Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
This inequality $a+b^2+c^3+d^4\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}+\frac{1}{d^4}$
let $0<a\le b\le c\le d$, and such $abcd=1$,show that
$$a+b^2+c^3+d^4\ge \dfrac{1}{a}+\dfrac{1}{b^2}+\dfrac{1}{c^3}+\dfrac{1}{d^4}$$
it seems harder than This inequality $a+b^2+c^3\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}$
| Suppose that it is true for $n=3$:
$$
\forall 0 \le a \le b \le c \wedge abc = 1 :
a - \frac{1}{a}
+ b^2 - \frac{1}{b^2}
+ c^3 - \frac{1}{c^3} \ge 0.
$$
Let $d \ge c \ge 1$, then it is clear that
$$
\forall 0 \le a \le b \le c \le d \wedge abcd = d :
a - \frac{1}{a}
+ b^2 - \frac{1}{b^2}
+ c^3 - \frac{1}{c^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
Evaluating the infinite product $ \prod\limits_{n=1}^{\infty} \cos(\frac{y}{2^n}) $ How does one evaluate
$ \displaystyle\prod\limits_{n=1}^{\infty} \cos(\frac{y}{2^n}) $?
Seems fairly straightforward, as I just plugged in some initial values $n = 1, 2, 3,\dotsc$
$n = 1$
$ \sin(y)= 2\sin(\frac{y}{2})\cos(\frac{y}{2})$... | Hint: Use the Taylor series and let $n\rightarrow \infty:$
$$2^n\sin\left(\frac{y}{2^n}\right)= 2^n\left(\frac{y}{2^n}-\left(\frac{y}{2^n}\right)^3/3! + \dots\right) = y - \frac{y^3}{3!2^{2n}} + \dots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Special Palindromic String A string of length $N$ can be made from $6$ characters $a$, $b$, $c$, $d$, $e$ and $f$. There are some rules to make such a string:
*
*$b$ can not come directly after $a$.
*$d$ can not come directly after $a$ or $f$
*$c$ can not come directly after $c$ or $e$
*The string is a palindro... | Letting $r=\lceil n/2 -1 \rceil$, we have
$$f(n)=\begin{cases}
\frac{2^{-r-1}}{\sqrt{17}} \left(\left(5 \sqrt{17}-21\right)
\left(5-\sqrt{17}\right)^r+\left(21+5 \sqrt{17}\right)
\left(5+\sqrt{17}\right)^r\right) & n \textrm{ even;}\\
\frac{3\ 2^{-r}}{\sqrt{17}} \left(\left(\sqrt{17}-4\right)
\left(5-\sqrt{17}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the Integral: $\int\ e^{x}(1+e^x)^{\frac{1}{2}}\ dx$ Evaluate the Integral: $\int\ e^{x}(1+e^x)^{\frac{1}{2}}\ dx$
$u=1+e^x$
$du=e^x\ dx$
$\int\ u^{\frac{1}{2}}\ du$
$u+c=1+e^x+c$
| $$\int\ e^{3}(1+e^x)^{\frac{1}{2}}\ dx=$$
$$e^3\int\sqrt{e^x+1} dx=$$
(substitute $u=e^x$ and $du=e^xdx$):
$$e^3\int\frac{\sqrt{u+1}}{u} du=$$
(substitute $s=\sqrt{u+1}$ and $ds=\frac{1}{2\sqrt{u+1}}du$):
$$2e^3\int\frac{s^2}{s^2-1} ds=$$
$$2e^3\int\left(-\frac{1}{2(s+1)}+\frac{1}{2(s-1)}+1\right)ds=$$
$$-e^3\int \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find lim$_{x \to 0}\left(\frac{1}{x} - \frac{\cos x}{\sin x}\right).$ $$\lim_{x \to 0} \left(\frac{1}{x} - \frac{\cos x}{\sin x}\right) = \frac{\sin x - x\cos x}{x\sin x}= \frac{0}{0}.$$
L'Hopital's: $$\lim_{x \to 0} \frac{f'(x)}{g'(x)} = -\frac{1}{x^2} + \frac{1}{\sin ^2x} = \frac{0}{0}.$$
Once again, using L'Hopital... | Use Taylor-MacLaurin at order $2$:
\begin{align*}\frac1x - \frac{\cos x}{\sin x}&=\frac1x - \frac{1-\dfrac{x^2}2+o(x^2)}{x-\dfrac{x^3}6+o(x^3)}=\frac1x-\frac1x\left(\frac{1-\smash[t]{\dfrac{x^2}2}+o(x^2)}{1-\smash[b]{\dfrac{x^2}6}+o(x^2)}\right)\\&=\frac1x-\frac1x\Bigl(1-\frac{x^2}3+o(x^2)\Bigr)= \frac x3+o(x)\underset... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
Prove by induction that $4$ divides $n^3+(n+1)^3+(n+2)^3+(n+3)^3$ Just looking for someone to check my work and for feedback, thanks!
Base case: $n=0$
$0+1+8+27 = 36$
$4$ divides $36.$
Inductive step: Assume $4$ divides $k^3+(k+1)^3+(k+2)^3+(k+3)^3$ for some number where $k$ is a natural number including zero. So $k^... | To go from
$n$ to $n+1$,
you subtract $n^3$
and add $(n+4)^3$.
This means that
the sum changes by
$(n+4)^3 - n^3$,
so if this is divisible by 4,
divisibility by 4 remains.
But
$(n+4)^3 - n^3
=(n^3+12n^2+48n+64)-n^3
=12n^2+48n+64
=4(3n^2+12n+16)
$
is divisible by 4.
Since the first sum
(for n=0)
is
$0^3+1^3+2^3+3^3
=1+8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Powers of a $2 \times 2$ matrix Let $A$ be a $2 \times 2$ matrix such that $$A = \begin{pmatrix} \sin\frac{\pi}{18} & -\sin\frac{4\pi}{9} \\ \sin\frac{4\pi}{9} & \sin\frac{\pi}{18} \end{pmatrix}$$ Find the smallest number $n \in N$ such that $A^n = I$, where $I$ is the identity matrix of order $2$.
| Using the identity $\sin\theta=\cos(\frac{\pi}{2}-\theta)$, you get $\sin\frac{\pi}{18}=\cos\frac{4\pi}{9}$.
Substitute this in and you get your standard rotation matrix:
$\begin{pmatrix}
\cos\frac{4}{9}\pi & -\sin\frac{4}{9}\pi \\
\sin\frac{4}{9}\pi & \cos\frac{4}{9}\pi
\end{pmatrix}$
This is equivalent to a counte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=3/2$
Prove that $$\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=\frac{3}{2}$$
I thought of rewriting $$\cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})$$ as $$\cos^2(90^{\circ}+ (\theta +30^{\circ... | As $4\cos^3x-3\cos x=\cos3x$
If $\cos3x=\cos3A, 3x=2n\pi\pm3A$ where $n$ is any integer
$\implies x=\dfrac{2n\pi}3\pm A$ where $n\equiv0,1,2\pmod3$
So, the roots of $4\cos^3x-3\cos x-\cos3A=0$ are $\cos\dfrac{2n\pi}3\pm A$ where $n\equiv0,1,2\pmod3$
$\implies\sum_{r=0}^2\cos\left(\dfrac{2n\pi}3+A\right)=0$
and $\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 4
} |
Show that $\left(1 + \frac{x}{n}\right)^{-n} \le 2^{-x}$, when $x,n \ge 0$, $x \le n$ Show that $\left(1 + \frac{x}{n}\right)^{-n} \le 2^{-x}$, when $x,n \ge 0$, $x \le n$. This is driving me crazy... I have plotted the graphs to be sure that the inequality is true, and it is, but I can't seem to show it. Here is what ... | I have already posted an answer where I considered asymptotic behavior of inequality as $n\to \infty$.
Below are more general conclusions.
Given $0\le x \le n$, show that $\left(1 + \dfrac{x}{n}\right)^{-n} \le 2^{-x}$.
First, note that
$$
0\le x \le n \implies 0 \le \frac{x}{n}\le 1
\implies 1\le \left(1 + \dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Find the values of $\cos(\alpha+\beta) $ if the roots of an equation are given in terms of tan It is given that $ \tan\frac{\alpha}{2} $ and $ \tan\frac{\beta}{2} $ are the zeroes of the equation $ 8x^2-26x+15=0$ then find the value of $\cos(\alpha+\beta$).
I attempted to solve this but I don't know if my solution is r... | Yes, your answer and method, all are right. Just $\tan(\frac{a+b}{2})=\frac{\tan(\frac a2)+\tan(\frac b2)}{1-\tan(\frac a2)\cdot\tan(\frac b2)}$.
But your values are right. And this was the slight correction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How to solve $\sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ$ Question:
$ \sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6° $
I have partially solved this:-
$$ \sin78^\circ-\sin42^\circ +\sin6^\circ-\sin66^\circ $$
$$ 2\cos\left(\frac{78^\circ+42^\circ}{2}\right) \sin\left(\frac{78^\circ-42^\circ}{2}\right) + 2\cos\le... | Angles that are multple of $18$ degrees occur in the regular pentagon with its diagonals and their trigonometric ratios are related to the golden ratio.
To find them, take an isosceles triangle with angles $A=36$ and $B=C=72$ degrees. Draw the bisector of $B$ that intersects $AC$ in a point $D$. Note that the triangles... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Contradiction between integration by partial fractions and substitution Integration by substitution:
$$\int \frac {dx}{x^2-1}$$
Let $x=\sec\theta$ and $dx=\sec\theta\tan\theta \,d\theta$
$$\int \frac {dx}{x^2-1} = \int \frac{\sec\theta\tan\theta \,d\theta}{\sec^2\theta-1} = \int \frac {\sec\theta\tan\theta\, d\theta}{\... | Let's try to clear out some of the confusion on what is going on here.
First of all, $x\mapsto \frac 1 {x^2 - 1}$ is continuous function everywhere except for $x=\pm 1$, so it is Riemann integrable on any segment not containing $\pm 1$. That said, we would very much like to find primitive function defined on $\mathbb R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
How To Tackle Trigonometric Proofs involving $4$th and $6$th powers? How do I prove that
$\cos^4A - \sin^4A+1=2\cos^2A$
$\cos^6A + \sin^6A =1-3\sin^2A\cdot\cos^2A$
I was going through a very old and very rich book of Plane Trigonometry to build a nice foundation for calculus , when I had came across these two rather in... | 1: Let's take $$LHS=\cos^4 A-\sin^4 A+1$$ $$=\cos^4 A+(1-\sin^4 A)$$ $$=\cos^4 A+(1-\sin^2 A) (1+\sin^2 A)$$ $$=\cos^4 A+\cos^2 A (1+\sin^2 A)$$ $$=\cos^2 A(\cos^2A+\sin^2 A+1)$$ $$=\cos^2 A(1+1)$$ $$=\color{blue}{2\cos^2 A=RHS}$$
2: Again let's take $$LHS=\cos^6 A+\sin^6 A$$ $$=(\cos^2 A)^3+(\sin^2 A)^3$$ $$=(\cos^2A+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Convergence of an oscillating recursive sequence Define the recursive sequence
$ q_{n+1} = \dfrac{q_n+2}{q_n+1};\;q_0=1 $
If we knew that
$ q_n \to q;\;n\to \infty $
then it's easy to show what follows
$ q_{n+1}\left(q_n+1\right) = q_n+2 $
$ q_{n+1}\cdot q_n+q_{n+1}-q{n}=2 $
$ q^2 = 2 \rightarrow q = \sqrt 2;\;n\to\i... | Your series, when written as a sequence of fractions, is:
$1, \frac{3}{2}, \frac{7}{5}, \frac{17}{12}, \frac{41}{29}...$
Notice that the denominator of term $n + 1$ is the sum of the denominator and numerator of term $n$, and the numerator of term $n + 1$ is twice the denominator of term $n$ plus its numerator.
So let'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove the inequality $\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$ with the constraint $abc=1$ If $a,b,c$ are positive reals such that $abc=1$, then prove that $$\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$$ I tried substituting $x/y,y/z,z/x$, but it didn't help(I got the r... | First let
$$ x = \sqrt{\frac{a}{a+8}}, \,\, y = \sqrt{\frac{b}{b+8}}, \,\, z = \sqrt{\frac{c}{c+8}} \,\, $$
Then $1 > x,y,z > 0$ and
$$ a = \frac{8x^2}{1 - x^2}, \,\, b = \frac{8y^2}{1 - y^2}, \,\, c = \frac{8z^2}{1 - z^2},\,\, $$
So the question transforms to this:
Given that $1 > x,y,z > 0, \, \, \frac{512x^2y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 6,
"answer_id": 4
} |
Find the sum: $\sum_{i=1}^{n}\dfrac{1}{4^i\cdot\cos^2\dfrac{a}{2^i}}$ Find the sum of the following :
$S=\dfrac{1}{4\cos^2\dfrac{a}{2}}+\dfrac{1}{4^2\cos^2\dfrac{a}{2^2}}+...+\dfrac{1}{4^n\cos^2\dfrac{a}{2^n}}$
| Hint:
$$\frac{1}{4^n\cos^2\frac{a}{2^n}}+\frac{1}{4^n\sin^2\frac{a}{2^n}}=
\frac{1}{4^{n-1}\sin^2\frac{a}{2^{n-1}}}.$$
Adding $\displaystyle \frac{1}{4^n\sin^2\frac{a}{2^n}}$ to the sum, the result thus telescopes to $\displaystyle\frac{1}{\sin^2a}$, and hence the initial sum is
$$\frac{1}{\sin^2a}-\frac{1}{4^n\sin^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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Power serie of $f'/f$ It seems that I'm [censored] blind in searching the power series expansion of $$f(x):=\frac{2x-2}{x^2-2x+4}$$ in $x=0$.
I've tried a lot, e.g., partiell fraction decomposition, or regarding $f(x)=\left(\log((x+1)^2+3)\right)'$ -- without success.
I' sure that I'm overseeing a tiny little missing... | The given $f(x)$ is given by
\begin{align}
f(x) = \frac{2(x-1)}{x^2 - 2x + 4}
\end{align}
and can be seen as, where $a = 1 + \sqrt{3} i$ and $b = 1 - \sqrt{3} i$, such that $ab=4$,
\begin{align}
f(x) = \frac{-2 \, (1 - x)}{ab \, \left(1 - \frac{x}{a}\right) \left( 1 - \frac{x}{b}\right)}
\end{align}
for which
\begin{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Why does the Pythagorean Theorem have its simple form only in Euclidean geometry? Below are the right-angled forms of the Pythagorean Theorem in elliptic, Euclidean, and hyperbolic geometry, respectively.
$$\cos\left(\frac{c}{R}\right) = \cos\left(\frac{a}{R}\right)\cos\left(\frac{b}{R}\right)$$
$$c^2 = a^2 + b^2$$
$$\... | You can even generalize the law of cosines.
$$
\cos\left(\frac{z}{r}\right) =
\cos\left(\frac{x}{r}\right) \cos\left(\frac{y}{r}\right)
+ \cos(\phi) \sin\left(\frac{x}{r}\right) \sin\left(\frac{y}{r}\right).
\tag 1
$$
What we have is
$$
\begin{array}{rcl}
r^2 > 0 &\rightarrow& \textrm{spherical}\\
&& \cos\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Why do remainders show cyclic pattern? Let us find the remainders of $\dfrac{6^n}{7}$,
Remainder of $6^0/7 = 1$
Remainder of $6/7 = 6$
Remainder of $36/7 = 1$
Remainder of $216/7 = 6$
Remainder of $1296/7 = 1$
This pattern of $1,6,1,6...$ keeps on repeating. Why is it so? I'm asking in general, that is for every case o... | Note that as you find more and more powers, you'll eventually run out of remainders. So you'll come back to a previous number, and since exponentiation is repeated multiplication, you'll generate the cycle again. As an example, let's look at powers of $2$ modulo $7$. We have
\begin{align*}
2^1 &\equiv 2 \pmod{7} \\
2^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 1
} |
how can i prove this trigonometry equation I need help on proving the following:
$$\frac{\cos {7x} - \cos {x} + \sin {3x}}{ \sin {7x} + \sin {x} - \cos {3x} }= -\tan {3x}$$
So far I've only gotten to this step:
$$\frac{-2 \sin {4x} \sin {3x} + \sin {x}}{ 2 \sin {4x} \cos {3x} - \cos {x}}$$
Any help would be appreciated... | The LHS can seen as $$\frac{\cos 7x -\cos x + \sin 3x}{\sin 7x + \sin x - \cos 3x} = \frac{-2\sin 4x \sin 3x + \sin 3x}{2\sin 4x\cos 3x - \cos 3x} = \tan 3x \frac{1-2\sin 4x}{-1+2\sin 4x} = -\tan 3x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove using mathematical induction that $x^{2n} - y^{2n}$ is divisible by $x+y$ Prove using mathematical induction that
$(x^{2n} - y^{2n})$ is divisible by $(x+y)$.
Step 1: Proving that the equation is true for $n=1 $
$(x^{2\cdot 1} - y^{2\cdot 1})$ is divisible by $(x+y)$
Step 2: Taking $n=k$
$(x^{2k} - y^{2k})$ is... | When $n=1$,
$$x^{2n} - y^{2n} =x^2-y^2=(x+y)(x-y)$$
which is divisible by $(x+y)$. Assume true for $n=k$. Then $x^{2k}-y^{2k}$ is divisible by $x+y$.
Putting $n=k+1$,
$$x^{2n}-y^{2n}=x^{2k+2}-y^{2k+2}=\left(x^{k+1}-y^{k+1}\right) \left(x^{k+1}+y^{k+1}\right)$$
When $k$ is an odd number, $x+y$ is a factor of $x^{k+1}-y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
How to integrate ${x^3}/(x^2+1)^{3/2}$? How to integrate $$\frac{x^3}{(x^2+1)^{3/2}}\ \text{?}$$
I tried substituting $x^2+1$ as t, but it's not working
| Let $x^2+1=t\implies 2xdx=dt$ or $xdx=\frac{dt}{2}$ $$\int \frac{x^3 dx}{(x^2+1)^{3/2}}=\frac{1}{2}\int \frac{(t-1)dt}{(t)^{3/2}}$$
$$=\frac{1}{2}\int \frac{(t-1)dt}{t^{3/2}}$$ $$=\frac{1}{2}\int (t^{-1/2}-t^{-3/2})dt$$ $$=\frac{1}{2}\left(2t^{1/2}+2t^{-1/2}\right)$$ $$=\sqrt{x^2+1}+\frac{1}{\sqrt{x^2+1}}+C$$ $$=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
$\int_{-\infty}^{\infty}e^{-\pi x^2}\cdot e^{-2\pi ix\xi}dx = e^{\pi\xi^2}$ Prove that for all $\xi \in \mathbb{C}$,
$$\int_{-\infty}^{\infty}e^{-\pi x^2}\cdot e^{-2\pi ix\xi}dx = e^{\pi\xi^2}$$
I don't really know how to compute this integral. Can you please help me?
| By noticing that
$$a x^{2} + b x = \left( \sqrt{a} x + \frac{b}{2 \sqrt{a}} \right)^{2} - \frac{b^{2}}{4 a}$$
then
\begin{align}
I &= \int_{-\infty}^{\infty} e^{-a x^{2} - b x} \, dx \\
&= e^{\frac{b^{2}}{4 a}} \, \int_{-\infty}^{\infty} e^{- \left( \sqrt{a} x + \frac{b}{2 \sqrt{a}} \right)^{2}} \, dx
\end{align}
Mak... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How to factor intricate polynomial $ ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c $ I would like to know how to factor the following polynomial.
$$ ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c $$
What is the method I should use to factor it? If anyone could help.. Thanks in advance.
| Let
$$
f(a)=ab^3-a^3b+a^3c-ac^3+bc^3-b^3c.
$$
If $a=b$, then $f(a)=0$. Therefore $(a-b)\mid f(a)$. Analogously $(a-c)\mid f(a)$; so, $(a-b)(a-c)\mid f(a)$. But we can take it directly:
$$
ab^3-a^3b+a^3c-ac^3+bc^3-b^3c = ab(b^2-a^2) + c^3(b-a) - c(b^3-a^3) =\\=
ab(b-a)(b+a)+c^3(b-a)-c(b-a)(b^2+ba+a^2) =\\= (b-a)[ab(b+a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1379045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integration by Partial Fractions $\int\frac{1}{(x+1)^3(x+2)}dx$ I'm trying to do a problem regarding partial fractions and I'm not sure if I have gone about this right as my answer here doesn't compare to the answer provided by wolfram alpha. Is it that I can't Seperate things that are raised to powers on the denominat... | Observe that:
$$x+2 = (x+1)+1$$
Rewrite the numerator as:
$$\left( (x+1)^3+1^3 \right) - (x+1)$$
And cancel using the sum of cubes identity
$$a^3+b^3\equiv(a+b)(a^2-ab+b^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1379702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Maximum value of expression Let the maximum value of the expression $y=\frac{x^4-x^2}{x^6+2x^3-1}$ for $x>1$ is $\frac p q$,where $p$ and $q$ are relatively prime positive integers.Find $p+q$.
My attempt: $y=\frac{x^2(x^2-1)}{(x^3+1)^2-2}$, then I could not think of any way to find maximum value of $f(x)$. Can someone ... | Let $t:=x-\frac{1}{x}$, which can take any value in $\mathbb{R}_{>0}$ Then, $y=\frac{t}{t^3+3t+2}$. To maximize $y$, we have to minimize $\frac{1}{y}=t^2+3+\frac{2}{t}$. By the AM-GM Inequality,
$$\frac{1}{y}=\left(t^2+\frac{1}{t}+\frac{1}{t}\right)+3\geq 3\sqrt[3]{t^2\cdot\frac{1}{t}+\frac{1}{t}}+3=6\,.$$
That is, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
find the total differential of this equation $ xyz + \sqrt{ x^2 + y^2 + z^2} = \sqrt 2 $ How to calculate the total differential of $ z= z(x,y)$, which is $ xyz + \sqrt{ x^2 + y^2 + z^2} = \sqrt 2 $ at point (1, 0, -1)？
The evaluation of mine seems wrong,
$ dz= \frac{\partial z}{\partial x} dx + \frac{\partial z}{\pa... | I have done the following workings for the question. I believe the answer is ok but please forgive if it is incorrect (and point out my oversight if possible).
As mentioned in the comments $z$ is implicitly a function of both $x$ and $y$. To be specific $z=z(x,y)$
First let's calculate $\frac{\partial z}{\partial x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Rearrangements that never change the value of a sum Which bijections $f:\{1,2,3,\ldots\}\to\{1,2,3,\ldots\}$ have the property that for every sequence $\{a_n\}_{n=1}^\infty$,
$$
\lim_{n\to\infty} \sum_{k=1}^n a_k = \lim_{n\to\infty} \sum_{k=1}^n a_{f(k)},
$$
where "$=$" is construed as meaning that if either limit exis... | The hypothesis is wrong.
Let $f$ be the following sequence:
$1,2 ; 3,5,4,6; 7,9,11,8,10,12; \cdots$
where each block has two more numbers than the previous block and goes through the odd numbers in order before the even ones.
Clearly $f(n) \in n + O(\sqrt{n})$ as $n \to \infty$.
Now consider the following series:
$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 0
} |
Solving simple mod equations Solve $3x^2 + 2x + 1 \equiv 0 \mod 11$
Additionally, I have an example problem, but a step in the middle has confused me:
$3x^2 + 5x - 7 \equiv 0 \mod 17$. Rearrange to get $3x^2 + 5x \equiv 7 \mod 17$.
$\implies 6\cdot 3x^2 + 6\cdot 5x \equiv 6*7 \mod 17$.
The next line reads $x^2 + 30x \e... | Note that $3 \equiv 5^2$, therefore $3x^2 +2x +1 \equiv 5^2x^2 + 2 \cdot 5 \cdot 5^{-1} x + 5^{-2} - 5^{-2} +1$ and since $5^{-1} \equiv 9$ this means $3x^2 +2x +1 \equiv 5^2x^2 + 2 \cdot 5 \cdot 9 x + 9^2 - 3 \equiv (5x + 9)^2 - 3$, so you want to solve $(5x + 9)^2 \equiv 3 \equiv 5^2$, therefore $(5x + 9) \equiv \pm ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Prove the series has positive integer coefficients How can I show that the Maclaurin series for
$$
\mu(x) = (x^4+12x^3+14x^2-12x+1)^{-1/4}
\\
= 1+3\,x+19\,{x}^{2}+147\,{x}^{3}+1251\,{x}^{4}+11193\,{x}^{5}+103279\,
{x}^{6}+973167\,{x}^{7}+9311071\,{x}^{8}+\cdots
$$
has positive integer coefficients? (I have others to d... | For the periodicity mod $5$, note that
$$(1 - x^4)^4 \equiv (1 + 3 x + 4 x^2 + 2 x^3)^4 (1-12 x+14 x^2+12 x^3+x^4) \mod 5 $$
and thus
in the field $\mathbb Z_5((x))$ of formal Laurent series in $x$ over the integers mod $5$ we have
$$ \eqalign{&\dfrac{1}{1-12 x+14 x^2+12 x^3+x^4} = \left(\dfrac{1+3x+4x^2+2x^3}{1-x^4}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
Proving that $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$ I have written the left side of the equation as $$\left(1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-1}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}\right).$$ I... | Denote $H_n = 1 + \frac{1}{2} + \cdots + \frac{1}{n}$. What you want to prove is exactly
$$
1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n-1}-\frac{1}{2n} = H_{2n} - H_n
$$
i.e.,
$$
1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n-1}-\frac{1}{2n} - H_{2n} = - H_n
$$
Note that the l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Solving algebraic equations with radicals: $x(x-\sqrt3)(x+1)+3-\sqrt3=0$ I have several problems requiring assistance.
Solve for $x$:
$x\left( x-\sqrt { 3 } \right) \left( x+1 \right) +3-\sqrt { 3 } \quad =\quad 0$
I've followed the suggestion to get x^2 - (√3 -1)x + (3-2√3) and shall proceed to the quadratic formula.... | The originally posted equation appeared to be $(x - \sqrt{3}) x (x+1) + 3 - \sqrt{3} = 0$ for which this polynomial can be seen in the form
\begin{align}
& x(x+1)(x - \sqrt{3}) + 3 - \sqrt{3} = 0 \\
& x^{2} (x + 1 - \sqrt{3}) - \sqrt{3} (x + 1 - \sqrt{3}) = 0 \\
& (x - \sqrt[4]{3})(x + \sqrt[4]{3})(x + 1 - \sqrt{3}) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A basic square numbers equation I couldn't solve this equation. Could you help me to solve this step by step.
Thank you.
$$\sqrt[4]{\frac{8^x}{8}}=\sqrt[3]{16}$$
| \begin{align}
\sqrt[4]{8^{x} \times \frac{1}{8}} &= \sqrt[3]{16} \\
\sqrt[4]{8^{x-1}} &= \sqrt[3]{16} \\
8^{x-1} &= \left( \sqrt[3]{16} \right)^{4} \\
8^{x-1} &= 16^{\frac{4}{3}} \\
2^{3x-3} &= 2^{\frac{16}{3}} \\
3x-3 &= \frac{16}{3} \quad \to \quad
3x = \frac{16}{3} +3 \quad \to \quad
x = 1+\frac{16}{9} \quad \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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$\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$ If $A_1A_2A_3.....A_n$ be a regular polygon and $\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$(number of vertices in the regular polygon).
I know that sides of a regular polygon are equal but i could not rel... | I was casting about for a method that didn't require a lot of work with multiple-angles and trigonometric identities. Up to now, I had a different -- though ultimately related -- argument (without using a circumscribed circle) which led me to the same equation, $ \ \frac{1}{\sin\theta} \ = \ \frac{1}{\sin2\theta} \ + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 0
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Give the equation of a line that passes through the point (5,1) that is perpendicular and parallel to line A. The equation of line $A$ is $3x + 6y - 1 = 0$. Give the equation of a line that passes through the point $(5,1)$ that is
*
*Perpendicular to line $A$.
*Parallel to line $A$.
Attempting to find the parallel,... | For the second case:
the line you want to find its equation is parallel to $A$, so it means they have the same slope$-\frac{1}{2}$, which yields:
$$
{y}_{1} = -\frac{1}{2}x+p
$$
Now you'll find $p$ from another condition you gave(the line passes from $(5,1)$), so we get :
$$
1 = \frac{-5}{2}+p\\
p = 1+\frac{5}{2} = \f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove for primes p $>2$ that $\sum_{k=1}^{p−1}{k^{2p−1}}\equiv\frac{1}{2}p(p+1)\pmod {p^2}$
Let $p$ be an odd prime. Prove that:
$$\sum_{k=1}^{p−1}{k^{2p−1}}\equiv\dfrac{p(p + 1)}{2}\pmod {p^2}$$
The problem is taken from the 2004 Canada National Olympiad.
I am only able to show that the sum is congruent to $0$ mo... | In general: $$(a+b)^n \equiv a^n + nba^{n-1}\pmod{b^2}.$$ Letting $a=-k, b=p, n=2p-1$, you get that $$(p-k)^{2p-1} \equiv (2p-1)pk^{2p-2} - k^{2p-1}\pmod{p^2}$$
And $k^{2p-2}=(k^2)^{p-1}\equiv 1\pmod p$, we have $$(2p-1)pk^{2p-2}\equiv -pk^{2p-2}\equiv -p\pmod {p^2}.$$
So $$(p-k)^{2p-1} \equiv -p -k^{2p-1}\pmod{p^2}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Perpendicular Bisector of Made from Two Points For a National Board Exam Review:
Find the equation of the perpendicular bisector of the line joining
(4,0) and (-6, -3)
Answer is 20x + 6y + 29 = 0
I dont know where I went wrong. This is supposed to be very easy:
Find slope between two points:
$${ m=\frac{y^2 - y^1}... | lol. a lot of confusion on this thread. When you are computing the midpoints, what you actually do is find the distance between them and divide by two. This isn't the coordinate of the midpoint; this is just the distance form one end to the midpoint.
What you meant to do is take the average of the x and y values, not t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Calculate the volume bounded by the surfaces Calculate the volume of the solid bounded by the surfaces $$\begin{aligned}z&=4x^2+4y^2, \\ z&=x^2+y^2, \\z&=4.\end{aligned}$$
I made an equation of $4x^2+4y^2=4-x^2+y^2$ and solved it to get $x^2+y^2=\dfrac{4}{5}$.
Then I did a double integration
$$\displaystyle
\iint_{x^2... | Let
$(x,y,z) \in \mathbb{R}^{3}$ and let $S$ be the region enclosed by the surfaces $z = x^{2}+y^{2}$ and $z=4$. Then $(x,y,z) \in S$ if and only if $|x| \leq 4, |y| \leq \sqrt{4-x^{2}}, 0 \leq z \leq 4$. Thus
$$4\int_{0}^{4}\int_{0}^{\sqrt{4-x^{2}}} x^{2} + y^{2} dy dx = 4\int_{0}^{4} x^{2}\sqrt{4-x^{2}} + \frac{(4-x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Where is my error in solving $y'' + y' + y = 0, y(0) = 1, y'(0) = 0$ with Laplace transform? Im trying to solve a laplace transoform question, but i am stuck.
The question is $y′′(t) + 2ζy′(t) + y(t) = 0, y(0) = 1, y′(0) = 0$ and $ζ = 0.5$.
I have so far done:
Laplace transform which gives
$s^2Y(s)-sy(0)-y'(0)+2ζ(sY(s)... | Since the denominator is not easily factorable, we avoid partial fractions and instead complete the square. Recall the following inverse Laplace transforms:
$$
Y(s) = \frac{b}{(s - a)^2 + b^2} \implies y(t) = e^{at}\sin bt \\
Y(s) = \frac{s - a}{(s - a)^2 + b^2} \implies y(t) = e^{at}\cos bt
$$
Indeed, observe that:
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Find $ax^5 + by^5$ by matrix from $ \left\{ \begin{aligned} ax+by&=3\\ ax^2+by^2&=7\\ ax^3+by^3&=16\\ ax^4+by^4&=42 \end{aligned} \right.$
Find $ax^5 + by^5$ if the real numbers $a$, $b$, $x$, and $y$ satisfy the equations
$$
\left\{
\begin{aligned}
ax+by&=3 \\
ax^{2}+by^{2}&=7 \\
ax^{3}+by^{3}&=16 \\
ax^{4}+by^{4... | The secuence $c_n=ax^n+by^n$ satisfies a linear two step recurrence relation of the form
$$c_{n+2}=Ac_{n+1}+Bc_n$$
for some constants $A,B$. This means that if we perform the row operations coming from that recurrence relation (subtract the second row multiplied by $A$ and the first row multiplied by $B$ from the third... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Integral of $\frac{1}{x^2+x+1}$ and$\frac{1}{x^2-x+1}$ How to integrate two very similar integrals. I am looking for the simplest approach to that, it cannot be sophisticated too much as level of the textbook this was taken from is not very high. $$\int \frac{1}{x^2+x+1} dx$$ and$$\int \frac{1}{x^2-x+1} dx$$
| For the first one:
$$
\int { \frac { 1 }{ 1+{ x }^{ 2 }+x } } dx\quad =\quad \quad \int { \frac { 1 }{ { x }^{ 2 }+x+\frac { 1 }{ 4 } +\frac { 3 }{ 4 } } dx } \\ \qquad \qquad \qquad \qquad =\quad \int { \frac { 1 }{ { \left( x+\frac { 1 }{ 2 } \right) }^{ 2 }+\frac { 3 }{ 4 } } dx } \\ \qquad \qquad \qquad \qquad... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
Find $x$ and $y$ by Cramer from $\left\{\begin{aligned} \frac{2x+2}{2y-3}&=\frac{x-2}{y+3}\\ \frac{x+2}{3y-2}&=\frac{x-1}{3y+8} \end{aligned}\right.$
Find $x$ and $y$ with Cramer rule from these equations
$$\left\{
\begin{aligned}
\frac{2x+2}{2y-3}&=\frac{x-2}{y+3} \\
\frac{x+2}{3y-2}&=\frac{x-1}{3y+8}
\end{aligned... | I will show you the first one:
$$\frac{2x+2}{2y-3}=\frac{x-2}{y+3} \implies\\
(2x+2)(y+3)=(x-2)(2y-3)\implies\\
2xy+6x+2y+6=2xy-3x-4y+6$$
Can you see how you can cancel and combine the terms now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find Sum of n terms of the Series: $ \frac{1}{1\times2\times3} + \frac{3}{2\times3\times4} + \frac{5}{3\times4\times5} + \cdots $ I want to find the sum to n terms of the following series.
$$ \frac{1}{1\times2\times3} + \frac{3}{2\times3\times4} + \frac{5}{3\times4\times5} + \frac{7}{4\times5\times6} + \cdots $$
I have... | $$\begin{align}
\sum\limits_{i=1}^n \frac{2i - 1}{i(i+1)(i+2)}
&=\sum_{i=1}^n\frac {Ai+B}{i(i+1)}-\frac{A(i+1)+B}{(i+1)(i+2)}\\
&=\sum_{i=1}^n\frac{2i-\frac12}{i(i+1)}-\frac{2(i+1)-\frac12}{(i+1)(i+2)}\\
&=\frac34-\frac{4n+3}{2(n+1)(n+2)}\qquad\text{by telescoping}\\
&=\frac{n(3n+1)}{4(n+1)(n+2)}\qquad\blacksquare
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$.
Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$.
Attempts so far:
Used Descartes signs stuff so possible number of real roots is $6,4,2,0$
tried differentiating the equation $4$ times and got an equation with no... | First, rescale by $720$ to get integer coefficients: $$x^6+6x^5+30x^4+120x^3+360x^2+720x+720$$
Now repeated completion of binomial powers:
$$\begin{align}
&\phantom{{}={}}(x+1)^6+15x^4+100x^3+345x^2+714x+719\\
&=(x+1)^6+15(x+5/3)^4+95x^2+\frac{3926}{9}x+\frac{16288}{27}\\
\end{align}$$
You could complete the square aga... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 4
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To evaluate $\int_{0}^\pi \frac {\sin^2 x}{a^2-2ab \cos x + b^2}\mathrm dx$? How do we evaluate $\displaystyle\int_{0}^\pi \dfrac {\sin^2 x}{a^2-2ab \cos x + b^2}\mathrm dx$ ? I tried substitution and some other methods , but its not working ; please help .
| Let $I$ be the integral given by
$$I=\int_0^\pi \frac{\sin^2x}{a^2+b^2-2ab\cos x}\,dx$$
Exploiting the even symmetry of the integrand, we have
$$I=\frac12\int_{-\pi}^\pi \frac{\sin^2x}{a^2+b^2-2ab\cos x}\,dx$$
Next, using $\sin^2x=\frac12(1-\cos 2x)$ we obtain
$$\begin{align}
I&=\frac14\int_{-\pi}^\pi \frac{1-\cos 2x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398196",
"timestamp": "2023-03-29T00:00:00",
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} |
Evaluate $\int e^x \sin^2 x \mathrm{d}x$ Is the following evaluation of correct?
\begin{align*} \int e^x \sin^2 x \mathrm{d}x &= e^x \sin^2 x -2\int e^x \sin x \cos x \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (\cos^2 x - \sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (1 -... | We have $$\int e^{x}\sin^{2}\left(x\right)dx=\frac{1}{2}\int e^{x}dx-\frac{1}{2}\int e^{x}\cos\left(2x\right)dx
$$ and $$\int e^{x}\cos\left(2x\right)dx=\textrm{Re}\left(\int e^{x+2ix}dx\right)
$$ then $$\int e^{x+2ix}dx=\frac{e^{x+2ix}}{1+2i}
$$ and so $$\int e^{x}\sin^{2}\left(x\right)dx=\frac{1}{2}e^{x}-\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Find $\int_0^\frac{\pi}{2} \frac {\theta \cos \theta } { \sin \theta + \sin ^ 3 \theta }\:d\theta$ My Calc 2 teacher wasn't able to solve this:
$$\int_0^\frac{\pi}{2} \frac {\theta \cos \theta } { \sin \theta + \sin ^ 3 \theta }\:d\theta$$
Can someone help me solve this?
| $$
\int \frac {\theta \cos \theta } { \sin \theta + \sin ^ 3 \theta} \,d\theta = \underbrace{\int \theta \, dx = \theta x - \int x\,d\theta}_{\text{integration by parts}}
$$
$$
dx = \frac { \cos \theta } { \sin \theta + \sin ^ 3 \theta} \,d\theta = \frac{du}{u+u^3} = \frac{du}{u(1+u^2)} = \left( \frac A u + \frac{Bu+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
Determine the equation of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ such that it has the least area but contains the circle $(x-1)^2+y^2=1$ Determine the equation of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ such that it has the least area but contains the circle $(x-1)^2+y^2=1$
Since the area of ellipse i... | Means ellipse and circle touches each other. So $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ and $(x-1)^2+y^2 = 1$
So Ellipse and Circle touches each other. So we will solve these two equations.
$\displaystyle \frac{x^2}{a^2}+\frac{1-(x-1)^2}{b^2} = 1\Rightarrow b^2x^2+a^2\left[1-(x-1)^2\right] = a^2b^2$
so $\di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $(a+b)(b+c)(c+a) \ge8$
Given that $a,b,c \in \mathbb{R}^{+}$ and $abc(a+b+c)=3$,
Prove that $(a+b)(b+c)(c+a)\ge8$.
My attempt: By AM-GM inequality, we have
$$\frac{a+b}{2}\ge\sqrt {ab} \tag{1}$$
and similarly
$$\frac{b+c}{2}\ge\sqrt {bc} \tag{2},$$
$$\frac{c+d}{2}\ge\sqrt {cd} \tag{3}.$$
Multiplying $(1)... | A geometric approach is to notice that $a,b,c>0$ and $abc(a+b+c)=3$ grant that there is a triangle with side lengths $a+b,a+c,b+c$ and area $\sqrt{3}$. Since the area of a triangle is given by the product of its side lengths divided by four times the length of the circumradius, the problem boils down to understanding w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Finding a perpendicular vector This is given: $P = (-1,1,0),\; Q = (1,5,6)\; \text{and} \; R = (3,-1,4).$
My question is:
Find the values of $x$ (where x is a real number) for which $PR + xQR\;$ is perpendicular to $PR$.
So far, what I have done is used the dot product of two vectors and equated
$\vec {u} \cdot \vec{v... | You are correct:
$$PR = (4, -2, 4)\\
QR = (2, -6, -2)\\
PR + xQR = (4 + 2x, -2-6x, 4-2x)\\
PR\cdot (PR + xQR) = 4(4+2x) - 2(-2 -6x) + 4(4-2x) = \\=16+8x+4+12x+16-8x=36+12x$$
So the solution is $28+12x=0$ or $x=-\frac{36}{12} = -3$.
Alternatively, you are solving the equation $$PR\cdot(PR-x\cdot QR) = 0$$
You can rewrit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $(1+ \tan{A}\tan{2A})\sin{2A} = \tan{2A}$. Prove the following statement
$$ (1+ \tan{A}\tan{2A})\sin{2A} = \tan{2A}. $$
On the left hand side I have put the value of $\tan{2A}$ and have then taken the LCM.
I got $\sin{2A}\cos{2A}$. How do I proceed?
Thanks
| \begin{align}
& (1 + \tan A \tan(2A))\sin(2A) \\
= & \left(1 + \frac{\sin A}{\cos A}\frac{\sin(2A)}{\cos(2A)}\right)\sin(2A)\\
= & \left(1 + \frac{\sin A}{\cos A}\frac{2\sin A\cos A}
{\cos(2A)}\right)\sin(2A) \\
= & \left(1 + \frac{2\sin^2 A}{\cos(2A)}\right)\sin(2A) \\
= & \frac{\cos(2A) + 2\sin^2 A}{\cos(2A)}\sin(2A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
An arctan integral $\int_0^{\infty } \frac{\arctan(x)}{x \left(x^2+1\right)^5} \, dx$ According to Mathematica, we have that
$$\int_0^{\infty } \frac{\arctan(x)}{x \left(x^2+1\right)^5} \, dx=\pi \left(\frac{\log (2)}{2}-\frac{1321}{6144}\right)$$
that frankly speaking looks pretty nice.
However Mathematica shows tha... | I'm not sure if this is nice and simple, but here is one way:
Substitute $y = \arctan x$ to rewrite as $$\int_0^{\pi/2} \frac{y}{\tan y \sec^{8} y} \, dy$$
Now write this as an integral in $y$ and $\sec y$:
$$\int_0^{\pi/2} y \cdot \frac{\sec y \tan y}{(\sec^2 y - 1) \sec^{9} y} \, dy$$
You can integrate this by parts.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 0
} |
$\int_{0}^{\pi}|\sqrt2\sin x+2\cos x|dx$ $\int_{0}^{\pi}|\sqrt2\sin x+2\cos x|dx$
MyAttempt
$\int_{0}^{\pi}|\sqrt2\sin x+2\cos x|=\int_{0}^{\pi/2}\sqrt2\sin x+2\cos x dx+\int_{\pi/2}^{\pi}|\sqrt2\sin x+2\cos x| dx$
I could solve first integral but in second one,i could not judge the mod will take plus sign or minus sig... | We Can Write $\displaystyle \sqrt{2}\sin x+2\cos x = \sqrt{\left(\sqrt{2}\right)^2+2^2}\left\{\frac{\sqrt{2}}{\sqrt{6}}\sin x+\frac{2}{\sqrt{6}}\cos x\right\}$
$\displaystyle = \sqrt{6}\left\{\sin x\cdot \frac{1}{\sqrt{3}}+\cos x\cdot \frac{2}{\sqrt{6}}\right\} = \sqrt{6}\left\{\sin x\cdot \cos \alpha+\cos x\cdot \sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Prove that $\int_{0}^{2\pi}\frac{x^2\sin x}{8+\sin^2x}=\frac{2\pi^2}{3}\ln\frac{1}{2}$ Prove that $$\int_{0}^{2\pi}\dfrac{x^2\sin x}{8+\sin^2x}=\frac{2\pi^2}{3}\ln\frac{1}{2}$$
My Attempt:
$$\int_{0}^{2\pi}\dfrac{x^2\sin x}{8+\sin^2x}=\int_{0}^{2\pi}x^2\frac{\sin x}{8+\sin^2x}$$
I applied integration by parts,consideri... | Let $$\displaystyle I = \int_{0}^{2\pi}\frac{x^2\cdot \sin x}{8+\sin^2 x}dx ...................(1)$$
Now Replace $x\rightarrow (2\pi-x)\;,$ We get
$$\displaystyle I = -\int_{0}^{2\pi}\frac{\left(2\pi-x\right)^2\cdot \sin x}{8+\sin^2 x}dx...........(2)$$
Now Add these Two equations, We get
$$\displaystyle 2I = \int_{0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to find the MacLaurin series of $\frac{1}{1+e^x}$ Mathcad software gives me the answer as: $$ \frac{1}{1+e^x} = \frac{1}{2} -\frac{x}{4} +\frac{x^3}{48} -\frac{x^5}{480} +\cdots$$
I have no idea how it found that and i don't understand. What i did is expand $(1+e^x)^{-1} $ as in the binomial MacLaurin expansion. I ... | Just a sketch of a proof to be expanded later (sorry, I am in a hurry): by taking the logarithmic derivative of the Weierstrass product of the $\cosh$ function we have the Taylor series of the $\tanh$ function, with the coefficients depending on values of the $\zeta$ function at even integers. With a little maquillage ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Find the range of the function,$f(x)=\int_{-1}^{1}\frac{\sin x}{1-2t\cos x+t^2}dt$ Find the range of the function,$f(x)=\int_{-1}^{1}\frac{\sin x}{1-2t\cos x+t^2}dt$
I tried to solve it,i got range $\frac{\pi}{2}$ but the answer is ${\frac{-\pi}{2},\frac{\pi}{2}}$
$f(x)=\int_{-1}^{1}\frac{\sin x}{1-2t\cos x+t^2}dt=\int... | Notice, second last line,
$$\frac{\sin x}{\sin x}\left[\tan^{-1}\left(\frac{t-cos x}{\sin x}\right)\right]_{-1}^{1}$$
$$=\tan^{-1}\left(\frac{1-cos x}{\sin x}\right)-\tan^{-1}\left(\frac{-1-cos x}{\sin x}\right)$$
$$=\tan^{-1}\left(\frac{1-cos x}{\sin x}\right)+\tan^{-1}\left(\frac{1+cos x}{\sin x}\right)$$
$$=\tan^{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Integrating $\int \sqrt{x+\sqrt{x^2+1}}\,\mathrm{d}x$ Integrating $$\int \sqrt{x+\sqrt{x^2+1}}\,\mathrm{d}x$$
Using substitution of $x=\tan \theta$, I got the required answer. But is there a more elegant solution to the problem?
| Let $$ I = \int \sqrt{x+\sqrt{x^2+1}} \, dx\;,$$ Let $$x+\sqrt{x^2+1} = t^2 \tag 1$$
Then $$ \left(1+\frac{x}{\sqrt{x^2+1}}\right) \, dx = 2t \, dt\Rightarrow t^2 \, dx = 2t\sqrt{x^2+1} \, dt$$
Now using $$\bullet\; \left(\sqrt{x^2+1}+x\right)\cdot \left(\sqrt{x^2+1}-x\right) = 1$$
So we get $$ \sqrt{x^2+1}-x = \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 0
} |
Rotation and translation of coordinate axes I am studying rotation and translation of conical but have no doubt in basic concept (Sorry, I know this is a very stupid question but I'm really struggling to understand). Especially in this equation:
$$
9x^2 - 4y^2 - 18x - 16y - 7 = 0
$$
$$
\begin{cases}
x = u + h\\
y = v +... | What happens when you substitute $u+h$ for $x$ in the equation
$9x^2 - 4y^2 - 18x - 16y - 7 = 0$?
First of all, $x^2 = (u + h)^2 = u^2 + 2hu + h^2$.
Therefore $$9x^2 = 9(u^2 + 2hu + h^2) = 9u^2 + 18hu + 9h^2.$$
And of course $-18x = -18u - 18h$. So
\begin{align}
9x^2 - 4y^2 - 18x - 16y - 7
& = 9u^2 + 18hu + 9h^2 - 18u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $\cos^2{\theta}-\sin{\theta} = 1$ Can someone please help me solve this?
$$\cos^{2}{\theta}-\sin{\theta} = 1, \quad\theta\in[0^{\circ}, 360^{\circ}]$$
| Notice, we have
$$\cos^2\theta-\sin\theta=1$$
$$1-\sin^2\theta-\sin\theta=1$$
$$-\sin^2\theta-\sin\theta=0$$
$$\sin^2\theta+\sin\theta=0$$
$$\sin\theta(\sin\theta+1)=0$$ $$\sin\theta=0\iff \theta=n(180^\circ)$$
Where, $n$ is any integer.
But for the given interval $[0^\circ, 360^\circ]$, substituting $n=0, 1, 2$, we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to integrate $\frac{\cos x-\cos2x}{1-\cos x}$? I want to find $$\int\dfrac{\cos x-\cos2x}{1-\cos x}\ dx$$
I have tried solving the question using substitution. how do I solve it?
| Notice, we have $$\int\frac{\cos x-\cos 2x}{1-\cos x}dx$$
$$=\int\frac{\cos x-2\cos^2x+1}{1-\cos x}dx$$
$$=\int\frac{\cos x-2\cos^2x+1}{1-\cos x}dx$$
$$=\int\frac{\cos x-\cos^2x+1-\cos^2 x}{1-\cos x}dx$$
$$=\int\frac{\cos x(1-\cos x)+(1-\cos^2 x)}{1-\cos x}dx$$
$$=\int\frac{\cos x(1-\cos x)}{1-\cos x}dx+\int\frac{(1+\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove the trigonometric identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$ While solving an equation i came up with the identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$.
Prove whether this is really true or not. I can add that $$\tan\left(\frac{x}{2}\right) = \sqrt{\frac{1-\cos x}{1+\cos x}}$$
| Let me try. You have: $$\cos x + \sin x \tan (\frac{x}{2}) = 1 - 2\sin^2 (\frac{x}{2}) + 2 \sin (\frac{x}{2}) \cos (\frac{x}{2}) \tan (\frac{x}{2}) = 1 -2\sin^2 (\frac{x}{2}) + 2\sin^2 (\frac{x}{2}) = 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 4
} |
I just need your approval Verify by susbtitution if the given functions are a solution to the next differential equations.
a)
$$x^2y''+xy'-y=\ln x \quad ,\quad y_p=x^{-1}-\ln x $$
simplifying:
$$ x^2y''+xy'-y=\ln x $$
$$ x^2y''=-xy'+\ln x+y $$
$$y''=\frac{-xy'+\ln x+y}{x^2} $$
$$y_p=x^{-1}-\ln x $$
$$y_p'= -\frac{1}... | You first idea is correct,
but you could do it simpler without first simplification.
Just "plug" it in...
$$
\begin{array}{ccc}
x^2 y'' & + & x y' & - & y & = & \ln(x)\\\\
\hline\\
\displaystyle x^2 \left( \frac{2}{x^3} + \frac{1}{x^2} \right) &+&
\displaystyle x \left( - \frac{1}{x^2} - \frac{1}{x} \right) &-&
\displ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1408330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the least positive residue of $10^{515}\pmod 7$. I tried it, but being a big number unable to calculate it.
| All answers here use Fermat's little theorem, which is cool, but there is another simpler way (simpler if you don't know Fermat's theorem, otherwise it's more complicated).
$$10^0\equiv 1\mod 7\\
10^1 \equiv 3\mod 7\\
10^2\equiv 10\cdot 10 \equiv 3\cdot 3\equiv 9\equiv 2\mod 7\\
10^3 \equiv 10\cdot 10^2\equiv 3\cdot 2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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If $f: \Bbb{N} \to \Bbb{N}$ is strictly increasing and $f(f(n))=3n$, find $f(2001)$. I have this question which seems a little harder than I thought. It has been about an hour for me hitting aimless thoughts on this one. I can really use a hint here if some one knows how to tackle it.
Let $f: \Bbb{N} \to \Bbb{N}$ such... | Part A: $f(1)=2$.
First $f(1)$ cannot be $1$. Otherwise $3=f(f(1)) = f(1) = 1$.
So $f(1)$ is at least $2$. But $3 = f(f(1)) > f(2)$ because $f$ is increasing and $f(1)>2$.
Then $f(2)$ can only be $2$ or $1$. But $f(1)$ is at least $2$ and $f(1)<f(2)$.
This is a contradiction. So $f(1)$ must be $2$.
Part B
Since $2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
$z=100^2-x^2$. Then, how many values of $x,z$ are divisible by $6$?
$z=100^2-x^2$. Then, how many values of $x,z$ are divisible by $6$?
My approach:
For $x=1$, $z$ is not divisible by $6$.
For $x=2$, $z$ is divisible by $6$.
For $x=3$, $z$ is not divisible by $6$.
For $x=4$, $z$ is divisible by $6$.
For $x=5$, $z$... | HINT:
$$z=100^2-x^2=(100-x)(100+x)$$
As $(100-x)+(100+x)=200,100\pm x$ have same parity and
if one is divisible by $3,$the other is not
So if $2|z,100\pm x$ must be even
If $3|z,3|(100-x)(100+x)\implies$
either $3|(100-x)\iff x\equiv1\pmod3$
or $3|(100+x)\iff x\equiv-1\pmod3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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How many terms required in $e =\sum^∞_{k=0}{1\over k!}$ to give $e$ with an error of at most ${6\over 10}$ unit in the $20$th decimal place? How many terms are required in the series $e =\sum^∞_{k=0}{1\over k!}$ to give $e$ with an error of at most ${6\over 10}$ unit in the $20$th decimal place?
Here is what I have:
$$... | Every tail of this series is bounded above by a geometric series. For example:
\begin{align}
& \frac 1 {0!} + \frac 1 {1!} + \cdots + \frac 1 {6!} + \overbrace{\frac 1 {7!} + \frac 1 {8!} + \frac 1 {9!} + \frac 1 {10!} + \cdots} \\[10pt]
\le {} & \frac 1 {0!} + \frac 1 {1!} + \cdots + \frac 1 {6!} + \overbrace{\frac 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Algebra Problem: Division Can someone help me with a problem involving the expression $$\frac{(2x^3-3x^2+b)}{(4-x^2)}?$$
The question is to find which values $b$ can be to simplify the expression, but I do not know how to begin.
Thanks for any help you can give me.
| $4-x^2=-(x-2)(x+2)$, so you are looking for values of $b$ that make $2x^3-3x^2+b$ divisible by either $(x-2)$ or $(x+2)$.
So if you substitute $x=2$ into $2x^3-3x^2+b$ it will be 0, and the same for $x=-2$. This give you two simple equations in $b$, which gives you two values for $b$.
For $x=2$, you get $b=-4$, so you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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"Trig Substitutions", I tried half- angle and trig indentity in this one, but doesn't work I´m really lost in this one.
$\int \sin^3 (2x) \cos^2 (2x) dx$
I know that the answer is:
$\frac{1}{10}cos^5(2x)-\frac{1}{6}cos^3(2x) + c$
Please help
| $$
I = \int sin^3 (2x) cos^2(2x) dx
$$
Substituting
$$
t = 2x
\\dt = 2 dx
\\dx = \frac{dt}{2}
$$
So,
$$
I = \frac{1}{2}\int sin^3(t) cos^2(t) dt
$$
I'm now going to factor out
$$
sin^3 t = sin(t) (1 - cos^2 t)
$$
Because the factor of $sin(t)$ will simplify the integral.
$$
I = \frac{1}{2}\int sin(t)\; (1 - cos^2 (t)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
eigenvalues of cycle graph and its complement graph I am trying to find the eigenvalue of cycle graph and its complement. How to simplify.Suppose $\omega^{1}+\omega^{n-1}=2\cos (2\pi/n) $, then, $\omega^{\frac{n-1}{2}}+\omega^{\frac{n+1}{2}}=\ ?$ Is it true that $\omega^{1}+\omega^{n-1}=\omega^{\frac{n-1}{2}}+\omega^{... | We have $\omega$ is a primitive $n$th root of unity, $\omega^{k}$ and $\omega^{n-k}$ are conjugates. This means that $(1/2)(\omega^{k}+\omega^{n-k}) = \mathrm{Re}(\omega^{k}) = \cos{\frac{2\pi k}{n}}$ (and so $\omega^{\frac{n-1}{2}} + \omega^{\frac{n+1}{2}} = 2 \cos{\frac{(n-1)\pi}{n}}$).
Also, $\omega$ is a root of
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove $\sum _{k=1}^{\infty} \frac{k-1}{2 k (1+k) (1+2 k)}=\log_e 8-2$? How to prove:
$$\sum _{k=1}^{\infty} \frac{k-1}{2 k (1+k) (1+2 k)}=\log_e 8-2$$
Is it possible to convert it into a finite integral?
| We can evaluate the series using partial fraction expansion. To that end, we have
$$\begin{align}
\sum_{k=1}^{\infty}\frac{k-1}{2k(k+1)(2k+1)}&=\frac12\sum_{k=1}^{\infty}\left(\frac{6}{2k+1}-\frac{3}{k+1}\right)+\frac12\sum_{k=1}^{\infty}\left(\frac{1}{k+1}-\frac{1}{k}\right)\\\\
&=-\frac12+\frac32\sum_{k=1}^{\infty}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1419166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Proof by induction for "sum-of" Prove that for all $n \ge 1$:
$$\sum_{k=1}^n \frac{1}{k(k+1)} = \frac{n}{n+1}$$
What I have done currently:
Proved that theorem holds for the base case where n=1.
Then:
Assume that $P(n)$ is true. Now to prove that $P(n+1)$ is true:
$$\sum_{k=1}^{n+1} \frac{1}{k(k+1)} = \frac{n}{n+1} + n... | This step is wrong:
$$\sum_{k=1}^{n+1} 1/k(k+1) = n/(n+1) + n+1$$
it should read:
$$\sum_{k=1}^{n+1} \dfrac{1}{k(k+1)} = \dfrac{n}{n+1} + \dfrac{1}{(n+1)(n+2)}$$
so we have:
$$\sum_{k=1}^{n+1} \dfrac{1}{k(k+1)} = \dfrac{n(n+2)+1}{(n+1)(n+2)}$$
$$= \dfrac{n^2+2n+1}{(n+1)(n+2)}$$
$$= \dfrac{(n+1)^2}{(n+1)(n+2)}$$
$$= \df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1419749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
How to evaluate $45^\frac {1-a-b}{2-2a}$ where $90^a=2$ and $90^b=5$ without using logarithm? Let $90^a=2$ and $90^b=5$, Evaluate
$45^\frac {1-a-b}{2-2a}$
I know that the answer is 3 when I used logarithm, but I need to show to a student how to evaluate this without involving logarithm. Also, no calculators.
| $$2=90^a=(2\cdot5\cdot3^2)^a\iff5^a3^{2a}=2^{1-a}$$
and similarly, $$5=90^b\iff5^{1-b}3^{-2b}=2^b$$
Equating the powers of $2,$
$$\implies(5^a3^{2a})^b=(5^{1-b}3^{-2b})^{1-a}\iff3^{2b}=5^{1-a-b}$$
$$45^{1-a-b}=(3^2\cdot5)^{1-a-b}=3^{2-2(a+b)}\cdot5^{1-a-b}=3^{2-2(a+b)}\cdot3^{2b}$$
$$\implies45^{1-a-b}=3^{2(1-a)}$$
$\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1421740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\int_{0}^{\frac{\pi}{2}} (\sin (x) +\cos (x))^4 - (\sin(x) - \cos (x))^4 dx$ I would like to evaluate $$\int_{0}^{\frac{\pi}{2}} (\sin (x) +\cos (x))^4 - (\sin(x) - \cos (x))^4 dx.$$ One approach I used was to split the integral into two pieces and evaluating each piece directly with the anti derivatives $$... | You didn't compute antiderivatives correctly. To check your work, simply differentiate your result:
$$\frac{\mathrm d}{\mathrm dx}\left[\frac {1}{4} (\sin(x) - \cos (x))^4 + C\right] = (\sin x + \cos x)(\sin x - \cos x)^3 \neq (\sin x + \cos x)^4$$
Please, observe that we used the chain rule to derive our result. The s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1422463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the sum of the following series to n terms $\frac{1}{1\cdot3}+\frac{2^2}{3\cdot5}+\frac{3^2}{5\cdot7}+\dots$ Find the sum of the following series to n terms $$\frac{1}{1\cdot3}+\frac{2^2}{3\cdot5}+\frac{3^2}{5\cdot7}+\dots$$
My attempt:
$$T_{n}=\frac{n^2}{(2n-1)(2n+1)}$$
I am unable to represent to proceed furthe... | You can compute the partial fraction decomposition of $T_n$;
$$T_n = \frac18 \left( \frac{1}{2n-1} - \frac{1}{2n+1} + 2 \right)$$
Then, you can separate the sum into three sum :
$$\sum_{n=1}^N T_n = \frac18 \left( \sum_{n=1}^N \frac{1}{2n-1} - \sum_{n=1}^N\frac{1}{2n+1} + \sum_{n=1}^N2 \right)$$
$$ = \frac18 \left( \su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1428163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Use De Moivre's Theorem to express $\cos(4θ)$ in terms of sums and differences of powers of $\sin(θ)$ and $\cos(θ)$ So far I have gotten this far using the binomial theorem.
$(x + iy)^4 = x^4 + 4x^3y + 6i^2x^2y^2 + 4i^3xy^3 + i^4y^4$=
$x^4 - 6x^2y^2 + y^4 + i(4x^3y - 4xy^3)$
$i^2 = -1 $ let $x = cosθ$ and y = $sinθ$
... | ${(\cos{\theta}+i\sin{\theta})^4}\to$
${(c+is)^4}\to$ (for brevity)
$\begin{matrix}
1&4&6&4&1\\
c^4&c^3&c^2&c&1\\
1&is&-s^2&-is^3&s^4\\
\hline
c^4&4ic^3s&-6c^2s^2&-4ics^3&s^4
\end{matrix}$
The terms pertaining to $\cos(4\theta)$ are the real ones; the imaginaries, after dispensing with $i$, pertain to $\sin(4\theta)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1429222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Computing $E(XY)$ for finding $Cov(X,Y)$ Consider tossing a cubic die once and let $n$ be the smallest number of dots that appear on top. Define two random variables $X$ and $Y$ such that:
*
*$X=1$ if $n \in \left \{1,2 \right \}$, $X=2$ if $n \in \left \{3,4 \right \}$ and $X=3$ if $n \in \left \{5,6 \right \}$... | Tabulate:
$\boxed{\begin{array}{r|r}
n & 1& 2& 3& 4& 5& 6 \\ \hline
X & 1& 1& 2& 2& 3& 3 \\ \hline
Y & 1& 2& 0& 1& 2& 0 \\ \hline
(X{-}\mathsf E(X))(Y{-}\mathsf E(Y)) & 0 & -1 & 0 &0 &1 &-1
\end{array}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Simple limit of a sequence
Need to solve this very simple limit $$ \lim _{x\to \infty
\:}\left(\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}\right) $$
I know how to solve these limits: by using
$a−b= \frac{a^3−b^3}{a^2+ab+b^2}$. The problem is that the standard way (not by using L'Hospital's rule) to solve this limit - ver... | You have $$f(x)=\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}=\sqrt[3]{3x^2}\left(\sqrt[3]{1+\frac{4}{3x}+\frac{1}{3x^2}}-\sqrt[3]{1+\frac{3}{x}+\frac{2}{3x^2}}\right)$$ Using Taylor expansion at order one of the cubic roots $\sqrt[3]{1+y}=1+\frac{y}{3}+o(y)$ at the neighborhood of $0$, you get: $$f(x)=\sqrt[3]{3x^2}\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Is there a closed-form of $\frac{\zeta (2)}{\pi ^2}+\frac{\zeta (4)}{\pi ^4}+\frac{\zeta (6)}{\pi ^6}+.....$ The value of $$\frac{\zeta (2)}{\pi ^2}-\frac{\zeta (4)}{\pi ^4}+\frac{\zeta (6)}{\pi ^6}-.....=\frac{1}{e^2-1}$$
Is there a closed-form of $$\frac{\zeta (2)}{\pi ^2}+\frac{\zeta (4)}{\pi ^4}+\frac{\zeta (6)}{\p... | $$\begin{align}
\sum_{n=1}^{\infty}\frac{\zeta(2n)}{\pi^{2n}}
&=\sum_{n=1}^{\infty}\frac{1}{\pi^{2n}}\sum_{k=1}^{\infty}\frac{1}{k^{2n}}\tag{1}\\
&=\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{(\pi^2 k^2)^n}\tag{2}\\
&=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{(\pi^2 k^2)^n}\tag{3}\\
&=\sum_{k=1}^{\infty}\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
} |
Condition for quartic polynomial coefficients given at least one real root
Find the minimum possible value of $a^2+b^2$ where $a$, $b$ are two real numbers such that the polynomial
$$x^4+ax^3+bx^2+ax+1,$$
has at least one real root.
My attempt: Let p be a real root.
Therefore $p^4 + a(p^3) + b(p^2) + ap + 1 = 0$... | Hint:
Put $t=x+\frac{1}{x}$ then:
$x^4+ax^3+bx^2+ax+1=0$ has a real root if and only if $t^2+at+(b-2)=0$ has a real root $t$ such that $|t| \ge 2$, which is equivalent to
\begin{equation}\tag{*}
\left[\begin{matrix}
\frac{-a-\sqrt{a^2-4(b-2)}}{2} \le -2 \\
\frac{-a+\sqrt{a^2-4(b-2)}}{2} \ge 2
\end{matrix}\right.
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1439501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How many ways to show $\sum_{n=1}^{\infty}\ln \left|1-\frac{x^2}{n^2\pi^2}\right|$ is pointwise convergence? $\sum_{n=1}^{\infty}\ln \left|1-\frac{x^2}{n^2\pi^2}\right|$. where $x\not= k\pi, k\in \mathbb{Z}$ is pointwise convergence.
Assume $n$ is sufficiently large($\geq N$),for a fixed point $x_0$,$-\ln \left|1-\fra... | Another way for math freaks: using the Euler formula for the sine
$$
\sum_{n=1}^\infty\ln\left|1-\frac{x^2}{n^2\pi^2}\right|=
\ln\prod_{k=1}^\infty \left|1-\frac{x^2}{n^2\pi^2}\right|=\ln\left|\frac{\sin x}{x}\right|.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1441883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Minimum value of $\frac{4}{x}+2x+10+\frac{3+x}{4x^2+1}$ If the minimum value of $\frac{4}{x}+2x+10+\frac{3+x}{4x^2+1}$ when $x>0$ is $\frac{p}{q}$ where $p,q\in N$ then find the least value of $(p+q).$
I can find the minimum value of $\frac{4}{x}+2x$ using $AM- GM$ inequality but that too comes irrational and when i tr... | HINT: There is another approach using calculus
let $$y=\frac{4}{x}+2x+10+\frac{3+x}{4x^2+1}$$
$$\frac{dy}{dx}=-\frac{4}{x^2}+2+\frac{(4x^2+1)(1)-(3+x)(8x)}{(4x^2+1)^2}$$
Now, for maxima or minima, put $$\frac{dy}{dx}=0$$
$$\frac{(4x^2+1)(1)-(3+x)(8x)}{(4x^2+1)^2}-\frac{4}{x^2}+2=0$$
Since, $x>0$ then find the positive ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1442089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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How to find triangle vertices given midpoints? I have a task to find vertices if midpoints are given: $M1(2;1)$, $M2(5;3)$, $M3(3;-4)$. I know one way to solve it through making a system of equations with three variables.
My teacher says there is faster way by using the midline of a triangle, and I can`t find this way ... | The matrix equation giving the midpoints from the vertices is
$$
\begin{bmatrix}
\frac12&\frac12&0\\
0&\frac12&\frac12\\
\frac12&0&\frac12
\end{bmatrix}
\begin{bmatrix}
a\vphantom{\frac12}\\b\vphantom{\frac12}\\c\vphantom{\frac12}
\end{bmatrix}
=
\begin{bmatrix}
\frac{a+b}2\\\frac{b+c}2\\\frac{c+a}2
\end{bmatrix}
$$
In... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
} |
If $\frac{x^2+y^2}{x+y}=4$, then what are the possible values of $x-y$?
If $\frac{x^2+y^2}{x+y}=4$,then all possible values of $(x-y)$ are given by
$(A)\left[-2\sqrt2,2\sqrt2\right]\hspace{1cm}(B)\left\{-4,4\right\}\hspace{1cm}(C)\left[-4,4\right]\hspace{1cm}(D)\left[-2,2\right]$
I tried this question.
$\frac{x^2+y^2... | $\frac{x^2+y^2}{x+y}=4$ is equivalent to $(x-2)^2+(y-2)^2 = 8$, hence $(x,y)$ lies on a circle centered at $(2,2)$ with radius $2\sqrt{2}$. The tangents at the points $(0,4)$ and $(4,0)$ are parallel to the $y=x$ line, so the right answer is $(C)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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How to solve $\ x^2-19\lfloor x\rfloor+88=0 $ I have no clue on how to solve this. If you guys have, please show me your solution as well.
$$\ x^2-19\lfloor x\rfloor+88=0 $$
| $$\ x^2-19\lfloor x\rfloor+88=0 \tag{*}$$
Start by solving $x^2-19x+88=0$. This factors as $(x-8)(x-11)=0$ giving solutions $x\in\{8,11\}$. Since these are integers, they are solutions of (*) as well.
There are no more solutions to (*) for $8<x<9$ since 8 was an ordinary root and $x^2$ is increasing.
For $9\le x<10$, w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1444045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
$\int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}}$ $\int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}}$
I was given this question by my senior.I tried to solve it but could not reach the answer.
Let $I= \int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}} $
$I=\int\frac{dx}{\sqrt{x^2+4x-6+\frac{4}{x}+\frac{1}{x^2}}}$
Then after repeated attem... | In case that there is a typo, as the comments suggest, and the function is:
$$\int\frac{x dx}{\sqrt{x^4+4x^3+6x^2+4x+1}}$$
Then the integral is fairly easy once noticing that:
$$x^4+4x^3+6x^2+4x+1=(x+1)^4$$
And thus, the integral can be simplified to:
$$\int\frac{x dx}{(x+1)^2}=\int\frac{(x+1) dx}{(x+1)^2}-\int\frac{1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving $\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$ $$\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$$
So
$$\frac{1-\sqrt[3]{4-3x}}{x-1} \cdot \frac{1+\sqrt[3]{4-3x}}{1+\sqrt[3]{4-3x}}$$
Then
$$\frac{1-(4-3x)}{(x-1)(1+\sqrt[3]{4-3x})}$$
That's
$$\frac{3\cdot \color{red}{(x-1)}}{\color{red}{(x-1)}... | Alternatively you may observe that, for any differentiable function $f$, you have
$$
\frac{f(x)-f(a)}{x-a} \to f'(a).
$$
Then use it with $a=1$ and
$$
f(x)=\sqrt[3]{4-3x},\quad \quad f'(x)=-\frac{1}{(4-3 x)^{2/3}}
$$ giving
$$
\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)=-f'(1)=1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1447217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Find the limit as $n$ approaches infinity: $\lim_{n\to \infty} \sqrt{3^n + 3^{-n}} - \sqrt{3^n + 3^{\frac{n}{2}}}$ $$\lim_{n\to \infty} \sqrt{3^n + 3^{-n}} - \sqrt{3^n + 3^{\frac{n}{2}}}$$
I am taking calculus in university and this is the problem I have been given. I haven't even seen limits involving a variable in th... | Hint. You may write
$$
\begin{align}
\sqrt{3^n + 3^{-n}} - \sqrt{3^n + 3^{0.5n}}&=\left(\sqrt{3^n + 3^{-n}} - \sqrt{3^n + 3^{0.5n}}\right)\dfrac{\sqrt{3^n + 3^{-n}} + \sqrt{3^n + 3^{0.5n}}}{\sqrt{3^n + 3^{-n}} + \sqrt{3^n + 3^{0.5n}}}\\\\
&=\dfrac{(3^n + 3^{-n})-(3^n + 3^{0.5n})}{\sqrt{3^n + 3^{-n}} + \sqrt{3^n + 3^{0.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1448690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Prove $\lim_{a \to \infty} \frac{3a+2}{5a+4}= \frac35$ using definition of limit Prove the limit using definition of limit$$ \ \lim_{a \to \infty}\ \frac{3a+2}{5a+4}= \frac{3}{5} $$
Answer: Let $\varepsilon \ >0 $. We want to obtain the inequality $$\left|\frac{3a+2}{5a+4}- \frac{3}{5}\right|< {\varepsilon}$$
$$ \Righ... | let $ζ>0$ be any arbitrary real number
consider,
$$\begin{align}
&|[(3a+2)/(5a+4)]-(3/5)| < ζ\\
\Leftrightarrow&|-2/5(5a+4)| < ζ\\
\Leftrightarrow& a > 1/5[(2/5ζ)-4]=∆
\end{align}$$
hence for every $ζ>0$ there exists $∆>0$ such that which satisfies the definition of convergence
hence $f(a)=[(3a+2)/(5a+4)]$ converges t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1451900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Three vertices of a triangle are (2,7),(4,-1),(0,Y).if the perimeter of the triangle is the least then express Y in the lowest rational form as (p/q). I don't know how to proceed in this question. What I know is that the perimeter is the sum of all the sides.
| Notice, the length of side with vertices $(2, 7)$ & $(4, -1)$, using distance formula, is $$=\sqrt{(2-4)^2+(7+1)^2}=\sqrt{68}$$
Similarly, the length of side with vertices $(2, 7)$ & $(0, y)$ is $$=\sqrt{(2-0)^2+(7-y)^2}=\sqrt{(y-7)^2+4}$$ the length of side with vertices $(4, -1)$ & $(0, y)$ is $$=\sqrt{(4-0)^2+(-1-y)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Complex numbers solving for all solutions
*
*Find all complex numbers $z$ such that $z + \frac{1}{z}$ is real
*Let $a \gt b \gt 0$ be integers. Find all the solutions in complex numbers to the equation $z^a = z^{-b}$
For the first one, I tried manipulating the equation $Re(z) = \frac{z + \bar{z}}{2}$ to equal $z + ... | Question 1
Let $z=x+iy$ for real numbers $x,y$. Clearly, they may not be both $0$ given we are dividing by $z$. $$z + z^{-1} = (x+iy) + \left(\frac{x}{x^2+y^2} - \frac{iy}{x^2+y^2}\right).$$
In order for $z+z^{-1}$ to be real, its imaginary part must be $0$. Therefore,
$$y- \frac{y}{y^2+x^2}=\frac{y(x^2+y^2-1)}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1454551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Recurrent integral formula $$ I_{m,n}= \int {\sin(x)^m}{\cos(x)^n} dx $$
Determine the recurrence formula for the given integral,where "$n$" and "$m$" are $2$ natural parameters.
EDIT: Can it be solved by using only definite integrals formulas or the integration by parts method?
| Use integration by parts on $I_{m,n}$ with:
$$\begin{array}{ll}
u=\sin^{m-1}x & dv=\sin x \cos^n x\,dx \\
du=(m-1)\sin^{m-2}x \cos x\,dx & v =\frac{-1}{n+1}\cos^{n+1}x
\end{array}$$
Then we have:
$$
I_{m,n} = \frac{-1}{n+1}\sin^{m-1}x\cos^{n+1}x + \frac{m-1}{n+1}\color{blue}{\int{\sin^{m-2}x\,\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1455601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Two numbers $a$ and $b$ are chosen at random from the set of first 30 natural numbers.Find the probability that $a^2-b^2$ is divisible by $3$ Two numbers $a$ and $b$ are chosen at random from the set of first 30 natural numbers.Find the probability that $a^2-b^2$ is divisible by $3$.
Total number of ways of choosing tw... | Note that $a^2-b^2$ is divisible by $3$ if and only if either (i) $a$ and $b$ are both divisible by $3$ or (ii) neither $a$ nor $b$ is divisible by $3$. This is because if $n$ is not divisible by $3$, then $n$ has remainder $1$ or $2$ on division by $3$. If $a$ and $b$ have the same remainder on division by $3$, then ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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trigonometry equation $3\cos(x)^2 = \sin(x)^2$ I tried to solve this equation, but my solution is wrong and I don't understand why. the answer in the book is: $x = \pm60+180k$. my answer is: $x= \pm60+360k$.
please help :)
3cos(x)^2 = sin(x)^2
3cos(x)^2 = 1 - cos(x)^2
t = cos(x)^2
3t=1-t
4t=1
t=1/4
cos(x)^2 = 1/4
co... | \begin{align*}
3\cos^2 x & = \sin^2x\\
3 & = \frac{\sin^2x}{\cos^2x}\\
\tan^2x & = 3\\
\tan x & = \sqrt{3}\\
\tan x & = \tan 60
\end{align*}
From there, you can use the tangent equation:
$$x= \pm 60+180k$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Where am I going wrong while trying to solve this logarithmic equation? $$\log _{ 0.2 }{ x } +\log _{ \sqrt { 5 } }{ x } =\log _{ 25 }{ x } +1$$
Steps I took:
$$\log _{ \frac { 1 }{ 5 } }{ x } +\frac { \log _{ \frac { 1 }{ 5 } }{ x } }{ \log _{ \frac { 1 }{ 5 } }{ \sqrt { 5 } } } =\frac { \log _{ \frac { 1 }{ 5 ... | at line 3: $\log _{ \frac { 1 }{ 5 } }{ \sqrt { 5 } }$ doesn't equal -2, but $-\frac {1}{2}$, so the third line need to be:
$\log _{ \frac { 1 }{ 5 } }{ x } +-2\log _{ \frac { 1 }{ 5 } }{ x } =-\frac { \log _{ \frac { 1 }{ 5 } }{ x } }{ 2 } +1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1463452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to sketch the subset of a complex plane? The question asks to sketch the subset of $\{z\ \epsilon\ C : |Z-1|+|Z+1|=4\}$
Here is my working:
$z=x+yi$
$|x+yi-1| + |x+yi+1|=4$
$\sqrt{ {(x-1)}^2 + y^2} + \sqrt{{(x+1)}^2+y^2}=4$
${ {(x-1)}^2 + y^2} + {{(x+1)}^2+y^2}=16$
$x^2 - 2x+1+y^2+x^2+2x+1+y^2=16$
$2x^2+2y^2+2=... | Or in another but similar parametrization to Yves':
\begin{align}
\sqrt{a+b}+\sqrt{a-b}&=4\\
\text{after squaring: }a+\sqrt{a^2-b^2}&=8\\
\text{rearrange and square: }a^2-b^2&=(8-a)^2=64-16a+a^2\\
16a-b^2&=64
\end{align}
where $a=x^2+y^2+1$ and $b=2x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1464152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
The number of solutions of the equation $4\sin^2x+\tan^2x+\cot^2x+\csc^2x=6$ in $[0,2\pi]$ The number of solutions of the equation $4\sin^2x+\tan^2x+\cot^2x+\csc^2x=6$ in $[0,2\pi]$
$(A)1\hspace{1cm}(B)2\hspace{1cm}(C)3\hspace{1cm}(D)4$
I simplified the expression to $4\sin^6x-12\sin^4x+9\sin^2x-2=0$
But i could not s... | Subtract $6$ from both sides of the original equation.
$$4\sin^2x-4+\csc^2x+\tan^2x-2+\cot^2x=0$$
$$(2\sin x-\csc x)^2+(\tan x-\cot x)^2=0$$
Since the square of a real number is non-negative, the only solution is if both squares are $0$. So we have $\tan x=\cot x$ and
$$2\sin x=\csc x$$
$$\sin^2x=\frac12$$
$$\left\{\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
$\int_{-1}^{+1}\sin(\sqrt{1-x^2})\cos(x)\,dx$ =? I am trying to find the value of the definite integral:
$$\int_{-1}^{+1}\sin\left(\sqrt{1-x^2}\right)\cos(x)\,dx$$
The answer from WolframAlpha is $1.20949$ but I can't solve it analytically.
| $\int_{-1}^1\sin\sqrt{1-x^2}\cos x~dx$
$=2\int_0^1\sin\sqrt{1-x^2}\cos x~dx$
$=2\int_0^\frac{\pi}{2}\sin\cos\theta\cos\sin\theta\cos\theta~d\theta$
$=\int_0^\frac{\pi}{2}\sin(\cos\theta+\sin\theta)\cos\theta~d\theta+\int_0^\frac{\pi}{2}\sin(\cos\theta-\sin\theta)\cos\theta~d\theta$
$=\int_0^\frac{\pi}{2}\sin\left(\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1472674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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A problem relating to triangles and progressions If $D,E,F$ are the points of contact of the inscribed circle with the sides $BC, CA, AB$ of a triangle $\triangle ABC$, we need to show that if the squares of $AD, BE, CF$ are in arithmetic progression, then the sides of a triangle are in harmonic progression.
I tried us... | $$AD^2=c^2+(s-b)^2-2c(s-b)cosB=(c-(s-b))^2+2c(s-b)(1-cosB)=(b+c-s)^2+4c(s-b)sin^2(\frac{B}{2})$$ But $$sin^2\left(\frac{B}{2}\right)=\frac{(s-a)(s-c)}{ac}$$ So
$$AD^2=(b+c-s)^2+4c(s-b)\frac{(s-a)(s-c)}{ac}=(s-a)^2+4\frac{\Delta^2}{as}=(s-a)^2+\frac{4r\Delta}{a}$$ Similarly
$$BE^2=(s-b)^2+\frac{4r\Delta}{b}$$ and
$$CF... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1474338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How do you factorize quadratics when the coefficient of $x^2 \gt 1$? So I've figured out how to factor quadratics with just $x^2$, but now I'm kind of stuck again at this problem:
$2x^2-x-3$
Can anyone help me?
| For quadratic polynomials of the form $f(x)=ax^2+bx+c$ with $a,b,c$ real numbers, the roots are known to be given by the following:
The Quadratic Formula: The roots of $f(x)=ax^2+bx+c$ are
$$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$$
and
$$x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$$
Note that when $b^2-4ac=0$, then the root is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1477307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
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Using arithmetic mean>geometric mean Prove that if a,b.c are distinct positive integers that
$$a^4+b^4+c^4>abc(a+b+c)$$
My attempt:
I used the inequality A.M>G.M to get two inequalities
First inequality
$$\frac{a^4+b^4+c^4}{3} > \sqrt[3]{a^4b^4c^4}$$
or
$$\frac{a^4+b^4+c^4}{3} > abc \sqrt[3]{abc}$$ or new --
first ine... | Here is a slightly more general approach to construct the right AM-GM. Note, for non-negative $x,y$ s.t. $x+y\le 1$,
$$x a^4+yb^4+(1-x-y)c^4\ge a^{4x}b^{4y}c^{4-4x-4y}$$
Clearly we are done if we have $4x=2,4y=1\implies 4-4x-4y=1$, so $x=\frac12, y=\frac14$ and what you need is to sum
$$\frac12a^4+\frac14b^4+\frac14c^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.