
$$I = \ldots = \int e^{ax} \cos bx \ \mathrm{d}x = \frac{1}{b} e^{ax} \sin bx + \frac{a}{b^{2}} \cos bx - \frac{a^{2}}{b^{2}} I.$$
Thus
$$\left( 1 + \frac{a^{2}}{b^{2}} \right) I = \frac{1}{b} e^{ax} \sin bx + \frac{a}{b^{2}} \cos bx + C... | The integral calculation is done by integration by parts:
\begin{align}
I = \int e^{ax} \cos bx \ dx = \frac{1}{b} e^{ax} \sin bx + \frac{a}{b^{2}} \cos bx - \frac{a^{2}}{b^{2}} I.
\end{align}
Solving for $I$, we have
\begin{align}
I + \frac{a^{2}}{b^{2}} I = \left( 1 + \frac{a^{2}}{b^{2}} \right) I = \frac{1}{b} e^{ax... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What will the value of following log expression What will be the value of the expression
$$\log_x \frac{x}{y} + \log_y \frac{y}{x}?$$
I tried:
$$\log_x x - \log_x y + \log_y y - \log_y x = 1 - \log_x y + 1 - \log_y x
= 2 - \log_x y - \log_y x.$$
Now what after this ?
| If $\log_a b = r$, this means that $a^r = b$, so $b = e^{\ln(a^r)} = e^{r\ln(a)}$. Therefore, $\ln(b) = r\ln (a)$, or
$$\log_a b = \frac{\ln(b)}{\ln a}.$$
Thus, for
$$\log_x y = \frac{\ln y}{\ln x}\quad\text{and}\quad \log_y x = \frac{\ln x}{\ln y},$$
so
$$\log_x y = \frac{1}{\log_y x}.$$
So:
$$\begin{align*}
\log_x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Where is my (algebra) mistake? Converting parametric to Cartesian equation I'm having a problem with my solution to a textbook exercise:
Find the Cartesian equation of the curve given by this parametric equation:
$$x = \frac{t}{2t-1}, y = \frac{t}{t+1}$$
The textbook's answer is $y = \frac{x}{3x-1}$
My solution is diff... | As you said Danny,
$$\frac{a}{b+c} \neq \frac{a}{b} + \frac{a}{c}$$
and Theo pointed out with a simple example why:
$$0.5 = \frac{1}{2} = \frac{1}{1+1} \neq \frac{1}{1} + \frac{1}{1} = 2 \; .$$
It is a common mistake however, so tempting that few people have resisted making it.
P.S: Note you made the mistake twice, on... | {
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Solve recursion $a_{n}=ba_{n-1}+cd^{n-1}$ Let $b,c,d\in\mathbb{R}$ be constants with $b\neq d$. Let
$$\begin{eqnarray}
a_{n} &=& ba_{n-1}+cd^{n-1}
\end{eqnarray}$$
be a sequence for $n \geq 1$ with $a_{0}=0$. I want to find a closed formula for this recursion. (I only know the german term geschlossene Formel and tran... | If $b=0$, then $a_n = cd^{n-1}$.
Now assume $b\neq 0$. You can first divide both sides of the equation by $b^n$, then you get
$$
\frac{a_n}{b^n} = \frac{a_{n-1}}{b^{n-1}} + \frac{cd^{n-1}}{b^n}
$$
Let $x_n = \frac{a_n}{b^n}$ and $q = \frac{d}{b}$, then $x_0 = a_0/b = 0$, $q\neq 1$, and we have
$$
x_n = x_{n-1} + (c/... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Interesting problem on "neighbor fractions" This is from I. M. Gelfand's Algebra book.
Fractions $\displaystyle\frac{a}{b}$ and $\displaystyle\frac{c}{d}$ are called
neighbor fractions if their difference $\displaystyle\frac{ad - bc}{bd}$ has numerator of $\pm 1$,
that is, $ad - bc = \pm 1$. Prove that
(a) in this cas... | Assume $\frac{e}{f}$ is (strictly) between $\frac{a}{b}$ and $\frac{c}{d}$. Then $\left|\frac{a}{b}-\frac{e}{f}\right| + \left|\frac{e}{f}-\frac{c}{d}\right| = \left|\frac{a}{b}-\frac{c}{d}\right| = \frac{1}{bd}$
But $\left|\frac{a}{b}-\frac{e}{f}\right| \geq \frac{1}{bf}$ and $\left|\frac{e}{f}-\frac{c}{d}\right|\geq... | {
"language": "en",
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Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$ I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:
$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$
I really have no idea why this statement is true. Can someone please explain ... | Proof 1. (Exercise 2.5.1 in Dias Agudo, Cândido da Silva, Matemáticas Gerais III). Let $S:=\sum_{k=1}^{n}k^{2}$. Consider $(1+a)^{3}=1+3a+3a^{2}+a^{3}$ and sum
$(1+a)^{3}$ for $a=1,2,\ldots ,n$:
$$\begin{eqnarray*}
(1+1)^{3} &=&1+3\cdot 1+3\cdot 1^{2}+1^{3} \\
(1+2)^{3} &=&1+3\cdot 2+3\cdot 2^{2}+2^{3} \\
(1+3)^{3} &... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$ Let $a,b,c$ be positive, not equal. Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$.
I know the proof by subtracting LHS by RHS and then doing some arrangement.
But isn't there any inequality which can be used in $(1+1+1)(a^3+b^3+c^3) >(a+b+c)(a^2+b^2+c^2)$, or an e... | By Holder and C-S for non-negatives $a$, $b$ and $c$ we obtain:
$$(3(a^3+b^3+c^3))^2=(1+1+1)^2(a^3+b^3+c^3)(a^3+b^3+a^3)\geq(a+b+c)^3(a^3+b^3+c^3)=$$
$$=(a+b+c)(a^3+b^3+c^3)(a+b+c)^2\geq(a^2+b^2+c^2)^2(a+b+c)^2$$
and we are done!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Is this Batman equation for real? HardOCP has an image with an equation which apparently draws the Batman logo. Is this for real?
Batman Equation in text form:
\begin{align}
&\left(\left(\frac x7\right)^2\sqrt{\frac{||x|-3|}{|x|-3}}+\left(\frac y3\right)^2\sqrt{\frac{\left|y+\frac{3\sqrt{33}}7\right|}{y+\frac{3\sqrt{... | Here's what I got from the equation using Maple...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
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The number of partitions of $n$ into distinct parts equals the number of partitions of $n$ into odd parts This seems to be a common result. I've been trying to follow the bijective proof of it, which can be found easily online, but the explanations go over my head. It would be wonderful if you could give me an understa... | I'll give a sketch of the bijective proof; ask me if there's some part you don't understand and need fleshed out (or maybe someone else will post a detailed version).
The important idea is that every number can be expressed uniquely as a power of 2 multiplied by an odd number. (Divide the number repeatedly by 2 till yo... | {
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"source": "stackexchange",
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How to solve these inequalities? How to solve these inequalities?
*
*If $a,b,c,d \gt 1$, prove that $8(abcd + 1) \gt
(a+1)(b+1)(c+1)(d+1)$.
*Prove that $ \cfrac{(a+b)xy}{ay+bx} \lt \cfrac{ax+by}{a+b}$
*Find the greatest value of $x^3y^5z^7$ when $2x^2+2y^2+2x^2=15$
Any hints/solution are welcome.
| Let $a=1+x$, $b=1+y$, $c=1+z$ and $d=1+t$. Hence,
$$8(abcd + 1)-(a+1)(b+1)(c+1)(d+1)=$$
$$=8(1+x)(1+y)(1+z)(1+t)+8-(2+x)(2+y)(2+z)(2+t)=$$
$$=4(xy+xz+yz+xt+yt+zt)+6(xyz+xyt+xzt+yzt)+7xyzt>0$$
Done!
| {
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"source": "stackexchange",
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Show that $PSL(3,4)$ has no element of order $15$.
$PSL(3,4)$ has no element of order $15$. Thus it is no isomorphic to $A_8$.
Here, $PSL(3,4)$ denotes the $3 \times 3$ projective special linear group on the field with $4$ elements.
As listing all the elements takes too much work, is there any better way to prove the... | I try to write out the proof sketched by user641.
Let $F$ denote the field with 4 elements. Speaking of a polynomial, "irreducible" will mean "irreducible over $F$".
We have to prove that $PSL(3, 4)$ has no element of order 6 and this amounts to prove that if $M$ is a matrix in $SL(3, 4)$ such that $M^{6}$ is scalar, ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Elementary central binomial coefficient estimates
*
*How to prove that $\quad\displaystyle\frac{4^{n}}{\sqrt{4n}}<\binom{2n}{n}<\frac{4^{n}}{\sqrt{3n+1}}\quad$ for all $n$ > 1 ?
*Does anyone know any better elementary estimates?
Attempt. We have
$$\frac1{2^n}\binom{2n}{n}=\prod_{k=0}^{n-1}\frac{2n-k}{2(n-k)}... | A way to get explicit bounds via Stirling's approximation is to use the following more precise formulation: $$n! = \sqrt{2\pi n} \left( \frac{n}{e} \right)^n e^{\alpha_n} $$ where $ \frac{1}{12n+1} < \alpha_n < \frac{1}{12n} $.
With this one arrives at $$ \binom{2n}{n} = \frac{4^n}{\sqrt{\pi n}} e^{\lambda_n} $$ where... | {
"language": "en",
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"source": "stackexchange",
"question_score": "39",
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What is the maximum value of this trigonometric expression What is the maximum value of the expression
$1/(\sin^2 \theta + 3\sin\theta \cos\theta+ 5\cos^2 \theta$).
I tried reducing the expression to $1/(1 + 3\sin\theta$ $\cos\theta + 4\cos^2 \theta)$.
How do I proceed from here?
| This is the same as minimizing the function $f(\theta)=\sin^2 \theta+3 \sin \theta \cos \theta+5\cos^2 \theta$ (subject to $f(\theta)>0$). We know such a minimum will occur only when $f'(\theta)=0$. We have
\begin{align*}
f'(\theta)&= 2 \sin \theta \cos \theta - 3 \sin^2 \theta +3 \cos^2 \theta-10 \sin \theta \cos \t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Number of positive integral solutions for $ab + cd = a + b + c + d $ with $1 \le a \le b \le c \le d$
How many positive integral solutions exist for: $ab + cd = a + b + c +
d $,where $1 \le a \le b \le c \le d$ ?
I need some ideas for how to approach this problem.
| The equation can be rewritten as
$$(a-1)(b-1)+(c-1)(d-1)=2.$$
Now there are not many possibilities to consider! If the first product is $0$, the second must be $2$, and if the first product is $1$, so is the second.
If $a=1$, then we need to have $(c-1)(d-1)=2$. Since $1\le c\le d$, this forces $c=2$, $d=3$. And $b$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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limit of $\frac{f'(x)}{x}$, $f$ convex of class $C^1$ Let $f:\mathbb R\to \mathbb R$ be a convex function of class $C^1$. Let us suppose that there exists the limit
$$L:=\lim_{x\to+\infty}\frac{f(x)}{x^2}$$ and that $0<L<+\infty$.
a) Prove that $$0<\liminf_{x\to+\infty}\frac{f'(x)}{x}\leq\limsup_{x\to+\infty}\frac{f'(... | Pick any $\epsilon>0$, then there is an $x_0$ big enough so that for $x>x_0$, $\left|\frac{f(x)}{x^2}-L\right|<\epsilon^2$. Then, for $x>y>x_0$,
$$
\begin{align}
\frac{f(x)-f(y)}{x^2-y^2}&\le\frac{(L+\epsilon^2)x^2-(L-\epsilon^2)y^2}{x^2-y^2}\\
&=L+\epsilon^2\frac{x^2+y^2}{x^2-y^2}
\end{align}
$$
and
$$
\begin{al... | {
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How many positive integer solutions to $a^x+b^x+c^x=abc$? How many positive integer solutions are there to $a^{x}+b^{x}+c^{x}=abc$? (e.g the solution $x=1$, $a=1$, $b=2$, $c=3$). Are there any solutions with $\gcd(a,b,c)=1$? Any solutions to $a^{x}+b^{x}+c^{x}+d^{x}=abcd$ etc. ?
| Unfortunately, I don't know how to make comments here, would be better to move it there. Anyways, consider for example, $a^x+b^x+c^x=abc$ with $a\le b\le c.$ Clearly, $c^x<LHS\le c^3,$ therefor $x<3.$ Case $x=1$ is fairly simple as you can bound $ab\le 3$ (following the same lines). So let's consider the case $x=2.$ Ou... | {
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Generalisation of the identity $\sum\limits_{k=1}^n {k^3} = \bigg(\sum\limits_{k=1}^n k\bigg)^2$ Are there any generalisations of the identity $\sum\limits_{k=1}^n {k^3} = \bigg(\sum\limits_{k=1}^n k\bigg)^2$ ?
For example can $\sum {k^m} = \left(\sum k\right)^n$ be valid for anything other than $m=3 , n=2$ ?
If not, i... | Here is a curious (and related) identity which might be of interest to you. Let $D_{k} = ${ $d$ } be the set of unitary divisors of a positive integer $k$, and let $\sigma_{0}^{*} \colon \mathbb{N} \to \mathbb{N}$ denote the number-of-unitary-divisors (arithmetic) function. Then it is relatively straightforward to prov... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 0
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Finding asymptotes to $y = \frac{2x^2 + 3x - 6}{2x + 1}$ I need to find the asymptotes of
$y = \frac{2x^2 + 3x - 6}{2x + 1}$.
The asymptote at $x = -1/2$ is clear. If one long divides they can easily see that there is an asymptote of $y = x + 1$ as $x$ goes to infinity.
However, what is wrong with this reasoning? I cl... | As $x \to \infty$, $2x^2$ is the dominant term in the numerator, while $2x$ is the dominant term in the denominator. So the leading term in $y$ is $\frac{2x^2}{2x} = x$. You still need the constant term, which you can do this way: $$\frac{2x^2 + 3x - 6}{2x+1} \approx
\frac{2x^2 (1 + 3/(2x))}{2x (1 + 1/(2x))} \appr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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System of $\sqrt{7x+y}+\sqrt{x+y}=6$ and $\sqrt{x+y}-y+x=2$ $$\begin{align*}\sqrt{7x+y}+\sqrt{x+y}=6\\\sqrt{x+y}-y+x=2\end{align*}$$
I have tried various things squaring, summing but nothing really helped, got some weird intermediate results which are probably useless such as:
$$(y-x)(y+x+4)+4-x-y=0$$
or
$$x_{1,2}=\fra... | When we see a square root, it is hard to resist the impulse to square. But resist we will, and such a decision is often useful.
The kind of solution we are looking is not specified, but the expressions $\sqrt{7x+y}$ and $\sqrt{x+y}$ suggest we are looking for real solutions. Let
$$7x+y=p^2 \qquad\text{and}\qquad x+y=q... | {
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Simplifying a simple fraction with exponent I am trying to simplify this fraction :
$
\dfrac{(3^2)(5^4)}{15^3}
$
The answer is : $ \dfrac{5}{3} $
I am trying to do the following: $ \dfrac{3^2}{15^3} \cdot \dfrac{5^4}{15^3} $ so ... $ \dfrac{1^{-3}}{3} \cdot \dfrac{1^1}{3} $
But that's not giving me the right answer, ... | Here it is explicitly. Notice $15^3=(3\cdot 5)^3=3^3\cdot 5^3$. So
$$
\dfrac{(3^2)(5^4)}{15^3}=\frac{3^2\cdot 5^4}{3^3\cdot 5^3}=\frac{5}{3}.
$$
| {
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Finding limit of $\lim\limits_{h\to 0} \frac1{h}\left(\frac1{\sqrt{x+h-2}}-\frac1{\sqrt{x-2}}\right)$ As expected, if you plug in 0 into the initial equation, the answer is undefined or indeterminate. I tried multiplying the conjugate $\frac1{\sqrt{x+h-2}}+\frac1{\sqrt{x-2}}$ to the numerator and the denominator, but i... | An alternative evaluation. Let $f(x)=\frac{1}{\sqrt{x-2}}$. Then, by definition of $f'(x)$
$$
\lim_{h\rightarrow 0}\frac{1}{h}\left( \frac{1}{\sqrt{x+h-2}}-\frac{1}{\sqrt{
x-2}}\right) =f^{\prime }(x),
$$
which is
$$
\begin{eqnarray*}
f^{\prime }(x) &=&\left( \frac{1}{\sqrt{x-2}}\right) ^{\prime }=\left( \sqrt{
x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/64112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solving Radical Equations $x-7= \sqrt{x-5}$ This the Pre-Calculus Problem:
$x-7= \sqrt{x-5}$
So far I did it like this and I'm not understanding If I did it wrong.
$(x-7)^2=\sqrt{x-5}^2$ - The Square root would cancel, leaving:
$(x-7)^2=x-5$ Then I F.O.I.L'ed the problem.
$(x-7)(x-7)=x-5$
$x^2-7x-7x+14=x-5$
$x^2-14x+1... | $(x-7)=\sqrt{x-5}$
Check the domain first
$x-5 \geq0 \cap x-7 \geq0$
$$x \geq7$$
$(x-7)^2=\sqrt{x-5}^2$
$(x-7)^2=x-5$
$(x-7)(x-7)=x-5$
$x^2-7x-7x+49=x-5$
$x^2 - 15x + 54 = 0$
$(x - 9)(x - 6) = 0$
$x - 9 = 0 $ or $x - 6 = 0$
$x = 9$ or $x = 6$
checking the domain x=6 is an extraneous root.
x = 9
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/66255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Special numbers Our teacher talked about some special numbers.
These numbers total of 2 different numbers' cube. For example :
$x^3+y^3 = z^3+t^3 = \text{A-special-number}$
What is the name of this special numbers ?
| Here's a sample identity by Ramanujan,
$(3x^2+5xy-5y^2)^3 + (4x^2-4xy+6y^2)^3 = (-5x^2+5xy+3y^2)^3 + (6x^2-4xy+4y^2)^3$
Let {x,y} = {-1,0} and you get the nice $3^3+4^3+5^3 = 6^3$. Or {x,y} = {-1,2} for $1^3+12^3 = 9^3+10^3 = 1729$, the smallest non-trivial "taxicab number" (after transposition and removing common fact... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/67406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
The $3 = 2$ trick on Google+ I found out this on Google+ yesterday and I was thinking about what's the trick. Can you tell?
How can you prove $3=2$?
This seems to be an anomaly or whatever you call in mathematics. Or maybe I'm just plain dense.
See this illustration:
$$ -6 = -6 $$
$$ 9-15 = 4-10 $$
Adding $\frac{25}{4... | It's simply not true in the reals that if $ x^2=y^2 $ then x=y. For example, if $ (-2)^2=2^2 $, but 2 does not equal -2. The scheme of inference used in the last step in general isn't valid.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/68913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
What is the asymptotic bound for this recursively defined sequence? $f(0) = 3$
$f(1) = 3$
$f(n) = f(\lfloor n/2\rfloor)+f(\lfloor n/4\rfloor)+cn$
Intuitively it feels like O(n), meaning somewhat linear with steeper slope than c, but I have forgot enough math to not be able to prove it...
| Suppose we first study $$g(n) = f(n)-3$$ which has $g(0)=g(1)=0$ and
$$g(n) = g(\lfloor n/2 \rfloor) + g(\lfloor n/4 \rfloor) + cn+3.$$
Then it is not difficult to see that for the binary representation of $n$ being
$$n = \sum_{k=0}^{\lfloor \log_2 n \rfloor} d_k 2^k$$
we have that for $n\ge 2,$
$$g(n) = \sum_{j=0}^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/69017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How do I show that $2\ln(1+1/x) > 1/x > \ln(1+1/x)$ for any positive integer $x$? How do I show that $$2\ln\left(1+\frac1x\right) > \frac1x > \ln\left(1+\frac1x\right)$$ for any positive integer $x$?
I know it's true but how do I show it?
| Let's observe: $\ln(1+\frac{1}{x})<\frac{1}{x}$
We can define $y=\ln(1+\frac{1}{x})-\frac{1}{x}$ ,Now we may find first derivative $y'$:
$y'=\frac{1}{1+\frac{1}{x}}(1+\frac{1}{x})'-(\frac{1}{x})' \Rightarrow y'=\frac{1}{x^2}(\frac{1}{x+1}) \Rightarrow y'>0$ for all positive integers..so function $y$ increases as $x$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/69974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding a Particular Coefficient Using Generating Functions I have a homework question to solve the number of ways to choose 25 ice creams of a selection of 6 types of ice creams and there are only 7 of each ice cream type.
The question requires the use of generating functions and I have gotten to this point:
$$({{x}^{... | It might be useful (or not) to note that the first factor is $(z^8 - 1)^6$. Try:
$\begin{align*}
[x^{25}] (x^{48}-6 x^{40} + 15 x^{32} - 20 x^{24} + 15 x^{16} -6 x^{8} + 1)
\cdot \left( \frac{1}{1 - x} \right)^6
&= [x^{25}] (1 - 6 x^8 + 15 x^{16} - 20 x^{24})
\cdot \sum_{k \ge 0} (-1)^k \bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/70374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Cauchy-Schwarz inequality and three-letter identities (exercise 1.4 from "The Cauchy-Schwarz Master Class") Exercise 1.4 from a great book The Cauchy-Schwarz Master Class asks to prove the following:
For all positive $x$, $y$ and $z$, one has
$$x+y+z \leq 2 \left(\frac{x^2}{y+z} + \frac{y^2}{x+z} + \frac{z^2}{x+y}\righ... | I think the symmetry here comes from being able to derive a similar set of inequalities incorporating the same splitting method. To build off of @Martin's method, we have
$$ (x+y+z) \leq 2\left(\frac{x^2}{y+z} + \frac{y^2}{x+z} + \frac{z^2}{x+y}\right) $$
And using the same splitting method we can derive
$$ (x+y+z) \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/71509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 5
} |
How to Evaluate $ \int \! \frac{dx}{1+2\cos x} $ ?
Possible Duplicate:
How do you integrate $\int \frac{1}{a + \cos x} dx$?
I have come across this integral and I tried various methods of solving. The thing that gets in the way is the constant $2$ on the $\cos(x)$ term. I tried the conjugate (works without the 2$\co... | Generalization:
Let's consider $\cos x = \frac{1-t^2}{1+t^2}; t = \tan\frac{x}{2}; dx=\frac{2}{1+t^2} dt.$ Then we get that our integral becomes:
$$J = \int \frac{2dt}{(a+b)+(a-b) t^2}$$
I. For the case $a>b$, consider $a+b=u^2$ and $a-b=v^2$, and obtain that:
$$J = 2\int \frac{dt}{u^2+v^2 t^2}=\frac{2}{uv} \arctan\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/73250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Find all integers $m$ such that $\frac{1}{m}=\frac{1}{\lfloor 2x \rfloor}+\frac{1}{\lfloor 5x \rfloor} $ How would you determine all integers $m$ such that the following is true?
$$\frac{1}{m}=\frac{1}{\lfloor 2x \rfloor}+\frac{1}{\lfloor 5x \rfloor} .$$
Note that $\lfloor \cdot \rfloor$ means the greatest integer fun... | If $k + j/10 \le x < k + (j+1)/10$ where $k$ and $j$ are nonnegative integers and $j \le 9$, $\lfloor 2x \rfloor = \begin{cases} 2k & 0 \le j \le 4 \\ 2k+1 & 5 \le j \le 9 \end{cases}$ while $\lfloor 5x \rfloor = \begin{cases} 5k + j/2 & j \ \text{even} \\ 5k + (j-1)/2 & j \ \text{odd} \end{cases}$.
Going over the var... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/77290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Expanding $(2y-2)^2$ by FOIL Expanding $(2y-2)^2$
Isn't this same as $$(2y-2)(2y-2)\ ?$$
$$4y^2-6y+4$$
This should be FOILd shouldn't it?
| Yes, $(2y-2)^2=(2y-2)(2y-2)$.
FOILing should work, but will get you $4y^2−8y+4$, rather than $4y^2−6y+4$, as shown:
$(2y-2)(2y-2)=(2y)(2y)+(2y)(-2)+(-2)(2y)+(-2)(-2)$
$=4y^2-4y-4y+4=4y^2−8y+4$
If you want to memorize the formula which will get you the same result, it is
$$(a+b)^2=a^2+2ab+b^2$$
In your example, $a=2y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/77606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the slope of the tangent line to $\frac{8}{\sqrt{4+3x}}$ at $(4,2)$ In order to find the slope of the tangent line at the point $(4,2)$ belong to the function $\frac{8}{\sqrt{4+3x}}$, I choose the derivative at a given point formula.
$\begin{align*}
\lim_{x \to 4} \frac{f(x)-f(4)}{x-4} &=
\lim_{x \mapsto 4}... | Use the L'Hopital method, several equations becomes easy to solve
$$\lim_{x \to 4}\frac{f(x)-f(4)}{x-4}=\lim_{x \to 4}\frac{f'(x)-0}{1-0}=\lim_{x \to 4}f'(x)$$
Where
$f(x)=\frac{8}{\sqrt{3 x+4}}$ and $f'(x)$, the derivative of $f(x)$ is defined by
$f'(x)=-\frac{12}{(3 x+4)^{3/2}}$
The final equation results, just d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/80010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Group under matrix multiplication I am trying to show that this set P={ $p(\alpha,\beta,\gamma)=\pmatrix{1&\alpha&\beta\\0&1&\gamma\\0&0&1}$ $|$ $\alpha,\beta,\gamma$ $\in R$} is a group under matrix multiplication. I have already proved the closure, identity and associative properties. But with the inverse, Im stuck a... | $$\begin{align}
\begin{pmatrix} 1 & x & y & \vdots & 1 & 0 & 0 \\
0 & 1 & z & \vdots & 0 & 1 & 0 \\
0 & 0 & 1 & \vdots & 0 & 0 & 1 \end{pmatrix} &\sim
\begin{pmatrix} 1 & x & 0 & \vdots & 1 & 0 & -y \\
0 & 1 & 0 & \vdots & 0 & 1 & -z \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/81268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $a^n+n^{a}$ is prime number and $a=3k-1$ then $n\equiv 0\pmod 3$? Is it true that :
If $a^n+n^{a}$ is prime number and $a=3k-1$ then $n\equiv 0\pmod 3$
where $a>1,n>1 ; a,n,k \in \mathbb{Z^+}$
I have checked statement for many pairs $(a,n)$ and it seems to be true.
Small Maple code that prints $(a,n)$ pairs :
An... | Since it is prime, we know that $a^n+n^a \equiv 1 \text{ or } 5 \mod 6$, and because $a \equiv 2 \mod 3$ we know that $a \equiv 2 \text{ or } 5 \mod 6.$
The case $a \equiv 2 \mod 6:$
Note that $n^2 \equiv n^8 \equiv n^{14} \dots \mod 6.$
When $n \equiv 0, 1, 2, 3, 4, 5 \mod 6, n^2 \equiv 0, 1, 4, 3, 4, 1 \mod 6$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/82631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Negative Exponents in Binomial Theorem I'm looking at extensions of the binomial formula to negative powers. I've figured out how to do $n \choose k$ when $n < 0 $ and $k \geq 0$:
$${n \choose k} = (-1)^k {-n + k - 1 \choose k}$$
So now let's look at one case for using the binomial coefficient:
$$(1+x)^n = \sum_{k=0}^n... | For $a=1$, the negative binomial series simplifies to
$(x+1)^{-n}=1-nx+\frac1{2!}n(n+1)x^2-\frac1{3!}n(n+1)(n+2)x^3+....$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/85708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 2,
"answer_id": 1
} |
Distance Between A Point And A Line Any Hint on proving that the distance between the point $(x_{1},y_{1})$ and the line $Ax + By + C = 0$ is,
$$\text{Distance} = \frac{\left | Ax_{1} + By_{1} + C\right |}{\sqrt{A^2 + B^2} }$$
What do I use to get started? All I know is the distance formula $\sqrt{(x_{2}-x_{1})^2+(y_{2... | Nothing to see here, just me preparing for my Calc 3 final by attempting this.
We can Lagrange this. Let $f(x) = \sqrt{(x - x_1)^2 + (y - y_1)^2}$ be the distance of $(x_1, y_1)$ from some $(x, y)$ on the line $Ax + By + C = 0$. Also, let $g(x) = Ax + By + C = 0$ be our constraint function, and $(f(x, y))^2 = (x - x_1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/85761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 9,
"answer_id": 7
} |
How to compute the characteristic polynomial of $A$ The matrix associated with $f$ is:
$$ \left(\begin{array}{rrr}
3 & -1 & -1 \\
-1 & 3 & -1 \\
-1 & -1 & 3
\end{array}\right) .
$$
First, I am going to find the characteristic equation of $\det(A- \lambda I)$. Please correct me if I'm wrong.
$$= (3-\... | Notice that your characteristic equation is
$$(3-\lambda)^3 - 3(3-\lambda) - 2 = 0.$$
Making the change of variable $x=3-\lambda$, we get the depressed cubic
$$x^3 - 3x - 2 = 0.$$
By the rational root theorem, you can test $1$, $-1$, $2$ and $-2$, which tells you that $x=2$ is a solution. Factoring out $x-2$ we have
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/88324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Right triangle where the perimeter = area*k I was doodling on some piece of paper a problem that sprung into my mind. After a few minutes of resultless tries, I advanced to try to solve the problem using computer based means.
The problem stated is
Does a right angle triangle with integer sides such that
$$P = A\c... | We have the following relations:
$$P=a+b+c,\quad 2A=ab,\quad a^2+b^2=c^2$$
In fact, there is one more that is implicit: the triangle inequality $a+b>c$ (which follows from $a^2+b^2=c^2$). We wish to relate the first and second relations, and so after staring at the problem for a while we realize the correct course of a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/90236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
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Solving for the smallest $x$ : $1! + 2! + \cdots+ 20! \equiv x\pmod 7$ I know the smallest $x \in \mathbb{N}$, satisfying $1! + 2! + \cdots + 20! \equiv x\pmod7$ is $5$. I would like to know methods to get to the answer.
| Note that each of $7!$, $8!$, $9!,\ldots, 20!$ is congruent to $0$ modulo $7$, since they are all divisible by $7$.
Note that $6!\equiv -1\pmod{7}$ by Wilson's Theorem, which cancels $1!$.
That leaves $2!+3!+4!+5! = 2! + 3!(1 + 4 + 20)$. But $3!\equiv -1\pmod{7}$, and $20\equiv -1\pmod{7}$, so $2!+3!+4!+5! \equiv 2-(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/91253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Is there a binomial identity for this expression $\frac{\binom{r}{k}}{\binom{n}{k}}$? What I'm trying to prove is this summation:
$$\sum_{i=0}^{k} \dfrac{\dbinom{r}{i} \cdot \dbinom{n - r}{k - i}}{\dbinom{n}{k}} \cdot i = \dfrac{r}{n} \cdot k$$
I used induction on $k$ as follows:
$$LHS = \sum_{i=0}^{k + 1} \dfrac{\dbin... | We don't need induction:
\begin{align*}
\sum_{i=0}^k\binom ri\binom{n-r}{k-i}i&=\sum_{i=1}^k\frac{r!}{(i-1)!(r-1-(i-1))}\binom{n-r}{k-i}\\
&=r\sum_{l=0}^{k-1}\binom{r-1}{l}\binom{n-r}{k-1-l}\\
&=r\binom{n-r+r-1}{k-1}\\
&=r\binom{n-1}{k-1}\\
&=\frac rn\frac{n!}{(k-1)!(n-k)!}\\
&=\frac {rk}n\binom nk,
\end{align*}
the t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/92719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Finding the $n$-th derivatives of $x^n \ln x$ and $\frac{\ln x}{x}$. How can I prove the following identities:
$$ \left( x^n \ln x \right)^{(n)}= n! \left(\ln x + 1 + \frac{1}{2} +\cdots +\frac{1}{n} \right), \quad x>0, \quad n\ge 1, \tag{a}$$
$$ \left( \frac{\ln x}{x} \right)^{(n)}= (-1)^n \ n! \ x^{-n-1} \left( \ln... | I prefer using induction to prove the statement. For (a), when $n=1$, by product rule
$$\left( x \ln x \right)'=\ln x+x\cdot\frac{1}{x}=1!(\ln x+1),$$
which shows that (a) is true for $n=1$. Now suppose that (a) is true for $n=k$, i.e.
$$\left( x^k \ln x \right)^{(k)}= k! \left(\ln x + 1 + \frac{1}{2} +\cdots +\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/94018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
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Can anyone prove this formula? I found formula below$$p_n=6\left \lfloor \frac{p_n}{6}+\frac{1}{2} \right \rfloor+\left ( -1 \right )^\left \lfloor \frac{p_n}{3} \right \rfloor$$ for $n>2$, $p_n$ is prime number sequence.
Can anyone prove this formula?
| For $n>2$ the prime $p_n$ is greater than $3$, so it must be of the form $6k+1$ or $6k+5$. If $p_n=6k+1$, then $$6\left\lfloor \frac{p_n}6+\frac12\right\rfloor= 6\left\lfloor k+\frac16+\frac12\right\rfloor=6k\;,$$ and $$\left\lfloor\frac{p_n}3\right\rfloor=\left\lfloor\frac{6k+1}3\right\rfloor=2k\;,$$ so $$6\left\lfloo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/95163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The Taylor series of $\int_0^x \operatorname{sinc}(t) dt$ I tried to find what is the Taylor series of the function $$\int_0^x \frac{\sin(t)}{t}dt .$$
Any suggestions?
| I just want to extend J.M.'s solution to a full solution of the exercise:
$$\int_0^x \frac{\sin(t)}{t} \,dt$$
Is with Maclaurin expansion ($\sin\,t=t\left(1-t^2/3!+t^4/5!-t^6/7!+\cdots\right)$) equals to
$$\int_0^x \frac{t}{t}\left(1-\frac{t^2}{3!}+\frac{t^4}{5!}-\frac{t^6}{7!}+\cdots\right) \,dt = \int_0^x \left(1-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/95395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Using Dyson's conjecture to give another proof of Dixon's identity. For natural numbers $a_1,\dots,a_n$, Freeman Dyson conjectured (and it was eventually proven) that the Laurent polynomial
$$
\prod_{i,j=1\atop i\neq j}^n\left(1-\frac{x_i}{x_j}\right)^{a_i}
$$
has constant term the multinomial coefficient $\binom{a_1... | Setting $a_1=a$, $a_2=b$, $a_3=c$, we can start with
$
\prod_{1\le i,j\le 3, \ i\ne j}\left(1-\frac{x_i}{x_j}\right)^{a_i}
$
and combine pairs of factors involving the same pair of variables to find that
$$
\prod_{1\le i,j\le 3, \ i\ne j}\left(1-\frac{x_i}{x_j}\right)^{a_i}
$$
$$
= (-1)^{a+b+c} x_1^{a-c} x_2^{b-a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/103764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
Trigonometric equality $\cos x + \cos 3x - 1 - \cos 2x = 0$ In my text book I have this equation:
\begin{equation}
\cos x + \cos 3x - 1 - \cos 2x = 0
\end{equation}
I tried to solve it for $x$, but I didn't succeed.
This is what I tried:
\begin{align}
\cos x + \cos 3x - 1 - \cos 2x &= 0 \\
2\cos 2x \cdot \cos x - 1 - \... | Hint: Can you write $\cos 2x$ and $\cos 3x$ in terms of $\cos x$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/104350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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A symmetric inequality
Possible Duplicate:
Cauchy-Schwarz inequality and three-letter identities (exercise 1.4 from “The Cauchy-Schwarz Master Class”)
Is it true for all $x, y, z > 0$ that
$$ x + y + z \leq 2 \left\{ \frac{x^2}{y+z} + \frac{y^2}{x+z} + \frac{z^2}{x+y} \right\} $$
This is an exercise (1.4) in "The Ca... | It is correct $ x + y + z = \frac{x}{\sqrt{y+z}}\sqrt{y+z} + \frac{y}{\sqrt{x+z}}\sqrt{x+z}
+ \frac{z}{\sqrt{x+y}}\sqrt{x+y}\le \left(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{y+x}\right)^{1/2}(2x+2y+2z)^{1/2}.$ Square both sides, and cancel out $x+y+z$.
| {
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$(\cos \alpha, \sin \alpha)$ - possible value pairs We introduced the complex numbers as elements of $ \mathrm{Mat}(2\times 2, \mathbb{R})$ with
$$
\mathbb{C} \ni x =
\left(\begin{array}{cc}
a & -b \\
b & a \\
\end{array}\right) =
\frac{1}{\sqrt{a^2+b^2}}
\left(\begin{array}{cc}
\frac{a}{\sqrt{a^2+b^2}} & \frac... | Find an $\alpha$ so that $\cos(\alpha)= \frac{a}{\sqrt{a^2+b^2}}$. This is possible because the fraction is between $-1$ and $1$.
Then note that
$$\sin^2(\alpha)= 1- \cos^2(\alpha)=1- \frac{a^2}{a^2+b^2}=\frac{b^2}{a^2+b^2} \,.$$
Thus either $\sin(\alpha)=\frac{b}{\sqrt{a^2+b^2}}$ or $\sin(\alpha)=\frac{-b}{\sqrt{a^... | {
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How can I find the area of the region bounded by the hyperbola now I got region question....
The question was "find the area of the region bounded by the hyperbola $9x^{2} - 4y^{2} = 36$ and the line $x = 3$
I drew the graph of $9x^{2} - 4y^{2} = 36$ and $x = 3$ and I got right side is bounded by $x=3$ but since the gr... | Solving for $y$ produces:
$$ y =\frac{1}{2} \sqrt{9x^2-36} $$
What we need to find is the area of $y$ from $x=2$ to $x=3$.
Thus we put:
$$A = \frac{1}{2}\int_2^3 \sqrt{9x^2-36} dx$$
Make an "hiperbolic" substitution:
$$3x = 6 \cosh u$$
$$3dx = 6 \sinh u du$$
We get the new limits are $0$ and $b = \cosh^{-1}\frac{3}{2}... | {
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Possibility to simplify $\sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}} = \frac{\pi }{{\sin \pi a}}} $ Is there any way to show that
$$\sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}} = \frac{1}{a} + \sum\limits_{k = 1}^\infty {{{\left( { - 1} \righ... | EDIT: The classical demonstration of this is obtained by expanding in Fourier series the function $\cos(zx)$ with $x\in(-\pi,\pi)$.
Let's detail Smirnov's proof (in "Course of Higher Mathematics 2 VI.1 Fourier series") :
$\cos(zx)$ is an even function of $x$ so that the $\sin(kx)$ terms disappear and the Fourier expans... | {
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Find smallest number when divided by $2,3,4,5,6,7,8,9,10$ leaves $1,2,3,4,5,6,7,8,9$ remainder Find smallest number when divided by $2,3,4,5,6,7,8,9,10$ leaves $1,2,3,4,5,6,7,8,9$ remainder.How to go about solving this problem??
| There is no smallest such number. But there is a smallest positive integer.
Note that $-1$ has the desired property. The integers that have the desired property are all the integers of the form $-1+kM$, where $M$ is the least common multiple of the integers $2,3,4,\dots, 10$ and $k$ ranges over the integers. So the sm... | {
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Showing $ \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\frac{\pi}{3\sqrt{3}}$ I would like to show that:
$$ \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\frac{\pi}{3\sqrt{3}} $$
We have:
$$ \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\sum_{n=0}^{\infty} \frac{1}{3n+1}-\frac{1}{3n+2} $$
I wanted to use the fact that $$\arcta... | There is another way to solve the problem. Note
$$ (3n+1)(3n+2)=9\left[(n+\frac{1}{2})^2+\left(\frac{i}{6}\right)^2\right] $$
and
and hence
\begin{eqnarray}
\sum_{n=0}^\infty\frac{1}{(3n+1)(3n+2)}&=&\frac19\sum_{n=0}^\infty\frac1{(n+\frac{1}{2})^2+\left(\frac{i}{6}\right)^2}\\
&=&\frac1{18}\sum_{n=-\infty}^\infty\frac1... | {
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Limit of the nested radical $x_{n+1} = \sqrt{c+x_n}$ (Fitzpatrick Advanced Calculus 2e, Sec. 2.4 #12)
For $c \gt 0$, consider the quadratic equation
$x^2 - x - c = 0, x > 0$.
Define the sequence $\{x_n\}$ recursively by fixing $|x_1| \lt c$ and then, if $n$ is an index for which $x_n$ has been defined, defining
$$x_{n+... | I am going to do this Ramanujan-style: pick some real positive $a$.
$$
a=\sqrt{a^2}=\sqrt{a^2-a+a}=\sqrt{a^2-a+\sqrt{a^2-a+a}}=\sqrt{a^2-a+\sqrt{a^2-a+\sqrt{a^2-a+a}}}=\dots=\sqrt{a^2-a+\sqrt{a^2-a+\sqrt{a^2-a+\sqrt{a^2-a+\dots}}}}
$$
What you do is keep replacing the last $a$ in the expression by $\sqrt{a^2-a+a}$.
Now... | {
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Finding the minimum value of $\left(\frac{a + 1}{a}\right)^2 + \left(\frac{b + 1}{b}\right)^2 + \left(\frac{c + 1}{c}\right)^2 $ Find the minimum value of
$$\left(\frac{a + 1}{a}\right)^2 + \left(\frac{b + 1}{b}\right)^2 + \left(\frac{c + 1}{c}\right)^2 $$
I tried to expand it and break it into individual terms and ... | Assuming $a,b,c$ are positive reals, there is no minimum!
Consider the individual term: $$\left(\frac{a + 1}{a}\right)^2 = \left(1 + \frac{1}{a}\right)^2$$
For positive real $a$ this is strictly greater than $1$, but as $a \to \infty$, this can be made as close to $1$ as we want.
Thus the total sum is always strictly $... | {
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Evaluate the integral $\int_{0}^1\frac{x-6}{x^2-6x+8}dx$ I used partial fractions to get:
$\displaystyle\frac{A}{x-4}+\frac{B}{x-2}= \frac{x-6}{(x-4)(x-2)}$
$A = -1$
$B = 2$
$\displaystyle\int_{0}^1\frac{-1}{x-4}+\frac{2}{x-2}dx$
Found the anti-derivative to be:
$(-\ln|x-4| + 2 \ln|x-2|)_0^{1}$
My answer came out to be... | In the interval $[0,1]$ we have $\frac{-5}{3}\leq\frac{x-6}{x^2-6x+8}\leq\frac{-3}{4}$, hence:
$$ \frac{-5}{3}\leq\int_0^1\frac{x-6}{x^2-6x+8}\leq\frac{-3}{4} $$
| {
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Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$ In Spivak's Calculus 3rd Edition, there is an exercise to prove the following:
$$x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$$
I can't seem to get the answer. Either I've gone wrong somewhere, I'm overlooking something, ... | Using Induction Hypothesis
Let n =k
Then ,
x^k -y^k = (x-y)(x^(k-1) +x^(k-2) y ....+y^k-1)
Let n= (k-1)
Then by Induction Hypothesis,
(x-y){x(x^(k-1) +x^(k-2) y) +y^k }
=» (x-y)(x^k +x^(k-1) y ... +y^k
Hence the formula is true for k+1 whenever true for k .
Therefore by Principle of Induction The formula is true for Po... | {
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For complex $z$, $|z| = 1 \implies \text{Re}\left(\frac{1-z}{1+z}\right) = 0$
If $|z|=1$, show that: $$\mathrm{Re}\left(\frac{1 - z}{1 + z}\right) = 0$$
I reasoned that for $z = x + iy$, $\sqrt{x^2 + y^2} = 1\implies x^2 + y ^2 = 1$ and figured the real part would be:
$$\frac{1 - x}{1 + x}$$
I tried a number of mani... | Let $z=a+bi$, where $a^2+b^2=1$ and $z\neq -1$. Then $$\begin{eqnarray}
\Re\left(\frac{1-z}{1+z}\right)
&=&\Re\left(\frac{(1-z)\overline{(1+z)}}{|1+z|^2}\right)\\
&=&\Re\left(\frac{(1-a-bi)(1+a-bi)}{(a+1)^2+b^2}\right)\\
&=&\Re\left(\frac{1-(a^2+b^2)-2bi}{(a+1)^2+b^2}\right)\\
&=&\Re\left(\frac{-2bi}{(a+1)^2+b^2}\... | {
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How does expanding $(2s+1)^2$ help us prove "$n^2$ even $\implies$ $n$ even"? How can $n^2 = (2s + 1)^2= 4s^2 + 4s + 1$ help in proving "If $n^2$ is even then $n$ is even"?
| Based on your comment on Lazar Ljubenović answer, here is an attempt to explain how to expand $(2s+1)^2$.
First let us rewrite $(2s+1)^2$ as $(2s+1)(2s+1)$. For the sake of this explanation let us rewrite $(2s+1)(2s+1)$ as $c(2s+1)$ where $c=2s+1$.
Using the distributive law $a(b+c)=ab+ac$ we know $$c(2s+1)=c\cdot 2s+... | {
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Show that $2^{3^k}+1$ is divisible by $3^k$ for all positive integers $k$ I attempted this by induction: Here is what I did
For $k=1, 2^{3^{1}}+1 = 2^{3}+1 = 9 = 3^{2}$ is divisible by $3^{1}$ , so the result is true for $k=1$
Now I assume the result to be true for $k=m$,
$2^{3^{m}}+1$ is divisible by $3^m$. To show t... | We show, as was observed by Sivaram Ambikasaran, that $3^{n+1}$ divides $2^{3^n}+1$. For $\varphi(3^{n+1})=2\cdot 3^n$. By Euler's Theorem, $2^{2\cdot 3^n}\equiv 1\pmod{3^{n+1}}$. Thus $2^{3^n}\equiv \pm 1 \pmod{ 3^{n+1}}$. But $2^{3^n}\not\equiv 1 \pmod{3}$, so $2^{3^n}\equiv -1\pmod{3^{n+1}}$.
Remark: It is not... | {
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Smith Normal Form Would the Smith Normal Form of the following matrix over $\mathbb Q[x]$
$$\begin{pmatrix}
(x+a)(x+b) & 0 & 0 &0 \\
0 & (x+c)(x+d) & 0 & 0 \\
0 &0 & x^3(x+a) & 0 \\
0 & 0 & 0& x^2(x+b)
\end{pmatrix}$$
simply be
$$\begin{pmatrix}
f(x) & 0 & 0 &0 \\
0 & f(x) & 0 & 0 \\
0 &0 & f(x... | No.
Let $s_k$ denote the $k$th entry on the diagonal of Smith form (i.e., the $k$th invariant factor). Then $$s_k = \frac{d_k}{d_{k-1}},$$ where $d_k = \gcd$ of all $k \times k$ minors of the original matrix (aka $k$th determinantal divisor). I'm assuming $a,b,c,d$ are distinct here. So the Smith form is:
$$
\begin{pma... | {
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What is the largest positive $n$ for which $n^3+100$ is divisible by $n+10$ What is the largest positive $n$ for which $n^3+100$ is divisible by $n+10$
I tried to factorize $n^3+100$, but $100$ is not a perfect cube. I wish it were $1000$.
| By division we find that $n^3 + 100 = (n + 10)(n^2 − 10n + 100)−900$.
Therefore, if $n +10$ divides $n^3 +100$, then it must also divide $900$.
Since we are looking for largest $n$, $n$ is maximized whenever $n + 10$ is, and since the largest divisor of $900$ is $900$, we must have $n + 10 = 900 \Rightarrow n = 890$
T... | {
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Plot $|z - i| + |z + i| = 16$ on the complex plane
Plot $|z - i| + |z + i| = 16$ on the complex plane
Conceptually I can see what is going on. I am going to be drawing the set of points who's combine distance between $i$ and $-i = 16$, which will form an ellipse. I was having trouble getting the equation of the ellip... | Sketching the graph gives a centrally symmetric ellipse with the imaginary axis as the major axis. The length of the half the major axis is clearly 8 (square 64), and pythagoras gives the square of half the minor axis as 63. This is reflected in the answer obtained by algebraic manipulation.
$$2x^2+2y^2 + 2\sqrt{x^2 + ... | {
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Proving a complicated inequality involving integers Let $a,b,c,d$ be integers such that $$\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) = \left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right) \mod 2$$ $$ ad-bc =1$$ $$\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \neq \left( \begin{matrix} \p... | In the case where $a$ and $d$ have the same sign, rewrite your expression as
$$ c^2(y^2 + x^2) + (cx + (d-a))^2 + 2ad + \frac{1}{y^2}(b-(d-a+3cx)x)^2$$
Since each term is non-negative your inequality follows easily. Perhaps you can adjust this expression so your inequality follows in the case where $a$ and $d$ have o... | {
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Help with basic high school math. What happens to $j$? I know my math is very rusty, actually, its always been that way. but I need help with this. The question below has me stumped. I've tried to show the steps I went through to get the answer. Please tel me where I made the mistake.
If x=a and x=b are two roots of a... | Here are your steps with mistakes highlighted in red:
$[-2 + j\sqrt5][-2 - j\sqrt5] = (-2)(-2) + (-2)(-j\sqrt5) + (-2)(+j\sqrt5) + (+j\sqrt5)(-j\sqrt5)$
$= 4 + (2j\sqrt5) - (2j\sqrt5) + \color{red}{(-j\sqrt5)^2}$
$ = 4 + (2j\sqrt5) - (2j\sqrt5) + \color{red}{(-j^2)(-\sqrt5)^2}$
$ = 4 + \color{red}{j^2 5}$
Here is fixed... | {
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What is the radius of the circle in cm? The rectangle at the corner measures 10 cm * 20 cm.
The right bottom corner of the rectangle is also a point on the circumference of the circle.
What is the radius of the circle in cm?
Is the data sufficent to get the radius of circle?
| $$r^2=x^2+y^2 \tag{1}$$
$r=y+20$ and $r=x+10$ therefore
$$y+20=x+10 \quad \mbox{then } \quad y=x-10 \tag{2}$$
Substitute $(2)$ into $(1)$
$$(r+10)^2=x^2+(x-10)^2$$
$$x^2+20x+100=x^2+x^2-20x+100$$
$$X^2=40x$$
$$x=40$$
Then substitute $x=40$ into $(2)$
$$y=40-10$$
$$y=30$$
substitute $x=40$ and $Y=30$ into $(1)$
$$r^2=(... | {
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Integrate $\sqrt{x^2 - 2x}$ $$\int{\sqrt{x^2 - 2x}}$$
I think I should be doing trig substitution, but which? I completed the square giving
$$\int{\sqrt{(x-1)^2 -1}}$$
But the closest I found is for
$$\frac{1}{\sqrt{a^2 - (x+b)^2}}$$
So I must add a $-$, but how?
| The standard way to solve these problems is indeed with trigonometric substitutions.
But it is not the only way to solve these.
Another way that was used more before is called Euler substitutions
if we are to integrate $\sqrt{ax^2+bx+c}$ that has real roots $\alpha$ and
$\beta$, then we can use the substitution $\... | {
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Extremely hard geometric problem Given a triangle ABC. BL is the bisector of angle ABC, H is the orthocenter and P is the mid-point of AC. PH intersects BL at Q. If $\angle ABC= \beta $, find the ratio $PQ:HQ$.If $QR\perp BC$ and $QS \perp AB$, prove that the orthocenter lies on $RS$.
| In the figures below, I have added the circumcenter, $U$, and the centroid, $E$. I have also placed $L$ on the circumcircle.
$\hspace{8mm}$
Note that since both are perpendicular to $\overline{AC}$, we have $\overline{BH}\,||\,\overline{UP}$; furthermore, $|\overline{BH}|=2|\overline{UP}|$. The latter is because $\tri... | {
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Partial fraction with a constant as numerator I am trying to express this as partial fraction:
$$\frac{1}{(x+1)(x^2+2x+2)}$$
I have a similar exaple that has $5x$ as numerator, it is easy to understand. I do not know what to do with 1 in the numerator, how to solve it?!
| You set it up in the usual way as $$\begin{align*}\frac1{(x+1)(x^2+2x+2)}&=\frac{A}{x+1}+\frac{Bx+C}{x^2+2x+2}\\
&=\frac{A(x^2+2x+2)+(Bx+C)(x+1)}{(x+1)(x^2+2x+2)}\;,
\end{align*}$$
so that $$A(x^2+2x+2)+(Bx+C)(x+1)=1\;.$$
Now multiply out the lefthand side to get $$(A+B)x^2+(2A+B+C)x+(2A+C)=1$$
and equate coefficient... | {
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Evaluate $(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011}$ Evaluate $$(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011}$$
So ...
$$(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011} = (-\sqrt{2}+\sqrt{2}i)^{-2011}$$
$$\theta=\pi - \arctan(\frac{\sqrt{2}}{\sqrt{2}}) = \frac{3\pi}{4}$$
$$-\sqrt{2}+\sqrt{2}i=\cos{\theta} + i \sin{\theta}$$
$$(-\sqrt{2}... | First off, let's get the signs correct:
$$\left(\frac{1}{-\sqrt{2}+\sqrt{2}~i}\right)^{\color{Red}+2011}=(-\sqrt{2}+\sqrt{2}~i)^{-2011}.$$
Second, if $z=-\sqrt{2}+\sqrt{2}i=\sqrt{2}(-1+i)$, then we do have a magnitude of two:
$$|z|=\sqrt{2}\cdot |-1+i|=\sqrt{2}\cdot\sqrt{(-1)^2+1}=2.$$
The angle $\theta=\frac{3}{4}\pi$... | {
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Equivalence of a Real Number to Itself If $\alpha$ and $\beta$ are two real numbers, we say that $\beta$ is equivalent to $\alpha$ if there are integers $a$, $b$, $c$, and $d$ such that $ad-bc=\pm1$ and $\beta=\frac{a\alpha+b}{c\alpha+d}$.
How can I show that a real number $\alpha$ is equivalent to itself?
What I have ... | $$ \alpha = \frac{ 1 \cdot \alpha + 0}{0\cdot \alpha + 1} .$$
| {
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How do you integrate $\int \frac{1}{a + \cos x} dx$? How do you integrate $\int \frac{1}{a + \cos x} dx$? Is it solvable by elementary methods? I was trying to do it while incorrectly solving a homework problem but I couldn't find the answer.
Thanks!
| Generalization:
Let's consider $\cos x = \frac{1-t^2}{1+t^2}; t = \tan\frac{x}{2}; dx=\frac{2}{1+t^2} dt.$ Then we get that our integral becomes:
$$J = \int \frac{2dt}{(a+b)+(a-b) t^2}$$
I. For the case $a>b$, consider $a+b=u^2$ and $a-b=v^2$, and obtain that:
$$J = 2\int \frac{dt}{u^2+v^2 t^2}=\frac{2}{uv} \arctan\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/134577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 4
} |
Solving the recursion $3a_{n+1}=2(n+1)a_n+5(n+1)!$ via generating functions I have been trying to solve the recurrence:
\begin{align*}
a_{n+1}=\frac{2(n+1)a_n+5((n+1)!)}{3},
\end{align*}
where $a_0=5$, via generating functions with little success. My progress until now is this:
Let $A(x)=\sum_{n=0} ^{\infty} a_nx^n$. ... | Being lazy and prove it by MI for fun.
Obviously, $a_0=5(0!)$, now assume $a_n=5n!$ for some natural number $n$, then
$$
\begin{aligned}
3 a_{n+1} &=2(n+1) 5 n !+5(n+1)_{0}^{1} \\
&=5 n !(2 n+2+n+1) \\
a_{n+1} &=5(n+1) !
\end{aligned}
$$
proved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/135803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 3
} |
Another evaluating limit question: $\lim\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}$ How do I begin to evaluate this limit: $$\lim_{n\to \infty}\ \frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}\;?$$
Thanks a lot.
| A simple, but famous trick works here: Observe that $(n-1)(n+1) = n^2 - 1 \leq n^2$. Thus we have
$$ \begin{align*}
& \left[ \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \right]^2 \\
&= \frac{1 \cdot 1}{2 \cdot 2}\cdot\frac{3 \cdot 3}{4 \cdot 4}\cdot\frac{5 \cdot 5}{6 \cdot 6}\cdots\frac{(2n-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/139494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
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