Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Multivariable limit which should be simple ! How to calculate the following limit WITHOUT using spherical coordinates?
$$
\lim _{(x,y,z)\to (0,0,0) } \frac{x^3+y^3+z^3}{x^2+y^2+z^2}
$$
?
Thanks in advance
| Since $x^2 \le x^2+y^2+z^2$, we have $|x| \le \sqrt{x^2+y^2+z^2}$, and similarly for $|y|$ and $|z|$. This gives
$$
\left|\frac{x^3+y^3+z^3}{x^2+y^2+z^2}\right|
\le \frac{|x|^3+|y|^3+|z|^3}{x^2+y^2+z^2}
\le \frac{(x^2+y^2+z^2)^{3/2}+(x^2+y^2+z^2)^{3/2}+(x^2+y^2+z^2)^{3/2}}{x^2+y^2+z^2}
.
$$
Can you take it from there?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1282418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
converge value of series $\sum_{n=0}^{\infty} \left( \frac{1}{n+d+1} - \frac{1}{n+5d+1} \right) $ \begin{align}
\sum_{n=0}^{\infty} \left( \frac{1}{n+d+1} - \frac{1}{n+5d+1} \right) = \sum_{n=0}^{\infty}\frac{4d}{(n+d+1)(n+5d+1)}=
?
\end{align}
I know from the $p$-test, ($i.e$ $\sum \frac{1}{n^p}$ : $p>1$ series conver... | One may recall the following series representation of the digamma function $\displaystyle \psi : = \Gamma'/\Gamma$,
$$
\psi(u+1) = -\gamma + \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{u+n}
\right), \quad u >-1, \tag1
$$ where $\gamma$ is the Euler-Mascheroni. From $(1)$ you get
$$
\sum_{n= 1}^{N}\frac{1}{n+u}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1283313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
how to evaluate the product $\prod _{n=2}^\infty (1+ \frac{1}{n^2}+\frac{1}{n^4}+\frac{1}{n^6}+\cdots )$? Evaluating the infinite product of $\prod _{n=2}^\infty (1+ \frac{1}{n^2}+\frac{1}{n^4}+\frac{1}{n^6}+\cdots )$. Please Help.
| Step $1$: Notice that $$\prod_{n=2}^{\infty} \bigg(\sum_{k=0}^{\infty} \frac{1}{n^{2k}}\bigg) = \prod_{n=2}^{\infty} \frac{1}{1-\frac{1}{n^2}} = \prod_{n=2}^{\infty}\frac{n^2}{n^2-1}$$
Step $2$: Use induction on $N$ to show that $$\prod_{n=2}^N \frac{n^2}{n^2-1} = \frac{2N}{N+1}$$
Step 3: Conclude that $$\prod_{n=2}^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1284512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
$ \lim_{x\to o} \frac{(1+x)^{\frac1x}-e+\frac{ex}{2}}{ex^2} $
$$ \lim_{x\to o} \frac{(1+x)^{\frac1x}-e+\frac{ex}{2}}{ex^2} $$
(can this be duplicate? I think not)
I tried it using many methods
$1.$ Solve this conventionally taking $1^\infty$ form in no luck
$2.$ Did this, expand $ {(1+x)^{\frac1x}}$ using binomial th... | We can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\dfrac{(1 + x)^{1/x} - e + \dfrac{ex}{2}}{ex^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + x)}{x}\right) - e + \dfrac{ex}{2}}{ex^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + x)}{x} - 1\right) - 1 + \dfrac{x}{2}}{x^{2}}\text{ (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1284801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
How to find cotangent? Need to find a $3\cot(x+y)$ if $\tan(x)$ and $\tan(y)$ are the solutions of $x^2-3\sqrt{5}\,x +2 = 0$.
I tried to solve this and got $3\sqrt{5}\cdot1/2$, but the answer is $-\sqrt{5}/5$
| \begin{align}
& x^2 - 3\sqrt{5}\,x+ 2 = (x-a)(x-b) \\[6pt]
= {} & x^2 - (a+b) x + ab.
\end{align}
Therefore $3\sqrt 5= a+b$ and $2=ab$.
Hence $3\sqrt 5 = \tan x + \tan y$ and $2 = \tan x\tan y$.
So
$$
\tan(x+y) = \frac{\tan x+\tan y}{1-\tan x\tan y} = \frac{a+b}{1-ab} = \frac{3\sqrt 5}{1-2},
$$
and finally,
$$
3\cot(x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Finding the surface area of the spheroid $\frac{x^2}{3} + \frac{y^2}{3} + \frac{z^2}{4} = 1$ I'm asked to evaluate this:
What is the surface area of the surface defined by $\frac{x^2}{3} + \frac{y^2}{3} + \frac{z^2}{4} = 1$?
I first parameterized it with spherical coordinates and then I took the cross product and the... | Note that two of the axes of the ellipsoid are the same, and the third is longer. Thus it is actually a prolate spheroid. The formula for the surface area of the prolate spheroid $\frac{x^2 + y^2}{a^2} + \frac{z^2}{c^2}=1$ is:
$$ S=2\pi a^2 + 2\pi\frac{ac}{e}\sin^{-1}e $$
where $e$ is the ellipticity $\sqrt{1-\frac{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $\gcd(a^2, b^2) = \gcd(a,b)^2$
Let $a$ and $b$ be two integers. Show that $\gcd(a^2, b^2) = \gcd(a,b)^2$.
This is what I have done so far:
Let $d = \gcd(a,b)$. Then $d=ax+by$ for some $x,y$. Then $d^2 =(ax+by)^2 = a^2x^2 + 2axby+b^2y^2$.
I am trying to create a linear combination of $a^2$ and $b^2$ but do n... | Suppose that $(a,b)=d$. Then Bezout's Identity says that we have some $x,y$ so that $ax+by=d$, and therefore
$$
a^3x^3+3a^2bx^2y+3ab^2xy^2+b^3y^3=d^3\tag{1}
$$
Dividing $(1)$ by $d$, remembering that both $d\mid a$ and $d\mid b$, we get
$$
a^2\left(\frac adx^3+3\frac bdx^2y\right)+b^2\left(3\frac adxy^2+\frac bdy^3\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
show that $\frac{1}{F_{1}}+\frac{2}{F_{2}}+\cdots+\frac{n}{F_{n}}<13$ Let $F_{n}$ is Fibonacci number,ie.($F_{n}=F_{n-1}+F_{n-2},F_{1}=F_{2}=1$)
show that
$$\dfrac{1}{F_{1}}+\dfrac{2}{F_{2}}+\cdots+\dfrac{n}{F_{n}}<13$$
if we use
Closed-form expression
$$F_{n}=\dfrac{1}{\sqrt{5}}\left(\left(\dfrac{1+\sqrt{5}}{2}\right... | Lemma
$$F_{n}\ge\dfrac{n(n+1)(n+2)}{42},n\ge 5$$
proof:use induction, since
$$\dfrac{1}{42}[k(k+1)(k+2)+(k+1)(k+2)(k+3)]=\dfrac{1}{42}(k+1)(k+2)(2k+3)$$
and
$$(k+1)(2k+3)\ge (k+3)(k+4)$$
so
$$\dfrac{n}{F_{n}}\le \dfrac{42}{(n+1)(n+2)}$$
so
$$\sum_{k=1}^{n}\dfrac{k}{F_{k}}\le 1+2+\dfrac{3}{2}+\dfrac{4}{3}+42\left(\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1287613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 2
} |
For which values of $\alpha \in \mathbb{R}$, does the series $\sum_{n=1}^\infty n^\alpha(\sqrt{n+1} - 2 \sqrt{n} + \sqrt{n-1})$ converge? How do I study for which values of $\alpha \in \mathbb{R}$ the following series converges?
(I have some troubles because of the form [$\infty - \infty$] that arises when taking the l... | Using Taylor expansion $\sqrt{1+x} =_{x\to 0} 1 + \frac x 2 - \frac {x^2}{8} + O(x^3)$, we obtain for $n \to \infty$:
$$\begin{eqnarray}
n^\alpha(\sqrt{n+1} - 2 \sqrt{n} + \sqrt{n-1}) & = & n^\alpha \sqrt{n} \left(\sqrt{1+\frac 1 n} - 2 + \sqrt{1-\frac 1 n}\right) \\
& = & n^{\alpha + 1/2} \left(1 + \frac 1 {2n} - \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Solving a special Quartic Equation.
Solve for $x$
$$(x^2-4)(x^2-2x)=2$$
I have tried the Rational Root Theorem and found that there are no rational roots. Further, the polynomial $p(x)=(x^2-4)(x^2-2x)-2$ is irreducible since when I tried expanding it and writing it as a product of two quadratics, there were no in... | $$(x^2-4)(x^2-2x)=2$$
$$\Rightarrow x^4-2x^3-4x^2+8x-2=0$$
$$\Rightarrow (x^2-x-1)^2-3(x-1)^2=0$$
$$\Rightarrow (x^2-x-1+\sqrt 3\ (x-1))(x^2-x-1-\sqrt 3\ (x-1))=0$$
$$\Rightarrow x^2+(\sqrt 3-1)x-1-\sqrt 3=0\ \ \text{or}\ \ x^2-(\sqrt 3+1)x-1+\sqrt 3=0$$
$$\Rightarrow x=\frac{-\sqrt 3+1\pm\sqrt{8+2\sqrt 3}}{2},\frac{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
How to prove $\int_0^1 \ln\left(\frac{1+x}{1-x}\right) \frac{dx}{x} = \frac{\pi^2}{4}$? Can anyone suggest the method of computing
$$\int_0^1 \ln\left(\frac{1+x}{1-x}\right) \frac{dx}{x} = \frac{\pi^2}{4}\quad ?$$
My trial is following
first set $t =\frac{1-x}{1+x}$ which gives $x=\frac{1-t}{t+1}$
Then
\begin{ali... | Setting the convergence issue aside, I'll compute the integral formally using power series.
Note that
\begin{eqnarray}\frac{\ln\frac{1+x}{1-x}}{x}=\frac{\ln(1+x)-\ln(1-x)}{x}=\sum_{n=1}^\infty\frac{2x^{2n-2}}{2n-1}\end{eqnarray}
Integrating termwise,
\begin{eqnarray}
\int_0^1\sum_{n=1}^\infty\frac{2x^{2n-2}}{2n-1}=2\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
$3^x + 4^y = 5^z$ This is an advanced high-school problem.
Find all natural $x,y$, and $z$ such that
$3^x + 4^y = 5^z$.
The only obvious solution I can see is $x=y=z=2$. Are there any other solutions?
| Note
$$2^z\equiv (3+2)^z=5^z=3^x+4^y\equiv 0+1=1\pmod 3$$
so
$z$ is even number,let $z=2z_{1}$,then we have
$$(5^{z_{1}}+2^y)(5^{z_{1}}-2^y)=3^x$$
so we have
$$5^{z_{1}}+2^y=3^x,5^{z_{1}}-2^y=1$$
then we have
$$(-1)^{z_{1}}+(-1)^y\equiv 0\pmod 3,(-1)^{z_{1}}-(-1)^y\equiv 1\pmod 3$$
so we have $z_{1}$ is odd,$y$ is eve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to solve the difference equation $u_n = u_{n-1} + u_{n-2}+1$ Given that:
$$
\begin{equation}
u_n=\begin{cases}
1, & \text{if $0\leq n\leq1$}\\
u_{n-1} + u_{n-2}+1, & \text{if $n>1$}
\end{cases}
\end{equation}
$$
How do you solve this difference equation?
Thanks
EDIT:
From @marwalix's answer:
$$
u_n=v_n... | Same (essentially) as marwalix ... Can you do the homogeneous equation? Can you find a particular solution to the inhomogeneous equation? Put them together to get the general solution. Then use your initial values to get your solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1302599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve Inequality for $ |x| $ Given $$\big|\frac{(x-2)}{(x+3)}\big| < 4,$$ solve for $x.$
\ My solution
$$|x - 2| < 4|x + 3|$$
Since,
$ |x - 2| \ge |x| - |2| $ and
$ |x + 3| \le |x| + |3| $ according to triangle inequality;
$|x| - |2| < 4|x| + 4|3| $
$-14 < 3|x|$
$|x| > \frac{-14}{3}$
Is this the final answer?
| $$\left|\frac{x-2}{x+3}\right|<4$$
$$-4<\frac{x-2}{x+3}<4~~ |\cdot (x+3)^2\ne 0$$
$$-4(x+3)^2<(x-2)(x+3)<4(x+3)^2$$
$$\begin{cases}
-4(x+3)^2<(x-2)(x+3)\\
(x-2)(x+3)<4(x+3)^2
\end{cases}$$
$$\begin{cases}
0<(x+3)((x-2)+4(x+3))\\
0<(x+3)(4(x+3)-x+2)
\end{cases}$$
$$\begin{cases}
0<(x+3)(5x+10)\\
0<(x+3)(3x+14)
\end{case... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1303632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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:
$$\begin{align}\lim \frac{(x+y)^2}{x^2+y^2} &= \lim \frac{x^2+y^2}{x^2+y^2} + \lim \fr... | If we approach $y=mx $ then $\lim_{(x,y)\to (0,0)}\frac{(x+y)^2}{x^2+y^2}
=\lim_{x\to 0}\frac{(x+xm)^2}{x^2+m^2x^2}=\lim_{x\to 0}\frac{(1+m)^2}{1+m^2}=\frac{(1+m)^2}{1+m^2} $ which is depends on $m$ so limit is not unique . So limit does not exist .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
Solution of a recurrence equations $T(1) = 1$
$T(n) = 2T(\frac{n}{3}) + n + 1$
How do you solve this equzione recurrence? I arrived at this point and then I don't know how to proceed...
$2^kT(\frac{n}{3^k}) + \frac{2^{k-1}n}{3^{k-1}} + 2^{k-1} + \frac{2^{k-2}n}{3^{k-2}} + 2^{k-2} + \frac{2n}{3} + 2 + 1$
Thanks a lot!
| Use the Master Method:
$T(n) = 2T(\frac{n}3) + n + 1$ falls into Case 3, because $c > \log_{b}(a)$, with $c=1$, $a=2$, $b=3$, $f(n) = n+1$. Additionally $af(\frac{n}{b}) \leq kf(n)$ is trivially satisfied with $k = \frac{2}{3}$.
Thus, $T(n) = \Theta(f(n)) = \Theta(n)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding $\lim\limits_{n\to\infty }\frac{1+\frac12+\frac13+\cdots+\frac1n}{1+\frac13+\frac15+\cdots+\frac1{2n+1}}$ I need to compute: $\displaystyle\lim_{n\rightarrow \infty }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}$.
My Attempt: $\dis... | If the numerator is always twice the denominator, then this works provided you include a proof of that. Let's try it when $n=2$:
$$
\frac{1+\frac12}{1+\frac13+\frac15} = \frac{30+15}{30+10+6}= \frac{45}{46}\ne 2.
$$
Let's try it when $n=3$:
$$
\frac{1+\frac12+\frac13}{1+\frac 13+\frac15+\frac17}= \frac{210+105+70}{210... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"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... | I just thought of this one:
Consider the equation $x^2-n=0$ for natural $n$. Evidently, $\sqrt n$ is a solution to the equation. Now the rational root theorem implies that for a root to be rational for that equation, it must be a factor of $n$ (up to sign). If $\sqrt n$ is a factor of $n$ ($n$ is a perfect square), the... | {
"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": 15
} |
Proving a formula using another formula These questions are from the book "What is Mathematics":
Prove
formula 1: $$1 + 3^2 + \cdots + (2n+1)^2 = \frac{(n+1)(2n+1)(2n+3)}{3}$$
formula 2: $$1^3 + 3^3 + \cdots + (2n+1)^3 = (n+1)^2(2n^2+4n+1)$$
Using formulas 4 and 5;
formula 4: $$1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n... | I think you're basically on the right track. Notice that
$$2^2 + 4^2 + 6^2 + \cdots + (2n)^2 = 4(1^2 + 2^2 + 3^2 + ... + n^2).$$
That should make things relatively easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Sum with Generating Functions Find the sum
$$\sum_{n=2}^{\infty} \frac{\binom n2}{4^n} ~~=~~ \frac{\binom 22}{16}+\frac{\binom 32}{64}+\frac{\binom 42}{256}+\cdots$$
How can I use generating functions to solve this?
| For an alternative technique, let
$$ S = \sum_{n\geq 2}\frac{\binom{n}{2}}{4^n}.$$
We have:
$$\begin{eqnarray*} 3S = 4S-S &=& \sum_{n\geq 2}\frac{\binom{n}{2}}{4^{n-1}}-\sum_{n\geq 2}\frac{\binom{n}{2}}{4^{n}}=\frac{1}{4}+\sum_{n\geq 2}\frac{\binom{n+1}{2}-\binom{n}{2}}{4^n}\\&=&\frac{1}{4}+\sum_{n\geq 2}\frac{n}{4^n}.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integral of Trigonometric Identities $$\int(\sin(x))^3(\cos(2x))^2dx$$
I can write $$\sin^3(x)=\sin(x)(1-\cos^2(x)=\sin(x)-\sin(x)\cos^2(x)$$
for $$\cos^2(2x)=(1-\sin^2(x))^2=1-4\sin^2(x)+4\sin^4(x)$$
after simplifying the Trig identities i get:
$$\int(sin^3(x)-4sin^5(x)+4sin^7(x))dx$$
so i need to know how to go furth... | Note $f(x)=\sin^3 x \cos^2 2x$. You have
$$\cos 2x = 2 \cos^2 x -1$$
Hence
$$(\sin^3 x)(\cos^2 2x)=\sin x \sin^2 x (2 \cos^2 x -1)=\sin x (1- \cos^2 x) (2 \cos^2 x -1)$$
As $(\cos x)^\prime = -\sin x$, we get by expending the right hand side $$f(x)=-(\cos x)^\prime (- 2 \cos^4 x +3 \cos^2 x -1)$$ and finally
$$\int f(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Finding the partial fraction decomposition of $\frac{4s^2 - 5s + 2}{s^2(s^2 +9)}$ I am trying to find the partial fraction decomposition of $\dfrac{4s^2 - 5s + 2}{s^2(s^2 +9)}$ into something of the form $A\dfrac{1}{s} + B\dfrac{1}{s^2} + C\dfrac{1}{s^2+9} + D\dfrac{s}{s^2 + 9}$.
I am unable to factor the numerator int... | $$\dfrac{4s^2 - 5s + 2}{s^2(s^2 +9)}=\frac{As+B}{s^2}+\frac{Cs+D}{s^2+9}$$
Multiply both sides by $s^2(s^2-9):$
$${4s^2 - 5s + 2}=({As+B})({s^2}+9)+{(Cs+D)}s^2$$
Combine terms:
$$4s^2-5s+2=(A+C)s^3+(B+D)s^2+(9A)s+9B$$
Solve for coefficients:
$$9B=2\implies B=\frac 29$$
$$\left(\frac 29+D\right)=4\implies D=\frac{34}9$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
with inequality $\frac{y}{xy+2y+1}+\frac{z}{yz+2z+1}+\frac{x}{zx+2x+1}\le\frac{3}{4}$ Let $x,y,z\ge 0$, show that
$$\dfrac{y}{xy+2y+1}+\dfrac{z}{yz+2z+1}+\dfrac{x}{zx+2x+1}\le\dfrac{3}{4}$$
I had solve
$$\sum_{cyc}\dfrac{y}{xy+y+1}\le 1$$
becasuse After some simple computations, it is equivalent to
$$(1-xyz)^2\ge 0$$
| Note that by CS inequality,$$\frac{4y^2}{xy^2+2y^2+y} =\frac{(y+y)^2}{(xy^2+y^2)+(y^2+y)}\le \frac{y^2}{xy^2+y^2}+\frac{y^2}{y^2+y}=\frac{1}{x+1}+\frac{y}{y+1}$$
Summing three such terms we get
$$4\sum_{cyc}\frac{y}{xy+2y+1}\le \sum_{cyc}\frac1{x+1}+\sum_{cyc}\frac{x}{x+1}=3$$
P.S. You will need to handle the case $xyz... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to find $\lim_{x \to 0}\frac{\cos(ax)-\cos(bx) \cos(cx)}{\sin(bx) \sin(cx)}$ How to find $$\lim_\limits{x \to 0}\frac{\cos (ax)-\cos (bx) \cos(cx)}{\sin(bx) \sin(cx)}$$
I tried using L Hospital's rule but its not working!Help please!
| One can use only the following standard limits (without L'Hospital's rule nor Taylor series)
\begin{equation*}
\lim_{u\rightarrow 0}\frac{1-\cos u}{u^{2}}=\frac{1}{2}\ \ \ \ \ \ \ \ ,\
\lim_{u\rightarrow 0}\cos u=\cos 0=1\ \ \ \ \ \ \ \ \lim_{u\rightarrow 0}%
\frac{\sin u}{u}=1
\end{equation*}
To this end, transform th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Proof by induction that $3^n - 1$ is an even number How to demonstrate that $3^n - 1$ is an even number using the principle of induction?
I tried taking that $3^k - 1$ is an even number and as a thesis I must demonstrate that $3^{k+1} - 1$ is an even number, but I can't make a logical argument.
So I think it's wrong... | First, $ 3^0 - 1 = 0 $ which is even so $3^n - 1 $ is even for $ n = 0 $
Suppose that
$3^k - 1 $ is even. We need to show that $3^{k+1}- 1 $ is even. Well,
$ 3^{k+1}-1 = 3(3^{k}) -1 = 3(3^k - 1) + 2 $
By our assumption, $ 3^k-1 $ is even so $ 3^{k} - 1 = 2m $, for some integer $m$
Thus, $ 3(3^k - 1) + 2 = 3(2m) + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 3
} |
Show that $30 \mid (n^9 - n)$ I am trying to show that $30 \mid (n^9 - n)$. I thought about using induction but I'm stuck at the induction step.
Base Case: $n = 1 \implies 1^ 9 - 1 = 0$ and $30 \mid 0$.
Induction Step: Assuming $30 \mid k^9 - k$ for some $k \in \mathbb{N}$, we have $(k+1)^9 - (k+1) = [9k^8 + 36k^7 + 84... | Here's an alternative way of doing it:
Since $\phi(30)=8$, we get that $m^9 \equiv m \pmod{30}$ whenever $(m,30)=1$.
Since the only factors of $30$ are $2$,$3$, and $5$, we write:
$$ n=2^a \cdot 3^b \cdot 5^c \cdot m$$
Where $(m,30)=1$. Now:
$$ n^9 = 2^{9a} \cdot 3^{9b}\cdot 5^{9c} \cdot m^9 \equiv (2^9)^a \cdot (3^9)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1320452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
trigonometric inequality bound? Suppose that one wants to determine an upper bound of the trigonometric expression
$$
a\sin(x)+b\cos(x),
$$
where $a,b\in\mathbb{R}$. My instinct is to proceed as follows:
$$
a\sin(x)+b\cos(x)\leq |a|+|b|,
$$
which is correct. If one were to drop the modulus signs, would that make the bo... | the conservative(naive) upper bound $|a| + |b|$ is too big because $\sin x$ and $\cos x$ cannot be both max or both min simultaneously. the reason is the ever present constraint $$\sin^2 x +\cos^2 x = 1.$$
what happens is $$\begin{align}a\sin x + b \cos x &= \sqrt{a^2 + b^2}\left(\frac a{\sqrt{a^2 + b^2}}\sin x + \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1320788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Evaluate $\int_{\pi/6}^{\pi/2} (\frac{1}{2} \tan\frac{x}{2}+\frac{1}{4} \tan\frac{x}{4}+ \cdots+\frac{1}{2^n} \tan\frac{x}{2^n}+\cdots) dx$ $$f(x)=\frac{1}{2} \tan\frac{x}{2}+\frac{1}{4} \tan\frac{x}{4}+....+\frac{1}{2^n} \tan\frac{x}{2^n}+...$$
Check the function $f(x)$ is continuous on $[\frac{\pi}{6},\frac{\pi}{2}]$... | $$\sum_{k=1}^{n}\dfrac{1}{2^k}\tan{\dfrac{x}{2^k}}=-\left(\sum_{k=1}^{n}\ln{\left|\cos{\dfrac{x}{2^k}}\right|}\right)'=-\left(\ln{\left|\dfrac{\sin{x}}{2^n\sin{(x/2^n)}}\right|}\right)'=\dfrac{1}{2^n}\cot{\dfrac{x}{2^n}}-\cot{x}$$
so
$$\sum_{k=1}^{+\infty}\dfrac{1}{2^k}\tan{\dfrac{x}{2^k}}=\dfrac{1}{x}-\cot{x}$$
then i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1321153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
square of a permutation cycle $$\sigma = \begin{bmatrix}
1 &2 &3 &4 &5 &6 &7 &8 &9 \\
1&5 &7 &4 &6 &9 &3 &2 &8
\end{bmatrix}$$
$$\sigma^{2} = \begin{bmatrix}
1 &2 &3 &4 &5 &6 &7 &8 &9 \\
1& 6&3 &4 &9 &8 &7 &5 & 2
\end{bmatrix}$$
Given $$\sigma$$ I found $$\sigma^2.$$
I want to find $$\s... | Permutations are represented in two ways. One is a verbose description of the mapping: $$\sigma = \begin{bmatrix}
1 &2 &3 &4 &5 &6 &7 &8 &9 \\
1&5 &7 &4 &6 &9 &3 &2 &8
\end{bmatrix}$$ means that $1$ goes to $1$, $2$ goes to $5$, and so on.
We abbreviate this to cycle notation, which is more compact a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1321311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Puiseux Series? WolframAlpha says that $$\sqrt{x^2-1}$$ expanded in Puiseux series near 1 is $\sqrt 2 \sqrt{x-1}$.
I don't know what a Puiseux series is; I have searched on the net but I haven't understood much... can you briefly explain it to me and how I can obtain this result?
| If you are familiar with Taylor series, in this case you easily can get the same expansion "for free," without having to sweat too much. Set $y=x-1$, so that you are looking at
$$
\sqrt{x^2-1} = \sqrt{(x+1)(x-1)} = \sqrt{(y+2)y} = \sqrt{2y}\sqrt{1+\frac{y}{2}}
$$
when $y\to 0$ (i.e., $x\to 1$). Recalling the Taylor ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1321992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 0
} |
Simplifying sum equation. (Solving max integer encoded by n bits) Probably a lack of understanding of basic algebra. I can't get my head around why this sum to N equation simplifies to this much simpler form.
$$
\sum_{i=0}^{n-2} 2^{-i+n-2} + 2^i = 2^n - 2
$$
Background
To give you some background I am trying to derive... | First notice that for any $m$, $\sum_{i=0}^{m-1} 2^i = 2^m-1$. To see this
(and really this is the geometric series proof applied to your situation), let $$S=\sum_{i=0}^{m-1} 2^i=1+2+4+8+\cdots+2^{m-1}.$$ Then \begin{eqnarray*} 2S&=&2+4+8+\cdots+2^{m-1}+2^m\\ &=& (-1+1)+2+4+\cdots+2^{m-1}+2^m\\&=& -1+(1+2+4+\cdots+2^{m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Summation of series in powers of x with certain combinations as coefficients How can I find the sum: $$\sum_{k=0}^{n} \binom{n-k}{k}x^{k}$$
Edit: The answer to this question is: $$\frac{{(1+\sqrt{1+4x})}^{n+1}-{(1-\sqrt{1+4x})}^{n+1}}{2^{n+1}\sqrt{1+4x}}$$ I don't know how to arrive at this answer.
| This answer is similar to this answer and this answer. That is, compute the generating function of the given sequence:
$$
\begin{align}
&\sum_{n=0}^\infty\sum_{k=0}^n\binom{n-k}{k}x^ky^n\\
&=\sum_{k=0}^\infty\sum_{n=k}^\infty\binom{n-k}{k}x^ky^n\tag{1}\\
&=\sum_{k=0}^\infty\sum_{n=0}^\infty\binom{n}{k}x^ky^{n+k}\tag{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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What does $\sum_{k=0}^\infty \frac{k}{2^k}$ converge to? This problem comes from another equation on another question (this one).
I tried to split it in half but I found out that
$$\sum_{k=0}^\infty \frac{k}{2^k}$$
can't be divided.
Knowing that $$\sum_{k=0}^\infty x^k=\frac{1}{1-x}$$
I wrote that
$$\sum_{k=0}^\infty \... | You can also view it this way, which is quite intuitive.
\begin{align*}
\sum_{k=0}^\infty \frac{k}{2^k}
&= \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \frac{5}{32} + \cdots \\
&= \frac12 + \Bigl(\frac14 + \frac14\Bigr) + \Bigl(\frac18 + \frac18 + \frac18\Bigr) + \Bigl(\frac1{16} + \frac1{16} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 2
} |
How prove $\frac{(a-b)^4+(b-c)^4+(c-a)^4}{(a+b+c)^4}=2$ Let $a,b,c\in R$,and such $ab+bc+ac=0,a+b+c\neq 0$
show that
$$\dfrac{(a-b)^4+(b-c)^4+(c-a)^4}{(a+b+c)^4}=2$$
| WLOG, assume $a+b+c=1 \Rightarrow (a-b)^2+(b-c)^2+(c-a)^2 = 2(a^2+b^2+c^2)=2$, since $1=(a+b+c)^2 = a^2+b^2+c^2$, and also $2((a-b)(b-c))^2 = 2(ab-ac-b^2+bc)^2=2(-2ac-b^2)^2=2(b^4+4acb^2+4a^2c^2)$. Thus:$\displaystyle \sum_{cyclic}2((a-b)(b-c))^2=2(a^4+b^4+c^4)+8abc(a+b+c)+8(a^2b^2+b^2c^2+c^2a^2)=2(1-2(a^2b^2+b^2c^2+c^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $x ^ 6 = x$, prove that $x ^ 2 = x$, in a ring I found a short and interesting problem:
Given a ring $(R, +, \cdot)$ and knowing that $x ^ 6 = x\ (\forall x\in R)$, prove that $x ^ 2 = x\ (\forall x \in R)$.
While it is short, I cannot figure out how to solve it. If it would be the reverse, then the solution were ... | Consider the case where the ring has a unit (if not, then one could consider $R$ as a $\mathbb{Z}$-algebra, but the details would change in that case).
Observe $2^6=2$ and $3^6=3$. In other words, $64=2$ and $729=3$. So $62=0$ and $726=0$. Since $\gcd(62,726)=2$, it follows that $2=0$ (by repeated subtraction).
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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Proving “The sum of $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.” This site
states:
Example $\boldsymbol 3$. The sum of consecutive cubes. Prove this remarkable fact of arithmetic: $$1^3 +2^3 +3^3 +\ldots +n^3 =(1 +2 +3 +\ldots +n)^2.$$
“The sum of $n$ consecutive cubes is equa... | No! Generally speaking, one shows by induction that $\,1^r+2^r+\dots+n^r\,$ has a closed form which is a polynomial in $n$ of degree $\color{red}{r+1}$.
Examples:
*
*$1 +2 +\dots+n =\dfrac{n(n+1)}2$.
*$1^2+2^2+\dots+n^2=\dfrac{n(n+1)(2n+1)}6$
*$1^0+2^0+\dots+n^0=\underbrace{1+1++\dots+1}_{n \ \text{times}}=n$
an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
} |
Rotation of a line by a matrix
Give the equation of the line $\ell'$ that is obtained by rotating $\ell$: $x+2y=5$ by an angle of $\theta=\frac{1}{2}\pi$ with center point $O(0,0)$.
The rotation matrix is $\left.\begin{pmatrix}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{pmatrix}\right|_{\alpha=\frac{1... | A shorter way:
Given line is $x+2y=5$, so we could say $x=r$ and $y=\dfrac{5-r}{2}$.
Now we multiply $\begin{pmatrix} 0&-1\\1&0\\ \end{pmatrix}\begin{pmatrix} r\\ \dfrac{5-r}{2}\\ \end{pmatrix}$, it gives us coordinates after rotation which are
$\begin{pmatrix} \dfrac{r-5}{2}\\ r\\ \end{pmatrix}$. namely, the new $x=\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$\frac{(-1)^n}{2\cdot 4\cdot \cdot\cdot2n}=\frac{(-1)^n}{2^n\cdot n!}$ $$\frac{(-1)^n}{2\cdot 4\cdot \cdot\cdot2n}=\frac{(-1)^n}{2^n\cdot n!}$$
$$\frac{(-1)^n}{3\cdot 5\cdot \cdot \cdot(2n+1)}=\frac{{(-2)^n} \cdot n! }{(2n+1)!}$$
can anyone tell me if these are true or false?
| Yes, both are true.
For the first, note that
$$ 2 \cdot 4 \dots 2n = (1 \cdot 2) \cdot (2 \cdot 2) \cdot (3 \cdot 2) \cdot (4 \cdot 2) \dots (n \cdot 2) = n! \cdot 2^n$$
For the second:
$$3 \cdot 5 \dots (2n + 1) = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \dots (2n + 1)}{2 \cdot 4 \dots 2n} = \frac{(2n+1)!}{n! \cdot 2^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1330229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Quick way to solve the system $\displaystyle \left( \frac{3}{2} \right)^{x-y} - \left( \frac{2}{3} \right)^{x-y} = \frac{65}{36}$, $xy-x+y=118$. Consider the system
$$\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}$$
I have solved... | Let $u=x-y$ and $a=\frac23$
\begin{align}
\left( \frac{3}{2} \right)^{x-y} - \left( \frac{2}{3} \right)^{x-y} & = a^{-u} - a^{u}\\
&=e^{-u\log a}-e^{u \log a}\\
&=-2\sinh (u\log a)
\end{align}
therefore \begin{align}
\log a^u &=\text{arcsinh} \frac{-36}{72}\\
&=\log\Big(-\frac{36}{72}+\sqrt{1+(-\frac{36}{72})^2}\B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find the cubic equation of $x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$ Find the cubic equation which has a root $$x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$$
My attempt is
$$x^3=2-\sqrt{3}+3\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(... | The general solution to a cubic:
$$\begin{align}
r & = \sqrt[3]{
\underbrace{ \left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right) }_{p}
+ \sqrt[2]{
\left(\underbrace{ \frac{c}{3a} - \frac{b^2}{9a^2} }_{q} \right)^3 + \left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right)^2
}
} \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 7
} |
Substituting the value $x=2+\sqrt{3}$ into $x^2 + 1/x^2$ My teacher gave me a question which I am not able to solve:
If $x=2+\sqrt{3}$ then find the value of $x^2 + 1/x^2$
I tried to substitute the value of x in the expression, but that comes out to be very big.
| By brute-force:
$$(2+\sqrt3)^2+\frac1{(2+\sqrt3)^2}=\frac{(2+\sqrt3)^4+1}{(2+\sqrt3)^2}.$$
Then
$$(2+\sqrt3)^2=7+4\sqrt3,\\
(2+\sqrt3)^4=(7+4\sqrt3)^2=97+56\sqrt3,$$
$$\frac{(2+\sqrt3)^4+1}{(2+\sqrt3)^2}=\frac{98+56\sqrt3}{7+4\sqrt3}=\frac{14\cdot7+14\cdot4\sqrt3}{7+4\sqrt3}=14.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 6
} |
Finding the Jacobian of a system of 1st-order ODEs I am trying to find the Jacobian matrix of the following system of 1st-order ODEs.
My system is: $\dfrac{dx}{dt} = \left(x-3\right)\!\left(y+x\right) \\ \dfrac{dy}{dt} = \left(x+4\right)\!\left(y-2x\right)$
Since $(x-3)(y+x) = xy+x^2 - 3y - 3x$ and $(x+4)(y-2x) = xy-2x... | Note that
$\dfrac{\partial (xy + x^2 - 3x - 3y)}{\partial y}$
$ = x - 3, \tag{1}$
not $x- 2$; the correct matrix is thus
$\begin{bmatrix} y + 2x -3 & x - 3 \\ y - 4x - 8 & x + 4 \end{bmatrix}. \tag{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1332311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding $ \lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x^2-2x} $ I'm kind of stuck on this problem, I could use a hint.
$$
\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x^2-2x}
$$
After some algebra, I get
$$
{\lim_{x\to 2}\frac{x+2 - 2x}{x(x-2)-\sqrt{x+2}+\sqrt{2x}}}
$$
EDIT above should be:
$$
\lim_{x\to 2}\frac{x+2 - 2x}{x(x... | You can notice that the denominator is $x(x-2)$, you have
$$
\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x^2-2x}=
\left(\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x-2}\right)
\left(\lim_{x\to 2}\frac{1}{x}\right)
$$
provided the limit exists. Thus we can concentrate on
$$
\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x-2}
$$
whi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Evaluate $\int^0_1 \frac{\ln(t)}{1-t^2}dt$
Evaluate: $$\int^0_1 \dfrac{\ln(t)}{1-t^2}dt$$
This actually came up while solving another integral. It was suggested that I use a binomial series, but unfortunately I do not understand how to use this. Can anyone help me out?
| First consider the operation
\begin{align}
\partial_{n} \, t^{n} = \frac{d}{dn} \, e^{n \ln(t)} = \ln(t) \, e^{n \ln(t)} = t^{n} \, \ln(t).
\end{align}
Now consider the integral, where the operation just presented will be used,
\begin{align}
I_{n} = \int_{0}^{1} \ln(t) \, t^{n} \, dt = \partial_{n} \, \int_{0}^{1} t^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Derivative of a lemniscate at the left hand side How does $${d \over{dx}}(3(x^2+y^2)^2)$$ turn into $$12y(x^2+y^2){{dy} \over{dx}}+12x(x^2+y^2)$$? I'm having a hard time solving it algebraically without it turning into a huge polynomial.
| Use the chain rule.
$$3\left({d\over{dx}}(x^2+y^2)^2\right)=3\cdot2{d\over{dx}}(x^2+y^2)(x^2+y^2)=6\left(2x+2{d\over{dx}}(y)y\right)(x^2+y^2)=6\left(2x+2y{dy\over{dx}}\right)(x^2+y^2)=12(x^2+y^2)\left(y{dy\over{dx}}+x\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1335731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve the system of equations with $x=y$ Solve the system of equations: $\left\{\begin{array}{l}\sqrt{x^2+(y-2)(x-y)}+\sqrt{xy}=2y\\\sqrt{xy+x+5}-\dfrac{6x-5}{4}=\dfrac{1}{4}\left(\sqrt{2y+1}-2\right)^2\end{array}\right.$
I used wolframalpha.com and got the only solution: $(x;y)=(4;4)$
And I guess that we can get $x=y... | If we allow the non-principal square root, you can actually have more than one solution to the system,
$$\left\{\begin{array}{l}\sqrt{x^2+(y-2)(x-y)}\color{red}\pm\sqrt{xy}=2y\\\sqrt{xy+x+5}-\dfrac{6x-5}{4}=\dfrac{1}{4}\left(\sqrt{2y+1}-2\right)^2\end{array}\right.$$
The positive case is solved by $x=y = 4$, but the ne... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1335998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Finding the basis of a subset of polynomials
Let $W$ be a subspace of the polynomials with maximum degree of $3$ and $p(1) = p(2) = 0$. Find the basis and the dimension of the subspace. The field is the real numbers.
My attempt:
My first thought is that it is the usual basis of $\mathbb{P}_3 = \{1, x, x^2, x^3\}$. H... | Note that $p(x)=ax^3+bx^2+cx+d$ satisfies $p(1)=p(2)=0$ if and only if
\begin{array}{rcrcrcrcrcrcrc}
a&+&b&+&c&+&d&=&0 \\
8\,a&+&4\,b&+&2\,c&+&d&=&0
\end{array}
But
$$
\DeclareMathOperator{rref}{rref}\rref
\begin{bmatrix}
1&1&1&1&0\\8&4&2&1&0
\end{bmatrix}
=
\begin{bmatrix}
1&0&-1/2&-3/4&0\\ 0&1&3/2&7/4&0
\end{bmatrix}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to integrate $\int^{\infty}_{-\infty} e^{-2\pi^2/x^2} dx$? I am wondering how can i integrate this quantity above?
Here it is again,
$$\int^{\infty}_{-\infty} e^{-2\pi^2/x^2}dx.$$
Thanks a lot.
| Since the function is even then
\begin{align}
I = \int_{ - \infty }^\infty {\exp \left( { - \frac{{2\pi ^2 }}{{x^2 }}} \right)dx} = 2\int_0^\infty {\exp \left( { - \frac{{2\pi ^2 }}{{x^2 }}} \right)dx}
\end{align}
Setting
\begin{align}
x = \frac{{\sqrt 2 }}{{\sqrt u }}\pi ,\,\,\,\text{i.e.,}\,\,\,u = \frac{{2\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1338795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the following definite and indefinite integrals I want to calculate the integral
$$\int_0^{\frac{\pi}{2}} e ^{ \sin t}\, dt.$$
Can we find a primitive function for $f(t) = e ^{\sin t}$?
| For $\int e^{\sin t}~dt$ ,
$\int e^{\sin t}~dt$
$=\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n}t}{(2n)!}dt+\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}t}{(2n+1)!}dt$
$=\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\sin^{2n}t}{(2n)!}\right)dt+\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}t}{(2n+1)!}dt$
For $n$ is any natu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$ \lim_{n\to+\infty} \frac{1\times 3\times \ldots \times (2n+1)}{2\times 4\times \ldots\times 2n}\times\frac{1}{\sqrt{n}}$
Knowing that :
$$I_n=\int_0^{\frac{\pi}{2}}\cos^n(t) \, dt$$
$$I_{2n}=\frac{1\times 3\times \ldots \times (2n-1)}{2\times 4\times \ldots\times 2n}\times\dfrac{\pi}{2}\quad \forall n\geq 1$$
$... | So we know that:
$$ I_{2n}=\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta = \frac{\pi}{2}\cdot\frac{(2n-1)!!}{(2n)!!}=\frac{\pi}{2}\cdot\frac{1}{4^n}\binom{2n}{n}.\tag{1}$$
If, in the same way, we prove:
$$ I_{2n+1}=\int_{0}^{\pi/2}\cos^{2n+1}(\theta)\,d\theta = \frac{(2n)!!}{(2n+1)!!}\tag{2} $$
then we have:
$$\lim_{n\to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Find the equation of base of Isosceles Traingle Given the two Legs $AB$ and $AC$ of an Isosceles Traingle as $7x-y=3$ and $x-y+3=0$ Respectively. if area of $\Delta ABC$ is $5$ Square units, Find the Equation of the base $BC$
My Try:
The coordinates of $A$ is $(1,4)$. Let the Slope of $BC$ is $m$. Since angle between $... | The problem can be solved using parametric equations. Since you know that both equations pass through the point, $(1,4)^T$, convenient parameterizations for the lines are:
$\left(\begin{array}{c}x_1\\y_1\end{array}\right) = \left(\begin{array}{c}1\\7\end{array}\right)s+\left(\begin{array}{c}1\\4\end{array}\right)$
and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the value of this series what is the value of this series $$\sum_{n=1}^\infty \frac{n^2}{2^n} = \frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+\frac{25}{32}+\cdots$$
I really tried, but I couldn't, help guys?
| Assuming as @Soke does, your series is $$\sum_{n=1}^\infty \frac{n^2}{2^n}$$ we can use the geometric series in order to find a closed form solution. The trick is to take derivatives and multiply by $x$.
Note that $$f(x) = \sum_{n=1}^\infty x^n = \frac{x}{1-x}$$ yields a derivative of $$\sum_{n=1}^\infty n x^{n-1} = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Exponentiation of Pascal's Triangle(in matrix form) I want to find a pattern in subsequent exponentiations of the pascal triangle shown in the form below
Matrix P[K+1][K+1]:
$$
\begin{matrix}
\binom{0}{0} & 0 & 0 & 0\cdots &0\\
\binom{1}{0} & \binom{1}{1} & 0 & 0\cdots &0 \\
\binom{2}{0}... | You can use a binomial expansion and rearrangement to show
$$(1+N)^{i-j}{i\choose j}=\sum_{n=0}^{i-j} N^{i-j-n}{i-j \choose n} {i\choose j} =\sum_{m=j}^i N^{i-m}{i\choose m} {m\choose j} $$
and then induction to show
$$P^N=\left(\begin{matrix}
N^{0-0}\binom{0}{0} & 0 & 0 & 0&\cdots &0\\
N^{1-0}\binom{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Approximating $\tan61^\circ$ using a Taylor polynomial centered at $\frac \pi 3$ : how to proceed? Here's what I have so far...
I wrote a general approximation of $f(x)=\tan(x)$ , which then simplified a bit to this:
$$\tan \left(\frac{61π}{180}\right) + \sec^2\left(\frac{61π}{180}\right)\left(\frac{π}{180}\right) + \t... | $$f(x)=\tan(x)\\x=60^\circ=\frac{\pi}{3} \\h=\Delta x=1^\circ=\frac{\pi}{180}$$now
$$ f(x+h)=f(x)+hf'(x)+\frac{h^2}{2!}f''(x)+\frac{h^3}{3!}f'''(x)+...\\f(\frac{\pi}{3}+\frac{\pi}{180})=f(\frac{\pi}{3})+hf'(\frac{\pi}{3})+\frac{h^2}{2!}f''(\frac{\pi}{3})+\frac{h^3}{3!}f'''(\frac{\pi}{3})+...\\f(\frac{\pi}{3})+(\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculating $\sum_{k=0}^{n}\sin(k\theta)$ I'm given the task of calculating the sum $\sum_{i=0}^{n}\sin(i\theta)$.
So far, I've tried converting each $\sin(i\theta)$ in the sum into its taylor series form to get:
$\sin(\theta)=\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\frac{\theta^7}{7!}...$
$\sin(2\theta)=2\theta... | $$
2\sin \frac{\theta}2 \sin k\theta = \cos \frac{(2k-1)\theta}2 -\cos \frac{(2k+1)\theta}2
$$
so (telescoping sum)
$$
2 \sin \frac{\theta}2 \sum_{k=1}^n\sin k\theta = \cos (\frac{\theta}2) -\cos \frac{(2n+1)\theta}2 \\
= 2 \sin \frac{n\theta}2 \sin \frac{(n+1)\theta}2
$$
giving
$$
\sum_{k=1}^n\sin k\theta = \frac{\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that in any triangle, we have $\frac{a\sin A+b\sin B+c\sin C}{a\cos A+b\cos B+c\cos C}=R\left(\frac{a^2+b^2+c^2}{abc}\right),$ Show that in any triangle, we have $$\frac{a\sin A+b\sin B+c\sin C}{a\cos A+b\cos B+c\cos C}=R\left(\frac{a^2+b^2+c^2}{abc}\right),$$
where $R$ is the circumradius of the triangle.
Here is... | Use Law of Sines for the numerator, $\displaystyle\sum_\text{cyc}a\sin A=\dfrac{\sum_\text{cyc} a^2}{2R}$
For the denominator, $\displaystyle\sum_\text{cyc}a\cos A=\sum_\text{cyc}(2R\sin A\cos A)=R\sum_\text{cyc}\sin2A$
Using Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a tr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Product of cosines: $ \prod_{r=1}^{7} \cos \left(\frac{r\pi}{15}\right) $
Evaluate
$$ \prod_{r=1}^{7} \cos \left({\dfrac{r\pi}{15}}\right) $$
I tried trigonometric identities of product of cosines, i.e, $$\cos\text{A}\cdot\cos\text{B} = \dfrac{1}{2}[ \cos(A+B)+\cos(A-B)] $$
but I couldn't find the product.
Any help w... | Note: Here's another variation inspired by an answer to this question.
We consider the roots of unity $e^{\frac{2\pi i k}{15}}, 0\leq k < 15$ of the polynomial
$$p(z)=z^{15}-1=\prod_{k=0}^{14}(z-e^{\frac{2\pi i k}{15}})$$
We obtain
\begin{align*}
-p(-z)=z^{15}+1&=\prod_{k=0}^{14}(z+e^{\frac{2\pi i k}{15}})\\
&=(z+1)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 5,
"answer_id": 3
} |
this inequality $\prod_{cyc} (x^2+x+1)\ge 9\sum_{cyc} xy$ Let $x,y,z\in R$,and $x+y+z=3$
show that:
$$(x^2+x+1)(y^2+y+1)(z^2+z+1)\ge 9(xy+yz+xz)$$
Things I have tried so far:$$9(xy+yz+xz)\le 3(x+y+z)^2=27$$
so it suffices to prove that
$$(x^2+x+1)(y^2+y+1)(z^2+z+1)\ge 27$$
then the problem is solved. I stuck in here
| It's enough to prove that
$$\prod_{cyc}\left(x^2+\frac{x(x+y+z)}{3}+\frac{(x+y+z)^2}{9}\right)-\frac{(xy+xz+yz)(x+y+z)^4}{9}\geq$$
$$\geq\frac{1}{2916}\left(\sum_{cyc}(5x^3+6x^2y+6x^2z-17xyz)\right)^2.$$
Let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative, and $xyz=w^3$.
Since $$\sum_{cyc}(5x^3+6x^2y+6x^2z-17x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Determine whether $\sum \frac{2^n + n^2 3^n}{6^n}$ converges For the series $$\sum_{n=1}^{\infty}\dfrac{2^n+n^23^n}{6^n},$$ I was thinking of using the root test? so then I would get $(2+n^2/n+3)/6$ but how do I find the limit of this?
| Hint: $\lim_{n \rightarrow \infty} \dfrac {n^4}{2^n} = 0$
$\implies $ After a particular value of $m \in \mathbb N : n^4 << 2^n ~\forall~ n \ge m$ or
$\dfrac {n^4}{2^n}<1 ~\forall~ n \ge m$.
Hence, $\dfrac {n^2}{2^n} < \dfrac {1}{n^2} \implies \sum \dfrac {n^2}{2^n} < \sum \dfrac {1}{n^2} $
and $\sum \dfrac {1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
If $\frac{a}{b}<\frac{c}{d}$ and $\frac{e}{f}<\frac{g}{h}$, then $\frac{a+e}{b+f} < \frac{c+g}{d+h}$.
If $a, b, c, d, e, f, g, h$ are positive numbers satisfying $\frac{a}{b}<\frac{c}{d}$ and $\frac{e}{f}<\frac{g}{h}$ and $b+f>d+h$, then $\frac{a+e}{b+f} < \frac{c+g}{d+h}$.
I thought it is easy to prove. But I could ... | The updated question (with the additional constraint $b+f>d+h$) is also false. For example,
$\frac{1}{1}<\frac{3}{2}$ and $\frac{9}{4}<\frac{5}{2}$, but $\frac{1+9}{1+4} = \frac{10}{5} = \frac{8}{4} = \frac{3+5}{2+2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to rewrite $\pi - \arccos(x)$ as $2\arctan(y)$? I get the following results after solving the equation $\sqrt[4]{1 - \frac{4}{3}\cos(2x) - \sin^4(x)} = -\,\cos(x)$, :
$$
x_{1} = \pi - \arccos(\frac{\sqrt{6}}{3}) + 2\pi n: n \in \mathbb{Z}\\
x_{2} = \pi + \arccos(\frac{\sqrt{6}}{3}) + 2\pi n: n \in \mathbb{Z}
$$
Wo... | Let $\arctan x=y\implies x=\tan y$
Using this, $0\le y\le\dfrac\pi2\iff0\le2y\le\pi$ and $\arccos$ lies in $[0,\pi]$
$\cos2y=\dfrac{1-\tan^2y}{1+\tan^2y}=\dfrac{1-x^2}{1+x^2}$
$\implies\arccos\dfrac{1-x^2}{1+x^2}=2y=2\arctan x$ if $0\le2\arctan x\le\pi\iff0\le\arctan x\le\dfrac\pi2\implies0\le x\le\infty$
For $x<0\imp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding period of $f$ from the functional equation $f(x)+f(x+4)=f(x+2)+f(x+6)$ How can I find the period of real valued function satisfying $f(x)+f(x+4)=f(x+2)+f(x+6)$?
Note: Use of recurrence relations not allowed. Use of elementary algebraic manipulations is better!
| You are given that $f(x)+f(x+4) = f(x+2)+f(x+6)$ for all $x \in \mathbb{R}$.
Replace $x$ with $x+2$ to get $f(x+2)+f(x+6) = f(x+4)+f(x+8)$ for all $x \in \mathbb{R}$.
Thus, $f(x)+f(x+4) = f(x+2)+f(x+6) = f(x+4)+f(x+8)$ for all $x \in \mathbb{R}$.
Subtract $f(x+4)$ from the left and right side of the last equation to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1357922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the roots of the summed polynomial
Find the roots of: $$x^7 + x^5 + x^4 + x^3 + x^2 + 1 = 0$$
I got that:
$$\frac{1 - x^8}{1-x} - x^6 - x = 0$$
But that doesnt make it any easier.
| $\dfrac{x^7 + x^5 + x^4 + x^3 + x^2 + 1}{x+1}=x^6-x^5+2x^4-x^3+2x^2-x+1$
$=x^6-x^5+x^4+(x^4-x^3+x^2)+(x^2-x+1)$
$=x^4(x^2-x+1)+x^2(x^2-x+1)+(x^2-x+1)$
$=(x^2-x+1)[x^4+x^2+1]$
Now $x^4+x^2+1=(x^2+1)^2-(x)^2=\cdots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Let $Y=1/X$. Find the pdf $f_Y(y)$ for $Y$. The Statement of the Problem:
Let $X$ have pdf
$$f_X(x) =
\begin{cases}
\frac{1}{4} & 0<x<1 \\
\frac{3}{8} & 3<x<5 \\
0 & \text{otherwise}
\end{cases}$$
(a) Find the cumulative distribution function of $X.$
(b) Let $Y=1/X$. Find the pdf $f_Y(y)$ for $Y$. Hint: Co... | $$
f_X(x) =
\begin{cases}
\frac{1}{4} & 0<x<1 \\
\frac{3}{8} & 3<x<5 \\
0 & \text{otherwise}
\end{cases}
$$
\begin{align}
f_Y(y) & = \frac d {dy} F_Y(y) = \frac d {dy} \Pr(Y\le y) = \frac d {dy} \Pr\left( \frac 1 X \le y \right) \\[10pt]
& = \frac d {dy} \Pr\left( X\ge \frac 1 y \right) \text{ (if }y>0) \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sum_{i=1}^{i=n} \frac{1}{i(n+1-i)} \le1$ $$f(n)=\sum_{i=1}^{i=n} \dfrac{1}{i(n+1-i)} \le 1$$
For example, we have $f(3)=\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot2}+\dfrac{1}{3\cdot1}=\dfrac{11}{12}\lt 1$
If true, it can be used to prove:
Proving $x\ln^2x−(x−1)^2<0$ for all $x∈(0,1)$
Also, can you prove $f(n)\g... | $\begin{array}\\
i(n+1-i)
&=i(n+1)-i^2\\
&=(n+1)^2/4-(n+1)^2/4+i(n+1)-i^2\\
&=(n+1)^2/4-((n+1)/2-i)^2\\
\end{array}
$
Therefore
$$i(n+1-i)
=(n+1)^2/4-((n+1)/2-i)^2
\le (n+1)^2/4
$$
and,
since $1 \le i \le n$,
so that
$(n-1)/2 \ge (n+1)/2-i
\ge -(n-1)/2$,
$$i(n+1-i)
=(n+1)^2/4-((n+1)/2-i)^2
\ge (n+1)^2/4-((n-1)/2)^2
=n
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Summation stuck under radical sign I am trying to evaluate the following sum, but I'm unable to solve it in any general way.
$$S=\sum_{k=1}^n\sqrt{1+\frac{1}{(k)^2}+\frac{1}{(k+1)^2} }$$
How can I do it?
| A different method:
\begin{align}1 + \frac{1}{k^2} + \frac{1}{(k+1)^2} &= 1 + \frac{2}{k(k+1)} + \left[\frac{1}{k^2} - \frac{2}{k(k+1)} + \frac{1}{(k+1)^2}\right]\\
&= 1 + 2\left(\frac{1}{k} - \frac{1}{k+1}\right) + \left(\frac{1}{k} - \frac{1}{k+1}\right)^2\\
&= \left(1 + \frac{1}{k} - \frac{1}{k+1}\right)^2,
\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to find the integral $\int \frac{\sqrt{1+x^{2n}}\left(\log(1+x^{2n}) -2n \log x\right)}{x^{3n+1}}dx$? How to evaluate the integral :
$$\int \frac{\sqrt{1+x^{2n}} \, \left(\ln(1+x^{2n}) -2n \, \ln x \right) \, dx}{x^{3n+1}}$$
I have attempted an evaluation, but I am at a loss as to a useful result. Thanks for any a... | Let
\begin{align}
I = \int \frac{\sqrt{1+x^{2n}} \, \left(\ln(1+x^{2n}) -2n \, \ln x \right) \, dx}{x^{3n+1}}
\end{align}
and make the transformation $x = t^{1/2n}$ to obtain
\begin{align}
I = \frac{1}{2n} \, \int t^{-5/2} \, \sqrt{1+t} \, \left(\ln(1+t) - \ln(t)\right) \, dt.
\end{align}
Let $t = \sinh^{2}\theta$ to o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1361575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Are the real parts of the vectors constituting the discrete Fourier transform matrix linearly independent? Let W denote the n- dimensional symmetric discrete Fourier transform matrix and $W_{i}$ denote its column vectors. Then, is the set { Re($W_{i}$) | i= 1... n } linearly independent? Or similarly, find det( Re ( W ... | The short answer is: No.
For example, consider $n=3$. You have $\omega=e^{-2\pi i/ 3} = -\frac{1}{2} - i \frac{\sqrt{3} }{2}$.
Then; the corresponging DTF matrix, $W$, becomes
$$ W=
\begin{pmatrix}
1 & 1 & 1 \\
1 & \omega &\omega^2 \\
1 & \omega^2 & \omega^4\\
\end{pmatrix}
=
\begin{pmatrix}
1 & 1 & 1 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1361793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that if $a^2+b^2$ is a multiple of three, then a and b are multiples of three I have attempted to prove the above. I am uncertain about the correctness of my proof:
Both numbers have to be multiples of three, i.e. $3a+3b=3n$, $\ 3(a+b)=3n$
It is not possible to arrive at an integer that is a multiple of three wit... | For any integer $n$, we have $n^2 \equiv 0 \bmod{3}$ or $n^2 \equiv 1 \bmod{3}$.
Since $a^2+b^2 \equiv 0 \bmod{3}$, by the above fact we must have $a^2\equiv 0 \bmod{3}$ and $b^2 \equiv 0 \bmod{3}$.
Since $3$ is a prime dividing $a^2$, $3$ divides $a$. Similarly $3$ divides $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find lim$_{x \to 0}\left(\frac{1}{x} - \frac{\cos x}{\sin x}\right).$ $$\lim_{x \to 0} \left(\frac{1}{x} - \frac{\cos x}{\sin x}\right) = \frac{\sin x - x\cos x}{x\sin x}= \frac{0}{0}.$$
L'Hopital's: $$\lim_{x \to 0} \frac{f'(x)}{g'(x)} = -\frac{1}{x^2} + \frac{1}{\sin ^2x} = \frac{0}{0}.$$
Once again, using L'Hopital... | Since no one mentioned it, I will go for the overkill. $\cot z$ is a meromorphic function on the complex plane and $\frac{1}{z}$ is exactly its singular part in the origin, since:
$$ \sin z = z \prod_{n\geq 1}\left(1-\frac{z^2}{n^2\pi^2}\right)\tag{1} $$
implies:
$$ \log \sin z = \log z + \sum_{n\geq 1}\log\left(1-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
} |
Bounds for $\frac{x-y}{x+y}$ How can I find upper and lower bounds for $\displaystyle\frac{x-y}{x+y}$? So I do see that
$$\frac{x-y}{x+y} = \frac{1}{x+y}\cdot(x-y) = \frac{x}{x+y} - \frac{y}{x+y} > \frac{1}{x+y} - \frac{1}{x+y} = \frac{0}{x+y} = 0$$
(is it correct?) but I don't get how to find the upper bound.
| so we have $f(x,y)=\frac{x-y}{x+y}$, we want to show that $f(x,y)$ is unbounded, therefore we fix $y=y_0>0$ and then have
$$
f(x,y_0)=f(x)=\frac{x-y_0}{x+y_0}
$$
now choose $x=x^*-y_0$ which gives us
$$
f(x)=\frac{x^*-2y_0}{x^*}
$$
and now observe that for $\lim_{x^*\to0+}\frac{x^*-2y_0}{x^*}=-\infty$ and for $\lim_{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$A$ and $B$ similar if $A^2=B^2=0$ and dimension of range $A$ and $B$ are equal Suppose $A$ and $B$ are linear transformations on finite dimensional vector space $V$,s.t. $A,B\neq 0$ and $A^2=B^2=0$. Suppose the dimension of range $A$ and $B$ are equal, can $A$ and $B$ be similar?
| If $A^2 = B^2 = 0$, then all the eigenvalues of $A$ and $B$ are identically equal to $0$. Now, if we put $A$ into Jordan normal form, the blocks can only be $1 \times 1$ or $2 \times 2$, else $A^2 \neq 0$; thus, $A$ is similar to $$ Q \left(\begin{array}{ccccccc}
0 & 1 & 0 & 0 & \cdots & 0 & 0\\
0 & 0 & 0 & 0 & \cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
finding a conditional expectation It is an old exam problem about conditional expectation:
Let $(\xi_1,\xi_2)$ be a Gaussian vector with zero mean and covariance matrix A with $A_{11}=A_{22}=1, A_{12}=A_{21}=1/2.$ What is $E(\xi_1^2\xi_2|2\xi_1-\xi_2)$?
With the condition I know $E(\xi_1\xi_2)=1/2$ and $E(\xi_1^2)=E(\x... | Make the transformation $\eta_1 = \xi_1, \eta_2 = 2\xi_1 - \xi_2$, since
$$\begin{pmatrix}
\xi_1 \\
\xi_2 \end{pmatrix} \sim \mathcal{N}\left(\begin{pmatrix} 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & 1\end{pmatrix}\right),$$
it follows that
$$\begin{pmatrix}
\eta_1 \\
\eta_2 \end{pmatrix}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Generalizing the Fibonacci sum $\sum_{n=0}^{\infty}\frac{F_n}{10^n} = \frac{10}{89}$ Given the Fibonacci, tribonacci, and tetranacci numbers,
$$F_n = 0,1,1,2,3,5,8\dots$$
$$T_n = 0, 1, 1, 2, 4, 7, 13, 24,\dots$$
$$U_n = 0, 1, 1, 2, 4, 8, 15, 29, \dots$$
and so on, how do we show that,
$$\sum_{n=0}^{\infty}\frac{F_n}{10... | That's because
$\sum_{n=0}^{\infty} F_nx^n
=\frac1{1-x-x^2}
$.
Putting
$x = \frac1{10^k}$
gives
$\sum_{n=0}^{\infty} \frac{F_n}{10^{kn}}
=\frac1{1-10^{-k}-10^{-2k}}
=\frac{10^{2k}}{10^{2k}-10^{k}-1}
$.
For the others,
the generating function is
$\sum_{n=0}^{\infty} G_n x^n
=\frac1{1-x-x^2-...-x^m}
$
where
$m=2$ for Fib... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
The maximum and minimum values of the expression Here is the question:find the difference between maximum and minimum values of $u^2$ where $$u=\sqrt{a^2\cos^2x+b^2\sin^2x} + \sqrt{a^2\sin^2x+b^2\cos^2x}$$
My try:I have just normally squared the expression and got
$u^2=a^2\cos^2x+b^2\sin^2x + a^2\sin^2x+b^2\cos^2x +2... | Expanding $u^2$ more:$$u^2=a^2+b^2 +2\sqrt{\sin^2x\cos^2x(a^4+b^4)+a^2b^2(\sin^4x+\cos^4x)}$$
Using trigonometric identity $\sin^2x+\cos^2x=1$ we can derive that:$$\sin^4x+\cos^4x=1-2\sin^2x\cos^2x$$
Rewrite $u^2$ again:$$u^2=a^2+b^2 +2\sqrt{\sin^2x\cos^2x(a^2-b^2)^2+a^2b^2}$$
The minimum value of $\sin^2x\cos^2x$ is $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Trigonometric Integrals $\int \frac{1}{1+\sin^2(x)}\mathrm{d}x$ and $\int \frac{1-\tan(x)}{1+\tan(x)} \mathrm{d}x$ Any idea of calculating this two integrals $\int \frac{1}{1+\sin^2(x)}\,dx$ and $\int \frac{1-\tan(x)}{1+\tan(x)} \mathrm{d}x$?
I found a solution online for the first one but it requires complex numbers w... | Notice, $$I_1=\int \frac{dx}{1+\sin^2 x}$$
$$=\int \frac{\sec^2 xdx}{\sec^2 x+\sec^2 x\sin^2 x}$$
$$=\int \frac{\sec^2 xdx}{1+\tan^2 x+\tan^2 x}$$
$$=\int \frac{\sec^2 xdx}{1+2\tan^2 x}$$
$$=\frac{1}{2}\int \frac{\sec^2 x dx}{\left(\frac{1}{\sqrt{2}}\right)^2+\tan^2 x}$$
Now, let $\tan x=t\implies \sec^2x dx=dt$
$$=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Value of an expression with cube root radical What is the value of the following expression?
$$\sqrt[3]{\ 17\sqrt{5}+38} - \sqrt[3]{17\sqrt{5}-38}$$
| $$\sqrt[3]{\ 17\sqrt{5}+38} - \sqrt[3]{17\sqrt{5}-38}=$$
$$\sqrt[3]{8+12\sqrt{5}+30+5\sqrt{5}} - \sqrt[3]{-8+12\sqrt{5}-30+5\sqrt{5}}=$$
$$\sqrt[3]{8+12\sqrt{5}+6\left(\sqrt{5}\right)^2+\left(\sqrt{5}\right)^3} - \sqrt[3]{-8+12\sqrt{5}-6\left(\sqrt{5}\right)^2+\left(\sqrt{5}\right)^3}=$$
$$\sqrt[3]{\left(\sqrt{5}+2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
} |
Finding $F(x)$ from $F(kx),$ where $F(x)$ is the antiderivative of the function $f(x)$. I have that $F(e^{x}x) = e^{x}x^{2} - e^{x}x + e^{x} - 1$, and I would like to find $F(x)$.
Attempt
Since $F(e^{x}x) = e^{x}x^{2} - e^{x}x + e^{x} - 1,$ $F(t) = \alpha_{1}t^{\beta_{1}} + \alpha_{2}t^{\beta_{2}} + \alpha_{3}t^{\beta_... | Ok, let's summarize the comments in an answer. Differentiating expression for $F(xe^x)$ we get $$(e^x+xe^x)f(xe^x) = x(e^x + xe^x)$$ and thus, for $x\neq -1$ we have $$f(xe^x) = x \implies f(x) = W(x)$$ and consequently, $$F(x) = \displaystyle\int_a^x W(t)\ dt$$ Now, to determine $a$:
$$\begin{align*}
F(xe^x) &= \disp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find product limit of this recursively-defined sequence?
Problem： if $a_1=3$, $a_n=2a_{n-1}^2-1$, $n\ge2$, find the limit of this expression:
$$\lim\limits_{n \to ∞} \prod\limits_{k=1}^{n-1} (1+\frac {1}{a_k})$$
The original problem asks to find this:
$$\lim\limits_{n \to ∞} \frac {a_n}{2^na_1a_2.....a_{n-1}}$$
Th... | Nice question, this following is my answer,(if I can't some wrong)
Let $a_{1}=\dfrac{1}{2}(x+\dfrac{1}{x})(x>1)\Longrightarrow x=3+2\sqrt{2}$
since
$$a_{2}=2a^2_{1}-1=\dfrac{1}{2}\left(x^2+\dfrac{1}{x^2}\right)$$
$$a_{3}=2a^2_{2}-1=\dfrac{1}{2}\left(x^4+\dfrac{1}{x^4}\right)$$
by indution you have
$$a_{n}=\dfrac{1}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Nature of the roots of quadratic equation Here is the problem that I need to prove:
If $x$ is real and $\displaystyle{\ p = \frac{3(x^2+1)}{(2x-1)}}$, prove that $\ p^2-3(p+3) \geq 0$
Here is what I did:
\begin{align*}
p(2x-1)=3(x^2+1) \\
3x^2 - 2px + (p+3)=0 \\
b^2 - 4ac = 4(p^2-3(p+3))
\end{align*}
By inspection I c... | $$3x^2-2px+(p+3) = 0 \\
9x^2-6px+3(p+3) = 0 \\
(3x-p)^2+(3(p+3)-p^2)=0 \\
p^2-3(p+3) = (3x-p)^2 \ge 0$$
Furthermore you can see that $p^2-3(p+3) = 0$ when $p=3x$ and $x^2 = x+1$, i.e. $x = \frac{1 \pm \sqrt 5}2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How to solve $\sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ$ Question:
$ \sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6° $
I have partially solved this:-
$$ \sin78^\circ-\sin42^\circ +\sin6^\circ-\sin66^\circ $$
$$ 2\cos\left(\frac{78^\circ+42^\circ}{2}\right) \sin\left(\frac{78^\circ-42^\circ}{2}\right) + 2\cos\le... | As $\sin(-A)=-\sin A,\sin(180^\circ-B)=\sin B$
$S=\sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ $
$=\sin(-138^\circ)+\sin(-66^\circ)+\sin6°+\sin78^\circ $
Observe that the angles are in Arithmetic Progression with common difference $=72^\circ$
Using How can we sum up $\sin$ and $\cos$ series when the angles are i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Find the number of sets of $(a,b,c)$ for $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{29}{72}$
If $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{29}{72},\ \ c<b<a<60,\ \ \{a,b,c\}\in\mathbb{N} $.
How many sets of $(a,b,c)$ exists ?
Options
$a.)\ 3 \quad \quad \quad \quad \quad b.)\ 4 \\
c.)\ 5 \quad \quad \quad \quad \... | In this type of problem,
you have to go through cases,
usually on the extreme variables.
Initially,
$\dfrac{1}{c} < \dfrac{29}{72}$,
so
$c > \dfrac{72}{29}
=2+\dfrac{14}{29}
$,
so $c \ge 3$.
In the other direction,
since
$\dfrac{1}{c}+\dfrac{1}{b}+\dfrac{1}{a}
< \dfrac{3}{c}
$,
$\dfrac{3}{c} > \dfrac{29}{72}$
or
$c < \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Integrate area of the shadow? Today I found an interesting article here. It computes (approximately) area of the shadow.
I was wondering what is exact value of the area. My first thought was to use integrals but it doesn't seem to be easy. How to compute this using integrals?
Only thing I know is that $$S\approx 2.92$... | There is of course a much easier approach that does not require any integrals whatsoever. Let the points of intersection of the circles be $A$ and $B$, the center of the small circle be $C=(1,1)$ and the center of the large circle, i.e., the lower left vertex of the square be $O=(0,0)$. (We will scale the resulting a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Finding $\frac {a}{b} + \frac {b}{c} + \frac {c}{a}$ where $a, b, c$ are the roots of a cubic equation, without solving the cubic equation itself Suppose that we have a equation of third degree as follows:
$$
x^3-3x+1=0
$$
Let $a, b, c$ be the roots of the above equation, such that $a < b < c$ holds. How can we find th... | We know that every symmetric function of the roots $a,b,c$ can be evaluated in terms of the elementary symmetric functions:
$$ e_1=a+b+c=0,\quad e_2=ab+ac+bc=-3,\quad e_3=abc=-1$$
or the power sums:
$$ p_1=e_1=0,\quad p_2=a^2+b^2+c^2 = 6,\quad p_3=a^3+b^3+c^3=3e_1-3=-3.$$
Now:
$$ g(a,b,c)=\frac{a}{b}+\frac{b}{c}+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
The expression $(1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64})$ where $q\ne 1$, equals The expression
$(1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64})$
where $q\ne 1$, equals
(A) $\frac{1-q^{128}}{1-q}$
(B) $\frac{1-q^{64}}{1-q}$
(C) $\frac{1-q^{2^{1+2+\dots +6}}}{1-q}$
(D) none of the fore... | $$(1+q)(1+q^2)=1+q+q^2+q^3=\frac{1-q^4}{1-q}$$
So,
$$(1+q)(1+q^2)(1+q^4)=\frac{(1-q^4)(1+q^4)}{1-q}=\frac{1-q^8}{1-q}$$
So,
$$(1+q)(1+q^2)(1+q^4)(1+q^8)=\frac{(1-q^8)(1+q^8)}{1-q}=\frac{1-q^{16}}{1-q}$$
Can you see the pattern?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find the $\int \frac{(1-y^2)}{(1+y^2)}dy$ $\int \frac{(1-y^2)}{(1+y^2)}dy$ first I tried to divide then I got 1-$\frac{2y^2}{1+y^2}$ and i still can't integrate it.
| Let $y = \tan \theta \Rightarrow I = \displaystyle \int \dfrac{\cos^2\theta - \sin^2 \theta}{\cos^2 \theta}d\theta= \displaystyle \int (1-\tan^2 \theta)d\theta = \displaystyle \int (2-\sec^2\theta)d\theta= 2\theta - \tan \theta + C=2\tan^{-1}y-y+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Range of an inverse trigonometric function Find the range of $f(x)=\arccos\sqrt {x^2+3x+1}+\arccos\sqrt {x^2+3x}$
My attempt is:I first found domain,
$x^2+3x\geq0$
$x\leq-3$ or $x\geq0$...........(1)
$x^2+3x+1\geq0$
$x\leq\frac{-3-\sqrt5}{2}$ or $x\geq \frac{-3+\sqrt5}{2}$...........(2)
From (1) and (2),
domain is $x\l... | Notice,
for the defined function $\cos^{-1}\sqrt{x^2+3x+1}$$$\implies -1\leq \sqrt{x^2+3x+1}\leq 1$$ $$\implies x^2+3x+1\geq 1$$ $$\implies x^2+3x\geq 0$$ $$\implies x(x+3)\geq 0$$ The above inequality holds for all $x$ such that $$x\in (-\infty, -3]\cup [0, \infty)\tag 1$$
Again, for the defined function $\cos^{-1}\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Power serie of $f'/f$ It seems that I'm [censored] blind in searching the power series expansion of $$f(x):=\frac{2x-2}{x^2-2x+4}$$ in $x=0$.
I've tried a lot, e.g., partiell fraction decomposition, or regarding $f(x)=\left(\log((x+1)^2+3)\right)'$ -- without success.
I' sure that I'm overseeing a tiny little missing... | Since
$$
x^2-2x+4=(x-1)^2+3=(x-2u)(x-2\bar{u}),
$$
with
$$
u=e^{i\frac\pi3},
$$
we have
\begin{eqnarray}
f(x)&=&\frac{2x-2}{(x-2u)(x-2\bar{u})}=\frac{1}{x-2u}+\frac{1}{x-2\bar{u}}=-\frac{\bar{u}}{2}\cdot\frac{1}{1-\frac{\bar{u}}{2}x}-\frac{u}{2}\cdot\frac{1}{1-\frac{u}{2}x}\\
&=&-\frac{\bar{u}}{2}\sum_{n=0}^\infty\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Counting with potency and simplifing So I have the question: Simplify $(6^{n+4}) / 2^{n+5} \cdot 3^{n+2}$
I tried to write the expresion as $6^{n+4-(2n+7)}/6$, but that is wrong.
So I guess I should factor it out. Perhaps $2^{2} + 2^{n+4}$ / $2^{n+5} \cdot 3^{n+2}$
Can you show me how to expand this expresion?
| $$\frac { { 6 }^{ n+4 } }{ { 2 }^{ n+5 }{ 3 }^{ n+2 } } =\frac { \left( 2\cdot 3 \right) ^{ n+4 } }{ { 2 }^{ n+5 }{ 3 }^{ n+2 } } =\frac { { 2 }^{ n+4 }{ 3 }^{ n+4 } }{ { 2 }^{ n+5 }{ 3 }^{ n+2 } } ={ 2 }^{ n+4-n-5 }{ 3 }^{ n+4-n-2 }={ 2 }^{ -1 }{ 3 }^{ 2 }=\frac { 9 }{ 2 } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
Evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
How to evalute this equation without using calculator?
| Another way : $$\begin{align}\\&\frac{1}{1+\sqrt 2+\sqrt 3}+\frac{1}{1-\sqrt 2+\sqrt 3}+\frac{1}{1+\sqrt 2-\sqrt 3}+\frac{1}{1-\sqrt 2-\sqrt 3}\\&=\left(\frac{1}{1+\sqrt 2+\sqrt 3}+\frac{1}{1+\sqrt 2-\sqrt 3}\right)+\left(\frac{1}{1-\sqrt 2+\sqrt 3}+\frac{1}{1-\sqrt 2-\sqrt 3}\right)\\&=\frac{1+\sqrt 2-\sqrt 3+1+\sqrt ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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remainder of $a^2+3a+4$ divided by 7
If the remainder of $a$ is divided by $7$ is $6$, find the remainder when $a^2+3a+4$ is divided by 7
(A)$2$ (B)$3$ (C)$4$ (D)$5$ (E)$6$
if $a = 6$, then $6^2 + 3(6) + 4 = 58$, and $a^2+3a+4 \equiv 2 \pmod 7$
if $a = 13$, then $13^2 + 3(13) + 4 = 212$, and $a^2+3a+4 \equiv 2 \pmod ... | The remainder of $a^2+3a+4$ divided by $7$ is sum of the remainder of each terms, modulo $7$.
So $a^2\equiv 1 \pmod{7}$ since $a=7k+6$ then $a^2=7l+1$; $\quad$
$3a\equiv 4 \pmod{7}$ since $3a=21k+18=21k+14+4$ and clearly $4\equiv 4 \pmod{7}$.
Finally $1+4+4 \equiv 2 \pmod{7}$ then the remainder is $2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is the inequality $| \sqrt[3]{x^2} - \sqrt[3]{y^2} | \le \sqrt[3]{|x -y|^2}$ true? I'm having some trouble deciding whether this inequality is true or not...
$| \sqrt[3]{x^2} - \sqrt[3]{y^2} | \le \sqrt[3]{|x -y|^2}$ for $x, y \in \mathbb{R}.$
| It is shown in Is $x^t$ subadditive for $t \in [0,1]$? that we have the inequality
$$(a+b)^t \le a^t + b^t \tag{1}$$
for all nonnegative $a,b$ and all $t \in [0,1]$.
For your inequality, let us assume without loss of generality that $x \ge y$. Applying the above inequality (1) with $a=y$, $b=x-y$, and $t=2/3$ we get
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the sum of series $\sum_{n=2}^\infty\frac{1}{n(n+1)^2(n+2)}$ How to find the sum of series $\sum_{n=2}^\infty\frac{1}{n(n+1)^2(n+2)}$ in the formal way? Numerically its value is $\approx 0.0217326$ and the partial sum formula contains the first derivative of the gamma function (by WolframAlpha).
| You may first write a partial fraction decomposition, giving
$$
\frac{1}{n(n+1)^2(n+2)}=\frac12\left(\frac{1}n-\frac1{n+2}\right)-\frac{1}{(1+n)^2}
$$ then obtain your sum, by telescoping for the first two terms above, then by identifying a celebrated series for the third term. You get
$$
\sum_2^N\frac{1}{n(n+1)^2(n+2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A basic root numbers question If $\sqrt{x^2+5} - \sqrt{x^2-3} = 2$, then what is $\sqrt{x^2+5} + \sqrt{x^2-3}$?
| $$\sqrt{x^2+5} - \sqrt{x^2-3} = 2\\ \sqrt{x^2+5} = \sqrt{x^2-3} + 2\\ (\sqrt{x^2+5})^2 = (\sqrt{x^2-3})^2 + 2^2\\x^2+5=x^2-3+4-4\sqrt{x^2-3} \\5-1=-4\sqrt{x^2-3}\\\sqrt{x^2-3}=1 \\x^2-3=1\\ \rightarrow x=\pm 2\\\sqrt{x^2+5} + \sqrt{x^2-3}=\sqrt{(\pm2)^2+5} - \sqrt{(\pm2)^2-3}=3+1=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove an identity (Trigonometry Angles--Pi/13) In this page http://mathworld.wolfram.com/TrigonometryAnglesPi13.html
I found equation (11) and (12).
$$\cos^2\frac{\pi}{13}+\cos^2\frac{3\pi}{13}+\cos^2\frac{4\pi}{13}=\frac{11+\sqrt{13}}{8}$$
$$\sin\frac{\pi}{13}+\sin\frac{3\pi}{13}+\sin\frac{4\pi}{13}=\sqrt{\frac... | For the first one, let $\zeta=e^{2\pi i/13}$. This is a thirteenth root of $1$, so $1+\zeta+\zeta^2+...+\zeta^{12}=0$.
\begin{align}
A &=\cos^2(\pi/13)+\cos^2(3\pi/13)+\cos^2(4\pi/13)\\
B &=2A-3=\cos(2\pi/13)+\cos(6\pi/13)+\cos(8\pi/13)\\
2B &=\zeta+\zeta^{-1}+\zeta^3+\zeta^{-3}+\zeta^4+\zeta^{-4}\\
C &=-1-2B=\zeta^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381294",
"timestamp": "2023-03-29T00:00:00",
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Area of a triangle with sides $\sqrt{x^2+y^2}$,$\sqrt{y^2+z^2}$,$\sqrt{z^2+x^2}$ Sides of a triangle ABC are $\sqrt{x^2+y^2}$,$\sqrt{y^2+z^2}$ and $\sqrt{z^2+x^2}$ where x,y,z are non-zero real numbers,then area of triangle ABC is
(A)$\frac{1}{2}\sqrt{x^2y^2+y^2z^2+z^2x^2}$
(B)$\frac{1}{2}(x^2+y^2+z^2)$
(C)$\frac{1}{2}... | Use cosine rule to find say $\angle C$ then use formula of area as follows
Area of $\triangle ABC$ $$=\frac{1}{2}(a)(b)\sin C=\frac{1}{2}(a)(b)\sqrt{1-(\cos C)^2}$$ $$=\frac{1}{2}(\sqrt{x^2+y^2})(\sqrt{y^2+z^2})\sqrt{1-\left(\frac{(\sqrt{x^2+y^2})^2+\sqrt{y^2+z^2})^2-(\sqrt{x^2+z^2})^2}{2\sqrt{x^2+y^2}\sqrt{y^2+z^2}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Resources on Variants of the Clausen Functions I am interested in locating more information about the Clausen functions. Specifically I am looking for the closed forms of the Gl-type (or Sl-type as they are sometimes called) and the alternating analogues of the Cl and Gl-type Clausen functions.
In other words, I am loo... | Here's what I found:
let's pick the case where $m\quad =\quad 2k\quad /\quad k\in Z\\ $
As you know:
$\sum _{ n=1 }^{ \infty }{ \frac { { t }^{ n } }{ { n }^{ m } } } \quad ={ Li }_{ m }(t)$
Let's substitute $t={ e }^{ ix }$:
$\sum _{ n=1 }^{ \infty }{ \frac { { { e }^{ inx } } }{ { n }^{ m } } } \quad ={ Li }_{ m ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim\limits_{x\to 0}\frac{\sqrt{2(2-x)}(1-\sqrt{1-x^2})}{\sqrt{1-x}(2-\sqrt{4-x^2})}$ I use L'Hospitals rule, but can't get the correct limit.
Derivative of numerator in function is
$$\frac{-3x^2+4x-\sqrt{1-x^2}+1}{\sqrt{(4-2x)(1-x^2)}}$$
and derivative of denominator is
$$\frac{-3x^2+2x-2\sqrt{4-x^2}+4}{2\sqrt{(... | $\lim\limits_{x\to 0}\left(\frac{\sqrt{2(2-x)}(1-\sqrt{1-x^2})}{\sqrt{1-x}(2-\sqrt{4-x^2})}\right) = \lim\limits_{x\to 0}\left(\frac{\sqrt{2(2-x)}(2 + \sqrt{4-x^2})}{\sqrt{1-x}(1+\sqrt{1-x^2})}\right) = 4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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last two digits of $14^{5532}$? This is a exam question, something related to network security, I have no clue how to solve this!
Last two digits of $7^4$ and $3^{20}$ is $01$, what is the last two digits of $14^{5532}$?
| The OP quickly realizes that we can't write $14^k \equiv 1 \pmod{100}$ with $k \gt 0$, but there are still relations to be found in (multiplicative) semigroups.
If the last digit of integers $a$ and $b$ end in $6$, then the last digit of the product ends in $6$. This motivates us to write
$\quad 14^2 \equiv 96 \equiv -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Radical under Radical expression how to find the sum of $\sqrt{\frac54 + \sqrt{\frac32}} + \sqrt{\frac54 - \sqrt{\frac32}} $ ? Is there a method to solve these kind of equations ?
| The general idea is to move along:
$$\sqrt{a}+\sqrt{r^2\,a} = \sqrt{a}+r\,\sqrt{a} = (1+r)\,\sqrt{a} = \sqrt{(1+r)^2\,a}$$
I am using the following systematic method as here, to first get the quotient:
$$r^2 = \frac{5/4+\sqrt{3/2}}{5/4-\sqrt{3/2}} = 49+20\sqrt{6} = (5+2\sqrt{6})^2$$
Therefore we have:
$$\sqrt{5/4-\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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.