Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Summation of general series One of the problems in Donald Knuth's Art of Programming is phrased as follows:
Find and prove a simple formula for the sum $$\sum\limits_{n=0}^k\frac{(-1)^n(2n+1)^3}{(2n+1)^4+4}.$$
I have very little experience with summations. My method was to try to find some function such that
$$\frac{... | Since
$$
x^4+4=(x^2+2-2x)(x^2+2+2x),
$$
we get
$$
f(x)=\frac{x^3}{x^4+4}=\frac{1}{4}\frac{(x^2+2-2x+x^2+2+2x)^2}{(x^2+2-2x)(x^2+2+2x)}.
$$
Let $a(x)=4x^2+1$. Then
$$
4f(2n+1)=\frac{(a(n)+a(n+1))^3}{a(n)a(n+1)}=\frac{a(n)^2}{a(n+1)}+\frac{a(n+1)^2}{a(n)}+3(a(n)+a(n+1)).
$$
Hence
$$
\sum (-1)^nf(2n+1)=\sum (-1)^n\left\{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/898733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Solve $y' = x^4y+x^4y^4$ Solve the differential equation $$y' = x^4y+x^4y^4.$$
I'm not sure how to deal with the $x^4y^4$ term. So far I have only encountered differential equations where the exponent of $y$ was at most one.
Could someone please share a general strategy/hint?
(This is test preparation, not homework.)
| $$\frac{y'}{y+y^4}=x^4\tag{1}$$
and since
$$\int\frac{dz}{z+z^4}=\frac{1}{3}\log\frac{z^3}{1+z^3}\tag{2}$$
by partial fraction decomposition, by integrating both terms in $(1)$ we get:
$$\log\left(1+\frac{1}{y^3}\right)=-\frac{3}{5}x^5+C,\tag{3}$$
hence:
$$ 1+\frac{1}{y^3}= K e^{-\frac{3}{5}x^5},$$
$$ y= \frac{1}{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/903814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
solution of $y' + y^2 = \varphi^2(x)$ I need to solve differential equation in the interval $[-\pi/2,\pi/2]$
\begin{eqnarray}
y''(x) = y(x)\sin^2x
\end{eqnarray}
Trying $y(x) = \exp(\psi(x))$ yields,
\begin{eqnarray}
\zeta'(x) + \zeta^2(x) = \sin^2x \hspace{2cm} \zeta(x) = \psi'(x)
\end{eqnarray}
This equation seems t... | Let
$$
v = y’
$$
then
$$
y’’ = \frac{dv}{dx} = \frac{dy}{dx} \frac{dv}{dy} = v \frac{dv}{dy}
\\ \frac{y''}{y} = \frac{v}{y} \frac{dv}{dy} = \frac{dv^2}{dy^2} = \sin^2{x}
$$
This might be used as the basis of a series solution. For example let
$$
v = (a_0 + a_1y + a_2y^2 + ...)^{-1}
$$
Then the LHS becomes
$$
\frac{dv^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How prove this inequality $(1+\frac{1}{16})^{16}<\frac{8}{3}$
show that
$$(1+\dfrac{1}{16})^{16}<\dfrac{8}{3}$$
it's well know that
$$(1+\dfrac{1}{n})^n<e$$
so
$$(1+\dfrac{1}{16})^{16}<e$$
But I found this $e=2.718>\dfrac{8}{3}=2.6666\cdots$
so how to prove this inequality by hand?
Thank you everyone solve it,I w... | \begin{align}
(1+\dfrac{1}{16})^{16} &= \sum_{k=0}^{16} {16 \choose k}(\frac{1}{16})^k \\
& = 2 + \frac{15}{32} + \frac{35}{256} + \sum_{k=4}^{16} {16 \choose k}(\frac{1}{16})^k \\
& \leq 2 + \frac{15}{32} + \frac{35}{256} +\sum_{k=4}^{16} \frac{1}{k!}\\
& \leq 2+ \frac{15}{32} + \frac{35}{256} + e - 1 - 1- \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 2
} |
Quadratic formula and factoring are leading to different answers $$x^{ 2 }-2x-15=0$$
By factoring, I get:
$$(x-5)(x+3)$$
Which has the solutions:
$$x=5, x=-3$$
However when I use the quadratic formula (which is what the book saids to use), I get
$$\frac { 2 \pm \sqrt { 4-(4\cdot1\cdot(-15)) } }{ -2 } =$$
$$\frac { 2... | $$ax^2+bx+c=0\implies x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$\implies x^2-2x-15=0\implies x=\frac{2\pm\sqrt{(-2)^2-4\cdot1(-15)}}{2\cdot1}=\frac{2\pm8}2=5,-3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/906818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Dividing by $\sqrt n$ Why is the following equality true?
I know I should divide by $\sqrt n$ but how is it done exactly to get the RHS?
$$ \frac{\sqrt n}{\sqrt{n + \sqrt{n + \sqrt n}}} = \frac{1}{\sqrt{1 + \sqrt{\frac{1}{n} + \sqrt{\frac{1}{n^3}}}}}$$
| Suppose that $n>0$.
$$\frac{\sqrt{n}}{ \sqrt{n+\sqrt{n+\sqrt{n}}} } = \frac{1}{ \frac{1}{\sqrt{n}} \left( \sqrt{n+\sqrt{n+\sqrt{n}}} \right) } = \frac{1}{ \sqrt{1+ \frac{1}{n} \left( \sqrt{n + \sqrt{n}} \right) } } = \frac{1}{\sqrt{1+ \sqrt{ \frac{1}{n} + \frac{1}{n^2} \sqrt{n} } } } = \frac{1}{\sqrt{1+ \sqrt{ \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
How does $2^n + 2^n = 2^{n+1}$? What property of exponents can be used to show that
$$2^n + 2^n = 2^{n+1}$$
Does this work for all constants raised to a variable exponent?
| I still remember trying to make my friends understand how we can take $2^n$ outside, when I was in middle school.
$$2^n + 2^n \\
= 2(2^{n-1} + 2^{n-1}) \\
= 2^2(2^{n-2} + 2^{n-2}) \\
= 2^3(2^{n-3} + 2^{n-3})\\
\dots\\ \dots
$$
$$=2^n(2^{n-n} + 2^{n-n})\\
= 2^n(2^0 + 2^0)\\
= 2^n(1 + 1)\\
= 2^n\cdot 2^1\\
= 2^{n+1}$$
Bu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/907297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Finding the derivative of $y=12x^4\sqrt[3]{x^2}-2e^x+9$ Let
$$
y=12x^4\sqrt[3]{x^2}-2e^x+9
$$
How can we find $y^\prime$?
| $$\require{cancel} y=12x^4\underbrace{\sqrt[3]{x^2}}_{x^\frac23}-2e^x+9=12x^\frac{14}3-2e^x+9\\\frac{dy}{dx}=\cancelto{4}{12}\left(\frac{14}{\cancel{3}}\right)(x^\frac{11}3)-2e^x+0\\=56x^\frac{11}3-2e^x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/908240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Tricky geometry proof If a,b,c belong to the interval $(0,1)$ and $ab + ca + bc = 1$, prove that
$$\frac{a}{1-a^2} + \frac{b}{1-b^2} + \frac{c}{1-c^2}\ge\frac{3^{3/2}}{2}$$
How would you go about solving such a problem?
| Another way: $(a+b+c)^2 \ge 3(ab+bc+ca) =3 \implies s=\dfrac{a+b+c}3 \ge \dfrac1{\sqrt3}$.
Now use $t \mapsto \dfrac{t}{1-t^2}$ is convex and Jensen's inequality to get
$$LHS \ge 3\frac{s}{1-s^2} \ge \frac{3\sqrt3}2=RHS$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/912120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Factorization of $z^4 +1 = (z^2 - \sqrt 2z+1)(z^2 + \sqrt 2 z+1)$ for complex $z$ How can I get this equation from LHS to RHS by using the four roots of $z^4 +1 = 0$ are $z=\pm\sqrt{\pm i}$
$$z^4 +1 = (z^2 - \sqrt2 z+1)(z^2 + \sqrt2 z+1)$$
| If the roots are known then $x^4+1=(x+\sqrt i)(x-\sqrt i)(x+\sqrt{-i})(x-\sqrt{-i})$. Now difference of two squares will mean that $(x+\sqrt i)(x-\sqrt i)=x^2-i$ therefore multiplying with the other possible choices is a better idea. $(x+\sqrt i)(x+\sqrt{-i})$ gives $x^2+\sqrt{2}x+1$ therefore pairing the roots that wa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/913531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How find the value $\beta$ such $\left|\frac{p}{q}-\sqrt{2}\right|<\frac{\beta}{q^2}$ Find all positive real number $\beta$,there are infinitely many relatively prime integers $(p,q)$ such that
$$\left|\dfrac{p}{q}-\sqrt{2}\right|<\dfrac{\beta}{q^2}$$
maybe this problem background is Hurwitz's theorem:
$$\left|\sqrt{2... | Proof sketch. First write
$$\left|\frac{p}{q}-\sqrt{2}\right| = \frac{|p^2-2q^2|}{q^2\left|\frac{p}{q} + \sqrt{2}\right|}$$
The Pell equation $$p^2 - 2q^2 = 1$$
is known to have infinitely many solutions (should be proven) so
$$\left|\frac{p}{q}-\sqrt{2}\right| = \frac{|p^2-2q^2|}{q^2\left|\frac{p}{q} + \sqrt{2}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/915540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
If $\sec \theta+\tan \theta= \sqrt{3}$ then the positive value of $\sin \theta$ If $\sec \theta+\tan \theta=\sqrt{3}$ then the positive value of $\sin \theta$
Note:
$1/\cos\theta+\sin\theta/\cos \theta=\sqrt{3}$
$\sin\theta=\sqrt{3}\cos \theta-1$
squaring on both sides we get
$\sin^2\theta=$
| Since $1 = \sec^2 \theta - \tan^2 \theta = (\sec \theta + \tan \theta)(\sec \theta - \tan \theta)$, we have $\sec \theta - \tan \theta = \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}$.
Now, we have two equations:
(1) $\sec \theta + \tan \theta = \sqrt{3}$
(2) $\sec \theta - \tan \theta = \dfrac{\sqrt{3}}{3}$
Adding the t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/915618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Prove $a^4+b^4+(a-b)^4=c^4+d^4+(c-d)^4$ if $a^2+b^2+(a-b)^2=c^2+d^2+(c-d)^2$
if $a,b,c,d$ are positive real numbers and $$a^2+b^2+(a-b)^2=c^2+d^2+(c-d)^2$$
Prove $a^4+b^4+(a-b)^4=c^4+d^4+(c-d)^4$
Things i have done: from assumption $a^2+b^2+(a-b)^2=c^2+d^2+(c-d)^2$ I can conclude that $$a^2+b^2-ab=c^2+d^2-cd$$
Powe... | Hint:
$$\left(a^2+b^2+(a-b)^2\right)^2=2\left(a^4+b^4+(a-b)^4\right).$$
I wish I knew a smart, non brute-force, way to show that.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/920104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Require help with Inequality problems I am unable to find the solution for below Inequality problems.
1) $2/x<3$
The answer seems to be x belong to $(-\infty,0)\cup (2/3,\infty)$
2) $\dfrac{x+4}{x-3}<2$
The answer seems to be x belongs to $(-\infty,3)\cup (10,\infty)$
Progress
This is how I solved the second problem, ... | 1) $$\frac{2}{x}<3$$
$\frac{2}{x}$ is not defined at $x=0$.
*
*For $x > 0$ :
$\frac{2}{x}<3 \Rightarrow 2<3x \Rightarrow x>\frac{2}{3}$
$\{x>0\} \cap \{x>\frac{2}{3} \}=\{x>\frac{2}{3}\}$
*
*For $x < 0$ :
$\frac{2}{x}<3 \Rightarrow 2>3x \Rightarrow x<\frac{2}{3}$
But since $\{x<\frac{2}{3}\} \cap \{x<0\}=\{x<... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/920505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Inequality: $x^2+y^2+xy\ge 0$ I want to prove that $x^2+y^2+xy\ge 0$ for all $x,y\in \mathbb{R}$.
My "proof": Suppose wlog that $x\ge y$, so $x^2\cdot x\ge x^2\cdot y\ge y^2\cdot y=y^3$ (because $x^2\ge 0$ so we can multiply both sides by it without changing the inequality sign) giving $x^3\ge y^3$. Substracting we ha... | Another approach: note that $-2|xy| \leq xy \leq 2|xy|$, so we have
$$x^2 - 2|xy| + y^2 \leq x^2 + xy + y^2 \leq x^2 + 2|xy| + y^2$$
or equivalently,
$$(|x|-|y|)^2 \leq x^2 + xy + y^2 \leq (|x| + |y|)^2$$
Since the left hand side is nonnegative, the result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/920605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 4
} |
What is the Hilbert curve's equation? The Hilbert curve has always bugged me because it had no closed equation or function that I could find. What is its equation or function? For example, if I wanted to find the Hilbert's curve point at 4/7, how would I find it?
| As pointed out by almagest, there is a formula for Hilbert's space filling curve in Space-Filling Curves by Hans Sagan. The following formula appears as formula 2.4.3 on page 18 of the text.
If we write $t\in[0,1)$ in its base four expansion,
$$t=0_{\dot 4}q_1q_2q_3\ldots,$$
then
$$h(t) = \sum_{j=1}^{\infty} \frac{(-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/921985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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How to simplify $\sqrt{\tan^2\theta}\sin\theta\cos^2\theta$? I am solving $\int_{7}^{14}\frac{\sqrt{x^2-49}}{x^4}$ and got the integral down to $\frac{1}{343}\int_{0}^{\frac{\pi}{3}}\sqrt{\tan^2\theta}\sin\theta\cos^2\theta$ and wolfram simplified $\sqrt{\tan^2\theta}\sin\theta\cos^2\theta$ to $\sin^2(\theta)\cos(\thet... | Note that $\tan^2 \theta = (\tan\theta)^2$, and likewise for $\cos^2 \theta$, and on the interval of integration, we have that $\tan\theta \geq 0$.
$$\sqrt{\tan^2 \theta}\sin \theta \cos^2 \theta = \tan\theta\cdot \sin \theta \cos^2 \theta$$
$$= \frac{\sin\theta}{\cos\theta}\cdot \sin\theta \cos^2 \theta $$
$$ = \sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/922832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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What is $\sqrt{-x^3}$? What is $\sqrt{-x^3}$, assuming $x \in \mathbb R$ and $x < 0$? It seems as if there are two possibilities:
$\sqrt{-x^3} = \sqrt{-x\times x \times x} = \sqrt{-x \times x^2} = x\sqrt{-x}$
$\sqrt{-x^3} = \sqrt{(-x)^3} = \sqrt{(-x)\times (-x)^2} = -x\sqrt{-x}$
But I get the feeling I'm not doing the ... | $\sqrt {-x^3} = \sqrt {-x}^3 = \sqrt {x(-1)}^3 = \sqrt {xi^2}^3 = \sqrt{x}^3 \times \sqrt{i^2}^3 = \sqrt {x}^3 \times i^3 = -i\sqrt{x}^3 = -i\sqrt{x^3}$
$\because i^3 = -\sqrt{-1} =$ {$-i : i = \sqrt {-1}$}, assuming we are proposing that $\sqrt {-x^3} = \sqrt {(-x)^3}$. Otherwise we would write the term as not $\sqrt ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/923199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Question about G.C.D. Let,
$$a_{n}=n^2+20$$
$$d_{n}=\gcd(a_{n},a_{n+1})$$
where $n$ is a positive integer. Find the set of all values attained by $d_{n}$
I tried,
$d_{n}=\gcd(n^2+2n+21,n^2+20)$
$=\gcd(n^2+2n+21,2n+1)$
$=\gcd(n^2+20+2n+1,2n+1)$
$=\gcd(n^2+20,2n+1)$
However, after this I'm stuck.
Please He... | If $d$ divides both $n^2+20,(n+1)^2+20$
$d$ must divide $(n+1)^2+20-(n^2+20)=2n+1$
$d$ must divide $2(n^2+20)-n(2n+1)=40-n$
$d$ must divide $2(40-n)+(2n+1)=81$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/926334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proof - Area of a cyclic quadrilateral So we have a cyclic quadrilateral, as depicted below:
I have a conjecture that the area of this cyclic quadrilateral equals $$ \dfrac{\sqrt{(a+b+c-d)(a+b+d-c)(a+c+d-b)(b+c+d-a)}}{4} $$
I want to prove this. I know that the area of triangle ABC equals $\dfrac{1}{2}ab\sin(B)$ and ... |
The Inscribed Angle Theorem shows that opposite angles of a cyclic quadrilateral are supplementary. Thus,
$$
D=\pi-B
$$
Therefore, $\cos(D)=-\cos(B)$ and $\sin(D)=\sin(B)$.
The Law of Cosines says
$$
e^2=\overbrace{a^2+b^2-2ab\cos(B)}^{\text{as a side of $\triangle ABC$}}=\overbrace{c^2+d^2+2cd\cos(B)}^{\text{as a sid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/926598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Integral $\int_{0}^1\frac{\ln\frac{3+x}{3-x}}{\sqrt{x(1-x)}}dx$ I have a problem with the following integral:
$$
\int_{0}^{1}\ln\left(\,3 + x \over 3 - x\,\right)\,
{{\rm d}x \over \,\sqrt{\,x\left(\,1 - x\,\right)\,}\,}
$$
The first idea was to use the integration by parts because
$$
\int{{\rm d}x \over \,\sqrt{x\... | Let $x = \sin(t)^2$ and $s = 2t$, we have
$$\int_0^1 \log\left(\frac{3+x}{3-x}\right)\frac{dx}{\sqrt{x(1-x)}}
= \int_0^{\pi/2} \log\left(\frac{3 + \sin(t)^2}{3-\sin(t)^2}\right)\frac{2\sin t\cos tdt}{
\sqrt{\sin(t)^2(1-\sin(t)^2)}}\\
= 2 \int_0^{\pi/2}\log\left(\frac{3 + \sin(t)^2}{3-\sin(t)^2}\right) dt
= \int_0^{\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/927656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 3,
"answer_id": 1
} |
Compound interest problem with increasing deposits An Investor starts with an initial investment : $A$
He earns a steady profit of 10 percent per year.
But every year he adds additional amount which increases by 15 percent every year.
At the end of first year he adds an amount $x$
That means he adds 1.15$x$ at the end... | If the initial amount is $A$, at the end of the first year the money amount is $\frac{11}{10}A+x$.
At the end of the second year, the money amount is:
$$\frac{11}{10}\left(\frac{11}{10}A+x\right)+\frac{23}{20}x,$$
while at the end of the $n$-th year it is:
$$\begin{eqnarray*}A\cdot\left(\frac{11}{10}\right)^n + x\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/929468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Rewriting $1/(v^2-(mg/k))$ as two fractions I'm looking at the solution someone gave to me and I'm having a bit of trouble following one of the steps. This step in particular,
$\frac{1}{k}\frac{1}{v^2-\frac{mg}{k}}=\frac{1}{2k\sqrt{mg/k}}(\frac{1}{v-\sqrt{mg/k}}-
\frac{1}{v+\sqrt{mg/k}})$. All letters are constants exc... | $$\frac{1}{k}\frac{1}{v^2-\frac{mg}{k}}=\frac{1}{k}\frac{1}{v+\sqrt\frac{mg}{k}}\frac{1}{v-\sqrt\frac{mg}{k}}=$$
$$\ =\frac{1}{k}(\frac{A}{v+\sqrt\frac{mg}{k}}+\frac{B}{v-\sqrt\frac{mg}{k}})$$
$$\text{now you bring every term in the brackets under a common denominator}$$
$$\text{so you get:}\begin{cases}
A+B=0\\
-A+B= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/931188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Other ways to evaluate $\lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1}} - 1\right ]$? Using the facts that:
$$\begin{align}
\sqrt{1 + x} &= 1 + x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt{1 - x} &= 1 - x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt[3]{1 + x} &= 1 + x/3 + \mathcal{o}(x)
\end{align... | Multiply $\frac{1-\sqrt{1-x}}{\sqrt{x+1}-1}$ by $\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}$ and simplify to get $\frac{1+\sqrt{x+1}-\sqrt{1-x}-\sqrt{x+1} \sqrt{1-x}}{x}$
Expanding this, you only need keep two terms and you get
$$ \frac{1+\sqrt{x+1}-\sqrt{1-x}-\sqrt{x+1} \sqrt{1-x}}{x} = \frac{x+x^2/2 + \mathcal{O}(x^3)}{x}$$
t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/932411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Sum of the infinite series $\frac16+\frac{5}{6\cdot 12} + \frac{5\cdot8}{6\cdot12\cdot18} + \dots$ We can find the sum of infinite geometric series but I am stuck on this problem.
Find the sum of the following infinite series:
$$\frac16+\frac{5}{6\cdot 12} + \frac{5\cdot8}{6\cdot12\cdot18} + \frac{5\cdot8\cdot11}{6\cdo... | Using binomial expansion, we have:
$(1-x)^{-2/3} = 1 + \dfrac{\frac{2}{3}}{1!}x + \dfrac{\frac{2}{3} \cdot \frac{5}{3}}{2!} x^2 + \dfrac{\frac{2}{3} \cdot \frac{5}{3} \cdot \frac{8}{3}}{3!}x^3 + \dfrac{\frac{2}{3} \cdot \frac{5}{3} \cdot \frac{8}{3} \cdot \frac{11}{3}}{3!}x^4 + \cdots$
$(1-x)^{-2/3} = 1 + \dfrac{2}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/936146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Is $5^2x^3-x^5 = x^3(x-5)(x+5)$ or $-x^3(5-x)(5+x)$ Geogebra's Factor function says that
$5^2x^3-x^5$
is
$-x^3(x-5)(x+5)$
but from what I do, it is positive, $x^3(5+x)(5-x)$
Note the x isnt in the same position
Am I wrong?
| No, you're not wrong (but the title doesn't reflect the question):
$$
5^2x^3-x^5=x^3(5^2-x^2)=x^3(5-x)(5+x)
$$
You can use $5-x=-(x-5)$ to rewrite it as
$$
-x^3(x-5)(x+5)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/940607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Looking for the logic of a sequence from convolution of probability distributions I am trying to detect a pattern in the followin sequence from convolution of a probability distribution (removing the scaling constant $\frac{6 \sqrt{3}}{\pi }$: $$\left(\frac{1}{\left(x^2+3\right)^2},\frac{2
\left(x^2+60\right)}{\left... | Given that $\frac{6\sqrt{3}}{\pi}\cdot\frac{1}{(x^2+3)^2}$ is the pdf of a random variable $X$, we have:
$$\varphi_X(t)=\mathbb{E}[e^{itX}]=(1+\sqrt{3}\,|t|)\,e^{-\sqrt{3}\,|t|}$$
hence the characteristic function of $Y=X_1+\ldots+X_n$, where $X_1,\ldots,X_n$ are independent random variables with the same distribution ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/942204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Is the union of all elliptic curves $\mathbb{R}^2$? An elliptic curve
could be written as
$$y^2 = x^3 + a x + b \;.$$
Q1. Is it the case that every point $p \in \mathbb{R}^2$ lies
on some elliptic curve?
Q2. And what is the natural generalization of this question to higher
dimensions $d$, $p \in \mathbb{R}^d$?
I ... | Not sure about the second question, but here is my take on the first.
Let $p = (p_1, p_2)$.
Suppose $p_2^2 \neq p_1^3$, then $p$ lies on $y^2 = x^3 + (p_2^2 - p_1^3)$; note, here $\Delta = -432(p_2^2 - p_1^3)^2 \neq 0$, so $y^2 = x^3 + (p_2^2 - p_1^3)$ is non-singular.
If $p_2^2 = p_1^3$, then $p$ lies on $y^2 = x^3 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/943683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Multiple answer for integration of a function? Q. $\int \left(\frac{sin2x}{sin^4x+cos^4x}\right)\:dx$
My method:
$$\int \:\left(\frac{sin2x}{sin^4x+cos^4x}\right)\:dx=\int \:\:\left(\frac{sin2x}{\left(cos^2x-sin^2x\right)^2+2sin^2\left(x\right)cos^2\left(x\right)}\right)\:dx\:$$
=> $$\int \:\left(\frac{2sin2x}{2\left(c... | $$\frac{\sin2x}{\sin^4x+\cos^4x}=2\frac{\tan x\sec^2x}{1+\tan^4x}$$
Set $u=\tan^2x$ to find $I=\arctan(\tan^2x)=\arctan\dfrac{1-\cos2x}{1+\cos2x}=\arctan1-\arctan(\cos2x)$
$$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-\frac{\sin^22x}2=1-\frac{1-\cos^22x}2$$
Set $\cos2x=u$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/945555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Equation simplification, can't get it right $$\frac{1}{\frac{x-1}{x+2}}-\frac{2}{x^2-1}$$ should be simplified into $$\frac{x^2+3x}{x^2-1} \quad .$$
However, when I try to do it (tried several times), I fail to get it done right:
$$\frac{1}{\frac{x-1}{x+2}}-\frac{2}{x^2-1} = 1 * \frac{x+2}{x-1} - \frac{2}{x^2-1} $$ $$... | $$\begin{align}\frac{x+2}{x-1}-\frac{2}{x^2-1}&=\frac{x+2}{x-1}-\frac{2}{(x-1)(x+1)}\\&=\frac{(x+2)(x+1)}{(x-1)(x+1)}-\frac{2}{(x-1)(x+1)}\\&=\frac{(x+2)(x+1)-2}{(x-1)(x+1)}\\&=\frac{x^2+3x}{x^2-1}.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/945805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Partitions without twice odd numbers and where every odd number appears at most once Let $A=\{2,6,10,14,\ldots\}$ be the set of integers that are twice an odd number.
Prove that, for every positive integer $n$, the number of partitions of $n$ in which no odd number appears more than once is equal to the number of part... | So I'm currently solving this problem with the AOPS intermediate counting and probability course, and if you are too, you should probably use the message board but here is an explanation for the two values:
We start by making a generating function for partitions when no odd number appears more than once.
For the even n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/947665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is it true that $\left\lfloor\sum_{s=1}^n\operatorname{Li}_s\left(\frac 1k \right)\right\rfloor\stackrel{?}{=}\left\lfloor\frac nk \right\rfloor$ While studying polylogarithms I observed the following.
Let $n>0$ and $k>1$ be integers. Is the following statement true?
$$\left\lfloor \sum_{s=1}^n \operatorname{Li}_s\left... | We use the series expansion for the polylogarithm.
Note that
$$\sum_{s=1}^{n}\operatorname{Li_{s}}\left(\frac{1}{k}\right) =
\sum_{s=1}^{n}\sum_{j=1}^{\infty}\frac{1}{j^sk^j} = \sum_{j=1}^{\infty}\sum_{s=1}^{n}\frac{1}{j^{s}k^j}$$
we are able to interchange the summations because each term is positive, which implies... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/950418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Evaluate this square root $\sqrt{6 + 2\sqrt{5}} + \sqrt{6 - 2\sqrt{5}}$
I have no clue where to begin. I would appreciate a hint, the answer should be
$2\sqrt{5}$
In general, how do you evaluate
$\sqrt{a + b} + \sqrt{a - b}$?
Thanks!
| Set $r=\sqrt{6 + 2\sqrt{5}} + \sqrt{6 - 2\sqrt{5}}$ and observe that $r>0$; then
\begin{align}
r^2
&=6+2\sqrt{5}+2\sqrt{(6 + 2\sqrt{5})(6 - 2\sqrt{5})} + 6 - 2\sqrt{5}\\
&=12+2\sqrt{36-20}\\
&=12+2\sqrt{16}\\
&=12+8=20
\end{align}
Thus $r=\sqrt{r^2}=\sqrt{20}=2\sqrt{5}$.
More generally, if $a>b$ and $r=\sqrt{a+b}+\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/951065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Trigonometric Integral $\int _0^{\pi/4}\:\frac{dx}{\cos^4x-\cos^2x\sin^2x+\sin^4x}$ Q. $$\int _0^{\pi/4}\:\frac{1}{\cos^4x-\cos^2x\sin^2x+\sin^4x}$$
My method:
$$\int_0^{ \pi /4} \frac{dx}{(\cos^2x+\sin^2x)^2-3\cos^2x \sin^2x}=\int_0^{ \pi /4} \frac{dx}{1-3\cos^2x\sin^2x} $$
Dividing numerator and denominator by $\cos^... | Basically, you make a change of variable $t=\tan(x)$; doing so, you have $$\int _0^{\frac{\pi }{4}}\:\left(\frac{1}{\cos^4x-\cos^2x\sin^2x+\sin^4x}\right)\:dx=\int _0^1\frac{t^2+1}{t^4-t^2+1}dt$$ and $$\int\frac{t^2+1}{t^4-t^2+1}dt=\tan ^{-1}\left(\frac{t}{1-t^2}\right)$$
I must say that I do not see where the $3$ dis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/952307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find $z^{10}+\frac{1}{z^{10}}$ given $z^2+z+1=0$ $z$ is a complex number and $z^2+z+1=0$.
$$z^{10}+\frac{1}{z^{10}}=?$$
For the solution:
*
*the roots of $z^2+z+1$ are: $z_1=-\frac12+\frac{\sqrt3}{2}i$ and $z_2=-\frac12-\frac{\sqrt3}{2}i$
*converting these to their trigonometrical forms, we get: $z_1=\cos\frac{2\p... | $z^2+z+1 = 0 \implies z^2+z = -1 $
$z^2= -z-1 \implies z^3 = -z^2-z = -(z^2+z)= 1$
$z^{10} = z^3\cdot z^3\cdot z^3\cdot z= z \implies$
$z^{10} + z^{-10} = z + 1/z = \frac{z^2 +1 }z$ but $z^2+1 = -z$ so this results in $-z/z = -1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/952968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Divisibility property of $(a+b)^n-a^n-b^n$ Let $n$ be a natural number of the form $n=6k+1$ (while $k$ is a positive integer).
Show that $(a^2+ab+b^2)^2$ divides $(a+b)^n-a^n-b^n$ for all integer numbers $a,b$ (such that $a^2+ab+b^2\ne0$).
| if $\omega$ is a complex cube root of unity, then both $\omega$ and $1+\omega$ are sixth roots of unity. hence for $k \ge 1$$\omega$ is a root of:
$$
(x+1)^{6k+1} - x^{6k+1} - 1 = 0 \tag{1}
$$
likewise for $\omega^2$
but these are exactly the roots of
$$
x^2+x+1=0
$$
this shows that $a^2+ab+b^2$ is a factor of $(a+b)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/956102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Theory number problem I need to prove that there are infinitely many natural numbers $n$ for which $2n^2+3$ and $n^2+n+1$ are relatively prime.
This is not true for every $n$ (for example, $n=4$), I tried to check for odd $n$ but can't find way to prove it (I tried induction). Any ideas?
| Any common divisor of $2n^2+3$ and $n^2+n+1$ divides $2(n^2+n+1)-(2n^2+3)$, which is $2n-1$.
Any common divisor of $2n^2+3$ and $2n-1$ divides $2n^2+3-n(2n-1)$, which is $n+3$.
And any common divisor of $2n-1$ and $n+3$ divides $7$.
Now produce infinitely many numbers $2n^2+3$ that are not divisible by $7$.
Remark: We... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/957040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Why does $ \frac{b^n-a^n}{b-a}=\sum_{k=1}^nb^{n-k}a^{k-1}$? Trying to work through the answer in this question:
The inequality $b^n - a^n < (b - a)nb^{n-1}$
| First consider a sample of the equation to be shown. Consider
\begin{align}
a^{2} + ab + b^{2} &= \frac{(b-a)(b^{2} + ab + a^{2})}{b-a} \\
&= \frac{1}{b-a} \, [ b^{3} -a b^{2} + a b^{2} - a^{2} b + a^{2} b - a^{3} = \frac{b^{3}- a^{3}}{b-a}.
\end{align}
Now apply the same logic to the form
\begin{align}
b^{n-1} + a b^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/958073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Is this the correct procedure for Integral Partial Fraction. $∫ (x^3+x^2+x+3)/((x^2+1)(x^2+3))$
The first step I did is distributed the denominator so that I can find out if I should use synthetic division. Which after doing this I discovered that the denominator's exponent is greater therefore I do not use synthetic d... | We do indeed have $$(Ax +B)(x^2+3)+( Cx+D)(x^2+1) = x^3 + x^2 + x + 3$$
If you put $x=0$, you get $$(A(0) +B)(0 + 3) + (C(0) + D)(0 + 1) = 3 \iff 3B + D = 3$$
Try using $x = (-1)$ and $x = 1$, and apply the same process.
Then use any other constant that hasn't been used yet, say $x = 2$, using the same process. That w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/958280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Mathematical Induction Question, Proof Help Prove using Mathematical Induction that for all natural numbers ($n>0$):
$$
\frac 1 {\sqrt{1}} + \frac 1 {\sqrt{2}} + \cdots + \frac 1 {\sqrt{n}} \ge \sqrt{n}.
$$
Proof by Induction:
Let P(n) denote 1/ √1 + 1/ √2 + … + 1/ √n ≥ √n
Base Case: n = 1, P(1) = 1/... | You know that
$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{k}} \geq \sqrt{k}$$$$
and want to prove that:
$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{k+1}} \geq \sqrt{k+1}$$$$
Add $\sqrt{k+1}-\sqrt{k}$ to both sides of first inequality, you get:
$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/959654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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A long nasty limit problem: $\lim_{x \to \infty}\left[8e\,\sqrt[\Large x]{x^{x+1}(x-1)!}- 8x^2-4x \ln x - \ln^2 x - (4x + 2 \ln x) \ln 2\pi\right]$ Does the following limit admit a closed-form?
$$\lim_{x \to \infty}\left[8e\,\sqrt[\Large x]{x^{x+1}(x-1)!}- 8x^2-4x \ln x - \ln^2 x - (4x + 2 \ln x) \ln 2\pi\right]$$
My p... | Taking the logarithm of the first term divided by $ 8e $:
$$ \frac 1 x\left[x\ln x + \ln\left(x!\right)\right], $$
Stirling-approximate the second term to get
$$ \ln\left(x!\right) = x\ln x - x + \frac 1 2\ln x + \frac 1 2 \ln 2\pi + \frac 1 {12x} + \mathcal{O}(x^{-1}). $$
The first term is therefore
$$\begin{align} &8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/960544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_{0}^{\infty} \mathrm{e}^{-x^2-x^{-2}}\, dx$ I have to find
$$I=\int_{0}^{\infty} \mathrm{e}^{-x^2-x^{-2}}\, dx $$
I think we could use
$$\int_{0}^{\infty} \mathrm{e}^{-x^2}\, dx = \frac{\sqrt \pi}{2} $$ But I don't know how.
Thanks.
| Consider
\begin{align}
x^{2} + \frac{1}{x^{2}} = \left( x - \frac{1}{x} \right)^{2} +2
\end{align}
for which
\begin{align}
I = \int_{0}^{\infty} e^{-\left(x^{2} + \frac{1}{x^{2}}\right)} \, dx = e^{-2} \, \int_{0}^{\infty} e^{-\left(x - \frac{1}{x}\right)^{2}} \, dx.
\end{align}
Now make the substitution $t = x^{-1}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/962420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
$ \lim\limits_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$ Find the limit: $$ \lim_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$$
I did the following:
\begin{align}
(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x}) = \frac{(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})}{1} \cdot \frac{(\sqrt{x^2 + 2x} + \sqrt... | $\bf{My\; Solution::}$ Given $\displaystyle \lim_{x\rightarrow \infty}\left(\sqrt{x^2+2x}-\sqrt{x^2-7x}\right) = \lim_{x\rightarrow \infty}x\left\{\left(1+\frac{2}{x}\right)^{\frac{1}{2}}-\left(1-\frac{7}{x}\right)^{\frac{1}{2}}\right\}$
Now Using Binomial Expansion::
$\displaystyle \lim_{x\rightarrow \infty}\left\{\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/962458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
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Evaluate $ \int_0^1 \sum_{k=0}^\infty (-x^4)^k dx = \int_0^1 \frac{dx}{1+x^4} $ I have read this thread and I found in some comments the above named equality. I couldn't follow the transformation, which are done to get from the left to the right side at that point in particular.
Can someone help me and show how it's do... | On the left, reverse the order of summation and integration to get
$$\sum_{k=0}^{\infty} (-1)^k \, \int_0^1 dx \, x^{4 k} = \sum_{k=0}^{\infty} \frac{(-1)^k}{4 k+1} = 1 - \frac15 + \frac19 - \frac1{13} + \cdots$$
This sum is equal to
$$\int_0^1 \frac{dx}{1+x^4}$$
To evaluate this, observe that
$$1+x^4 = (1+\sqrt{2} x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/962658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Calculating Moments The following problem is from a Schaum book on statistics. While I thought I did it right, I did not come up with the right answer. Therefore, I am thinking I did something wrong.
Problem: Find (a) the moment generating function of the random variable
$ x = \begin{cases}
\, \frac{1}{2} &\text{ p... | You work seems correct. For example, without using the moment generating function
$$
E[X^2] = (1/2)^2 (1/2) + (-1/2)^2(1/2) = 1/8+1/8 = 1/4.
$$
It seems the book chose to change the random variable to $\tilde{x}$ where $$\tilde{x} = \begin{cases} 1 & \text{ with probability $1/2$} \\ -1 & \text{ with probability $1/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/964511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$\Sigma_{k=m+1}^{\infty}\frac{1}{10^k} \leq \frac{1}{10^m}$ Is it true that $\Sigma_{k=m+1}^{\infty}\frac{1}{10^k} \leq \frac{1}{10^m}$ for $m \in \mathbb{N}$?
| $$\sum\limits_{k=m+1}^\infty \dfrac{1}{10^k} = \dfrac{1}{10^{m+1}}\sum\limits_{\ell=0}^\infty \dfrac{1}{10^\ell} = \dfrac{1}{10^{m+1}} \cdot \frac{1}{1-1/10} = \dfrac{1}{10^{m+1}}\cdot \frac{10}{9} = \frac{1}{9 \cdot 10^m} < \dfrac{1}{10^m}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/966046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solving $2\cos2x-4\sin x\cos x=\sqrt{6}$ How I solve the following equation for $0 \le x \le 360$:
$$
2\cos2x-4\sin x\cos x=\sqrt{6}
$$
I tried different methods. The first was to get things in the form of $R\cos(x \mp \alpha)$:
$$
2\cos2x-2(2\sin x\cos x)=\sqrt{6}\\
2\cos2x-2\sin2x=\sqrt{6}\\
R = \sqrt{4} = 2 \\
\alph... | You have made a mistake in the third step.
$$
2\cos2x-2(2\sin x\cos x)=\sqrt{6}\\
2\cos2x-2\sin2x=\sqrt{6}\\
R = \sqrt{8} = 2\sqrt{2} \\
\alpha = \arctan -\frac{2}{2} = -\frac{\pi}{4}\\
\therefore \cos(2x - \frac{\pi}{4}) = \frac{\sqrt3}{2}
$$
Hope this help.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/966798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to simplify $\sqrt[3]{29\sqrt{2}-45}-\sqrt[3]{29\sqrt{2}+45}$ I in trouble simplifying this:
$$\sqrt[3]{29\sqrt{2}-45}-\sqrt[3]{29\sqrt{2}+45}$$
couldn't find a solution. Can you help?
| HINT:
Let $$\sqrt[3]{29\sqrt{2}-45}-\sqrt[3]{29\sqrt{2}+45}=a$$
$$a^3=29\sqrt{2}-45-(29\sqrt{2}+45)-3a(-7)$$
as $(29\sqrt{2}-45)(29\sqrt{2}+45)=-7^3$
$$\iff a^3-21a+90=0$$ whose only real root is $-6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/970165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Calculate this sum $\sum_{x=1}^{{n}/{4}} \frac{1}{x}$ How can I calculate this sum
$$\sum_{x=1}^{{n}/{4}} \frac{1}{x}$$
for $n$ an odd integer?
Added later: For example, WolframAlpha outputs 4 - Pi/2 - Log[8] for HarmonicNumber[1/4], what are the steps that produce that? Can they be generalized for any odd number?
| The series could be defined by
\begin{align}
S_{m} = \sum_{n=1}^{\lfloor m/4 \rfloor} \frac{1}{n}
\end{align}
where $\lfloor x \rfloor$ represents the largest integer of $x$. Now, consider the series
\begin{align}
H_{n} = \sum_{k=1}^{n} \frac{1}{k}
\end{align}
which are known as the Harmonic numbers. From this it is se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/970732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Code is not cyclic for any q I have code $C$ over $F_p$ with generator matrix which looks like
$G =
\begin{pmatrix}
0 &0& 0& 1& 0& 1& 1 &1\\
1& 0 &0& 0 &1 &0 &1& 1\\
1& 1& 0& 0& 0& 1& 0& 1\\
1 &1& 1& 0 &0 &0& 1 &0\end{pmatrix}$
I need to show that this code is not cyclic for any $p$.
I constructed vector which looks... | A base for your code is a base for $\ker G$. So the code has dimension 4, and a base is
$$
v_1=\begin{pmatrix}0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ -1\end{pmatrix}
\quad v_2=\begin{pmatrix}-1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0\end{pmatrix}
\quad v_3=\begin{pmatrix}0 \\ -1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0\end{pmatrix}
\quad v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/972961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
how to solve $\int\frac{1}{1+x^4}dx$ i want find the answer and method of solve of $\int\frac{1}{1+x^4}dx$.
I know $$\int\frac{1}{a^2+x^2}dx=\frac{1}{a}\arctan\frac{x}{a}+C$$,
How I can use this to solve of that integration.
| Note the following:
*
*$2 = (1 + x^{2}) + (1-x^{2})$.
*$\displaystyle \int \frac{1}{1+x^{4}} = \frac{1}{2} \int\frac{2}{1+x^{4}} = \frac{1}{2} \int\frac{(1+x^{2})+(1-x^{2})}{1+x^{4}} = \frac{1}{2} \int\frac{1+x^{2}}{1+x^{4}} + \frac{1}{2}\int\frac{1-x^{2}}{1+x^{4}} = \frac{1}{2} I_{1} + \frac{1}{2}I_{2}$.
*$\displ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/973822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How do I solve this fraction addition problem? $4\frac{2}{9} + -9\frac{1}{2}$ yeilds result of $-5\frac{13}{18}$ but WolframAlpha says the answer is $-5\frac{5}{18}$
fixed.
| In mixed fractions (just as in decimal ones), the sign applies to both whole and fractional parts, so
$$4\frac{2}{9} + -9\frac{1}{2}=4\frac{2}{9}-\left(9+\frac{1}{2}\right)=(4-9)+\left(\frac{2}{9}-\frac{1}{2}\right)$$
Since
$$4-9=-5$$
and
$$\frac{2}{9}-\frac{1}{2}=\frac{2\times2}{9\times2}-\frac{1\times9}{2\times9}=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/981671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How to find inverse laplace transform $$
F(s) = \dfrac{6s+9}{s^2-10s+29}
$$
How do you solve the inverse Laplace transform of this above equation?
| $$
\underbrace{s^2-10s+29 = (s-5)^2+2^2}_{\text{completing the square}} = t^2+2^2
$$
So
$$
\frac{6s+9}{s^2-10s+29} = \frac{6(s-5)+39}{(s-5)^2+2^2} = \frac{6t+39}{t^2+2^2} = 6\frac{t}{t^2+2^2} + \frac{39}{2}\cdot\frac{2}{t^2+2^2}
$$
Now look at each term separately.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/983386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve for $\int \sqrt{x}(\sqrt{x}-2x)^2 dx $ $\int \sqrt{x}(\sqrt{x}-2x)^2 dx $ so I solved this using U-substitution where $u= \sqrt{x}$ so my $du2\sqrt{x}=dx$ then it will be $2 \int u^2(u-2u^2)^2$ and just expand then distribute the $u^2$ so i got $\frac{8u^7}{7}-\frac{4u^6}{3}+\frac{2u^5}{5}+c$ then just substitue ... | Not answering your question, but another way to solve is with partial integration, is to observe that $$\sqrt{x}(\sqrt{x}-2x)^2=\sqrt{x}\left(\sqrt{x}(1-2\sqrt{x})\right)^2=\sqrt{x}\cdot x\cdot(\sqrt{x}-2x)^2=x\dfrac{(1-2\sqrt{x})^2}{\sqrt{x}}$$
Now $$\left((1-2\sqrt{x})^3\right)'=3(1-2\sqrt{x})^2\dfrac{-2}{2\sqrt{x}}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/985811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Integration $I_n=\int_{0}^{1}\frac{dx}{(x^n+1)(\sqrt[n]{x^n+1})}$ $$I_n=\int_{0}^{1}\frac{dx}{(x^n+1)\large\sqrt[n]{\normalsize x^n+1}}$$
Could someone help me through this problem?
| Let $\displaystyle v=v(x)=\left(\frac{x^n+1}{x^n}\right)^{\Large \frac{1}{n}}$.
Then we have: $\displaystyle v(0)=+\infty \ , \ v(1)=2^{\large \frac{1}{n}} \ , \ \frac{1}{x^n+1} = 1 - v^{-n} \ , \ x=\frac{1}{\left(v^n-1\right)^{\large \frac{1}{n}}}$ and $$\displaystyle dx = - \frac{v^{n-1}}{\left(v^n-1\right)^{1+\larg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/987337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Trigonometry graphs sinusoidal waves i need help on this questions. I couldn't figure how to determine for both question A and B.
But i have the answers for them, i just don't understand how the amplitude is 3 and so on.
| To find the amplitude, take half the difference of the maximum and minimum values. In this case, the amplitude is
$$a = \frac{1 - (-5)}{2} = \frac{6}{2} = 3$$
If you subtract the amplitude from the maximum value, you will find the average value. In this case, the average value of the function is $1 - 3 = -2$. If the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/987495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $\frac{x^2+2xy+y^2}{x^2-y^2} >x+y$
Find the set of integer solutions $(x,y)$ to
$$\frac{x^2+2xy+y^2}{x^2-y^2} >x+y$$
I can't seem to multiply both sides by the expression in the denominator.
Nor can I simplify and cancel any terms. How should I it?
| $$\frac{x^2+2xy+y^2}{x^2-y^2} >x+y \iff \frac{x+y}{x-y}> x+y$$
Case 1: $x+y > 0$
In this case, we have $\dfrac1{x-y}> 1 \iff 0 < x-y< 1$ which is not possible.
Case 2: $x+y < 0$
In this case, we have $\dfrac1{x-y} < 1 \iff x-y < 0$ or $x-y > 1$.
Now $x+y < 0, \; x-y < 0$ gives $x < 0, \;x < y < -x$
and $x+y < 0, \; ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/988198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove $\sqrt{ \frac{2x^2 - 2x + 1}{2} } \geq \frac{1}{x + \frac{1}{x}}$ for $0 < x < 1$ I stumbled upon this question while doing practice inequalities questions, and I do not know how to start...
Problem: Prove that
\begin{align*}
\sqrt{ \frac{2x^2 - 2x + 1}{2} } \geq \frac{1}{x + \frac{1}{x}}
\end{align*}
for $0 < ... | This inequality is true for all $x>0$. Indeed$,$
$$ \frac{(x+\frac{1}{x})^2(2x^2-2x+1)}{2}-1$$
$$={\frac {2 \ \left( 2\ x-1 \right) ^{2}{x}^{2}+ \left( x-1
\right) ^{4} \left( x+1 \right) ^{2}+{x}^{2} \left( x-1 \right) ^{2}
\left( x+1 \right) ^{2}}{2{x}^{2}}} \geqq 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/988546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
For the series $S = 1+ \frac{1}{(1+3)}(1+2)^2+\frac{1}{(1+3+5)}(1+2+3)^2$...... Problem :
For the series $$S = 1+ \frac{1}{(1+3)}(1+2)^2+\frac{1}{(1+3+5)}(1+2+3)^2+\frac{1}{(1+3+5+7)}(1+2+3+4)^2+\cdots $$ Find the nth term of the series.
We know that nth can term of the series can be find by using $T_n = S_n -S_{n-1... | Consider the series
\begin{align}
S_{n} = 1 + \frac{(1+2)^{2}}{1+3} + \frac{(1+2+3)^{2}}{1+3+5} + \cdots + \frac{(1+2+\cdots+n)^{2}}{1+3+\cdots+(2n-1)}.
\end{align}
This series is seen as
\begin{align}
S_{n} &= 1 + \frac{1}{2^2}\binom{3}{2}^{2}+ \frac{1}{3^{2}} \binom{4}{2}^{2}+ \cdots + \frac{1}{n^{2}} \binom{n+1}{2}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/988628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\int_0^\infty \frac{\ln x}{x^n-1}\,dx = \Bigl(\frac{\pi}{n\sin(\frac{\pi}{n})}\Bigr)^2$ This question inspired me to ask the following.
Prove that
$$I_n = \int_0^\infty \frac{\ln x}{x^n-1}\,dx = \left(\frac{\pi}{n\sin\left(\frac{\pi}{n}\right)}\right)^2,$$
for $\Re(n)>1$.
For some cases there is a nice spec... | $$
\begin{align}
\int_0^\infty\frac{\log(x)}{x^n-1}\mathrm{d}x
&=\int_{-\infty}^\infty\frac{x}{e^{nx}-1}e^x\,\mathrm{d}x\\
&=\int_0^\infty\frac{x}{e^{nx}-1}e^x\,\mathrm{d}x+\int_0^\infty\frac{x}{1-e^{-nx}}e^{-x}\,\mathrm{d}x\\
&=\int_0^\infty x(e^{(1-n)x}+e^{(1-2n)x}+e^{(1-3n)x}+\dots)\,\mathrm{d}x\\
&+\int_0^\infty x(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/989504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 3
} |
Proving $a^ab^bc^c\ge(abc)^{(a+b+c)/3}$ for positive real numbers. Prove that
$$a^ab^bc^c\ge(abc)^{(a+b+c)/3}$$
where $a,b,c\in\mathbb{R^+}$
I tried using powered AM-GM but didn't get anything. please give me a hint to solve it.
| Take the $\log$ of both sides. This is equivalent to:
$$
a\log a + b\log b + c\log c \ge \frac{a+b+c}3 (\log a + \log b + \log c)
$$
now use the rearrangement inequality twice to get:
$$
a\log a + b\log b + c\log c \ge b\log a + c\log b + a\log c ;\\
a\log a + b\log b + c\log c \ge c\log a + a\log b + b\log c;\\
a\log ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/990103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Number of $0$ in the end of $11^n-1$ $n$ is integer, calculate number of $0$ in the end of $11^n-1$(i.e. largest integer $m$ such that $10^m|11^n-1$).
The original question was $n=100$ and I could only choose $m$ from 1 to 5. I calculated the $11^{100}-1$ mod $100000$ and got the answer, I'm looking for much better and... | The binomial expansion used for $n=100$ in the answer to the linked question can also be applied to general $n$. Let us set
$$11^n-1= (10+1)^n-1= \sum_{i=0}^n {n \choose i} 10^i -1$$
The sum can be expanded as
$$ 11^n-1= 10 \cdot n+ 100 \cdot \frac{n(n-1)}{2!}+ 1000 \cdot \frac{n(n-1)(n-2)}{3!}....$$
Now, as correctly ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/990221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that complex numbers are vertices of equilateral triangle 1)Show if $|z_1|=|z_2|=|z_3|=1$ and $z_1+z_2+z_3=0$ then $z_1,z_2,z_3$ are vertices of equilateral triangle inscribed in a circle of radius.
I thought I can take use from roots of unity here, since $|z_1|=|z_2|=|z_3|=1$ they lie at circle at radius $1$ but ... | Let: $z_1 =e^{ia} ; z_2 = e^{ib}; z_3 = e^{ic}$
$ z_1 +z_2 = e^{i\frac{a+b}{2}}*(e^{i\frac{a-b}{2}} + e^{-i\frac{(a-b)}{2}}) = e^{i\frac{a+b}{2}}*2*cos(\frac{a-b}{2}) = -z_3 $
=> $|2*cos(\frac{a-b}{2})| = |-z_3| = |z_3| = 1$ ,
If $ cos(\frac{a-b}{2}) =\frac{1}{2} $ -> $a = b \pm \frac{2\pi}{3}$ $mod(2\pi)$
here withou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/992262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Prove that $1+4+7+...+(3n-2) = \frac{n}{2}(3n-1)$ Using induction
prove that $1+4+7+...+(3n-2) = \frac{n}{2}(3n-1) \forall n \in \mathbb{N}$
Attempt:
Let $n =1$ so $3(1)-2 = 1$ and $\frac{1}{2}(3(1)-1)=1$
Assume true at $n=k$ so $3k-2 = \frac{k}{2}(3k-1)$
What do I do next? Here's where I'm stuck:
Let $n=k+1$ So $3(k+... | Notice that you have an aritmetic progression with initial term $1$ and difference $3$. So, $a_1 = 1$, and $a_k = 1 + 3(k-1)$. Then, $3n - 2 = 1 + 3(k-1) \implies k = n$, so $3n - 2$ is the $n$th term of the progression. This way: $$S_n = n\frac{a_1 + a_n}{2} = \frac{n}{2}(3n-1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/992827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\dots+\frac{1}{1331}=\frac{p}{q}$; is $p$ divisible by $1997$? if $p,q\in \mathbb{N}$ and
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\dots+\frac{1}{1331}=\frac{p}{q}$$
why is $p$ divisible by $1997$?
| Write $$1-\frac 12+\frac 13- \dots+\frac 1{1331}=\left(1+\frac 12+\frac 13+ \dots +\frac 1{1331}\right)-\left(1+\frac 12+ \dots +\frac 1{665}\right)$$
Using $1=2\cdot \frac 12$, and $ \frac 12=2\cdot \frac 14$, and $ \frac 13=2\cdot \frac 16$ etc to eliminate the negative fractions.
Then the sum becomes $$\frac 1{666}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/993852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Using Dirichlet's hyperbola method and Dirichlet's formula Dirichlet Hyperbola Method. For $x \geq 2$:
$$ \sum_{n \leq x} \frac{d(n)}{n} = \frac{1}{2} \log^2 x + 2\gamma \log x + \gamma^2 + O(\frac{\log x}{\sqrt{x}})$$
I know already that the summation of 1/n can be written $\log x + \gamma + O(1/x)$ and summation of ... | Following the notation from the
Planetmath Article
we note that
$$\frac{\tau(n)}{n} = 1/n * 1/n =
\sum_{d|n} \frac{1}{d} \left(\frac{n}{d}\right)^{-1}.$$
Substituting this into the formula for the method we obtain
$$\sum_{n\le x} \frac{\tau(n)}{n} =
\sum_{a\le\sqrt{x}} \sum_{b\le x/a}
\frac{1}{a} \frac{1}{b}
+ \sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/994179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
derivative of $y=x^3\sqrt{x}-\frac{1}{x^2\sqrt{x}}$ $y=x^3\sqrt{x}-\dfrac{1}{x^2\sqrt{x}}$
Both terms require the product rule, right? My try:
$x^3\dfrac{1}{2}x^{-1/2}+3x^2x^{1/2}-\dfrac{-1}{2}x^{-3/2}x^{-2}--2x^{-3}x^{-1/2}$
What am I doing wrong? The correct answer is: $y\;'=3.5x^2\sqrt{x}+\dfrac{2.5}{x^3\sqrt... | $$\ y=x^3\sqrt x-\frac{1}{x^2\sqrt x}=x^{\frac{7}{2}}-x^{-\frac{5}{2}}$$
$$\ y'=\frac{7}{2}x^{\frac{5}{2}}+\frac{5}{2}x^{-\frac{3}{2}}=$$
$$=\frac{7}{2}x^2\sqrt x+\frac{5}{2}\cdot\frac{1}{x\sqrt x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How to prove $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} <2$? Prove the inequality for a triangle with sides $a,b,c$ we have $$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} <2$$
Trial: Since $a,b,c$ are sides of a triangle I know $a+b>c,b+c>a,a+c>b$
| That $a,b,c$ are sides of a triangle is equivalent to $a=x+y$, $b=y+z$ and $c=z+x$. Therefore we can rewrite our inequality as follows:
$$
\dfrac{x+y}{x+y+2z}+\dfrac{y+z}{2x+y+z}+\dfrac{z+x}{x+2y+z} <2
$$
Now, we have $\dfrac{x+y}{x+y+2z}<\dfrac{x+y}{x+y+z}$, $\dfrac{y+z}{2x+y+z}<\dfrac{y+z}{x+y+z}$ and $\dfrac{z+x}{x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Proving Identity of Combination The lecturer had given two questions of proving that are
$$\binom{r}{r}+\binom{r+1}{r}+...+\binom{n}{r}=\binom{n+1}{r+1}\text{for }n\geq{r}\geq{1} $$
$$\binom{r}{0}+\binom{r+1}{1}+...+\binom{r+k}{k}=\binom{r+k+1}{k}\text{for }r,k\geq{1}$$
I tried to use the induction to prove these two... | $\bf{My\; Solution::}$ Given $\displaystyle \binom{r}{r}+\binom{r+1}{r}+\binom{r+2}{r}+\cdot \cdot \cdot \cdot \cdot \cdot\cdot +\binom{n}{r} = \binom{n+1}{r+1}\;,$
Where $n\geq r \geq 1.$
Coefficeint of $x^r$ in
$\displaystyle \left\{(1+x)^{r}+(1+x)^{r+1}+(1+x)^{r+1}\cdot \cdot \cdot \cdot \cdot \cdot \cdot+(1+x)^{n}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Poisson Integral is equal to 1
Show
$$
\int_{-\pi}^{\pi}P(r, \theta)d\theta = 1
$$
Let $\alpha(r) = \frac{r^2 - 1}{2r}$ and $\gamma(r) = -\big(\frac{r^2 + 1}{2r}\big)$.
Then
$$
\frac{1}{2\pi}
\int_{-\pi}^{\pi}\frac{1 - r^2}{1 - 2r\cos(\theta) + r^2}d\theta
= \frac{\alpha}{2\pi}
\int_{-\pi}^... | From your definition, $\gamma = -\frac{r^2+1}{2r}$ so $-\gamma>1$ for all positive $r$. So $-\gamma+\sqrt{\gamma^2-1}>-\gamma>1$ so that root cannot possibly lie in $|z|<1$. Are you sure you do not have a sign error somewhere between lines 3 and 4 of your derivation?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Can two pythagoras triplet have a common number If I have a pythagoras triplet $(a,b,c)$ such that
$$a^2+b^2=c^2$$
then is there another triplet $(a,d,e)$ possible such that
$$a^2+d^2=e^2, \; b\neq d$$
| You can easily find nonprimitive triples that share an element: start with your two favorite triples (say $3^2 + 4^2 = 5^2$ and $5^2 + 12^2 = 13^2$). We'll get triples with $15$ in each: multiply the first triple by $5^2$ and the second by $3^2$. Then we have $5^2(3^2 + 4^2) = 5^2\cdot5^2$ and $3^2(5^2 + 12^2) = 3^2\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1006351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Proving $\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\ldots\left(1+\frac{1}{n^3}\right)<3$ for all positive integers $n$
Prove that $\left(1+\dfrac{1}{1^3}\right)\left(1+\dfrac{1}{2^3}\right)\ldots\left(1+\dfrac{1}{n^3}\right)<3$ for all positive integers $n$
This problem is copied from Math Olympiad Tr... | Since:
$$ 1+\frac{1}{k^3}=\left(1+\frac{1}{k}\right)\left(1-\frac{1}{k}+\frac{1}{k^2}\right) = \left(1-\frac{1}{k^2}\right)\left(1+\frac{1}{k(k-1)}\right) $$
and:
$$ \prod_{k=2}^{+\infty}\left(1-\frac{1}{k^2}\right)=\frac{1}{2} $$
we have:
$$ \prod_{k=1}^{+\infty}\left(1+\frac{1}{k^3}\right)=\prod_{k=2}^{+\infty}\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 1
} |
Derive an explicit formula for a power series $\sum_{n=1}^\infty n^2x^n$ Could anyone help me find an explicit formula for:
$$
\sum_{n=1}^\infty n^2x^n
$$
We're supposed to use: $$\sum_{n=1}^\infty nx^n = \frac{x}{(1-x)^2} \qquad |x| <1 $$
| What if we didn't know
$$ \sum_{n=1}^\infty n x^n = \frac{x}{(1-x)^2}, |x| < 1 \text{?} $$
Note that $n^2 = (n+2)(n+1)-3(n+1) + 1$. (Where did this come from? We'll get to that.) So
$$ S(z) = \sum_{n=1}^\infty \left( (n+2)(n+1)-3(n+1) + 1 \right) z^n \text{.} $$
Assuming each of these converges on the same i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1010403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Number of real roots of the equation $(6-x)^4+(8-x)^4 = 16$
Find number of real roots of the equation $(6-x)^4+(8-x)^4 = 16$
$\bf{My\; Try::}$ Let $f(x) = (6-x)^4+(8-x)^4\;,$ and we have to find real values of $x$ for
which $f(x) = 16$. Now we will form Different cases.
$\bf{\bullet \; }$ If $x<6$ or $x>8\;,$ Then $... | I guess by Algebraic you mean this way
$$(6-x)^4+(8-x)^4 = 16$$
put $7-x=t$
(reason behind this substitution is not just because the later is easy to expand using binomial formula but the magic it does).
$$(t-1)^4+(t+1)^4 = 16$$
$$(t^4-4t^3+6t^2-4t+1)+(t^4+4t^3+6t^2+4t+1)=16$$
$$2(t^4+6t^2+1)=16$$
$$t^4+6t^2+1=8$$
Well... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1010622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Simplifying the derivative of $f(x)= \frac{e^x - e^{-x}}{e^x+e^{-x}}$ I was having some trouble on simplifying the derivative because I didn't know if it's correct. The original function is $$f(x)= \frac{e^x - e^{-x}}{e^x+e^{-x}}$$ What would the simplified derivative be with no negative exponents?
So far I got
$$f'(... | What you have thus far is correct.
Observe that the numerator of
$$f'(x) = \frac{(e^x + e^{-x})^2 - (e^x - e^{-x})^2}{(e^x + e^{-x})^2}$$
is a difference of squares. Thus,
\begin{align*}
f'(x) & = \frac{[(e^x + e^{-x}) + (e^x - e^{-x})][(e^x + e^{-x}) - (e^x - e^{-x})]}{(e^x + e^{-x})^2}\\
& = \frac{(2e^x)(2e^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Solving trigonometric system of equations What are the solutions for this system of equations when $\alpha \in \mathbb{R}$ is considered a constant and $0 \leq x < 2\pi$.
$$ I) \ (y - \cos x)\sin x + (\alpha - \sin x) (-\cos x) = 0$$
$$ II) \ (y - \cos x) = 0$$
My attempt:
From II) we know that $y = \cos x$.
If we appl... | $$ \begin{cases}
(y−\cos{x})\sin{x}+(α−\sin{x})(−\cos{x}) = 0 \\
y − \cos{x} = 0
\end{cases} \tag{1} $$
First, substitute all instances of $(y−\cos{x})$ with $0$.
$$ \begin{cases}
\color{red}{(0)}\sin{x}+(α−\sin{x})(−\cos{x}) = 0 \\
\color{red}{(0)} = 0
\end{cases} \tag{2} $$
Now, we are left with one equation.
$$ (α−... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Give a combinatorial argument Give a combinatorial argument to show that $$\binom{6}{1} + 2 \binom{6}{2} + 3\binom{6}{3} + 4 \binom{6}{4} + 5 \binom{6}{5} + 6 \binom{6}{6} = 6\cdot2^5$$
Not quite where to starting proving this one. Thanks!
| For whatever it's worth, here's a standard probabilistic argument: The number of ways to get $k$ heads in six trials when tossing a coin is $\dbinom 6 k$. If all $2^6=64$ sequences are equally likely (as they are when the coin is "fair", i.e. gives heads and tails equally often) then the probability of exactly $k$ su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
$a^2+ab+b^2 \equiv 0 \pmod n$ if and only if $ a\equiv b\equiv 0 \pmod n$ Let $n$ be a prime number with $n \equiv -1 \mod 6$ and $a,b$ be positive integers. I want to prove:
$$a^2+ab+b^2 \equiv 0 \mod n \iff a\equiv b\equiv 0 \mod n$$
| $a^2+ab+b^2\equiv0 \pmod n\to a^3-b^3\equiv0 \pmod n\to (ab')^3\equiv1\pmod n\to 3\mid n-1\; or\; a\equiv b \pmod 3$ where $b'$ is the inverse of $b$ $\pmod n$. Because $3\nmid n-1$ we have $a\equiv b \pmod n$ so we have $n\mid 3a^2$ we have $(n,3)=1$ so we have $n\mid a^2$ and $a\equiv 0 \pmod n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Calculate the area between the function $\sqrt{x}$ and $x^2$ in the interval $[0 , 2]$. Calculate the area between the function
$\sqrt{x}
$and
$x^2$
in the interval [0 , 2].
Ok, $0<x<1$, $\sqrt{x}$$>x^2$
and $1<x<2$, $\sqrt{x}$$<x^2$
$$ Area = \int_{0}^1(\sqrt{x}-x^2)dx + \int_{1}^2(x^2-\sqrt{x})dx$$
I know that $$... | To begin to answer your question one can first check have a graph of the equations handy, or understand that $y=\sqrt{x}$ raises faster than $y=x^2$ in the interval from $[0,1]$, after that $y=x^2$ is above the $y=\sqrt{x}$ from $[1,2]$. So know to setup the integrals:
\begin{align*}A_{\mathrm{tot}}&=A_1+A_2\\A_1&=\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the probability of two out of three events happening? All events are independent.
$$\Pr(A) = \frac{9}{10}$$
$$\Pr(B) = \frac{9}{10}$$
$$\Pr(C) = \frac{6}{10}$$
What is the probability of at least two events happening?
I'd like to use negation, to negate the possibility that event no event happen plus the probab... | Can confirm that is the right answer, calculated it directly and obtained the same number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1017787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Find the maximum value of the expression Let a,b,c,d be positive real numbers. Show that
$$\frac{(ab+bc+ca)}{(a^3+b^3+c^3)}+\frac{(ab+bd+da)}{(a^3+b^3+d^3)}+\frac{(ac+cd+da)}{(a^3+c^3+d^3)}+\frac{ (bc+cd+db)}{(b^3+c^3+d^3)} \le \frac{(a^2+b^2)}{(ab)^{3/2}} + \frac{(c^2+d^2)}{(cd)^{3/2}} +\frac{(a^2+c^2)}{(ac)^{3/2}} +... | Note that $\displaystyle \frac{a^2+b^2}{(ab)^{3/2}} \ge 2\frac{a^2+b^2}{a^3+b^3}$ by AM-GM. Using that, we get
$$\frac{a^2+b^2}{(ab)^{3/2}}+\frac{b^2+c^2}{(b c)^{3/2}}+\frac{c^2+a^2}{(c a)^{3/2}} \ge 2\left(\frac{a^2+b^2}{a^3+b^3} + \frac{b^2+c^2}{b^3+c^3} + \frac{c^2+a^2}{c^3+a^3} \right) \tag{1}$$
Using Cauchy-Schw... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate $\lim\limits_{x\to 0} \frac{\sin x - x + x^3/6}{x^3}$ I'm unsure as to how to evaluate:
$$\lim\limits_{x\to 0} \frac{\sin x - x + \frac{x^3}{6}}{x^3}$$
The $\lim\limits_{x\to 0}$ of both the numerator and denominator equal $0$. Taking the derivative of both ends of the fraction we get:
$$\lim\limits_{x... | Using L'Hospital twice,
$$
\lim_{x\to 0} \frac{\sin x - x + \frac{x^3}{6}}{x^3}=\lim_{x\to 0} \frac{x^2 + 2\cos x -2}{6x^2}=\lim_{x\to 0} \frac{x - \sin x }{6x}=\frac{1}{6}\lim_{x\to 0} \left(1-\frac{\sin x}{x}\right)=0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
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Manipulating roots of a cubic Given that $A,B,C$ are the roots of the equation $x^3-5x^2+x+1$, how do I find the value of $$\dfrac{A}{B+C}+\dfrac{B}{A+C}+\dfrac{C}{A+B}$$ I know the Vieta's formulas but I am not able to manipulate the above expression into something known. And taking the LCM doesn't help. Please help m... | By Vieta's, you know that $A+B+C = 5$. Therefore, we have
$\dfrac{A}{B+C}+\dfrac{B}{A+C}+\dfrac{C}{A+B} = \dfrac{A}{5-A}+\dfrac{B}{5-B}+\dfrac{C}{5-C}$
Also, you perform the following manipulation:
$x^3-5x^2+x+1 = 0$
$5x^2-x^3 = x+1$
$\dfrac{5-x}{x} = \dfrac{x+1}{x^3}$
$\dfrac{x}{5-x} = \dfrac{x^3}{x+1}$
$\dfrac{x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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can have solution of $x^4-3x^3+2x^2-3x+1=0$ using only high school methods can have solution of $x^4-3x^3+2x^2-3x+1=0$ using only high school methods???
i only know quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ i tried many algebraic manipulations and i get $(x^2+1)^2=3(x^3+x)$, so can we have solution ... | Hint: $$\color{#C00}{x^4}\color{darkmagenta}{-3x^3}\color{royalblue}{+2x^2}\color{darkmagenta}{-3x}\color{#C00}{+1},\tag1$$ is a symmetric polynomial of degree $4$. I will describe here the basic procedure to solve such polynomial equations. Plugging in $x=0$ yields $1$, thus $0$ isn't a root and therefore $x\neq0$ whi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Evaluating $\int e^{x\sin x+\cos x}\left(\frac{x^4\cos^3 x-x\sin x+\cos x}{x^2\cos^2 x}\right)dx$ Evaluate
$$\displaystyle \int e^{x\sin x+\cos x}\left(\frac{x^4\cos^3 x-x\sin x+\cos x}{x^2\cos^2 x}\right)dx$$
$\bf{My\; Try::}$
Let $$\begin{align}I &= \int e^{x\sin x+\cos x}\left(\frac{x^4\cos^3 x-x\sin x+\cos x}{x... | Let $$\begin{align}I&= \int e^{x\sin x+\cos x}\left[\frac{x^4\cos^3 x-x\sin x+\cos x}{x^2\cos^2 x}\right]\mathrm d x\\
&= \int e^{x\sin x+\cos x}\left[\frac{\cos x - x \sin x + x^4 \cos^3 x}{x^2\cos^2x}\right]\mathrm d x\tag{1}\\
&=\int e^{x\sin x+\cos x} \left[ \frac{1}{x^2 \cos x} -\frac{\sin x}{x\cos ^2x}+x^2\cos x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1024099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Reduce $\frac{dy}{dx}=\frac{4x-y+7}{2x+y-1}$ to a homogenous equation by substituting $x=X-1$ and $y=Y+3$
By substituting $x=X-1$ and $y=Y+3$ reduce the differential equation
$$
\frac{dy}{dx}=\frac{4x-y+7}{2x+y-1}
$$
to a homogeneous equation and hence find the general solution in terms of $x$ and $y$.
All I have... | Notice that $$\frac{dY}{dX}=\frac{4X-Y}{2X+Y}= \frac{4 -\frac{Y}{X}}{2 + \frac{Y}{X}}.$$
So, let $V= Y/X$, so $Y=VX\implies dY/dX = V + X(dV/dX)$. With these, we substitute:
$$V + X\frac{dV}{dX} = \frac{4 -V}{2 +V}\implies \frac{2+V}{4-3V-V^2}dV = \frac{dX}{X}.$$
This can be broken up using partial fractions. After we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1024522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Optimization problem - Trapezoid under a parabola recently I've been working on a problem from a textbook about Optimization. The result that I get is $k = 8$, even thought the answer from the textbook is $k = \frac{32}{3}$
The problem follows:
--
The x axis interepts the parabola $12-3x^2$ at the points $A$ and $B$, a... | The formula for the area should read
$$A_{T} = \frac{(B+b) \cdot h}{2} = \frac{\left(4+2 \cdot \sqrt{\frac{12-k}{3}}\right) \cdot k}{2}=\left(2+\sqrt{\frac{12-k}{3}}\right) \cdot k.$$
Now do the same you have done with this function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1025897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Proving a Binomial Sum How do I prove that $$\sum_{r=0}^{n-1}\left[ r \binom{n}{r} \binom{n}{r+1}\right]=n \binom{2n-1}{n-2}$$ without induction?
I've tried manipulating $(1+x)^n$ and the binomial coefficients, but to no avail. Also, how can one explain this identity combinatorially?
| By way of enrichment here is another algebraic proof using basic
complex variables.
We seek to compute
$$\sum_{r=0}^n r {n\choose r} {n\choose r+1}.$$
Introduce the integral representation
$${n \choose r+1}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{r+2}} \; dz.$$
We use this to obtain an int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Integration of $1/(x^4 \sin x +2x)$ $$\int\frac{1}{x^4 \sin x +2x} dx\ $$ How to evaluate this integral. How to go about evaluating these integrals?
| Hint:
$\int\dfrac{1}{x^4\sin x+2x}dx$
$=\int\dfrac{1}{2x\left(1+\dfrac{x^3\sin x}{2}\right)}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{3n-1}\sin^nx}{2^{n+1}}dx$
$=\int\left(\dfrac{1}{2x}+\sum\limits_{n=1}^\infty\dfrac{(-1)^nx^{3n-1}\sin^nx}{2^{n+1}}\right)dx$
But this approach is only suitable for the very limit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1028766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Area of $\triangle ABC$ whose sides $a,b,c$ satisfy $0\leq a \leq1;1\leq b \leq2;2\leq c \leq3$ is The maximum area of a triangle whose sides $a,b,c$ satisfy $0\leq a \leq1;1\leq b \leq2;2\leq c \leq3$ is
$\bf{My\; Try::}$ Area of $\displaystyle \triangle ABC = \frac{1}{2}ab\sin C = \frac{1}{2}ab\cdot \sqrt{1-\cos^2 C}... | As the above user said a better approach would be to look at the area of a triangle as $\frac12 ab \sin{x}$, where $a$ and $b$ are the lengths of two adjacent sides and $x$ is the measure of the angle between these two consecutive sides. The maximum value of $\sin{x}$ occurs when $x=\pi/2$ which gives a value of 1. We ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Improper Integral of $\int\frac{dx}{(2x-1)^3}$ Improper Integral of
$$\int_{-\infty}^0\frac{dx}{(2x-1)^3}$$
from Anton Calculus 8th Edition, page 576, question 9. Answer is $-\frac{1}{4}$ but I'm finding $-1$
The integral, substituting $u= (2x-1)$:
$$\frac{1}{2}\cdot\frac{-2}{(2x-1)^2}+C$$
Definite solution to use ... | You made correct substitution
$$u= 2x-1\iff du=2dx$$
$$\int\frac{dx}{(2x-1)^3}dx=\frac12\int\frac{du}{u^3}dx=-\frac{1}{4u^2}+C$$
$$\int\frac{dx}{(2x-1)^3}=-\frac{1}{4(2x-1)^2}+C$$
I hope you can take it from here!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Does $x,y,z>0$ and $x+y+z=1$ imply $\left(1+\frac 1x\right)\left(1+\frac 1y \right)\left(1+\frac 1z \right)\ge 64$? If $x,y,z$ are positive real numbers such that $x+y+z=1$ then is it true that
$\left(1+\dfrac 1x\right)\left(1+\dfrac 1y \right)\left(1+\dfrac 1z \right)\ge 64$ ?
| By AM-GM, we have
$$1+x=\frac{1}3+\frac13+\frac13+x\ge 4\sqrt[4]{\frac{x}{3^3}}.$$
So
$$(1+x)(1+y)(1+z)\ge 4^3\sqrt[4]{\frac{xyz}{3^9}}.$$
We then need only show that
$$\sqrt[4]{\frac{xyz}{3^9}}\ge xyz,$$
or that
$$1\ge 3^3xyz.$$
The last inequality is clear thanks to AM-GM.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1034802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Maximisation problem I am trying the following question:
If$$a+b+c+d=0,\;a^2+b^2+c^2+d^2=1$$ Then what is the maximum value of
$ab+bc+cd+da?$
By the rearrangement inequality I can get $ab+bc+cd+da\leq 1$ but I am having trouble bringing the first condition into this. I have tried squaring it but that does not look ... | $ab+bc+cd+da = (a+c)(b+d) = -(a+c)^2 \leq 0$, and $\text{max} = 0$ when $a = -c$,
and so $b = -d$. Thus $a^2+b^2+c^2+d^2 = 2(a^2+b^2) = 1 \to a^2+b^2 = \dfrac{1}{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Evaluation of $\lim_{x\rightarrow \infty}\sqrt{x^2+x+1} -\lfloor \sqrt{x^2+x+1 }\rfloor\;,$ Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\sqrt{x^2+x+1} -\lfloor \sqrt{x^2+x+1 }\rfloor\;,$ where $\lfloor x \rfloor$ represent floor function of $x$.
$\bf{My\; Try}::$
$\bullet\; $If $x\in \mathbb{Z}\;,$ and $x\r... | As you mention, when $x$ is a large integer, the difference is roughly $1/2$. On the other hand, when $\sqrt{x^2+x+1}$ is an integer, which happens for unboundedly large $x$, the difference is $0$. This gives two different sequences tending to infinity with two different limits of your function $\sqrt{x^2+x+1}-\lfloor\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1036936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Can -3 and 2 be eigenvalues of the following matrix? Can $-3$ and $2$ be eigenvalues of and nxn matrix B such that $A = B^{2}+B-6I$ and A's determinant is $0$?
So this is what I concluded:
At first glance, it can be seen that the matrix $A$ can be factored into two different terms.
\begin{align*} A = B^{2}+B-6I = (B+3I... | Your reasoning starts out fine, up to "In turn, this suggests that either
$$
|C|,|D|, \textrm{ or } |C| \textrm{ and } |D| = 0.''
$$
After that, I'm not sure what you're trying to do, but it is something circular/unnecessary.
To start over from your last correct assertion, you now know that
$$
\det[(B + 3I)(B-2I)] = 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1037001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Inequality: $(a^3+3b^2+5)(b^3+3c^2+5)(c^3+3a^2+5) \ge 27(a+b+c)^3$ Proving inequality for positive real $a,b,c > 0$: $$ (a^3+3b^2+5)(b^3+3c^2+5)(c^3+3a^2+5) \ge 27(a+b+c)^3$$
| Use AM-GM inequality we have $$a^3+2=a^3+1+1\ge 3a$$
so
$$a^3+3b^2+5\ge 3a+3b^2+3=3(a+b^2+1)$$
so
$$LHS\ge 27(a+b^2+1)(1+b+c^2)(a^2+1+c)$$
Use Holder inequality we have
$$(a+b^2+1)(1+b+c^2)(a^2+1+c)\ge (a+b+c)^3$$
By done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1039510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Proving the sum of the reciprocals squared converges I'm investigating the Basel Problem, and the sum to consider is:
$\dfrac{1}{1^2}+\dfrac{1}{2^2}+...+\dfrac{1}{n^2}$
How can I show this converges? Using graphs/computer software is also fine, but how would I do it? Is there a way using calculus?
I've looked on Google... | Note that
$$
\begin{alignat}{8}
&\frac{1}{1^2}+&&\frac{1}{2^2}+&&\frac{1}{3^2}+&&\frac{1}{4^2}+&&\frac{1}{5^2}+&&\frac{1}{6^2}+&&\frac{1}{7^2}+&&\dots\\
\leq &\frac{1}{1^2}+&&\frac{1}{2^2}+&&\frac{1}{2^2}+&&\frac{1}{4^2}+&&\frac{1}{4^2}+&&\frac{1}{4^2}+&&\frac{1}{4^2}+&&\dots
\end{alignat}
$$
where there are $2^n$ copi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1039699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Compute $\lim\limits_{n \to \infty }\frac{981}{n+5}\sum_{i=1}^{n} \left (\frac{i^2}{n^2} \right)$ Compute the given limit
$$
\lim_{n \to \infty }\frac{981}{n+5}\sum_{i=1}^{n} \left (\frac{i^2}{n^2} \right)
$$
The sum is:
Can someone please show me the steps to complete this problem? The answer I arrived at was 0 but ... | Hint1:
$$\begin{split}
\lim_{n \to \infty }\frac{981}{n+5}\sum_{i=1}^{n} \left(\frac{i^2}{n^2} \right)
&= \lim_{n \to \infty} \frac{981}{n+5} \cdot \frac{\sum_{i=1}^{n}i^2}{n^2}\\
\end{split}$$
Hint2:
$$ \sum_{i=1}^{n}i^2 = \frac{n\cdot (n+1) \cdot 2n+1)}{6}$$
$$\begin{split}
\lim_{n \to \infty }\frac{981}{n+5}\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Why is $(1-\cot 37^\circ)(1-\cot 8^\circ)=2.00000000\cdots$? Apparently,
$$(1-\cot 37^\circ)(1-\cot 8^\circ)=2.00000000000000000\cdots$$
Since it is a $2.0000000000\cdots$ instead of $2$, it isn't exactly $2$.
Why is that?
| Alternatively,
$$\cot(x+y) = \frac{\cos(x+y)}{\sin(x+y)} = \frac{\cos x \cos y - \sin x \sin y}{\sin x \cos y + \cos x \sin y} = \frac{\cot x \cot y - 1}{\cot x + \cot y}$$
Now if $x = 37^\circ$, $y = 8^\circ$, then $\cot(x + y) = \cot 45^\circ = 1$ and
$$\begin{align} \ \ \ \ 1 & = \frac{\cot x \cot y - 1}{\cot x + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Evaluating $\int [(5x^3-3x^2+7x-3)/(x^2+1)^2] dx$ Can I get hints on how to factor?
Should I break down by term, i.e. $$\frac{5x^3}{(x^2+1)^2}\text{...?}$$
| Since the degree of the numerator is less than the degree of the denominator, you can immediately apply the method of partial fractions to decompose the integrand: You're looking for numbers $A, B, C, D$ such that,
$$\frac{5x^3 - 3x^2 + 7x - 3}{(x^2 + 1)^2} = \frac{Ax + B}{(x^2 + 1)^2} + \frac{Cx + D}{x^2 + 1}$$
which ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1043735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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.