Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
If $a^2 + b^2 + c^2 = 2$, find the maximum of $\prod(a^5+b^5)$ Given that $a, b, c > 0$ and $a^2 + b^2 + c^2 = 2$, what is the maximum value of $(a^5 + b^5)(a^5 + c^5)(b^5 + c^5)$?
Normally when I encounter a problem like this, I seem to be able to push through with AM-GM. This one seems a little problematic since I c... | Let $c=\min\{a,b,c\}$, $a^2+\frac{c^2}{2}=x$ and $b^2+\frac{c^2}{2}=y$.
Hence, $x+y=2$, $xy\leq1$, $\sqrt{x}+\sqrt{y}\leq\sqrt{(1+1)(x+y)}=2$ and
$a^5+c^5\leq\left(a^2+\frac{c^2}{2}\right)^{\frac{5}{2}}=x^{\frac{5}{2}}$,$b^5+c^5\leq\left(b^2+\frac{c^2}{2}\right)^{\frac{5}{2}}=y^{\frac{5}{2}}$ and
$$a^5+b^5\leq\left(a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/51835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Find the area of the region inside: $r= 6\sin(\theta)$ but outside of $r = 1$ How do we find the area of the region inside $r = 6 \sin(\theta)$, but outside $r = 1$?
So, here's my work thus far:
First off, we know:
$r^2 = x^2 + y^2$
and $\mathrm{sin}(\theta) = y/r$
Therefore,
$r = \sqrt{x^2+y^2}$
and $6\mathrm{\sin}(\... | You want the region where $r>1$. That's the same as $\sin\theta > 1/6$. The boundary of that region is where $\sin\theta=1/6$. That happens when $\theta = \arcsin(1/6)$ and when $\theta = \frac\pi 2 - \arcsin(1/6)$. So you want
$$
\int_{\arcsin(1/6)}^{\pi/2-\arcsin(1/6)} \frac{r^2}{2} \, d\theta = \int_{\arcsin(1/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/54351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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What is the maximum value of this trigonometric expression What is the maximum value of the expression
$1/(\sin^2 \theta + 3\sin\theta \cos\theta+ 5\cos^2 \theta$).
I tried reducing the expression to $1/(1 + 3\sin\theta$ $\cos\theta + 4\cos^2 \theta)$.
How do I proceed from here?
| There are as usual many approaches. We describe one that uses pure trigonometry (and in particular no calculus), in line with the tag on the question.
Trigonometry can be bypassed in favour of algebra, as we show in the second part of this answer. But it would not be a good idea to omit the trigonometric approach alto... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/59049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Trigonometric system I would like to solve:
$ x +y+z=\frac{11\pi}{6} $
$ \sin(x)+\sin(y)+\sin(z)= \frac{\sqrt{3}}{2} $
$ \cos(x)+\cos(y)+\cos(z)=\frac{1}{2} $
After eliminating $ z $ I get:
$ 2\sin(x)+2\sin(y)-\cos(x+y)-\sqrt{3}\sin(x+y)=\sqrt{3}\tag{1}$
$ 2\cos(x)+2\cos(y)+\sqrt{3}\cos(x+y)-\sin(x+y)=1\tag{2}$
Also: ... | Multiply the 2nd equation by $i$ and add the 3rd equation to get $$e^{ix}+e^{iy}+e^{iz}=e^{i\pi/3}$$ Now you can argue geometrically that for three numbers of modulus 1 to sum to a number of modulus 1, one of the summands must be the negative of another summand (and so the third summand must equal the lone number on th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/60904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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Proving the identity $\sum_{k=1}^n {k^3} = \big(\sum_{k=1}^n k\big)^2$ without induction I recently proved that
$$\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2$$
using mathematical induction. I'm interested if there's an intuitive explanation, or even a combinatorial interpretation of this property. I would also l... | $f(n)=1^3+2^3+3^3+\cdots+n^3$
$f(n-1)=1^3+2^3+3^3+\cdots+(n-1)^3$
$f(n)-f(n-1)=n^3$
if $g(n)= (1+2+3+4+\cdots+n)^2$ then
$$g(n)-g(n-1)=(1+2+3+4+\cdots+n)^2-(1+2+3+4+\cdots+(n-1))^2$$
using $a^2-b^2=(a+b)(a-b)$
$$\begin{align}g(n)-g(n-1)&=\\
&=[(1+\dots+n)+(1+\dots +(n-1))][(1+\dots+ n)-(1+\dots+(n-1)]\\
&=[2(1+2+3+4+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/61482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "184",
"answer_count": 28,
"answer_id": 22
} |
Is this asymptotic equation correct? Is this equation correct?
$$ \frac {1 + \Theta(\frac 1 {2n})} {(1 + \Theta(1/n))^2} = 1 + O(1 / n) $$
I need this equation to prove that
$$ \binom {2n} n = \frac {2 ^ {2n}} {\sqrt {\pi n}} (1 + O(1 / n)) $$
which could be calculated from Stirling's Approximation:
$$ n! = \sqrt {2\p... | Yes. One way to see it is to do the long division; i.e., divide the denominator directly into the numerator. The numerator is $1 + \Theta(\frac{1}{2n}) = 1 + \Theta(\frac{1}{n})$, and the denominator is $\left(1 + \Theta(\frac{1}{n})\right)^2 = 1 + \Theta(\frac{1}{n}) + \Theta(\frac{1}{n^2}) = 1 + \Theta(\frac{1}{n})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/62996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Difficult Integral: $\int\frac{x^n}{\sqrt{1+x^2}}dx$ How to calculate this difficult integral: $\int\frac{x^2}{\sqrt{1+x^2}}dx$?
The answer is $\frac{x}{2}\sqrt{x^2\pm{a^2}}\mp\frac{a^2}{2}\log(x+\sqrt{x^2\pm{a^2}})$.
And how about $\int\frac{x^3}{\sqrt{1+x^2}}dx$?
| Recall the hyperbolic functions
$$\cosh t= \frac{e^t + e^{-t}}{2} = \cos(it)$$
and $$\sinh t=\frac{e^t - e^{-t}}{2} = i\sin(-it).$$
Note that $\frac{d}{dt}\sinh t = \cosh t$, $\frac{d}{dt}\cosh t = \sinh t$ and also $\cosh^2 t -\sinh^2 t = 1$.
Making the substitution $\sinh t=x $ we see that
$$\frac{x^n\, dx}{\sqrt{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/64450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solving Radical Equations $x-7= \sqrt{x-5}$ This the Pre-Calculus Problem:
$x-7= \sqrt{x-5}$
So far I did it like this and I'm not understanding If I did it wrong.
$(x-7)^2=\sqrt{x-5}^2$ - The Square root would cancel, leaving:
$(x-7)^2=x-5$ Then I F.O.I.L'ed the problem.
$(x-7)(x-7)=x-5$
$x^2-7x-7x+14=x-5$
$x^2-14x+1... | $x-7= \sqrt{x-5}$
$(x-7)^2=\sqrt{x-5}^2$
$(x-7)^2=x-5$
$(x-7)(x-7)=x-5$
$x^2-7x-7x+49=x-5$
$x^2 - 15x + 54 = 0$
$(x - 9)(x - 6) = 0$
$x - 9 = 0 $ or $x - 6 = 0$
$x = 9$ or $x = 6$
Now check for extraneous solutions...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/66255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to show a scalar inequality? Let $a_1,\cdots, a_n\ge0$ and $S=a_1+\cdots+a_n$ with $k\ge 2, k\in\mathbb{Z}$. How to show
$$\sum_{i=1}^n(S-a_i)^k\le 2^{k-2}\left(\sum\limits_{i=1}^n a_i ^k+(n-2)S^k\right)?$$
| There must be a slick way to do this, but here's one way. We will split the problem into different cases; the inequality is quite obvious in each of the cases. (Convince yourself that the cases exhaust all the possibilities. :))
Case 1: $n = 2$. For $n=2$, we have $S = a_1 + a_2$. In this case the inequality is obvious... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a^3 =a$ for all $a$ in a ring $R$, then $R$ is commutative. Let $R$ be a ring, where $a^{3} = a$ for all $a\in R$. Prove that $R$ must be a commutative ring.
| I happen to have come across a recent set of exercises on many of the small-$n$ cases of Jacobson's Theorem. It also happens that my solution is different than those contained in the link of @lhf above.
So we have that $a^3 = a \quad\forall a \in R$, and so $2a = (a+a)^3 =8a$, thus $6a = 0$.
Now consider the ideals $2R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/67148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "57",
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Convergence of Series, problems with intermediate steps I have problems with two exercises. I know the answer of those two but both of them have a step which I don't understand.
1) I have to prove that $\prod_{k=2}^n \frac{k^3-1}{k^3+1}$ is convergent.
My first steps were:
$\prod_{k=2}^n \frac{k^3-1}{k^3+1}= \prod_{k=2... | 1) Factoring further
$$
\begin{align}
k^3-1&=(k-1)(k^2+k+1)=(k-1)(k+\alpha)(k+\bar{\alpha})\\
k^3+1&=(k+1)(k^2-k+1)=(k+1)(k-\alpha)(k-\bar{\alpha})
\end{align}
$$
where $\alpha+\bar{\alpha}=1$. It is easy to see that $\prod\frac{k-1}{k+1}$ is telescoping. However, when we write $\prod\frac{(k+\alpha)(k+\bar{\alpha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/67399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $a$ and $r$ such that $\lim\limits_{n\to \infty} n^r \cdot \frac12 \cdot \frac34 \cdots \frac{2n-1}{2n}=a$ Find $a,r>0$ such that
$$\lim_{n\to \infty} n^r \cdot \frac12 \cdot \frac34 \cdots \frac{2n-1}{2n}=a$$
I don't have any idea to solve it. How can I solve it?
| I'll start out from a celebre limit, namely Wallis product that states that:
$$ \frac{\pi}{2} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdot \cdot \cdot $$
Without loss of generality, we consider an even factors number of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/67592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Integrate $\int \sin (2x) \cos (2x)\,dx$ I have
$$\int \sin(2x) \cos (2x)\,dx = \frac12 \int \sin(4x)\,dx = -\frac18 \cos(4x),$$
but I also have
$$\int \sin(2x) \cos (2x)\,dx = \frac12 \int \sin 2x \cdot 2 \cos 2x \, dx = \frac14 \sin^2(2x).$$
Which one is correct, and why is the other method wrong?
| To illustrate the phenomena in your calculation, I've chosen the number $\pi/8$ as the lower boundary.
\begin{align*}
\int_{\frac{\pi}{8}}^{x} \sin(4t) \,\mathrm{d}t &= \int_{\frac{\pi}{8}}^{x} 2 \sin(2t) \cos(2t) \,\mathrm{d}t \\
&= \int_{\frac{\pi}{8}}^{x} \sin(2t) \,\mathrm{d}(\sin(2t)) \\
\left . -\frac{\cos(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/67793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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The $3 = 2$ trick on Google+ I found out this on Google+ yesterday and I was thinking about what's the trick. Can you tell?
How can you prove $3=2$?
This seems to be an anomaly or whatever you call in mathematics. Or maybe I'm just plain dense.
See this illustration:
$$ -6 = -6 $$
$$ 9-15 = 4-10 $$
Adding $\frac{25}{4... | HINT $\ $ You erroneously inferred $\rm\ x^2 =\: (-x)^2\ \Rightarrow\ x\: =\: -x\:,\ $ for $\rm\ x\:=\:1/2\:.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/68913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Getting the root out of the denominator Alright quick question,
If I have
$$\frac{1}{2^{1/3} - 2},$$
how do i change it to not have the root in the denominator? I know the trick with the root of 2, but this seems a bit more complicated. Thanks in advance :)
| If you mean
$$\frac{1}{2^{1/3}} - 2,$$
then just multiply the numerator and denominator by $2^{2/3}$; we get
$$\frac{2^{2/3}}{2} - 2.$$
If you mean
$$\frac{1}{2^{1/3} - 2}$$
then you can use the formula
$$(a-b)(a^2+ab+b^2) = a^3-b^3$$
so
$$\frac{1}{2^{1/3}-2} = \frac{2^{2/3} + 2(2^{1/3}) + 4}{(2^{1/3}-2)(2^{2/3} + 2(2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/69335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why doesn't this method of calculating Gaussian curvature of a sphere work? Suppose we have a sphere defined as $x^2 + y^2 + z^2 = 9$. With radius of 3, we know the Gaussian Curvature should be $\frac{1}{9}$ at every point on the surface.
Since this defines a level set, the gradient gives the normals. So, the normals s... | First observe that your normalization simplifies greatly since $x^2 + y^2 + z^2 = 9$. So really the unit normal vector is just $\frac{1}{3} (x,y,z)$. Now the Gauss map takes the sphere to the unit sphere and the differential of this in coordinates should be a two by two matrix since this is a map between surfaces. L... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/72161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Maximize the distance between a line normal to an ellipse and its center My friend sent me this problem, which (upon Googling) seems to be from a Cornell class (1220?). Anywho.
My advice to him was to parametrize the ellipse (say, in the first quadrant) with
$x = a \cos(t); y = b \sin(t)$, find normal lines, then use... | The same maximal distance occurs in each quadrant, so we can restrict attention to $t\in[0,\pi/2]$. The tangent vector at $t$ is $(-a\sin t,b\cos t)$. This vector is normal to the line, so we just have to take the scalar product of a unit vector in this direction with the position vector in order to find the distance o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/75600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Problem with generating functions and binary recurrences I am considering the following recurrence:
$a_0 = 1$;
$a_1 = 2$
$a_{n} = 2 (a_{n - 1} + a_{n - 2})$
Then I proceeded with the generating function:
$F(x) = \displaystyle\sum_{n = 0}^\infty a_n x^n = 1 + 2x + \displaystyle\sum_{n = 2}^\infty a_{n} x^{n} = 1 + 2x +... | A simpler way to handle such is to write the recurrence as:
$$
a_{n + 2} = 2 (a_{n + 1} + a_n)
$$
multiply by $x^n$, sum over $n \ge 0$. Recognize the sums that result:
$$
\frac{A(x) - a_0 -a_1 z}{z^2}
= 2 \left( \frac{A(x) - a_0}{x} + A(x) \right)
$$
Plugging in $a_0 = 1$ and $a_1 = 2$, solving for $A(z)$:
$$
A(z) =... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find all integers $m$ such that $\frac{1}{m}=\frac{1}{\lfloor 2x \rfloor}+\frac{1}{\lfloor 5x \rfloor} $ How would you determine all integers $m$ such that the following is true?
$$\frac{1}{m}=\frac{1}{\lfloor 2x \rfloor}+\frac{1}{\lfloor 5x \rfloor} .$$
Note that $\lfloor \cdot \rfloor$ means the greatest integer fun... | You can solve it by cases. Let $x=n+r$, where $n$ is an integer and $r=\lfloor x\rfloor$. Then $\lfloor 2x\rfloor = \lfloor 2n+2r\rfloor = 2n+\lfloor 2r\rfloor$ and $\lfloor 5x\rfloor = \lfloor 5n+5r\rfloor = 5n+\lfloor 5r\rfloor$, and the original equation can be written as $$\frac1m = \frac1{2n+\lfloor 2r\rfloor}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/77290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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How to prove a fact about the sum of three squares? How would I go about proving the following?
If $a$, $b$, $c$, $n$ are positive integers, then
$a^2+b^2+c^2 \neq 2^nabc$
I tried doing something similar to the proof for Adrien-Marie Legendre's Three Square theorem: $a^2+b^2+c^2=n$ iff there are not integers $k$, and ... | To go from $a^2+b^2+c^2$ even to $a,b,c$ even in this case, you need an argument, although it is true for $n$ strictly positive.
You should try to express your equation for only $a_k$, $b_k$ and $c_k$, then you will see that your argument does not end after $n$ steps. Either, you argue with infinite descent or equivale... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/78122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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How to solve $z^4 +6z^2 +25 = 0$ into complex conjugate? How to solve $z^4 +6z^2 +25 = 0$ into complex conjugate?
I started with
$$(z^2 + 3)^2 + 16 = 0$$
$$(z^2 + 3)^2 = - 16$$
$$z^2 + 3 = \pm 4i$$
Is this the way to start solving the equation or am I completely of?
| You have $z^2=-3\pm4i$. You need the two square roots of $-3+4i$ and the two square roots of $-3-4i$. I can think of two ways to do this. One is polar coordinates:
$$
-3+4i = \sqrt{3^2+4^2}\left(\cos\theta+i\sin\theta\right) = 5\left(\cos\theta+i\sin\theta\right)
$$
where $\theta$ is an angle between $\pi/2$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/78931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Finding the slope of the tangent line to $\frac{8}{\sqrt{4+3x}}$ at $(4,2)$ In order to find the slope of the tangent line at the point $(4,2)$ belong to the function $\frac{8}{\sqrt{4+3x}}$, I choose the derivative at a given point formula.
$\begin{align*}
\lim_{x \to 4} \frac{f(x)-f(4)}{x-4} &=
\lim_{x \mapsto 4}... | Just note that
$$\begin{align}
\frac{8}{\sqrt{4 + 3x}} - 2 &= \frac{8 - 2 \sqrt{4 + 3x}}{\sqrt{4 + 3x}}
\\
&= \frac{8 - 2 \sqrt{4 + 3x}}{\sqrt{4 + 3x}} \cdot \frac{8 + 2\sqrt{4 + 3x}}{8 + 2 \sqrt{4 + 3x}}
\\
&= \frac{64 - 4(4 + 3x)}{\sqrt{4 + 3x}(8 + \sqrt{4 +3x})}.
\end{align}$$
I'll leave the rest up to you.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\sum\limits_{n=0}^{20} \frac{(-1)^{n}2^{n+1}}{3^{n}},$ I know the sum of the series $$2 - \frac{4}{3} + \frac{8}{9} - \cdots + \frac{(-1)^{20}2^{21}}{3^{20}}$$ is equal to $$\sum\limits_{n=0}^{20} \frac{(-1)^{n}2^{n+1}}{3^{n}},$$ but I don't know how to calculate the sum without manually entering it into th... | $$\sum_{n=1}^{20}\frac{(-1)^n2^{n+1}}{3^n}=2\sum_{n=1}^{20}\left(-\frac{2}{3}\right)^n=2\left(-\frac{2}{3}\right)\frac{\left(-\frac{2}{3}\right)^{20}-1}{\left(-\frac{2}{3}-1\right)}=$$
$$=\left(-\frac{4}{3}\right)\left(-\frac{3}{5}\right)\frac{2^{20}-3^{20}}{3^{20}}=\frac{4}{5}\frac{2^{20}-3^{20}}{3^{20}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/82200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Expected value and Variance of $Y=\frac{1}{a} X-b$ where $X \sim N(\mu, \sigma^2)$ I absolutely know I am not doing this right. :[
Could I get some input or point back in the right direction?
My work done so far is shown below.
Let $X$ be a normal random variable with parameters $N(\mu, \sigma^2)$. Please find the E... | Since expectation is linear, you have that $\mathsf{E}(Y)=\mathsf{E}(1/a\:X-b)=1/a\:\mathsf{E}(X)-b$. Furthermore,
$$
\begin{align}
\mathsf{Var}(Y)
&=\mathsf{E}(Y^{\;2})-\mathsf{E}(Y)^2\\
&=\mathsf{E}(1/a^2X^2-2b/aX+b^2)-(1/a\:\mathsf{E}(X)-b)^2\\
&=1/a^2\mathsf{E}(X^2)-2b/a\mathsf{E}(X)+b^2-1/a^2\mathsf{E}(X)^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/85658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
What is the least positive integer $n$ for which $(-\sqrt{2}+i\sqrt{6})^n$ is an integer?
Compute the least positive integer $n$ for which $(-\sqrt{2}+i\sqrt{6})^n$ will be an integer, where $i$ is the imaginary unit.
I did the binomial expansion and just plugged in numbers for $n$ starting from $1$ to see any patte... | $$
\begin{align}
(-\sqrt{2} + i \sqrt{6})^n
& = \left(\sqrt{2} \left(-1 + i \sqrt{3} \right)\right)^n
& = 2^{n/2} \left(-1 + i \sqrt{3} \right)^n\\
& = 2^{n/2} \left( 2 \left( - \frac12 + i \frac{\sqrt{3}}{2} \right) \right)^n
& = 2^{n/2} 2^n \left( - \frac12 + i \frac{\sqrt{3}}{2} \right)^n \\
& = 2^{3n/2} \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/87292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
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Four kissing circles How can one go about solving the following problem?
Inscribe a circle in an arbitrary triangle. Call it's radius $r_1$.
Inscribe three more circles so that each one is tangent to two sides of
the triangle and the first circle (i.e., each at a different corner). Call the
radii $r_2, r_3, r_4$... | I'm using the same notation as in the link mentioned by David Mitra. By similarity we have
$$
\frac{r}{OA}=\frac{r_a}{OA-r-r_a}=\sin\left(\frac{\alpha}{2}\right)
$$
where $\alpha=\angle A$. Then
$$
\frac{r_a}{r} = \frac{1-\sin(\alpha/2)}{1+\sin(\alpha/2)}
= \frac{1-\cos((\beta+\gamma)/2)}{1+\cos((\beta+\gamma)/2)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/87458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Help with a geometry problem The problem says:
A triangle has its lengths in an arithmetic progression,
with difference d. The area of the triangle is t. Find the dimensions.
the solution says:
the notation can be even better if we make it more symmetrical, by making the
side lengths $b − d, b,$ and $b + d$ .
by Heron... | Yes, you know that $b^2$ and $b$ are both positive real numbers, so you select the positive sign twice while taking square root.
If you want to know why the inequality is true, note that both sides of it are positive, so you can just square both sides to get an equivalent inequality which is easily seen to be correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/88274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Proof of an inequality: $\sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$
Possible Duplicate:
Proving $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$ with induction
How do I prove the following?
$$\sqrt{n} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \cdots + \dfrac{1}{\sqrt{... | We can even do better:
$$
\sqrt{n}-\sqrt{n-1}=\frac{1}{\sqrt{n}+\sqrt{n-1}}<\frac{1}{2\sqrt{n-1}}\tag{1}
$$
Summing, we get
$$
\sqrt{n}-1<\frac{1}{2}\left(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{n-1}}\right)\tag{2}
$$
So, for $n\ge2$,
$$
\sqrt{n}<1+\frac{1}{2}\left(\frac{1}{\sqrt{1}}+\frac{1}{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/88695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
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How can we prove the locus is a circle? Given two fixed points A and B, find the locus of the point P, satisfying PA=2PB. Of course we can use Cartesian geometry to find the equation of the curve.
Let the midpoint of A and B be the origin, and the line AB be the x-axis, then the coordinates of A and B is (a,0), (-a,0... | Without loss of generality, let the distance between $A$ and $B$ equal 3 (this is for convenience in calculations) and suppose that $\overline{AB}$ is horizontal (this just makes it easier to talk about where things are).
Now, the point $P_1$ that is on $\overline{AB}$ a distance of 2 from $A$ and 1 from $B$ is in the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/89473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Rectifiability of a curve Let $f$ be a function defined on $[0,1]$ by
$$f(x) = { 0, \text{ if } x = 0} $$
$$f(x) = { x \sin \frac 1 x , \text{ if } 0 < x \leq 1} $$
Prove that the curve $\{(x, f(x)) : x \in [0,1]\}$ is not rectifiable.
I'm not sure how to approach this. The general idea seems logical, we're proving t... | The arclength $L$ of a curve $(x,f(x))$ from $x=a$ to $x=b$ is defined as
$$L=\int_a^b\sqrt{1+(f'(x))^2}dx.$$
Therefore, in this case, $f(x)=\displaystyle x\sin(\frac{1}{x})$, which implies that
$$(f'(x))^2=\Big[\sin(\frac{1}{x})-\frac{1}{x}\cos(\frac{1}{x})\Big]^2\geq-\frac{2}{x}\sin(\frac{1}{x})\cos(\frac{1}{x})+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/90699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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A quartic diophantine equation Here is the statement: Let $a,b \in \mathbb{Z}$ positive integers such that $a^2=b^4+b^3+b^2+b+1.$ Prove $b=3.$
I've tried is the following: Let $\Sigma=b^4+b^3+b^2+b+1$. If $a\equiv 0\mod 3$, then $a^2\equiv 0 \mod 3$. If $a\equiv 1\mod 3$ or $a\equiv 2\mod 3$, then $a^2\equiv 1\mod 3$.... | If $b$ is even then $b^4+b^3+b^2+b+1$ is caught between $$(b^2+(b/2))^2=b^4+b^3+(1/4)b^2$$ and $$b^4+b^3+(9/4)b^2+b+1=(b^2+(b/2)+1)^2$$
If $b$ is odd, have a look at $(b^2+(b-1)/2)^2$ and $(b^2+(b+1)/2)^2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How to find $\sqrt[3]{8i}$ How do I find the following cube root?
$$\sqrt[3]{8i} = ?$$
I tried by adding $\sqrt[3]{i^3 + 8i + i}$ but that is where my imagination quits.
| Here is another way that avoids exponential functions
for this problem but is not a method that I would
recommend in general, e.g. to find $\sqrt[10]{8i}$,
for which the method described Andre Nicolas's and
Agusti Roig's answers work much better.
Suppose $a$ and $b$ are real numbers such that $(a+bi)^3 = 8i$. We hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/93130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Dice Question Probability A fair die is rolled until a $6$ appears. What is the probability that is must be cast more than 5 times?
So this is $1- P(\text{dice has to be cast less than or equal to}\ 5 \ \text{times})$. So this probability is equal to $$ P =\frac{1}{6}+ \frac{5}{6} \frac{1}{6}+\left(\frac{5}{6} \right)... | The easiest way to do this is simply to raise the probability that a six is not rolled to the fifth power, as these probabilities are independent, which gives $$P(\text{cast more than five times}) = P(\text{individual roll is not 6})^5 = \left(\frac{5}{6}\right)^5$$
This is equivalent to $1 - P$ where $P$ is what you w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/94079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Number of possible triangle for the given triangle. An equilateral triangle has n equally spaced dots on each side. How many triangles can be formed (of any size)?
Analysis
If there are n dots on each side then total no of dots =n(n+1)/2
Number of combination using 3 at a time =${n(n+1)/2 \choose 3}$
but all set of ... | This question has previously been considered for triangles created by points in a square lattice; see here.
As with the square case, the challenge on the triangular lattice is determining the number of collinear triples. A000938 in OEIS gives the answer for the square lattice as
$$\left(2 \;\sum _{m=2}^n\; \sum _{k=2}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/94684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How can I solve a complicated induction exercise? (formula for sum of fourth powers) I have an induction exercise:
It is for $n \in \mathbb{N_0}$. Show:
$$ \sum\limits_{k=1}^{n} k^4 = \frac 1 {30} n(n+1)(2n+1)(3n^2+3n-1) $$
As far as I understand it, you have to put in $(n+1)^4$ at the end and you have to resolve it to... | There is one way to more or less predict the coefficients in one of these sums. It is not a use of induction as such, but I think it is worth knowing when you are beginning the study of induction. It is just Pascal's triangle.
$$ \sum_{k=0}^n \left( \begin{array}{c} k \\ 0 \end{array} \right) = \left( \begin{array}{c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/98321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculating this residue (residue theorem) Given $$\int_{\gamma}\frac{1}{(z-a)(z-\frac{1}{a})}dz,$$ and $0<a<1$, where $\gamma(t)=e^{it}$ and $0\le t \le 2\pi$
I am trying to find the residue of$$f(z)=\frac{1}{(z-a)(z-\frac{1}{a})}$$
The answer says $\operatorname{res}(f,\mathbb C)=\frac{1}{(a-\frac{1}{a})}$?
Any help ... | $$ f(z)=\frac{1}{(z-a)(z-\frac{1}{a})}=\frac{\frac{1}{a-\frac{1}{a}}}{z-a}+\frac{\frac{-1}{a-\frac{1}{a}}}{z-\frac{1}{a}} $$
$$ \gamma \text{ is unit circle} \hspace{20mm} \text{Poles of Function} \begin{cases} z_1=a &\text{placed inside of unit circle} \\
z_2=\frac{1}{a} &\text{placed outside of unit circle} \e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/100316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Showing $24|(n+1)\implies24|\sigma(n)$ Question:
Show that if $n$ is a positive integer such that $24$ divides into $n + 1$, then $24$ divides the sum of all divisors of $n$ (denoted in number theory by $\sigma(n)$ or $\sigma_1(n)$).
For example if $n = 95$, then $n + 1 = 96 = 4 \times 24$ and the sum of the divisor... | We're looking for divisors $a, b$ such that $a \times b = n \equiv - 1 \pmod{24}$. In order for this to be possible, $a$ and $b$ must both be coprime to 24. Now note the following:
$a \times b \equiv -1 \pmod{24}$
$a^2 + a \times b \equiv a^2 -1 \pmod{24}$
$a(a + b) \equiv (a-1)(a+1) \pmod{24}$
Since $a$ is coprime to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/103379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving the inequality: $[x-(x^2)/2 < \ln(1+x) < x]$ , $x>0$ Prove the inequality:
$[x-(x^2)/2 < \ln(1+x) < x]$ , $x>0$
The right side is easy, I used taylor expansion to show that $e^x > 1+x$ since
$e^x = 1 + x + x^2/2 + x^3/3! +\cdots $
The left side I expanded $\ln(1+x)$ to $x-x^2/2 + x^3/3 - x^4/4 +\cdots$
m... | For the left inequality, since you know that $x-\frac{x^2}{2}=\ln(1+x)$ when $x=0$ you need only check the derivatives: $\frac{d}{dx}(x-\frac{x^2}{2})=1-x$ while $\frac{d}{dx}(\ln(1+x))=\frac{1}{1+x}$, and so
$$\frac{d}{dx}(\ln(1+x))-\frac{d}{dx}(x-\frac{x^2}{2})=\frac{1}{1+x}-(1-x)=\frac{1-(1+x)(1-x)}{1+x}=\frac{x^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/105281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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differential equation nondevelopable I try to solve this differential equation whose solution seems not to be constructable in power series
$y''+(x+a/x^2+b)y=0$, where $a$ and $b$ are some positive real numbers.
If one can help me please?
| Another way of solving this equation will be a series expansion in the parameter $b$. We know that for $b=0$ the two independent solutions can be easily found by means of the series expansion method.Those solutions are related to Bessel functions. Let us therefore assume that the whole solution reads:
\begin{equation}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/105422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Lim $x\to 0$ of $\frac{\sin(\pi x)}{\tan(\sqrt{3}x)}$ $$
\lim_{x \to 0} \frac{\sin(\pi x)}{\tan(\sqrt{3} x)}
$$
I need a step by step explanation. Thank you.
| Hint:
First, let's expand what we are looking at:
$$\begin{align}
\frac{\sin(\pi x)}{\tan(\sqrt{3}x)} = \frac{\sin(\pi x)}{\frac{\sin(\sqrt{3} x)}{\cos(\sqrt{3} x)}} = \sin(\pi x) \cdot \frac{\cos(\sqrt{3}x)}{\sin(\sqrt{3}x)} = \sin(\pi x)\cdot\frac{1}{\sin(\sqrt{3}x)}\cdot \cos(\sqrt{3}x).
\end{align}$$
Next, we wan... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find$\int x^2 \arctan(x) \;dx $ I was solving $\int x^2 \arctan(x) \;dx $
I set $u=x^2$, $dv= \arctan(x)$, so I could get $du=2x$, $v=x\arctan(x)-\frac12\ln(1+x^{2})$.
From $\int x^2 \arctan(x)\;dx = uv - \int v \; du$
I got
$$\begin{align*}&\int x^2 \arctan(x) \; dx =\\
&x^3\arctan(x) -\frac12\ln(1+x^{2})-\in... | We have
$$
\int x^2 \arctan x\;dx.
$$
Let
$$
\begin{align}
u & = \arctan x, \\ \\
du & = \frac{dx}{x^2+1}, \\ \\
dv & = x^2 \;dx, \\ \\
v & = \frac{x^3}{3}.
\end{align}
$$
Then
$$
\begin{align}
\int u\;dv & = uv - \int v\;du = \frac{x^3}{3}\arctan x - \int \frac{x^3\;dx}{3(x^2+1)} = \frac{x^3}{3}\arctan x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/106827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Steps in evaluating $\int \frac{1}{x^4-c^4} dx$? $$ \int\ \frac {dx} {x^4 - c^4} \\ $$
is equal to (from integral tables)
$$ \frac {1} {4 c^3} \ln \frac {x-c} {x+c} - \frac {1} {2 c^3} \tan^{-1}\frac {x} {c}$$
If I let $\frac{x}{c}=\tan u$, then $dx/c= \sec^2 u du$, and the integral becomes
$$ \frac {1} {c^3} \int\ \fr... | You’d be better off writing factoring $x^4-c^4$ as $(x^2-c^2)(x^2+c^2)$ and then as $(x-c)(x+c)(x^2+c^2)$ and decomposing $\frac1{x^4-c^4}$ into partial fractions:
$$\frac1{x^4-c^4}=\frac{A}{x-c}+\frac{B}{x+c}+\frac{Cx+D}{x^2+c^2}\;.$$
Once you’ve done the algebra to find $A,B$, and $C$, the integrations should be pret... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/107015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $\lim\limits_{(x,y) \to (0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4} $ I'm trying to find $$\lim\limits_{(x,y) \to (0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4} .$$
After I tried couple of algebraic manipulation, I decided to use the polaric method.
I choose $x=r\cos \theta $ , $y=r\sin \theta$, and $r= \sqrt{x^... | We have for $t>0$ that $e^t\geq\frac{t^3}3$ so
$$\frac{\exp\left(-\frac 1{x^2+y^2}\right)}{x^4+y^4}\leq \frac 3{\frac 1{(x^2+y^2)^3}}\frac 1{x^4+y^4}=\frac{3(x^2+y^2)^3}{x^4+y^4}=3(x^2+3y^2+3x^2+y^2)=12(x^2+y^2),$$
and you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/107171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Computing $ \sum_{n=0}^{\infty} (-1)^{n+1}\int_{0}^{1}\frac{t^{3n+3}}{1+t^3}\mathrm{d} t$ I would like to find the exact value of the following series:
$$ \sum_{n=0}^{\infty} (-1)^{n+1}\int_{0}^{1}\frac{t^{3n+3}}{1+t^3}\mathrm{d} t$$
We can easily show that the series converges using the alternating series test:
$$ 0 \... | We have
$$ \begin{align*}
\sum_{n=0}^{\infty} (-1)^{n+1} \int_{0}^{1} \frac{t^{3(n+1)}}{t^3 + 1} \; dt
& = - \int_{0}^{1} \frac{t^3}{(t^3 + 1)^2} \; dt \\
&= \left[ \frac{t}{3} \frac{1}{t^3 + 1}\right]_{0}^{1} - \frac{1}{3} \int_{0}^{1} \frac{dt}{t^3 + 1} \\
&= \frac{1}{6} - \frac{1}{3} \int_{0}^{1} \left( \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/112400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Try to prove $\lim_{n \to \infty}n(\ln 2-A_n) = \frac{1}{4}$ $$A_n=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$$
Try to prove $$\lim_{n \to \infty}n(\ln 2-A_n) = \frac{1}{4}$$
I try to decompose $\ln 2$ as $$\ln(2n)-\ln(n)=\ln\left(1+\frac{1}{2n-1}\right)+\dots+\ln\left(1+\frac{1}{n}\right)\;,$$ but I can't continu... | Trying to follow the idea of the OP, write $$ \log 2 = \log (2n+2) - \log (n+1) = \log \left( 1+ \frac{1}{2n+1} \right) + \log \left( 1+ \frac{1}{2n} \right) + \cdots + \log \left( 1+ \frac{1}{n+1} \right) $$
so then $$ n( \log 2 - A_n) = n\log \left(1+ \frac{1}{2n+1} \right) + n\sum_{k=1}^n \left( \log \left( 1+ \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/114831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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What is the least value of the function $y= (x-2) (x-4)^2 (x-6) + 6$? What is the least value of function:
$$y= (x-2) (x-4)^2 (x-6) + 6$$
For real values of $x$ ?
Does $\frac{dy}{dx} = 0$, give the value of $x$ which will give least value of $y$?
Thanks in advance.
| $$\frac{dy}{dx} = (x-4)^2 (x-6)+(x-4)^2 (x-2)+2(x-2)(x-4)(x-6) = 0 $$
$$ \Rightarrow (x-4) \left[ (x-4)(2x-8)+2(x-2)(x-6) \right] = 0$$
$$ \Rightarrow (x-4) \left[ 2x^2-16x+32+2x^2-16x+24 \right] = 0$$
$$ \Rightarrow (x-4) \left[ 4x^2-32x+56 \right] = 0$$
$$ \Rightarrow 4(x-4) \left[ (x-4+\sqrt{2})(x-4-\sqrt{2}) \right... | {
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"timestamp": "2023-03-29T00:00:00",
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If $f(x)f(y)=f(\sqrt{x^2+y^2})$ how to find $f(x)$ As we know, for the $$f(x)f(y)=f(x+y)$$ $f(x)=\mathrm e^{\alpha x}$ is a solution.
What about
$f(x)f(y)=f(\sqrt{x^2+y^2})$?
Does anybody know about the solution of the function equation?
I tried to find $f(x)$.
See my attempts below to find $f(x)$.
$$f(x)=a_0+a_1x+\... | Let' assume you want solution to $f:\mathbb{R}\mapsto\mathbb{R}$ that satisfies this equation $$\forall x,y\in\mathbb{R},f(x)f(y)=f\left(\sqrt{x^2+y^2}\right)\tag{1}$$
Let's plug in $(x,y)=(0,0)$ in the equation $(1)$,
$$f(0)^2=f(0)\Rightarrow f(0)\in\{0,1\}$$
Case-1 : $f(0)=0$
For $x>0, f(x)=f\left(\sqrt{x^2+0^2}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/115784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Help to evaluate determinant I want to evaluate the determinant of the $n \times n$ matrix
$\left|\begin{array}{ccccc} 1 & 0 & \ldots & 0& 0 \\
0 & 0 & \ldots & 0 & -a\\
0 & 0 & \ldots & -a & 0\\
&&&\vdots \\
0 & -a & 0 &\dots & 0
\end{array}\right|.$
So I try to say that it is $(-1)^{ n + (n-1) + \ldots n-(n-2)}(... | Using cofactor expansion, we have:
$$
\left|\begin{array}{ccccc} 1 & 0 & \ldots & 0& 0 \\
0 & 0 & \ldots & 0 & -a\\
0 & 0 & \ldots & -a & 0\\
&&&\vdots \\
0 & -a & 0 &\dots & 0
\end{array}\right|
=
\left|\begin{array}{ccccc}
0 & \ldots & 0 & -a\\
0 & \ldots & -a & 0\\
&&&\vdots \\
-a & 0 &\dots & 0
\end{array}\right|
.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/116240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int \cos(\cos x)~dx$
Evaluate $\int \cos(\cos x)~dx$
I tried to use chain rule but failed. Can anyone help me please?
| $\int\cos\cos x~dx=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\cos^{2n}x}{(2n)!}~dx=\int\left(1+\sum\limits_{n=1}^\infty\dfrac{(-1)^n\cos^{2n}x}{(2n)!}\right)~dx$
For $n$ is any natural number,
$\int\cos^{2n}x~dx=\dfrac{(2n)!x}{4^n(n!)^2}+\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin x\cos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/117536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 3
} |
Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$ In Spivak's Calculus 3rd Edition, there is an exercise to prove the following:
$$x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$$
I can't seem to get the answer. Either I've gone wrong somewhere, I'm overlooking something, ... | Here is the inductive step, presented more conceptually
$$\rm\frac{x^{n+1}-y^{n+1}}{x-y}\: =\ x^n\: +\ y\ \frac{x^n-y^n}{x-y}$$
So, intuitively, proceeding inductively yields
$$\rm\:x^n + y\: (x^{n-1} + y\: (x^{n-2} +\:\cdots\:))\ =\ x^n + y\: x^{n-1} + y^2\: x^{n-2} + \:\cdots $$
Use this intuition to compose a formal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/117660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 9,
"answer_id": 6
} |
Fitting a sine wave of known frequency through three points We have a computationally expensive function of a large set of data and an angle, that is known to result in a sine wave:
$$ f(\text{data}, \theta) \approx a \sin (\theta + b) + c$$
We want to find the constants $a$, $b$, and $c$, executing the function as few... | Yes, you can do it in 3 points, considering $f(0)$, $f(\pi)$, and $f(\frac{\pi}{2})$.
$$
\begin{aligned}
f(0) &= a\sin b + c & \\
f(\tfrac{\pi}{2}) &= a\sin(b + \tfrac{\pi}{2}) + c & &= a\cos b + c \\
f(\pi) &= a\sin(b + \pi) + c & &= - a\sin b + c
\end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/118526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Sums of Double series where $a_{m,n}=\frac {m-n} {2^{m+n}}\frac {\left( m+n-1\right) !} {m!n!}$ I am trying to show if $$a_{m,n}=\dfrac {m-n} {2^{m+n}}\dfrac {\left( m+n-1\right) !} {m!n!}$$ such that $$\left( m,n>0\right) $$ that $\sum _{m=0}^{\infty }\left( \sum _{n=0}^{\infty }a_{m,n}\right) =-1$, $\sum _{n=0}^{\inf... | Here is a possible approach:
$$\frac{1}{(1-x)^{m+1}} = \sum_{n=0}^{\infty} \binom{m+n}{n} x^n$$
Write $\displaystyle a_{m,n}$ as $\displaystyle \frac{m}{(m+n)2^{m+n}} \binom{m+n}{n}$ - $\displaystyle \frac{n}{(m+n)2^{m+n}} \binom{m+n}{n}$
For the first term, multiply by $x^m$ and integrate between $0$ and $1/2$.
For th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/119063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\displaystyle\int {1\over (1+kx^2)^{3/2}}dx$ What change of variable should I use to integrate $$\displaystyle\int {1\over (1+kx^2)^{3/2}}dx$$
I know the answer is $$\displaystyle x\over \sqrt{kx^2+1}.$$ Maybe a trig or hyperbolic function?
| $\int \dfrac{1}{(1+kx^2)^{3/2}}dx$
Put $1+kx^2=t$, Then,
$$2kx\cdot dx = dt \text{. Also, } x = \sqrt{t-1\over k}$$
$$2kx\cdot dx = dt$$
$$dx = \frac{dt}{2kx} = \frac{\sqrt{k}dt}{2k\sqrt{t-1}}= \frac{dt}{2\sqrt{k}\sqrt{t-1}}$$
$\int \dfrac{1}{(1+kx^2)^{3/2}}dx=$ $\int \dfrac{1}{2t^{1.5}\sqrt{k}\sqrt{t-1}}dt$
$$\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/119184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find a three digit number ($\overline{xyz}$)? (Excuse me for my english: I'm spanish speaker)
"Find a three digit number $\overline{xyz}$ such that $x^2 +y^2 + z^2$ is equal to the number (xyz). "
I have this equation:
$$x^2 -100x +(y^2-10y+z^2-z)=0$$
And the discriminant (in $x$):
$$\Delta_x = -4y^2 +40y-4z^2+4z+100... | It seems that solving it for $y$ is a better idea, as we are going to have $y=5\pm\sqrt{something}$.
Let us solve it for a number with any number of digits: $\overline{...vwxyz}$.
$y^2-10y+(z^2-z)+(x^2-100x)+(w^2-1000w)+...=0$
$y=5\pm\sqrt{25-(z^2-z)-(x^2-100x)-(w^2-1000w)-...}$.
Note, that $0\le y\le9$, therefore, $0\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/119710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Show that $2^{3^k}+1$ is divisible by $3^k$ for all positive integers $k$ I attempted this by induction: Here is what I did
For $k=1, 2^{3^{1}}+1 = 2^{3}+1 = 9 = 3^{2}$ is divisible by $3^{1}$ , so the result is true for $k=1$
Now I assume the result to be true for $k=m$,
$2^{3^{m}}+1$ is divisible by $3^m$. To show t... | The last step you did can be corrected as
$$2^{3^{m+1}}+1 = (2^{3^m})^3+1$$ because if you look at it this way
$$(2^{3^m})^3 = (2^{3^m}) \times (2^{3^m}) \times (2^{3^m}) = 2^{3 \times 3^m} = 2^{3^{m+1}}$$
Using the fact that the result is true for $k=m$
$$2^{3^{m}}+1 = x \times 3^{m} \implies 2^{3^{m}} = x3^{m}-1 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/119890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Calculate the integral of $1/(2x^2+2x+1)$ I need to calculate the integral of $1 / (2x^2+2x+1)$.
I used WolframAlpha and get this answer: $$\tan^{-1}(2x+1)$$ but I don't understand how to get there.
Can you help?
| First note that $\displaystyle \int \frac{dy}{a^2+y^2} = \frac1a \tan^{-1}\left(\frac{y}{a} \right)$. This is obtained by substituting $y = a \tan( \theta)$ in the integrand and integrating.
For your problem, $I = \displaystyle \int \frac{dx}{2x^2 + 2x+1}$. First complete the square in the denominator i.e. rewrite $2x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/120329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Please help me prove by induction that $n^n>1\cdot3\cdot5\cdot\ldots\cdot(2n-1)$ Please help me prove by induction that
$$
n^n>1\cdot3\cdot5\cdot\ldots\cdot(2n-1)
$$
| We need to prove $1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)< n^n$
For $n=1$ we have $1 < 1$, what is wrong.
For $n=2$ we have $3 < 4$ what is true.
Note that $1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1) (2\cdot (n+1)-1)=n^n(2(n+1)-1)$
then prove $(n+1)^{n+1} - n^n\cdot (2\cdot (n+1)-1) > 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/120442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 3
} |
Limit of sequence $x_n^n$ Let $x_1=2$, $x_{n+1}=\sqrt{x_n+\frac{1}{n}}$ for all $n\geq 1$. Prove that $\lim\limits_{n\to\infty}x_n=1$ and evaluate $\lim\limits_{n\to\infty}x_n^n$.
| For $1$, first observe that if $x_n>1$ then $x_{n+1}=\sqrt{x_n+\frac{1}{n}}>\sqrt{1+\frac{1}{n}}>1$, so the sequence $(x_n)$ is bounded below by $1$. Note that
$$\begin{eqnarray}x_n>x_{n+1}&\iff&x_n>\sqrt{x_n+\frac{1}{n}}\\
&\iff&x_n^2>x_n+\frac{1}{n}\\
\end{eqnarray}$$
and that this is clearly true when $n=1$. If it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/120867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Polynomial equation Is it possible to find polynomials with rational coefficients $P(x),Q(x)$ such that $Px^3+Px^2+Qx+2Q=1$? I have trying in vain to find one by inspection, but that might just be me.
| Just in case the OP does not know the Euclidean algorithm, I'll compute it.
Divide $(x^3 + x^2)$ by $x+2$. This is just ordinary high school long division. So we find that
$$\begin{equation}(x^3 + x^2) = (x+2)q(x) + r(x)\end{equation}$$
where $q(x) = x^2 - x$ and $r(x) = 2x$. Now by the Euclidean algorithm we then divi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/121653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Computation of a residue Consider the function $\cos\left(\frac{z}{z+1}\right)$, which has an essential singularity at the point $z=-1$.
How does one compute its residue at $z=-1$.
| $$ A=\cos\left(\frac{z}{z+1}\right) = \cos\left(\frac{z\color{red}{+1-1}}{z+1}\right)=\cos \left( 1- \frac{1}{z+1} \right) $$
$$ \cos(a\pm b)=\cos(a)\cos(b) \mp \sin(a)\sin(b)$$
$$A=\cos(1)\cos\left(\frac{1}{z+1}\right)+\sin(1)\sin\left(\frac{1}{z+1}\right) $$
$$ \cos(z)=1-\frac{z^2}{2!}+\frac{z^4}{4!}-... \hspace{10mm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/121903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Sum of the reciprocal of sine squared I encountered an interesting identity when doing physics homework, that is,
$$ \sum_{n=1}^{N-1} \frac{1}{\sin^2 \dfrac{\pi n}{N} } = \frac{N^2-1}{3}. $$
How is this identity derived? Are there any more related identities?
| Let's denote $ S_N=\sum_{n=1}^{N-1} \frac{1}{\sin^2 \dfrac{\pi n}{N} } $
Consider an equality:
$$\frac{\pi^2}{\sin^2(\pi s)}=\int_0^{\infty}\frac{x^{s-1}}{1-x}\ln\frac{1}{x}dx;0<s<1$$
Because of $0<\frac{n}{N}<1$, the integral form applies. Thus:
$$S_N=\frac{1}{{\pi}^2} \int_0^{\infty}\frac{\ln\frac{1}{x}}{1-x}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/122836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 1
} |
How can I determinate the bases for the most simple representation of a linear transformation? Imagine a linear transformation $\Phi : \mathbb{R}^4 \rightarrow \mathbb{R}^3$ with the ordered standard basis:
$B = (\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmat... | Try the singular value decomposition, which gives orthonormal bases in the domain and in the codomain with respect to which the linear transformation is given by a diagonal matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/123495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Let $a,b$ be positive real numbers. Prove $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}} \geq \frac{2}{\sqrt{1+ab}}$ Let $a,b$ be positive real numbers. Prove $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}} \geq \frac{2}{\sqrt{1+ab}}$
if either
$(1) 0 \leq a,b \leq 1$
OR
$(2) ab \geq 3$
Since this question was under ... | Let $f(x)=\frac{1}{\sqrt{1+e^{2x}}}$.
Hence, $f''(x)=\frac{e^{2x}(e^{2x}-2)}{\sqrt{(e^{2x}+1)^5}}\geq0$ for all $x\geq\frac{1}{2}\ln2$.
Id est, for all $\{a,b\}\subset[\sqrt2,+\infty)$ by Jensen we obtain:
$$\sum_{cyc}\frac{1}{\sqrt{1+a^2}}=\sum_{cyc}f(\ln{a})\geq2f\left(\frac{\ln{a}+\ln{b}}{2}\right)=\frac{2}{\sqrt{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/124926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Inter-causal reasoning: How to solve probability with two conditions? Below is the scheme of conditional dependence and the probabilities of events:
P(A=1) = 0.01
P(A=0) = 0.99
P(B=1) = 0.1
P(B=0) = 0.9
P(C=1|A=0,B=0) = 0.1
P(C=1|A=0,B=1) = 0.5
P(C=1|A=1,B=0) = 0.6
P(C=1|A=1,B=1) = 0.9
Given the probabilities above I... | The typical way I do inter-causal reasoning is to flip the conditional probabilities around --
$$
\begin{align}
P(B = 1 \vert C = 1) & = \frac{P(B = 1, C = 1)}{P(C = 1)} \\
& = \frac{P(C = 1 \vert B = 1)P(B = 1)}{P(C = 1)} \\ \\
P(B = 1 \vert C = 1, A = 1) & = \frac{P(B = 1, C = 1, A = 1)}{P(C = 1, A = 1)} \\
& = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/126216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Plot $|z - i| + |z + i| = 16$ on the complex plane
Plot $|z - i| + |z + i| = 16$ on the complex plane
Conceptually I can see what is going on. I am going to be drawing the set of points who's combine distance between $i$ and $-i = 16$, which will form an ellipse. I was having trouble getting the equation of the ellip... | Maybe it's quicker way.
Equation $|z-i| + |z+i| = 16$ is equivalent to
$$
\sqrt{x^2 + (y-1)^2} = 16 - \sqrt{x^2 + (y+1)^2}.
$$
Squaring both sides you obtain
$$
x^2 + (y-1)^2 = 256 - 32\sqrt{x^2+(y+1)^2} + x^2 + (y+1)^2.
$$
Some terms cancel out hence you get
$$
8\sqrt{x^2 + (y+1)^2} = 64 + y,
$$
and
$$
64(x^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/126518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
Proof- central term of recursive pyramid
Moderator Note: At the time that this question was posted, it was from an ongoing contest. The relevant deadline has now passed.
How can we prove $4^{-n}(n+1)\left(2^{2n+1}-{2n+1\choose n+1}\right)$ produces the central term in the $2n+1$th row of the following pyramid?
The ... | It appears that joriki has answered the original question. I got bogged down on that when I got sick, but I did manage to establish that $4^{-n}(n+1)\left(2^{2n+1}-{2n+1\choose n+1}\right)$ gives the expected number of tosses of a fair coin required to get either $n+1$ heads or $n+1$ tails, as I mentioned in the commen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/127039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Is this limit correct: $\lim_{x \to+\infty} \frac{\log_{2}(x-1)}{x} = 0$?
Find $\space\ \begin{align*} \lim_ {x \to+\infty} \left [ \frac{\log_{2}(x-1)}{x}\right]
\end{align*}$.
After some minutes around this limit I did it this way:
$\log_{2}(x-1)=y \Leftrightarrow 2^y=x-1$
So,$\space x=2^y+1$.
When $x \to +\inft... | Your way is fine, as an alternative we have that
$$ \frac{\log_{2}(x-1)}{x}= \frac{\log_{2}(x-1)}{x-1}\frac{x-1}{x}\to 0\cdot 1 =0$$
indeed eventually $\log_{2}(x-1) \le \sqrt{x-1}$ and then
$$\frac{\log_{2}(x-1)}{x-1} \le \frac{\sqrt{x-1}}{x-1} \to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/127750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Linear Algebra Working with Linear Transformations Let $v_1=[-3;-1]$ and $v_2= [-2;-1]$
Let $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ be the linear transformation satisfying:
$T(v_1)=[15;-6]$ and $T(v_2)=[11;-3]$
Find the image of an arbitrary vector $[x;y]$
| Note sure if (homework) yet. So hint:
Let
$$
T =
\begin{pmatrix}
a & b \\ c & d
\end{pmatrix}
$$
We can re-interpret the given $T(v_1)$ and $T(v_2)$ as:
$$
\begin{pmatrix}
a & b \\ c & d
\end{pmatrix}
\begin{pmatrix}
-3 \\ -1
\end{pmatrix}
=
\begin{pmatrix}
15 \\ -6
\end{pmatrix} ,
\\
\begin{pmatrix}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/128221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Differentiate respect to $x$ $(x^2+2x+1)^3$
let u=$x^2+2x+1$ $\frac{du}{dx} = 2x$ or $2x+2$ $\frac{dy}{dx}=3u^2 $
if $\frac{du}{dx} = 2x$ then
$3(x^2+2x+1)^2 (2x)$
answer is $6x(x^2+2x+1) $
Or if $\frac{du}{dx} = 2x+2$ then
$3(x^2+2x+1)^2 (2x+2)$
$6x(x^2+2x+1)+2$
however the right answer is $6(x+1)^5$
can ple... | by Chain rule :
$f'(x)=3(x^2+2x+1)^2 \cdot(x^2+2x+1)'=3\cdot(x+1)^4\cdot(2x+2)=$
$=6\cdot(x+1)^4\cdot(x+1)=6(x+1)^5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/128624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
First Four Nonzero Terms of Taylor Series I'm not acing Calculus II and while this problem might be easy to some, it's not to me. Any help will do.
Problem: Find the first nonzero terms of the Taylor Series for $f(x) = \cos x$ where
$$a = {\pi\over 6}$$
My work:
\begin{array}{rlrll}
f(x) =& \cos x & f(a) =& \cos {... | Recall that $f(h+a)=\cos(h+(\pi/6))=\cos(h)\cos(\pi/6)-\sin(h)\sin(\pi/6)$. Hence, $$2f(h+a)=\sqrt3\cos(h)-\sin(h).
$$
Now, $\cos(h)=1-\frac12h^2+O(h^4)$ and $\sin(h)=h-\frac16h^3+O(h^4)$, hence
$$
2f(h+a)=\sqrt3(1-\tfrac12h^2)-(h-\tfrac16h^3)+O(h^4).
$$
The first four terms of the expansion of $f(h+(\pi/6))$ are the $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/130205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding $\int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\,dx$ Finding $$\int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\,dx$$
I suppose I need integration by parts and trigo substitution
Let $u=\frac{1}{x} \Rightarrow du = -\frac{1}{x^2} dx$
Let $dv = \sqrt{1+(\frac{1}{x^2})^2}$, $\frac{1}{x^2} = \tan{\theta}$. Is my substitution OK?
S... | Usually we want to get rid of the square root. One way to do so is to bring in into the form $\sqrt{1+\cosh^2(u)}$ so lets try this in two steps
$$\frac{1}{x^2}=v \Rightarrow dv = \frac{-1}{2x^3}dx \Leftrightarrow dx = -2x^3 \, dv$$
$$ \int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\,dx = \int \frac{-2x^3}{x}\sqrt{1+v^2}\,dv = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/130394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove via mathematical induction that $4n < 2^n$ for all $ n≥5$. I did the following base case $n = 5$
$$\begin{align*}
4(5) &\lt 2 ^5\\
20 &\lt 32
\end{align*}$$
So true.
$$\begin{align*}
4n &\lt 2^n\\
n &\lt 2^{n-2}\\
\log_2(n)+2 &\lt n
\end{align*}$$
But I don't think this is right. Where do I add in the... | Given $5\leq{n}$
Assume $4n<2^n$
Prove $4(n+1)<2^{n+1}$:
*
*$4(n+1)=4n+4$
*$4n+4<2^n+4$ assumption used here
*$2^n+4<2^n+32$
*$2^n+32=2^n+2^5$
*$2^n+2^5\leq2^n+2^n$ given fact used here
*$2^n+2^n=2^{n+1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/131036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Fallacy in Fermat's Little Theorem? Fermat's Little Theorem: If $p$ is prime, then for every $1 ≤ a < p$,
$a^{p-1} ≡ 1$ $(mod$ $p)$
Let $p$ be 9 (a composite number), and let $a$ be 2.
Let $S$ be the nonzero integers modulo $9$
$S = (1, 2, 3, 4, 5, 6, 7, 8)(mod$ $9)$
$2^8S$
$=2^8(1, 2, 3, 4, 5, 6, 7, 8)$$=
(1*2,2*2,... | i) No see Fermat pseudoprime.
ii) $2^8=256$ and $2+5+6=13$ so that $2^8= 1+3 \pmod{9}$ (the other one is right)
$2^{340} = 1\pmod{341}$ and $341=11\cdot 31$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/132695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
why does $\frac{1}{z\cdot \sin{z}} $ only have pole when clearly its undefined at $n\pi$ I am having trouble with a specific problem actually. I have a function $$f(z) = \frac{1}{z\cdot \sin{z}}$$
Now I want to find the residues of this. The Laurent series expanded about $0$ shows that $0$ is a pole of order $2$. The ... | Rather than expanding the function $f(z)$ around the point $z=2\pi$, let's rearrange $f(z)$ instead by looking at the Taylor series expansion of $\sin(z)$
$$ f(z) = \frac{1}{(z)(z-\frac{z^3}{3!}+\frac{z^5}{5}+\cdots)}$$
$$= \frac{1}{(z^2)(1-\frac{z^2}{3!}+\frac{z^4}{5}+\cdots)}$$
$$ f(z) = \frac{1}{(z^2)}\left(1+\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/133218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
How do you integrate $\int \frac{1}{a + \cos x} dx$? How do you integrate $\int \frac{1}{a + \cos x} dx$? Is it solvable by elementary methods? I was trying to do it while incorrectly solving a homework problem but I couldn't find the answer.
Thanks!
| This doesn't help you to evaluate the indefinite integral, but I though I would add that the definite integral $\int_0^{2 \pi
} \frac{1}{a + \cos x} \ dx$ can also be evaluated using methods from complex analysis. Let us make the simplifying assumption that $a > 1$ to avoid a blow up in the integral. Other values of $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/134577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 2
} |
Linear algebra - Dual, functionals Given three functionals
*
*$f_1(p) = \int_0^1 p(t)\,dt$
*$f_2(p) = \int_0^2 p(t)\, dt$
*$f_3(p) = \int_0^{-1} p(t)\, dt$
defined on $V = P_2$, the space of all polynomials over $\mathbb R$ of degree not greater than 2. Asking to prove that
$\{f_1, f_2, f_3\}$ is a basis for $V^... | You want to find polynomials $p_1,p_2,p_3$ of degree at most $2$ with the property that $f_i(p_j) = 1$ if $i=j$, and $f_i(p_j) = 0$ if $i\neq j$.
For instance, you want to find $p_1(x) = a+bx+cx^2$ such that
$$\begin{align*}
1=f_1(p_1) &=\int_0^1(a+bx+cx^2)\,dx \\
&= \left.\left( ax + \frac{b}{2}x^2 + \frac{c}{3}x^3\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/135372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solving the recursion $3a_{n+1}=2(n+1)a_n+5(n+1)!$ via generating functions I have been trying to solve the recurrence:
\begin{align*}
a_{n+1}=\frac{2(n+1)a_n+5((n+1)!)}{3},
\end{align*}
where $a_0=5$, via generating functions with little success. My progress until now is this:
Let $A(x)=\sum_{n=0} ^{\infty} a_nx^n$. ... | Exponential generating function of sequence $\{a_n\}$ is $f(x) = \sum_{n=0}^\infty a_n \frac{x^n}{n!}$. Rewriting the recurrence equation as
$$
\frac{a_{n+1}}{n+1} = \frac{2}{3} a_n + \frac{5}{3} n!
$$
Now multiplying both sides by $\frac{x^n}{n!}$ and using recurrence relation for factorial:
$$
a_{n+1} \frac{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/135803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 1
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A 'should be simple' derivative Help! I can't seem to prove that $$\Bigl[\log(x+\sqrt{x^2 + 1})\Bigr]' = \frac{1}{\sqrt{x^2 + 1}}$$ I keep getting some horrible answer namely $$\frac{x + \sqrt{x^2 + 1}}{x\sqrt{x^2 + 1} + 1 + x^2}$$ does this cancel down at all?
| The other answers are quite what's necessary, but note that working tidely you get
$$\frac{d}{{dx}}\log \left( {x + \sqrt {1 + {x^2}} } \right) = \frac{{\frac{d}{{dx}}\left( {x + \sqrt {1 + {x^2}} } \right)}}{{x + \sqrt {1 + {x^2}} }}=$$
$$\frac{{1 + \frac{x}{{\sqrt {1 + {x^2}} }}}}{{x + \sqrt {1 + {x^2}} }}$$
Now, bef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/135920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Summation by parts of $\sum_{k=1} ^n \frac{2k+1}{k(k+1)}$ Let $\{f_k\}$ and $\{g_k\}$ be sequences of real numbers. The formula for summation by parts is given by:
$\sum_{k=m} ^n f_k \Delta g_k=(f_{n+1}g_{n+1}-f_mg_m)-\sum_{k=m} ^n g_{k+1}\Delta f_k$,
where $\Delta f_k=f_{k+1}-f_k$.
Letting $f_k=2k+1$ and $g_k=-\frac{... | It does look to be your final conversion to harmonic numbers is at fault. In particular,
$$\begin{align*}
\sum_{k=1}^n \frac1{k+1}&=\sum_{k-1=1}^n \frac1{k-1+1}\\
&=\sum_{k=2}^{n+1} \frac1{k}\\
&=H_{n+1}-1\\
&=H_n+\frac1{n+1}-1\\
&=H_n-\frac{n}{n+1}
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/138454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Absolute max for $f(x,y,z)=x^ay^bz^c$, with constraint $g(x,y,z)=x+y+z-1$ I need to show absolute max for $f(x,y,z)=x^ay^bz^c$, with constraint $g(x,y,z)=x+y+z-1$ is $$\frac{a^ab^bc^c}{(a+b+c)^{a+b+c}}$$
So I I do have $$ax^{a-1}y^bz^c = bx^ay^{b-1}z^c = cx^ay^bz^{c-1} =\lambda$$
then I went on to equating each of the... | When you have
$$ax^{a-1}y^bz^c = bx^ay^{b-1}z^c = cx^ay^bz^{c-1} =\lambda$$
$ax^{a-1}y^bz^c =\lambda \hspace{4pt}$ will give you $\hspace{4pt} \lambda x = a x^a y^b z^c$
Similarly
$bx^ay^{b-1}z^c =\lambda \hspace{4pt} \Rightarrow \hspace{4pt} \lambda y = b x^a y^b z^c$
$cx^ay^bz^{c-1} =\lambda \hspace{4pt} \Rightarrow ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/139251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int \frac{x+8}{\sqrt {x+12}}dx$ How should I go about integrating the function $$\frac{x+8}{\sqrt {x+12}}$$
I have tried substituting $u = \sqrt{ x + 12 }$, but that leads me nowhere...
Could somebody possibly just tell me which steps have to be followed in order to evaluate this integral?
| Let $u = \sqrt{x+12}$ and $du = \frac{1}{2\sqrt{x+12}} dx$
$$
\begin{align*}
\int \frac{x+8}{\sqrt{x+12}} \hspace{3pt}dx &= \int {\hspace{3pt}\frac{u^2-12+8}{u}}\cdot2u\hspace{3pt} du\\
&= \int {\hspace{3pt}2u^2-8}\hspace{3pt} du\\
&= 2\int {\hspace{3pt}u^2-4}\hspace{3pt} du\\
&= 2\cdot \frac{u^3}{3} -8u {\hspace{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/139598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to prove that $\int_0^1\left(\sum\limits_{k=n}^\infty {x^k\over k}\right)^2\,dx = \int_0^1 2x^{n-1}\log\left(1+{1\over\sqrt{x}}\right)\,dx$ American Mathematical Monthly problem 11611 essentially asks you to show that
$$\lim_n\ n \int_0^1\left(\sum_{k=n}^\infty {x^k\over k}\right)^2\,dx=2\log(2).\tag1$$
This wou... | Here's a way to do it by brute force. First write
$$
\begin{align}
\int_0^1\left(\sum_{k = n}^\infty \frac{x^k}{k}\right)^2\,dx & = \sum_{k,m\geq n}\frac{1}{km}\int_0^1x^{k+m}\,dx \\
& = \sum_{k,m \geq n} \frac{1}{km(k+m+1)}.
\end{align}
$$
Put $r = k+m$, so that $m = r-k$, and transform the sum:
$$
\begin{align}
\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/139729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 4,
"answer_id": 2
} |
How Can One Prove $\cos(\pi/7) + \cos(3 \pi/7) + \cos(5 \pi/7) = 1/2$ Reference: http://xkcd.com/1047/
We tried various different trigonometric identities. Still no luck.
Geometric interpretation would be also welcome.
EDIT: Very good answers, I'm clearly impressed. I followed all the answers and they work! I can only... | Hint: start with $e^{i\frac{\pi}{7}} = \cos(\pi/7) + i\sin(\pi/7)$ and the fact that the lhs is a 7th root of -1.
Let $u = e^{i\frac{\pi}{7}}$, then we want to find $\Re(u + u^3 + u^5)$.
Then we have $u^7 = -1$ so $u^6 - u^5 + u^4 - u^3 + u^2 -u + 1 = 0$.
Re-arranging this we get: $u^6 + u^4 + u^2 + 1 = u^5 + u^3 + u$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/140388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 9,
"answer_id": 5
} |
Finding where the slope of tangent line is = 1 I am suppose to find the points on the curve $x^2+2y^2 = 1$ where the tangent line has a slope of 1.
I am not sure how to do this but I guessed that if I set it is
$y = \sqrt{\frac{1-x^2}{2}}$ then I can find that derivative and work from there.
I find the derivative as $y... | Don't solve for $y$ first; use implicit differentiation!
If $x^2+2y^2=1$, then
$$\begin{align*}
\frac{d}{dx}(x^2+2y^2) &= \frac{d}{dx} 1\\
2x + 4y\frac{dy}{dx} &= 0\\
4y\frac{dy}{dx} &= -2x\\
\frac{dy}{dx} &= -\frac{2x}{4y}\\
\frac{dy}{dx} &= -\frac{x}{2y}.
\end{align*}$$
If the tangent has slope $1$, then $\frac{dy}{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/140985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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A trigonometric identity: $(\sin x)^{-2}+(\cos x)^{-2}=(\tan x+\cot x)^2$ I've been trying to prove it for a while, but can't seem to get anywhere.
$$\frac{1}{\sin^2\theta} + \frac{1}{\cos^2\theta} = (\tan \theta + \cot \theta)^2$$
Could someone please provide a valid proof?
I am not allowed to work on both sides of t... | Hint:
$$\tan(\theta) + \cot(\theta)= \frac{\sin(\theta)}{\cos(\theta)}+\frac{\cos(\theta)}{\sin(\theta)}=\frac{\sin^2(\theta)}{\sin(\theta)\cos(\theta)}+\frac{\cos^2(\theta)}{\sin(\theta)\cos(\theta)} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/142252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Multiplication convention rules Student was asked to convert the following statement into multiplication format
$$7+7+7+7+7+7$$
She wrote the answer as $7\times 6=42$ and was marked wrong as the teacher expected $6\times 7=42$.
Is there any rule that can clarify the answer format?
The same with converting a multiplicat... | I suspect there is no commonly accepted convention on whether $2+2+2$ is $2\times 3$ or $3\times 2$. But note that $2\times 2\times 2$ is $2^3$, so I personally prefer to say that $2+2+2=2\times 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/143763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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Polynomial inequality I found the following problem on a website and would be curious to find a solution.
Let $a_1\ge a_2\ge\cdots\ge a_n$ be real numbers such that for all integer $k>0$:
$$a_1^k+a_2^k+\cdots+a_n^k\ge 0$$
Let $p=\max\{|a_1|,\ldots,|a_n|\}$. Show that $p=a_1$, and that
$$(x-a_1)(x-a_2)\cdots(x-a_n)\le ... | Part a) is indeed simple. Let me give a proof for part b). For $k=1$ our condition gives $a_1\ge -\sum_{k=2}^{n}a_k=S.$ Now we apply AM-GM to estimate $$\prod_{k=2}^{n}(x-a_k)\le \left(\frac{(n-1)x-S}{n-1}\right)^{n-1}\le \left(x+\frac{a_1}{n-1}\right)^{n-1}.$$
So we are left to show that $$(x-a_1)\left(x+\frac{a_1}{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/144383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Logarithms - Get the second solution to a logarithm equation Here's the original equation:
$\ln(11x-10) + \Big(\ln(11x-10)\Big)^2$ = 6
I've managed to obtain one solution: $x = \frac{e^2 + 10}{11}$
through those steps:
*
*$\ln\Big((11x-10)(11x-10)^2\Big) = 6$
*$\ln\Big((11x-10)^3\Big) = 6$
*$(11x-10)^3 = e^6$
... | Your solution is incorrect because :
$\ln(11x-10) +(\ln(11x-10))^2$ is not equal to $\ln(11x-10 \times {(11x-10)}^2)$
To solve
$$\ln(11x-10) + (\ln(11x-10))^2 = 6$$
Let $t=\ln(11x-10)$
$$t^2 +t -6=0$$
this gives , $t=2$ and $t=-3$
when $t=2 \longrightarrow \ln(11x-10)=2 \longrightarrow x=\frac{e^2+10}{11}$
when $t=-3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/146670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Using contour integration, or other means, is there a way to find a general form for $\int_{0}^{\infty}\frac{\sin^{n}(x)}{x^{n}} \, dx$? While studying some CA, I ran across methods of evaluating $$\int_0^\infty \frac{\sin x}{x} \, dx, \;\ \int_0^\infty \frac{\sin^2 x}{x^2} \, dx, \;\ \text{and} \ \int_0^\infty \frac{\... |
$$
\int_0^\infty \frac{\sin^n(x)}{x^n} \mathrm{d} x = \frac{\pi}{2^{n+1} \cdot (n-1)!} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} (2k-n)^{n-1} \operatorname{sign}(2k-n)
$$
where $\operatorname{sign}(x) = \cases{ 1 & $x > 0$ \\ 0 & $x = 0$\\ -1 & $x < 0$}$.
As to the (probabilistic) proof, notice that $\frac{\sin(t)}{t}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/146741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
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How to show that $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $ using induction? I am attempting a question, where I have to show $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $.
I have managed to solve the base case, which gives 9, which is a multiple of 3.
From here on,
I have $(n+1)((n+1)^2 + 8)$
$n^3 + 3n^2 + 1... | Without induction:
Since $8=-1\pmod{3}$, $n(n^2+8)=n(n^2-1)=(n-1)n(n+1)\pmod{3}$. Since $n-1$, $n$ and $n+1$ are three consecutive integers, at least (and in fact, exactly) one of them is a multiple of $3$, hence their product is a multiple of $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/150425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 1
} |
Evaluating $\int \sqrt{\frac{x^a}{(x+b)^c}} dx$ for $a,b,c\in\mathbb N$ and $a \lt c$ For $a,b,c\in\mathbb N$ are they any procedures how to calculate the following indefinite integral?
$$\int \sqrt{\frac{x^a}{(x+b)^c}} dx$$
it is also possible to assume that $a \lt c$ but it would be great if it works more general.
| It seems in any particular case, Maple knows how to do this.
$$
\int \frac{x^{\frac{3}{2}}}{(x + 5)^{\frac{5}{2}}} d x = \frac{-2\sqrt{5} x^{\frac{5}{2}}
\Biggl(4 x^{\frac{7}{2}} + 15 x^{\frac{5}{2}} - 15 \sqrt{5} \operatorname{arcsinh} \biggl(\frac{\sqrt{5} \sqrt{x}}{5}\biggr) x^{2} \biggl(\frac{x}{5} + 1\biggr)^{\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/150721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is $n = k \cdot p^2 + 1$ necessarily prime if $2^k \not\equiv 1 \pmod{n}$ and $2^{n-1} \equiv 1 \pmod{n}$? $p$ is an odd prime and $k$ is a positive integer.
Let $n=k \cdot p^2+1$.
If $2^k \not\equiv 1 \pmod n$ and $2^{n-1} \equiv 1 \pmod n$, is $n$ prime? If it is, why?
| I'm able to find a proof of the $k=2$ case, which is the first non-trivial case.
Result: Let $p$ be an odd prime and $n=2p^2+1$ such that $2^{n-1} \equiv 1 \pmod n$. Then $n$ is prime.
This proof splits into four steps:
*
*We will prove that if $p$ divides $\varphi(n)$, then $n$ is prime.
*We will prove that if $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/150853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Triangle related question My question is:
In $\Delta ABC$, let $AE$ be the angle bisector of $\angle A$. If $\frac{1}{AE} = \frac{1}{AC} + \frac{1}{AB}$ then prove that $\angle A = 120^\circ$.
What I tried: I extended side $AB$ and took a point $M$ on it such that $AC$ is congruent to $AM$. Then I proved that $AE$ is p... | Mark point $D$ on side $AC$ such that $AE=AD$.
Rewrite the given relation as follows:
$$\frac{1}{AE}-\frac{1}{AC}=\frac{1}{AB}$$
$$\frac{AC-AE}{AE\cdot AC}=\frac{1}{AB}$$
$$\frac{AC-AD}{AD\cdot AC}=\frac{1}{AB}$$
$$\frac{DC}{AD}=\frac{AC}{AB}$$
Now by the angle bisector theorem:
$$\frac{BE}{EC}=\frac{AB}{AC}$$
Combini... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/154432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to find the closed form formula for this recurrence relation $ x_{0} = 5 $
$ x_{n} = 2x_{n-1} + 9(5^{n-1})$
I have computed: $x_{0} = 5, x_{1} = 19, x_{2} = 83, x_{3} = 391, x_{4} = 1907$, but cannot see any pattern for the general $n^{th}$ term.
| Note that Ayman’s technique of ‘unwinding’ the recurrence works even without the preliminary division by $2^n$:
$$\begin{align*}
x_n&=2x_{n-1}+9\cdot5^{n-1}\\
&=2\left(2x_{n-2}+9\cdot5^{n-2}\right)+9\cdot5^{n-1}\\
&=2^2x_{n-2}+2\cdot9\cdot5^{n-2}+9\cdot5^{n-1}\\
&=2^2\left(2x_{n-3}+9\cdot5^{n-3}\right)+2\cdot9\cdot5^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/154667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
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Prove that the product of four consecutive positive integers plus one is a perfect square I need to prove the following, but I am not able to do it. This is not homework, nor something related to research, but rather something that came up in preparation for an exam.
If $n = 1 + m$, where $m$ is the product of four co... | Recursion on the square root, k(n)
n = lead integer
k(n) = square root [n(n+1)(n+2)(n+3) + 1]
k(1)^2 = 25
k(2)^2 = 121
k(3)^2 = 361
k(4)^2 = 841
k(5)^2 = 1681
k(1) = 5
k(2) = 11 = 5+ 6 = 5 + 2*(2+1)
k(3) = 19 = 11+ 8 = 11 + 2*(3+1) = 5 + 2 * ((2+1) + (3+1))
k(4) = 29 = 19+10 = 19 + 2*(4+1) = 5 + 2 * ((2+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/155040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "43",
"answer_count": 16,
"answer_id": 13
} |
Evaluating $\int \frac{x^3}{(x^2 + 1)^\frac{1}{3}}dx$ $$\int \frac{x^3}{(x^2 + 1)^\frac{1}{3}}dx$$
I am suppose to make a $u$ substitution and to make this a rational integral and then evaluate it from there but I have no idea how to do that. There aren't any good examples of this in the book and I can not find any us ... | Let $u = x^2 + 1$, $du = 2x\,dx$:
\begin{align*}
\int \frac{x^3\,dx}{(x^2 + 1)^\frac{1}{3}} &= \int \frac{x^2\,x\,dx}{(x^2 + 1)^\frac{1}{3}} = \frac{1}{2} \int \frac{u-1}{u^\frac{1}{3}}\,du = \frac{1}{2} \int \left(u^\frac{2}{3} - u^{-\frac{1}{3}}\right) \,du \\
&= \frac{3}{10} u^{\frac{5}{3}} - \frac{3}{4} u^{\frac{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/156553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Simplifying a radical? How would I simplify the following two radicals.
$$\sqrt{\frac{X^3}{50}}$$
For my answer I got $\frac{X^2}{50X}$ but I am not sure if this is correct.
My second question is how would I simplify $\sqrt{\frac{1}{12}}$
| For your first one,
$$\sqrt\frac{x^3}{50} = \frac{\sqrt{x^2}\sqrt{x}}{\sqrt{25}\cdot\sqrt{2}} = \frac{x\sqrt{x}}{5\sqrt{2}}$$
At this point, you could multiply top and bottom through by $\sqrt{2}$ (to make the $\sqrt{2}$ part of the bottom "go away"):
$$\frac{x\sqrt{x}\cdot\sqrt{2}}{5\sqrt{2}\cdot\sqrt{2}} = \frac{x\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/157948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$1 +1$ is $0$ ?
Possible Duplicate:
-1 is not 1, so where is the mistake?
$i^2$ why is it $-1$ when you can show it is $1$?
So:
$$
\begin{align}
1+1 &= 1 + \sqrt{1} \\
&= 1 + \sqrt{1 \times 1} \\
&= 1 + \sqrt{-1 \times -1} \\
&= 1 + \sqrt{-1} \times \sqrt{-1} \\
&= 1 + i \times i \\
&= 1 + (... | $$\sqrt{ab} = \sqrt{a} \times \sqrt{b}$$ is valid only for non-negative real numbers $a$ and $b$. Hence, the error is in the step $$\sqrt{(-1) \times (-1)} = \sqrt{-1} \times \sqrt{-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/158409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Positive definite quadratic form $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ Is $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ positive definite?
Approach:
The matrix of this quadratic form can be derived to be the following
$$M := \begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{2} & \cdots & \frac... | A direct approach would note that your form is
$${1\over2}\left[\left(\sum_{i=1}^n x_i\right)^2+\sum_{i=1}^n x_i^2\right].$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/159506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
} |
Integration of $\int\frac{1}{x^{4}+1}\mathrm dx$ I don't know how to integrate $\displaystyle \int\frac{1}{x^{4}+1}\mathrm dx$. Do I have to use trigonometric substitution?
| HINT :
Use
$$x^4+1=(x^2+1)^2-(\sqrt 2x)^2=(x^2+\sqrt 2x+1)(x^2-\sqrt 2x+1)$$
to get
$$\begin{align}\int\frac{dx}{1+x^4}&=\int\frac{\frac{1}{2\sqrt 2}x+\frac 12}{x^2+\sqrt 2x+1}dx+\int\frac{-\frac{1}{2\sqrt 2}x+\frac 12}{x^2-\sqrt 2x+1}dx\\&=\frac{1}{2\sqrt 2}\int\frac{x+\sqrt 2}{x^2+\sqrt 2x+1}dx-\frac{1}{2\sqrt 2}\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/160157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 20,
"answer_id": 4
} |
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Fractions in Questions and Answers
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