Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Complex analysis computation of $Im(f(z)-f(z_0))$ I have a computation that I wanted to carry out and know the answer. My issue is that when I did the computation I did not arrive at the answer that I know to be true. I would really appreciate a fresh set of eyes to comment on my work.
For $f(z)= kz- \sqrt{z}$ we want ... | A mistake I see is that the imaginary part of $r^{1/2}e^{\theta/2}$ is $r^{1/2}\sin\theta/2$ (and not cosine). The change is minor, though: your last equality changes to
$$
k^2y^2= \frac{1}{2}(\sqrt{y^2+x^2} -x).
$$
So
$$
\sqrt{x^2+y^2}=x+2k^2y^2.
$$
Squaring,
$$
x^2+y^2=x^2+4xk^2y^2+4k^4y^4,
$$
so, after subtracting... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Problem in partial fraction decomposition Consider the following rational function which we want to decompose to partial fractions $$\frac{x}{x^3-3x+2}=\frac{x}{(x-1)(x^2+x-2)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x-2}$$ Now when we multiply out we get the following expression $$x=A(x^2+x-2)+Bx(x-1)+C(x-1)$$ If we consider $x... | $$\frac{x}{x^3-3x+2}=\frac{x}{(x-1)(x^2+x-2)}=\frac{x}{(x-1)(x-1)(x+2)} = \frac{x}{(x-1)^2(x+2)}$$ $$= \frac{A}{x-1}+\frac B{(x-1)^2} + \frac{C}{x+2}$$
$$x = A(x-1)(x+2) + B(x+2) + C(x-1)^2$$
Now, when we test with $x=1$, we get $3B= 1 \iff B=\frac 13$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\sum_{k=0}^{\infty}\frac{1}{1-16k^2}$ I could use some help on calculating this infinite sum: $\sum_{k=0}^{\infty}\frac{1}{1-16k^2}$. Included was that I had to start with a Fourier series for the function $f:\Re \to \Re: x \mapsto \sin(x)$ for $x\in[0, \frac{\pi}{2}[$, so let's start with that.
Let
\begin{eqna... | Maybe it is interesting to see that this series can be easily calculated with the residue theorem. Observing that $$S=\sum_{n\geq0}\frac{1}{1-16n^{2}}=\frac{1}{2}\sum_{n\in\mathbb{Z}}\frac{1}{1-16n^{2}}+\frac{1}{2}$$ and using $$\sum_{n\in\mathbb{Z}}f\left(n\right)=-\sum\left\{ \textrm{Res}\left(\pi\cot\left(\pi z\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
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Proving $\sec^3\frac{2\pi}{7}+\sec^3\frac{4\pi}{7}+\sec^3\frac{6\pi}{7}=-88$ and $\sec^2\frac{2\pi}{7}+\sec^2\frac{4\pi}{7}+\sec^2\frac{6\pi}{7}=24$
Prove:
$$\begin{align}
\sec^3\frac{2\pi}{7}+\sec^3\frac{4\pi}{7}+\sec^3\frac{6\pi}{7} &=-88 \tag{1} \\[6pt]
\sec^2\frac{2\pi}{7}+\sec^2\frac{4\pi}{7}+\sec^2\frac{6\pi}{... | you do need to know how to work with symmetric polynomials. The basic fact is that the roots of
$$ x^3 + x^2 - 2x-1 $$ are
$$ 2 \cos \frac{2 \pi}{7}, \; \; \; 2 \cos \frac{4 \pi}{7}, \; \; \; 2 \cos \frac{6 \pi}{7}. $$
This tells you the sum $(-1)$, the sum of pairwise products$(-2)$, and the product$(1)$. The obser... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can the following integral $I(c): = \int_{|c-1|}^{c+1} \frac{2x\ln(x)}{\pi\sqrt{-x^4 + 2(1+c^2)x^2-(1-c^2)^2}} \: dx$ be computed? Let $c > 0$ be a real number and consider the following integral:
\begin{equation} I(c): = \int_{|c-1|}^{c+1} \frac{2x\ln(x)}{\pi\sqrt{-x^4 + 2(1+c^2)x^2-(1-c^2)^2}} \: dx \end{equation}
... |
Any help is appreciated.
Since $~2x=\big(x^2\big)',\quad\ln x=\dfrac{\ln\big(x^2\big)}2,\quad$ and $x^4=\big(x^2\big)^2,\quad$ substituting $x^2\mapsto x$ seems like the
natural course of action. We then have $I(c)=\dfrac1{2\pi}\displaystyle\int_{(1-c)^2}^{(1+c)^2}\frac{\ln x~\cdot~dx}{\sqrt{-x^2+2(1+c^2)~x-(1-c^2)^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the largest power $n$ of 3 such that ${3}^n$ divides $2012^4-2011^4$? What is the largest power $n$ of 3 such that ${3}^n$ divides $2012^4-2011^4$?
[\begin{align}
& \text{My try follows : } \\
& \text{ 201}{{\text{2}}^{\text{4}}}\text{-201}{{\text{1}}^{\text{4}}}= \\
& =\left( \text{201}{{\text{2}}^{\te... | Just to give another approach, note that $81\cdot25=2025$, so that working mod $81$ (from which we can draw conclusions mod $27$), we have
$$\begin{align}
2012^4-2011^4&\equiv(-13)^4-(-14)^4\\
&\equiv169^2-196^2\\
&\equiv7^2-34^2\\
&=(7-34)(7+34)\\
&=-27\cdot41\mod81\\
&\equiv0\mod27
\end{align}$$
Since $3\not\mid41$, ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find locus of intersection of two lines with parameters $\require{cancel}$
I have a triangle
where $$DE || BC, A(a,0), B(0,b), b>0, a>0$$
I need to find the locus of the intersection between $BE$ and $CD$ (in terms of $a$ and $b$).
They didn't tell me more about this parallel line that cuts AB y and AC, therefore I th... | Parameterize the point $D$ as $(1-t)A+tB=((1-t),bt)$, $t\in[0,1]$. An equation for the line $\overline{CD}$ is then $btx-(1-t)ay=0$. The point $E$ is $((1-t)a,0)$, so an equation for $\overline{BE}$ is $bx+(1-t)ay=(1-t)ab$ and the intersection of these two lines is $x={1-t\over1+t}a$, $y={t\over1+t}b$. From here, it’s ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2220757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a six-digit perfect square of a particular form – BMO 1993 P1
1993 British Mathematical Olympiad, Round 1, Question 1
Find, showing your method, a six-digit integer $n$ with the following properties:
*
*$n$ is a perfect square,
*the number formed by the last three digits of $n$ is exactly one greater than the ... | Hint The condition $n = 1001 r + 1$ does not express one of our conditions, namely that $n$ is a square, say, $m^2$ for some positive integer $m$. In terms of $m$, our condition is $m^2 = 1001 r + 1$, and the key observation is that this equation can be rearranged and then factored:
$$(m + 1) (m - 1) = 1001 r .$$
Now, ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Integrate: $\int (x^2+a^2)^{-3/2} \cdot dx$
Integrate: $\int (x^2+a^2)^{-3/2} \cdot dx$
My Approach:
$\int (x^2+a^2)^{-3/2} \cdot dx$
$\int (x^2+a^2)^{-3/2} \cdot d(a^2+x^2)\cdot \frac{dx}{d(x^2+a^2)}$
But this doesn't give the right answer. I showed this to my friend and he said $d(x^2+a^2)$ is not possib... | Here is the correct solution. Note for anything of the form $({x^2+a^2)^{n/2}}$ for n odd, consider using the following trigonometric substituion.
So, let $x=atan(u) \ (or \ x=asinh(u)$ works aswell$)$ $\Rightarrow dx=(a)sec^2(u)du$
Thus we have:
$$\int \frac{1}{(x^2+a^2)^{3/2}}dx=\int \frac {asec^2u}{(a^2tan^2(u)+a^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2222429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Maclaurin Series : $f(x) = 3\sin^2(x)$ Use a Maclaurin series in this table to obtain the Maclaurin series for the given function.
$$f(x) = 3\sin^2(x)$$
[Hint: Use $\sin^2(x) = \frac{1}{2}\cdot(1-\cos(2x))$]
Series Sum from $0$ to $\infty$ = ?
| Alternatively, consider that:
$f(x) = 3\sin^2(x) \implies f'(x) = 6\sin(x)\cos(x) = 3\sin(2x)$
The series for $f'(x)$, we can easily find:
$\displaystyle f'(x) = 3\cdot\sum_{k=0}^{\infty} \frac{(-1)^{k}2^{2k + 1}x^{2k + 1}}{(2k + 1)!}$
Thus $\displaystyle f(x) = \int f'(x) dx = \int 3\cdot\sum_{k=0}^{\infty} \frac{(-1)... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Kindly provide the stepwise solution for this integral. $$\int_{-b}^b\int_{-a}^a\frac{1}{(x^2+y^2+h^2)^{3/2}}dxdy$$
First I put x=$\sqrt{y^2+h^2}tan\theta$ and arrive at:
$$\int_{-b}^b\frac{2a}{(y^2+h^2)(y^2+h^2+a^2)^{1/2}}dy$$
Can you please tell me what to do after this?
| If we substitute
$$y=\sqrt{a^2+h^2}\tan u$$
$$dy=\sqrt{a^2+h^2}\sec^2u \text{ }du$$
We get the integral
$$\int \frac{2a(\sqrt{a^2+h^2}\sec^2u)}{((a^2+h^2)\tan^2u+h^2)((a^2+h^2)\tan^2u+a^2+h^2)^{1/2}}du$$
Remembering that $\tan^2x+1=\sec^2x$, we get
$$\int \frac{2a\sqrt{a^2+h^2}\sec^2u}{((a^2+h^2)\sec^2u -h^2)((a^2+h^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Which of the following circles has the greatest number of points of intersection with the parabola $x^{2} = y + 4$ Which of the following circles has the greatest number of points of intersection with the parabola $x^2 = y + 4$?
(A) $x^2 + y^2=1$
(B) $x^2 + y^2=2$
(C) $x^2 + y^2=9$
(D) $x^2 + y^2=16$
(E) $x^2 + y^2=25$... | Eliminate $x$ between them leading to quadratic equation
$$ x^2+y^2= R^2,\, x^2= y+4 ,\, \rightarrow y^2+ y + 4- R^2 =0 $$
If its discriminant is $>0$ then real roots occur when
$$ R>\frac{\sqrt15} {2} \approx 1.9365$$
Repeated roots occur when $ R = \dfrac{\sqrt15}{2} $
But graphically repeated roots occur also for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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$\int_0^1 \frac{1}{1-t} ( \frac{1}{\sqrt{1-4 x}}-\frac{1}{\sqrt{1-4 t x}}) \, dt=\frac{2}{\sqrt{1-4x}}\log (\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}})$ How to prove
$$\int_0^1 \frac{1}{1-t} \bigg( \frac{1}{\sqrt{1-4 x}}-\frac{1}{\sqrt{1-4 t
x}}\bigg) \, dt=\frac{2}{\sqrt{1-4x}}\log \bigg(\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}... | This is a weird one. Put $t = \frac{1}{4x}(1-u^2)$. The limits become $1$ and $\sqrt{1-4x}$. Then
$$ \frac{dt}{1-t} = \frac{2u \, du}{1-4x-u^2}, $$
while $1/\sqrt{1-4tx} = 1/u$. But $1-4x-u^2 = (\sqrt{1-4x}+u)(\sqrt{1-4x}-u)$, so we can cancel part of this with the numerator of
$$\frac{1}{\sqrt{1-4x}}-1/u = \frac{u-\s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove by Induction $1\cdot 2+2\cdot 5+3\cdot 8+4\cdot 11+...+ n(3n-1) = n^2(n+1)$ Prove by induction that the following equality holds true for all n that's an element of a natural number.
$$1\cdot 2+2\cdot 5+3\cdot 8+4\cdot 11+...+ n(3n-1) = n^2(n+1)$$
My work:
Base Case: $n = 1$
l.s = 2
r.s = 2
True
Induction Hypothe... | Your proof is incredibly confusing (not least due to lack of proper type-setting).
As you know, you really should just start on one side and try to get to the other side:
$$\sum_{n=1}^{k+1}(n(3n-1)) = $$
$$\sum_{n=1}^{k}(n(3n-1)) + (k+1)(3(k+1)-1)= $$ (Inductive Hypothesis)
$$k^2(k+1) + (k+1)(3(k+1)-1)= $$
$$(k+1)(k^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2229745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of $S_n = \frac{1}{2n + 1} + \frac{1}{2n + 3} + ... + \frac{1}{4n - 1}$ when $n\to\infty$ $$S_n = \frac{1}{2n + 1} + \frac{1}{2n + 3} + ... + \frac{1}{4n - 1}$$
the task is to find $$\lim_{n \to \infty} S_n$$
I've tried different ways, but all I could do is to make an estimation that the limit is somewhere betwee... | Consider the following inequalities:
$$ \begin{align}
\frac{1}{2n+1+2x} &\le \frac{1}{2n+1} \le \frac{1}{2n-1+2x}, \quad 0\le x\le 1, \quad (1) \\
\frac{1}{2n+1+2x} &\le \frac{1}{2n+3} \le \frac{1}{2n-1+2x}, \quad 1\le x\le 2 , \quad (2)\\
\frac{1}{2n+1+2x} &\le \frac{1}{2n+5} \le \frac{1}{2n-1+2x}, \quad 2\le x\le ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove this $(ab)^2+(bc)^2+(ca)^2\le ab+bc+ac$
Let$a$, $b$ and $c$ be non-negative numbers such that
$$a^2+b^2+c^2=a+b+c.$$
Show that
$$(ab)^2+(bc)^2+(ca)^2\le ab+bc+ac.$$
I tried the uvw's technique and BW and more but without some success.I think can use C-S solve it?
| Another way:
$$\sum_{cyc}(ab-a^2b^2)=\frac{1}{2}\sum_{cyc}(2ab-2a^2b^2)=$$
$$=\frac{1}{2}\sum_{cyc}(a^2+2ab-a^4-2a^2b^2+a^4-a^2)=\frac{1}{2}\sum_{cyc}(a^4-a^2)=$$
$$=\frac{1}{2}\sum_{cyc}(a^4-a^2-2(a^2-a))=\frac{1}{2}\sum_{cyc}(a^4-3a^2+2a)=\frac{1}{2}\sum_{cyc}a(a+2)(a-1)^2\geq0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Analyzing quadratic forms $x^2-3xy+y^2$, $2xy+yz-3xz$ and $x^2+y^2+2xy-xt+2yt$. Positive, negative or indefinite?
By means of successive coordinate changes, write each one of the
quadratic forms below as a sum of terms of the type $\pm u^2$ and
decide wich ones are positive, negative or indefinite:
$$A(x,y) = x^2-... | If you want to work directly with the quadratic expressions, you should just "complete" the square. For $C$, we have
$$ C(x,y,z,t) = x^2 + y^2 + 2xy - xt + 2yt = \left( x + y - \frac{1}{2}t \right)^2 - \frac{1}{4}t^2 + 3yt = \\ \left( x + y - \frac{1}{2}t \right)^2 - \left(\frac{1}{2}t - 3y \right)^2 - 9y^2.$$
I've man... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to compute the determinant of $D_n$? Let
\begin{align*}
D_n = \begin{pmatrix}
a & 0 & b & 0 & 0 & \cdots & b & 0 \\
0 & a & 0 & b & 0 & \cdots & 0 & b \\
b & 0 & a & 0 & b & \cdots & 0 & 0 \\
0 & b & 0 & a & 0 & \ddots & 0 & 0 \\
0 & 0 & b & 0 & a & \ddots & b & 0 \\
\vdots & \vdots & \vdots & \ddots & \ddots ... | This is a circulant matrix with associated polynomial $f(x)=a+bx^2+bx^{n-2}$; the definition, as well as the result used below, can be found in the Wikipedia link. The eigenvectors/eigenvalues of a circulant matrix can be written down explicitly, and from the latter the determinant can be found. In the present case, we... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A point is reflected in the three sides of a triangle; when do the lines connecting the reflected points to corresponding vertices concur?
Let $P$ be a point in $\triangle ABC$, and let $P_A$ be its reflection about $\overline{BC}$, and similarly define $P_B$ and $P_C$. When do $\overleftrightarrow{AP_A}, \overleftri... | I'll derive a condition based on the distances from $P$ to the edges of $\triangle ABC$. For simplicity, I'll assume that $P$ lies in the interior of the triangle; some tweaks to the intermediate may be required otherwise, but the final condition is (probably) universal.
Let the midpoints of $\overline{PP_A}$, $\overl... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is it a coincidence that the 12th Fibonacci number is 12 squared? The first $12$ Fibonacci numbers are:
$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
n & 0& 1& 2& 3& 4& 5& 6& 7& 8 & 9& 10& 11& 12\\
\hline
f_{n} & 0& 1& 1& 2& 3& 5& 8& 13& 21& 34& 55& 89& 144\\
\hline
\end{array}$$
When you get to $f_{12}$ you ... | This is a very nice manifestation of the Strong Law of Small Numbers formulated by Richard K. Guy:
"There aren't enough small numbers to meet the many demands made of them."
A more striking manifestation of the same "law" of small numbers is this:
$$
\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
n & 0& 1& 2& 3&... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Better way of finding $\int {x^2 + 1 \over x^4 + x^2 + 1} dx$.
$$\int {x^2 + 1 \over x^4 + x^2 + 1} dx$$
By separating partial fractions,
$${Ax + B \over x^2 - x + 1} + {Cx + D \over x^2 + x + 1} = {x^2 + 1 \over x^4 + x^2 + 1} \\ \implies (Ax + B)(x^2 + x + 1) + (Cx + D)(x^2 - x + 1) = x^2 + 1$$
I get $$\begin{... | Notice that if you plug $x=i$ into $$(Ax+B)(x^2+x+1)+(Cx+D)(x^2-x+1)=x^2+1$$
You get $$Bi-A+C-Di=(C-A)+(B-D)i=0$$
or $C=A,B=D$ so
$$(Ax+B)(x^2+x+1+x^2-x+1)=2(Ax+B)(x^2+1)=x^2+1$$
From this it's easy to see that $A=C=0$ and $B=D=1/2$
| {
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"url": "https://math.stackexchange.com/questions/2240614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simple inequality $\left|\frac{3x+1}{x-2}\right|<1$
$$\left|\frac{3x+1}{x-2}\right|<1$$
$$-1<\frac{3x+1}{x-2}<1$$
$$-1-\frac{1}{x-2}<\frac{3x}{x-2}<1-\frac{1}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2} \text{ , }x \neq 2$$
$${-x+1}<{3x}<{x-3} \text{ , ... | We can not multiply by $x-2$ even if it is $\ne0$
In fact the inequality sign remains valid only if $x-2>0$
$$-1<\dfrac{3x+1}{x-2}\iff0<\dfrac{3x+1-(x-2)}{x-2}=\dfrac{x+3}{x-2}$$
So, we need $(x-2)(x+3)>0$
Now if $(x-a)(x-b)>0;a\le b$ we can prove either $x>b$ or $x<a$
| {
"language": "en",
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"answer_id": 0
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How to solve the equation $x^2y-y^2 = e^ax^2+x^2$ in terms of $x$? Please help me solve the following equation in terms of $x$, I have difficulty with this:
$$x^2y-y^2 = e^ax^2+x^2$$
Edit with my solution:
Rewrite the equation
$$0 = y^2 - x^2y + (e^a+1)x^2$$
Using the quadratic formula we obtain
$$y = \frac{x^2±\sqrt{x... | Rewrite your equation as
$$y^2 - (x^2)y + (e^ax^2+x^2) = y^2 -by +c$$
where $b$ and $c$ correspond to the terms in brackets. Then apply the standard technique to solve a quadratic equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is $1+x+x^2+x^3...$? What is the difference between these two series?
$$
\begin{align}
1+x+x^2+x^3+...+x^n+\mathcal O(x^{n+1})&=\frac{1}{1-x}\\
\\
1+r+r^2+r^3+...+r^{n-1}&=\frac{r^n-1}{r-1}\\
\end{align}
$$
I can't wrap my head around it; they both start with $1+x+x^2+x^3...$? Of course the first is the maclaurin... | Beware that "$=$" is not a symmetric relation in the presence of asymptotic notation. So while it is true that
$$\frac{1}{1-x} = 1+x+x^2+x^3+...+x^n+O(x^{n+1})$$
as $x\to 0$, we do not usually say that
$$1+x+x^2+x^3+...+x^n+O(x^{n+1})=\frac{1}{1-x}$$
because the left-hand side of that can be a lot of things other than ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Why is $\int \frac{1}{\sqrt{y}\sqrt{1 - y}} dy = \frac{2\sqrt{y - 1}\sqrt{y} \log(\sqrt{y - 1} + \sqrt{y})}{\sqrt{(-(y - 1) y)}} $? Fairly self-explanatory question title. Why is
$$\int \frac{1}{\sqrt{y}\sqrt{1 - y}} dy = \frac{2\sqrt{y - 1}\sqrt{y} \log(\sqrt{y - 1} + \sqrt{y})}{\sqrt{-(y - 1)}\sqrt{y}}\ ? $$
I'm as... | Is the answer correct? Notice $y\in (0,1)$, as both $y> 0$, $1-y>0$. But why answer has $\sqrt{y-1}$?
Let $y=\sin ^2x$, $x\in (0,\frac{\pi}{2})$
$$\int \frac{1}{\sqrt{y}\sqrt{1 - y}} dy = \int \frac{2\sin x \cos x}{\sin x\cos x} dx =2x +C=2\arcsin \sqrt{y}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2246475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\int_C\frac{z+1}{z^2-2z}dz$, where $C$ is the circle $|z|=3$
Evaluate the contour integral $\int_C\frac{z+1}{z^2-2z}dz$ using Cauchy's residue theorem, where $C$ is the circle $|z|=3$.
I see that the function has 2 singularities, at 0 and 2, so I need to find the residue of each. By examining the Laurent ... | You can calculate the partial fraction decomposition to avoid calculating the Laurent's series :
$$\int_C\frac{z+1}{z^2-2z}dz=\frac32\int_C\frac{1}{z-2}dz-\frac12\int_C\frac{1}{z}dz$$
Then we apply Cauchy formula :
$$\int_C\frac{1}{z-2}dz=2\pi i=\int_C\frac{1}{z}dz$$
Or with Cauchy residue's theorem, $\frac{1}{z-2}$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Determine the unique vector x in the row space for A, for which Ax = b Given the Vector $b = (9, -6, 3)^T$
$A =\begin{pmatrix}
1 & 2 & 0 & 1\\
-1 & -1 & -1 & 0 \\
1 & 0 & 2 &-1
\end{pmatrix}$
I know that the row space is $ \begin{pmatrix} 1 & 0 & 2& -1\end{pmatrix}$ , $ \begin{pmatrix... | As far as I can see, this system of equations doesn't even a have a solution, for we have:
$$
\begin{align}
\left[
\begin{array}{cccc|c}
1&2&0&1&6\\
-1&-1&-1&0&-6\\
1&0&2&-1&3
\end{array}
\right]
&\iff
\left[
\begin{array}{cccc|c}
1&2&0&1&6\\
0&1&-1&1&0\\
0&-2&2&-2&-3
\end{array}
\right]
\iff
\left[
\begin{array}{cccc|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2251271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A question regarding Czarnowski's cat function The Czarnowski cat function $c(x, n) = \sin^n(x) + \cos(x)$ gives the so-called Czarnowski's cats. There are infinitely many of these cats, and legend has it that Czarnowski created a clever mathematical naming scheme to give them all unique names.
The problem: for Czarno... | We seek a reduction formula for integrands of the form $$\sin^{2n} x.$$ Integration by parts with the choice $$u = \sin^{2n-1} x, \quad du = (2n-1) \sin^{2n-2} x \cos x \, dx, \\ dv = \sin x \, dx, \quad v = -\cos x, $$ gives
$$\begin{align*} I_n (x) &= \int \sin^{2n} x \, dx = -\sin^{2n-1} x \cos x + (2n-1) \int \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Error of approximation for Euler's constant Is there a simple way to show this bound for Euler’s constant:
$$ \frac{1}{2n+2} < \sum_{k=1}^n\frac{1}{k} - \ln(n) - \gamma < \frac{1}{2n}$$
For proving convergence, I showed that $\sum_{k=1}^n 1/k - \ln(n)$ is bounded and monotonic, using the property
$$\frac{1}{k+1} < \int... | Use convexity.
Since $f(x) = 1/x$ is both decreasing and convex we have for $k \leqslant x \leqslant k+1$,
$$\frac{f(k+1) - f(k+2)}{k + 2 - (k+1)} \leqslant \frac{f(x) - f(k+1)}{k + 1 - x} \leqslant \frac{f(k) - f(k+1)}{k + 1 - k} $$
Hence,
$$\tag{*}\left(\frac{1}{k+1} - \frac{1}{k+2}\right)(k + 1 - x) \leqslant \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2253605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$ then the value of $\alpha^6+\beta^6$ is
If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$ then the value
of $\alpha^6+\beta^6$ is
I know that, here, $\alpha\beta=4$ and $\alpha + \beta = 2$ and use that result to find $\alpha^2 + \beta^2$ using the expansion... | From $\alpha^2-2\alpha+4=0$ we get that for $n\geq 0$ we have $\alpha^{n+2}-2\alpha^{n+1}+4\alpha^n=0$, and we have the same relation for $\beta$. Putting $u_n=\alpha^n+\beta^n$ and adding, we get that $u_{n+2}-2u_{n+1}+4u_n=0$ for all $n$. We have $u_0=2$, $u_1=2$, and now it is easy to compute $u_6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Prove that there are infinite sets of $(a, b, c)$ such that $ab + 1$, $ac + 1$ and $bc + 1$ are perfect square Prove that there are infinite sets of integer numbers $(a, b, c)$ that $ab + 1$, $ac + 1$ and $bc + 1$ are perfect square. ($a$, $b$, $c$ are different numbers)
Any hints how to prove the statement? I'm trying... | If $a=4n$ , $b=n+1$ and $c=n-1$ then
\begin{eqnarray*}
ab+1&=&(2n+1)^2 \\
bc+1&=& n^2 \\
ca+1&=&(2n-1)^2
\end{eqnarray*}
EDIT : for Ovi.
We $ab+1,bc+1,ca+1$ are perfect squares. So there exist $x,y,z$ such that
\begin{eqnarray*}
z^2=ab+1 \\
x^2=bc+1 \\
y^2=ca+1
\end{eqnarray*}
Now subtract these equations pairwise (dif... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Analytical evaluation of double integral I'm looking to evaluate the following integral:
$$\int^a_{-a}\int^a_{-a}\frac{\sqrt{x^2+y^2}}{4a^2} \,\mathrm{d}x \mathrm{d}y,$$
where $a> 0$.
Mathematica gives the following answer:
$$ \frac{a}{3} (\sqrt{2} + \sinh^{-1}(1)).$$
This goes well beyond my basic calculus 101 trainin... | $$
\frac 1 {4a^2} \int^a_{-a}\int^a_{-a} \sqrt{x^2+y^2} \, dx \, dy = \text{what?}
$$
$$
\int_{-a}^a (\text{an even function}) = 2\times\int_0^a (\text{the same function}), \text{ so}
$$
\begin{align}
& \int_{-a}^a \sqrt{x^2+y^2} \ dx = 2\int_0^a \sqrt{x^2+y^2}\ dx = 2\int_0^{\arctan a} \sqrt{y^2\tan^2\theta + y^2} (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Find the sum of all values of $ f ( 2017 ) $ given $ f ^ { f ( a ) } ( b ) f ^ { f ( b ) } ( a ) = f ( a + b ) ^ 2 $.
Let $ f :\mathbb N \to \mathbb N $ be an injective function such that
$$ f ^ { f ( a ) } ( b ) f ^ { f ( b ) } ( a ) = f ( a + b ) ^ 2 $$
for all $ a , b \in \mathbb N $. Let $ S $ be the sum of all po... | It's easy to see that $ f ( n ) = n + 1 $ satisfies the functional equation
$$ f ^ { f ( a ) } ( b ) f ^ { f ( b ) } ( a ) = f ( a + b ) ^ 2 \text , \tag 0 \label 0 $$
since then for every positive integers $ m $ and $ n $, one will have $ f ^ m ( n ) = n + m $. It can be shown that this is the only injective solution ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2257621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Solving Recurrence Relation with substitution How to solve T(n) = T(n-2) + n using iterative substitution
Base case:
T(0) = 1
T(1) = 1
Solve:
T(n) = T(n-2) + n
Currently I have:
T(n) = T(n-2) + n
= T(n-4) + n - 2 + n = T(n-4) + 2n - 2
= T(n-6) + n - 4 + n - 2 + n = T(n-6) + 3n - 6
= T(n-8) + n - 6 + n ... | $$\begin{align}
T(n) &= T(n-2) + n \\
T(n) &= T(n-4) + (n-2) + (n-0) \\
T(n) &= T(n-6) + (n-4) + (n-2) + (n-0) \\
T(n) &= T(n-8) + (n-6) + (n-4) + (n-2) + (n-0) \\
\vdots\\
T(n) &= T(n-2r) + \sum_{k=0}^{r-1}(n-2k)\\
T(n) &= T(n-2r) + \left(n\sum_{k=0}^{r-1}(1)\right)-2\left(\sum_{k=0}^{r-1}k\right)\\
T(n) &= T(n-2r) + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2259243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Given $x,y,z >0$ and $xy^2z^3 = 108 $, what is the minimum value of $x+y+z$? Given $x,y,z >0$ and $xy^2z^3 = 108 $, what is the minimum value of $x+y+z$ ?
This is a homework problem, so if someone could just give me an outline/hint of the method used to solve this it would be much appreciated.
Thanks
| By AM-GM Inequality,$$x+y+z = x+\frac{y}{2}+\frac{y}{2}+\frac{z}{3}+\frac{z}{3}+\frac{z}{3}\geq6\left(\frac{xy^2z^3}{2^2\cdot3^3}\right)^\frac{1}{6}=6\left(\frac{108}{2^2\cdot3^3}\right)^\frac{1}{6}=6$$
Equality holds when $x=\frac{y}{2}=\frac{z}{3}$, so plugging this into $xy^2z^3=108$ we know the equality occurs when... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2261563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many twin primes are of the form $2^n-1$ and $2^n+1$? The first pair is $(3,5)$ for $n=2$. Is there any other pair beside this?
| No, there are no more. A number of the form $2^n-1$ can only be prime if $n$ is prime (although there are prime $n$ which doesn't work, like $n = 11$, so $2^{11}-1$ is not prime), while a number of the form $2^n+1$ can only be prime if $n$ is a power of $2$ (again, there are power-of-two's which do not work, like $32$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2262070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Integral $\int_0^\pi \big( (1+\alpha \cos x) \cos x \big)^n dx $ I have been struggling with the integral
$$ I_n(\alpha) = \frac{1}{\pi} \int_0^\pi \big( (1+\alpha \cos x) \cos x \big)^n dx,$$
where $\alpha$ is real and $n$ is a non-negative integer.
It is relatively easy to get the values for specific $n$;
$$I_0(\al... | I am not entirely sure the binomial theorem can be avoided, but also the Weierstrass substitution ($(t = \tan x/2)$ transforms your integral into an integral of a rational function from $0$ to infinity, at which point the residue theorem is your friend.
EDIT
A similar method is the following: first note that your integ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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proving $ \lim_{x \rightarrow 1} \frac{1-\sqrt{x}}{1-x} = \frac{1}{2}$. I'm having trouble with proving $ \lim_{x \rightarrow 1} \frac{1-\sqrt{x}}{1-x} = \frac{1}{2}$.
So far I have:
$\frac{1-\sqrt{x}}{1-x} =\frac{1}{1+\sqrt{x}} $
If $x\in dom (\frac{1}{1+\sqrt{x}}$) and $|x-1| <\delta$ then $ |\frac{1}{1+\sqrt{x}}- \f... | Hint: for $x>0$:
$|\frac{1-x}{2(1+2\sqrt(x)+x)}| \le |\frac{1-x}{2}|$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If $\cos A+\cos B=p$ and $\sin A+\sin B=q$, then find $\cos\left( \frac {A+B}{2}\right)$ in terms of $p$ and $q$ If $\cos A+\cos B=p$ and $\sin A+\sin B=q$ then find $\cos \left( \dfrac {A+B}{2}\right)$ in terms of $p$ and $q$.
My Attempt:
$$\cos A+\cos B=p$$
$$2\cos \left( \dfrac {A+B}{2}\right)\cos \left( \dfrac {A-... | here is a geometric interpretation of your work: think of the points $(\cos A, \sin A)$ and $(\cos B , \sin B)$ as two points also called $A$ and $B$ on the unit circle. the arc length is measured from the initial point $O = (1,0).$ now,
the mid point of the chord $AB$ has the coordinate $(p/2, q/2).$ we push this poin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2265706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Parameterize Intersection of Surfaces I need to parameterize the intersection of $$4x^2 + y^2 + z^2 = 9\tag{1}$$ and $$z=x^2+y^2\tag{2}$$.
First, I'll solve (2) for $y^2$ and substitute the result into (1):
$$3x^2+z+z^2 = 9 \tag{3}$$
Next, I'll make the substitution $u=\sqrt{3}x$, such that we can complete the square ... | It is not necessary to convert to polar coordinates, really. You have
$$z = x^2 + y^2 \tag{1}\label{1}$$
$$4 x^2 + y^2 + z^2 = 9 \tag{2}\label{2}$$
Equation $\eqref{1}$ can be written as
$$x^2 = z - y^2$$
and substituting it into equation $\eqref{2}$ yields
$$4 (z - y^2) + y^2 + z^2 = -3 y^2 + z^2 + 4 z = 9$$
which we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2265819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Prove by induction that $2^n + 4^n \leq 5^n$ I'm trying to prove by induction that $2^n + 4^n \leq 5^n$. Through some value plugging I've established that the induction must start from $n = 2$ because $2^2 + 4^2 \leq 5^2 \equiv 20 \leq 25$; for $n = 1$ it doesn't hold since $2 + 4 \geq 5$.
Now I assume that $2^k + 4^... | You already have established from the induction hypothesis that
$$2^{k+2} + 4^{k+1} \leq 4 \cdot 5^{k}$$
So:
$$ 2^{k+1} + 4^{k+1} < 2^{k+2} + 4^{k+1} \leq 4\cdot5^k < 5 \cdot 5^k =5^{k+1}$$
Which already shows the required inequality for $k+1$.
It's simpler than you think...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2266291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Find $b$ in $2\sin(\frac{x}{2}+b)$ given the graph Following is the graph of the function. $2\sin \left(\dfrac{x}{2}+b\right)$
I tried solving the equation using the point $\left (\dfrac{\pi}{2},2\right)$:
$$2 = 2\sin\left(\dfrac{x}{2}+b\right) \Leftrightarrow \\
1 = \sin\left(\dfrac{x}{2}+b\right)\Leftrightarrow \\
\... | $\sin\left( \dfrac x 2 + b \right) = 1$ when $\dfrac x 2 + b = \dfrac \pi 2,$ and that implies $x=\pi-2b.$
Since it appears that the point where the curve reaches its maximum is half-way between $0$ and $\pi/2,$ we have $\dfrac\pi 2 = \pi - 2b,$ and from that we get $b=\dfrac \pi 4.$
A bit of an optical illusion seems ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2267715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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What's the limit of $\sqrt{2 + \sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2 + ...}}}}}} $? Let's look at the continued radical
$ R = \sqrt{2 + \sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2 + ...}}}}}} $
whose signs are defined as $ (+, -, +, -, -, + ,-, -, -,...)$, similar to the sequence $101001000100001...$, where
$1 = +$
$0 =... | Let $R$ be the iterated square root in question
One easily check that any finite expression of the form
$$\sqrt{2\pm\sqrt{2\pm\sqrt{2\pm\cdots}}}$$
is between $0$ and $2$. So $0\le R\le 2$.
Suppose that $a=2\cos t$. Then
$$\sqrt{2+a}=\sqrt{2(1+\cos t)}=\sqrt{4\cos^2(a/2)}=2\cos\frac a2$$
and
$$\sqrt{2-a}=\sqrt{2(1-\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2270730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
"answer_count": 3,
"answer_id": 1
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System of Linear Congruences Find all $x$ such that
\begin{align}
x&\equiv 1 \pmod {12}\\
x&\equiv 4 \pmod {21}\\
x&\equiv 18 \pmod {35}
\end{align}
Im not quite sure if this system of linear congruence is solvable. Since
$\gcd(12,21) =3$, $\gcd (12,35)=1$ and $\gcd(21,35) = 7$, and the CRT states that "If(m1, m2) = ... | \begin{align}
x&\equiv 1\phantom {8} \pmod {{\color{blue}{12}}}\tag {1}\\
x&\equiv 4\phantom {8} \pmod {{\color{teal}{21}}}\tag {2}\\
x&\equiv 18 \pmod {{\color{blue}{35}}}\tag{3}
\end{align}
Since $\gcd(12,35)=1 $ you can first apply the Chinese Remainder Theorem to find the solution to the two simultaneous congruence... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2272593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Find the arc length of a path I'm trying to solve the arc length for the following. $c(t) = (2, 6t^2, 4t^3)$ from $0\le t\le 1.$
I've checked on WolframAlpha and I get the answer $8\sqrt2 - 4$ but when I work it out I get $16\sqrt2$ where am I going wrong?
\begin{align*}
L&= \int_0^1 \sqrt{(12t)^2 + (12t^2)^2}\, dt\\
&... | From $$\int_{0}^{1}12t\sqrt{1+t^2}dt$$
Use $u=1+t^2\implies \frac{1}{2}du=tdt$ to get $$\int_{0}^{1}12t\sqrt{1+t^2}dt=\int12\sqrt{u}\cdot\frac{1}{2}du=6u^{3/2}\cdot\frac{2}{3}=4(1+t^2)^{3/2}\bigg\vert_{0}^{1}=4\cdot2^{3/2}-4\cdot1$$ $$=4\cdot2\cdot2^{1/2}-4\cdot1=8\sqrt{2}-4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Find parameter of inequality Find the parameter $m \in \mathbb R$ so that $$x^2 - 2xy + 2y^2 + 2x - 6y + m \ge 0, \forall x, y \in \mathbb R$$
I tried forming some squares but it seems it doesn't help me
$$(x-y)^2 + y^2 + 2x -6y + m \ge 0$$
| $y^2+x^2+1+2x-2y-2xy=(y-x-1)^2$ and $y^2-4y+4=(y-2)^2$. Total of these, we yields
$$x^2 - 2xy + 2y^2 + 2x - 6y + m = (y-x-1)^2 + (y-2)^2 + m - 5 \geq 0$$
Hence, we find that $ m\geq 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve for xy in terms of a and b $$x^2 + xy + y^2 = a$$
$$x + y = b$$
$xy = ?$
I tried this and did this:
$xy = a - x^2 - y^2$
$xy = a - (x^2 + y^2)$
$xy = a - (x + y)(x - y)$
$xy = a - b(x - y)$
At this point I can't think of anything to do to represent the $x - y$ part in terms of $a$ and $b$.
Any help would be appre... | We are given \begin{align}x^2+xy+y^2&=a\tag{1}\\
x+y&=b\tag{2}\end{align}
We can square the equation $(2)$ to give us \begin{align}b^2&=(x+y)^2\\
&=x^2+2xy+y^2\end{align}
Therefore, we now have \begin{align}a&=x^2+xy+y^2\tag{1}\\
b^2&=x^2+2xy+y^2\tag{3}\end{align}
We can rearrange equation $(3)$ and substitute equation... | {
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"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Functions and Sequences Problem
The function $F(k)$ is defined for positive integers as $F(1) = 1$,
$F(2) = 1$, $F(3) = -1$ and $F(2k) = F(k)$, $F(2k + 1) = F(k)$ for $k \geq
2$. Then $$F(1) + F(2) + \dotsb + F(63)$$ equals
$\begin{array}{lr}
(\text{A}) & 1 \\
(\text{B}) & -1 \\
(\text{C}) & -32 \\
(\text{D})... | With induction it can be shown that:$$F(2^k)+\cdots+F(2^{k+1}-1)=0$$
for $k=1,2,\dots$
This because $F(2^1)+F(2^2-1)=F(2)+F(3)=1+(-1)=0$ and:$$F(2^{k+1})+F(2^{k+1}+1)+\cdots+F(2^{k+2}-2)+F(2^{k+2}-1)=2F(2^k)+\cdots+2F(2^{k+1}-1)$$
Then $$F(1)+\cdots+F(63)=F(1)+\sum_{k=1}^5\left[F(2^k)+\cdots+F(2^{k+1}-1)\right]=F(1)=1$... | {
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"url": "https://math.stackexchange.com/questions/2278951",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Identify the plane defined by $|z-2i| = 2|z+3|$ I tried:
$$|z-2i| = 2|z+3| \Leftrightarrow \\
|x+yi-2i|=2|x+yi+2|\Leftrightarrow \\
\sqrt{x^2+(y-2)^2}=\sqrt{4((x+2)^2+y^2)} \Leftrightarrow \\
\sqrt{x^2+y^2-4y+4} = \sqrt{4x^2+24x+36+4y^2} \Leftrightarrow \\
x^2+y^2-4y+4 = 4x^2+24x+36+4y^2 \Leftrightarrow \\
y^2-4y-4y^2=... | You can square without adding spurious solutions, because $|w|\ge0$ by definition. So $|z-2i|^2=4|z+3|^2$ and, using $|w|^2=w\bar{w}$,
$$
(z-2i)(\bar{z}+2i)=4(z+3)(\bar{z}+3)
$$
This simplifies to
$$
3z\bar{z}+12(z+\bar{z})-2i(z-\bar{z})+32=0
$$
Now recall that, for $z=x+yi$, $z\bar{z}=x^2+y^2$, $z+\bar{z}=2x$ and $z-\... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Can this ratio be an integer? Is it true that the number
$$\frac{b^3+1}{ab-1}$$
is never an integer for positive integers $a$ and $b$ such that $a\geq 6$ and $a\geq b$?
| Here is my approach:
By definition you can say that $b^2+{2 \over b} \ge a \ge b$ because $(b^2+{2 \over b})b-1 = b^3+1$ . But because ${2 \over b}$ is only an integer for $b=1$ or $b=2$ your interval for $a$ is effectively $[b, b^2]$ . This means you can write $a=b\cdot b^x$ with $0 \le x \le 1$ or $a=b \sqrt[y]{b}$ ... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Help solving the PDE $u_x^2 + u_y^2 = u$ I'm trying to solve this Cauchy problem:
$$ u_x^2 + u_y^2 = u \\
u(x,0) = x^2+1$$
Here's what I've tried so far:
Letting $p(r,s) = u_x$, $q(r,s) = u_y$ and $z(r,s) = u$, we have:
$$ F = p^2 + q^2 - z = 0 $$
and thus:
$$ \begin{align} \frac{dx}{ds} &= F_p = 2p \\
\frac{dy}{ds} &=... | Obviously, the main difficulty comes from the boundary condition $u(x,0)=x^2+1$ because it involves to solve a polynomial equation leading to huge formula. Even if the solving is theoretically possible, one have to merely accept a result on implicit form.
To make more clear where the difficulty arises, we will avoid t... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Find the straight line from two plane vectors Two planes have equations $x + 3y - 2z = 4$ and $2x +y +3z = 5$. The planes intersect in the straight line $l$.
Find a vector equation for the line $l$.
How can I do this type of question?
| Let $\vec{l}(a,b,c)$.
Hence, $a+3b-2c=0$ and $2a+b+3c=0$, which gives $b=-\frac{7}{11}a$ and $c=-\frac{5}{11}a$.
Let $a=11$. Hence, $b=-7$ and $c=-5$ and $\vec{l}(11,-7,-5)$.
But for $z=0$ we get $x=\frac{11}{5}$ and $y=\frac{3}{5}$, which gets an answer:
$$\left\{\left(\frac{11}{5}+11t,\frac{3}{5}-7t,-5t\right)|t\in\m... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Bound on $x^TAx$ for spd matrix $A$ in terms of diagonal Let $A\in\mathbb{R}^{n\times n}$ be a symmetric positive definite matrix and $x\in\mathbb{R}^n$ some vector. I want to find a bound of the form
$$x^T Ax \leq c \sum_{i=1}^n a_{ii} x_i^2$$
with a constant $c>0$.
If $\lambda_\text{min}$ and $\lambda_\text{max}$ den... | You can substitute $y_i = \sqrt{a_{ii}}\;x_i$. Then you are looking for the smallest $c$ with
$$
y^T
\begin{pmatrix} \frac{1}{\sqrt{a_{11}}} & & & 0 \\
& \frac{1}{\sqrt{a_{22}}} & & \\
& & \ddots & \\
0 & & & \frac{1}{\sqrt{a_{nn}}}
\end{pmatrix}
A
\begin{pmatrix} \frac{1}{\sqrt{a_{11}}} & & & 0 \\
& \frac{1}{\sqrt{a_{... | {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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find oblique asymptotes : $f(x) =\sqrt[q]{\frac{a_mx^m+a_{m-1}x^{m-1}+\cdots+a_0}{b_nx^n+b_{n-1}x^{n-1}+\cdots+b_0}}$ let : $$f(x) =\sqrt[q]{\frac{a_mx^m+a_{m-1}x^{m-1}+\cdots+a_0}{b_nx^n+b_{n-1}x^{n-1}+\cdots+b_0}}$$
shuch that : $$m-n=q$$
then : fine oblique asymptotes :
i know that : $y=mx+h$ is oblique asymptotes ... | For the slope, you can do the same calculation you mentioned:
$$\begin{align}
\displaystyle\lim_{x\to\infty}\frac{f(x)}{x} &= \displaystyle\lim_{x\to\infty} \sqrt[q]{\frac{a_mx^m+a_{m-1}x^{m-1} + \cdots+a_0}{b_nx^n+b_{n-1}x^{n-1}+\cdots+b_0}}\cdot\frac{1}{x}\\
&= \displaystyle\lim_{x\to\infty} \sqrt[q]{\frac{a_mx^m+a_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2283767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
partially differentiate $\frac{x-y}{\sqrt{x^2-y^2}}$ How do you partially differentiate $\frac{x-y}{\sqrt{x^2-y^2}}$, with respect to $y$ then $x$?
I have done it and got the answer $$\frac{y^2+yx}{\sqrt[3]{x^2-y^2}}$$
However, the answer given in my paper gives it with a $-y^2$ on the top with all the rest the same. ... | We should get the following steps:
\begin{align}\frac{\partial}{\partial y}\left(\frac{x-y}{\sqrt{x^2-y^2}}\right)&=\frac{-x}{(x+y)\sqrt{x^2-y^2}}\\
\frac{\partial}{\partial x}\left(\frac{-x}{(x+y)\sqrt{x^2-y^2}}\right)&=\frac{x^2-xy+y^2}{(x-y)(x+y)^2\sqrt{x^2-y^2}}\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2285032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Counting sets which can be partition in two subsets satisfying certain criterion I have a question that involves some ability in combinatorics that I don't have. Any hint could be extremely appreciated.
Take $n>1$ with $n\in \mathbb{N}$.
Let $\mathcal{G}$ be the set of all possible binary matrices of dimension $n\times... | This is not an answer but an input regarding the maximum of $M$.
Claim:
For $M \gt 2^{n(n-1)}-n(2^{n-1}-1)$ there is no $C$ satisfying the criterion.
Argument:
When pairing matrices, at least two rows must differ. This means the only pairing which isn't allowed is the pairing of matrices which differ in exactly one ro... | {
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"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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Help in evaluating integral using elementary methods I have the following integral:
$\int\limits_{0}^{\pi/2}\frac{x\sin x \cos x}{a^2 \cos^2x+ b^2\sin^2x} \ dx$, for $a,b \geq 0$, both not zero.
I have tried several substitutions without any success at all. How can this be tackled?
Thanks.
| First we deal with the case $b= a$, then:
$$I(a,a)=\int\limits_{0}^{\pi/2}\frac{x\sin x \cos x}{a^2 \cos^2x+ a^2\sin^2x} \ dx=\frac{1}{a^2} \int\limits_{0}^{\pi/2} x\sin x \cos x dx$$
Now we use integration by parts with:
$$u=x,\qquad dv=\sin x \cos x dx$$
$$v=\int \sin x \cos x dx=\frac{1}{4} \int \sin 2x d2x=-\frac{\... | {
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Limit with 2 variables: $\frac{x^2y}{x^4 +y^2}$, $\frac{e^{xy^3}-1}{x^2 +y^4}$ and $(x^2 +y^2)^{xy}$ at $(0,0)$ I am new to this topic so I appreciate any help on this.
a)$$\lim_{(x,y) \rightarrow (0,0)} f(x,y) = \frac{x^2y}{x^4 +y^2}$$
For $x=0 \lor y=0 : f(x,y) \longrightarrow 0$ but for $y = x^2$ the limit is $\fra... | You've made a mistake in c): $(1/y,y)$ does not approach $(0,0)$ as $y\to 0.$
To do it, we can apply $\ln$ to get $xy\ln (x^2+y^2).$ Recall $|xy|\le (x^2+y^2)/2.$ Thus
$$\tag 1 |xy \ln (x^2+y^2)| \le \frac{(x^2+y^2)\ln (x^2+y^2)}{2}.$$
Now $u\ln u \to 0$ as $u\to 0^+.$ Thus the limit of $(1)$ as $(x,y) \to (0,0)$ is $0... | {
"language": "en",
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"source": "stackexchange",
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"answer_id": 2
} |
Confusion about proving $\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{n(n+1)} = \frac{n}{n+1}$ by induction I know what the answer to this question is, but I am not sure how the answer was reached and I would really like to understand it! I am omitting the base case because it is not re... | "What I am confused about is where the 1/(k+1)*(k+2) comes from in the first line of the inductive step."
It comes because you are trying to evaluate for $n = k+1$
You want to prove the statement $\frac 1{1*2} + .....+ \frac 1{(n-1)(n)} + \frac 1{n*(n+1)} = \frac n{n+1}$.
For $n = k$ if you replace $n$ with $k$ you a... | {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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How to take $\int_0^{+\infty} \frac{x^2+1}{x^4+1}dx$? The integral:
$$\int_0^{+\infty} \frac{x^2+1}{x^4+1}dx$$
If num were greater than denum I would just devide it normally with long division, but it is not, how should I handle it then?
| By symmetry,
$$2\int_0^{+\infty}\frac{x^2+1}{x^4+1}\ dx=\int_{-\infty}^{+\infty}\frac{x^2+1}{x^4+1}\ dx$$
By taking a semi-circle contour in the upper half of the complex plane and applying Cauchy's residue formula, we find that
$$\begin{align}\int_{-\infty}^{+\infty}\frac{x^2+1}{x^4+1}\ dx&=2\pi i\left[\lim_{a\to e^{\... | {
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Find the last digit of $43^{43^{43}}$ Find the last digit of $43^{43^{43}}$. My attempt:
$$(40+3)^3 \equiv 3\cdot 40\cdot 9+27 \equiv 1080+27\equiv 07\pmod{100}$$
Am I right or is there any other way to find the answer?
| There will come some number $n $ where the last digit of $43^n $ is $1$.( $43^{n}\equiv 1 \mod 10$) Then any $43 ^{kn+m}=(43^n)^k43^m\equiv 1^k43^m \mod 10$ which will probably be easier to solve.
So our first step is to find that $n $.
Since $43=40+3$ then $43^n=40^n+a_140^{n-1}*3+......+a_{n-1}40*3^{n-1}+3^n $ and ... | {
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"answer_count": 4,
"answer_id": 3
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Without the aid of a calculator, compute $\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}$
Without the aid of a calculator, compute $\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}$
The only method in doing this that I know was to just plug it in into the calculator, but I really had no clue on how to do this without the aid of a calc... | One more method:
\begin{align*}
\left(\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}\right)^2&=\left(\cos\frac{\pi}{12}+\sin\frac{\pi}{12}\right)^2\\
&=\cos^2\frac{\pi}{12}+\sin^2\frac{\pi}{12}+2\sin\frac{\pi}{12}\cos\frac{\pi}{12}\\
&=1+\sin\frac{\pi}{6}\\
&=\frac{3}{2}
\end{align*}
As $\displaystyle \sin\frac{5\pi}{12}>0$ an... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Solve: $(i+i^2+i^3+...+i^{102})^2$
Solve $$(i+i^2+i^3+...+i^{102})^2$$
I have tried 2 approaches:
*
*to look at power of $2 (=-1)$ and power of $4 (=1)$ and there are $102$ elements which is even number, but because there are $4$ "types" $\{i,-1,-i,1\}$ it did not work it out
*to look at
$$(i+i^2+i^3+...+i^{102... | $$a + a^2 + a^3 + \dots a^n = a\frac{1-a^{n}}{1-a}$$
Therefore,
\begin{align*}
(i+i^2 + \cdots + i^{102}) &= i \frac{1-i^{102}}{1-i}\\
& = i \frac{1-i^{100}i^2}{1-i} \\
&= i \frac{1+(i^{4})^{25}}{1-i}\\
&= \frac{2i}{1-i}\\
&= -1 + i
\end{align*}
Squaring yields $-2i$.
| {
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"answer_count": 5,
"answer_id": 2
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Find all complex roots for $z^6 + (1 + i)z^3 + i = 0$ Find all complex roots of $z^6 + (1 + i)z^3 + i = 0$.
I've tried by first settings $w = z^3$, so then I get $$w^2 + (1 + i)w + i = 0$$
Now I rewrite it as $$(w+(\frac{1+i}{2}))^2 = -i + (\frac{1+i}{2})^2$$
Set $w+(\frac{1+i}{2}) = a+bi$
$$(a+bi)^2 = -i + 2i=i$$
Whic... | It's $$(z^3+i)(z^3+1)=0.$$
We have two cases.
*
*$z^3+i=0$ or $z^3-i^3=0$ or
$$(z-i)(z^2+iz-1)=0,$$ which gives
$$\left\{i,\pm\frac{\sqrt3}{2}-\frac{1}{2}i\right\}$$
*$z^3+1=0$ or
$$(z+1)(z^2-z+1)=0,$$ which gives
$$\left\{-1,\frac{1}{2}\pm\frac{\sqrt3}{2}i\right\}$$
Done!
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Find the power series representation of the function I have to find $a_n$ so, my first attempt was to do partial fraction, but with no real solution. Any recommendation will be appreciate $$\frac{1}{(2x-3)(x^2-x+1)}=\sum_{n=0}^\infty a_n (x-1)^n$$
| Let us replace $x$ with $z+1$ for the sake of simplicity. We have to find the sequence $\{a_n\}_{n\geq 0}$ given by:
$$ f(z)=\frac{1}{(2z-1)(z^2+z+1)}=\sum_{n\geq 0}a_n z^n \tag{1} $$
but $f(z)$ is a meromorphic function with simple poles at $z=\frac{1}{2},\,z=\omega,\,z=\omega^2$, where $\omega=e^{\frac{2\pi i}{3}}$.
... | {
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"question_score": "1",
"answer_count": 2,
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Prove $(1-\frac{8}{\pi^2})x^2\cot^2(x)+\frac{8}{\pi^2}x\cot(x)+\frac{4}{\pi^2}x^2-1\ge0$ for $x\in[0,\pi/2]$
How to prove the following inequality:
$$\left(1-\frac{8}{\pi^2}\right)x^2\cot^2(x)+\frac{8}{\pi^2}x\cot(x)+\frac{4}{\pi^2}x^2-1\ge0,\quad x\in[0,\pi/2]?$$
I evaluated the LHS of the above inequality numerical... | A partial answer - I am going to show that the given inequality holds for every $x\in(0,1]$.
Since the cotangent is the logarithmic derivative of the sine function we have
$$ \cot x = \frac{1}{x} + \sum_{n\geq 1}\left(\frac{1}{x-n\pi}+\frac{1}{x+n\pi}\right) \tag{1} $$
and
$$ \frac{1-\pi z\cot(\pi z)}{2} = \sum_{n\geq ... | {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
fermat's last theorem by elementary methods I was wondering about correctness of the following theorem/proof.
Theorem:
Suppose $p$ is prime in:
$ \quad \quad p^k | x - y$
$ \quad \quad p^{k + 1} \not | x - y$
$ \quad \quad gcd(x,y,z) = 1$
$ \quad \quad n > 2$
$ \quad \quad x^n = y^n + z^n$
Then:
$ \quad \quad p | n$
P... | Let $n$ be a positive integer, $n > 1$, and suppose $b,c,d$ are positive integers with $\gcd(b,c,d)=1$ such that $d^n = b^n + c^n$.
Next, suppose $p$ is a prime such that $d=ap+b$.
\begin{align*}
\text{Then}\;\;&(ap + b)^n = b^n + c^n\\[4pt]
\implies\;&p\mid c^n\\[4pt]
\implies\;&p\mid c\\[4pt]
\end{align*}
So you can... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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every integer can be represented in the form $x^2+y^2-z^2$ Every integer can be represented in the form $x^2+y^2-z^2$ and show that $6$ actually requires all three terms.
I put
*
*$z=y+1$
*$x=n^2+3$
*$y=3n^2+4+(n^4-n)/2$
what does it mean that $6$ actually requires all three terms?
| It means that if $x,y,z$ are such that
$$x^2+y^2-z^2=6$$
then $x,y,z$ are all non-zero.
You can prove this by looking at the equation mod $4$.
You know that a square is always $0$ or $1$ mod $4$, and $6\equiv 2\pmod 4$.
So if $x^2+y^2-z^2=6$, then $z$ must be $0$ mod $4$, otherwise you are not able to get to $2$ mod $4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
How multiply Blocked Matrices? I am having a hard time understanding how to multiply blocked matrices with rectangle matrices and blocking into non-square matrices. Can someone please explain me how that works?
| I hope it can help you
Example:
$A=
\begin{bmatrix}
2 & 3 & 2\\
4 & 1 & 3\\
3 & 2 & 1\\
\end{bmatrix}
$$\to$
$ \left[
\begin{array}{cc|c}
2&3&2\\
4&1&3 \\
\hline
3 &2 &1
\end{array}
\right] $
$ B=
\begin{bmatrix}
1 & 3 \\
2 & 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to prove that $\sqrt{-3-2i}+\sqrt{-3+2i} = \sqrt{2(\sqrt{13}-3)}$? Is there a trick to show that
$$\sqrt{-3-2i}+\sqrt{-3+2i} = \sqrt{2(\sqrt{13}-3)}$$
is true ?
| Start by squaring the LHS and RHS
\begin{align*}
(\sqrt{-3-2i} + \sqrt{-3+2i} )^2 &= \sqrt{2(\sqrt{13}-3)}^2 \\
(-3-2i) + 2\sqrt{(-3-2i)(-3+2i)} + (-3 + 2i) &= 2(\sqrt{13}-3) \\
-6 + \sqrt{4 (9 + 4)} &= 2(\sqrt{13}-3) \\
-6 + 2\sqrt{13} &= 2(\sqrt{13}-3) \\
2(\sqrt{13}-3) &= 2(\sqrt{13}-3)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2310053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
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Evaluate series $\sum\limits_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}}$
Determine the value of
$$\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}}$$
or $$\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+\cdots$$
for $x\in\mathbb{R}$.
The answer is $\dfrac{x}{1-x}$ for $x\in(0,1)$. To prove this, notice
$$\f... | As soon as $|x|<1$ we have
$$ \frac{x^{2^{n-1}}}{1-x^{2^n}} = x^{2^{n-1}}+x^{3\cdot 2^{n-1}}+x^{5\cdot 2^{n-1}}+\ldots \tag{1}$$
hence:
$$ \sum_{n\geq 1}\frac{x^{2^{n-1}}}{1-x^{2^n}} = \sum_{m\geq 0}\sum_{h\geq 0} x^{(2h+1) 2^m} = \sum_{n\geq 1} x^n = \frac{x}{1-x}\tag{2} $$
since every positive integer can be represen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2310696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
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Prove that $\frac{(\sin 20^\circ + \cos 20^\circ)^2}{\cos 40^\circ} = \cot 25^\circ$ So I'm trying to come up with an answer to this question for hours now. I don't know what I'm doing wrong and none of the calculators on the internet couldn't help so I figured I should ask people.
What have I done so far:
$\frac{(\sin... | Generalization:
$$\dfrac{(\cos x+\sin x)^2}{\cos2x}=\dfrac{(\cos x+\sin x)^2}{\cos^2x-\sin^2x}=\dfrac{\cos x+\sin x}{\cos x-\sin x}$$ provided $\cos x+\sin x\ne0\iff\tan x\ne-1$
Now $$\dfrac{\cos x+\sin x}{\cos x-\sin x}=\dfrac{1+\tan x}{1-\tan x}=\tan\left(45^\circ+x\right)$$
Can you identify $x$ here?
Alternatively,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Which way is correct to represent this differentiation? $$3x^2+x^3$$
Option 1
$$\frac{d}{dx}(3x^2+x^3)=6x+3x^2$$
Option 2
$$\frac{d(3x^2+x^3)}{dx}=6x+3x^2$$
My student wrote the Option 2 style, not sure whether I can consider is ok or not.
| Both of the given representations are equivalent.
For:
$$y(x)=3x^2+x^3$$
$$\frac{dy}{dx} \equiv \frac{d(3x^2+x^3)}{dx} \equiv \frac{d}{dx}(3x^2+x^3) \equiv \frac{d(3x^2)}{dx}+\frac{d(x^3)}{dx} \equiv \frac{d}{dx}(3x^2)+\frac{d}{dx}(x^3)
=6x+3x^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2313391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Confusion about infinitely many bounded prime gaps? I was working at something and I had this idea: Let $f(x) = \frac{1}{x}$ for all $x > 0$. Let $p_n$ denote the $n$'th prime number. Then by mean value theorem $$ f(p_{n+1}) - f(p_n) = f'(c_n) (p_{n+1} - p_n)$$ where $c_n \in (p_n,p_{n+1})$. Hence $$ \frac{1}{p_{n+1}} ... | You conclude at some point:
Therefore $$\frac{1}{p_{n+1}}<−D\cdot Z_p(2,n+1)+D\cdot C_2+\frac{1}{p_N}$$
which basically is the same as;
$$\color{red}{0<}\frac{1}{p_{n+1}}<\color{red}{D\cdot \left(C_2- Z_p(2,n+1)\right)+\frac{1}{p_N}}$$
and the next thing you do is:
We take $D$ such that $D\cdot \left(C_2−Z_p(2,n+1)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2313772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Manipulating an algebraic equation This came in a competition, I went back home and tried it but was unsuccessful.
If $$\frac{a^3+3ab^2}{3a^2b+b^3} = \frac{x^3+3xy^2}{3x^2y+y^3}$$ then -
$$1) ~ax =by
~~2)~ xy = ab
~~3)~ ay = bx
~~4) ~ax = b$$
It is clear that if the following given condition holds for the variables th... | Rewrite $$\frac{a^3+3ab^2}{3a^2b+b^3} = \frac{x^3+3xy^2}{3x^2y+y^3}\tag 1$$ as
$$\frac{ ({a \over b})^3+3{a \over b}}{3({a\over b})^2+1} = \frac{({x \over y})^3+3{x\over y}}{3({x\over y})^2+1} \tag 2$$ and set $t={a\over b},\; s={x\over y}.$ Then
$$0=\frac{t^3+3t}{3t^2+1}-\frac{s^3+3s}{3s^2+1}=\frac{(t-s)\left[(t-s)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Point out my fallacy, in sequence and series.
The sum of the first $n$--terms of the series $1^2+2\cdot2^2+3^2+2\cdot4^2+\cdots$ is $\dfrac{n(n+1)^2}{2}$, when $n$ is even. When $n$ is odd, the sum is?
I got the correct answer when is replaced $n\rightarrow (n+1)$ to make above valid for odd, but when I tried the dif... | For $n$ odd case: $$\big(1^2+3^2+\cdots +n^2\big)+2\big(2^2+4^2+\cdots+(n-1)^2\big) \\= \big(1^2+3^2+\cdots +n^2\big)+2\big(2^2+4^2+\cdots+(n+1)^2\big) - 2(n+1)^2 \\= \frac{(n+1)(n+2)^2}{2} - 2(n+1)^2 \\= \frac{n^2(n+1)}{2}.$$
Let $T1$ denote number of terms contributed by odd indexes (for example $1^2, 3^2$ etc) and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
$a_1+a_2+…+a_n=2008$ where all $a_i$ are positive integers. If $A_k = a_1 a_2 … a_k$ , what is the largest possible value of $A_1+A_2+…+A_n$? The sum of a set of positive integers = $2008$.
$a_1+a_2+…+a_n=2008$ where all $a_i$ are positive integers.
If $A_k=a_1 a_2 … a_k$ , what is the largest possible value of $A_1+A_... | All of the $a_i$ must come in weakly decreasing order.
If $a_1+\dots+a_n=2008$ with $a_1$ an even number bigger than $2$, then consider $$(a_1/2)+2+a_2+\dots+a_n+1+\dots+1=2008.$$
If $a_1+\dots+a_n=2008$ with $a_1$ odd and at least $5$, say $a_1=2k+1$, consider
$$
(a_1-2)+2+a_2\dots+a_n=2008
$$
(Clearly $a_1=1$ cannot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2316434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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How to find eigenvector of large matrix without substitution directly. \begin{bmatrix}a&b&b&0&0&0\\b&a&b&0&0&0\\b&b&a&0&0&0\\0&0&0&a&b&b\\0&0&0&b&a&b\\0&0&0&b&b&a\end{bmatrix}
I have a large matrix as above which I already know that
\begin{bmatrix}a&b&b\\b&a&b\\b&b&a\end{bmatrix} have three eigenvectors equal to \begi... | In general, if $A$ and $B$ are diagonalizable as $A=MD_AM^{-1}$ and $B=ND_BN^{-1}$ ( where the columns of $M$ are the eigenvectors of $A$ and the columns of $N$ are the eigenvectors of $B$) than we have:
$$
C=\begin{pmatrix}A&0\\0&B
\end{pmatrix}=\begin{pmatrix}MD_AM^{-1}&0\\0&ND_BN^{-1}
\end{pmatrix}=
$$
$$
=\begin{pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2318474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find closed formula and limit for $a_1 =1$, $2a_{n+1}a_n = 4a_n + 3a_{n+1}$ Tui a sequence $(a_n)$ defined for all natural numbers given by
$$a_1 =1, 2a_{n+1}a_n = 4a_n + 3a_{n+1}, \forall n \geq 1$$
Find the closed formula for the sequence and hence find the limit.
Here, what I have done:
$$2a_{n+1}a_n = 4a_n + 3a_{... | Enforcing $u_n=1/a_n$ gives
$$2a_{n+1}a_n = 4a_n + 3a_{n+1}\Longleftrightarrow 4u_{n+1}+ 3u_n =2\Longleftrightarrow u_{n+1}=- \frac34u_n +\frac12$$
Now set $x_n = u_n-\frac27$ then you can easily check that we have $x_{n}= -\frac34x_{n-1}$ which is a geometric sequence
$$x_{n}= -\frac34x_{n-1}\implies x_n =\left(-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find Jordan's normal form It is first time I try normally to figure out how it works.
So there's the matrix:
$$A=\begin{pmatrix} 0 & 6 & 4 & 0 \\ 0 & 4 & 0 & 2 \\ 0 & -1 & 2 & -1 \\1 & -3 & -2 & 2 \end{pmatrix}$$
1.) First of all we have to find eignevalue:
$$\mathrm{det}|A-\lambda{I}|= \begin{vmatrix} 0 - \lambda & 6 ... | You can easily, though annoyingly carefully, check that $\;(A-2I)^3=0\;$ and that means the minimal polynomial of your matrix is $\;(x-2)^3\;$ , which means its Jordan form must have at least (and, in this case, obviously also at most) a Jordan block of size three, so
$$J_A=\begin{pmatrix}
2&1&0&0\\
0&2&1&0\\
0&0&2&0\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2320749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove or disprove that where $a, b \in Z$, the number a+b is even if and only if a-b is even.
Prove or disprove that where $a, b \in Z$, the number a+b is even if and only if a-b is even.
Can some help me improve on this or check it please?
If $a + b$ is even, then there is some number $k \in Z$ such that $a + b = 2k... | We have $a+b = a-b +2b$ and so $a+b$ and $a-b$ differ by an even number. Therefore, they have the same parity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Find $(x + 1)^{−3}$ in $\mathbb{Z}_2[x]/(x^5 + x^4 + 1)$ In a previous question, I established that $x^5 + x^4 + 1$ is irreducible in $\mathbb{Z}_2[x]/(x^5 + x^4 + 1)$, so it will not be divisible by $x+1$. Hence $x^5+x^4+1$ will be coprime to $(x+1)^3 = x^3+x^2+x+1$
So far I have used the Euclidean algorithm to get:
... | Using the algorithm detailed in this answer with polynomials (and writing top to bottom instead of left to right),we get
$$
\begin{array}{c|c|c|c}
x^5+x^4+1&0&1&\\
x^3+x^2+x+1&1&0&\\
x&x^2+1&1&x^2+1\\
1&\color{#C00}{x^4+x^3+x}&x^2+x+1&x^2+x+1\\
0&x^5+x^4+1&x^3+x^2+x+1&x
\end{array}
$$
which says that in $\mathbb{Z}_2$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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$8\sin x\cos x-\sqrt{6} \sin x- \sqrt{6} \cos x+1=0$, Solve for $x$
Solve for $x:0\leq x \leq \frac{\pi}{2}$
$$8\sin x\cos x-\sqrt{6} \sin x- \sqrt{6} \cos x+1=0$$
My attempt,
I changed it into $$1+4 \sin 2x-2\sqrt{3} \sin (x+\frac{\pi}{4})=0$$
| Hint: Let $t=\sin x+ \cos x$ and use $1=\sin^2 x+\cos^2 x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Proving a relation related to quadratic equation Question:If $α$ and $β$ be the roots of $ax^2+2bx+c=0$ and $α+δ$, $β+δ$ be those of $Ax^2+2Bx+C=0$, prove that, $\frac{b^2-ac}{a^2}=\frac{B^2-AC}{A^2}$.
My Attempt: Finding the sum of roots and product of roots for both the equations we get,
$α+β=\frac{-2b}{a}$
$αβ=\... | Alternative hint: we can assume WLOG that $\,a=A=1\,$, since both the roots and the equality to be proved are homogeneous in the respective coefficients.
Then, if $\alpha, \beta$ are the roots of $x^2+2bx+c=0$, the polynomial with roots $\alpha+\delta, \beta+\delta$ is:
$$(x-\delta)^2+2b (x-\delta)+c=0 \;\;\iff\;\; x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
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Multiplied X in Quadratic Equation - Delta always lower than 0 I'm new to Quadratic Equations and I am following the two formulas to find $x_1$ and $x_2$ and they are:
$\Delta = b^2 - 4ac$
$x = \dfrac{-(b) \pm \sqrt\Delta}{2a}$
but delta always gives a negative number;
for example I have the following equation:
$6x^2 +... | $6x^2+11x−35=0$
$a =6; b= 11; c= -35$
$b^2 = 121$.
$4ac = 4*6*(-35) =-840$
$b^2 - 4ac = 121 - (-840) = 121 + 841 = 961=31$.
So $\Delta = 961=31^2 > 0$ and $\sqrt{961} = 31$
$x = \frac {-11 \pm 31}{2*6}$.
$x = \frac {-42}{12}$ or $x = \frac {20}{12}$
$x = -\frac 72$ or $x = \frac 53$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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Infinitely many positive integers $n$ such that $n^2+1 \mid 1 \cdot 2 \cdot 5 \cdot 10 \cdots ((n-1)^2+1)$
Prove that there exist infinitely many positive integers $n$ such that $$n^2+1 \mid 1 \cdot 2 \cdot 5 \cdot 10 \cdots ((n-1)^2+1).$$
We need to prove that there are infinitely many $n$ such that $\prod_{m=0}^{n-... | I conjecture that $n=2^{2k+1}$ is always a solution. We have that $$2^{4k+2}+1=(2^{2k+1}+1)^2-2^{2k+2}=(2^{2k+1}-2^{k+1}+1)(2^{2k+1}+2^{k+1}+1)$$
Now we can pick $m=2^{k+1}+1$ then $$m^2+1=2^{2k+2}+2^{k+2}+2=2(2^{2k+1}+2^{k+1}+1)$$ and $m=2^{k+1}-1$ then $$m^2+1=2^{2k+2}-2^{k+2}+2=2(2^{2k+1}-2^{k+1}+1)$$
And both $m$'s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
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$\arctan(\tan(-\frac{3\cdot \pi}{5}))+\operatorname{arccot}(\cot(-\frac{3\cdot \pi}{5}))=?$ We know that $$\DeclareMathOperator{\arccot}{arccot}\arctan(-x)=-\arctan(x)$$ $$\arccot(-x)=\pi-\arccot(x)$$ $$\cot(-x)=-\cot(x)$$ $$\tan(-x)=-\tan(x)$$So I have figured out that $$\arctan(\tan(-\frac{3\cdot \pi}{5}))=\arctan(-\... | For all real $x$ we have $-\frac{\pi}{2}<\arctan{x}<\frac{\pi}{2}$ and $0<\operatorname{arccot}{x}<\pi$.$\DeclareMathOperator{\arccot}{arccot}$
$$\arctan\tan\left(-\frac{3 \pi}{5}\right)+\arccot\cot\left(-\frac{3\pi}{5}\right)=$$
$$=\arctan\tan\frac{2 \pi}{5}+\arccot\cot\frac{2\pi}{5}=\frac{4\pi}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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A simple trigonometric inequality How to sole this inequalities:
$$2\cos\left(\frac{x}{2}\right) \leq -\cos(x).$$
I use the equality $$\cos\left(\frac{x}{2}\right)=\pm\sqrt{\frac{1+\cos x}{2}}$$
to find the solution for the equality
$$2\cos\left(\frac{x}{2}\right) = -\cos(x).$$
I get a second degree equation $t^2-2t-... | \begin{align}
2\cos\frac{x}{2}&\le-\cos x\\
\cos x+2\cos\frac{x}{2}&\le0\\
2\cos^2\frac{x}{2}-1+2\cos\frac{x}{2}&\le0\\
\left(2\cos\frac{x}{2}+1\right)^2&\le3\\
\frac{-1-\sqrt{3}}{2}\le\cos\frac{x}{2}&\le\frac{-1+\sqrt{3}}{2}
\end{align}
So, $\displaystyle x\in\bigcup_{n\in\mathbb{Z}}\left[4n\pi+2\cos^{-1}\left(\frac{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Arithmetic-Geometric mean sequences limit If $a,b$ are positive quantities such that $(a<b)$ and if
\begin{align}
a_1 &= \frac{a+b}2 & b_1 & = \sqrt{a_1 b} \\
a_2 &= \frac{a_1+b_1}2 & b_2 &= \sqrt{a_2 b_1} \\
&\phantom{36pt} \vdots & & \\
a_n &= \frac{a_{n-1}+b_{n-1}}2 & b_n & = \sqrt{a_n b_{n-1}} \\
&\phantom{... | If we assume that $b=1$ and $a=\cos\theta$ with $\theta\in\left(0,\frac{\pi}{2}\right)$ we get:
$$ a_1 = \frac{1+\cos\theta}{2} = \cos^2\frac{\theta}{2},\qquad b_1=\cos\frac{\theta}{2}$$
$$ a_2 = \cos\frac{\theta}{2}\left(\frac{1+\cos\frac{\theta}{2}}{2}\right),\qquad b_2=\cos\frac{\theta}{2}\cos\frac{\theta}{4}$$
so b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Somebody help me please. I have a difficult inequality. Let $ab+bc+ca=1$. Prove that $2 \ge \sqrt{1+a^2} + \sqrt{1+b^2}+\sqrt{1+c^2}-a-b-c \geq \sqrt3 $.
| The right inequality.
We need to prove that
$$ \sum\limits_{\text{cyc}}\sqrt {\left(a+c\right)\left(a+b\right)}\geq a+b+c + \sqrt3\cdot\sqrt {ab+ac+bc}$$ or
$$2\left(a+b+c - \sqrt {3(ab+ac+bc)}\right) - \sum_{cyc}\left(\sqrt {a+c} - \sqrt {b+c}\right)^{2}\geq0.$$
But $$ 2\left(a+b+c- \sqrt {3(ab+ac+bc)}\right) - \sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Directional derivatives of a function with partial derivatives The prompt is to 1. find the directional derivatives of the function $f$ at a point $P(1, -1,2)$ in the direction towards origin and we are given $$f(1, -1, 2) = 5$$ $$f_x(1, -1, 2) = 1$$ $$f_y(1, -1, 2) = 3$$ $$f_z(1, -1, 2) = -1$$
*Find the plane tangen... | The unit vector at $P(1, -1,2)$ in the direction of the origin is $\mathbf{u}=\left(-\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}}\right)$ and the gradient is $\nabla f = (1, 3, -1)$ therefore the directional derivative at $P$ in the direction $\mathbf{u}$ is
$$ D_{\mathbf{u}}f(1,-1,2)=\nabla f\cdot\mathbf{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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sum of integer parts of roots
If $\alpha,\beta,\gamma$ are real roots of the equation $\displaystyle \frac{1}{x}+\frac{1}{x-1}+\frac{1}{x-2} = \frac{11}{54}x^2.$ Then $\lfloor \alpha \rfloor + \lfloor \beta \rfloor +\lfloor \gamma \rfloor $
$\bf{Attempt}$ From equation $$\frac{x^2-3x+2+x^2-2x+x^2-x}{x(x-1)(x-2)} = ... | If $x\ge 3$, then the LHS is $\le \frac13+\frac12+1=\frac{11}{6}$ and the RHS is $\ge \frac{11}6$ with equality iff $x=3$. Thus we have found our first real root and shown that it is the largest: $x=3$.
If $x<0$, the LHS is negative, the RHS positive, hence we cannot find a real solution there.
If $2<x<3$, then the LHS... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a,b,c$ be roots of $x^3+px+r=0$. Find the cubic whose roots are $(a-b)^2$,$ (b-c)^2$ and $(c-a)^2$ Question
Let $a,b,c$ be roots of $x^3+px+r=0$. Then find the cubic whose roots are $(a-b)^2, (b-c)^2$ and $ (c-a)^2$
Attempt
I have tried using Vieta's formulas to compute coefficients of the sought cubic.
For sum of... | Hint: assume WLOG that the three roots $a,b,c$ are all distinct (the case of multiple roots follows by continuity, or could be handled separately). Then subtracting the equations $a^3+pa+r=0\,$ and $b^3+pb+r=0\,$ gives:
$$\require{cancel}
a^3-b^3 + p(a-b) = 0 \iff \cancel{(a-b)}\big(a^2+ab+b^2+p\big) = 0 \iff (a-b)^2=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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BMO2 1992 - Triangle Circumscribed in Circle The circumcircle of triangle $ABC$ has radius $R$ satisfying
$$AB^2+AC^2=BC^2-R^2$$
Prove that the angles in the triangle are uniquely determined, and state the values for the angles.
So far I have used the Extended Law of Sines and the fact that $\alpha+\beta+\gamma=180$ ... | $b^2+c^2=a^2-R^2$ leads to
$$ 4\sin^2 B+4\sin^2 C=4\sin^2(B+C)-1 \tag{1}$$
that can be written as
$$ \left(2\cos(B+C)-\cos(B-C)\right)^2 + \sin^2(B-C) = 0.\tag{2} $$
It follows that $B=C$ and $2\cos(2B)-1=0$, from which
$$ A=\frac{2\pi}{3},\quad B=C=\frac{\pi}{6}.\tag{3} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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What is the probability of the sum of four dice being 22? Question
Four fair six-sided dice are rolled. The probability that the sum of the results being $22$ is $$\frac{X}{1296}.$$ What is the value of $X$?
My Approach
I simplified it to the equation of the form:
$x_{1}+x_{2}+x_{3}+x_{4}=22, 1\,\,\leq x_{i} \,\,\le... | It is the coefficient of $x^{22}$ in the expansion of the generating function $$(x^6+x^5+x^4+x^3+x^2+x^1)^4=\left(\frac{x(1-x^6)}{1-x}\right)^4$$ which is $$1\cdot{{x}^{24}}+4\cdot {{x}^{23}}+10\cdot {{x}^{22}}+20\cdot {{x}^{21}}+35\cdot {{x}^{20}}+56\cdot {{x}^{19}}+80\cdot {{x}^{18}}\\+104\cdot {{x}^{17}}+125\cdot {{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 11,
"answer_id": 1
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$\text{If } x(x+y+z)=20,y(x+y+z)=30 \text{ and } z(x+y+z)=50 \text{ then the value of } 2(x+y+z) is:$
If $x(x+y+z)=20$, $y(x+y+z)=30$ and $z(x+y+z)=50$ then what is the value of $2(x+y+z)$?
Ans. $20$
I have tried the following:
$$ \frac{20}{x}=\frac{30}{y}=\frac{50}{z}$$
From which I get:
$$x:y:z=2:3:5$$
Now,
$$2(x+y... | $$x(x+y+z)=20,\\ y(x+y+z)=30\\ z(x+y+z)=50\\ $$ by summing all these three eqatuations$$\left( x+y+z \right) \left( z+y+x \right) =100\\ { \left( x+y+z \right) }^{ 2 }=100\\ x+y+z=\pm 10$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Does $\lim_{n\to \infty} \left(\frac{a}{a+bn}\right)^{\frac{1}{n}}$ exist? Let $a, b >0$ and $a \ll b$, $a<1$. Does $\lim_{n\to \infty} \left(\frac{a}{a+bn}\right)^{\frac{1}{n}}$ exist ?
| Using
\begin{align}
\frac{a}{a + b \, n} &= 1 - \frac{1}{1 + \frac{a}{b \, n}} \\
&= 1 - \left( 1 - \frac{a}{b \, n} + \frac{a^{2}}{b^{2} \, n^{2}} + \mathcal{O}\left(\frac{1}{n^{3}}\right) \right) \\
&= \frac{a}{b \, n} - \left(\frac{a}{b \, n}\right)^{2} + \mathcal{O}\left(\left(\frac{a}{b \, n}\right)^{3}\right) \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2354323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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solve trigonometric lim with tan Hi guys I tried to solve the following problem lately and got stucked.
I'd love to get some help and guidance.
consider this:
$$\lim_{x \rightarrow\infty}\tan\left(\frac{\pi x}{2x+1}\right)^{\frac{1}{x}}
$$
I did the following:
1. break to sin and cos :
$$
\lim_{x \rightarrow\infty}\ta... | Considering $$A=\tan\left(\frac{\pi x}{2x+1}\right)^{\frac{1}{x}}$$ first $$\tan\left(\frac{\pi x}{2x+1}\right)=\tan\left(\frac{\pi} 2-\frac{\pi }{2(2 x+1)}\right)=\cot \left(\frac{\pi }{2(2 x+1)}\right)$$ So $$\log(A)=\frac 1 x \log\left(\cot \left(\frac{\pi }{2(2 x+1)}\right)\right)$$ Now, using Taylor series startin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
if a,b,c are roots of a cubic equation then for the following question... If $a, b, c$ are roots of $x^3 -3x^2 + 2x +4 = 0$ and
$$y= 1 + \frac{a}{x-a} + \frac{bx}{(x-a)(x-b)} + \frac{cx^2}{(x-a)(x-b)(x-c)}$$
then value of $y$ at $x=2$ is:
| We can reduce this to all be over the denominator $(x-a)(x-b)(x-c)$:
\begin{align}y&=1 + \frac{a}{x-a} + \frac{bx}{(x-a)(x-b)} + \frac{cx^2}{(x-a)(x-b)(x-c)}\\\\
&=\frac{(x-a)(x-b)(x-c)}{(x-a)(x-b)(x-c)}+\frac{a(x-b)(x-c)}{(x-a)(x-b)(x-c)}\\
&\qquad\qquad+\frac{bx(x-c)}{(x-a)(x-b)(x-c)}+\frac{cx^2}{(x-a)(x-b)(x-c)}\\\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How to calculate $\sum_{r=1}^\infty\frac{8r}{4r^4+1}$?
Calculate the following sum: $$\frac{8(1)}{4(1)^4+1} + \frac{8(2)}{4(2)^4+1} +\cdots+ \frac{8(r)}{4(r)^4+1} +\cdots+ \text{up to infinity}$$
MY TRY:- I took $4$ common from the denominator. and used $a^2+b^2=(a+b)^2-2ab$. It gave me two brackets, whose subtractio... | Hint. The sum is telescopic. Note that $4r^4+1=(2r^2-2r+1)(2r^2+2r+1)$, and
$$\frac{8r}{4r^4+1}=\frac{2}{2r(r-1)+1}-\frac{2}{2(r+1)r+1}.$$
Hence
$$\sum_{r=1}^n\frac{8r}{4r^4+1}=\sum_{r=1}^n\left(\frac{2}{2r(r-1)+1}-\frac{2}{2(r+1)r+1}\right)=\frac{2}{2\cdot 1(1-1)+1}-\frac{2}{2(n+1)n+1}.$$
Can you find the sum of the s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2357340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
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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.