Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Multivariable limit exists? Does the limit
$$\lim_{(x,y)\rightarrow (0,0)} \frac{y^4}{x^\beta(x^2+y^4)}$$ exists for $\beta>0$? I don't think it exists but how do you prove it rigorously. Thanks
| $$\lim\limits_{(x,y)\to (0,0)} \frac{y^4}{x^\beta\left(x^2+y^4\right)}$$
Using polar coordinates, we have
$$\lim\limits_{r\to 0^+} \frac{(r\sin\phi)^4}{(r\cos\phi)^\beta\left((r\cos\phi)^2+(r\sin\phi)^4\right)}$$
$$=\lim\limits_{r\to 0^+} \frac{r^4\sin^4\phi}{r^\beta\left(\cos^\beta\phi\right)\left(r^2\cos^2\phi+r^4\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1187338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the derivative using the chain rule and the quotient rule
$$f(x) = \left(\frac{x}{x+1}\right)^4$$ Find $f'(x)$.
Here is my work:
$$f'(x) = \frac{4x^3\left(x+1\right)^4-4\left(x+1\right)^3x^4}{\left(x+1\right)^8}$$
$$f'(x) = \frac{4x^3\left(x+1\right)^4-4x^4\left(x+1\right)^3}{\left(x+1\right)^8}$$
I know the fin... | Apparently you complicated things too much.
Starting from $$f(x) = \left(\frac{x}{x+1}\right)^4$$ with the chain rule we get first
\begin{align}
f'(x) &= 4\left(\frac{x}{x+1}\right)^3\cdot \left(\frac{x}{x+1}\right)^\prime\notag\\
&= 4\left(\frac{x}{x+1}\right)^3\cdot \frac{x'(x+1)-x(x+1)'}{(x+1)^2}\notag\\
&= 4\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1189494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Generating Functions - Extracting Coefficients In many counting problems, we find an appropriate generating function which allows us to extract a given coefficient as our answer.
In cases where the generating function is not one that is easily used as an infinite sum, how does one alter the generating function for sim... | You are looking for the coefficient of $x^{24}$ in:$$x^{12}\left(\frac{1-x^6}{1-x}\right)^4$$
Which is the same as the coefficient of $x^{24-12}=x^{12}$ in $$\left(\frac{1-x^6}{1-x}\right)^4\tag{1}$$
Now, $$(1-x^6)^4 = 1-4x^6+6x^{12}-4x^{18}-x^{24}$$
And $$\frac{1}{(1-x)^4} = \sum_{n=0}^{\infty} \binom{n+3}{3}x^n$$
So ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Does every $9 \times 9$ Latin square contain a $3 \times 3$ submatrix containing each symbol in $\{1,2,\ldots,9\}$?
Q: Does every $9 \times 9$ Latin square on the symbol set $\{1,2,\ldots,9\}$ contain a $3 \times 3$ submatrix containing each symbol in $\{1,2,\ldots,9\}$?
This one has $1728$ such submatrices, which is... | We might consider the general concept of a saturated sub-matrix, or $SSM$, being an $M \times K$ sub-matrix within a Latin square of of rank $N = M \times K$, that contains all $N$ values.
We are looking for Latin squares that are completely void of $SSM$s, so they might well be called poly-unsaturated!
For example $6 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Posed in regional mathematics Olympiad 1995 Call a positive integer $n$ good if there are $n$ integers, positive or negative, and not necessarily distinct, such that their sum and product are both equal to $n$
For example, $8$ is good since
$$\begin{align}
8 = 4\times2\times1\times1&\times1\times1\times(-1)\ti... | As “Your Ad Here” wrote, if $n=4k+1$, take the following numbers:
$$4k+1,\underbrace{1,\dots,1}_{2k\,\text{times}},\underbrace{-1,\dots,-1}_{2k\,\text{times}}.$$
If $n=4k$, where $k>1$, write $n=2^j q$, where $q$ is odd and $j\ge2$, and proceed by induction on $j$.
If $j=2$, $n=4k$ with $k$ odd. Then take these numbers... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1194020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Taylor series for the function $f(z) = \frac{1}{(z-5)(z-7)}$ on a disc centered at point $z_0=3$ I started by expressing the function as sum of two fractions using partial fraction decomposition to get $\frac{-1}{2(z-5)} + \frac{1}{2(z-7)}$
However I could only then end up writing that as the sum of two power series :
... | If we are looking to construct the series around $z_0 = 3$, we could have $\lvert z - 3\rvert < 1$ or $\lvert z - 3\rvert > 1$ which would be the series inside and outside a disc of radius one. In the first case,
\begin{align}
\frac{1/2}{z - 3 - 4} - \frac{1/2}{z - 3 - 2}&= \frac{-1/8}{1 - (z-3)/4} + \frac{1/4}{1-(z-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Length and breadth of a rectangle enclosed between two semi-circles of given radii
34. It is required to take a rectangular frame in a horizontal position along a corridor bounded by vertical walls of which a horizontal cross-section is two concentric semicircles of radii $r$ and $r\ \sqrt{3}$; the frame is of length ... | Place the circle centers at the origin of a Cartesian grid. Without loss of generality, we can assume optimal rectangle is oriented with its base parallel to the $X$ axis, touching the smaller circle at the base midpoint at $(r,0)$.
Then the coordinates of the upper right hand corner can be taken to be $(rp,rq)$ where,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Aproximation of $a_n$ where $a_{n+1}=a_n+\sqrt {a_n}$ Let $a_1=2$ and we define $a_{n+1}=a_n+\sqrt {a_n},n\geq 1$.
Is it possible to get a good aproximation of the $n$th term $a_n$?
The first terms are $2,2+\sqrt{2}$, $2+\sqrt{2}+\sqrt{2+\sqrt{2}}$ ...
Thanks in advance!
| Here is a method to develop the asymptotic behaviour of $a_n$ with a precision of two terms.
First, simply remark that $a_n$ grows to infinity since $a_n \geq n$ (by induction).
Then remark that $\sqrt{a_{n+1}} - \sqrt{a_n} = \frac{a_{n+1}-a_{n}}{\sqrt{a_{n+1}} + \sqrt{a_n}} = \frac{1}{1+\sqrt{1+\sqrt{\frac{1}{a_n}}}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1196979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Integrating reciprocals of functions with known antiderivatives If
$$\int_{}^{} f(x)\,dx$$
is known, is there a way to directly find
$$\int_{}^{} \frac{1}{f(x)}\,dx$$
| Consider the function $f(x)$ in a power series. Let
\begin{align}
f(x) = \sum_{n=0}^{\infty} a_{n} \, x^{n}
\end{align}
then it can be seen that
\begin{align}
\frac{1}{f(x)} = \frac{1}{a_{0}} - \frac{a_{1}}{a_{0}^{2}} \, x + \frac{a_{1}^{2} - a_{2}}{a_{0}^{3}} \, x^{2} + \cdots
\end{align}
Integration of each shows
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1198520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Radius of convergence of the power series $\sum_{n=1}^{\infty}a_nz^{n^2}$
Find the radius of convergence of the power series $$\sum_{n=1}^{\infty}a_nz^{n^2}$$ where , $a_0=1$ and $a_n=\frac{a_{n-1}}{3^n}$.
My Work :
We, have, $$\frac{a_1}{a_0}.\frac{a_2}{a_1}...\frac{a_n}{a_{n-1}}=\frac{1}{3}.\frac{1}{3^2}...\frac{1}... | Just to add, if $\sum_n b_nz^n$ is a power series, then there is an explicit formula for the radius of convergence: $$R = \frac{1}{\limsup_n |b_n|^{\frac{1}{n}}}$$ In this case we have that $b_{n^2}=a_n=3^{-\frac{n(n+1)}{2}}$. Hence $b_{n^2}^{\frac{1}{n^2}}=3^{-\frac{1}{2}(1+\frac{1}{n})}$. Thus, substituting $k$ for $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1199359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Let $∑_{n=0}^∞c_n z^n $ be a representation for the function $\frac{1}{1-z-z^2 }$. Find the coefficient $c_n$ Let $∑_{n=0}^∞c_n z^n $ be a power series representation for the function $\frac{1}{1-z-z^2 }$. Find the coefficient $c_n$ and radius of convergence of the series.
Clearly this is a power series with center $z_... | Using the factorization $1 - z - z^2 = -(z + \varphi)(z + \bar{\varphi})$, where $\varphi = (1 + \sqrt{5})/2$ and $\bar{\varphi} = (1 - \sqrt{5})/2$, we write
\begin{align}\frac{1}{1 - z - z^2} &= \frac{1}{\sqrt{5}}\left(\frac{1}{z + \varphi} - \frac{1}{z + \bar{\varphi}}\right) \\
& = \frac{1}{\sqrt{5}}\left(\frac{1/\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1199807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Sum $\pmod{1000}$
Let $$N= \sum_{k=1}^{1000}k(\lceil \log_{\sqrt{2}}k\rceil-\lfloor \log_{\sqrt{2}}k \rfloor).$$ Find $N \pmod{1000}$.
Let $\lceil x \rceil$ be represented by $(x)$ and $\lfloor x \rfloor$ be represented by $[x]$.
Consider $0 < x < 1$ then:
$$(x) - [x] = 1 - 0 = 1$$
Consider $x=0$ then:
$$(x) - [x] ... | Note that $(\lceil \log_{\sqrt{2}}k\rceil-\lfloor \log_{\sqrt{2}}k \rfloor)=1$ unless $k$ is an integer power of $\sqrt 2$, which is when $k$ is a power of $2$. In that case it is zero. So your sum is $$N= \sum_{k=1}^{1000}k-\sum_{m=0}^92^m$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1204494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A problem I didn't know since high school algebra Determine all positive integers which can be written as a sum of two squares of integers.
This is a problem I saw when I was in high school...
*
*sum of two squares of integers can be (4k, 4k+1, 4k+2, but no 4k+3?)
| Every integer $n$ can be expressed as $n=4k+r$ for $r=0,1,2,3$. $r$ is called the remainder after division by 4. Now $n^2 = (4k+r)^2 = 4^2k^2 + 8kr + r^2$.
If $r = 0$ then $r^2=0$. So $n^2=4k'$ for some $k'$.
If $r=1$ then $r^2=1$. So $n^2=4k'+1$ for some $k'$.
If $r=2$ then $r^2=4$. So $n^2=4k'$ for some $k'$.
If $r=3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1204641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Use logarithmic differentiation to find $\frac{dy}{dx}$ Here is the problem as well as my work so far. Any advice or hints regarding where I should go from here would be appreciated.The arrow indicates where I got stuck.
Where do I go from here?
EDIT
I realize my mistake in not using logarithms. Here is my second at... | here is a way to do check your i will do it little differently. i will start with
$$\ln y = 5 \ln (x^2 + 3) - \frac 12 \ln (x+1) $$ differencing this we get $$\frac{dy}{y} = 5 \frac{d(x^2 + 3)}{x^2 + 3} - \frac{d(x+1)}{2(x+1)}=\frac{10x\, dx}{x^2 + 3} - \frac{dx}{2(x+1)} $$ therefore
$$\frac{dy}{dx} = y\left(\frac{10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1205328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Showing $1+2+\cdots+n=\frac{n(n+1)}{2}$ by induction (stuck on inductive step) This is from this website:
Use mathematical induction to prove that
$$1 + 2 + 3 +\cdots+ n = \frac{n (n + 1)}{2}$$
for all positive integers $n$.
Solution to Problem 1: Let the statement $P(n)$ be
$$1 + 2 + 3 + \cdots + n = \frac{n... | We know that
$P(k) = 1 + 2 + 3 + ... + k$
Therefore:
$P(k+1) = 1 + 2 + 3 + ... + k + (k+1)$
By induction hypothesis we have:
$1 + 2 + 3 + ... + k + (k+1) = \frac{(k+1)(k+2)}{2}$
so
$P(k+1) = 1 + 2 + 3+...+k+(k+1) = \frac{(k+1)(k+2)}{2}$
so
$P(k+1) = \frac{(k+1)(k+2)}{2}$
By induction we now know that since this is tr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1206645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
If $A,B>0$ and $A+B = \frac{\pi}{3},$ Then find Maximum value of $\tan A\cdot \tan B$. If $A,B>0$ and $\displaystyle A+B = \frac{\pi}{3},$ Then find Maximum value of $\tan A\cdot \tan B$.
$\bf{My\; Try::}$ Given $$\displaystyle A+ B = \frac{\pi}{3}$$ and $A,B>0$.
So we can say $$\displaystyle 0< A,B<\frac{\pi}{3}$$. No... | $$(\tan A+\tan B)^2-4\tan A\tan B=(\tan A-\tan B)^2\ge0$$
$$\iff4\tan A\tan B\le(\tan A+\tan B)^2$$
Also by $\bf{A.M\geq G.M},$ $\dfrac{\tan A+\tan B}2\ge\sqrt{\tan A\tan B}$
By Jensen's inequality, $$\frac{\tan A+\tan B}2\le\tan\frac{A+B}2=\frac1{\sqrt3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1207197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Apostol (3.8.22): Finding $f \circ g (x)$ $$
f(x)=
\begin{cases}
1 &\text{if} & |x|\leq 1,\\
0 &\text{if} & |x|>1,\\
\end{cases}
\qquad
g(x)=
\begin{cases}
2-x^2 &\text{if} & |x|\leq 2,\\
2 &\text{if} & |x|>2.
\end{cases}
$$
I need to find the composition $h(x)$ of $f(x)$ and $g(x)$ such that $h(x)=f(g(x))$. My answer ... | $$
f(g(x)) = \begin{cases} 1 & \left|g(x)\right| \leq 1 \\
0 & \left|g(x)\right| > 1
\end{cases}
$$
So you need to find those cases:
First, it's clear that when $|x| > 2$, $g(x) = 2 > 1$ and thus when $|x| > 2$, $f(g(x)) = 0$. So now we need to investigate when $|x| \leq 2$ and $g(x) = 2 - x^2$. It might help to righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1209828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Does this series $\sum_{i=0}^n \frac{4}{3^n}$ diverge or converge? I a newbie to series, and I have not done too much yet. I have an exercise where I have basically to say if some series are convergent or divergent. If convergent, determine (and prove) the sum of the series.
This is the first series:
$$\sum_{i=0}^n \fr... |
A proof that, if the left-hand-side converges:
$$a+ar+ar^2+\dotsb=\frac a{1-r}$$
Call the sum $S$. Thus, we have $S=a+ar+ar^2+\dotsb$.
$\phantom rS=a+ar+ar^2+ar^3+\dotsb$
$rS=\phantom{a+{}}ar+ar^2+ar^3+\dotsb$ (using the distributive property)
Thus:
\begin{align}
S&=a+rS\\
S-rS&=a\\
S(1-r)&=a\\
S&=\frac a{1-r}
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1210267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
$(\delta,\varepsilon)$ Proof of Limit I wish to prove that $\lim_{x\to 2} {x+1 \over x+2} = {3 \over 4} $.
The $(\delta,\varepsilon)$ limit definition in this case is:
$\forall \epsilon >0, \exists \delta >0$ such that $0<|x-2|<\delta \Rightarrow |{x+1 \over x+2} - {3 \over 4}| < \epsilon.$
Thus, I need to provide a $\... | You want to prove
$$\lim_{x\to 2}\frac{x+1}{x+2}=\frac{3}{4}.$$
By the definition of limit you need to find, for all $\epsilon>0$, a $\delta$ such that $0<|x-2|<\delta$ implies $$\left|\frac{x+1}{x+2}-\frac{3}{4}\right|<\epsilon.$$ Your best bet is to work back from the complicated expression. We can simplify,
$$\left|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1212672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Find the derivative of the function $F(x) = \int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt$. $$\begin{align}
\left(\int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt\right)' &= \frac{1}{\sqrt{2+t^4}}2x - \frac{1}{\sqrt{2+t^4}}\sec^2{x} \\
&= \frac{2x}{\sqrt{2+t^4}} - \frac{\sec^2{x}}{\sqrt{2+t^4}} \\
&= \frac{2x-\sec^2{x... | No. $t$ is a mute variable. The answer must have $x$ as the only variable. Yor computations are similar to
$$
\int_1^2t\,dt=2\cdot t-1\cdot t=t,
$$
which is obviously wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1213093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Limit as $x$ tend to zero of: $x/[\ln (x^2+2x+4) - \ln(x+4)]$ Without making use of LHôpital's Rule solve:
$$\lim_{x\to 0} {x\over \ln (x^2+2x+4) - \ln(x+4)}$$
$ x^2+2x+4=0$ has no real roots which seems to be the gist of the issue.
I have attempted several variable changes but none seemed to work.
| In the eternal words of Claude Leibovici, love Taylor Series.
We have
\begin{align*}
\ln(x^{2}+2x+4) &= \ln(4) + \frac{x}{2} + \frac{x^{2}}{8} + \mathcal{O}(x^{3})\\
\ln(x+4) &=\ln(4) + \frac{x}{4} -\frac{x^{2}}{32} + \mathcal{O}(x^{3}).
\end{align*}
So
$$\lim_{x\to 0} {x\over \ln (x^2+2x+4) - \ln(x+4)} = \lim_{x\to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1213655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Find $z$ s.t. $\frac{1+z}{z-1}$ is real I must find all $z$ s.t. $\dfrac{1+z}{z-1}$ is real.
So, $\dfrac{1+z}{1-z}$ is real when the Imaginary part is $0$.
I simplified the fraction to $$-1 - \dfrac{2}{a+ib-1}$$
but for what $a,b$ is the RHS $0$?
| Hint:
$$\begin{align}\frac{1+z}{z-1} \,\,\, \text{is real} \ &\iff \frac{1+z}{z-1} = \overline{\frac{1+z}{z-1}} = \frac{\overline{1+z}}{\overline {z-1}} \\&\iff (1+z)(\overline {z}-1) = (1+\overline {z})(z-1)\\&\iff \mathfrak {Im} (z) = 0\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1216378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
A sequence is defined by $f(x) = 4^{x-1}$, find the sum of the first $8$ terms. A sequence is defined by $f(x) = 4^{x-1}$, find the sum of the first $8$ terms.
$\dfrac{a(1-r^n)}{1-r}$
$\dfrac{1(1-4^7)}{1-4} = 5461$.
The answer in the book is $21845$. How is this so?
Thank you
| For $a_n = 4^{n-1}$,
$$
a\frac{r^n-1}{r-1} = \frac{4^8-1}{4-1} = \frac{65536 -1}{3} = 21845.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1216672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that equation $x^6+x^5-x^4-x^3+x^2+x-1=0$ has two real roots Prove that equation
$$x^6+x^5-x^4-x^3+x^2+x-1=0$$ has two real roots
and $$x^6-x^5+x^4+x^3-x^2-x+1=0$$
has two real roots
I think that:
$$x^{4k+2}+x^{4k+1}-x^{4k}-x^{4k-1}+x^{4k-2}+x^{4k-3}-..+x^2+x-1=0$$
and
$$x^{4k+2}-x^{4k+1}+x^{4k}+x^{4k-1}-x^{... | I'm not sure of the general case; but for the first polynomial divide through by $x^3$, since $x\neq0$, and substitute $y=x-1/x$. You then obtain $y^3+y^2+2y+1=0$, which must have just one real root, in the interval $(-1,0).$ Call this root $y=a=x-1/x$. Then $x^2-ax-1=0$, which has a positive discriminant ($a^2+4$), ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
How to factorize $(x-2)^5+x-1$? This is a difficult problem.
How to factorize this?
$$(x-2)^5+x-1$$
we can't do any thing now and we should expand it first:
$$x^5-10x^4+40x^3-80x^2+81x-33$$
but I can't factorize it.
| finally found how looking in my previous questions.
first substitute $n=x-2$ then
$$(x-2)^5+x-1=n^5+n+1$$
$$n^5+n+1=n^5-n^2+n^2+n+1=n^2(n^3-1)+n^2+n+1$$
$$=n^2(n-1)(n^2+n+1)+n^2+n+1$$
$$=(n^2+n+1)(n^3-n^2+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1222420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Proof of sum results I was going through some of my notes when I found both these sums with their results
$$
x^0+x^1+x^2+x^3+... = \frac{1}{1-x}, |x|<1
$$
$$
0+1+2x+3x^2+4x^3+... = \frac{1}{(1-x)^2}
$$
I tried but I was unable to prove or confirm that these results are actually correct, could anyone please help me conf... | Others have answered on how to justify that $S=1+x+x^2+\cdots+x^n+\cdots$ equals $\frac{1}{1-x}$ for $|x|<1$. To get the second identity without differentiation, note that
\begin{align*}
1+2x+3x^2+4x^3+\cdots&=1+(x+x)+(x^2+x^2+x^2)+(x^3+x^3+x^3+x^3)+\cdots\\
&=(1+x+x^2+x^3+\cdots)+(x+x^2+x^3+\cdots)+(x^2+x^3+\cdots)+\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1223811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Minimising the distance covered I am trying to solve the question: You are trying to get to go from A to B 10 times. At each journey, a coin is flipped and if its heads, a wall appears in the middle as in scenario 2. If tails, no wall appears. Construct a method to minimize the distance covered. What is the expected di... | I assume that the coin is tossed each trip only upon reaching the wall position, and also that the diagram is a top view so we are walking around, not climbing over, the wall.
Firstly, the default expected distance should probably assume, in the case of tails, that you walk diagonally on the second part of the trip, so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1226646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to calculate $\lim_{n \to \infty}\sqrt[n]{\frac{2^n+3^n}{3^n+4^n}}$ I came across this strange limit whilst showing convergence of a series:
$$\lim_{n \to \infty}\sqrt[n]{\frac{2^n+3^n}{3^n+4^n}}$$
How can I calculate this limit?
| Use the sandwich/squeeze theorem. Since
$$\frac{3^n}{2\cdot 4^n}<\frac{2^n + 3^n}{3^n + 4^n} < \frac{2\cdot 3^n}{4^n}$$
for all $n$,
$$\frac{1}{2^{1/n}} \cdot \frac{3}{4} < \sqrt[n]{\frac{2^n + 3^n}{3^n + 4^n}} < 2^{1/n}\cdot \frac{3}{4}$$
for all $n$. Since $2^{1/n} \to 1$ as $n\to \infty$, by the squeeze theorem, you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1229117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
What will be the sum of the series of binomial co-efficients? What will be the sum of the following binomial co-efficent series
$$\binom{z+1}{z} + \binom{z+2}{z} + \binom{z+3}{z} + \dots + \binom{z+r}{z} = \sum\limits_{i=1}^r \binom{z+i}{z}$$
Thank you
| $\dbinom{z+1}{z} +\dbinom{z+2}{z} +\cdots+\dbinom{z+r}{z}$
= Coeff of $x^{z}$ in $(1+x)^{z+1}$+ Coeff of $x^{z}$ in $(1+x)^{z+2}$+$\cdots$+ Coeff of $x^{z}$ in $(1+x)^{z+r}$
= Coeff of $x^{z}$ in $((1+x)^{z+1}+ (1+x)^{z+2}+\cdots+ (1+x)^{z+r})$
= Coeff of $x^{z}$ in $((1+x)^{z+1}\frac{(1+x)^{r}-1}{1+x-1})$
= Coeff of $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can you derive $\sin(x) = \sin(x+2\pi)$ from the Taylor series for $\sin(x)$? \begin{eqnarray*}
\sin(x) & = & x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots\\
\sin(x+2π) & = & x + 2\pi - \frac{(x+2π)^3}{3!} + \frac{(x+2π)^5}{5!} - \ldots \\
\end{eqnarray*}
Those two series must be equal, but how can you show that by ... | With series manipulation, we can get
$$
\begin{align}
\sin(x+y)
&=\sum_{k=0}^\infty(-1)^k\frac{(x+y)^{2k+1}}{(2k+1)!}\tag{1a}\\
&=\sum_{k=0}^\infty\sum_{j=0}^{2k+1}\frac{(-1)^k}{(2k+1)!}\binom{2k+1}{j}x^jy^{2k+1-j}\tag{1b}\\
&=\sum_{k=0}^\infty\sum_{j=0}^{2k+1}(-1)^k\frac{x^j}{j!}\frac{y^{2k+1-j}}{(2k+1-j)!}\tag{1c}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 5,
"answer_id": 0
} |
If the equation $|x^2+4x+3|-mx+2m=0$ has exactly three solutions then find value of m. Problem :
If the equation $|x^2+4x+3|-mx+2m=0$ has exactly three solutions then find value of $m$.
My Approach:
$|x^2+4x+3|-mx+2m=0$
Case I : $x^2+4x+3-mx+2m=0$
$\Rightarrow x^2+ x (4-m) + 3+2m=0 $
Discriminant of above qudratic ... | The parabola $x^2 + 4x + 3 = (x+3)(x+1)$ is upward facing and has roots at $x = -3, x = -1$, so it is negative in $(-3, -1)$. Break it into cases. One of these cases has two solutions, the other has exactly one (i.e. discriminant = 0).
Case 1: $-3 \leq x \leq -1$. The equation is $-x^2 - 4x - 3 - mx + 2m = 0$ so $x^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1234482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
substitution integration question I want to integrate $\int \sqrt{1 - x^2} dx $.
When I substitute $x = \sin θ$ , I get the right answer. ( $ \cos^2\theta$ before integration)
But when I substitute $x = \cos θ$ , I don't get the right answer. ( $ -\sin^2\theta$ before integration).
What step is wrong here? If I proceed... | The short answer is they become the same answer once you back-substitute $\theta$ for $x$.
It makes more sense if you actually do the integrals.
$$\begin{align} \int \cos^2 \theta \,d\theta &= \frac{1}{2}\int (1 +\cos 2\theta)\,d\theta \\&= \frac{1}{2}\theta + \frac{1}{4}\sin 2\theta + C \\&= \frac{1}{2}\theta + \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1235574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Logarithm multivariable limit $\frac{\ln(1+x^3+y^3)}{\sqrt{x^2+y^2}}$ Find multivariable limit $$\lim_{\left( x,y \right) \rightarrow (0,0)}\frac{\ln(1+x^3+y^3)}{\sqrt{x^2+y^2}}$$
I was trying to find and inequality i've found out that:
$$\frac{\ln(1+x^3+y^3)}{\sqrt{x^2+y^2}} \le \frac{\sqrt{x^2+y^2} \ln(1+x^3+y^3)}{2x... | Using Polar coordinates.
Let $x = r\cos(\theta)\ and \ y=r\sin(\theta)$ hence the function becomes $\rightarrow$ $\frac{\ln( 1 +r^3\cos^3(\theta)sin^3(\theta))}{r}$
.
As $r^3\cos^3(\theta)\sin^3(\theta)$ $\rightarrow$ $0$ as $r$ $\rightarrow$ $0$ hence $\ln(1 + r^3\cos^3(\theta)\sin^3(\theta))$ $\sim$ $r^3\cos^3(\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1236147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove the set $\mathcal{O}_d := \left\{\frac{a + b\sqrt{d}}{2}:a,b \in \mathbb{Z}, a \equiv b \mod 2 \right\}$ is a ring.
Prove that if $d$ is a non-square integer with $d \equiv 1 \mod 4$ then the set $$\mathcal{O}_d := \left\{\frac{a + b\sqrt{d}}{2}:a,b \in \mathbb{Z}, a \equiv b \mod 2 \right\}$$ is a ring, and in ... | Hint: $\mathcal O_d$ is a subset of real numbers, so it is enough to show that it is closed under $+,-,\cdot$ and that it contains $0$ and $1$. I'm pretty sure the operations are (implicitly) the standard addition and multiplication.
For the closure under multiplication, just consider the four cases (depending on the r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1237953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find generating functions for the Perrin and Padovan sequences
The Perrin sequence is defined by $a_0 = 3, a_1 = 0, a_2 = 2$ and $a_k = a_{k-2}+a_{k-3}$ for $k \ge 3$. The Padovan sequence is defined by $b_0 = 0, b_1=1, b_2=1$ and $b_k=b_{k-2}+b_{k-3}$ for $k\ge 3$.
Find generating functions in the form of rational f... | Denote the generating function for the Perrin sequence as $A(x) = \sum_{n=0}^\infty a_n x^n$, where $\{a_k\}_{k=0}^\infty$ is the Perrin sequence.
We get
$$\begin{align}
A(x) &= \sum_{n=0}^\infty a_n x^n \\
&= \sum_{n=3}^\infty a_n x^n + a_2x^2+a_1x^1+a_0x^0\\
&= \sum_{n=3}^\infty \left(a_{n-2} + a_{n-3}\right)x^n+ a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1238699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving $6^n - 1$ is always divisible by $5$ by induction I'm trying to prove the following, but can't seem to understand it. Can somebody help?
Prove $6^n - 1$ is always divisible by $5$ for $n \geq 1$.
What I've done:
Base Case:
$n = 1$: $6^1 - 1 = 5$, which is divisible by $5$ so TRUE.
Assume true for $n = k$, whe... | We can show by induction that $6^k$ has remainder $1$ after division by $5$.
The base case $k=1$ (or $k=0$) is straightforward, since $6=5\cdot 1+1$.
Now suppose that $6^k$ has remainder $1$ after division by $5$ for $k\ge 1$. Thus $6^k = 5\cdot m+1$ for some $m \in \mathbb{N}$. We can then see that $$6^{k+1}=6\cdot 6^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1239106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Use the generating function to solve a recurrence relation We have the recurrence relation $\displaystyle a_n = a_{n-1} + 2(n-1)$ for $n \geq 2$, with $a_1 = 2$.
Now I have to show that $\displaystyle a_n = n^2 - n +2$, with $n \geq 1$ using the generating function.
The theory in my book is scanty, so with the help o... | Note: Here is another approach using generating functions.
According to the recurrence relation
\begin{align*}
a_1&=2\\
a_n&=a_{n-1}+2(n-1)\qquad\qquad n\geq 2
\end{align*}
we set
\begin{align*}
A(x):=\sum_{n=1}^{\infty}a_nx^n
\end{align*}
and use the Ansatz
\begin{align*}
\sum_{n=2}^{\infty}a_nx^n&=\sum_{n=2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $(a-1)(x^4+x^2+1)+(a+1)(x^2+x+1)^2 = 0$ are real and distinct, Then set of all values of $a$
If the two roots of the equation $(a-1)(x^4+x^2+1)+(a+1)(x^2+x+1)^2 = 0$ are
real and distinct, Then the set of all values of $a$ is.
$\bf{Options::}$ $(a)\;\; \displaystyle \left(0,\frac{1}{2}\right)\;\;\;\; (b)\;\; \displ... | A slightly different method. Seemed easier.
$$(a-1)(x^4+x^2+1)+(a+1)(x^2+x+1)^2=0$$
$$(x^2+x+1)\left[(a-1)(x^2-x+1)+(a+1)(x^2+x+1)\right]=0$$
$x^2+x+1=0$ has no real roots. So
$$(a-1)(x^2-x+1)+(a+1)(x^2+x+1)=0$$
$$2ax^2+2x+2a=0$$
$$ax^2+x+a=0$$
For this to have real and distinct roots, we have $\mathrm{Discriminant}>0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1244377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Let $p_n$ denote the $n$th prime and $c_n$ the $n$th composite number. Determine all positive integers $n$ such that $|p_n − c_n| = 1$. The only ones with the property in title I could find are $n = 5$ and $n = 6$: $|11 − 10| = 1$ and $|13 − 12| = 1$. Past the $20$th prime, the list of primes grows too fast for compos... | If $|p_n-c_n| = 1$ then there must exist a $N$ such that the set $\{2,3,\ldots,N\}$ contains an equal number of primes and composite numbers.
The number of composite numbers in the $N-1$ element set $\{2,\ldots, N\}$ is larger than (numbers divisible by $2$ + numbers divisible by $3$ - numbers divisible by $2\cdot 3$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1245145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is this proof for 1/4 mod 9 = x, correct? Find an integer x so (1/4) mod 9 = x
Proof:
> 1/4 mod 9 = x
> 1 mod 9 = 4 * x
> - using x = 7 -
> 1 mod 9 = 28
> 28 mod 9 = 1 (to validate)
> - using Euler division theorem m = nq + r -
> 28 = 9(3) + 1
> r = 1, so 1 = 1
I appreciate the feedback.
| Here is a systematic approach to finding the inverse.
Since $\gcd(4, 9) = 1$, $4$ has a multiplicative inverse modulo $9$. To find it, we must solve the equivalence
$$4x \equiv 1 \pmod{9}$$
To find the solution, we can use the Euclidean Algorithm.
$$9 = 2 \cdot 4 + 1$$
Solving for $1$ yields
$$9 - 2 \cdot 4 = 1$$
Hen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1246092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to Find the Function of a Given Power Series? (Please see edit below; I originally asked how to find a power series expansion of a given function, but I now wanted to know how to do the reverse case.)
Can someone please explain how to find the power series expansion of the function in the title? Also, how would you... | $x = (1-x-x^3) x + (x^2 + x^4)$
$x^2 + x^4 = (1-x-x^3) * x^2 + (x^3 + x^4 + x^5)$
$(x^3 + x^4 + x^5) = (1-x-x^3) * x^3 + (2*x^2+x^5 + x^6)$
$(2*x^2+x^5 + x^6) = (1-x-x^3) * 2x^4 + (x^5+3*x^6+x^7)$ then
$\frac{x}{1-x-x^3} = 1 + x + x^2 + x^3 + 2x^4 +\frac{(x^5+3*x^6+x^7)}{1-x-x^3}$
if you continue you all the coefficien... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1249107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
How does $\frac{-1}{x^2}+2x=0$ become $2x^3-1=0$? Below is part of a solution to a critical points question. I'm just not sure how the equation on the left becomes the equation on the right. Could someone please show me the steps in-between? Thanks.
$$\frac{-1}{x^2}+2x=0 \implies 2x^3-1=0$$
| For $x\neq 0$, we have $0= - \frac{1}{x^2} + 2x =- \frac{1}{x^2} + \frac{x^2}{x^2} \times 2x = - \frac{1}{x^2} + \frac{2x^3}{x^2} = \frac{2x^3 -1}{x^2}$.
So we have $\frac{2x^3 -1}{x^2} = 0$ which will give us $2x^3 -1 = 0.$
And we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1250132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Does $2^2=4$ imply $2=\pm \sqrt{4}$? I read the square root property from the book, College Algebra by Raymond A Barnett and Micheal R Ziegler that,
The square root property says,
If $A^2=C$ then $A=\pm \sqrt{C}$
I took the equality,
$2^2=4$
$\implies 2=\pm \sqrt{4}$
$\implies 2=\pm {2}$
but, How $2$ can be equal to... | Let's see how "If $A^2=C$ then $A=\pm \sqrt{C}$" is obtained:
$A^2=C \Rightarrow A^2-C=0 \Rightarrow (A-\sqrt{C})(A+\sqrt{C})=0 \Rightarrow$
$A-\sqrt{C}=0$ or $A+\sqrt{C}=0 \Rightarrow A_1=\sqrt{C}$ or $A_2=-\sqrt{C} \Rightarrow A=\pm \sqrt{C}$.
Now let's see what happens to "$2^2=4$":
$2^2=4 \Rightarrow 2^2-4=0 \Righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1250325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How Deficient a Number is? (Finding numbers having a certain deficiency) This question was edited, in particular equations were corrected:
A number N is said to be deficient by an integer $d$ if:
$\sigma(N)=2N-d$
Note that powers of 2 are deficient by 1.
While a prime $p$ is deficient by $p-1$
My question will be this:... | Let us try using the following criterion:
Let $1 < d = 2N - \sigma(N)$. Then
$$\frac{2N}{N + d} < I(N) < \frac{2N + d}{N + d}$$
where $I(N)=\sigma(N)/N$ is the abundancy index of $N$.
Since you require
$$\frac{2}{3}N < d < N - \sqrt{N},$$
then we have
$$\frac{2}{3} < \frac{d}{N} = 2 - I(N) < 1 - \sqrt{\frac{1}{N}}$$
s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1251097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to calculate this multivariable limit? $$
\lim_{(x,y,z)\to (0,0,0) } \frac{\sin(x^2+y^2+z^2) + \tan(x+y+z) }{|x|+|y|+|z|}
$$
I know the entire limit should not exist. In addition, the limit:
$$
\lim_{(x,y,z)\to (0,0,0) } \frac{\tan(x+y+z) }{|x|+|y|+|z|}
$$
does not exist and it seems like the limit:
$$
\lim_{(x,y,z... | Write $\frac{\sin(x^2+y^2+z^2)}{|x|+|y|+|z|}=\frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2}\frac{\sqrt{x^2+y^2+z^2}}{|x|+|y|+|z|}\sqrt{x^2+y^2+z^2}$. The first factor tends to $1$, the second is bounded and the last tends to zero.
To answer your question in your comment:
$$x^2+y^2+z^2\leq x^2+y^2+z^2+2|x||y|+2|x||z|+2|y||z|=(|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1251299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Proving convergence/divergence via the ratio test Consider the series
$$\sum\limits_{k=1}^\infty \frac{-3^k\cdot k!}{k^k}$$
Using the ratio test, the expression $\frac{|a_{k+1}|}{|a_k|}$ is calculated as:
$$\frac{3^{k+1}\cdot (k+1)!}{(k+1)^{k+1}}\cdot \frac{k^k}{3^k\cdot k!}=\frac{3}{(k+1)^{k}}\cdot {k^k}=3\cdot \frac{... | $$\lim_{k\to\infty} 3\cdot \bigg(\frac{k}{k+1}\bigg)^k = \lim_{k\to\infty} 3\cdot \bigg(1 - \frac{1}{k+1}\bigg)^k = 3\cdot \lim_{k\to\infty}\bigg(1-\frac{1}{k}\bigg)^k = 3\cdot e^{-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1253061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Prove that $f : [-1, 1] \rightarrow \mathbb{R}$, $x \mapsto x^2 + 3x + 2$ is strictly increasing.
Prove that $f : [-1, 1] \rightarrow \mathbb{R}$, $x \mapsto x^2 + 3x + 2$ is strictly increasing.
I do not have use derivatives, so I decided to apply the definition of being a strictly increasing function, which should ... | Seems good to me. I would be lazier, though. Pick $a,b\in[-1,1]$, with $a<b$. Since $a \geq -1$, then $a+1 \geq 0$. Since $b > a$, we have also that $b+2,b+1,a+2 > 0$. So: $$\begin{cases} b+2>a+2 >0 \\ b+1 > a+1 \geq 0 \end{cases} \implies (b+2)(b+1) > (a+2)(a+1) \implies f(b) > f(a).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1253120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Finding equation of hyperbola with only foci and asymptote This is a concept we learned in class today, which I still can't seem to grasp. I have no specific question that necessarily has to be done, so I will use one of the examples my book gives me:
Given the foci are: (6√5, 10) and (-6√5,10)
The asymptote is y = 1/... | It looks like you know all of the equations you need to solve this problem. I also see that you know that the slope of the asymptote line of a hyperbola is the ratio $\dfrac{b}{a}$ for a simple hyperbola of the form $$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$$
So, given $y=\dfrac{1}{2}x+10$, we can see that the ratio $\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1255277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\int \limits_0^1 \int \limits_x^1 \arctan \bigg(\frac yx \bigg) \, \, \, dx \, \, dy$ Find $$\int \limits_0^1 \int \limits_x^1 \arctan \bigg(\frac yx \bigg)dx \, \, dy$$
So obviously using cylindrical is the way to go to give $\theta r$ inside the integral (after considering the jacobian term).
But how can the li... | Let $I=\int_{0}^{1} \int_{x}^{1} arctan(\frac{y}{x}) dy dx $, then considering polar coordinates we have the foolowing region :
so as we see from the picture, the region is bounded by $\theta$ between $\frac{\pi}{4} $ and $\frac{\pi}{2} $. Now taking a ray from the origin crossing the region, we recignize that it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1256465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Positivity of power function. Prove that
$6^a-7^a+2\cdot 4^a-3^a-5^a\ge0$ for $-\frac{1}{2}\le a\le0$.
I tried to do it by first derivative test but derivative become more complicated (same with 2nd derivative for convexity).
Here is it's plotting.
| We want to show that
$$\left((3^x+7^x+5^x)/3\right)^{1/x} > \left((6^x+4^x+4^x)/3\right)^{1/x}\hbox{ for }-0.5\le x<0$$ "as it may be seen from the plot".
It's not too hard to test that
$$\hbox{(1) }4.57 < \left((3^{-\frac{1}{2}}+7^{-\frac{1}{2}}+5^{-\frac{1}{2}})/3\right)^{-2} $$
$$\hbox{(2) }\left((6^{-\frac{1}{4}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1257102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show that only one prime can be expressed as $n^3-1$ for some positive integer $n$
Show that only one prime can be expressed as $n^3-1$ for some positive integer $n$
$n^3-1=(n-1)(n^2+n+1)$. If $n^3-1$ is prime then $n-1=1$ or $n=2$ so $n^3-1=7$.
But I need a concrete proof of it, not from intuition.
| $$n^3-1=(n-1)(n^2+n+1)$$ Suppose that $n^2+n+1\le n-1\implies n^2\le -2$ which leads to contradiction.
Therefore $n^2+n+1\gt n-1$ and since $n^3-1$ is prime only possibility is $$n-1=1\implies \color{Green}{n=2}\,\,\,\,\,\text{and} \,\,\,\,\,\,\color{Red}{n^3-1=7}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1257198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Find the equation whose roots are each six more than the roots of $x^2 + 8x - 1 = 0$
Find the equation whose roots are each six more than the roots of $x^2 + 8x - 1 = 0$
I must use Vieta's formulas in my solution since that is the lesson we are covering with our teacher.
My solution:
Let p and q be the roots of the q... | The original quadratic equation is given to you as $x^2 + 8x - 1 = 0$ . Let it's roots be $(x_1 , x_2)$ . The roots of required equation be $(\alpha ,\beta)$ (say).
Now, $$x_1 + x_2 = -8 \\ x_1 x_2 = -1$$
And $$\alpha = x_1 + 6 \\ \beta = x_2 + 6 $$
So, $$\alpha + \beta = x_1 + x_2 + 12 \\ \alpha \beta = (x_1 + 6)(x_2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1259949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What is the connection between the discriminant of a quadratic and the distance formula? The $x$-coordinate of the center of a parabola $ax^2 + bx + c$ is $$-\frac{b}{2a}$$
If we look at the quadratic formula
$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
we can see that it specifies two points at a certain offset from the cen... | Without loss of generality, we can find the horizontal distance between $x=\dfrac{-b}{2a}$ and $x=\dfrac{-b+\sqrt{b^2-4ac}}{2a}.$ So inputting these coordinates into the distance formula, with both $y-$coordinates equal to zero, we have
\begin{align}
d&=\sqrt{\left(\dfrac{-b+\sqrt{b^2-4ac}}{2a}-\dfrac{-b}{2a}\right)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1264091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 8,
"answer_id": 6
} |
I cannot solve this limit $$
\lim_{n\to\infty}\frac{(\frac{1}{n}+1)^{bn+c+n^2}}{e^n}=e^{b-\frac{1}{2}}
$$
I am doing it like this, and I cannot find the mistake:
$$
\lim_{n\to\infty}\frac{1}{e^n}e^{n+b+c/n}=
\lim_{n\to\infty}e^{n+b-n+c/n}=e^b
$$
| You could use L'Hospital's rule and $x=\frac{1}{n}$ to get
$\displaystyle\lim_{n\to\infty}\ln\bigg[\frac{(1+\frac{1}{n})^{bn+n^2+c}}{e^n}\bigg]=\lim_{n\to\infty}(bn+n^2+c)\ln\big(1+\frac{1}{n}\big)-n$
$=\displaystyle\lim_{x\to0}\bigg(\frac{b}{x}+c+\frac{1}{x^2}\bigg)\ln(1+x)-\frac{1}{x}=\lim_{x\to0}\frac{(bx+cx^2+1)\ln... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1264971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Finding the inverse of a number under a certain modulus How does one get the inverse of 7 modulo 11?
I know the answer is supposed to be 8, but have no idea how to reach or calculate that figure.
Likewise, I have the same problem finding the inverse of 3 modulo 13, which is 9.
| To find the inverse of $7$ modulo $11$, we must solve the equivalence $7x \equiv 1 \pmod{11}$. To do this, we use the Extended Euclidean Algorithm to express $1$ as a linear combination of $7$ and $11$. The coefficient of $7$ will be the inverse modulo $11$. By the Euclidean Algorithm,
\begin{align*}
11 & = 1 \cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1266282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
For $n>2, n\in\mathbb{Z}$, why is this true: $\left\lfloor 1/\left(\frac{1}{n^2}+\frac{1}{(n+1)^2}+\cdots+\frac{1}{(n+n)^2}\right)\right\rfloor=2n-3$ Let $n>2$ be a positive integer, prove that
$$\left\lfloor \dfrac{1}{\dfrac{1}{n^2}+\dfrac{1}{(n+1)^2}+\cdots+\dfrac{1}{(n+n)^2}}\right\rfloor=2n-3?$$
before I use hand C... | Let $f(n) = \dfrac1{n^2} + \dfrac1{(n+1)^2} + \cdots + \dfrac1{(2n)^2}$.
We then have
$$\int_{n-1}^{2n} \dfrac{dx}{x^2} > \dfrac1{n^2} + \dfrac1{(n+1)^2} + \cdots + \dfrac1{(2n)^2} \geq \int_n^{2n+1} \dfrac{dx}{x^2}$$
This gives us
$$\dfrac1n-\dfrac1{2n+1} < f(n) < \dfrac1{n-1} - \dfrac1{2n} \implies \dfrac{n+1}{n(2n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1268447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding the evolute of a parabola I previously tried to find the evolute of a parabola by using parameterisation by arc length. It didn't work. While I was hoping for an answer I kept working on it and came up with the following method (unfortunately, something is not quite right as it leads to a different result than ... | Right, your method is not inherently flawed, but you have calculational mistakes: they first appear in the normal vector: differentiating the tangent vector should give
$$ -\frac{4t}{(1+4t^2)^{3/2}} \begin{pmatrix} 1 \\ 2t \end{pmatrix} + \frac{1}{(1+4t^2)^{1/2}}\begin{pmatrix} 0 \\ 2 \end{pmatrix} = \frac{1}{(1+4t^2)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}$, prove that $\frac{a}{d}=\sqrt{\frac{a^5+b^2c^2+a^3c^2}{b^4c+d^4+b^2cd^2}}$ What I've done so far;
$$\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\\
a=bk, b=ck, c=dk\\
a=ck^2, b=dk^2\\
a=dk^3$$
I tried substituting above values in the right hand side of the equation to get $\frac{a}{d}$... | Having $a=dk^3,b=dk^2,c=dk$ gives you$$\begin{align}\sqrt{\frac{a^5+b^2c^2+a^3c^2}{b^4c+d^4+b^2cd^2}}&=\sqrt{\frac{d^5k^{15}+d^2k^4\cdot d^2k^2+d^3k^9\cdot d^2k^2}{d^4k^8\cdot dk+d^4+d^2k^4\cdot dk\cdot d^2}}\\&=\sqrt{\frac{dk^{15}+k^6+dk^{11}}{dk^9+1+dk^5}}\\&=\sqrt{\frac{k^6(dk^9+1+dk^5)}{dk^9+1+dk^5}}\\&=\sqrt{k^6}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1271114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that number of zeros at the right end of the integer $(5^{25}-1)!$ is $\frac{5^{25}-101}{4}.$
Prove that number of zeros at the right end of the integer $(5^{25}-1)!$ is $\frac{5^{25}-101}{4}.$
Attempt:
I want to use the following theorem:
The largest exponent of $e$ of a prime $p$ such that $p^e$ is a divisor... | Since $e_1\le e_2$, all you need is $e_1$.
$$\begin{align}e_1&=\sum_{k=1}^{25}\left\lfloor\frac{5^{25}-1}{5^k}\right\rfloor\\&=\sum_{k=1}^{25}\left\lfloor 5^{25-k}-\frac{1}{5^k}\right\rfloor\\&=\sum_{k=1}^{25}5^{25-k}+\sum_{k=1}^{25}\left\lfloor-\frac{1}{5^k}\right\rfloor\\&=\frac{5^{25}-1}{4}+25\times(-1)\\&=\frac{5^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1272660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How does $-\frac{1}{x-2} + \frac{1}{x-3}$ become $\frac{1}{2-x} - \frac{1}{3-x}$ I'm following a solution that is using a partial fraction decomposition, and I get stuck at the point where $-\frac{1}{x-2} + \frac{1}{x-3}$ becomes $\frac{1}{2-x} - \frac{1}{3-x}$
The equations are obviously equal, but some algebraic mani... | $$
\frac{1}{x-a} = \frac{1}{-(a - x)} = - \frac{1}{a - x}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 4
} |
How to compute $\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}}$? I have to compute the series $\displaystyle\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}}$.
$$\displaystyle\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}} = \sum_{n=0}^{\infty}{\frac{3^n\frac{1}{2}}{n!}} + \sum_{n=0}^{\infty}{\frac{3^nn}... | We have
$$
x(e^x)'=xe^x=\sum_{n=0}^\infty \frac{nx^n}{n!}
$$
Now take $x=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Find the domain of $x$ in $4\sqrt{x+1}+2\sqrt{2x+3}\leq(x-1)(x^2-2)$
Solve this equation for $x$: $4\sqrt{x+1}+2\sqrt{2x+3}\leq(x-1)(x^2-2)$
I have no idea to solve that, but I know solutions are $x=-1$ or $x\ge 3$.
| After more than two years I managed to find a more simple and creative solution of this homework-looking inequality. :-)
Define the function from the left hand side of the inequality by $f(x)$ and the function from the right hand side of the inequality by $g(x)$. Draw the graphs of these functions
at a segment contai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1276510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Calculating all possible sums of the numbers $2^0, 2^1, \ldots, 2^{(n-1)}$ Using the simple equation $2^{n-1}$ you get values such as:
$1,2,4,8,16,32,64,128,256,$ etc.
How can I find all possible number combinations within this range? For example, the numbers $1,2,4,8$ give:
\begin{align}
1 &= 1 \\
2 &= 2 \\
1 + 2 &= ... | Every integer from $1$ to $2^n-1$ is expressible using the sums of the numbers $1,2,4,8, ... 2^{n-2}, 2^{n-1}$, and no other numbers. This can be proven using induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1278579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $\int \frac{1}{x^4+x^2+1} \,\, dx$ Find $$\int \frac{1}{x^4+x^2+1} \,\, dx$$
I tried to find like that:
$\int \frac{1}{x^4+x^2+1} = \int \frac{\frac{1}{2}x + \frac{1}{2}}{x^2+x+1} \,\, dx + \int \frac{-\frac{1}{2}x + \frac{1}{2}}{x^2-x+1} \,\, dx = \frac{1}{2} \Big(\int \frac{2x + 1}{x^2+x+1} - \int \frac{x}{x^2+... | $$\int\frac{xdx}{x^2-x+1}=\int\frac{(x-\frac12)dx}{x^2-x+1}+\int\frac{\frac12dx}{(x-\frac12)^2+\frac34}$$. I'm guessing you know how to solve that. The other half is very similar.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1278706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Show that $f:= \frac{1 - 2x}{x^2 - 1}$ is monotonically increasing in the open interval $(-1, 1)$ Show that $f:= \frac{1 - 2x}{x^2 - 1}$ is monotonically increasing in the open interval $(-1, 1)$
To show a function is monotonically increasing, I started by saying that:
A function $f$ is monotonically increasing in an ... | A differentiable function in some interval is monotonically aascending if its derivative is positive:
$$f'(x)=\frac{-2(x^2-1)-2x(1-2x)}{(x^2-1)^2}=\frac{-2x^2+2-2x+4x^2}{(x^2-1)^2}=$$
$$=2\frac{x^2-x+1}{(x^2-1)^2}$$
Now just prove the above indeed is positive for all $\;x\in\Bbb R\setminus\{-1,1\}\;$ , in particular in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1279206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the value of this sum. $$\sum_{n=5}^\infty \frac {2^n-4}{3^n}$$
I tried treating it as a geometric sum but I get 2 as the value, when it should be $\frac {10}{27}$
| $\sum=\sum_{n=0}^{\infty} (\frac{2}{3})^n-4\sum_{n=0}^\infty (\frac{1}{3})^n-\sum_{n=0}^4 \frac{2^n-4}{3^n}$
$=\frac{1}{1-\frac{2}{3}}-\frac{4}{1-\frac{1}{3}}+\frac{91}{27}$
$=3-6+\frac{91}{27}$
$=\frac{10}{27}$
QED.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1279309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
complex nos in ellipse. I was practising some ques on ellipses when I came a criss this question:
If normal at four points $(x_1,y_1)$..... on the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ are concurrent then find the value of $$(x_1+x_2+x_3+x_4)\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}\right)... | Equation of tangent at $(x_k,y_k)$,
$$\frac{x_kx}{a^2}+\frac{y_ky}{b^2}=1$$
Equation of normal at $(x_k,y_k)$,
$$\frac{y_k}{b^2}(x-x_k)-\frac{x_k}{a^2}(y-y_k)=0$$
If $(X,Y)$ is the common point of the four normals, then
$$\frac{y_k}{b^2}(X-x_k)-\frac{x_k}{a^2}(Y-y_k)=0$$
Hence, all the points $(x_k,y_k)$ lie on another... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1281470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Roots of a polynomial whose coefficients are ratios of binomial coefficients
Prove that $\left\{\cot^2\left(\dfrac{k\pi}{2n+1}\right)\right\}_{k=1}^{n}$ are the roots of the equation
$$x^n-\dfrac{\dbinom{2n+1}{3}}{\dbinom{2n+1}{1}}x^{n-1} + \dfrac{\dbinom{2n+1}{5}}{\dbinom{2n+1}{1}}x^{n-2} - \ldots \ldots \ldots + \d... | Like Sum of tangent functions where arguments are in specific arithmetic series,
$$\displaystyle\tan(2n+1)x=\dfrac{\binom{2n+1}1\tan x-\binom{2n+1}3\tan^3x+\cdots}{1-\binom{2n+1}2\tan^2x+\cdots}$$
Multiplying the denominator & the numerator by $\cot^{2n+1}x$ we get,
$$\displaystyle\tan(2n+1)x=\dfrac{\binom{2n+1}1\cot^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Proving simple trigonometric identity I need help with this one
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \mathrm{tan}^2\alpha} - \cos\alpha = \sin \alpha
$$
I tried moving sin a on the other side of the eqation
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\... | $$LHS=\frac{\sin^2 \alpha}{\sin \alpha-\cos \alpha}+\frac{\sin\alpha+\cos\alpha}{1-\tan^2\alpha}-\cos\alpha\\=\frac{\sin^2 \alpha}{\sin \alpha-\cos \alpha}-\frac{\cos^2\alpha}{\sin\alpha-\cos\alpha}-\cos\alpha\\=\sin\alpha+\cos\alpha-\cos\alpha=\sin\alpha$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Understanding density of irrational numbers and Archemedian property From
Density of irrationals
I know this much of the proof of the density of irrational numbers
"We know that $y-x>0$.
By the Archimedean property, there exists a positive integer $n$ such that $n(y-x)>1$ or $1/n < y-x$. There exists an integer $m$ ... | If $M$ is the least $m$
such that
$m\left(\frac{\sqrt{2}}{n}\right)\gt x
$,
then
$(M-1)\left(\frac{\sqrt{2}}{n}\right)
\le x
\lt M\left(\frac{\sqrt{2}}{n}\right)
$.
Since
$\frac{\sqrt{2}}{n}\lt \frac{y-x}{2}
$,
$\begin{array}\\
y
&\gt 2\frac{\sqrt{2}}{n}+x\\
&\ge 2\frac{\sqrt{2}}{n}+(M-1)\left(\frac{\sqrt{2}}{n}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving the following number is real Let $z_i$ be complex numbers such that $|z_i| = 1$ .
Prove that :
$$ z\, :=\, \frac{z_1+z_2+z_3 +z_1z_2+z_2z_3+z_1z_3}{1+z_1z_2z_3} \in \mathbb{R} $$
This problem was featured on my son's final exam today, I tried helping him with this, but I guess I got a bit rusty myself.
Any idea... | If $|z|=1$, then $z+\frac1z=z+\bar{z}=2\mathrm{Re}(z)$. Therefore, the following are both real:
$$
\begin{align}
\small\left(z_1+\frac1{z_1}\right)\left(z_2+\frac1{z_2}\right)\left(z_3+\frac1{z_3}\right)
&\small=\frac{z_1^2z_2^2z_3^2+z_1^2z_2^2+z_2^2z_3^2+z_3^2z_1^2+z_1^2+z_2^2+z_3^2+1}{z_1z_2z_3}\tag{1}\\
z_1z_2z_3+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 5,
"answer_id": 4
} |
Matrix Problem of form Ax=B The matrix $A$ is given by
$$\left(\begin{array}{ccc}
1 & 2 & 3 & 4\\
3 & 8 & 11 & 8\\
1 & 3 & 4 & \lambda\\
\lambda & 5 & 7 & 6\end{array} \right)$$
Given that $\lambda$=$2$, $B$=$\left(\begin{array}{ccc}
2 \\
4 \\
\mu \\
3 \end{array} \right)$ and $X$=$\left(\begin{array}{ccc}
x \\
y \\
z ... | First, note that
$$
\det A=2\,(\lambda-2)^2
$$
so $A$ is invertible if and only if $\lambda\neq 2$. In this case we can use Cramer's rule to solve for $X$. For example, we have
$$
x=
\frac{
\begin{vmatrix}
2 & 2 & 3 & 4 \\
4 & 8 & 11 & 8 \\
\mu & 3 & 4 & \lambda \\
3 & 5 & 7 & 6
\end{vmatrix}
}
{
\begin{vmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to compute $\int_0^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx =\frac{\pi}{48\sqrt{3}}$? $$\int_0^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx =\frac{\pi}{48\sqrt{3}}$$
I have difficulty to evaluating above integrals.
First I try the substitution $x^4 =t$ or $x^4 +x^2+1 =t$ but it makes integral worse.
Using Mathematica I fou... | For the antiderivative, you could also "simplify" the problem using partial fraction decomposition since $$\frac{x^4}{\left(x^4+x^2+1\right)^3}=-\frac{3 (x-1)}{16 \left(x^2-x+1\right)}+\frac{3 (x+1)}{16
\left(x^2+x+1\right)}-\frac{3}{16 \left(x^2-x+1\right)^2}-\frac{3}{16
\left(x^2+x+1\right)^2}+\frac{x}{8 \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 0
} |
Derivative of $f(x) = \frac{\cos{(x^2 - 1)}}{2x}$ Find the derivative of the function $$f(x) = \frac{\cos{(x^2 - 1)}}{2x}$$
This is my step-by-step solution: $$f'(x) = \frac{-\sin{(x^2 - 1)}2x - 2\cos{(x^2 -1)}}{4x^2} = \frac{2x\sin{(1 - x^2)} - 2\cos{(1 - x^2)}}{4x^2} = \frac{x\sin{(1 - x^2)} - \cos{(1 - x^2)}}{2x^2} ... | Differentiating $\cos(x^2-1)$ gives you $-2x\sin(x^2-1)$, using the Chain Rule
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Do the spaces spanned by the columns of the given matrices coincide? Reviewing linear algebra here.
Let $$A = \begin{pmatrix}
1 & 1 & 0 & 0 \\
1 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 1 & 1
\end{pmatrix} \qquad \text{and} \qquad
B = \begin{pmatrix}
1 & 0 & 0 \\
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmat... | You are correct. One can also verify this by checking that $A, B$ are column-equivalent, or equivalently, that is, that $A^T, B^T$ are row-equivalent. One can check the latter algorithmically by checking that the reduced row echelon form of the $A^T, B^T$ coincide (at least after padding with an appropriate number of z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving $3x^3\equiv 7\pmod{925}$ I am trying to solve $3x^3\equiv 7\pmod{925}$.
I thought of using brute force, but $925$ is too big for that.
I also tried raising both sides of the equation to the power of $3$, but it didn't help.
How can I solve it?
| With these, it's often easier to consider the problem after breaking the moduli down and then recombining the results using the Chinese Remainder Theorem.
So we consider the pair of problems
$$ \begin{align}
3x^3 &\equiv 7 \pmod {37}\\
3x^3 &\equiv 7 \pmod {25}.
\end{align}$$
In the first, it's easy to check that the i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding a nullspace of a matrix - what should I do after finding equations? I am given the following matrix $A$ and I need to find a nullspace of this matrix.
$$A =
\begin{pmatrix}
2&4&12&-6&7 \\
0&0&2&-3&-4 \\
3&6&17&-10&7
\end{pmatrix}$$
I have found a row reduced form of this matrix, which is:
$$A' =
\beg... | \begin{cases} x_1+2x_2+\frac{23}{10}x_5=0 \\ x_3+\frac{13}{10}x_5=0 \\ x_4+\frac{22}{10}x_5=0 \end{cases}
$$
x_1=-2x_2-\dfrac{23}{10}x_5
$$
$$
x_3=-\dfrac{13}{10}x_5
$$
$$
x_4=-\dfrac{11}{5}x_5
$$
Therefore,basis of null space=
$$
\begin{pmatrix}
-2 \\
1 \\
0 \\
0 \\
0
\end{pmatrix},
\begin{pmatrix}
-\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solutions of $\sqrt{x+4+2\sqrt{x+3}}-(x^2+4x+3)^{1/3}=1$ $\sqrt{x+4+2\sqrt{x+3}}-(x^2+4x+3)^{1/3}=1$
I get that $-3$ as a solution, but apparently 1 is as well a solution, and I don't see a mechanism through which I could find it. Any help would be appreciated.
| we have $$\sqrt{x+4+2\sqrt{x+3}}-(x^2+4x+3)^{1/3}=1 $$ let us make a change of variable $u = x + 3 \ge 0, x = u - 3.$ with that we have
$$\sqrt{u+1+2\sqrt u}-(u(u-2))^{1/3}=1 \to 1+\sqrt u=(u(u-2))^{1/3}+1$$ this gives us $$\sqrt u = (u(u-2))^{1/3}\tag 1 $$
now, exponentiating $(1)$ implies $$ 0=u^3 - u^2(u-2)^2 \to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the solutions of $x^2\equiv 9 \pmod {256}$. Find the solutions of $x^2\equiv 9 \pmod {256}$. I try to follow an algorithm shown us in class, but I am having troubles doing so. First I have to check how many solutions there are. Since $9\equiv 1 \pmod {2^{\min\{3,k\}}}$ where k fulfills $256=2^k$, then since $k\... | This is equivalent to finding all $0\leq x \leq 255$ such that $256\mid x^2-9 $. But $256 = 2^8$ and $x^2-9 = (x-3)(x+3)$. But the greatest common divisor of $x-3$ and $x+3$ must divide $(x+3)-(x-3) = 6$. Hence, if $x+3$ is divisible by $4$, then $x-3$ is only divisible by $2$, and vice versa.
The statement is thus equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Having problem in last step on proving by induction $\sum^{2n}_{i=n+1}\frac{1}{i}=\sum^{2n}_{i=1}\frac{(-1)^{1+i}}{i} $ for $n\ge 1$ The question I am asked is to prove by induction $\sum^{2n}_{i=n+1}\frac{1}{i}=\sum^{2n}_{i=1}\frac{(-1)^{1+i}}{i} $ for $n\ge 1$
its easy to prove this holds for $n =1$ that gives $\fra... | $$\sum_{i=n+2}^{2n+2} \frac{1}{i}=\sum_{i=n+1}^{2n}\frac{1}{i}+ \frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}=\sum_{i=1}^{2n}\frac{(-1)^{i+1}}{i}+\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}$$
It remains to prove that:
$$ \sum_{i=1}^{2n}\frac{(-1)^{i+1}}{i}+\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}=\sum_{i=1}^{2n+2}\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1298016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Reducing the form of $2\sum\limits_{j=0}^{n-2}\sum\limits_{k=1}^n {{k+j}\choose{k}}{{2n-j-k-1}\choose{n-k+1}}$. I've been toying around with simplifying the expression $2\sum\limits_{j=0}^{n-2}\sum\limits_{k=1}^n {{k+j}\choose{k}}{{2n-j-k-1}\choose{n-k+1}}$ (for integer only $n$) for a while, as I was hoping it would h... | Alternate solution.
As before we start trying to evaluate
$$S(n) =
\sum_{q=0}^{n-2}
\sum_{k=1}^n {k+q\choose k} {2n-q-k-1\choose n-k+1}$$
which we re-write as
$$-\sum_{q=0}^{n-2} {2n-q-1\choose n+1}
-\sum_{q=0}^{n-2} {n+1+q\choose n+1}
+ \sum_{q=0}^{n-2}
\sum_{k=0}^{n+1} {k+q\choose k} {2n-q-k-1\choose n-k+1}.$$
Cal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1299123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Finding two solutions to $x^2 - 6y^2 = 1$ using continued fractions Can anyone show me how to find the solutions to $x^2-6y^2=1$ by using continued fractions? I know how to find the fractions for $\sqrt6$ but do not know how to proceed. THANK YOU!!!
| Suppose you are able to find a solution to $x^2-6y^2=1$ like $u_1=5$ and $v_1=2$ as indicated in the answer given by user17762. Then ALL other positive integer solutions $(u_n,v_n)$ can be found by
$$u_n+\sqrt{6}v_n=(u_1+\sqrt{6}v_1)^n.$$
Thus another solution will be
$$u_2+\sqrt{6}v_2=(u_1+\sqrt{6}v_1)^2=(5+2\sqrt{6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1299954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Help with Lagrange multipliers on an intresting function Hi guys I am trying to do Lagrange multipliers to figure out $\lambda$
$$F=a \log(x^2-y)+b\log(x^3-z)-\lambda (x^2-y+x^3-z -1)$$
Where a and b are constants and we have the constraint $x^2-y+x^3-z =1$
What I did was take the partials and added them together and ... | Suppose your $\log$ means natural logarithm. Then set the partial derivatives to $0$:
$$\frac{2ax}{x^2-y}+\frac{3bx^2}{x^3-z}-2x\lambda+3x^2\lambda=0 \\
\frac{-a}{x^2-y}+\lambda=0 \\
\frac{-b}{x^3-z}+\lambda=0\\
x^2-y+x^3-z =1$$
From equation 2 and 3:
$$\lambda=\frac{a}{x^2-y}=\frac{b}{x^3-z}$$
Plug these information ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving differential equation $x'=\frac{x+2t}{x-t}$ I am trying to solve the following differential equation:
$$x'=\frac{x+2t}{x-t}$$
with initial value condition: $x(1)=2$
This is what I have so far:
Substitution: $u=\frac{x}{t}$
$$\implies u't+u=\frac{2t+tu}{-t+tu}$$
Separation of variables:
$$\implies \frac{u'(u-1)}... | it must be $$\frac{u-1}{-u^2+2u+2}du=\frac{dt}{t}$$
we get
$$-\frac{1}{2}\ln|-u^2+2u+2|=\ln|t|+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limit as $x$ tends to zero of $\frac{\csc(x)}{x^3} - \frac{\sinh(x)}{x^5}$ How would I find the $$\lim_{x\to0}\left(\frac {\csc(x)}{x^3} - \frac{\sinh(x)}{x^5}\right)$$ The only way I know how to do this is with l'hopitals rule but I don't see it helping here as we have x's in our denominator.
| you have $$ \left(\frac {\csc(x)}{x^3} - \frac{\sinh(x)}{x^5}\right)= \frac{2x^2 - (e^x - e^{-x})\sin x}{2x^5\sin x}$$ we will look at maclaurin expansion of $$\begin{align} (e^x - e^{-x})\sin x - 2x^2 &=2\left(x+ \frac16 x^3+\frac1{120}x^5 + \cdots\right)\left(x - \frac16x^3 + \frac1{120}x^5+\cdots\right)-2x^2 \\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Matrices $\begin{pmatrix} a &b \\ c &d \end{pmatrix}\begin{pmatrix} x\\y \end{pmatrix}=k\begin{pmatrix}x\\y \end{pmatrix}$ If $\begin{pmatrix}
a &b \\
c &d
\end{pmatrix}\begin{pmatrix}
x\\y \end{pmatrix}=k\begin{pmatrix}x\\y \end{pmatrix}$, prove that $k$ satisfies the equation $k^2-(a+d)k+(ad-bc)=0$.
If the roots of... | we can rewrite the matrix equation as s system of two linear equations. they are $$\begin{align} (a-k)x + by &= 0\\cx + (d-k)y &=0\end{align}$$
multiply the first equation by $(d-k)$, the second one by $b$, and subtracting gives $$\left((a-k)(d-k)-bc\right)x = 0.$$
we have two choices: (a) $x = 0$ or (b) $$(a-k)(d-k)-b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Showing that $x^3+2y^3+4z^3=2xyz$ has no integer solutions except $(0,0,0)$. Let $x,y,z\in \mathbb{Z}$ satisfy the equation:
$$
x^3+2y^3+4z^3=2xyz
$$
How do I prove that the only solution is $x=y=z=0$?
| If following equation has a non-trivial solution
$$x^3 + 2y^3 + 4z^3 = 2xyz\tag{*1}$$
there must be one whose $x^2 + y^2 + z^2$ is minimal. Let $(x_1,y_1,z_1)$ be such a minimal non-trivial solution. We have
$$x_1^3 + 2y_1^3 + 4z_1^3 = 2x_1y_1z_1 \implies x_1^3 \equiv 0 \pmod 2 \implies 2|x_1$$
Let $(x_2,y_2,z_2) = (y_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Remainder of $(1+x)^{2015}$ after division with $x^2+x+1$
Remainder of $(1+x)^{2015}$ after division with $x^2+x+1$
Is it correct if I consider the polynomial modulo $5$
$$(1+x)^{2015}=\sum\binom{2015}{n}x^n=1+2015x+2015\cdot1007x^2+\cdots+x^{2015}$$
RHS stays the same and then The remainder must be of the form $Ax+B... | Recall that $x^2+x+1=\Phi_3(x)$. If we set $\omega=\exp\left(\frac{2\pi i}{3}\right),\bar\omega=\left(\frac{2\pi i}{3}\right)$, we must have:
$$ (1+x)^{2015} = q(x)\cdot \Phi_3(x)+(Ax+B), \tag{1}$$
and evaluating the previous identity for $x\in\{\omega,\bar{\omega}\}$ we get:
$$ (1+\omega)^{2015} = A\omega+B,\quad (1+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Determine whether or not the limit exists: $\lim_{(x,y)\to(0,0)}\frac{(x+y)^2}{x^2+y^2}$
Determine whether or not the limit $$\lim_{(x,y)\to(0,0)}\frac{(x+y)^2}{x^2+y^2}$$
exists. If it does, then calculate its value.
My attempt:
$$\begin{align}\lim \frac{(x+y)^2}{x^2+y^2} &= \lim \frac{x^2+y^2}{x^2+y^2} + \lim \fr... | Consider $$f(x,y)=\frac{(x+y)^2}{x^2+y^2} .$$
If you take the path $(0,y)$, then:
$$\displaystyle\lim_{y\to0} f(0,y) =\lim_{y\to0} \frac{y^2}{y^2}=1$$
If you take the path $(x,x)$, then:
$$ \lim_{x\to0}f(x,x) =\lim_{x\to0} \frac{(2x)^2}{2x^2}=2$$
So, the limit doesn't exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
} |
find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$ Question:
Given $x + y = 12$, find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$?
Key:
I use $y = 12 - x$ and substitute into the equation, and derivative it.
which I got this
$$\frac{x}{\sqrt{x^2+4}} + \frac{x-12}{\sqrt{x^2-24x+153}} = f'(x).$$
However, aft... | $$\begin{align} f'(x)=0 &\iff \frac{x}{\sqrt{x^2+4}} = \frac{12-x}{\sqrt{x^2-24x+153}}\\
&\implies \frac{x^2}{{x^2+4}} = \frac{x^2 - 24x + 144}{{x^2-24x+153}}\\
&\iff 1 - \frac{4}{{x^2+4}} = 1 - \frac{9}{{x^2-24x+153}}\\
&\iff 4(x^2 -24x + 153) = 9(x^2+4) \end{align}$$
from which you find the roots (if I'm not mistaken... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
What's the relation between different antiderivatives? If a function $f(x)$ has different forms of antiderivatives:
$\frac { d }{ dx } { F }_{ 1 }(x)=f(x)$
$\frac { d }{ dx } { F }_{ 2 }(x)=f(x)$
What's the relationship between $F_1$ and $F_2$, is that ${F}_{1}(x)-{F}_{2}(x)=constant$ correct?
For example, question fin... | Your method 1 is correct.
Your method 2 is not:
$$\int\frac{dx}{x^4-1} \neq \frac12\int\frac{d(x^2)}{(x^2)^2-1}$$
If you're going to substitute for $x^2$, $d(x^2) = 2x dx$, so the equality would be
$$\int\frac{dx}{x^4-1} = \frac12\int\frac{du}{\sqrt{u}(u^2-1)}$$
which gets you nowhere.
But had both methods been correc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\lim\limits_{x\to\infty}(\sin\sqrt{x+1}-\sin\sqrt{x})$ When using Maclaurin series, the limit is
$$\lim\limits_{x\to\infty}\frac{1}{\sqrt{x+1}+\sqrt{x}}=0$$
If we expand the expression with two limits
$$\lim\limits_{x\to\infty}\sin\sqrt{x+1}-\lim\limits_{x\to\infty}\sin\sqrt{x}$$
it diverges.
Which solution i... | Using the prosthaphaeresis formulae,
$$ \sin{A}-\sin{B} = 2\sin{\tfrac{A+B}{2}} \cos{\tfrac{A-B}{2}}, $$
which gives you
$$ 2\sin{\left( \frac{\sqrt{x+1}-\sqrt{x}}{2} \right)} \cos{\left( \frac{\sqrt{x+1}+\sqrt{x}}{2} \right)} $$
Then you have
$$ (\sqrt{x+1}-\sqrt{x})(\sqrt{x+1}+\sqrt{x}) = 1+x-x=1, $$
so we have
$$ 2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
} |
What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer... | Consider $\mathbb Z[\sqrt{2}]= \{a + b\sqrt 2 ; a,b \in \mathbb Z\}$. Take $\alpha = \sqrt 2 - 1 \in \mathbb Z [\sqrt 2]$, then $$0 < \alpha < 1 \implies \alpha ^k \to 0 \,\,\text{as} \,\, k \to \infty \tag {*}$$
Say $\sqrt 2 = \frac{p}{q}$, since $\mathbb Z[\sqrt 2]$ is closed under multiplication and addition we hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "114",
"answer_count": 19,
"answer_id": 12
} |
solving difficult complex number proving if $z= x+iy$ where $y \neq 0$ and $1+z^2 \neq 0$, show that the number $w= z/(1+z^2)$ is real only if $|z|=1$
solution :
$$1+z^2 = 1+ x^2 - y^2 +2xyi$$
$$(1+ x^2 - y^2 +2xyi)(1+ x^2 - y^2 -2xyi)=(1+ x^2 - y^2)^2 - (2xyi)^2$$
real component
$$(1+ x^2 - y^2)x - yi(2xyi) = x + x^3... | HINT:
Let $z=r(\cos A+i\sin A)$ where $r\ge0,A$ are real
As $y\ne0,\sin A\ne0$
$1+z^2=1+r^2\cos2A+i(r^2\sin2A)$
$\dfrac1{1+z^2}=\dfrac{1+r^2\cos2A-ir^2\sin2A}{(1+r^2\cos2A)^2+(r^2\sin2A)^2}$
$r(\cos A+i\sin A)\cdot(1+r^2\cos2A-ir^2\sin2A)$
$=\cdots+ir[\sin A(1+r^2\cos2A)-r^2\cos A\sin2A]=\cdots+ir[\sin A-r^2\sin(2A-A)]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Laurent Series of $\frac{z+1}{z(z-4)^3}$ in $0<|z-4|<4$ Find the Laurent Series of $\displaystyle \frac{z+1}{z(z-4)^3}$ in $0<|z-4|<4$
I thought about doing partial fraction decomposition first, so I'd have
$\displaystyle \frac{A}{z}+\frac{B}{z-4}+\frac{C}{(z-4)^2}+\frac{D}{(z-4)^3}$, but that would take awhile...so I... | Setting $\xi=\frac{z-4}{4}$, i.e. $z=4\xi+4$ we have $0<|\xi|<1$ and therefore
\begin{eqnarray}
\frac{z+1}{z(z-4)^3}&=&\frac{4\xi+5}{(4\xi+4)(4\xi)^3}=\frac{4\xi+5}{4^4\xi^3}\cdot\frac{1}{\xi+1}=\left(\frac{1}{4^3\xi^2}+\frac{5}{4^4\xi^3}\right)\cdot\frac{1}{1+\xi}\\
&=&\left(\frac{1}{4^3\xi^2}+\frac{5}{4^4\xi^3}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve $p_{n+1} + \frac 16 p_n = \frac 1 2 (\frac 5 6 ) ^{n-1}$ I'm trying to solve:
$$p_{n+1} + \frac 16 p_n = \frac 1 2 \left(\frac 5 6 \right) ^{n-1}$$
with initial condition: $p_1 = 1$.
First, I search particular solution of the form $p_n^* = \lambda (\frac 5 6 ) ^n$. I found $\lambda = \frac 3 5$.
Next, I know tha... | An alternate method of finding the solution of the difference equation is through generating functions. The following illustrates this method.
For the difference equation
\begin{align}
p_{n+1} + \frac{1}{6} \, p_n = \frac{1}{2} \, \left(\frac{5}{6} \right) ^{n-1}
\end{align}
it is seen that
\begin{align}
\sum_{n=0}^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1320529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding a polynomial by divisibility
Let $f(x)$ be a polynomial with integer coefficients. If $f(x)$ is divisible by $x^2+1$ and $f(x)+1$ by $x^3+x^2+1$, what is $f(x)$?
My guess is that the only answer is $f(x)=-x^4-x^3-x-1$, but how can I prove it?
| In fact, there are infinitely many solutions. If $f(x)=-x^4-x^3-x-1$ and $g(x)$ is any polynomial, then
$$f(x)+g(x)(x^2+1)(x^3+x^2+1)$$
is a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1322741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find all the natural numbers which are coprimes to $n$ and are not a fermat witness to compositeness of $n$. The number $n=35$ is given. Find all the natural numbers $1 \leq a \leq n-1$ which are coprimes to $n$ and are not a fermat witness to compositeness of $n$.
Is it enough to say that we are looking for these num... | We want to find the $a$ in the interval $1\le a\le 34$ such that $a^{34}\equiv 1\pmod{35}$.
Note that since $a$ is relatively prime to $5$ and $7$, we have $a^4\equiv 1\pmod{5}$ and $a^{6}\equiv 1\pmod{7}$. It follows that $a^{12}\equiv 1$ modulo each of $5$ and $7$, and hence modulo $35$.
We have $a^{12}\equiv 1\pmo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
System of equations with complex numbers-circles The system of equations
\begin{align*}
|z - 2 - 2i| &= \sqrt{23}, \\
|z - 8 - 5i| &= \sqrt{38}
\end{align*}
has two solutions $z_1$ and $z_2$ in complex numbers. Find $(z_1 + z_2)/2$.
So far I have gotten the two original equations to equations of circles,
$(a-2)^2 +(b-2... | We can also treat this in a way that has little to do with complex numbers. The line connecting the centers of the circles at $ \ (2, \ 2) \ $ and $ \ (8, \ 5 ) \ $ has a slope of $ \ \frac{5 - 2}{8 - 2} \ = \ \frac{1}{2} \ $ and has the equation
$$ y \ - \ 2 \ = \ \frac{1}{2} \ (x - 2) \ \ \Rightarrow \ \ y \ = \ \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Quick way to solve the system $\displaystyle \left( \frac{3}{2} \right)^{x-y} - \left( \frac{2}{3} \right)^{x-y} = \frac{65}{36}$, $xy-x+y=118$. Consider the system
$$\begin{aligned} \left( \frac{3}{2} \right)^{x-y} - \left( \frac{2}{3} \right)^{x-y} & = \frac{65}{36}, \\ xy -x +y & = 118. \end{aligned}$$
I have solved... | If $\left(\dfrac32\right)^u=a$
If $a+\dfrac1a=\dfrac{65}{36}<2$
and for $a>0, a+\dfrac1a\ge2\sqrt{a\cdot\dfrac1a}=2,$
I assume $a-\dfrac1a=\dfrac{65}{36}$
$\iff36a^2-65a-36=0$
$a=\dfrac{65\pm\sqrt{65^2+4\cdot36^2}}{2\cdot36}=\dfrac94,-\dfrac49$
For real $u,$
$\left(\dfrac32\right)^u=\dfrac94=\left(\dfrac32\right)^2\imp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"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.