Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to factorize $2ab^2+2b^2c-ac^2-bc^2-a^2b-a^2c$? If $a,b,c$ are in AP $\implies \frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}$ are in AP $\implies
\frac{ab+bc+ca}{bc}, \frac{ab+bc+ca}{ca}, \frac{ab+bc+ca}{ab}$ are in AP $\implies \frac{bc+ca}{bc},\frac{ab+bc}{ca},\frac{bc+ca}{ab}$ are in AP $\implies \frac{bc}{bc+ca}, \f... | If rearranging terms as suggested in the previous answer is difficult to imagine here you have another approach but this has much simplification work.
Let the given polynomial be denoted by f(a,b,c)
Since this is homogeneous polynomial of a,b,c when you replace a by x root of x of f(x,b,c) =0 must be in the form of m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4441291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Fourier series of $\sqrt[3]{\sin x}$ So I've done some experiments with how to add distortion to audio, and one of the methods I'm proposing is to take the cube root of an audio signal as a way to add overdrive. As the waveform that you're supposed to start with is a sine wave, I decided to tackle the problem of findin... | The solution to the problem is straighforward, and I'm going to present it here with a minor generalisation.
Remember that the Fourier series to order $n$ of a function $f(x)$ is defined as
$$f_s(x,n) = \sum_{k=-n}^{n}c(k) e^{i k x}\tag{1a}$$
where the Fourier coefficients are defined as
$$c(k) = \frac{1}{2\pi} \int_{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4441796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Algebra of o-symbols In theorem $7.8$, Tom Apostol in his Calculus Vol. $1$ gave and proved the following basic rules for algebra of o-symbols
*
*$o(g(x)) \pm o(g(x)) = o(g(x)) $
*$o(cg(x)) = o(g(x)) = o(g(x)) $, if $c \ne 0$
*$f(x) \cdot o(g(x)) = o(f(x)g(x)) $
*$o(o(g(x))) = o(g(x)) $
*$\frac{1}{1 + g(x)} = 1 -... | Please correct me if I'm wrong, but to prove $o(x^2) + o(x^3) = o(x^2)$, when $x \to 0$ we can take a simpler route, than what was explained in the answer (the edited part).
Namely, because of property $4$ of theorem $7.8$, we have the following:
\begin{equation}
\tag{By the definition of the $o$ notation}
x^3 = o(x^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4442024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving $ab^n + ba^n < a^{n+1} + b^{n+1}$ by Induction Given $0<a<b , ab \in \mathbb{R}$ , $n \geq 1$
By induction, Assuming $ab^n + ba^n < a^{n+1} + b^{n+1}$. - $(1)$
To prove $ab^{n+1} + ba^{n+1} < a^{n+2} + b^{n+2}$
Multiplying (1) by ab we get
$a^2b^{n+1} + b^2a^{n+1} < a^{n+2}b + b^{n+2}a$
How to proceed ?
Tha... | Your equation is equivalent to
$$0 < aa^n-a^nb-ab^n+bb^n$$
$$0 < (a-b)(a^n-b^n)$$
Since $0 < a < b,$ where $ a,b \in \mathbb{R}$,
$$0>(a-b)$$
that leaves
$$0 > (a^n-b^n)$$
Working on $n$, we have
$$(a^n-b^n) < 0$$
$$n\ln\frac{a}{b} < 0.$$
Again, $\ln\frac{a}{b}<0$ when $a<b$ so
$$n > 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4442251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Convergence or divergence of $\int_2^\infty \frac{1-\cos(4-x^2)}{x-2} \, \mathrm{d}x$ Is the integral
$$ \int_2^\infty \frac{1-\cos(4-x^2)}{x-2} \, \mathrm{d}x $$
convergent or divergent? As the numerator goes to $0$ infinite times I can't apply the comparison test, and I can't seem to find any function to compare the ... | Alternative solution:
Let
$$I_n := \int_3^{\sqrt{2n\pi + 4}} \frac{1 - \cos(4 - x^2)}{x - 2}\mathrm{d} x
\overset{x = \sqrt{y + 4}} = \int_5^{2n\pi} \frac{1 - \cos y}{\sqrt{y + 4} - 2}\, \frac{1}{2\sqrt{y + 4}}\mathrm{d} y.$$
We have
\begin{align*}
I_n &= \int_5^{2n\pi} \frac{1 - \cos y}{y}\, \frac{\sqrt{y + 4} + 2}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4444670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove the independence of a certain segment in a special triangle Let $\triangle ABC$ be a triangle with obtuse angle $\angle A$ and $\overline{AB} = 1$. Also, let $\angle C = \gamma$ and $\angle B = 2\gamma$. If $E$ and $F$ are intersection points of perpendicular bisector of $\overline{BC}$ and circle $(A, \overline{... | Note that if we fix the length $AB$, then the length of the chord $EF$ depends only on the distance from $A$ to $\overline{EF}$. Let $\overline{AD}$ be an altitude of $\triangle ABC$. Then
\begin{align*}
BD &= AB \cos(2\gamma) & BC &= AD \cot(2\gamma) + AD \cot(\gamma) = AB \sin(2\gamma) (\cot(2\gamma) + \cot(\gamma))
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
One of the angles of a trapezoid $ABCD$ is $30^\circ$ and its diagonals are perpendicular to each other One of the angles of a trapezoid $ABCD (AB\parallel CD)$ is $30^\circ$ and its diagonals are perpendicular to each other. If the midsegment is $10$ and one of the bases is $8$, find the other base, the diagonals and ... | Hint
The triangles $\Delta OCD$ and $\Delta OAB$ are similars with raio $8/12 = 2/3$. It means we can write $OD=2x$, $OB=3x$, $OC=2y$ and $OA=3y$.
We can also write $$\tan (OBA) = \frac yx \quad (1)$$ and $$\tan (OBC)=\tan (30-OBA)=\frac{\tan (30)-\tan(OBA)}{1+\tan(30)\tan(OBA)}=\frac 23 \frac yx.\quad (2)$$
Solve the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4448361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Determine the geometric centre of a circle with a quarter missing The question
I have a circle of radius $a$, (where $a$ is a known positive constant), centred at Cartesian coordinates $(a,a)$. The bottom left quarter of the circle is missing.
Let the two-dimensional region $R$ be formed by this circle with a missing q... | Choose polar coordinates centered at the center of your circle so the change-of-variables looks like
\begin{cases}
x = a + r \cos \theta, \\
y = a + r \sin \theta.
\end{cases}
Notice that when $r=0$, we're at the center $(x, y) = (a, a)$. As you correctly observed, the region of interest is defined by the inequalitie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4448485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Calculate: $\Delta=\left|\begin{array}{ccc} b c & c a & a b \\ a(b+c) & b(c+a) & c(a+b) \\ a^{2} & b^{2} & c^{2} \end{array}\right|$ Calculate:
$$\Delta=\left|\begin{array}{ccc}
b c & c a & a b \\
a(b+c) & b(c+a) & c(a+b) \\
a^{2} & b^{2} & c^{2}
\end{array}\right|$$
Does anyone know any easy way to calculate this dete... | If you add the first row to the second, the second row becomes three copies of $ab+bc+ca$, and you can write
$$\Delta=(ab+bc+ca)\begin{vmatrix}bc&ca&ab\\1&1&1\\a^2&b^2&c^2\end{vmatrix}.$$
Now, subtract the first column from the second and third to get
$$\Delta=(ab+bc+ca)\begin{vmatrix}bc&c(a-b)&b(a-c)\\1&0&0\\a^2&b^2-a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4450509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Solving $\tan ^{-1}(\frac{1-x}{1+x})=\frac{1}{2}\tan ^{-1}(x)$ I was solving the following equation,$$\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\tan ^{-1}\left(x\right)$$
But I missed a solution (don't know where's the mistake in my work).
Here's my work:
$$\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\ta... | Apply $tan$ to both sides
$ \dfrac{1 - x}{1 + x} = \tan \left( \dfrac{1}{2} \tan^{-1}(x) \right) \\
= \dfrac{ x }{ \sqrt{1 + x^2} + 1 } $
From this,
$ (1 - x) (\sqrt{1 + x^2} + 1 ) = x (1 + x) $
$ (1 - x) \sqrt{1 + x^2} + 1 - x = x + x^2 $
$ \sqrt{ 1 + x^2} = \dfrac{(x^2 + 2 x - 1)}{ ( 1 - x)} $
$ 1 + x^2 = \dfrac{ (x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4450671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to solve $'''+ ''+ = 0$? $$'''+ ''+ = 0$$
What method can I use to further solve the equation?
| We can write the first as
\begin{align}
(xy'')^{'} + \left(\dfrac{x^2}{2}\right)' = 0
\end{align}
so it becomes a problem of integration, since of course the integrating process can be applied linearly. As such
\begin{align}
xy'' + \dfrac{x^2}{2} = C_1 &\implies y'' = \dfrac{C_1}{x} - \dfrac{x}{2}\\
&\implies \int y''(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4450843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I prove this these inequalities, preferably without calculus? I'm trying to solve an inequality which stems from this problem:
If x, y, z are positive real numbers and $x^5+y^5+z^5=3$, prove $\frac{x^8}{y^3}+\frac{y^8}{z^3}+\frac{z^8}{x^3} \ge 3$
I thought that solving this directly would require an inequality I... | Note that by the rearrangement inequality for triples $(x^3,y^3,z^3)$ and $(x^2,y^2,z^2)$ we can conclude that $$y^3x^2+z^3y^2+x^3z^2 \le x^5+y^5+z^5$$
Now by Cauchy-Schwarz inequality
$$(\frac{x^8}{y^3}+\frac{y^8}{z^3}+\frac{z^8}{x^3})(y^3x^2+z^3y^2+x^3z^2) \ge (x^5+y^5+z^5)^2 \implies$$ $$\frac{x^8}{y^3}+\frac{y^8}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4451000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to find $\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{-x}$? By factorization:
$$\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{-x}\tag{1}$$
$$=\lim_{x\to-\infty} \frac{x\sqrt{1+\frac{2}{x}}}{-x}$$
$$=\lim_{x\to-\infty}-\sqrt{1+\frac{2}{x}}$$
If I input $x=-\infty$, the limiting value seems to be $-1$. But according to desmos, ... | Your approach is almost correct, you've just made a common mistake regarding square roots.
Writing out your manipulation of the numerator, you did $\sqrt{x^2 + 2x} = \sqrt{x^2(1 + \frac2x)} = \sqrt{x^2}\sqrt{1 + \frac2x} = x\sqrt{1 + \frac2x}.$ However, recall that because the principal square root is always positive (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4451437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Restoring third coordinate for triangle by its orthogonal projection and similar triangle Suppose we have triangle $\Delta OAB$ lying on plane $z=0$ with coordinates $O(0,0,0), A(x_a,y_a,0), B(x_b,y_b,0)$
Also there is triangle $\Delta EFG$, but we know only coordinates of its orthogonal projection on plane $z=0$. $E(x... | For $\triangle OAB$ we have
$ o^2 = \overline{AB}^2 , a^2 =\overline{OB}^2 , b^2 = \overline{OA}^2 $
And for $\triangle EFG$ we have
$ e^2 = \overline{FG}^2 = (x_f - xg)^2 + (y_f - y_g)^2 + (z_f - z_g)^2 = K_1 + (z_f - z_g)^2 $
$ f^2 = \overline{EG}^2 = (x_e - x_g)^2 + (y_e - y_g)^2 + (0 - z_g)^2 = K_2 + z_g^2 $
$ g^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4453168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find $\sum_{k=1}^\infty\frac{1}{x_k^2-1}$ where $x_1=2$ and $x_{n+1}=\frac{x_n+1+\sqrt{x_n^2+2x_n+5}}{2}$ for $n \ge 2$ Given $x_1=2$
and $x_{n+1}=\frac{x_n+1+\sqrt{x_n^2+2x_n+5}}{2}, n\geq 2$
Prove that $y_n=\sum_{k=1}^{n}\frac{1}{x_k^2-1}, n\geq 1$ converges and find its limit.
*
*To prove a convergence we can jus... | Using the relation $x_{n+1}^2 - 1 = x_{n+1}(x_n + 1)$, we find that
\begin{align*}
\frac{1}{x_n + 1} - \frac{1}{x_{n+1} + 1}
&= \frac{x_{n+1}}{x_{n+1}^2 - 1} - \frac{1}{x_{n+1} + 1} \\
&= \frac{1}{x_{n+1}^2 - 1}.
\end{align*}
So it follows that
\begin{align*}
y_n
&= \frac{1}{x_1^2 - 1} + \sum_{k=1}^{n-1} \left( \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4455721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Under what conditions is $\vert x-y \vert = \vert x \vert -\vert y \vert$? This is a fairly basic question, however, I don't know what I am missing here. Solving the equation $\vert x- y\vert = \vert x \vert -\vert y \vert$ yields:
\begin{align}
\vert x- y\vert &= \vert x \vert -\vert y \vert \\
(x-y)^2&=x^2+y^2-2\vert... | You have that
$$
|x-y| = |x| - |y| \qquad\implies\qquad
(x-y)^2 = x^2 + y^2 - 2|xy|
$$
which is an implication but not an equivalence. The problem is that squaring adds solutions, for example $1\neq-1$ but $1^2=(-1)^2$.
If you'd take the square root of the right equation, you'd get $$|x-y| = \big|\,|x|-|y|\,\big|$$
The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4472227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
To prove an Inequality: $ ( x^2 +2x)e^x + (x^2-2 x)e^{-x} \ge 0$
$ \left(x^2 +2x\right)e^x + \left(x^2-2 x\right)e^{-x} \ge 0$.
I used photomath to plot its graph: $y=(x^{2}+2x))e^{x} + \frac{{x}^{2}-2x}{{e}^{x}}$
But how do I prove it without an image? Should I take the derivative of it and reason, please tell me ... | Rearranging gives
$$2x^2\frac{(e^x+e^{-x})}{2}+4x\frac{e^x-e^{-x}}{2}=2f(x)+4g(x)$$
Now we will use Taylor expansion !!!
note $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$
$-e^{-x}=-1+x-\frac{x^2}{2!}+\frac{x^3}{3!}-....$
So,
$$\frac{e^x-e^{-x}}{2}=x+\frac{x^3}{3!}+...$$
So $$x\frac{(e^x-e^{-x})}{2}=x^2+\frac{x^4}{3!}+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4472916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Sameness of Riemann surfaces I need to show that the equations $x^3+y^3=1$ and $y^2=4x^3-1$ are the same Riemann surface in $\mathbb{CP}^2$ and as a consequence to show that there two meromorphic functions $f,g$ such that $f^3+g^3=1$.
I tried using Able's theorem but didn't get much
Edit: Using compactification I manag... | I'm thinking if this proof is correct or not:
$x^3+y^3=1$ is the Riemann surface in $\Bbb{CP}^2$ define by:
$$
\{[X,Y,Z] \in \Bbb{C}^3| X^3+Y^3=Z^3, (X, Y, Z) \neq (0, 0, 0)\}
$$
Let
$$
\begin{aligned}
X &= U \\
Y &= \frac{\sqrt{3}}{6}V -\frac{1}{2}W \\
Z &= \frac{\sqrt{3}}{6}V +\frac{1}{2}W \\
\end{aligned}
$$
Then $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Answer:
$a+b = 7, ab = 2$
$$\begin{align}
(a+b)^6 &= a^6 + 6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6 \\[4pt]
a^6 + b^6 &= (a+b)^6 - (6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5) \... | What the hey, I'll toss in one more (there are plenty of ways to work with the binomials, and some of the other answers already given are ones I'd have tried first):
We need $$ \ a + b \ = \ 7 \ \ , \ \ a - b \ = \ \sqrt{\Delta} \ = \ \sqrt{41} \ \ , \ \ ab \ = \ 2 \ \ , $$ $$ a^2 + b^2 \ = \ (a + b)^2 - 2ab \ = \ 49 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 8
} |
Find $\int \sin^4x dx$ In my textbook, I came upon the following problem:
Problem: Find $\int \sin^4x dx$
Since there was only given a solution sketch and I came to a different conclusion in one step, I ask if my following solution is correct.
Solution: In my solution, I use the following identities:
$$\sin^2x = \frac... | An alternative approach: A complex variable identity reduces the problem to an exercise in the Binomial Theorem.
$$\begin{align}
\sin^4 x &= \left( \frac{1}{2i} (e^{ix}-e^{-ix})\right)^4 \\
&= \frac{1}{16} \left( e^{4ix} -4 e^{2ix} +6 -4 e^{-2ix} + e^{-4ix} \right)
\end{align}$$
so
$$\begin{align}
\int \sin^4 x \;dx &=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
help me to evaluate these integrals Find the value of $$\int\frac{x}{x^2-x+1} dx$$ and $$\int\frac{1}{x^2-x+1} dx$$ This was not the original question. Original question was tougher and I have simplified that to these two integrals. I am having a hard time evaluating these two integrals but what I know is that, in thes... | For the second one: A common method would be: complete the square in the denominator...
$$
x^2-x+1 = x^2 - x + \frac14+\frac34
=\left(x-\frac12\right)^2+\frac34
$$
Substitute $x-\frac12 = y$.
$$
\int\frac{1}{x^2-x+1}\;dx = \int\frac{1}{y^2+(\sqrt{3}/2)^2}\;dy
$$
recognize an arctangent integral
$$
=\frac{2}{\sqrt{3}}\a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Integral $\int_0^1 \frac{\ln(1-\sqrt{3}t+t^2)}{t} dt$ $$I=\int_0^1 \frac{\ln(1-\sqrt{3}t+t^2)}{t} dt=-\frac{13}{72}\pi^2$$
Here is what I tried: $1-\sqrt{3}t+t^2=(z_1-t)(z_2-t)$ where $z_1=e^{\frac{\pi}{6}i}, z_2=e^{-\frac{\pi}{6}i}$
$$I=\int_0^1 \frac{\ln(z_1)+\ln(1-\frac{t}{z_1})}{t}+\frac{\ln(z_2)+\ln(1-\frac{t}{z_2... | Continue with the identity $\operatorname{Li}_2(z)+\operatorname{Li}_2(1/z)=-\frac{\pi^2}6-\frac12\ln^2(-z)$
\begin{align}
I= & \int_0^1 \frac{\ln(1-e^{-i\frac{\pi}6}{t})}{t}+\frac{\ln(1-e^{i\frac{\pi}6}t)}tdt\\
= & -\operatorname{Li}_2\left(e^{-i\frac{\pi}6}\right)-\operatorname{Li}_2\left(e^{i\frac{\pi}6}\right)\\
= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4486282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Expand the function $f(z) = \frac{1+2z^2}{z^2+z^4} $ into power series of $z$ in all areas of convergence. I'm studying for my upcoming complex analysis qualifying exam by working through problems in past exams. For this problem, I'd like to know (1) if my answer is correct and complete (i.e. whether I've made any erro... | Yes, that's perfect. Note that you can even simplifie more your expression:
$$\sum_{k=0}^{\infty} \left(z^{-2}+2\right)(-1)^k z^{2k}=\sum_{k=0}^{\infty}(-1)^kz^{2(k-1)}+2\sum_{k=0}^{\infty}(-1)^kz^{2k}=\\\sum_{k=-1}^{\infty}(-1)^{k+1}z^{2k}+2\sum_{k=0}^{\infty}(-1)^kz^{2k}\\=\frac{1}{z^2}+\sum_{k=0}^{\infty}(-1)^{k+1}z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4486804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $a > 0$, $b > 0$ and $b^{3} - b^{2} = a^{3} - a^{2}$, where $a\neq b$, then prove that $a + b < 4/3$. If $a > 0$, $b > 0$ and $b^{3} - b^{2} = a^{3} - a^{2}$, where $a\neq b$, then prove that $a + b < 4/3$.
Now what I thought is to manipulate given result somehow to get something in the form of $a + b$:
\begin{align... | Refer to 'Inequality of arithmetic and geometric means' in wiki
$\frac{{a + b}}{2} \geqslant \sqrt {ab} \Rightarrow \frac{{{{(a + b)}^2}}}{4} \geqslant ab$
Next,
$\begin{gathered}
{a^2} + ab + {b^2} = a + b \hfill \\
{(a + b)^2} = a + b + ab < a + b + \frac{{{{(a + b)}^2}}}{4} \hfill \\
\frac{{3{{(a + b)}^2}}}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4487764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$T(n) = 2T\left(\lfloor n/2\rfloor +1\right) + T(n-1) + 1$, $T(3) = 1$ I am looking for asymptotic bound on $T$, where $T$ satisfies the following recurrence:
$$T(n) = 2T\left(\lfloor{\frac{n}{2}\rfloor} +1\right) + T(n-1) + 1\text{ , }T(3) = 1.$$
I plotted the graph for $T$, and it seems like $T$ grows somewhere betwe... | Your prediction is correct, $T(n)$ has growth rate between polynomial and exponential. Let $F(x)$ be defined by $$
F(x) = \sum_{n=0}^\infty \frac{c_n (x-3)^n}{n!}
$$
where $c_n$ is defined by $c_0=1$ and $c_n = 3 \cdot 2^{-\binom{n-1}2}$ for $n\ge 1$. This series converges (quite rapidly, I would add) for all $x\in \ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4489415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Irrational equation $\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$ I saw the problem from one math competition:
$$\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$$.
I tried to solve it this way:
\begin{align*}
& \left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0\\
\Leftrightarrow\,\, & \left... | Of course, $x= \pm \sqrt{2}$ are two roots of this equation. Assume $x^2 \ne 2$, we have:
\begin{align*}
&(x^{3} - 3x + 1)(\sqrt{x^{2} - 1} + x) = -1 &\Longleftrightarrow & x^{3} - 3x + 1 = \sqrt{x^{2} - 1}-x \\\\
\Longleftrightarrow & x^{3} - 2x + 1 = \sqrt{x^{2} - 1} & \Longleftrightarrow & x^{3} - 2x = \sqrt{x^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4490883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
Prove $\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\ge \sqrt{7\left(a^3+b^3+c^3+a^2b+b^2c+c^2a+b^2a+a^2c+c^2b\right)-3}$ with $a^2+b^2+c^2=1$ and $a,b,c\ge 0$ With $a,b,c\ge 0$ and $a^2+b^2+c^2=1$, prove $\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\ge \sqrt{7\left(a^3+b^3+c^3+a^2b+b^2c+c^2a+b^2a+a^2c+c^2b\right)-3}$.
I don't know when equali... | It suffices to prove that
$$(\sqrt{a + b} + \sqrt{b + c} + \sqrt{c + a})^2 \ge 7(a + b + c) - 3$$
or
$$2a + 2b + 2c + \sum_{\mathrm{cyc}} 2\sqrt{(a + b)(b + c)} \ge 7(a + b + c) - 3.$$
Using Cauchy-Bunyakovsky-Schwarz inequality and GM-HM inequality, we have
$$\sqrt{(a + b)(b + c)} \ge b + \sqrt{ca} \ge b + \frac{2ca}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4491232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Analytical approach to this 2nd order ODE? Is there any analytical approach to solving an ODE of the form below? It could be more elegant than numerical approximation...
$$y''-\frac{a}{y^3}+b=0,\quad\textrm{where $a$ and $b$ are constants.}$$
If anyone has any pointers it would be greatly appreciated!
| Another option is
Solve
\begin{gather*}
\boxed{y^{\prime \prime}-\frac{a}{y^{3}}+b=0}
\end{gather*}
Writing the ode as
\begin{align*}
y^{\prime \prime}&=\frac{a -y^{3} b}{y^{3}}
\end{align*}
Multiplying both sides by $y^{\prime}$ gives
\begin{align*}
y^{\prime} y^{\prime \prime}&=\frac{\left(a -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4491652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $\int_{t}^1 \frac{\sin(\pi x)^2}{(\sin(\pi x/2)^2-a^2)^2} \ dx. $ Let $a \in (0,1)$, I would like to integrate
$$\int_{t}^1 \frac{\sin(\pi x)^2}{(\sin(\pi x/2)^2-a^2)^2} \ dx. $$
Now $\sin(\pi x/2)^2$ is a monotonically increasing function from $0$ to $1$, therefore there exists a unique $t^*$ such that $\sin(\pi ... | Use the result in this post:
Note that
\begin{align}
\frac{\frac14\sin^2\pi x}{\sin^2\frac{\pi x}2-a^2}
=-a^2 + \cos^2\frac{\pi x}2 + \frac{a^2(1-a^2)}{\sin^2\frac{\pi x}2-a^2}
\end{align}
Integrate respectively to obtain
\begin{align}
&\int_ 0^{x_{\downarrow}} \frac{\sin^2\pi x}{\sin^2\frac{\pi x}2-a^2}dx\\
=& \ 2(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4494665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}$ $$
\begin{align*}
&\text { Let } \mathrm{x}=2 \mathrm{y} \quad \because x \rightarrow 0 \quad \therefore \mathrm{y} \rightarrow 0\\
&\therefore \lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}\\
&=\lim _{y \rightarrow 0} \frac{12-6(2 y)^{... | Use $\cos z=1-z^2/2!+z^4/4!+...$
$$L=\lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}$$$$=\displaystyle\lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 (1-\frac{x^2}{2}+\frac{x^4}{24}+...)}{x^{4}}=-\frac{1}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
How to calculate $\sum _{n=1}^{\infty }\:\left(\frac{2}{3}\right)^n\sin\left(\frac{\pi }{3}n\right)$ I believe that Complex numbers should be used in order to calculate this.
Let $z = \frac{2}{3}e^{\frac{i\pi }{3}}$,
So,
$$\sum _{n=1}^{\infty }\left(\frac{2}{3}\right)^n\sin\left(\frac{\pi }{3}n\right)=\sum _{n=1}^{\inf... | It is also possible to do it directly in the real domain.
Since the series is absolutely convergent, we can rearrange terms to group the same values of the sinus.
*
*$n\equiv 0,3\pmod 6\implies \sin(\frac{n\pi}3)=0$
*$n\equiv 1,2\pmod 6\implies \sin(\frac{n\pi}3)=\frac 12\sqrt{3}$
*$n\equiv 4,5\pmod 6\implies \sin(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4498220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$\int_0^1 \frac{\arcsin x\arccos x}{x}dx$ Someone on Youtube posted a video solving this integral.
I can't find on Math.stack.exchange this integral using search engine https://approach0.xyz
It is related to $\displaystyle \int_0^\infty \frac{\arctan x\ln x}{1+x^2}dx$
Following is a solution that is not requiring the u... | Use the relation: $$\arcsin(x)+\arccos(x)=\frac{\pi}2$$
So,
$$J=\frac{\pi}2\int_0^1 \frac{\arcsin(x)}{x}dx-\int_0^1\frac{\arcsin^2(x)}{x} dx=\frac{\pi}2I_1-I_2$$
Let $t=\arcsin(x)$
$$I_1=\int_0^{\pi/2} \frac{t}{\tan(t)} dt,~~~~I_2=\int_0^{\pi/2} \frac{t^2}{\tan(t)}dt$$
For each integral use series:
$$\frac{1}{\tan(t)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4500073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
$a, b, c, d, e$ are distinct real numbers. Prove that this equation has distinctive 4 real roots.
$a, b, c, d, e$ are distinct real numbers. Prove that this equation has distinctive four real roots.
$$(x-a)(x-b)(x-c)(x-d) \\ +(x-a)(x-b)(x-c)(x-e) \\ +(x-a)(x-b)(x-d)(x-e) \\ +(x-a)(x-c)(x-d)(x-e) \\ + (x-b)(x-c)(x-d)... | This can also be approached without theorems from calculus, though much of it uses ideas from pre-calculus that lead into limits and the Intermediate Value Theorem. For this argument, we will use a simpler function with just three terms, $ \ g(x) \ = \ (x - a)·(x - b) + (x - a)·(x - c) + (x - b)·(x - c) \ \ , $ from ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4500296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find all $(p,n)$ such that $p^n+1\mid n^p+1$ where $p$ is prime. Let $p$ be a prime number. Find all $(p,n)$ such that $$p^n+1\mid n^p+1.$$
Where $n$ is a positive integer.
Here is some obvious solutions $(p,p)$ for any prime $p$ and $(2,4)$.
If $p=2$ then we get $2^n+1\mid n^2+1$ therefore $2^n\le n^2$ but for $n>4$ w... | If $p = 2$, then $n^2 + 1 \geq 2^n + 1$ is required (since $2^n + 1 \mid n^2 + 1$ and both sides are positive), which implies $n \leq 4$, which we can see only gives solutions $(p, n) = (2, 2), (2, 4)$.
Now if $p \neq 2$, then $p$ is an odd prime, and so $n$ must be odd since $p^n + 1$ is even. Note that $n \geq 3$ sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Summation of reciprocal products When studying summation of reciprocal products I found some interesting patterns.
$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)}=\frac{1}{1\cdot1!}-\frac{1}{1\cdot(N+1)}$$
$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)}=\frac{1}{2\cdot2!}-\frac{1}{2\cdot(N+1)(N+2)}$$
$$\sum_{k=1}^{N} \frac{1}{k\cdo... | I think I've posted this before,
but what the heck.
The telescoping occurs not only
for the product of consecutive integers
but for its reciprocal.
Here's the proofs.
$p_m(x)
=\prod_{k=0}^{m-1} (x+k)
$.
$\begin{array}\\
p_m(x+1)-p_m(x)
&=\prod_{k=0}^{m-1} (x+1+k)-\prod_{k=0}^{m-1} (x+k)\\
&=\prod_{k=1}^{m} (x+k)-\prod_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4506521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\lim_{n \to \infty} \frac{a_n}{2 ^ {n - 1}}$ if $a_n = a_{n - 1} + \sqrt{a_{n - 1}^2 + 1}$
Let $a_i (i \in \mathbb{N}_{0})$ be a sequence of real numbers such that $a_0 = 0$ and $$a_n = a_{n - 1} + \sqrt{a_{n - 1}^2 + 1} \text{ } \forall n \geq 1$$ Evaluate the limit $$\lim_{n \to \infty} \frac{a_n}{2 ^ {n -... | This is a general method that you may be able to use.
Suppose given a function
$$f(x) := x + \sqrt{x^2+1}. $$
Given a number $\,a_0\ge 0,\,$ define the sequence
$$a_n := f(a_{n-1}). $$
Note the series expansion $$ f(x) = 2x + \frac1{2x} - \frac1{8x^3} + \frac1{16x^5} + \cdots. $$
Iterate to get
$$ f^2(x) := f(f(x)) = 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4507788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
How many methods to tackle the integral $\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x ?$ $ \text{We are going to evaluate the integral}$
$\displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x \tag*{} \\$
by letting $ y=\frac{\pi}{4}-x. $ Then
$$\begin{aligned} \displaystyle ... | Using the famous tangent half-angle substitution,
$$I=\int \frac{\sin (x)+\cos (x)}{9+16 \sin (2 x)}\,dx=\int \frac{-2 t^2+4 t+2}{9 t^4-64 t^3+18 t^2+64 t+9}\,dt$$
$$\frac{-2 t^2+4 t+2}{9 t^4-64 t^3+18 t^2+64 t+9}=\frac{-2 t^2+4 t+2}{\left(t^2-8 t+9\right) \left(9 t^2+8 t+1\right)}=$$
$$\frac{4-t}{20 \left(t^2-8 t+9\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
If $\vec a,\vec b,\vec c$ be three vectors such that $|\vec a|=1,|\vec b|=2,|\vec c|=4$ and then find the value of $|2\vec a+3\vec b+4\vec c|$ If $\vec a,\vec b,\vec c$ be three vectors such that
$\vert \vec a\vert =1,\vert \vec b\vert =2,\vert \vec c\vert=4$
and
$\vec a \cdot \vec b+\vec b \cdot \vec c+\vec c \cdot\ve... | You have
$$2a + 3b +4c = 3(a+b+c) -a+c.$$ therefore
$$\begin{aligned}
\lVert 2a + 3b +4c \rVert^2 &= 9 \lVert a + b +c \rVert^2 + \lVert a \rVert^2 + \lVert c \rVert^2 - 6 \lVert a \rVert^2 - 6 a \cdot b - 6 a \cdot c + 6 \lVert c \rVert^2 + 6 b \cdot c + 6 a \cdot c\\
&=9 + 1 + 16 -6 + 96 - 6 a \cdot b + 6 a \cdot c\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Implicit Differentiation - $x^my^n = (x+y)^{m+n}$
Use implicit differentiation to find $\frac{\mbox{dy}}{\mbox{dx}}$ if $$x^my^n = (x+y)^{m+n}$$
Differentiation both sides with respect to $x$:
$$mx^{m-1}y^n + x^mny^{n-1}y' =(m+n)(x+y)^{m+n-1}(1 + y')$$
$$y' = \frac{(m+n)(x+y)^{m+n-1} - mx^{m-1}y^n}{x^mny^{n-1}-(m+n)(... | For any homogeneous relation of $x$ and $y$, we always have $$\frac{dy}{dx}=\frac{y}{x}.$$ Because the relationship is always satisfied by $y=vx$ meaning $$\frac{dy}{dx}=v=\frac{y}{x}.$$ The relation $$x^my^n=(x+y)^{m+n}$$ is a homogeneous function of degree $m+n$.
For $x^2+xy+y^2=0$ and $4x^2y+5y^3+6x^3-7xy^2=0.$
Afte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int_{0}^{\pi}\sqrt\frac{1+\cos2x}{x^2+1}dx$.
Evaluate$$\int_{0}^{\pi}\sqrt\frac{1+\cos2x}{x^2+1}\,\mathrm dx$$
We have $\sqrt{1+\cos2x}=\sqrt{2\cos^2x}=\sqrt{2} |\cos x|$. Then we need to solve $\int_{0}^{\pi}\frac{\sqrt{2} |\cos x|}{\sqrt{x^2+1}}dx$. Using symmetry of $\cos x$ from $0$ to $\pi$ the integr... | Just for the fun !
Being skeptical about a possible closed form, starting from
$$I=\sqrt2\left(\int_{0}^{\frac \pi 2} \frac{\cos (x)}{\sqrt {x^2+1}}\, dx-\int_{\frac \pi 2}^{\pi} \frac{\cos (x)}{\sqrt {x^2+1}}\, dx\right)$$ I rewrote it as
$$I=\sqrt2\left(\int_{0}^{\frac \pi 2} \frac{\cos (x)}{\sqrt {x^2+1}}\, dx+\int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4513126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Find the last two non-zero digits of $70!$ The question itself is quite straightforward, however, I am unable to get an exact answer to the problem. I have narrowed it down to four possibilities one from $\{18, 43, 68, 93\}$.
The approach
We begin by finding the number of zeros that are contained in $70!$. This is done... |
From here we will find the remainder of $N$ modulo $4$ and $25$
respectively so we can use the Chinese Remainder Theorem to finish the
problem.
$$ N \equiv 0 \pmod 4 $$
$$ N \equiv \frac{M}{2^{16}} \equiv \frac{23}{11} \pmod {25} $$
$$\color{red}{\text{THIS IS WHERE I GOT STUCK!}}$$
To simplify $\frac{23}{11} \pmod {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Find the sum of radicals without squaring, Is that impossible?
Find the summation:
$$\sqrt {3-\sqrt 5}+\sqrt {3+\sqrt 5}$$
My attempts:
\begin{align*}
&A = \sqrt{3-\sqrt{5}}+ \sqrt{3+\sqrt{5}}\\
\implies &A^2 = 3-\sqrt{5} + 3 + \sqrt{5} + 2\sqrt{9-5}\\
\implies &A^2 = 6+4 = 10\\
\implies &A = \sqrt{10}
\end{align*}
S... | Use Vieta's formulas:
Let, $$x_1=\sqrt {3-\sqrt 5},\,\,\,x_2=\sqrt {3+\sqrt 5}$$
$$x^2-px+2=0\\p=\frac{x^2+2}{x}$$
and we have,
$$p=\frac{5+\sqrt 5}{\sqrt {3+\sqrt 5}}=\frac{5-\sqrt 5}{\sqrt {3-\sqrt 5}}$$
Then, using the rule
$$p=\frac ab=\frac cd\implies p=\frac{a+c}{b+d}$$
We get
\begin{aligned}p&= \frac{5+\sqrt 5+5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 2
} |
Finding the equation of a plane given three points Below is a problem I did from a Calculus text book. My answer matches the
back of the book and I believe my answer is right. However, the method
I used is something I made up. That is, it is not the method described
in the text book.
Is my method correct?
Problem:
Find... | Yes, your method is perfectly fine, and nicely laid out, though at the very end in the original $x,y,z$ were accidentally replaced by $A,B,C$. :)
In terms of numerical computation, this is a reasonably efficient algorithm.
In terms of formulaic or abstract presentation, (conceivably what the book did), we realize that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4517005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How do we prove that the simple continued fraction for $e^{2/n}=[1;\frac{n-1}{2},6n,\frac{5n-1}{2},1,1,...]$? Motivation: remarkably, the simple continued fraction - which is unique - for $e^{1/n},e^{2/n}$ is known for every $n\in\Bbb N$. The expansions for $e^{3/n}$, or even $e^{p/q}$, are not known in general, and it... | If you have prior knowledge of the answer, you can figure it out. I don’t know how you might derive it “from first principles” though! I’m a little surprised by the lack of publicly available information about continued fractions - you’d think ones concerning notable values, such as $e$, would have well-documented expl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4519805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Find the derivative of $y=\sqrt{\ln\left(4x-x^2\right)}$ Find the derivative of $$y=\sqrt{\ln{\left(4x-x^2\right)}}$$
So we can rewrite the function as $$y=\left[\ln\left(4x-x^2\right)\right]^\frac12$$ Let's try to break it down a bit. So let's set $$a(x)=x^\frac12$$ and $$b(x)=\ln(4x-x^2)$$ then $$y=a(b(x))$$ The chai... | $y'(x) =\frac{(\ln{{(4x-x^2)}})'}{2\times\sqrt{\ln{{4(x-x^2)}}}}=\frac{\frac{(4x-x^2)'}{4x-x^2}}{2\times\sqrt{\ln{{(4x-x^2)}}}}=\frac{\frac{4-2x}{x(4-x)}}{2\times\sqrt{\ln{{(4x-x^2)}}}}=\frac{\frac{x-2}{x(x-4)}}{\sqrt{\ln{{(4x-x^2)}}}}$.
Now,
write $\frac{x-2}{x(x-4)}$ such that $\frac A{x}+\frac B{x-2}=\frac{A(x-4)+B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4520056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
If $g(x)=ax^2+bx+c$ and $f(x)= \begin{cases} {g(x)} & {x\ge k} \\ {g'(x)} & {x< k} \end{cases} $ , $\max (k)=?$ if $f$ is a differentiable function
If $g(x)=ax^2+bx+c$ and $f(x)= \begin{cases} {g(x)} & {x\ge k} \\
{g'(x)} & {x< k} \end{cases} $. If $f(x)$ is a differentiable
function, what is the maximum value of $k... | (A solution with no square roots needed.)
$g'(k) = g''(k)$ gives
$$ 2ak+b = 2a $$
If $a=0$ then this forces $b=0$, $0 = g'(k) = 2a$ so $a=0$, and finally $g(k)=c=0$. In the degenerate case where $f$ and $g$ are zero everywhere, $k$ can be any number at all. The problem should have eliminated this case, to have an answe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4523509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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How did people come up with the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$? Every resource that I've read proves the formula
$$ a^3 + b^3 = (a+b)(a^2-ab+b^2) \tag1$$
by just multiplying $(a+b)$ and $(a^2 - ab + b^2)$.
But how did people come up with that formula? Did they think like, "Oh, let's just multiply these polynomials... | $«$Who first factored the expression $a^3+b^3$ and what was the method used?$»$ - I don't know the exact answer to this question, but the similar question I'm trying to answer is:
$«$How can we factor $a^3+b^3$ using the most basic algebraic techniques?$»$
It seems to me that, this formula basically comes from the Bino... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Calculate $\sin^5\alpha-\cos^5\alpha$ if $\sin\alpha-\cos\alpha=\frac12$ Calculate $$\sin^5\alpha-\cos^5\alpha$$ if $\sin\alpha-\cos\alpha=\dfrac12$.
The main idea in problems like this is to write the expression that we need to calculate in terms of the given one (in this case we know $\sin\alpha-\cos\alpha=\frac12$).... | Another approach: Solve $ \sin{z}-\cos{z}=\frac{1}{2}$
This results in $e^{iz}=\frac{\sqrt{7}-i}{2(i-1)}$
Plugging this into the complex sin and cos definition results in:
$\sin{z}=\frac{1-1\sqrt{7}}{4}$ and $\cos{z}= \frac{-1-1\sqrt{7}}{4}$
When applying this in $sin^5z-cos^5z$ all uneven powers of $\sqrt(7)$ cancel o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 4
} |
Line of best fit for $\{(n,n+\sin n) : n \in \mathbb{Z}\}$ It seems intuitive that the line of best fit for $\{(n,n+\sin n) : n\in \mathbb{Z}\}$ should be $y=x$.
More concretely, it seems like a reasonable conjecture would be:
If $y = m_k x + b_k$ is the line of best fit for the set of points $$\{ (n,n+\sin n) : n\in ... | To start with, we're going to need a few identities:
$$\begin{eqnarray} \sum_{j = 1}^n j & = & \frac{n(n+1)}{2} \\
\sum_{j = 1}^n j^2 & = & \frac{n(n+1)(2n+1)}{6} \\
\sum_{j = 1}^n \sin j & = & \frac{\sin n - \sin (n+1) + \sin 1}{2(1 - \cos 1)} \\
\sum_{j = 1}^n j \sin j & = & \frac{(n + 1) \sin n - n \sin (n + 1)}{2(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4526759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Limit for $(x,y) \to (0,0)$ of $f(x,y)=\frac{x^2+y^4}{x}$ I tried to solve this limit, but I'm not sure whether it's correct. The candidate limit is $l=0$.
From $ | \frac{x^2+y^4}{x} | $, since $\sqrt{x^2-y^2} \leq \sqrt{x^2} = |x| $, we then have:
$$| \frac{x^2+y^4}{x} | \leq | \frac{x^2+y^4}{\sqrt{x^2-y^2}} | \leq ..... | I suggest consider curve $y=\sqrt[8]{x}=x^{\frac{1}{8}}$. Then we have on this curve
$$\frac{x^2+y^4}{x}=x + \frac{1}{\sqrt{x}}$$
limit of first summand exists, while second tends to infinity.
On another hand (accordingly comments) along the curve $y=x$ limit is obviously zero. Thus, we can postulate, that limit does n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4527261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\int_{0}^{\infty}\tan^{-1}\left(\frac{2x}{1+x^2}\right)\frac{x}{x^2+4}dx$ $$\int_{0}^{\infty}\tan^{-1}\left(\frac{2x}{1+x^2}\right)\frac{x}{x^2+4}dx$$
I am given solution for this definite integral is $\frac{\pi}{2}\left(\ln\frac{\sqrt2+3}{\sqrt2+1}\right)$. Any idea or approach you would use to solve this?
| Using integration by parts, we find
$$\int_0^\infty \tan^{-1}\left(\frac{2 x} {x^2 +1}\right)\frac{x}{x^2 +4}dx=\int_0^{\infty}\frac{(x^2-1)\ln(x^2+4)}{x^4+6x^2+1} dx.$$
Factoring the denominator as
$$x^4+6x^2+1=(x^2+3+2\sqrt{2})(x^2+3-2\sqrt{2})\\
=(x^2+(\sqrt{2}+1)^2)(x^2+(\sqrt{2}-1)^2)$$ and then applying the part... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4529185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Showing $\int_1^\infty\left(\sqrt{\sqrt{x}-\sqrt{x-1}}-\sqrt{\sqrt{x+1}-\sqrt{x}}\right)dx=\frac4{15}\left(\sqrt{26\sqrt2-14}-2\right)$ A Putnam problem asked to show that some improper integral is convergent, but I was curious to see if it can be computed in closed form and Mathematica came up with this:
$$\int_1^{\i... | This is a partial answer.
\begin{align}
a_n&=\int_n^{n+1} \left(\sqrt{\sqrt{x}-\sqrt{x-1}} -\sqrt{\sqrt{x+1}-\sqrt{x}} \right)\mathbb dx\\
&=\int_n^{n+1} \sqrt{\sqrt{x}-\sqrt{x-1}}\,\mathbb dx-\int_{n+1}^{n+2} \sqrt{\sqrt{x}-\sqrt{x-1}}\mathbb\,dx\\
\sum_{n=1}^{\infty}a_n&=\int_1^{2} \sqrt{\sqrt{x}-\sqrt{x-1}}\,\ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4529613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 0
} |
FOPDE: $2x z_x + 3y z_y = x + y$ Solving the First Order PDE:$$2x z_x + 3y z_y = x + y$$
for $c_1$ I get: $$ c_1 = \frac{x^3}{y^2} $$
I do not know how to go about solving for $c_2$
The correct general solution should be:
$$z = \frac{x}{2} + \frac{y}{3} + f\bigg(\frac{x^3}{y^2}\bigg)$$
How would one come to this calcu... | $$2x z_x + 3y z_y = x + y$$
Charpit-Lagrange characteristic ODEs :
$$\frac{dx}{2x}=\frac{dy}{3y}=\frac{dz}{x+y}$$
A first characteristic equation from solving $\frac{dx}{2x}=\frac{dy}{3y}$ that you found correctly :
$$\frac{x^3}{y^2}=c_1$$
A second characteristic equation from
$$\frac{dx}{2x}=\frac{dy}{3y}=\frac{dz}{x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4530604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there any closed form for the integral $\int_0^{\frac{\pi}{2}} \frac{\ln ^n(\sin x) \ln ^m(\cos x)}{\tan x} d x?$ Inspired by the question in the post, I started to generalise the integral
$$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \frac{\ln^n (\sin x) \ln (\cos x)}{\tan x} d x =& \frac{1}{n+1} \int_0^{\frac{\pi}{2}}... | After submitting the post, I suddenly recognise that $I(n,m) $ is actually a high derivative of a Beta function as
$$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \frac{\sin ^a x \cos ^b x}{\tan x} d x
&=\int_0^{\frac{\pi}{2}} \sin ^{a-1} x \cos ^{b+1} x d x
=\frac{1}{2} B\left(\frac{a}{2} , \frac{b}{2} +1\right)
\end{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4535182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Is there any natural number between $(N^2 + N)$ and $(N^2 + 1)$ that can divide $(N^2 + N)\times(N^2 + 1)$ My question is fairly simple and explained in title. I'm trying to prove that there are no natural number(s) between $(n^2+n)$ and $(n^2+1)$ that can divide $(n^2+1) \times (n^2 + n)$
[EDIT]
I was trying to solve ... | Let $N=(n^2+1)(n^2+n)$. Assume that there exists an integer $A$ such that $A\mid N$ and $(n^2+1)<A<(n^2+n)$. Then the complementary factor $B=N/A$ is also in that interval.
Because $A$ and $B$ are closer to each other than $n^2+1$ and $n^2+n$, and yet $AB=(n^2+1)(n^2+n)$, we must have
$A+B<(n^2+1)+(n^2+n),$ or $A+B\le ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4535406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Big-O analysis of a combination function Prove that $$\binom{2n}{n} \in O\left(4^n/\sqrt{n}\right) $$
I am not familiar with markdown syntax so sorry for the formula display
| In the comments you can find links to an argument using Stirling's approximation and an argument using an integral formula; here's a much more elementary argument (which gives a worse bound). Write $a_n = {2n \choose n}$. We have
$$\frac{a_{n+1}}{a_n} = \frac{(2n+2)! n!^2}{(2n)! (n+1)!^2} = \frac{4n+2}{n+1} = 4 - \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4536033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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prove that $\{\det(A) : A\in S\} = \{-4,-3,-2,-1,0,1,2,3,4\}$
Let $S$ be the set of all $3\times 3$ matrices with entries in $\{0,1,-1\}$. Prove that $\{\det(A) : A\in S\} = \{-4,-3,-2,-1,0,1,2,3,4\}$.
Let $R = [-4,4]\cap \mathbb{Z}$, $T =\{\det(A) : A\in S\}$. To show $R\subseteq T,$ it suffices to show that $R\cap ... | The main task is to prove that $T\subseteq R$.
$\det A$ is the sum of six elements of $\{0,1,-1\}$ but if some entry of $A$ is $0$, at least two of these six elements are $0$ and you have won.
If all entries of $A$ are $\pm1$ then $\det A$ is even and wlog (up to changing some rows to their opposite), $A = \begin{pmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4538325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Characterize all primes p such that 15 is a square modulo p I was having some difficulty understanding the following parts of this proof:
Proof:
Obviously, 15 is a square mod 2, 3, 5. So suppose p > 5. We compute the Jacobi
symbol:
($\frac{15}{p}$) = ($\frac{3}{p})(\frac{5}{p}$) = $(-1)^\frac{p-1}{2}(\frac{p}{3})(\frac... | This is a fairly standard exercise. As you note, using the Legendre (or Jacobi) symbol and Quadratic Reciprocity, we have
$$\begin{align*}
\left(\frac{15}{p}\right) &= \left(\frac{3}{p}\right)\left(\frac{5}{p}\right)\\
&=(-1)^{(\frac{p-1}{2})(\frac{3-1}{2})}\left(\frac{p}{3}\right)(-1)^{(\frac{p-1}{2})(\frac{5-1}{2})}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4547542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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A smarter (not bashy) way to solve this roots of unity problem? (Mandelbrot) Let $\xi = \cos \frac{2\pi}{5} + i \sin \frac{2pi}{5}$ be a complex fifth root of unity. Set $a = 20\xi^2 + 13 \xi, b = 20\xi^4 + 13\xi^2, c = 20\xi^3 + 13\xi^4, \text{and } d = 20\xi + 13\xi^3$. Find $a^3 + b^3 + c^3 + d^3$
Immediately what c... | Let $\varphi:\mathbb{Q}[\xi]\to\mathbb{Q}[\xi]$ be the field homomorphism which is the identity on $\mathbb{Q}$ and $\varphi(\xi)=\xi^2$.
Since $\xi+\xi^2+\xi^3+\xi^4=-1$, we have
$$
\begin{align}
a^3&=\left(20\xi^2+13\xi\right)^3&&=8000\xi+15600+10140\xi^4+2197\xi^3\\
b^3&=\varphi\!\left(a^3\right)&&=8000\xi^2+15600+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4549205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln(x)}\,dx = \ln\big(\frac mn \frac{n+p}{m+p} \frac{m+2p}{n+2p} \frac{n+3p}{m+3p}\dots\big)$ page 371 in ‘Synopsis Of Elementary Results In Pure Mathematics’ contains the following result,
$$\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln(x)}\,dx = \ln \left(\frac{m}{n} ... | I am going to evaluate the integral by differentiating its partner
$$
\begin{aligned}
I(a) &=\int_0^1 \frac{x^a-x^{n-1}}{\left(1+x^p\right) \ln x} dx\\
I^{\prime}(a) &=\int_0^1 \frac{x^a}{1+x^p} d x \\
&=\sum_{k=0}^{\infty} \frac{(-1)^k}{a+pk+1}
\end{aligned}
$$
Integrating back gives our integral
$$
\begin{aligned}
I ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4553487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Compute $I=\int_0^2(3x^2-3x+1)\cos(x^3-3x^2+4x-2)\,dx$ If the value of $I = \int\limits_0^2 {\left( {3{x^2} - 3x + 1} \right)\cos \left( {{x^3} - 3{x^2} + 4x - 2} \right)dx} $ can be expressed as $p(\sin q)$ , where $p,q\in \mathbb N$, then $p+q-1=$?
My approach is as follow ${x^3} - 3{x^2} + 4x - 2 = {x^3} - {x^2} - 2... | Let $y=x-1$, then $3x^2 - 3x + 1=3y^2+3y+1$, and $x^3-3x^2+4x-2=y^3+y$. Hence
$$I=\int_0^2(3x^2-3x+1)\cos(x^3-3x^2+4x-2)\,dx=\int_{-1}^1(3y^2+3y+1)\cos(y^3+y)\,dy.$$
Since $\frac{d}{dy}(y^3+y)=3y^2+1$, we have
$$\int_{-1}^1(3y^2+1)\cos(y^3+y)\,dy=\sin(y^3+y)\Big|_{-1}^1=2\sin 2.$$
Since $3y\cos(y^3+y)$ is an odd functi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4554753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Solve the equation $(x-1)^5+(x+3)^5=242(x+1)$ Solve the equation $$(x-1)^5+(x+3)^5=242(x+1)$$ My idea was to let $x+1=t$ and use the formula $$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4),$$ but I have troubles to implement it. The equation becomes $$(t-2)^5+(t+2)^5=242t\\(t-2+t+2)\left[(t-2)^4-(t-2)^3(t+2)+(t-2)^2(t+2)^2-\... | We have
$$0=(x-1)^5+(x+3)^5-242(x+1)=2(x^2 + 2x + 42)(x + 2)(x + 1)x
$$
by applying the rational root theorem to the polynomial equation
$$
2x^5 + 10x^4 + 100x^3 + 260x^2 + 168x=0,
$$
which yields the linear factors $x$, $x+1$ and $x+2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Number of strings of length $n$ with no consecutive $y$'s Suppose we have a set $S$ such that $\lvert S\rvert=k+1$. Fix an element $y$ in $S$. We want to find the recurrence relation on the number of $S$-strings of length $n$ that don't have two consecutive $y$'s, namely $yy$. Use $f(n)$ to denote the answer.
I already... | The empty string does not contain two consecutive $y$s, so $f(0) = 1$. None of the $k + 1$ strings of length $1$ contain two consecutive $y$s, so $f(1) = k + 1$. Every string of length $2$ is admissible except $yy$, so $f(2) = (k + 1)^2 - 1 = k^2 + 2k + 1 - 1 = k^2 + 2k$.
An admissible string of length $n \ge 2$ must... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4557916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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In how many ways can points be assigned to five questions with one to four points per question so that the total is $14$? A professor must write an exam with $5$ questions. Question number i should give $p_i \in \mathbb{Z}$ points. The sum of the points must be $14$ and each question must give at least $1$ point and a ... | If $p_i$ points are assigned to the $i$th question, with each $p_i$ a positive integer satisfying $1 \leq p_i \leq 4$, and the examination has a total of $14$ points, $1 \le i \le 5$, then
$$p_1 + p_2 + p_3 + p_4 + p_5 = 14 \tag{1}$$
is an equation in the positive integers subject to the restrictions $p_i \leq 4$, $1 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4558085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show $\sum_{n=0}^\infty (-1)^n\, \frac{\Gamma \left(\frac{n}{2}+1\right)}{n! \,\Gamma \left(2-\frac{n}{2}\right)}=\frac{3-\sqrt{5}}{2}$ The following sum was used in an unrelated answer on math.stackexchange.com.
$$\sum_{n=0}^\infty (-1)^n \frac{\Gamma \left(\frac{n}{2}+1\right)}{n! \Gamma \left(2-\frac{n}{2}\ri... | We will use the facts that
$$
\frac{1}{{\Gamma\! \left( {2 - \frac{n}{2}} \right)}} = \frac{{\Gamma \!\left( {\frac{n}{2} - 1} \right)}}{{\Gamma \!\left( {2 - \frac{n}{2}} \right)\Gamma\! \left( {\frac{n}{2} - 1} \right)}} = - \frac{1}{\pi }\sin \left( {\frac{{\pi n}}{2}} \right)\Gamma \!\left( {\frac{n}{2} - 1} \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4561897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Calculate $\sum_{n=2}^{\infty}\left (n^2 \ln (1-\frac{1}{n^2})+1\right)$ I am interested in evaluating
$$\sum_{n=2}^{\infty}\left (n^2 \ln\left(1-\frac{1}{n^2}\right)+1\right)$$
I am given the solution for the question is $\,\ln (\pi)-\frac{3}{2}\,.$
$$\sum_{n=2}^{\infty}\left(n^2\ln\left(\!1\!-\!\frac{1}{n^2}\!\right)... | (I have partially copied @Hamdiken's answer to make it more complete)
Define the sum
$$S(x):=\sum_{n=2}^{\infty}\left(n^2\ln\left(1-\frac{x^2}{n^2}\right)+x^2\right)$$
and differentiate with respect to $x$ to obtain
\begin{align*}
S'(x)=\sum_{n=2}^{\infty}\left(-2x{\frac{n^2}{n^2-x^2}}+2x\right)
&=2x\sum_{n=2}^{\infty}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4565763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 3
} |
Better approach to evaluate the limit $\lim_{x\to0^+}(\cot x-\frac{1}{x})(\cot x+\frac{1}{x})$ I solved it by rewriting the limit as indeterminate form $0/0$, then apply L'Hopital's rule 4 times, It was really lengthy and easy to make mistakes, If anyone got a better approach, please tell me!
$$
\begin{align}
\lim_{x\t... | Using $$\tan(x)=x+\frac{x^3}{3}+o(x^3)\quad \text{and}\quad \frac{1}{1+x}=1-x+o(x),$$ yields
\begin{align*}
\left(\cot(x)-\frac{1}{x}\right)\left(\cot(x)+\frac{1}{x}\right)&=\cot^2(x)-\frac{1}{x^2}\\
&=\frac{1}{\tan^2(x)}-\frac{1}{x^2}\\
&=\frac{1}{x^2+\frac{2}{3}x^4+o(x^4)}-\frac{1}{x^2}\\
&=\frac{1}{x^2}\left(\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4567326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
On an infinite harmonic series. Let $n \in \mathbb{N}$ and let $\mathcal{H}_m$ denote the $m$-th harmonic number. Evaluate in a closed form the sum
$$\mathcal{S}_n = \sum_{m=1}^{\infty} \mathcal{H}_m \left ( \frac{1}{m+1} - \frac{1}{m+n+2} \right )$$
I do not know how to tackle this. My instict tells me that there shou... | It seems I got preempted by a comment, but since nobody has posted an answer, I see no reason to not post my own answer, though it does use a very similar method.
First, we let
$$T_n=\frac{1}{2}[H_{n+1}^2+H^{(2)}_{n+1}]$$
We see that for $n\geq 0$,
\begin{equation}
\begin{split}
T_n-T_{n-1}&=\frac{1}{2}[H_{n+1}^2-H_n^2... | {
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"answer_count": 1,
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Prove that $\frac{\ln x}{x}+\frac{1}{e^x}<\frac{1}{2}$ for $x > 0$
Given $x>0$ , prove that $$\frac{\ln x}{x}+\frac{1}{e^x}<\frac{1}{2}$$
I have tried to construct $F(x)=\frac{\ln x}{x}+\frac{1}{e^x}$ and find the derivative function of $F(x)$ to find the maximun value, but I can't solve the transcendental equation.
... | Let
\begin{align*}
f(x) &= e^{-x} + \frac{\log (x)}{x}, \\
f'(x) &= -e^{-x} + \frac{1 - \log (x)}{x^2}, \\
f''(x) &= e^{-x} + \frac{2 \log (x) - 3}{x^3}.
\end{align*}
We would like to show that $f(x) < 1/2$ for all $x > 0$.
Case 1. $0 < x \le 1$
Using the inequalities $e^x > x + 1$ and $\log(x) \le x - 1$,
$$ \begin{sp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4573486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
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short weierstrass form for cubic I have some question about the derivation of the short Weierstrass form. In https://www.staff.uni-mainz.de/dfesti/EllipticCurvesNotes.pdf this note, I follow the derivation till the point $y^2 = β_0x^3 + β_1x^2 + β_2x + β_3$, but then it says that using the transformation $x' = x + β_1/... | You can get rid of the $\beta_0$ by multiplying both sides by $\beta_0^2$ and then letting $u = \beta_0 x$ and $v = \beta_0 y$. So in your example $y^2 = 5x^3+x+1$, we have
\begin{align*}
y^2 &= 5x^3+x+1\\
5^2 y^2 &= 5^3 x^3 + 5^2 x + 5^2\\
(5y)^2 &= (5x)^3 + 5(5x) + 25\\
v^2 &= u^3 + 5 u + 25 \, .
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/4574929",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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What is the value of $\int_{-1}^{2} 4 \, x^2 \, (2+x-x^2) \, dx$? $$\int_{-1}^{2} 4 \, x^2 \, (2+x-x^2) \, dx$$
*
*When I solved this took the constant out of the integral and then multiplied $x^2$
by the bracket and evaluated the integral to get:
$$4 \, \left[ \frac{x^4}{4} +\frac{2x^3}{3}-\frac{x^5}{5} \right]^2_{-... | The first step should be simplifying the integral, like this:
$$\displaystyle \int -4x^{4} + 4x^{3} + 8x^2 \,dx$$
$$-4\int x^4dx + 4\int x^3 dx + 8\int x^2 dx$$
use the power rule, $$\int x^n \,dx = \frac{x^{n+1}}{n+1}$$
and apply to the problem
$$-4 \times\frac{x^5}{5} + 4\times \frac{x^4}{4} + 8 \times\frac{x^3}{3}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4575117",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How is $\ln (x-2) - \frac{1}{2} \ln (x-1) = \frac{1}{2} \ln \frac{(x-2)^2}{x-1}$
How is $\ln (x-2) - \frac{1}{2} \ln (x-1) = \frac{1}{2} \ln \frac{(x-2)^2}{x-1}$
Can someone enlighten me on how is these 2 actually equals and the steps taken? the left hand side is actually the answer for $\int \frac{x}{2 (x-2)(x-1)} d... | HINT:
Multiply both side by $2.$ You need to show that:
$$2 \ln (x-2) -2 \cdot \frac{1}{2} \ln (x-1) = 2\cdot\frac{1}{2} \ln \frac{(x-2)^2}{x-1}$$
Use the rule
$$ \ln x - \ln y= \ln~(x/ y)$$
Next get rid of $\ln ..$ ( by exponentiation of both sides). What went before integration is not asked, but only about last step ... | {
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"question_score": "3",
"answer_count": 4,
"answer_id": 3
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How do you solve these kinds of system of equations? Recently, I came across this system of equations,
\begin{align}
\begin{cases} \dfrac{2x}{p(p+4)}=\dfrac{2x}{q(q-4)}=3 & (1) \\ \dfrac{x}{p+q} = 4 & (2) \end{cases}
\end{align}
or
\begin{align}
\begin{cases} \dfrac{2x}{p(p+4)}=\dfrac{2x}{q(q-4)}=3 & (1) \\ \dfrac{y}{p... | Consider the first set of fractions:
\begin{align}
\frac{2 \, x}{p (p+4)} &= \frac{2 \, x}{q (q-4)} = 3 \\
\frac{x}{p+q} &= 4
\end{align}
it would seem that the relations lead to an equation for $p$ and $q$. This follows from solving for $x$ in the second equation, $x = 4 \, (p+q)$, and using it in the first which give... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4579774",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find limit of trigonometric function with indeterminacy Find limit of the given function:
$$\lim_{x\rightarrow0} \frac{(4^{\arcsin(x^2)} - 1)(\sqrt[10]{1 - \arctan(3x^2)} - 1)}{(1-\cos\tan6x)\ln(1-\sqrt{\sin x^2})}
$$
I tried putting 0 instead of x
$$\lim_{x\rightarrow0} \frac{(4^{\arcsin(x^2)} - 1)(\sqrt[10]{1 - \arct... | $$\lim_{x\to 0} \frac{(4^{\arcsin(x^2)} - 1)(\sqrt[10]{1 - \arctan(3x^2)} - 1)}{(1-\cos\tan6x)\ln(1-\sqrt{\sin x^2})}
$$
$4^{\arcsin(x^2)} - 1=e^{\arcsin(x^2) \log 4}-1\sim x^2\log 4 $
$\sqrt[10]{1 - \arctan(3x^2)} - 1\sim 1-\dfrac{1}{10}\, \arctan(3x^2)-1\sim -\dfrac{3}{10}x^2 $
$1-\cos\tan6x \sim 1-1+\dfrac{\tan^2 6x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4582099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Prove $^{6n+2} − ^{6n+1} + 1$ is always divisible by $^2 − + 1$; = 1, 2, 3,... How can we prove that
$^{6n+2} − ^{6n+1} + 1$ is always divisible by $^2 − + 1$; = 1, 2, 3,...
I attempted to solve this with Mathematical Induction as follows:
Let s(n) = $x^2 - x + 1$ | $^{6n+2} − ^{6n+1} + 1$; = 1, 2, 3,..
Basic Ste... | Define the polynomial
$$P_n(x):=x^{6n+2}−x^{6n+1} + 1$$
Observe that, $x=-1$ is not a root of $x^2-x+1=0$, then multiplying both sides of the equation by $(x+1)$, yields:
$$(x+1)(x^2-x+1)=0$$
This implies that, $x^3= -1,\;x≠-1$.
Therefore, making $x^3\equiv -1$, by $\mod x^2-x+1$, we have:
$$
\begin{align}P_n(x)&\equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4584894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Equation has exactly one real solution I am trying to show that the equation
$$
4x^2y^4+12x^2y^2+4x^2+4xy^2+4x+1=0
$$
has exactly one real solution $x,y$ and to determine it.
The first observation is: If $y=0$, we are left with
$$
4x^2+4x+1=0
$$
which is solved by $x=-1/2$.
Thus a real solution is given by
$$
x=-\frac{... | I do not know if it helps, but you can rearrange the proposed equation as follows:
\begin{align*}
4x^{2}y^{4} + 12x^{2}y^{2} + 4x^{2} + 4xy^{2} + 4x + 1 = 0 & \Longleftrightarrow (4x^{2}y^{4} + 4xy^{2} + 1) + (12x^{2}y^{2} + 4x^{2} + 4x) = 0\\\\
& \Longleftrightarrow (2xy^{2} + 1)^{2} + 12x^{2}y^{2} + (4x^{2} + 4x + 1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4585033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Cyclic inequality. $\sum_{cyc} \frac{x+2y}{\sqrt{z(x+2y+3z)}}\geq \frac{3\sqrt{6}}{2}$ Let $x,y,z>0$. Show that
$$\sum_{cyc} \frac{x+2y}{\sqrt{z(x+2y+3z)}}\geq \frac{3\sqrt{6}}{2}$$
Equality case is for $x=y=z$.
A hint I have is that I have to amplify with something convenient, and then apply some sort of mean inequali... | By AM-GM and Cauchy-Schwarz we obtain:
$$\sum_{cyc}\frac{x+2y}{\sqrt{z(x+2y+3z}}=\sum_{cyc}\frac{2\sqrt{6}(x+2y)}{2\sqrt{6z(x+2y+3z})}\geq2\sqrt6\sum_{cyc}\frac{x+2y}{x+2y+9z}=$$
$$=2\sqrt6\sum_{cyc}\frac{(x+2y)^2}{(x+2y)^2+9(x+2y)z}\geq2\sqrt6\cdot\frac{\left(\sum\limits_{cyc}(x+2y)\right)^2}{\sum\limits_{cyc}((x+2y)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4586365",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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The parabola with the equation $y=-x^2+4x+8$ is shifted so that it passes through the points (1,1) and (3,5). Find the equation of the new parabola. Given the points $(1,1)$ and $(3,5)$, the vertex form would be:
$1=(1-h)^2+k$ for $(1,1)$ and
$5=(3-h)^2+k$ for $(3,5)$.
With a system of equations, I obtain that $h = ... | Here's a slightly different approach.
Assuming that a "shift" strictly means a translation, then the vector between points $A = (1, 1)$ and $B = (3, 5)$, which is $\begin{pmatrix} 2 \\ 4 \end{pmatrix}$ is preserved. Thus the following system of equations holds:
$$\begin{cases} b = -a^2 + 4a + 8 \\ b + 4 = -(a + 2)^2 +... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Evaluate $\int\frac{1}{x-\sqrt{1-x^2}}dx$ The integral $I$ in question is defined as follows
$$
I \equiv \int\frac{1}{x-\sqrt{1-x^2}}dx
$$
To solve this, I tried the trig substitution $x = \sin\theta$, with $dx = \cos\theta d\theta$, and rewrote the integral as follows
$$
\int\frac{\cos\theta}{\sin\theta-\sqrt{1-\sin^2... | You could also proceed in the following way.
By letting $\;t=x-\sqrt{1-x^2}\;,\;$ we get that
$x=\dfrac12\left(t\pm\sqrt{2-t^2}\right)\;,\quad\mathrm dx=\dfrac12\left(\!\!1\mp\dfrac t{\sqrt{2-t^2}}\!\!\right)\mathrm dt\;\;.$
$\displaystyle\int\frac1{x-\sqrt{1-x^2}}\,\mathrm dx=\frac12\!\int\frac1t\left(\!\!1\mp\frac t{... | {
"language": "en",
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"source": "stackexchange",
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"answer_count": 3,
"answer_id": 1
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A closed form expression of $\sum_{n \ge 0} \biggl( \sum_{k=1}^n \frac{1}{k} \biggr)z^n$
I am working on the following exercise: Use the identity $\frac{1}{1-z} = \sum_{n \ge 0} z^n$ and elementary operations on power series (addition, multiplication, integration, differentiation) to find closed form expressions of th... | Expanding the sum: $$\sum_{n\ge 0}H(n)z^n=z+\left(1+\dfrac{1}{2}\right)z^2+\left(1+\dfrac{1}{2}+\dfrac{1}{3}\right)z^3+...=\frac{z}{1-z}+\frac{1}{2}z^2+\left(\dfrac{1}{2}+\dfrac{1}{3}\right)z^3+...=\frac{z}{1-z}+\frac{z^2}{2(1-z)}+...$$Where $H(n)$ is the $n$th harmonic number. Using induction, you could prove that: $$... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Plot of function $(x^\frac{1}{2} + y^\frac{1}{2})^2 = 5$ Can someone explain me how can we make a plot of function:
$$ (x^\frac{1}{2} + y^\frac{1}{2})^2 = 5 $$
I tried to find a first derivative and a second derivative to understand something about this plot of function but it was useless as I understand. What should I... | Since the equation implies $\sqrt{x}+\sqrt{y}\geq 0$, therefore it is equivalent to $\sqrt{x}+\sqrt{y}=\sqrt{5}$. Therefore because $y\geq 0$, $y=(\sqrt{5}-\sqrt{x})^2=x+5-2 \sqrt{5}\cdot \sqrt{x}$. And we know from the condition of $\sqrt{x}, \sqrt{y}\geq 0$ that $0\leq x, y\leq 5$. Therefore, finally, the original st... | {
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"url": "https://math.stackexchange.com/questions/4595481",
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Show that $b^2+c^2-a^2\leq bc$. Let $a,b,c>0$ such that $b<\sqrt{ac}$, $c<\frac{2ab}{a+b}$. Show that $b^2+c^2-a^2\leq bc$.
I tried to construct a triangle with $a,b,c$ and to apply The cosine rule, but I am not sure that it's possible to construct it and also I have no idea how to prove that an angle it's greater than... | Another solution.
First, we find $c<\frac{2ab}{a+b}\leq \frac{2\left(\frac{a+b}{2}\right)^2}{a+b}=\frac{a+b}{2}$ by applying AM-GM to $ab$. We also find $b<\sqrt{ac}\leq \frac{a+c}{2}$.
Now if $(b+c)\leq a$, this implies $(b+c)^2\leq a^2 \Rightarrow b^2+c^2-a^2\leq -2bc<bc$, and we are done.
If not, then $b+c>a$ we'll ... | {
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"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Inequality, $\frac{u+v}{2}-\frac{1}{4}$ versus $uv$
On the unit square, where is $f(u,v) = uv$ greater than $g(u,v) = \frac{u+v}{2}-\frac{1}{4}$?
I know that
*
*$f \left( \frac{1}{2} , \frac{1}{2}\right ) = \frac{1}{4} = g \left( \frac{1}{2} , \frac{1}{2}\right ) $
From playing with numbers,
*
*on $\left(0, \f... | $$\begin{align}
f(u, v) - g(u, v) &= uv - \left(\frac{u+v}{2} - \frac14\right) \\
&= uv - \frac u2 - \frac v2 + \frac 14 \\
&= \left(u - \frac12\right)\left(v - \frac12\right)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4600323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How do we prove without calculus that $\forall \ x,y \ge 0$, we have $\ 1+x^3+y^3\ge x+x^2y+y^2$ I've been trying to prove an inequality I was given by a friend, but so far my only progress has been calculus bashing:
$$LHS \ge RHS\iff1+x^3+y^3 - x-x^2y-y^2 \ge0$$
Letting $f(x,y) = 1+x^3+y^3 - x-x^2y-y^2$, we want $\fra... | Note that
$$
\begin{align}
\frac{1}{3}(1 + 1+x^3) &\ge x \\
\frac{1}{3}(x^3 + x^3 + y^3) &\ge x^2y \\
\frac{1}{3}(1 + y^3 + y^3) &\ge y^2
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4602946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of $~\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta$ $$
I:=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta
$$
My tries
$$\begin{align}
s&:=\sin\theta\\
c&:=\cos\theta\\
I&=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\the... | $$
\begin{aligned}
\int_0^{2 \pi} \frac{\cos ^2 \theta-\sin ^2 \theta}{\sin ^4 \theta+\cos ^4 \theta} d \theta
= & \int_0^{2 \pi} \frac{\cos 2 \theta}{\left(\sin ^2 \theta+\cos ^2 \theta\right)^2-4 \sin ^2 \theta \cos ^2 \theta} d \theta \\
= & \int_0^{2 \pi} \frac{\cos 2 \theta}{1-\sin ^2 2 \theta} d \theta \\
= & \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4603090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
Finding the angles of a right triangle if $\frac{\text{area of triangle}}{\text{area of incircle}}=\frac{2\sqrt{3}+3}{\pi}$
Let $ABC$ be a right triangle with $\measuredangle ACB=90^\circ$. If $k(O;r)$ is the incircle of the triangle and
$$\dfrac{S_{ABC}}{S_k}=\dfrac{2\sqrt3+3}{\pi}$$ find the angles of the triangle.
... | Rewrite $$\dfrac{\sin\alpha+\cos\alpha+1}{\sin\alpha+\cos\alpha-1}$$
as
$$\dfrac{2}{\sin\alpha+\cos\alpha-1}+1.$$
Then
$$\dfrac{2}{\sin\alpha+\cos\alpha-1}+1=2\sqrt3+3\implies \dfrac{1}{\sin\alpha+\cos\alpha-1}=\sqrt3+1.$$
Rearranging, we get
$$\sin\alpha + \cos\alpha =\frac{1}{\sqrt{3}+1}+1=\frac{\sqrt{3}-1}{2}+1=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4603334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Contest Math Question: simplifying logarithm expression further I am working on AoPS Vol. 2 exercises in Chapter 1 and attempting to solve the below problem:
Given that $\log_{4n} 40\sqrt{3} = \log_{3n} 45$, find $n^3$ (MA$\Theta$ 1991).
My approach is to isolate $n$ and then cube it. Observe:
\begin{align*}
\frac{\l... | In this answer, we provide another method that uses the definition and basic properties of logarithm.
The original equation states :
$$
\begin{align}&\log_{4n}{40}\sqrt{3}=\log_{3n}{45}\\
\implies &{40}\sqrt 3=\left(4n\right)^{\log_{3n}45}\\
&\thinspace\thinspace\thinspace\thinspace \thinspace\thinspace\thinspace \thin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4610313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$ Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$.
My effort: I tried using the fact $9^{\sqrt{2}}<9^{1.5}=27.$
Also We have $512 <729 \Rightarrow 2^9<27^2 \Rightarrow 2^{\frac{9}{2}}<27 \Rightarrow \sqrt{2}^9=2^{4.5}<27$. But both are below $27$.
| It boils down to the comparison of some powers of $2$ and $3$, if I didn't do any mistake, in the following way: $9^{\sqrt{2}}<\sqrt{2}^9$ if $9<2^{\frac{9}{2\sqrt{2}}}$, by using the fact that $\sqrt{2}<1.415$, if $9<2^{\frac{9}{2\times 1.415}}$ if $9<2^{3.18}$ if $3^{100}<2^{159}$ which is true since:
*
*$2^{159}>\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 1
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residue $\frac{z}{(z-1)(z+1)^2}$ at $z = -1$ My attempt: $$f(z) = \frac{z}{(z-1)(z+1)^2} = \frac{1}{(z+1)^2}\cdot -z\frac{1}{1-z}$$ So at $D(-1;1)$ we have $$-z\frac{1}{1-z} = \sum_{k=0}^\infty-(z+1)^{k+1}$$ and this gives $$f(z) = \sum_{k=-1}^\infty-(z+1)^{k}.$$ So the residue at $z = -1$ is negative one but I feel li... | For each $z\in\Bbb C$ such that $|z+1|<2$, you have\begin{align}\frac z{z-1}&=1+\frac1{z-1}\\&=1-\frac1{2-(z+1)}\\&=1-\frac12\cdot\frac1{1-\frac{z+1}2}\\&=1-\sum_{n=0}^\infty\frac{(z+1)^n}{2^{n+1}}\\&=\frac12-\sum_{n=1}^\infty\frac{(z+1)^n}{2^{n+1}}\end{align}and therefore, if $z\ne-1$,\begin{align}\frac z{(z-1)(z+1)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4613060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Prove $\displaystyle\frac{H(x^2)}{H(x)}$ increases. For $x\in[0,1]$, let $f(x):=-x\ln x$ and the two-sample entropy function $H(x)=f(x)+f(1-x)$. Prove $h(x):=\displaystyle\frac{H(x^2)}{H(x)}$ increases.
Here is my proof which is a bit cumbersome. I am seeking a much more elegant approach.
The numerator of the derivati... | Let
$$u = \ln x, \quad v = \ln(1-x), \quad w = \ln(1 + x).$$
We have
\begin{align*}
&[H(x)]^2h'(x)\\[6pt]
=\,& [(1-x)^2v - (1+x^2)u]w + 2u^2x^2 - (1-x)(1-3x)uv + (1-x)^2v^2\\[6pt]
\ge\,& [(1-x)^2v - (1+x^2)u]\cdot (x - x^2/2)\\[6pt]
&\qquad + 2u^2x^2 - (1-x)(1-3x)uv + (1-x)^2v^2 \tag{1}\\[6pt]
=\,& 2u^2x^2 - [x(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4613282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 1
} |
Limit of $\lim_{(x,y)\to (0,0)}\frac{3x^3y^2+xy^4}{(x^2+y^2)^2}$ I've run across a particular limit in my multivariable calculus class, that being:
\begin{equation}
\lim_{(x,y)\to (0,0)}\frac{3x^3y^2+xy^4}{(x^2+y^2)^2}.
\end{equation}
I'm having quite a bit of trouble finding this limit using the squeeze theorem. Using... | Using polar coordinates is perfectly alright and does not require any special knowledge. The formal argument can be presented as follows:
Let $\varepsilon > 0$. Take $\delta = \varepsilon / 4$ and assume $|(x,y)| = (x^2+y^2)^{1/2} < \delta$. We can always find $r, \theta$ such that $x = r\cos\theta $ and $y = r\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4613396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to solve this functional equation and find the integral of $f(x)$? How to solve this equation and find the integral from $0$ to $1$ of $f(x)$ if
$$f(1-x)=6x^2f(x^3)-\dfrac{6}{\sqrt{3x+1}}?$$
I've tried take integration for two sides of the equation after switching $f$ to one side and $x$ to one side. After that I u... | Substituting $t$ = $x^3$, we get
$\int{6 x^2 f(x) dx} = \int{2 f(t) dt}$ --- (1)
Using $\int_{a}^{b}f(x)dx = \int_{a}^{b}f(b+a-x)dx\quad$, we get
$\int_{0}^{1}f(x)dx = \int_{0}^{1}f(1-x)dx\quad$ --- (2)
Using (1),(2) and integrating the equation in the question from 0 to 1, we get
$\int_{0}^{1}f(x)dx = 2 \int_{0}^{1}f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4613660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\lim\limits_{x\to \infty}\frac{3x^3+x^2+\sin(e^x)}{5x-8x^3+\arctan(\log x)} $ Compute the limit:
$$\lim_{x\to \infty}\frac{3x^3+x^2+\sin(e^x)}{5x-8x^3+\arctan(\log x)} $$
I worked out this to be $\frac{-3}{8}$. I believe this is correct. I used the sandwich theorem but my issue is that the denominator is sometimes neg... | Your method is correct, but you can find the limit instantly after reading the text reasoning in term of asymptotic behaviour.
$$\lim_{x\to \infty}\frac{3x^3+x^2+\sin(e^x)}{5x-8x^3+\arctan(\log x)} $$
Note that both $\sin e^x$ and $\tan^{-1}(\log x)$ are bounded, so when $x\to +\infty$ larger powers dominate:
$$\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4614829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $\frac{a+b}{10}$
If $\sin x+\cos x+\tan x+\cot x+\sec x +\csc x=7$, then assume that $\sin(2x)=a-b\sqrt7$, where $a$ and $b$ are rational numbers. Then find the value of $\frac{a+b}{10}$.
How to solve these kind of problems. I can make substitutions and convert all of them to $\sin$ and then solve f... | $\displaystyle \sin x+\cos x+\tan x+\cot x+\sec x+\text{cosec } x=7$
$\iff\displaystyle \sin x+\cos x+\frac{\sin x}{\cos x} +\frac{\cos x}{\sin x} +\frac{1}{\cos x} +\frac{1}{\sin x} =7$
$\displaystyle ( \sin x+\cos x)\left( 1+\frac{1}{\sin x\cos x}\right) +\frac{1}{\sin x \cos x} =7$
$\iff\displaystyle ( \sin x+\cos x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4615546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Composing function The function is $f: \Bbb{R}\rightarrow\Bbb{R}$ defined as $f(x)= 2/ (x -3).$
I need to find $(f o f)(1).$
I would like to ask which of the following answers are the right one for writing this function.
$( f o f) ( 1 ) = ( f ( f ( 1 ) ) )= ( f ( 2/ 1 - 3) = ( 2 / 1 - 3 - 2 ) = -1/2 $
or
$( f o f ) ( 1... | If $f(x) = \frac{2}{x} -3 $, then
\begin{align*}
(f\circ f)(x) = f(f(x)) = \frac{2}{f(x)}-3 = \frac{2}{\frac{2}{x}-3}-3
\end{align*}
And thus
\begin{align*}
(f\circ f)(1) = \frac{2}{\frac{2}{1}-3}-3 = \frac{2}{-1}-3 = -5
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4618947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Solve the diophantine equation involving floor function ${\text{Solve for } \forall n \in \mathbb{N}, n \in (1,10^9) \text{ and primes } p \text{ that hold this equation:}}$
$${\sqrt{\lfloor \sqrt{ n^2 }\rfloor+\lfloor \sqrt{ n^2+1 }\rfloor+\lfloor \sqrt{ n^2+2 }\rfloor} = p}$$
Since ${ \forall k \in \mathbb{N}:}$
${0<... | You've made a good start, but there's a simpler way to show there are no solutions for most of your cases. First, your equation is
$$\sqrt{\lfloor\sqrt{n^2}\rfloor+\lfloor\sqrt{n^2+1}\rfloor+\lfloor\sqrt{n^2+2}\rfloor} = p \tag{1}\label{eq1A}$$
With the case of $(n,n,n)$, you've found only $p = 3$ is possible, with $n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4619623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
I came across this fun pattern, can anyone give me a proof for this? The Pattern
$\begin{align*}1(8)&=(3^2)-1\\
2(8)&= (3+1)^2 \\
3(8)&= (3+2)^2-1\\
4(8)&= (3+3)^2 - 4\\
5(8)&= (3+4)^2- 9\\
6(8)&= (3+5)^2-16\end{align*}$
Conjecture
I think the pattern is that the numbers appear in the following form:
$$8(n+1) = (3+n)^2... | If you are just learning algebra, this might not be obvious. But, the rule to keep in mind is that multiplication distributes over addition. On the left-hand side, you have $8(n+1)$ which after distribution equals $8n+8.$ Now we have to show that the right-hand side gives the same thing.
Let's work out the two terms... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Can a "sum of two squares" solution be the average of two other solutions? If $n = a^2 + b^2$ and $n$ has more than two unique solutions for $a$ and $b$ (where unique means that $a$ and $b$ are unsigned and unordered), then is it possible, impossible, or unknown whether one solution can be the average of two other solu... | Observations towards a solution
*
*$ s^2 + x^2 = 2j^2$ has parametrized solutions for integers $a, b, k$ of
$$ s = k(a^2 + 2ab -b^2), x =k( -a^2 + 2ab + b^2), j = k(a^2 + b^2).$$
*
*This can be algebraically verified.
*In the event that the values are negative, just take the absolute value.
*Note: This is a compl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4624334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Let $A $ and $B$ be real symmetric matrices of order $n$, satisfying $A^2B=ABA.$ Proof that $AB=BA.$ Let $A $ and $B$ be real symmetric matrices of order $n$, satisfying $A^2B=ABA.$ Proof that $AB=BA.$
I only find that $A^2B=BA^2=ABA.$ Then I don't know what to do next.Can you help me solve this puzzle?
| If $A$ is invertible or $A=0$ then it is obvious. Assume now that $A\ne 0$ and $\det A=0$.
$A$ is symmetric and hence diagonalizabe by an orthnormal matrix $U$. Set $A=U^{-1}LU$, where$L$ is diagonal. Expressing $L$ and $\hat B=UBU^{-1}$ (which is also symmetric) in block form,
$$
L=\left(
\begin{array}{cc}D & 0 \\ 0 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4624706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\int_0^{\frac{\pi}{2}}x^2 \cot x\ln(1-\sin x)\mathrm{d}x$ I was able to find
$$\int_0^{\frac{\pi}{2}}x^2 \cot x\ln(1-\sin x)\mathrm{d}x=-\frac14\sum_{n=1}^\infty\frac{4^n}{{2n\choose n}}\frac{H_{2n}}{n^3}$$
$$=5\operatorname{Li}_4\left(\frac12\right)-\frac{65}{32}\zeta(4)-2\ln^2(2)\zeta(2)+\frac5{24}\ln^4(2... | Here is one way to break up the integral
\begin{align}
I=\int_0^{\frac{\pi}{2}}x^2 \cot x\ln(1-\sin x)\ dx
=I_1-2(I_2 -I_3)
\end{align}
where, with $\ln(1-\sin x)=\ln (\cos x) -2\tanh^{-1}(\tan\frac x2 )$ and $ \cot x = \csc x-\tan\frac x2$
\begin{align}
I_1=& \int_0^{\frac{\pi}{2}}x^2 \cot x \ln(\cos x)dx\\
=&\ \text... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4625939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.