Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Subsitution - integrals I am attempting to solve this integral using substituion
$\int (x^2 +1) (x^3 +3x)^4dx$ I make $u=x^3+3x$ and then made $dx=du/(3x^2 + 3)$
I then got $1/3 \int (x^3+3x)^4$ I have no idea what to do now.
| Your substitution $u=x^3+3x$ is a good idea as it yields $\frac{du}{dx}=3x^2 + 3$
Substituting in you get:
$\int (x^2 +1) (x^3 +3x)^4dx$
$= \frac{1}{3} \int (3x^2 + 3)(x^3 + 3x)^4 dx$
$= \frac{1}{3} \int u^4 du$ (this is where you seemed to have gone wrong by converting to du yet leaving the integral in terms of x)
$= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/80994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is there an easy way to determine when this fractional expression is an integer? For $x,y\in \mathbb{Z}^+,$ when is the following expression an integer?
$$z=\frac{(1-x)-(1+x)y}{(1+x)+(1-x)y}$$
The associated Diophantine equation is symmetric in $x, y, z$, but I couldn't do anything more with that. I tried several fact... | We write out a solution, to check whether something like it can be "borrowed" for a problem set.
The numerator is $1-x-y-xy$, which can be written as $-[(x+1)(y+1)-2]$. The denominator is $-[(x-1)(y-1)-2]$. So our fraction is
$$\frac{(x+1)(y+1)-2}{(x-1)(y-1)-2}.$$
We can take $x=1$ or $y=1$. The ratio is then an integ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/82087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solving $A(x) = 2A(x/2) + x^2$ Using Generating Functions Suppose I have the recurrence:
$$A(x) = 2A(x/2) + x^2$$ with $A(1) = 1$.
Is it possible to derive a function using Generating Functions? I know in Generatingfunctionology they shows show to solve for recurrences like $A(x) = 2A(x-1) + x$. But is it possible to s... | As per Qiaochu's comment on the answer so far, consider
$$a_n = 2 a_{\lfloor n/2 \rfloor} + n^2$$
with $a_1 = 1$ and $a_n = 1, 6, 11, 28, 37, 58, 71, 120,\dots$ for $n = 1,2,\dots$. Then
$$a_{2n} = 2 a_n + 4n^2 \quad\quad\text{and}\quad\quad a_{2n+1} = 2 a_n + 4n^2 + 4n + 1$$
where both recurrences are valid for $n\ge ... | {
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"url": "https://math.stackexchange.com/questions/85415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Trick to find multiples mentally We all know how to recognize numbers that are multiple of $2, 3, 4, 5$ (and other). Some other divisors are a bit more difficult to spot. I am thinking about $7$.
A few months ago, I heard a simple and elegant way to find multiples of $7$:
Cut the digits into pairs from the end, multipl... | I will try to explain a general rule using modular congruence. We can show that any integer in base $10$ can be written as $$ z = a_0 + a_1 \times 10 + a_2 \times 10^2 + a_3 \times10^3 + \cdots + a_n \times 10^n$$
Lets say we have to find a divisibility rule of $7$,Hence for congruence modulo $7$ have,
$$ 10 \equiv 3,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove that $2 \arctan\sqrt{x} = \arcsin \frac{x-1}{x+1} + \frac{\pi}{2}$ I want to prove that$$2 \arctan\sqrt{x} = \arcsin \frac{x-1}{x+1} + \frac{\pi}{2}, x\geq 0$$
I have started from the arcsin part and I tried to end to the arctan one but I failed.
Can anyone help me solve it?
| It's kind of fun to "unveil" these mysterious formulas using only trigonometry, by which they all appear to be very simple angle relationships.
For $x>1$ consider the Figure below.
$\hskip1.5in$
Start with a right-angled triangle with hypothenuse $\overline{AC} = x+1$ and side $\overline{BC}=x-1$, so that
$$\alpha = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/89759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
} |
Need help finding limit $\lim \limits_{x\to \infty}\left(\frac{x}{x-1}\right)^{2x+1}$ Facing difficulty finding limit
$$\lim \limits_{x\to \infty}\left(\frac{x}{x-1}\right)^{2x+1}$$
For starters I have trouble simplifying it
Which method would help in finding this limit?
| If you know that
$$\lim_{x\to\infty}\left(1 + \frac{a}{x}\right)^x = e^{a},$$
so that
$$\lim_{x\to\infty}\left(1 - \frac{1}{x}\right)^x = e^{-1},$$
then you can try to rewrite your limit into something involving this limit.
So try rewriting it; perhaps as a product,
$$\begin{align*}
\left(\frac{x}{x-1}\right)^{2x+1} &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/90324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Simplifying trig expression I was working through some trig exercises when I stumbled upon the following problem:
Prove that: $ \cos(A+B) \cdot \cos(A-B)=\cos^2A- \sin^2B$.
I started out by expanding it such that
$$ \cos(A+B) \cdot \cos(A-B)=(\cos A \cos B-\sin A \sin B) \cdot (\cos A \cos B+ \sin A \sin B),$$
which s... | Here is a detailed answer.Let's rock!
$$
\require{cancel}\begin{align}
\cos\left(A-B\right)\cdot\cos\left(A+B\right)&=\left(\cos A\cos B-\sin A\sin B\right)\left(\cos A\cos B+\sin A\sin B\right)\\
&=\cos^2A\cos^2B-\sin^2A\sin^2B\\
&=\cos^2A\left(1-\sin^2B\right)-\sin^2A\sin^2B\\
&=\cos^2A-\sin^2B\cos^2A-\sin^2A\sin^2B\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/92159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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How many ordered triple $ (p,a,b) $ is possible such that $p^a=b^4+4$? If we have a prime number $p$ and two natural numbers $a$ and $b$ such that $p^a=b^4+4$,
then how many such ordered triplets $(p,a,b)$ exist?
What should be the strategy to solve this one? The only I can see is $(5,1,1)$, is this the only one? if ye... | This is an old contest problem, I wish I could remember where I first saw it. Anyway, André's comment that $$b^4+4=(b^2-2b+2)(b^2+2b+2)$$ is the key to a solution.
Looking modulo $16$, we see that $b^4+4$ cannot be a power of $2$. For $b>1 $, both factors will be strictly greater then $1$, so that if $p^k|(b^4+4)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/93746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solving quadratic equation $$\frac{1}{x^2} - 1 = \frac{1}{x} -1$$
Rearranging it I get: $1-x^2=x-x^2$, and so $x=1$. But the question Im doing says to find 2 solutions. How would I find the 2nd solution?
Thanks.
| $$\frac{1}{x^2} - 1 = \frac{1}{x} -1$$
$$\frac{1}{x^2} = \frac{1}{x} $$
$$x^2-x=0$$
$$x(x-1)=0$$
The solutions are $x=1$ or $x=0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is $\sin^3 x=\frac{3}{4}\sin x - \frac{1}{4}\sin 3x$? $$\sin^3 x=\frac{3}{4}\sin x - \frac{1}{4}\sin 3x$$
Is there any formula that tells this or why is it like that?
| \begin{align}
\sin(3x) &= \sin(x+2x) \tag{1} \\
\sin(\alpha+\beta) &= \sin \alpha \cdot \cos \beta + \cos \alpha \cdot \sin \beta \tag{2} \\
\sin 2\alpha &= 2\cdot \sin \alpha \cdot \cos \alpha \tag{3} \\
\cos 2\alpha &= \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/97654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 5
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Solutions of some Diophantine equations Respected Mathematicians,
The Diophantine equation
$$2^x + 5^y = z^2$$
has solutions $$x = 3, y = 0, z = 3$$ and $$x = 2, y = 1, z = 3$$ I got these solutions by trial and error method. To be honest, these solutions are below the number $5$. So, I easily verified them by trial ... | I'll do a piece of it to show you some methods you can try on the other pieces.
$2^x+5^y=z^2$. Let's do the case where $y=2s$ is even.
$2^x=z^2-(5^s)^2=(z+5^s)(z-5^s)$, so $z+5^s=2^m$ and $z-5^s=2^n$ with $m+n=x$. Eliminating $z$, $2\times5^s=2^m-2^n$, so $5^s=2^{m-1}-2^{n-1}$. The left side is odd, so the right side... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/100709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Compute: $\int_{0}^{1}\frac{x^4+1}{x^6+1} dx$ I'm trying to compute: $$\int_{0}^{1}\frac{x^4+1}{x^6+1}dx.$$
I tried to change $x^4$ into $t^2$ or $t$, but it didn't work for me.
Any suggestions?
Thanks!
| Edited Here is a much simpler version of the previous answer.
$$\int_0^1 \frac{x^4+1}{x^6+1}dx =\int_0^1 \frac{x^4-x^2+1}{x^6+1}dx+ \int_0^1 \frac{x^2}{x^6+1}dx$$
After canceling the first fraction, and subbing $y=x^3$ in the second we get:
$$\int_0^1 \frac{x^4+1}{x^6+1}dx =\int_0^1 \frac{1}{x^2+1}dx+ \frac{1}{3}\int_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/101049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 3
} |
Find $DF$ in a triangle $DEF$ Consider we have a triangle $ABC$ where there are three points $D$, $E$ & $F$ such as point $D$ lies on the segment $AE$, point $E$ lies on $BF$, point $F$ lies on $CD$. We also know that center of a circle over ABC is also a center of a circle inside $DEF$. $DFE$ angle is $90^\circ$, $DE/... | I refer to the diagram of Victor Liu's answer. This is an analytical
verification that $DF=8$ (which means that $u=2$) but omits some details. Using the equation of the circle centered at $(0,0)$ with radius $14$, the equation of $AE$ (tangent to the small circle at the point $(x,y)=(-8/5,6/5)$)
$$
y=\frac{4}{3}\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/102406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to analyze (sum and convergence) step-by-step this series: $\sum_{n=1}^\infty\frac{4}{(n+1)(n+2)}$? I have this series:
$$\sum_{n=1}^\infty\frac{4}{(n+1)(n+2)}$$
and $$\sum_{n=1}^\infty\frac{4}{(n+1)(n+3)}$$
and I would like to know how to analyze, very step-by-step.
thanks
| One way to show convergence of the first series, is to first note that
$$ \frac{4}{(n+1)(n+2)} < \frac{4}{(n+1)^2}$$ for all $n$. Then write
$$ \sum_{n = 1}^{\infty} \frac{4}{(n+1)^2} = 4\sum_{n=2}^{\infty} \frac{1}{n^2}. $$ Presumably, you already know that the latter series converges, and so you can apply the compar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/104964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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show if $n=4k+3$ is a prime and ${a^2+b^2} \equiv 0 \pmod n$ , then $a \equiv b \equiv 0 \pmod n$ $n = 4k + 3 $
We start by letting $a \not\equiv 0\pmod n$ $\Rightarrow$ $a \equiv k\pmod n$ .
$\Rightarrow$ $a^{4k+2} \equiv 1\pmod n$
Now, I know that the contradiction will arrive from the fact that if we can show $a^2 \... | Assume that $a$ and $b$ are coprime. If $a$ and $b$ are odd, replace
$a$ and $ b$ by $(a-b)/2$ and $(a+b)/2$. Then $a^2 + b^2 = pm$, and by reducing $a$ and $b$
modulo $p$ you can make sure that $m < p$. Since $p = 4n+3$ and $a^2 + b^2 = 4k+1$,
we must have $m = 4j+3$. But then $m$ is divisible by a prime number $q = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/105034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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roots of complex polynomial - tricks What tricks are there for calculating the roots of complex polynomials like
$$p(t) = (t+1)^6 - (t-1)^6$$
$t = 1$ is not a root. Therefore we can divide by $(t-1)^6$. We then get
$$\left( \frac{t+1}{t-1} \right)^6 = 1$$
Let $\omega = \frac{t+1}{t-1}$ then we get $\omega^6=1$ which b... | Note that
$$(t+1)^6 - (t-1)^6=((t+1)^3-(t-1)^3)((t+1)^3+(t-1)^3)$$
(difference of squares).
When you simplify the first term in the product on the right, there is no $t^3$ term and no $t$ term! The second term in the product simplifies to $2t^3+6t$.
Remark: The solution by Arhabhata is the right one, it works if we re... | {
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"url": "https://math.stackexchange.com/questions/105129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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How to integrate $1-\tan^4(x)$ without using secant? How to integrate such function:
$\int_{-\pi/3}^{\pi/3}1-\tan^4(x)$
I already found a solution using the trigonometric function secant. It looks like this:
$$u=\tan(x),\quad \frac{du}{dx}=\sec^2(x)$$
$$
\begin{align}
\int 1-\tan^4(x) & = \int 1 \;dx - \int\tan^4(x) ... | Sasha's solution is very elegant. However, if you can't see a trick like this, you can always mechanically integrate any rational function of sine and cosine using the Weierstraß substitution.
With
$$\tan x=\frac{2t}{1-t^2}\;,$$
$$\frac{\mathrm dx}{\mathrm dt}=\frac2{1+t^2}\;,$$
$$t=\tan\frac x2\;,$$
the integral of $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/107195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Confused by Calc II question regarding derivative of rational integrals So here's the question:
If $f$ is a quadratic function such that $f(0) = 1$ and $\int \frac{f(x)}{x^2(x+1)^3}\,dx$ is a rational function, find the value of $f’(0)$.
What I've done so far is try to solve the integral using partial fractions i.e.
... | You have
$f(x)=ax^2+bx +c$. That $f(0)=1$ gives you $c=1$. We have $f'(x)=2ax+b$; and so $f'(0)=b$.
The integrand can be written as
$$
{ax^2+ f'(0)x+1\over x^2(x+1)^3} = {A\over x\vphantom{ )^2}}+{B\over x^2\vphantom{ )^2}}+ {C\over (x+1)\vphantom{ )^2}}+{D\over (x+1)^2}+{E\over (x+1)^3}.
$$
Here's the important ob... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/107829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Showing $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$ Given that n is a positive integer show that $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$.
I'm thinking that I should be using the property of gcd that says if a and b are integers then gcd(a,b) = gcd(a+cb,b). So I can do things like decide that $\gcd(n^3 + 1, n^2 + 2) = ... | As you note, $\gcd(n^3+1,n^2+2) = \gcd(1-2n,n^2+2)$.
Now, continuing in that manner,
$$\begin{align*}
\gcd(1-2n, n^2+2) &= \gcd(2n-1,n^2+2)\\
&= \gcd(2n-1, n^2+2+2n-1)\\
&= \gcd(2n-1,n^2+2n+1)\\
&= \gcd(2n-1,(n+1)^2).
\end{align*}$$
Consider now $\gcd(2n-1,n+1)$. We have:
$$\begin{align*}
\gcd(2n-1,n+1) &= \gc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/109876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Simple algebra. How can you factor out a common factor if it is $\frac{1}{\text{factor}}$ in one of the cases? I'm sure this is simple. I want to pull out a factor as follows...
I have the expression
$$\frac{a(\sqrt{x}) - (b + c)(\frac{1}{\sqrt{x}})c}{x}.$$
It would be useful for me to pull out the $\sqrt{x}$ from the... | In general, if you have an expression $a+\frac{1}{c}b$ and you want to factor out the $\frac{1}{c}$, multiply the expression by $\frac{c}{c}$ like this:
$$
a+\frac{1}{c}b \\
\frac{c}{c}\left(a+\frac{1}{c}b\right) \\
\frac{1}{c}\left(c\cdot a + c\cdot \frac{1}{c}b\right) \\
\frac{1}{c}\left( ca+b \right)
$$
I find ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/110669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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A minimization problem for a function involving maximum Let $a,b,c,d$ be constants in the interval $[-1,1]$. Define $$f(x,y)=\max\{|y-a|,1-b\}+\max\{1-x,1-y\}+\max\{|x-c|,1-d\}$$ for $ -1\le x\le 1, -1\le y\le 1.$
Prove, or disprove, that the minimum value of $f$ is $$\max\{2-b-c,2-a-d, 2-b-d\}.$$
Numerical evidence se... | That is true. Let $A=2-b-c, B=2-a-d, C=2-b-d$. Note that $f(x,y)\ge \max\{A,B,C\}$ for all $(x,y)$ in the domain. To prove that there exist $(x,y)$ in the domain such that $f(x,y)= \max\{A,B,C\}$ consider six cases: $A\le B\le C, A\le C\le B,\cdots$. For example in the case $C\le B\le A$, we have $$a\le b, c-d\le a-b.$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/110808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to get from $a\sqrt{1 + \frac{b^2}{a^2}}$ to $\sqrt{a^2 + b^2}$ I have the following expression: $a\sqrt{1 + \frac{b^2}{a^2}}$. If I plug this into Wolfram Alpha, it tells me that, if $a, b$ are positive, this equals $\sqrt{a^2 + b^2}$.
How do I get that result? I can't see how that could be done. Thanks
| If $a\ge0$,
$$\begin{align}
a\sqrt{1 + \frac{b^2}{a^2}}
&=\sqrt{a^2}\sqrt{1 + \frac{b^2}{a^2}}
\\
&=\sqrt{a^2\left(1 + \frac{b^2}{a^2}\right)}
\\
&=\sqrt{a^2 + b^2}.
\end{align}$$
($\sqrt{a^2}=|a|$ for all $a\in\mathbb{R}$ and $|a|=a$ when $a\ge0$.)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Notation of indexers with multiples in a series If $\sigma _{n}=1+\dfrac {1} {2}+\dfrac {1} {3}+\ldots +\dfrac {1} {n}$ what series is given by $\sigma _{2n}$ ? Does that mean we only take the even terms now or does it mean every term is multiplied by 2 ?
| Since
$$\sigma_{n} = 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1}+\frac{1}{n}$$
is the sum of the reciprocals of $1$ up to $n$, we have that $\sigma_{2n}$ is
$$\sigma_{2n} = 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n-1}+\frac{1}{2n}$$
That is, we sum up to $2n$.
If w... | {
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"url": "https://math.stackexchange.com/questions/115811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $ \sqrt{2-2\cos x}+\sqrt{10-6\cos x}=\sqrt{10-6 \cos 2x} $ $$ \sqrt{2-2\cos x}+\sqrt{10-6\cos x}=\sqrt{10-6 \cos 2x} $$
I tried squaring and/or using $1-\cos x=2\sin^2{\frac{x}2}$, but no luck.
| If $t = \cos(x)$, we have $\sqrt{2-2t} + \sqrt{10-6t} = \sqrt{16-12 t^2}$. Square both sides, isolate the term with square roots, square again, and factor. The result should be equivalent to $(t+1)(t-1)(3t-2)^2=0$. $t=-1$ does not work, but the other factors do.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/117807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Negative fractions - what's the difference? What's the difference between the following fractions:
$ \frac{-4}{-5}$
$ \frac{4}{-5}$
$ \frac{-4}{5}$
$ - \frac{4}{5}$
| Another way to think about this would be in terms of equivalence classes. If you are not familiar with this, it is pretty much how mathematicians say in the rationals that
$$\frac{1}{2} = \frac{2}{4} = \frac{4}{8}.$$
In fact one says that given two fractions
$$x = \frac{m}{n} \hspace{2mm} \text{and} \hspace{2mm} y= \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/118203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Numbers are too large to show $65^{64}+64^{65}$ is not a prime I tried to find cycles of powers, but they are too big. Also $65^{n} \equiv 1(\text{mod}64)$, so I dont know how to use that.
| $$64^{65}+65^{64} = 6^{65}+7^{64} \pmod{29}$$
$65=2 \times 28+9, 64 = 2 \times 28 +8$, and also gcd$(29,36)$ = gcd$(29,49) = 1$
Therefore by Fermat's Little Theorem
If gcd$(a,p)= 1$, and $p$ is a prime then $a^{(p-1)} \hspace{3pt}\equiv \hspace{3pt}1 \pmod{p}$
$36^{29-1} \equiv 1 \pmod{29}, \hspace{5pt}49^{29-1} \equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/119798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 1
} |
Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$ Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$
Use proof by induction. I tried for $n=1$ and got $\frac{27}{9}=3$, but if I assume for $n$ and show it for $n+1$, I don't know what... | Here is a simple "direct proof":
\begin{align*}
10^n+3 \times 4^n + 5&=10^n-1 +3 \times 2^{2n}+6 =9999..9+6 \times [2^{2n-1}+1] \\
&=9999..9+6 \times (2+1)(2^{2n-2}-2^{2n-3}+\cdots-2+1) \\
&= 9 \times [1111...1+2 \times (2^{2n-2}-2^{2n-3}+\cdots-2+1)]
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/120649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 2
} |
Compute the Unit and Class Number of a pure cubic field $\mathbb{Q}(\sqrt[3]{6})$ Find a unit in $\mathbb{Q}(\sqrt[3]{6})$ and show that this field has class number $h=1$.
I am done with the first part which is relatively simple:
Suppose that $\varepsilon$ is a unit in $\mathbb{Q}(\sqrt[3]{6})$. Then we have $\varepsil... | The first idea for computing units in such fields is finding a generator of a purely ramified prime. Here $2 - \sqrt[3]{6}$ has norm $2$, hence
$$ (2 - \sqrt[3]{6})^3 = 2(1 - 6\sqrt[3]{6} + 3\sqrt[3]{6}^2) $$
is $2$ times a unit.
Finding an element generating the prime ideal above $3$ is more difficult, but it turns o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/125291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
Show $ I = \int_0^{\pi} \frac{\mathrm{d}x}{1+\cos^2 x} = \frac{\pi}{\sqrt 2}$ Show $$ I = \int_0^{\pi} \frac{\mathrm{d}x}{1+\cos^2 x} = \frac{\pi}{\sqrt 2}$$
| If we make the standard "Weierstrass" $t=\tan(x/2)$ substitution, we get $\cos t=\frac{1-t^2}{1+t^2}$ and $dx=\frac{2\,dt}{1+t^2}$. We end up quickly with
$$\int_0^\infty \frac{1+t^2}{1+t^4}\,dt.$$
But $1+t^4=(1-\sqrt{2}t+t^2)(1+\sqrt{2}t +t^2)$, so by partial fractions our integrand is
$$\frac{1}{2-2\sqrt{2}t+2t^2} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/125637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Proving an asymptotic lower bound for the integral $\int_{0}^{\infty} \exp\left( - \frac{x^2}{2y^{2r}} - \frac{y^2}{2}\right) \frac{dy}{y^s}$ This is a follow up to the great answer posted to https://math.stackexchange.com/a/125991/7980
Let $ 0 < r < \infty, 0 < s < \infty$ , fix $x > 1$ and consider the integral
$$ I... | Let
$$
\phi_{r,x}(y)=-\frac{x^2}{2y^{2r}}-\frac{y^2}{2}\tag{1}
$$
Taking the first and second derivatives of $\phi_{r,x}(y)$ yields
$$
\phi_{r,x}^\prime(y)=r\frac{x^2}{y^{2r+1}}-y\tag{2}
$$
and
$$
\phi_{r,x}^{\prime\prime}(y)=-(2r+1)r\frac{x^2}{y^{2r+2}}-1\tag{3}
$$
Using $(2)$, $\phi_{r,x}(y)$ reaches a maximum ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/127177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
inequality $(a+c)(a+b+c)<0$, prove $(b-c)^2>4a(a+b+c)$
If $(a+c)(a+b+c)<0,$
prove $$(b-c)^2>4a(a+b+c)$$
I will use the constructor method that want to know can not directly prove it?
| Because
$$(b-c)^2-4a(a+b+c)=(b-c)^2+(4a+8c)(a+b+c)-8(a+c)(a+b+c)=$$
$$=(2a+b+3c)^2-8(a+c)(a+b+c)>0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/128898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
On solvable quintics and septics Here is a nice sufficient (but not necessary) condition on whether a quintic is solvable in radicals or not. Given,
$x^5+10cx^3+10dx^2+5ex+f = 0\tag{1}$
If there is an ordering of its roots such that,
$x_1 x_2 + x_2 x_3 + x_3 x_4 + x_4 x_5 + x_5 x_1 - (x_1 x_3 + x_3 x_5 + x_5 x_2 + x_... | It turns out there is an infinite number of such septics, such as the Hashimoto-Hoshi septic,
$$\small x^7 - (a^3 + a^2 + 5a + 6)x^6 +
3(3a^3 + 3a^2 + 8a + 4)x^5 + (a^7 + a^6 + 9a^5 - 5a^4 - 15a^3 - 22a^2 -
36a - 8)x^4 - a(a^7 + 5a^6 + 12a^5 + 24a^4 - 6a^3 + 2a^2 - 20a - 16)x^3 + a^2(2a^6 + 7a^5 + 19a^4 + 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/129655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
Critical points of $f(x,y)=x^2+xy+y^2+\frac{1}{x}+\frac{1}{y}$ I would like some help finding the critical points of $f(x,y)=x^2+xy+y^2+\frac{1}{x}+\frac{1}{y}$. I tried solving $f_x=0, f_y=0$ (where $f_x, f_y$ are the partial derivatives) but the resulting equation is very complex. The exercise has a hint: think of $f... | Here $f_{x}=2x+y-\frac{1}{x^{2}}$ and $f_{y}=2y+x-\frac{1}{y^{2}}$
For critical point
$$f_{x}=0,\ f_{y}=0$$
$$2x+y-\frac{1}{x^{2}}=0\ ,\ 2y+x-\frac{1}{y^{2}}=0$$
$$2x^{3}+x^{2}y-1=0\ ,\ 2y^{3}+xy^{2}-1=0$$
Substracting this two equations we get,
$$ 2(x^{3}-y^{3})+xy(x-y)=0$$
$$ 2(x-y)(x^{2}+xy+y^{2})+xy(x-y)=0$$
$$(x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/130277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
rotating a matrix Given a rectangular matrix $A$, what is the general form to rotate the matrix about the center term, e.g. such that
$$\pmatrix{a_{0,0} & a_{0,1} & a_{0,2} \\ a_{1,0} & a_{1,1} & a_{1,2} \\ a_{2,0} & a_{2,1} & a_{2,2}}\longrightarrow\pmatrix{a_{0,2} & a_{1,2} & a_{2,2} \\ a_{0,1} & a_{1,1} & a_{2,1} \\... | $$\pmatrix{0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0}A^T = \pmatrix{a_{0,2} & a_{1,2} & a_{2,2} \\ a_{0,1} & a_{1,1} & a_{2,1} \\ a_{0,0} & a_{1,0} & a_{2,0}} =A_r $$
Edit: reverse is,
$$A_r^T \pmatrix{0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0} = \pmatrix{a_{0,0} & a_{0,1} & a_{0,2} \\ a_{1,0} & a_{1,1} & a_{1,2} \\ a_{2,0} & a_{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/133156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Taking the derivative of $y = \dfrac{x}{2} + \dfrac {1}{4} \sin(2x)$ Again a simple problem that I can't seem to get the derivative of
I have $\frac{x}{2} + \frac{1}{4}\sin(2x)$
I am getting $\frac{x^2}{4} + \frac{4\sin(2x)}{16}$
This is all very wrong, and I do not know why.
| Jordan, The derivative of your function is $\frac{1}{2} + \frac{\cos 2x}{2}$. Now note that $\cos 2x = \cos^2 x -\sin ^2 = \cos^2 x -1 -\cos^2 x =2\cos^2x -1$. Rearranging, you get $$\cos^2 x =\frac{\cos 2x}{2} + \frac{1}{2}.$$
$$
\begin{align*}
\cos 2x = \cos(x+x) & =\cos x \cos x -\sin x \sin x \\
& = \cos^2x -\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/134855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
integration of fractions i am trying to integrate following equation
$$ \int\frac 1{(x^2-1)\cdot (x+2)}\,dx$$
i can represent $(x^2-1)=(x-1)(x+1)$ so,it would be converted in the following form
$$\int\frac1{(x^2-1)(x+2)}\,dx=\int \frac1{(x-1)(x+1)(x+2)}\,dx$$
or it is equal $$\int \frac1{(x-1)(x^2+3x+2)}\,dx$$ last ... | I think you can decompose it like this:
$$
\frac{1}{(x^2-1)\cdot(x+2)}=\frac{a}{x-1}+\frac{b}{x+1}+\frac{c}{x+2}
$$
Thus we can solve the following equations:
$$
a+b+c=0\\3a+b=0\\2a-2b-c=1
$$
getting $a=1/6,b=-1/2,c=1/3$.
Therefore,
$$
\int\frac{dx}{(x^2-1)\cdot(x+2)}\\=\int\frac{1}{6}\cdot\frac{dx}{x-1}-\int\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/135155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is so interesting about the zeroes of the Riemann $\zeta$ function? The Riemann $\zeta$ function plays a significant role in number theory and is defined by $$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} \qquad \text{ for } \sigma > 1 \text{ and } s= \sigma + it$$
The Riemann hypothesis asserts that all the non-tri... | the HIlbert Polya operator which would prove Riemann Hypothesis is the Wu-Sprung genralzed model with potential
$$ f^{-1} (x)=\frac{4}{\sqrt{4x+1} } +\frac{1}{4\pi } \int\nolimits_{-\sqrt{x} }^{\sqrt{x}}\frac{dr}{\sqrt{x-r^2} } \left( \frac{\Gamma '}{\Gamma } \left( \frac{1}{4} +\frac{ir}{2} \right) -\ln \pi \right) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/136417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "72",
"answer_count": 5,
"answer_id": 3
} |
Show that $ a，b，c, \sqrt{a}+ \sqrt{b}+\sqrt{c} \in\mathbb Q \implies \sqrt{a},\sqrt{b},\sqrt{c} \in\mathbb Q $ Assume that $a，b，c, \sqrt{a}+ \sqrt{b}+\sqrt{c} \in\mathbb Q$ are rational，prove $\sqrt{a},\sqrt{b},\sqrt{c} \in\mathbb Q$,are rational.
I know that can be proved, would like to know that there is no easier... | Maybe not easier, but quite elegant :
Suppose that $a,b,c$ are all non zero. Let $K=\mathbb{Q}(\sqrt{a},\sqrt{b},\sqrt{c})$ and $n = [K: \mathbb{Q}]$. Then since $Tr_{K/\mathbb{Q}}(\sqrt{a}) = Tr_{\mathbb{Q}(\sqrt{a})/\mathbb{Q}} \circ Tr_{K/\mathbb{Q}(\sqrt{a})} (\sqrt{a})$, we have
$$ Tr_{K/\mathbb{Q}}(\sqrt{a}) = \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/136556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Showing a series is a solution to a differential equation I am attempting to show that the series $y(x)\sum_{n=0}^{\infty} a_{n}x^n$ is a solution to the differential equation $(1-x)^2y''-2y=0$ provided that $(n+2)a_{n+2}-2na_{n+1}+(n-2)a_n=0$
So i have:
$$y=\sum_{n=0}^{\infty} a_{n}x^n$$
$$y'=\sum_{n=0}^{\infty}na_{n}... | You are right till the last step.
You have $$\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n}-2\sum_{n=0}^{\infty}n(n+1)a_{n+1}x^{n}+\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n}-2\sum_{n=0}^{\infty} a_{n}x^n=0$$
which gives us
$$\sum_{n=0}^{\infty} \left((n+2)(n+1) a_{n+2} - 2 n (n+1) a_{n+1} + (n^2-n-2)a_n \right)x^n = 0$$
and not
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/138520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Using modular arithmetic, how can one quickly find the natural number n for which $n^5 = 27^5 + 84^5 + 110^5 + 133^5$? Using modular arithmetic, how can one quickly find the natural number n for which $n^5 = 27^5 + 84^5 + 110^5 + 133^5$?
I tried factoring individual components out, but it seemed really tedious.
| Tabulating the expression with respect to low primes:
$\bmod 2: 27^5 + 84^5 + 110^5 + 133^5 \equiv 1^5 + 0^5 + 0^5 + 1^5 \equiv 0 \implies n\equiv 0$
$\bmod 3: 27^5 + 84^5 + 110^5 + 133^5 \equiv 0^5 + 0^5 + -1^5 + 1^5 \equiv 0 \implies n\equiv 0$
$\bmod 5: 27^5 + 84^5 + 110^5 + 133^5 \equiv 2^5 + (-1)^5 + 0^5 + (-2)^5 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/140659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Determinant of symmetric Matrix with non negative integer element Let \begin{equation*}
M=%
\begin{bmatrix}
0 & 1 & \cdots & n-1 & n \\
1 & 0 & \cdots & n-2 & n-1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
n-1 & n-2 & \cdots & 0 & 1 \\
n& n-1 & \cdots & 1 & 0%
\end{bmatrix}%
\end{equation*}
How can you prove ... | Let's take a $4\times 4$ matrix (I don't want to type much).
$$\begin{vmatrix}
0 & 1 & 2 & 3 \\
1 & 0 & 1 & 2 \\
2 & 1 & 0 & 1 \\
3 & 2 & 1 & 0
\end{vmatrix} $$
Since adding a row into another does not change determinant values. Add $-i'th$ row into $i+1$'th row.
$$\begin{vmatrix}
0 & 1 & 2 & 3 \\
1 & -1 & -1 & -1 \\
1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/144902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Approximate a series with finite number of terms How it is possible to approximate:
$$\sum_{i=1}^{NR}{i\cdot \left( \dfrac{1}{1-p} \right)^i} $$
| Let $x = \dfrac1{1-p}$ and let $n = NR$. Then we are interested in the sum $\displaystyle \sum_{i=1}^{n} i x^i$.
\begin{align}
\sum_{i=1}^{n} i x^i & = x \left(\sum_{i=1}^{n} i x^{i-1} \right)\\
& = x \left( \sum_{i=1}^{n} \frac{d x^i}{dx} \right)\\
& = x \frac{d}{dx} \left( \sum_{i=1}^{n} x^i\right)\\
& = x \frac{d}{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/146196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Surprising approximation of weighted sum of binomial coefficients The following sum appeared in connection to the problem addition of angular momentum in physics:
$$
\frac{1}{2^{n+3}}\sum_{k=0}^n \left(\frac{n-2k-1}{\sqrt{k+1}}+\frac{n-2k+1}{\sqrt{n-k+1}}\right)^2 {n\choose k}
$$
The intriguing thing about this sum is ... | Notice that the sum can be given a probabilistic interpretation, since $\frac{1}{2^n} \binom{n}{k}$ is the point mass function of a symmetric Binomial random variable.
Let $X \sim \operatorname{Binom}\left(n, \frac{1}{2}\right)$. Then the sum in question equals
$$
S = \frac{1}{8} \mathbb{E} \left( \left( \frac{n-2X ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/146261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Something is wrong with this proof, limit $\lim\limits_{(x,y) \to (0,0)} \frac{xy^3}{x^4 + 3y^4}$ Could someone please tell me what is wrong with this proof?
Show that $\lim\limits_{(x,y) \to (0,0)} \dfrac{xy^3}{x^4 + 3y^4}$ does not have a limit or show that it does and find the limit.
I know it is wrong because th... | The statement $x^4>x$ is clearly wrong especially when $x$ is in the neighborhood of $0$.
Also, you need to look at cases $x>0,y<0$ and $x<0,y>0$. But the main error is the statement $x^4 > x$.
EDIT
The answer is that the function is not continuous at the origin. This can be seen as follows. Remember that in two dimen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/149509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
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Calculate the volume between $z=x^2+y^2$ and $z=2ax+2by$ I'm trying to calculate the volume between the surfaces $z=x^2+y^2$ and $z=2ax+2by$ where $a>0,b>0$. Here's what I've tried:
First I noticed the projection of the volume to the xy plane is a circle: $(x-a)^2+(y-b)^2\leq a^2+b^2$. Using this I simplified the calcu... | Integrate in $y$ first; this gives the additional factor of $2\sqrt{(a^2+b^2)-(x-a)^2}$, so that the question reduces to
$$\int_{a-\sqrt{a^2+b^2}}^{a+\sqrt{a^2+b^2}} (x-a)^2 \sqrt{(a^2+b^2)-(x-a)^2}\,dx$$
which may look scary, but is in fact a typical trigonometric substitution problem. Namely, $x=a+\sqrt{a^2+b^2}\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/150251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Need help with the integral $\int \frac{2\tan(x)+3}{5\sin^2(x)+4}\,dx$ I'm having a problem resolving the following integral, spent almost all day trying.
Any help would be appreciated.
$$\int \frac{2\tan(x)+3}{5\sin^2(x)+4}\,dx$$
| If you convert everything to sines and cosines, you get $$\frac{2\tan x+3}{5\sin^2x+4}=\frac{2\sin x+3\cos x}{5\sin^2x\cos x+4\cos x}\;,\tag{1}$$ which probably doesn’t look very promising. However, you can rewrite it as $$\frac{2\sin x+3\cos x}{\cos x(9-5\cos^2x)}=\frac{2\sin x}{\cos x(9-5\cos^2x)}+\frac3{9-5\cos^2x}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/155022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $4^{2n} + 10n -1$ is a multiple of 25 Prove that if $n$ is a positive integer then $4^{2n} + 10n - 1$ is a multiple of $25$
I see that proof by induction would be the logical thing here so I start with trying $n=1$ and it is fine. Then assume statement is true and substitute $n$ by $n+1$ so I have the follo... | Another solution, via congruences mod. $25$. First note $16\bmod 25$ generates a cyclic group of order $5$:
$$16^2\equiv 6,\quad16^3\equiv 6\cdot 16\equiv-4,\quad16^4\equiv6^2\equiv 11, \quad 16^5\equiv6\cdot-4\equiv1\mod25.$$
So let's examine each case:
*
*If $n\equiv 0\mod 5$, $\;16^n+10n-1\equiv1+0-1=0$.
*If $n\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/155637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 3
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How do I show that this function is always $> 0$
Show that $$f(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +
\frac{x^4}{4!} > 0 ~~~ \forall_x \in \mathbb{R}$$
I can show that the first 3 terms are $> 0$ for all $x$:
$(x+1)^2 + 1 > 0$
But, I'm having trouble with the last two terms. I tried to show that the followin... | You have had some good ideas so far. You tried to see when this was true: $$\frac{x^3}{3!} \leq \frac{x^4}{4!}.$$
You rearranged this to $4x^3\leq x^4$ but you made an incorrect conclusion when you divided by $x^3$ (if $x<0$ then the inequality sign should flip). Instead, lets divide by $x^2$ to get $4x \leq x^2$ or $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/156117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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Evaluating $\int \frac{dx}{x^2 - 2x} dx$ $$\int \frac{dx}{x^2 - 2x}$$
I know that I have to complete the square so the problem becomes.
$$\int \frac{dx}{(x - 1)^2 -1}dx$$
Then I set up my A B and C stuff
$$\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{-1}$$
With that I find $A = -1, B = -1$ and $C = 0$ which I know is w... |
My book is telling me that I have to complete the square
$$I=\begin{eqnarray*}
\int \frac{dx}{x^{2}-2x} &=&\int \frac{dx}{\left( x-1\right) ^{2}-1}\overset{
u=x-1}{=}\int \frac{1}{u^{2}-1}\,du=-\text{arctanh }u+C
\end{eqnarray*},$$
$$\tag{1}$$
where I have used the substitution $u=x-1$ and the standard derivative $$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/156566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Combining a radical and simplifying? How would I combine and simplify the following radical:
$$\sqrt {\frac{A^2}{2}} - \sqrt \frac{A^2}{8}$$
| $\sqrt{\frac{A^2}{2}} - \sqrt{\frac{A^2}{8}} = \frac{A}{\sqrt{2}} - \frac{A}{2\sqrt{2}} = \frac{2A}{2\sqrt{2}} - \frac{A}{2\sqrt{2}} = \frac{A}{2\sqrt{2}}$
Assuming $A \geq 0$. If $A < 0$, you can replace with $|A|$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/157967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
What is the result of sum $\sum\limits_{i=0}^n 2^i$
Possible Duplicate:
the sum of powers of $2$ between $2^0$ and $2^n$
What is the result of
$$2^0 + 2^1 + 2^2 + \cdots + 2^{n-1} + 2^n\ ?$$
Is there a formula on this? and how to prove the formula?
(It is actually to compute the time complexity of a Fibonacci recur... | Let $S = 2^0 + 2^1 + 2^2 + \cdots + 2^{n}$.
Then $2S = 2^1 + 2^2 + 2^3 + \cdots + 2^{n} + 2^{n+1}$.
Then
$$\begin{align*}
S = 2S - S &= & & 2^1 &+& 2^2 & + & 2^3 & + & 2^4 &+&\cdots &+& 2^{n} &+& 2^{n+1}\\
&& -2^0 -& 2^1 & - & 2^2 & - & 2^3 & - & 2^4 & - & \cdots & - & 2^n
\end{align*}$$
How much is that?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/158758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 4
} |
Proof about $z\cot z=1-2\sum_{k\ge1}z^2/(k^2\pi^2-z^2)$ In Concrete Mathematics, it is said that
$$z\cot z=1-2\sum_{k\ge1}\frac{z^2}{k^2\pi^2-z^2}\tag1$$
and proved in EXERCISE 6.73
$$z\cot z=\frac z{2^n}\cot\frac z{2^n}-\frac z{2^n}\tan\frac z{2^n}+\sum_{k=1}^{2^{n-1}-1}\frac z{2^n}\left(\cot\frac{z+k\pi}{2^n}+\cot\fr... | This identity is also proven in this answer, but the limit of the trigonometric identity is a cute trick, too.
Concrete Mathematics claim:
For the limit claimed in Concrete Mathematics, we need a few things.
First, by inspecting the graph of $\frac{1-x\cot(x)}{x^2}$ for $-\frac{3\pi}{4}\le x\le\frac{3\pi}{4}$, we have
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/159752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
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?
| There are two (three) ways to go. One, assume
$$x^4+1=(x^2+ax+1)(x^2-ax+1)$$
You'll get that
$${x^4} + 1 = {x^4} + \left( {2 - {a^2}} \right){x^2} + 1$$
Then $a=\sqrt 2$ (or the other, by symmetry)
$${x^4} + 1 = {x^4} + 1 = \left( {{x^2} + \sqrt 2 x + 1} \right)\left( {{x^2} - \sqrt 2 x + 1} \right)$$
The other ${x^2}... | {
"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": 13
} |
What transforms under SU(2) as a matrix under SO(3)? A vector $\boldsymbol{r}$ in $\mathbb{R}^3$ transforms under rotation $\boldsymbol{A}$ to $\boldsymbol{r}'=\boldsymbol{Ar}$. It is equivalent to an SU(2) "rotation" as
$$\left( \boldsymbol{r}'\cdot\boldsymbol{\sigma} \right) = \boldsymbol{h} \left( \boldsymbol{r}\cdo... | Firstly, we need to map $\mathbb{R}^3$ to the representation space $V$ for $\mathrm{SU}(2)$. One possible map is given by the following formula:
$$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \mapsto
x \mathbf{I} +
y \mathbf{J} +
z \mathbf{K}$$
where
\begin{align}
\mathbf{I} & = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/160662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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System of equations of 3rd degree I need help with the following system of equations:
$
2y^3 +2x^2+3x+3=0 $
$
2z^3 + 2y^2 + 3y + 3= 0 $
$2x^3 + 2z^2 + 3z + 3 = 0$
| The only real solution is $x = y = z = -1$.
Claim 1: $x,y,z \ge -1$.
Proof. Suppose that $x < -1$. Then $0 = 2y^3 + 2x^3 + 3x + 3 > 2y^3 + 2$, so that $y < -1$ also. Similarly it follows that $z < -1$. Hence if one of $x,y,z$ is smaller than $-1$, all of them are. But then if for example $x<z$, we have
$$0 = 2x^3 + 2z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/160824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Series Expansion An old problem from Whittaker and Watson I'm having issues with. Any guidance would be appreciated.
Show that the function
$$
f(x)=\int_0^\infty \left\{ \log u +\log\left(\frac{1}{1-e^{-u}} \right) \right\}\frac{du}{u}e^{-xu}
$$
has the asymptotic expansion
$$
f(x)=\frac{1}{2x}-\frac{B_1}{2^2x^2}+\frac... | Note that:
$$\frac{d}{du} \left\{ \ln u + \ln \left( \frac{1}{1- e^{-u}}\right) \right\} = \frac{1}{u} - \frac{1}{e^u - 1} = -\sum_{n = 1}^{+\infty} \frac{B_n u^{n-1}}{n!}$$
then:
$$\ln u + \ln \left( \frac{1}{1- e^{-u}}\right) = -\sum_{n=1}^{+\infty} \frac{B_n}{n! \cdot n} u^n $$
so we can rewrite the integral as:
$$-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/162469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Calculate the limit at x=0 Find the limit of $f(x)=\frac{\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2}}{\sqrt{a-x}-\sqrt{a+x}}$ (at x=0) so that $f(x)$ becomes continuous for all $x$. My answer is $2\sqrt{a}$. Am I right?
Sorry to state the question incorrectly. We have to define $f(x)$ at $x=0$ such that $f(x)$ is continuous fo... | EDIT::
$$ \frac{\sqrt{a^2}-\sqrt{a^2}}{\sqrt a - \sqrt a} = \frac{0}{0}\neq 2\sqrt a $$
You have to multiply by conjugate of both terms (in numerator and denominator) and get the following.
$$ \frac{\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2}}{\sqrt{a-x}-\sqrt{a+x}} \\
= \frac{(a^2-ax+x^2)-(a^2+ax+x^2)}{(a-x)-(a+x)} \times \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/162531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Finding solutions to $(4x^2+1)(4y^2+1) = (4z^2+1)$ Consider the following equation with integral, nonzero $x,y,z$
$$(4x^2+1)(4y^2+1) = (4z^2+1)$$
What are some general strategies to find solutions to this Diophantine?
If it helps, this can also be rewritten as $z^2 = x^2(4y^2+1) + y^2$
I've already looked at On the equ... | Let $a$ be a positive integer.
Then
\begin{align}
(4a^2+1)(4((2a)^2)^2+1) &= 256a^6 + 64a^2 + 4a^2 + 1 \\
& = 4(64a^6 + 16a^4 + a^2) + 1 \\
&= 4(a^2(8a^2+1)^2)+1 \\
&= 4((8a^2+1)a)^2+1
\end{align}
so $(a, (2a)^2, (8a^2+1)a)$ is always a solution.
There are others as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/162862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Find all ordered pair of integers $(x,y)$
Obtain all ordered pair of integers $(x,y)$ such that $$x(x + 1) = y(y + 1)(y + 2)(y + 3)$$
I'm getting 8,
(0, 0), (0, -1), (0, -2), (0, -3)
(-1, 0), (-1, -1), (-1, -2), (-1, -3)
Please confirm my answer.
| Hint: It is easily proved that the product of four consecutive integers, plus $1$, is a perfect square. But $x(x+1)+1$ is hardly ever a perfect square!
Added: To prove that $y(y+1)(y+2)(y+3)+1$ is a perfect square, note that
$$y(y+1)(y+2)(y+3)=y(y+3)(y+1)(y+2)=(y^2+3y)(y^2+3y+2)=z(z+2),$$
where $z=y^2+3y$. And clearly... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/163902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Test for convergence $\sum_{n=1}^{\infty}\{(n^3+1)^{1/3} - n\}$ I want to expand and test this $\{(n^3+1)^{1/3} - n\}$ for convergence/divergence.
The edited version is: Test for convergence $\sum_{n=1}^{\infty}\{(n^3+1)^{1/3} - n\}$
| By direct inspection, for every pair of real numbers $A$ and $B$,
$$
A^3 - B^3 = (A-B) (A^2+AB+B^2).
$$
Choose now $A=\sqrt[3]{n^3+1}$ and $B=n$. Then
$$
(n^3+1)^{1/3} - n = \frac{n^3+1-n^3}{(n^3+1)^{2/3} + n (n^3+1)^{1/3}+n^2} \sim \frac{1}{n^2}
$$
as $n \to +\infty$. The limit is therefore zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/164495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Determine whether $\sum\limits_{i=1}^\infty \frac{(-3)^{n-1}}{4^n}$ is convergent or divergent. If convergent, find the sum. $$\sum\limits_{i=1}^\infty \frac{(-3)^{n-1}}{4^n}$$
It's geometric, since the common ratio $r$ appears to be $\frac{-3}{4}$, but this is where I get stuck. I think I need to do this: let $f(x) = ... | Let $q = \frac{-3}{4}$, $a_n = \frac{(-3)^{n-1}}{4^n}$, $b_0 = 0$, $b_n = b_{n-1} + a_n$.
$a_n = -\frac{1}{3} (\frac{-3}{4})^n = -\frac{1}{3} q^n$, hence $b_n = -\frac{1}{3} c_n$, where $c_0 = 0$, $c_n = c_{n-1} + q^n$.
The $c_n$ limit is equal to $q + q^2 + q^3 + \ldots = \frac{q}{1-q} = \frac{\frac{-3}{4}}{1-\frac{-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/166097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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proving :$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}\le\frac{3}{4}$. Let $a,b,c>0$ how to prove that :
$$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}\le\frac{3}{4}$$
I find that
$$\ \frac{ab}{a^{2}+3b^{2}}=\frac{1}{\frac{a^{2}+3b^{2}}{ab}}=\frac{1}{\frac{a}{b}+\frac{3b}{a}} $$
By AM-GM
$... | By AM-Gm $$\sum_{cyc}\frac{ab}{a^2+3b^2}=\sum_{cyc}\frac{ab}{a^2+b^2+2b^2}\leq\sum_{cyc}\frac{ab}{2\sqrt{2b^2(a^2+b^2)}}=\frac{1}{2\sqrt2}\sum_{cyc}\sqrt{\frac{a^2}{a^2+b^2}}.$$
Thus, it remains to prove that $\sum\limits_{cyc}\sqrt{\frac{x}{x+y}}\leq\frac{3}{\sqrt2}$, which followos from C-S.
Indeed, $$\left(\sum\limi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/167855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
} |
Polynomial-related manipulation My question is:
Factorize: $$x^{11} + x^{10} + x^9 + \cdots + x + 1$$
Any help to solve this question would be greatly appreciated.
| $$
\begin{align}
& {}\quad (x^{11} + x^{10}) + (x^9 + x^8)+(x^7+x^6)+(x^5+x^4)+(x^3+x^2 )+( x + 1)\\[8pt]
& =x^{10}(x+1)+x^8(x+1)+x^6(x+1)+x^4(x+1)+x^2(x+1)+(x+1)\\[8pt]
& =(x+1)(x^{10}+x^8+x^6+x^4+x^2+1)\\[8pt]
& =(x+1)(x^8(x^2+1)+x^4(x^2+1)+x^2+1)\\[8pt]
& =(x+1)((x^2+1)(x^8+x^4+1))\\[8pt]
& =(x+1)(x^2+1)(x^4+1-x^2)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/167981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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$(x^n-x^m)a=(ax^m-4)y^2$ in positive integers How do I find all positive integers $(a,x,y,n,m)$ that satisfy $ a(x^{n}-x^{m}) = (ax^{m}-4) y^{2} $ and $ m\equiv n\pmod{2} $, with $ax$ odd?
| Some partial results.
Case 1) $n<m$.
Then $ax^m<4$.
The solutions of this inequality are:
$$
a=1, m=1, x=1,2,3,\\
m\geq2, x=1,
$$
and
$$
a=2,3, x=1, m\geq 1.
$$
From these we obtain two candidates (i) $x=1$ and (ii) $x=3$.
(i) Substituting into the original equation we get
$$
0=(a-4)y^2.
$$
Since $y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/169863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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how to solve system of linear equations of XOR operation? how can i solve this set of equations ? to get values of $x,y,z,w$ ?
$$\begin{aligned} 1=x \oplus y \oplus z \end{aligned}$$
$$\begin{aligned}1=x \oplus y \oplus w \end{aligned}$$
$$\begin{aligned}0=x \oplus w \oplus z \end{aligned}$$
$$\begin{aligned}1=w \oplus... | As I wrote in my comment, you can just use any method you know for solving linear systems, I will use Gauss:
$$
\begin{array}{cccc|c||l} \hline
x & y & z & w &\ & \\ \hline\hline
1 & 1 & 1 & 0 & 1 & \\
1 & 1 & 0 & 1 & 1 & \text{$+$ I}\\
1 & 0 & 1 & 1 & 0 & \text{$+$ I}\\
0 & 1 & 1 & 1 & 1 & \\ \hline
1 & 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/169921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 2
} |
Prove $(-a+b+c)(a-b+c)(a+b-c) \leq abc$, where $a, b$ and $c$ are positive real numbers I have tried the arithmetic-geometric inequality on $(-a+b+c)(a-b+c)(a+b-c)$ which gives
$$(-a+b+c)(a-b+c)(a+b-c) \leq \left(\frac{a+b+c}{3}\right)^3$$
and on $abc$ which gives
$$abc \leq \left(\frac{a+b+c}{3}\right)^3.$$
Since ... | Here I give a detailed proof. Though steps could have been jumped to keep it short.
Without loss of generality we can assume that $a\ge b \ge c$
Let $$(-a+b+c)(a-b+c)(a+b-c)=S$$
$$\Rightarrow S=(-a+b+c)\{a-(b-c)\}\{a+(b-c)\}$$ $$\Rightarrow S=(-a+b+c)\{a^2-(b-c)^2\} $$ $$\Rightarrow S= (-a+b+c)\{a^2-b^2-c^2+2bc\}$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/170813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 7,
"answer_id": 0
} |
Trigonometric Identities: $\frac{\sin^2\theta}{1+\cos\theta}=1-\cos\theta$ $\dfrac{\sin^2\theta}{1+\cos\theta}=1-\cos\theta$
Right Side:
$1-\cos\theta$ either stays the same, or can be $1-\dfrac{1}{\sec\theta}$
Left Side:
$$\begin{align*}
&= \dfrac{\sin^2\theta}{1+\cos\theta}\\
&= \dfrac{1-\cos^2\theta}{1+\cos\... | Perhaps slightly simpler and shorter (FYI, what you did is correct):
$$\frac{\sin^2x}{1+\cos x}=1-\cos x\Longleftrightarrow \sin^2x=(1-\cos x)(1+\cos x)\Longleftrightarrow \sin^2x=1-\cos^2x$$
And since the last equality is just the trigonometric Pytahgoras Theorem we're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/170947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve for $x$; $\tan x+\sec x=2\cos x;-\infty\lt x\lt\infty$
Solve for $x$; $\tan x+\sec x=2\cos x;-\infty\lt x\lt\infty$
$$\tan x+\sec x=2\cos x$$
$$\left(\dfrac{\sin x}{\cos x}\right)+\left(\dfrac{1}{\cos x}\right)=2\cos x$$
$$\left(\dfrac{\sin x+1}{\cos x}\right)=2\cos x$$
$$\sin x+1=2\cos^2x$$
$$2\cos^2x-\sin x... | This is more a comment than an answer, but it is an important comment, since it affects the list of solutions. We seem to have solutions that come from $\sin x=\frac{1}{2}$, and solutions that come from $\sin x=-1$.
No problem with the $\sin x=\frac{1}{2}$ stuff. For the record, this gives the solutions $x=\frac{\pi}{6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/171792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
multiple choice summation problem
Let
$$X = \frac{1}{1001} + \frac{1}{1002} + \frac{1}{1003} + \cdots + \frac{1}{3001}.$$
Then
(A) $X < 1$
(B) $X > 3/2$
(C) $1 < X < 3/2$
(D) none of the above holds.
I assume that the answer is the third choice $1<X<3/2$. I integrate out $1/x$ in the interval $(1001, 3001)$ and... | With respect to your Riemann sum approach: the idea is that for positive, decreasing functions, the Riemann sum and the integral carefully approximate each other, more or less as in the proof of the integral test of convergence.
If you'd like another sort of approach, we could approach it naively. Separate the sum into... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/171999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Proving:$\tan(20^{\circ})\cdot \tan(30^{\circ}) \cdot \tan(40^{\circ})=\tan(10^{\circ})$ how to prove that : $\tan20^{\circ}.\tan30^{\circ}.\tan40^{\circ}=\tan10^{\circ}$?
I know how to prove
$ \frac{\tan 20^{0}\cdot\tan 30^{0}}{\tan 10^{0}}=\tan 50^{0}, $
in this way:
$ \tan{20^0} = \sqrt{3}.\tan{50^0}.\tan{10^0}$
... | Due to $\tan 40^\circ*\tan 20^\circ$
$$=\frac{\sin 40^\circ*\sin 20^\circ}{\cos 40^\circ*\cos 20^\circ}$$
$$=\frac{4\sin 40^\circ*(\sin 20^\circ)^2}{4\cos 40^\circ*\cos 20^\circ*\sin 20^\circ}$$
$$=\frac{2\sin 40^\circ*(1-\cos 40^\circ)}{2\cos 40^\circ*\sin 40^\circ}$$
$$=\frac{2\sin 40^\circ-2\sin 40^\circ*\cos 40^\ci... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/172182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Prove $ (r\sin A \cos A)^2+(r\sin A \sin A)^2+(r\cos A)^2=r^2$ How would I verify the following trig identity?
$$(r\sin A \cos A)^2+(r\sin A \sin A)^2+(r\cos A)^2=r^2$$
My work thus far is
$$(r^2\cos^2A\sin^2A)+(r^2\sin^2A\sin^2A)+(r^2\cos^2A)$$
But how would I continue? My math skills fail me.
| Oh I didn't read robjohn's answer carefully before making this colourful answer.. I will leave it here anyways.
To continue on what you have:
$$ (\color{red}{r^2}\cos^2A\sin^2A)+(\color{red}{r^2}\sin^2A\sin^2A)+(\color{red}{r^2}\cos^2A) \\
= \color{red}{r^2}( \cos^2A\color{blue}{\sin^2A}+\sin^2A\color{blue}{\sin^2A}+\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/172607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating the product $\prod\limits_{k=1}^{n}\cos\left(\frac{k\pi}{n}\right)$ Recently, I ran across a product that seems interesting.
Does anyone know how to get to the closed form:
$$\prod_{k=1}^{n}\cos\left(\frac{k\pi}{n}\right)=-\frac{\sin(\frac{n\pi}{2})}{2^{n-1}}$$
I tried using the identity $\cos(x)=\frac{\sin(... | If $n$ is even, then the term with $k=n/2$ makes the product on the left $0$ and $\sin\left(\frac{n}{2}\pi\right)=0$. So assume that $n$ is odd.
$$
\begin{align}
\prod_{k=1}^n\cos\left(\frac{k\pi}{n}\right)
&=-\prod_{k=1}^{n-1}\cos\left(\frac{k\pi}{n}\right)\tag{1}\\
&=-\prod_{k=1}^{n-1}\frac{\sin\left(\frac{2k\pi}{n}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/173238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 4,
"answer_id": 1
} |
Trigonometry proof involving sum difference and product formula How would I solve the following trig problem.
$$\cos^5x = \frac{1}{16} \left( 10 \cos x + 5 \cos 3x + \cos 5x \right)$$
I am not sure what to really I know it involves the sum and difference identity but I know not what to do.
| Applying the same approach as that of this,
let $A\cos5x+B\cos3x+C\cos x=\cos^5x$
As $\cos 3x = 4\cos^3x-3\cos x$ and
$\cos 5x = 16\cos^5 x-20\cos^3 x+5\cos x$
$A(16\cos^5 x-20\cos^3 x+5\cos x) + B( 4\cos^3x-3\cos x)+ C\cos x=\cos^5x$
Comparing the coefficients of different powers of cosx,
5th power=>16A=1=>$A=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/179584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluation of $\int_0^{2\pi} \frac{1}{1+3\cos^2(\theta)} \, d\theta$ For a test I had to evaluate $\int_0^{2\pi} \frac{1}{1+3\cos^2(\theta)}d\theta$. First I used substitution, with $z=e^{i\theta}$ and $d\theta=\frac{1}{iz}dz$, as shown:
$$\int_0^{2\pi} \frac{1}{1+3\cos^2(\theta)}d\theta=\int_{|z|=1}\frac{1}{1+\frac{3}... | Right after the first appearance of $f(z)$, when you factored the denominator, you forgot a factor $3$.
Note: Complex analysis might not be the simplest approach here, since the change of variables $t=\tan(θ)$ yields directly that, for every $a\gt-1$,
$$
\int_0^{2\pi}\frac{\mathrm d\theta}{1+a\cos^2(\theta)}=\frac{2\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/186025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Infinite descent This Wikipedia article of Infinite Descent says:
We have
$ 3 \mid a_1^2+b_1^2 \,$.
This is only true if both $a_1$ and $b_1$ are divisible by $3$.
But how can this be proved?
| Well, we have two possibilities, first if neither $a_1$ nor $b_1$ is divisible by $3$ then that means that they have the form $a_1 = 3k \pm 1$ and $b_1 = 3t \pm 1$. Then $$a_1^2 + b_1^2 = (3k \pm 1)^2 + (3t \pm 1)^2 = 9k^2 \pm 6k + 1 + 9t^2 \pm 6t + 1 = 3A + 2$$
where $A = 3k^2 \pm 2k + 3t^2 \pm 2t$ and thus since $a_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/186989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
how can one find the value of the expression, $(1^2+2^2+3^2+\cdots+n^2)$
Possible Duplicate:
Proof that $\sum\limits_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$?
Summation of natural number set with power of $m$
How to get to the formula for the sum of squares of first n numbers?
how can one find the value of the expressio... | If you can find or sketch some 3D blocks, there is a fun geometric proof.
Fix some $n$. If you are doing this with real blocks, $n=3$ or $4$ should be convincing. Let's take $n=4$ for now.
Make a $4\times 4\times 1$ base, laid flat, which of course has volume $4^2$.
Now make a $3\times 3 \times 1$ brick and place it, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/188602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
how to evaluate the integral $\frac{64}{\pi^3} \int_0^\infty \frac{ (\ln x)^2 (15-2x)}{(x^4+1)(x^2+1)}\ dx$ How to evaluate the 59 integral, possibly using real method?
$$\frac{64}{\pi^3} \int_0^\infty \frac{ (\ln x)^2 (15-2x)}{(x^4+1)(x^2+1)}\ dx$$
| Start by evaluating, for $-1<\Re(s)<4$, the following integral:
$$
I(s) = \int_0^\infty \frac{x^s}{(1+x^2)(1+x^4)} \mathrm{d}x = \frac{1}{2} \int_0^\infty x^s \left( \frac{1}{1+x^2} + \frac{1-x^2}{1+x^4} \right) \mathrm{d}x
$$
When $-1<\Re(s)<1$ we can rewrite this as a sum of integrals:
$$
I(s) = \frac{\pi}{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/188856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Proving inequality $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c} \geq 4$ I started to study inequalities - I try to solve a lot of inequlites and read interesting .solutions . I have a good pdf, you can view from here . The inequality which I tried to solve and I didn't manage to find a solution can... | Following above motivations and applying AM-GM three times:
\begin{align}
&\frac13\left(\frac ab+\frac ab+\frac bc\right)+\frac13\left(\frac bc+\frac bc+\frac ca\right)+\frac13\left(\frac ca+\frac ca+\frac ab\right)+\frac{3\sqrt[3]{abc}}{a+b+c}\\
&\ge \frac{a}{\sqrt[3]{abc}}+\frac{b}{\sqrt[3]{abc}}+\frac{c}{\sqrt[3]{ab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/189143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Plane intersection by a mapping and different cases In $\mathbb{R}^3$ say we have the 2 planes $A=\{z=1\}$ and $B=\{x=1\}$. A line through 0 meeting $A$ at $(x,y,1)$ meets $B$ at $(1,y/x,1/x).$ Consider the map $\phi: A \rightarrow B$ defined by $(x,y) \mapsto (y' = y/x, z' = 1/x)$.
I'm trying to figure out the image u... | To answer your last question first, notice that a line through $0$ meeting $A$ at $(0,y,1)$ does not meet $B$ at all. This explains why $\phi$ is undefined in such cases. Correspondingly, pick any point on $B$ with $z = 0$ and any line through the origin and that point is wholly within the $xz$-plane, so will never hit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/189924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Complex numbers Here's the question:
z is a complex, and if $z^5 + z^4 + z^3 + z^2 + z + 1 = 0$ then $z^6=1$.
use this fact to calculate how many answers is there for:
$$z^5 + z^4 + z^3 + z^2 + z + 1 = 0$$
Thanks.
| $z^6 = 1$ if and only if $z^6 - 1 = (z-1)(z^5+z^4+z^3+z^2+z+1) = 0$. So the roots of $z^5+z^4+z^3+z^2+z+1$ consist of the roots of $z^6 = 1$ excluding the root $z = 1$, which leaves $5$ roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/191826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving differential equation for an expanding bubble I need to solve the equation
\begin{eqnarray}
R^3 \frac{d } {dt} \left [ \frac{4}{3} \rho_{\rm ext} \left ( \frac{dR}{dt} \right )^2 \right ]+ 4 p R^2 \frac{d R} {dt} =\frac{F_E}{4\pi}
\end{eqnarray}
Could you please help in this regard?
| Let $X=\dfrac{dR}{dt}$ ,
Then $R^3\dfrac{d}{dt}\left[\dfrac{4}{3}\rho_{\rm ext}X^2\right]+4pR^2X=\dfrac{F_E}{4\pi}$
$R^3\dfrac{d}{dR}\left[\dfrac{4}{3}\rho_{\rm ext}X^2\right]\dfrac{dR}{dt}+4pR^2X=\dfrac{F_E}{4\pi}$
$\dfrac{8\rho_{\rm ext}R^3X^2}{3}\dfrac{dX}{dR}=\dfrac{F_E}{4\pi}-4pR^2X$
Let $Y=\dfrac{1}{R^2}$ ,
Then ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/191987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How may I prove this inequality? Let $a, b, c$ be positive real, $abc = 1$. Prove that:
$$\frac{1}{1+a+b} + \frac{1}{1+b+c} + \frac{1}{1+c+a} \le \frac{1}{2+a} + \frac{1}{2+b}+\frac{1}{2+c}$$
I thought of Cauchy and AM-GM, but I don't see how to successfully use them to prove the inequality. Any hint, suggestion will b... | $\frac{2}{1+a+b}-(\frac{1}{2+a}+\frac{1}{2+b})$
$=\frac{1}{1+a+b}-\frac{1}{2+a}+\frac{1}{1+a+b}-\frac{1}{2+b}$
$=\frac{1}{1+a+b}(\frac{1-b}{2+a}+\frac{1-a}{2+b})$
$=\frac{1}{(1+a+b)(2+a)(2+b)}((1-b)(2+b)+(1-a)(2+a))$
$≤\frac{1}{1\cdot 2\cdot 2}(4-(a+b+a^2+b^2))$ as $a,b>0,2+a>2$ and $a+b+1>1$
$\sum(\frac{2}{1+a+b}-(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/192371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Sum of the sequence What is the sum of the following sequence
$$\begin{align*}
(2^1 - 1) &+ \Big((2^1 - 1) + (2^2 - 1)\Big)\\
&+ \Big((2^1 - 1) + (2^2 - 1) + (2^3 - 1) \Big)+\ldots\\
&+\Big( (2^1 - 1)+(2^2 - 1)+(2^3 - 1)+\ldots+(2^n - 1)\Big)
\end{align*}$$
I tried to solve this. I reduced the equation into the followi... | Others have given the correct answer; here’s how you could have simplified your incorrect expression.
$$\begin{align*}
n(2^1) + (n-1)\cdot2^2 + (n-2)\cdot2^3 +\ldots&=\sum_{k=1}^n(n-k+1)2^k\\
&=(n+1)\sum_{k=1}^n2^k-\sum_{k=1}^nk2^k\\
&=(n+1)\left(2^{n+1}-2\right)-\sum_{k=1}^n\sum_{i=1}^k2^k\\
&=(n+1)\left(2^{n+1}-2\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/192520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How to solve $x^3=-1$? How to solve $x^3=-1$? I got following:
$x^3=-1$
$x=(-1)^{\frac{1}{3}}$
$x=\frac{(-1)^{\frac{1}{2}}}{(-1)^{\frac{1}{6}}}=\frac{i}{(-1)^{\frac{1}{6}}}$...
| $$x^3=-1$$
$$x^3+1=0$$
$$(x+1)(x^2+1-x)=0$$
$$x=-1 \quad\text{or}\quad x^2-x+1=0$$
When $$x^2-x+1=0, x= \frac{-1(\pm\sqrt{-3})}{2}$$
$$x= \frac{-1+\sqrt{3}i}{2} \quad\text{and}\quad \frac{-1-\sqrt{3}i}{2}$$
Which is equal to $e^{−i2π/3}$ and $e^{i2π/3}$ respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/192742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 8,
"answer_id": 3
} |
Solve sum for theta Is there any way to solve the following sum of trigonometric functions for theta without using a solver?
$$25\sin(\theta)-1.5\cos(\theta)=20$$
| $$
25\sin\theta-1.5\cos\theta = \sqrt{25^2+1.5^2}\left( \frac{25}{\sqrt{25^2+1.5^2}}\sin\theta - \frac{1.5}{\sqrt{25^2+1.5^2}}\cos\theta \right)
$$
$$
= \sqrt{25^2+1.5^2}(\cos\varphi\sin\theta-\sin\varphi\cos\theta) = \sqrt{25^2+1.5^2} \sin(\varphi-\theta).
$$
So you want
$$
\sin(\varphi-\theta)=\frac{20}{\sqrt{25^2+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/192988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Probability of throwing multiple dice of at least a given face with a set of dice I know how to calculate the probability of throwing at least one die of a given face with a set of dice, but can someone tell me how to calculate more than one (e.g., at least two)?
For example, I know that the probability of throwing at ... | The probability of no 4 with four 6-sided dice$(p_1)=(\frac{5}{6})^4$
The probability of exactly one 4 with four 6-sided dice$(p_2)$
$=4\frac{1}{6}(\frac{5}{6})^3$ as here the combinations are $4XXX$ or $X4XX$ or $XX4X$ or $XXX4$ where $X$ is some other face$≠4$
So, the probability of at least two 4s with four 6-sided ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/193050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$ Please help me for prove this inequality:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$$
| This inequality can also be rewritten as
$$\frac{a+b+c}{3} \geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \,,$$
which is just the AM-HM inequality.
A more direct proof would be to simply multiply:
$$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=3+ (\frac{a}{b}+\frac{b}{a})+(\frac{a}{c}+\frac{c}{a})+(\frac{c}{b}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/193771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
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Inequality. $\sum{(a+b)(b+c)\sqrt{a-b+c}} \geq 4(a+b+c)\sqrt{(-a+b+c)(a-b+c)(a+b-c)}.$ Let $a,b,c$ be the side-lengths of a triangle. Prove that:
I.
$$\sum_{cyc}{(a+b)(b+c)\sqrt{a-b+c}} \geq 4(a+b+c)\sqrt{(-a+b+c)(a-b+c)(a+b-c)}.$$
What I have tried:
\begin{eqnarray}
a-b+c&=&x\\
b-c+a&=&y\\
c-a+b&=&z.
\end{eqnarray}
... | I couldn't solve this with elementary means.
Take $\sum_{cyc}{\sqrt{x}(S+y)(S+z)} \geq 16S\cdot \sqrt{xyz} $ and note that it is enough to prove it for $S=1$ because it is homogeneous. Substituting $x\to x^2$, etc. consider the function $f(x,y,z)=(x^2+1) (y^2+1) z+(x^2+1) y (z^2+1)+x (y^2+1) (z^2+1)-16 x y z$. It is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/195185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove $\frac{1}{a(1+b)} + \frac{1}{b(1+c)} + \frac{1}{c(1+a)} \geq \frac{3}{ (abc)^{\frac{1}{3}}\big( 1+ (abc)^{\frac{1}{3}}\big) }$ using AM-GM I need to proof this inequality by AM-GM method.
Any ideas how to do it?
$$\frac{1}{a(1+b)} + \frac{1}{b(1+c)} + \frac{1}{c(1+a)} \geq \frac{3}{ (abc)^{\frac{1}{3}}\big( 1+ ... | We may assume without loss of generality that $abc = k^3$, which enables us to make the substitution $\displaystyle a = \frac{kq}{p}, b = \frac{kr}{q}, c = \frac{kp}{r}$.
Now, $\displaystyle a(1+b) = \frac{kq}{p} \left(1+ \frac{kr}{q} \right) = \frac{k(q+kr)}{p}$.
Thus, the inequality reduces to proving (after cancelli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/196176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Generating integral triangles with two equal sides How can I generate all triangles which have integral sides and area, and exactly two of its three sides are equal?
For example, a triangle with sides ${5,5,6}$ satisfies these terms.
| Heron's formula says the area of a triangle with sides $a, b, c$ is
$$Area = \sqrt{s(s-a)(s-b)(s-c)}$$
where $s$ is the semiperimeter, $s = \frac{a+b+c}{2}$.
Now, your assumption is two sides are equal, so $a, b, b$. The area is now
$$Area = \sqrt{s(s-a)(s-b)^2} = (s-b)\sqrt{s(s-a)}$$
Now $s-b$ might not be an integer... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/198034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve a simultaneous equation. How do we solve $|b-y|=b+y-2\;and\;|b+y|=b+2$? I have tried to square them and factorize them but got confused by and and or conditions.
| $b+2=|b+y|$ which is real, so is $b$
$y+b-2=|b-y|$ which is real, so is $y+b-2$ and $y$
(1)If $b \ge y, b-y=b+y-2\implies y=1 \implies |b+1|=b+2$ and $b \ge y=1$
So, $b+1 >0\implies |b+1|=b+1=b+2$ which has no finite solution.
(2) If $b<y, y-b=b+y-2\implies b=1, y>b=1$
So, $|1+y|=3\implies y+1=3\implies y=2$
The only ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/202791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Help me prove $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$ Please help me prove this Leibniz equation: $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$. Thanks!
| $\sqrt{1 + i \sqrt 3} + \sqrt{1 - i\sqrt 3} = \sqrt 6$ ?
$1 + i \sqrt 3 = 2 \exp \left( \dfrac {\pi}{3}i + 2 \pi n i \right)
\quad \{ n \in \mathbb Z \}$
$\sqrt{1 + i \sqrt 3} = \sqrt 2 \exp \left( \dfrac {\pi}{6}i + \pi n i \right)
\quad \{ n \in \mathbb Z \}$
$\sqrt{1 + i \sqrt 3} =
\pm \left( \dfrac{\sqrt 6}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/203462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 2
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difficulty working out equation $\int_{-a}^{a}2(\sqrt{a^2-x^2})dx $ =$\left.2[\frac{1}{2}\sqrt{a^2-x^2}+a^2\arcsin(\frac{x}{a})]\right|_{-a}^{a}$ I encountered the below formula in my text.
I know the author is using integration by substitution and double angle formula
but for some reason, every working on paper that i... | This can be done with a standard $\sin$ substitution: $(1)\ x=a\sin(\theta)$ followed by a change of variables: $(2)\ \phi=2\theta$.
$$
\begin{align}
\int_{-a}^a2\sqrt{a^2-x^2}\,\mathrm{d}x
&=\int_{-\pi/2}^{\pi/2}2a\cos(\theta)\,\mathrm{d}a\sin(\theta)\tag{1}\\
&=2a^2\int_{-\pi/2}^{\pi/2}\cos^2(\theta)\,\mathrm{d}\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/205535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Unsure about applying series comparison test Does this converge or diverge?
$$
\sum\limits_{n=1}^\infty (a_{n} = \frac{1}{2\sqrt{n} + \sqrt[3]{n}})
$$
The answer is: diverges by limit comparison to $\sum (b_{n} = \frac{1}{\sqrt{n}})$
If I look at $\lim_{n \to \infty}\frac{a_{n}}{b_{n}}$ I get
$$
\frac{\sqrt{n}}{2\sqrt{... | There seems to be a mistake in the algebra -- namely, you took $\sqrt[3]{n} = n^{1/2} n^{2/3}$. But the right-hand side of this is equal to $n^{1/2 + 2/3} = n^{3/6 + 4/6} = n^{7/6} \ne \sqrt[3]{n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/207663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $5a^2 - 4ab - b^2 + 9 = 0$, $ - 21a^2 - 10ab + 40a - b^2 + 8b - 12 = 0$
Solve $\left\{\begin{matrix} 5a^2 - 4ab - b^2 + 9 = 0\\ - 21a^2 - 10ab + 40a - b^2 + 8b - 12 = 0. \end{matrix}\right.$
I know that we can use quadratic equation twice, but then we'll get some very complicated steps. Are there any elegant wa... | You can notice that many terms of
$$(b+5a-4)^2=b^2+10ab-8b-40a+25a^2+16$$
appear in the first equation. Similarly, in the first one, you can notice $(b+2a)^2$.
By algebraic manipulation you get that the original equations are equivalent to
$$
\begin{align}
(b+5a-4)^2&=4(a^2+1)\\
(b+2a)^2&=9(a^2+1)
\end{align}
$$
which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/210454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
How to find the eigenvalues and eigenvector without computation? The given matrix is
$$
\begin{pmatrix}
2 & 2 & 2 \\
2 & 2 & 2 \\
2 & 2 & 2 \\
\end{pmatrix}
$$
so, how could i find the eigenvalues and eigenvector without computation?
Thank you
| Try to solve the following matrix equation, taking into account that your matrix is singular:
$$\begin{pmatrix}2&2&2\\2&2&2\\2&2&2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2(x+y+z)\\2(x+y+z)\\2(x+y+z)\end{pmatrix}=2\begin{pmatrix}x+y+z\\x+y+z\\x+y+z\end{pmatrix}=\lambda\begin{pmatrix}x\\y\\z\end{p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/211865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving trigonometric equations of the form $a\sin x + b\cos x = c$ Suppose that there is a trigonometric equation of the form $a\sin x + b\cos x = c$, where $a,b,c$ are real and $0 < x < 2\pi$. An example equation would go the following: $\sqrt{3}\sin x + \cos x = 2$ where $0<x<2\pi$.
How do you solve this equation wi... | This is little known, but you can solve the equation without much trigonometry.
WLOG, we can assume that $a^2+b^2=1$ (the coefficients can be normalized). Write
$$a\,S+b\,C=c,\\b\,C=c-aS,\\b^2(1-S^2)=(c-a\,S)^2,\\S^2-2ac\,S+c^2-b^2=0.$$
The solution of the quadratic equation is
$$S=ac\pm bd$$ where $d=\sqrt{1-c^2}$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/213545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 6,
"answer_id": 3
} |
Prove that $\prod_{i=2}^n (1-1/i^2) = {n+1\over 2n}$ prove the $$\prod_{i=2}^n (1-1/i^2) = {n+1\over 2n}$$ for all n greater or equal to 2.
$\pi$ should be a big pi from $i=2$ to $n$ for $(1-1/i^2)$. I'm really confused about the $\prod$ function.
UPDATE:
Suppose$$\prod_{i=2}^n (1-1/i^2) = {n+1\over 2n}$$
then $$\prod_... | $$\displaystyle \prod_{k=2}^n f(k)$$ is a short hand notation for $$f(2) \times f(3) \times f(4) \times \cdots \times f(n-1) \times f(n)$$
For instance, $$\displaystyle \prod_{k=2}^{4} \left(1 - \dfrac1{k^2} \right) = \left(1 - \dfrac1{2^2} \right) \times \left(1 - \dfrac1{3^2} \right) \times \left(1 - \dfrac1{4^2} \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/215259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to get rid of a cubic root in a logarithmic limit? The formula:
$$\lim_{n \to \infty}\dfrac{\log (1 - n + n^2)}{\log (1 + n + n^{10})^{1/3}}.$$
Thanks for any advice!
| Hint:
\begin{eqnarray*}
\frac{\log \left( 1-n+n^{2}\right) }{\log \left( 1+n+n^{10}\right) ^{1/3}}
&=&\frac{\log \left( 1-n+n^{2}\right) }{\frac{1}{3}
\log \left( 1+n+n^{10}\right) } \\
&=&\frac{\log \left( n^{2}\left( 1/n^{2}-1/n+1\right) \right) }{\frac{1}{3}
\log \left( n^{10}\left( 1/n^{10}+1/n^{9}+1\right) \right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/215914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Continued Fractions Approximation I have come across continued fractions approximation but I am unsure what the steps are.
For example How would you express the following rational function in continued-fraction form:
$${x^2+3x+2 \over x^2-x+1}$$
| Start out writing
$$\frac{x^2+3x+2}{x^2-x+1}=\frac{1}{\frac{(x^2+3x+2)-(4x+1)}{x^2+3x+2}}=\frac{1}{1-\frac{4x+1}{x^2+3x+2}}$$
and then iterate doing the same with the fraction in the denominator.
EDIT: Complete solution:
By polynomial long division we have $\frac{x^2+3x+2}{4x+1}=\frac{x}{4}+\frac{11}{16}+\frac{21}{16}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/218884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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.