Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove that $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ for $n \in \mathbb{N}$. Problem: Prove that $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ for $n \in \mathbb{N}$.
My work: So I think I have to do a proof by induction and I just wanted some help editing my proof.
My attempt:
Let $P(n)=1^2+2^2+\cdots+n^2=\frac{n(n+1)... | Consider any natural number $r$. You have $$r^3-(r-1)^3=3r^2-3r+1.$$
Now telescope it:
$$
1^3-0^3=3-3+1
$$
$$2^3-1^3=3\cdot2^2-3\cdot2+1
$$
$$\vdots
$$
$$
n^3-(n-1)^3=3n^2-3n+1
$$ Now add, and see them cancel out. You are left with $$n^3=3(1^2+2^2+\cdots+ n^2)-3(1+2+3+\cdots+n)+n$$ You must know
$$
1+2+3+\cdots+n=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1166027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
How to decompose the polynomial $x^{21} + 1$ into a product of irreducible factors over $\mathbb Z/2\mathbb Z$? This is a problem from a past abstract algebra exam, the degree $21$ was too high for me to solve it. No Wolfram Mathematica please!
| $$x^{21} + 1 = x^{21} - 1 = (x^7)^3 - 1 = (x^7 - 1)(x^{14} + x^7 + 1) = (x - 1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)(x^{14} + x^7 + 1) = (x -1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)(x^2 + x + 1)(x^{12} - x^{11} + x^9 - x^8 + x^6 - x^4 + x^3 - x +1)$$
Now: The $2$nd one can be shown to be irreducible in $\mathbb Z[x]$; I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1169222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Use $\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$ to compute $\sum_{n=1}^\infty \frac{(-1)^n}{n^4}$ Is it possible to use the fact that $\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$ to compute $\sum_{n=1}^\infty \frac{(-1)^n}{n^4}$?
| Note
$$\sum_{n=1}^\infty\frac{1}{n^4}+\sum_{n=1}^\infty\frac{(-1)^n}{n^4}=2\left(\frac{1}{2^4}+\frac{1}{4^4}+\dots+\right)=\frac{1}{8}\left(\frac{1}{1^4}+\frac{1}{2^4}+\dots\right)=\frac{1}{8}\sum_{n=1}^\infty\frac{1}{n^4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1169861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Using the limit of $ (1+1/n)^n$ to find the limit of $((n^2+2)/(n^2+1)) ^ {3n^2+1/n}$ Use: $ \lim_{n \to \infty} (1+1/n)^n = e$
To find: $ \lim_{n \to \infty} (\frac{n^2+2}{n^2+1}) ^ {3n^2+1/n}$
Any help is appreciated. Not looking for the complete answer since it's looked down upon here. So strong hints would be appre... | Hint: ${n^2 +2 \over n^2 +1}=1+{1 \over n^2 +1}$
It is convenient to write the exponent as $3n^2 + 1/n= 3(n^2 + 1) -3 + 1/n$ to get:
$${n^2 +2 \over n^2 +1}^{3n^2 + 1/n}=\left (\left (1+{1 \over n^2 +1} \right )^{n^2+1} \right )^3\left (1+{1 \over n^2 +1} \right )^{-3}\left (1+{1 \over n^2 +1} \right )^{1/n}$$
Now $n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1171242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
For what values of k is the vector u in the span of the columns of A? Let $$u = \begin{pmatrix} -1 \\1 \\-1 \end{pmatrix}$$
and let$$A = \begin{pmatrix} 1&0&1 \\1&1&0 \\0&1&k \end{pmatrix}$$
For what values of k is the vector u in the span of the columns of A?
Here is my work so far:
$$Au = \begin{pmatrix} 1&0&1&-1 \... | $(-1, 1, -1) = x (1, 1, 0) + y (0, 1, 1) + z (1, 0, k) $
\begin{equation*}
\left\{\begin{aligned}
-1 = x 1 + y 0 + z 1 \\
1 = x 1 + y 1 + z 0 \\
-1 = x 0 + y 1 + z k
\end{aligned}
\right.
\end{equation*}
\begin{equation*}
\left\{\begin{aligned}
-1 - x &= z(x) \\
1 - x &= y(x) \\
-1 &= y + z k = 1-x + (-1-x) k
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1171379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to prove $x^3-y^3 = (x-y)(x^2+xy+y^2)$ without expand the right side? I can prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$ by expanding the right side.
*
*$x^3-y^3 = (x-y)x^2 + (x-y)(xy) + (x-y)y^2$
*$\implies x^3 - x^2y + x^2y -xy^2 + xy^2 - y^3$
*$\implies x^3 - y^3$
I was wondering what are other ways to prove... | Another way is as follows:
$$
(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3 = (x^3 - y^3) -3xy(x - y) \quad \Rightarrow
$$
$$
x^3 - y^3 = (x - y)^3 + 3xy(x - y) = (x - y)[(x - y)^2 + 3xy] \quad \Rightarrow
$$
$$
x^3 - y^3 = (x - y)(x^2 + xy + y^2)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1172119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 12,
"answer_id": 6
} |
Does the series of $\sum_{n=2}^{\infty} \frac{(-1)^n(n+1)}{n-1}$ converge? Does the following sum converge?
$$
\sum_{n=2}^{\infty} \frac{(-1)^n(n+1)}{n-1}
$$
Have tried ratio test and root test, inconclusive. alternating series test not useful.
| Let
$s(m)
=\sum_{n=2}^{m} \frac{(-1)^n(n+1)}{n-1}
$.
$\begin{array}\\
s(2m+1)
&=\sum_{n=2}^{2m+1} \frac{(-1)^n(n+1)}{n-1}\\
&=\sum_{n=1}^{m} (\frac{(-1)^{2n}(2n+1)}{2n-1}+\frac{(-1)^{2n+1}(2n+1+1)}{2n+1-1})\\
&=\sum_{n=1}^{m} (\frac{(2n+1)}{2n-1}-\frac{(2n+2)}{2n})\\
&=\sum_{n=1}^{m} (\frac{2n-1+2}{2n-1}-\frac{n+1}{n})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1177171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Trying to determine the Inverse function of $\sinh$ and $\cosh$ I'm trying to find out how to determine the inverse function in order to develop the $$ \sinh(x).$$
I tried to expand to its exponential form $$\sinh(x) = \frac{1}{2} (e^x-e^{-x}) .$$
So I wrote
$$ \frac{1}{2}(e^x-e^{-x}) = y . $$
So we find ourselves w... | If you multiply both sides of the equation
$$y = \frac{e^x - e^{-x}}{2}$$
by $2e^x$, you obtain
\begin{align*}
2e^{x}y & = e^{2x} - 1\\
0 & = e^{2x} - 2e^{x}y - 1
\end{align*}
so you should have obtained the equation $t^2 - 2ty - 1 = 0$ when you made the substitution $y = e^x$.
To find the inverse, we interchange $x$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1180824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Maximum number of prime sums Zach thinks of six different positive integers, and for each pair of numbers, he adds them. This gives him $\binom{6}{2}=15$ sums. Among these sums, find the maximum number that can be prime numbers. I deducted that there has to be atleast one even number among the $6$ numbers. But will the... | The sum of two different positive integers is always at least $3$, and so in order for it to be prime, it has to be odd. So one summand has to be even, the other odd. If there are $a$ even numbers and $b = 6-a$ odd numbers, then the potential number of primes is at most $ab \leq 9$, attained when $a=b=3$. Can this boun... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proof of no primes such that $x^2 + y^2 = z^2$ I'm in a pretty simple "CS Math" course for year 1 Comp Sci, and I came across this:
Disprove, $x^2 + y^2 = z^2$, such that $x, y, z$ are primes
I thought of this as, if n is a prime, then prime factorization of n must be:
$n = z*1$
$n^2 = (z*1)(z*1) = (z)(z)$
So I tried t... | I think this proof is sufficient:
If x, y, z are primes, $x^2 + y^2 \ne z^2$
There exists x, y, z that are primes such that $x^2 + y^2 = z^2$, derive contradiction
Then: $x^2 = z^2 - y^2$
Then: $x^2 = (z-y)(z+y)$
Then: $z - y = 1$ and $z + y = x^2$, because by the Fundamental Theorem of Arithmetic, the prime factoriza... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1182494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Normalizer and Centralizer of Upper Triangular Matrix Consider the group $H_3(\mathbb{Z})=\left\{\begin{pmatrix}1 & a & c\\0 & 1 &
b\\0 & 0 & 1\end{pmatrix}:a, b, c\in\mathbb{Z}\right\}$.
Find $Z\left(\begin{pmatrix}1 & 1 & 0\\0 & 1 & 1\\0 & 0 &
1\end{pmatrix}\right)$.
Find $N_{H_3(\mathbb{Z})}\left(\left\{\begin{pmatr... | To find the centralizer of
$$
A = \begin{bmatrix}1 & 1 & 0\\0 & 1 & 1\\0 & 0 &
1\end{bmatrix}
$$
you need to solve the equation
$$
A X = X A,
$$
where
$$
X = \begin{bmatrix}1 & a & c\\0 & 1 & b\\0 & 0 &
1\end{bmatrix}.
$$
This will yield three linear equations which, as you will see, admit the immediate solution $a = b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1182857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve $\int\frac{dx}{\sqrt{4-x^2}}$ with trigonometric substitution? The integral
$$\int\frac{dx}{\sqrt{4-x^2}}$$
I've found the variable
$$x=2\sin\theta$$
$$x^2=4\sin^2\theta$$
$$dx=2\cos\theta\,d\theta$$
Which gave me by substitution
$$\int\frac{2\cos\theta}{4-\sqrt{4\sin^2\theta}}\,d\theta$$
$$\int\frac{\cos\... | Substitution of $x=2\sin\theta$ implies that $\sqrt{4-x^2}=\sqrt{4-4\sin^2\theta}=2\cos\theta$. Then,
$$\int \frac{dx}{\sqrt{4-x^2}}=\int \frac{\cos\theta d\theta}{\cos\theta}=\theta+C$$
where $C$ is a constant of integration. But, $\theta=\arcsin\left(\frac{x}{2}\right)$. So,
$$\int \frac{dx}{\sqrt{4-x^2}}=\arcsin\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1183095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
differential equation; how to find the general solution. how do i get the general solution of $$\frac{dy}{dx} = \frac{(10x^2-1)y}{ x(10x^2+7x+1)}$$ please? i have been trying for a while now. I have been doing it as a separable DE and I get $$\frac {dy}y = \frac{(10x^2-1)\, dx}{x(10x^2+7x+1)} $$I can not seem to carr... | The solution is as follows
\begin{align}
\frac{dy}{dx} = \frac{(10x^2-1)y}{x(10x^2+7x+1)}
\end{align}
can be seen to be of the form
\begin{align}
\frac{dy}{dx} = y \left( \frac{2}{2x+1} + \frac{5}{5x+1} - \frac{1}{x} \right)
\end{align}
or
\begin{align}
\frac{dy}{y} = \left(\frac{2}{2x+1} + \frac{5}{5x+1} - \frac{1}{x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1183189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Definition of multiplication of complex numbers I know that given two complex numbers $z_1 = a + bi$ and $z_2 = c + di$, the multiplication of these two numbers is defined as
$$
z_1*z_2 = (ac - bd) + i(ad + cb)
$$
I also know that I can easily derive this formula by applying the distributive property of multiplication ... | Indeed, $i^2=-1$ is coherent with the definition of multiplication, but things appeared in the reverse order.
While working on the resolution of algebraic equations (roots of polynomials), the need arose to consider square roots of negative numbers. Hence the "imaginary" symbol $i=\sqrt{-1}$, which is enough to represe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1185291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 3
} |
Evaluation of $\int\frac{1}{x^4-5x^2+16}dx$
Evaluation of $\displaystyle \int\frac{1}{x^4-5x^2+16}\,dx$
$\bf{My\; Try::}$ Given $$\displaystyle \int\frac{1}{x^4-5x^2+16}dx = \frac{1}{8}\int\frac{\left(x^2+4\right)-\left(x^2-4\right)}{x^4-5x^2+16}\,dx$$
So We get $$\displaystyle = \frac{1}{8}\int\frac{x^2+4}{x^4-5x^2+... | Use hyperbolic functions. The integrand can be stated as $ \frac {4}{4x^4 - 20x^2 + 64} = \frac {4}{(2x^2 - 5)^2 + 39} $
Let $ x = \sqrt{ \frac{5}{2} }\cosh y $, then $ \mathrm{d} x = \sqrt{ \frac {5}{2} } \sinh y \space \mathrm{d} y $
The integral becomes
$ \displaystyle 4 \cdot \sqrt{ \frac {5}{2} } \int \frac {\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1186356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How can we solve $y^2=x^3+23$ without trial and error? $$y^2=x^3+23$$
Are there any easy ways to solve this problem with number theory, abstract algebra, etc.?
(trial and error for mods by the way)
| $$ y^2 + 4 = x^3 + 27 = (x+3)(x^2 - 3 x + 9)$$
If $x$ were even, we would have $y^2 \equiv 3 \pmod 4,$ impossible.
So, $x$ is odd, $y$ is even. Note that $y^2 + 4 $ is divisible by $4.$ Since $x^2 - 3 x + 9$ is odd, this means $x+3$ is divisible by $4.$ That is $x \equiv 1 \pmod 4.$
As a result, since $-3 \equiv 1 \pmo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1187327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Center of mass of Right angle Trapezoid Given bases $a$, $b$, and the height $h$.
Get the $M(x,y)$ coordinates formula from point $O(0,0)$, where $M$ is center of mass. Wiki has a formula for $M(y) = \frac{h}{3}\frac{2a+b}{a+b}$. And I'm interested in how to find that formula (also for $M(x)$).
The median $c$ is divid... | The formula for the distance from the base $b$ to the center of mass of a trapezoid
is
$$\bar y = \frac{b+2a}{3(a+b)} h.$$
You can find this in many on-line sources such as Wolfram Mathworld.
You can prove this by integrating $\int_0^h y(b + (a-b)\frac yh) dy$
and dividing by the area of the trapezoid. But here's a de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1190922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Having trouble solving $\int\frac{5x^2+3x+2}{x(x+1)^2}$ I've first transformed the integral to
$$\int\frac{5x^2+3x+2}{x(x^2+2x+1)}dx$$
Which gave me
$$\frac{A}{x}+\frac{Bx+C}{x^2+2x+1}$$
$$=\frac{A(x^2+2x+1)+Bx^2+Cx}{x(x^2+2x+1)}$$
$$\frac{5x^2+3x+2}{x(x^2+2x+1)}=\frac{(A+B)x^2+(2A+C)x+A)}{x(x^2+2x+1)}$$
So I've found ... | you can see that $$\frac{5x^2 + 3x+2}{x(x+1)^2} = \frac 2 x + \frac{B}{x+1} - \frac 4{(x+1)^2}$$ by looking at the behavior of the two sides near the singularities $x = 0, x = -1.$ to find $B,$ put $x = 1.$ you will find $B = 3.$
you can now integrate $$ \int \frac{5x^2 + 3x+2}{x(x+1)^2} \, dx = 2 \ln x + 3 \ln(x+1) +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Indefinite integral of $ \int \frac{x^3}{\sqrt{x^2+1}} \,\text{d}x$. Can you please provide any sort of hint or suggestion on how to find the following indefinite integral?
$$\int\frac{x^3}{\sqrt{x^2+1}}\text{d}x$$
I tried substituting everything but it didn't work. I also tried trigonometric substitution but I couldn'... | $$
\begin{aligned}
\int \frac{x^{3}}{\sqrt{x^{2}+1}} d x &=\int x^{2} d\left(\sqrt{x^{2}+1}\right) \\
&=x^{2} \sqrt{x^{2}+1}-2 \int x \sqrt{x^{2}+1} d x \\
&=x^{2} \sqrt{x^{2}+1}-\frac{2}{3}\left(x^{2}+1\right)^{\frac{3}{2}}+C \\
&=\frac{\sqrt{x^{2}+1}}{3}\left[3 x^{2}-2\left(x^{2}+1\right)\right]+C \\
&=\frac{\sqrt{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Locus of a midpoint Let $Γ_1$ be a circle of radius $4$, and let $Γ_2$ be a circle of radius $14$. The distance between the centers of $Γ_1$ and $Γ_2$ is $25$. Let $A$ be a variable point on $Γ_1$, let $B$ be a variable point on $Γ_2$, and let $M$ be the midpoint of $AB$. Let $S$ be the set of all possible locations of... | We can write
$$A = A_0 + 2 a (\cos\theta, \sin\theta) \qquad B = B_0 + 2 b (\cos\phi, \sin\phi)$$
where $A_0$ and $B_0$ are the centers of the circles, and $2a$ and $2b$ are the radii. Then
$$M = C_0 + (a\cos\theta + b \cos\phi, a\sin\theta+b\sin\phi)$$
where $C_0 = \frac12(A_0+B_0)$. Note that
$$\begin{align}
|\overli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
$ \sum _{n=1}^{\infty} \frac 1 {n^2} =\frac {\pi ^2}{6} $ then $ \sum _{n=1}^{\infty} \frac 1 {(2n -1)^2} $ If $ \sum _{n=1}^{\infty} \frac 1 {n^2} =\frac {\pi ^2}{6}
$ then $ \sum _{n=1}^{\infty} \frac 1 {(2n -1)^2} $
Dont know what kind of series is this. Please educate.
How to do such problems?
| $$\sum_{n=1}^\infty \frac1{n^2} =\sum_{n=1}^\infty \frac1{(2n)^2} + \sum_{n=1}^\infty \frac1{(2n-1)^2}=\frac14 \sum_{n=1}^\infty \frac1{n^2} + \sum_{n=1}^\infty \frac1{(2n-1)^2}$$
So, $\sum\limits_{n=1}^\infty \frac1{(2n-1)^2}= \frac34 \sum\limits_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}{8}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1194009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is there another way to solve this integral? My way to solve this integral. I wonder is there another way to solve it as it's very long for me.
$$\int_{0}^{\pi}\frac{1-\sin (x)}{\sin (x)+1}dx$$
Let $$u=\tan (\frac{x}{2})$$
$$du=\frac{1}{2}\sec ^2(\frac{x}{2})dx $$
By Weierstrass Substitution
$$\sin (x)=\frac{2u}{u^2+1... | The question asked "is there another way", so here's another one. Start by removing the trig from the numerator:
$$ \int_0^{\pi} \frac{1-\sin{x}}{1+\sin{x}} \, \mathrm{d}x =
\int_0^{\pi} \frac{2 - (1 +\sin{x})}{1+\sin{x}} \, \mathrm{d}x =
\int_0^{\pi} \left( \frac{2}{1+\sin{x}} - 1 \right) \, \mathrm{d}x =
-\pi + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1194139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 9,
"answer_id": 3
} |
Various evaluations of the series $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$ I recently ran into this series:
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$$
Of course this is just a special case of the Beta Dirichlet Function , for $s=3$.
I had given the following solution:
$$\begin{aligned}
1-\frac{1}{3^3}+\frac... | Method by Fourier Series
Consider the function $f(x) = x(1 - x)$, $0 \le x \le 1$. It has Fourier sine series expansion
$$f(x) = \frac{8}{\pi^3}\sum_{n = 1}^\infty \frac{1}{(2n-1)^3}\sin{(2n-1)\pi x}.$$
Setting $x = \frac{1}{2}$ results in
$$\frac{1}{4} = \frac{8}{\pi^3}\sum_{n = 1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3},... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
A geometric proof about decagons. Let us have a regular decagon. Prove that the radius of the circle that can be drawn around it (if I have drawn correctly, that's the blue line) equals $|ad|-|ab|$.
How can I prove this statement, any ideas? :)
| WLOG you can take the radius to be unity. Using the Cosine Law we can easily say that $AD=2\sin\dfrac{3\pi}{10}$ and $AB=2\sin\dfrac{\pi}{10}$ . Then the difference $AD-AB:$
$$2\left(\sin\dfrac{3\pi}{10}-\sin\dfrac{\pi}{10}\right)=4\cos\dfrac{\pi}{5}\sin\dfrac{\pi}{10}={4\cos\dfrac{\pi}{5}\sin\dfrac{\pi}{10}\cos\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Finf $f(x)$ which is a second degree polynomial, such that $f(1)=0$ and $f(x) = f(x-1)$ I must find a function $f(x) = ax^2+bx+c$ such that:
$$f(1) = a+b+c=0\\f(x)=f(x-1)\implies ax^2+bx+c = a(x-1)^2+b(x-1)+c\implies\\ax^2+bx+c = ax^2+(-2a+b)x+a-b+c\implies\\a = a, b = -2a+b, c = a-b+c$$
but this results fo $a=b=c=0$. ... | If $f$ is a polynomial and $f(1)=0$ together with $f(x)=f(x-1)$, then $$0=f(1)=f(2)=f(3)=f(4)=\ldots $$
and $f$ has an infinite number of real roots. That implies $f\equiv 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Finding the limit $\lim_{x\rightarrow \infty} \sqrt[3]{x+1}-\sqrt[3]{x}$ I am trying to find this limit
$$\lim_{x\rightarrow \infty} \sqrt[3]{x+1}-\sqrt[3]{x}$$
My so far method is this
*
*$f(x)>0.$
*$f^{\prime}(x)=\frac{1}{3\sqrt[3]{(x+1)^2}}-\frac{1}{3\sqrt[3]{x^2}}<0.$
*For every $0<y<1$ the equation $f(x)=y$ ... | use the binomial theorem to expand $$(x+1)^{1/3} = x^{1/3} + \frac 13 x^{-2/3}+\cdots $$ so $$ (x+1)^{1/3} - x^{1/3} = \frac 13 x^{-2/3}+\cdots \rightarrow 0\text{ as } x \to \infty.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
Solve this system of equation Solve this system of equations for real $x$ and $y$:
*
*$5x\left(1+\dfrac{1}{x^2+y^2}\right)=12$
*$5y\left(1-\dfrac{1}{x^2+y^2}\right)= 4$
I juggled with those equations and got $x-y+\dfrac{x+y}{x^2+y^2}=\dfrac{8}{5}$, from where I guessed a solution $(2,1)$.
But I don't know how to ... | \begin{align}
5x\left(1+\frac{1}{x^2+y^2}\right)&=12\\
5y\left(1-\frac{1}{x^2+y^2}\right)&=4\\
\end{align}
Obviously, $x\neq0$ and $y\neq0$.
\begin{align}
\left(1+\frac{1}{x^2+y^2}\right)&=\frac{12}{5x}\\
\left(1-\frac{1}{x^2+y^2}\right)&=\frac{4}{5y}\\
1&=\frac{6}{5x}+\frac{2}{5y}&=\frac{6y+2x}{5xy}\\
\frac{1}{x^2+y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove that the algebraic expressions are equivalent.
$$\frac{3^{k+1}-1}{2} + 3^{k+1} = \frac{3^{k+2}-1}{2}$$
with steps make left hand side = right hand side by modifying one or both expressions
Thanks for your help guys, I solved it like this:
$$\frac{3^{k+1}-1}{2} + 3^{k+1} = \frac{3^{k+1}-1}{2} + \frac{2*3^{k+1}}... | Hint. $$
2\cdot 3^{k+1}=(3-1)\cdot 3^{k+1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1201612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proof by Induction Problem I need to prove that for natural numbers $a$, and positive integers $n$, the number $a^{2n+1}-a$ is divisible by $6$.
I have proved the case when $n=1$, that $a^3-a$ is divisible by $6$. I'm having trouble proving that if $a^{2n+1}-a$ is divisible by $6$, then $a^{2(n+1)+1}-a$ is also divisi... | hint: $a^{2n+1} - a = a(a^{2n}-1) = a(a^2-1)(a^{2n-2} + a^{2n-4} + \cdots + 1) = a(a-1)(a+1)(.....)$ is divisible by $6$ since it has a product of $3$ consecutive integers: $a-1,a,a+1$.
To do an induction proof.
The base case is done. Assume $6\mid a^{2n+1} - a \Rightarrow a^{2n+3} - a = (a^{2n+3} - a^{2n+1}) + (a^{2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1202923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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calculating an integral $\int\frac{x \ln x}{(1+x^{2})^{2}}dx $ I tried to do subtition and to integrate by parts, unsuccessfully.
$\int\frac{x \ln x}{(1+x^{2})^{2}}dx
$
Thanks in advanced.
| $$\int \frac{x}{(1+x^2)^2}=\frac{1}{2}\int\frac{2x}{(1+x^2)^2}=\frac{1}{2}\int\frac{1}{(1+y)^2}=-\frac{1}{2}\frac{1}{1+y}=-\frac{1}{2}\frac{1}{1+x^2},$$ with $y=x^2$, was easy to integrate and its integral is rational.
On the other hand, $(\ln(x))'=\frac{1}{x}$ is rational.
Therefore, integration by parts rationalizes ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1205812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove this limits with $\sin{(\tan{x})}-\tan{(\sin{x})}$ How Find limit
$$\lim_{x\to 0}\dfrac{\sin{(\tan{(\sin{(\tan{x})})})}-\tan{(\sin{(\tan{(\sin{x})})})}}
{\sin{(\tan{x})}-\tan{(\sin{x})}}$$
My approach is the following: I use wolframalpha found this limits is $2$
| For this kind of problems, Taylor series are very useful. Just start from the series of the most inner terms, cascade (replacing and simplifying) and be patient !$$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+\frac{17 x^7}{315}+O\left(x^9\right)$$ $$\sin(\tan(x))=x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{55 x^7}{1008}+O\left(x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1207074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Application of Implicit Function Theorem for Constrained Optimization Here's the problem:
Consider the subset $S \subset \mathbb R$ defined by
$$ x^4+2xy+y^4+yz+z^3 = 2 $$
Show that there exists a $C^1$ function $g: \mathbb R^2 \to \mathbb R$ defined near $(1,1)$ such that for $(x,y,z)$ near $(1,1,-1)$, then $(x,y,z) ... | From the equation:
$${x^4} + 2xy + {y^4} + yz + {z^3} = 2$$
We define a function:
$$F(x,y,z)= {x^4} + 2xy + {y^4} + yz + {z^3} - 2$$
Equation reads now:
$$F(x,y,z) = 0$$
Because of
$$dF(x,y,z) = (4{x^3} + 2y)dx + (2x + 4{y^3} + z)dy + (y + 3{z^2})dz$$
we have:
$$\nabla F(x,y,z) = \left( {\begin{array}{*{20}{c}}
{4{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1207948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to solve for $x$ in $2(x-5) + 4 (x-3) = -30$ In $2(x-5) + 4 (x-3) = -30$... I'm very confused as to how to solve for $x$, the correct response is $-2$, but I keep getting $4/3$.
| Assuming you transcribed the problem correctly, your answer is correct.
\begin{align*}
2(x - 5) + 4(x - 3) & = -30\\
x - 5 + 2(x - 3) & = -15 && \text{divide both sides of the equation by $2$}\\
x - 5 + 2x - 6 & = -15 && \text{distribute}\\
3x - 11 & = -15 && \text{combine like terms}\\
3x & = -4 && \text{add $11$ to e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1208440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Closed form for an almost-elliptic integral Does
$$\int_0^{2 \pi} \log\left(\frac{1}{2}[1+\sqrt{1-(a \sin\phi)^2}]\right) d\phi $$ have a closed form ?
An approximation for small $a$ is $2E-\pi$, but it is the exact form that is needed for any $|a|<1$.
The integration standing to Jack is
$$ -\frac{\pi a^2}{4}\p... | For any $b\in(0,1)$ we have:
$$I(b)=\int_{0}^{2\pi}\log\left(1+\sqrt{1-b^2\sin^2 x}\right)\,dx=4\int_{0}^{1}\frac{\log\left(1+\sqrt{1-b^2 t^2}\right)}{\sqrt{1-t^2}}\,dt$$
hence:
$$ I(b)=\frac{1}{b}\int_{0}^{b}\frac{\log(1+\sqrt{1-u^2})}{\sqrt{1-\frac{u^2}{b^2}}}\,du=\frac{1}{b}\int_{0}^{\arcsin b}\frac{\cos\theta}{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1208727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Probability that minimum of two numbers is less than 4 Suppose I have to choose two numbers from set $$S=\{1,2,3,4,5,6 \}$$ without a replacement , then what is the probability that minimum of two is less than $4$?
I made two groups for this problem $A= \{1,2,3 \}$ and $B=\{4,5,6 \}$
.There are two possibilities , eith... | It is easy as P.I.E.. Use the Principle of Inclusion and Exclusion.
$$\begin{align}
\mathsf P(\min(X,Y) < 4) & = \mathsf P(X < 4)+\mathsf P(Y < 4)- \mathsf P(X< 4 \cap Y< 4)
\\ &= \frac 3 6 + \frac 3 6 - \frac 3 6\cdot\frac 2 5
\\ & = \frac 4 5
\end{align}$$
Or you can use the Law of Complements
$$\begin{align}
\maths... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1213148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Eritrea's Theorem According to this newspaper, an Eritrean high school student named Saied Mohammed Ali has discovered a new geometric theorem. Another source seems to say that it's the following:
Say you have a triangle, with sides of length $a$, $b$, and $c$. Draw the medians (lines $\overline{AG}$, $\overline{BI}$, ... | Express α, β, γ in terms of a, b, c
According to the Encyclopedia of Triangle Centers, the orthocenter $X(4)$ has barycentric coordinates $[\tan A:\tan B:\tan C]$. From the cosine law you have $\cos C=\frac{a^2+b^2-c^2}{2ab}$ and likewise for the other angles. So you get
\begin{align*}
\tan C&=\frac{\sin C}{\cos C}=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1215178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 1
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Reverse engineering a Taylor expansion 2 So there is the sum:
$$S(x) = \frac{x^3}{3(1!)} + \frac{x^6}{6(2!)} + \frac{x^9}{9(3!)} \text{ }...$$
and we are instructed to find the sum of the series in a small expression.
I took the derivative to cancel out the increasing terms on the bottom:
$$S'(x) = \frac{x^2}{(1!)} + ... | You properly identified that $$S'(x) = \frac{x^2}{1!} + \frac{x^5}{2!} + \frac{x^8}{3!}+ \cdots=\frac 1x\Big(\frac{x^3}{1!} + \frac{x^6}{2!} + \frac{x^9}{3!}+ \cdots\Big)=\frac{e^{x^3}-1}{x}$$ So, this write $$\frac{dS}{dx}=\frac{e^{x^3}-1}{x}$$ Now, changing variable $x=\sqrt[3]y$, this leads to $$\frac{dS}{dy}=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1215242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to differentiate $y=(x+1)^3/x^{3/2}$ and $y=2x^4/(b^2-x^2)$ I need to solve a list of derivatives to help me on an exam; however, I'm in doubt when they use another variable (constant) or when I have a fraction with functions that use the power rule.
For example:
$$y = \frac{(x + 1)^3}{x^{3/2}}.$$
This one I tried ... | You really only need the quotient rule and the power rule.
For the quotient rule, when $y=\frac{f(x)}{g(x)}$ (where $g(x)\neq 0$), we have $y'=\frac{f'g-fg'}{g^2}$.
Problem 1: Differentiate $y = \frac{(x + 1)^3}{x^{3/2}}$ with respect to $x$.
Solution. We have $y=\frac{f(x)}{g(x)}$ where $f(x)=(x+1)^3$ and $g(x)=x^{3/2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1219363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The equality $\frac{3}{4}\zeta(2) = \sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n = 1}^\infty \frac{1}{(2n)^2} = \sum_{n = 0}^\infty \frac{1}{(2n + 1)^2}$ I was reading an article and the author immediately states that the following identity is clear
$$\frac{3}{4}\zeta(2) = \sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n = 1}^\... | It's because
$$\sum_{n = 1}^\infty \frac{1}{n^2} - \sum_{n = 1}^\infty \frac{1}{(2n)^2} = \sum_{n = 1}^\infty \frac{1}{n^2} - \sum_{\underset{n\,even}{n = 1}}^\infty \frac{1}{n^2} = \sum_{\underset{n\, odd}{n = 1}}^\infty \frac{1}{n^2} = \sum_{n = 0}^\infty \frac{1}{(2n+1)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1220571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that the digits in $\frac{n}{2^k}$ (where $n$ is an integer divisible by 9 and $k$ is any integer) sum to a multiple of 9? I don't know where to start. Here are two examples of the statement:
$36/8=4.5$ and $(4+5)\mod9=0$
$9/4=2.25$ and $(2+2+5)\mod9=0$
| Hint $\ $ Multiply by a power of $10$ to get an integer multiple of $9$ with the same digits.
$\qquad x= \dfrac{9j}{2^k}\ \Rightarrow\ 10^k x = 5^k(9j)\ $ has same digits as $\,x\, $ (shifted left by $\,k$), except $0$'s at end
E.g. in your case $\ x = \dfrac{9}{2^2} = \color{#c00}{2.25}\ $ so $\,10^2 x = 5^2\cdot 9 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1221846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$A$ is a sum of two postive integer squares? if $x,y,z,w$ be postive integer,and such $x^2+y^2$ is prime number,and $A=\dfrac{w^2+z^2}{x^2+y^2}\in N^{+}$
show that
$A$ is a sum of two postive integer squares?
maybe
$$A=\dfrac{(w^2+z^2)(x^2+y^2)}{(x^2+y^2)}=\dfrac{(wx+zy)^2}{(x^2+y^2)^2}+\dfrac{(wy-xz)^2}{(x^2+y^2)^... | Approach 1
In this answer, it is shown that a number, $n$, can be written as the sum of two squares if and only if each prime in the prime factorization of $n$ which is $\equiv3\pmod4$ appears with even exponent. Thus, if
$$
\frac{w^2+z^2}{x^2+y^2}\in\mathbb{Z}
$$
each prime $\equiv3\pmod4$ in $w^2+z^2$ appears with ev... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1225309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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How to prove this limit using epsilon delta method? i know how to construct a proof using epsilon delta method but as much as i have done i have to just algebraically manipulate and i got the value but i cant get this one.
$$\lim_{x \to c} \frac{x}{1+x^2} = \frac{c}{1+c^2}$$
My try: $$\left|\frac{x^2}{1+x^2} - \frac{c^... | If $c=0,$ take $\delta = \epsilon.$ If $c\ne 0,$ take $\delta = \min (\epsilon/(1+2c^2),|c|).$ Why? We have
$$\left |\frac{(x-c)(1-xc)}{(1+x^2)(1+c^2)} \right | \le |x-c|\,|1-xc|.$$
If $c=0,$ then we're left with $|x-c|$ on the right, perfect. If $c\ne 0,$ $|x-c|<|c| \implies |x|\le |x-c|+|c|<2|c|\implies |1-xc| < 1+2c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1225673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Laurent series of $\frac1{\sin^2z}$ around 0 I tried to expand $\frac{z^2}{\sin(z)^{2}}$ using Taylor expansion, but the coefficient involved some limit of $\frac{0}{0}$ and was really difficult to calculate. (I tried to convince myself the constant term of the Taylor expansion is 1 because $\lim_{x\rightarrow 0}(\fra... | You may just write, as $z \to 0$:
$$\begin{align}\frac{1}{\sin^2z}&=\frac{1}{\left(z-\dfrac{z^3}{3!}+\dfrac{z^5}{5!}+\mathcal{O}(z^7)\right)^2}\\\\&=\frac{1}{z^2\left(1-\dfrac{z^2}{3!}+\dfrac{z^4}{5!}+\mathcal{O}(z^6)\right)^2}
\\\\&=\frac{1}{z^2}\left( 1-2\times \left(-\frac{z^2}{3!}+\frac{z^4}{5!}\right)+3\times \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1227783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluate: $\int \frac{1}{\sqrt{x}\sqrt{1-x}}\:dx$ Evaluate: $\int \frac{1}{\sqrt{x}\sqrt{1-x}}\:dx$
I solved this question using u-substitution.
But I got two answers:
1) When $x=u^2 \implies \int \frac{1}{\sqrt{x}\sqrt{1-x}}\:dx =
$2arcsinx $\left(\sqrt{x}\right)$ +c and
2) When $\sqrt{1-x}=u$ ===> $\int \frac{1}{\... | another way is :
$$\int \frac{dx}{\sqrt{x}\sqrt{1-x}}\\=\int \frac{dx}{\sqrt{x-x^2}}\\=\int \frac{dx}{\sqrt{\frac{1}{4}-\frac{1}{4}+x-x^2}}\\=\int \frac{dx}{\sqrt{\frac{1}{4}-(x-\frac{1}{2})^2}}\\=\int \frac{2dx}{\sqrt{1-(2x-1)^2}}\\$$now use this substittuion $u=2x-1$ so $du=2dx$ $$\int \frac{du}{\sqrt{1-u^2}}=arcsin(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1228641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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limit using taylor series I keep getting an error in the expansion
$$\lim_{x\to 0 }\frac{2\exp(\sin(x))-2-x-x^2-\arctan (x) }{x^3}$$
The numerator works out as
$$\approx 2 (1+\sin(x) + 1/2 \sin^2(x)) -2-x-x^2 - (x-x^3/3+x^5/5)$$
$$\approx 2(1+x-x^3/3!+1/2(x-x^3/3!)^2 -2-2x+x^3/3-x^2 -x^5/5 $$
$$ = o(x^4)$$
so that the ... | $$\newcommand{\b}[1]{\left(#1\right)}
\newcommand{\O}{{\mathbb O}}
\sin x=x-\frac16x^3+\O(x^5)\\
e^x=1+x+\frac12x^2+\frac16x^3+\O(x^5)\\
\arctan x=x-\frac13x^3+\O(x^5)$$
Now:
$$e^{\sin x}=1+\sin x+\frac12\sin^2 x+\O(x^3)\\
=1+\b{x-\frac16x^3+\O(x^5)}+\frac12\b{x-\frac16x^3+\O(x^5)}^2+\frac16\b{x-\frac16x^3+\O(x^5)}^3+O... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1229968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Convergence of the series: $\sum_{n\geq1}\frac{(-1)^n\arctan (n)}{n+n^{1/2}}$ $$\sum_{n=1}^{\infty} (-1)^n \tan^{-1}(n)/(n+(n)^{1/2}).$$
I know that the series is not absolutely converges.
I want to prove using Alternative test. I don't know how to prove that sequence $ tan^{-1}n/(n+(n)^{1/2})$ is decreasing sequenc... | Certainly using a comparison test is easier, but here's what the Alternating Series test requires.
1) Since the arctangent function has an upper bound of $ \ \frac{\pi}{2} \ $ , it is reasonably clear that $$ \ \lim_{n \rightarrow \infty} \ \ \frac{\arctan \ n}{n \ + \ n^{1/2}} \ \ = \ \ 0 \ \ , $$
so the series passe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1230612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If the distance between two lines is $ \frac{1}{\sqrt{3}} $, then $ \alpha $ is...
If the shortest distance between the lines $\displaystyle \frac{x-1}{\alpha}=\frac{y+1}{-1}=\frac{z}{1}\;, (\alpha \neq - 1)$ and
$x+y+z+1=0 = 2x-y+z+3 = 0$ is $\displaystyle \frac{1}{\sqrt{3}}\;,$ Then $\alpha = $
$\bf{My\; Solution::... | I would go about doing it like this: put your lines into vector notation, in this case, $$(x_1,y_1,z_1)=(0,1,0)+t_1(2,1,0)$$ $$(x_2,y_2,z_2)=(1,-1,0)+t_2(\alpha,-1,1)$$ Then solve the messy system of equations yielded by$$\frac{\partial}{\partial t_1 }||(x_1,y_1,z_2)-(x_2,y_2,z_2)||^2=0$$ $$\frac{\partial}{\partial t_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1232568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Matrix addition/multiplication with different sizes I have the following two matrices:
$$A=\begin{pmatrix}1 & -2\\3 & 1\end{pmatrix}\text{ and }B=\begin{pmatrix}1 & 3 &
2\\-1 & 0 & 2\end{pmatrix}$$
So I have two matrixes with different sizes. Multiple sources tell me that I can't do multiplication or addition with ma... |
So I have two matrixes with different sizes. Multiple sources tell me that I can't do multiplication or addition with matrix of different sizes.
You can't add matrixes of different sizes as stated by @meshal. Hence, A+B or B+A can't be performed. However, you can multiply them. Here's how:
$$A=\begin{pmatrix}1 & -2\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1232835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Trigonometric integrals How do I evaluate this indefinite integral ? Integral $$\int\frac{x^2+n(n-1)}{(x\sin(x)+n\cos(x))^2}dx$$ What type of integral is it ? Is there any intuition involved in the approach to solve it?
Edit: The complete term in denominator has 2 as exponent. There was a typo previously.
| $\bf{My\; Solution:}$ Using $$\displaystyle (x\cdot \sin x+n\cdot \cos x) = \sqrt{x^2+n^2}\left\{\frac{x}{\sqrt{x^2+n^2}}\cdot \sin x+\frac{n}{\sqrt{x^2+n^2}}\cdot \cos x\right\}$$
$$\displaystyle = \sqrt{x^2+n^2}\cdot \cos\left(x-\phi\right)\;,$$ where $$\displaystyle \sin \phi = \frac{x}{\sqrt{x^2+n^2}}$$ and $$\disp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Rotations of complex graphs Let $c_1 = -i$ and $c_2 = 3$. Let $z_0$ be an arbitrary complex number. We rotate $z_0$ around $c_1$ by $\pi/4$ counter-clockwise to get $z_1$. We then rotate $z_1$ around $c_2$ by $\pi/4$ counter-clockwise to get $z_2$.
There exists a complex number $c$ such that we can get $z_2$ from $z_0... | Geometrically, this is simply stating that we wish to define a circle with center $c$ such that the points $z_0$ and $z_2$ lie on this circle.
In the complex plane, a point can be describe in polar coordinates by $re^{i\theta}+c$ for $0 \leq \theta < 2 \pi$ where $c$ is the center and $r$ is the radius.
The first poi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1238448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluate $\lim_{n \to \infty} \int_{0}^1 \frac{n+1}{2^{n+1}} \left(\frac{(t+1)^{n+1}-(1-t)^{n+1}}{t}\right) \mathrm{d}t$
Evaluate
$$\lim_{n \to \infty} \int_{0}^1 \frac{n+1}{2^{n+1}} \left(\frac{(t+1)^{n+1}-(1-t)^{n+1}}{t}\right) \mathrm{d}t$$
For this integral, I have tried using integration by parts and then e... | First, consider the following two Lemmas,
Lemma $1$: $$\lim_{n \to \infty} \sum_{r=0}^n \left(\dfrac{1}{\displaystyle\binom{n}{r}}\right) =2$$
Proof : First of all, note that the limit exists, since, if we let
$$\text{S}(n)=\displaystyle \sum_{r=0}^n \left(\dfrac{1}{\dbinom{n}{r}} \right)$$
then $\text{S}(n+1)<... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1238873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Proving $\int_{0}^{1}\sqrt{\frac{1-x}{1+x}}dx=\frac{π}{2}-1$ Proving $$\int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx= \frac{π}{2}-1$$
My attempt is:
I assumed the $1-x=u$
$du =-dx$
$$\int_0^1 \sqrt{\frac{u}{2+u}}\,(-du)$$
| Let $u = \sqrt{\dfrac{1-x}{1+x}} \to u^2 = \dfrac{1-x}{1+x} \to x = \dfrac{1-u^2}{1+u^2} \to dx = \dfrac{-4u}{(1+u^2)^2}du \to I = \displaystyle \int_{0}^1\dfrac{4u^2}{(1+u^2)^2}du$. Next let $u = \tan \theta \to du = \sec^2 \theta d\theta \to I = \displaystyle \int_{0}^{\pi/4} \dfrac{4\tan^2 \theta}{\sec^4 \theta}\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1239885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Contour integration
Consider the real-valued function
$$u(t) = \frac{1}{13-12\cos(t)}$$
By converting it to a contour integral along the unit circle in $\mathbb{C}$, evaluate
$$\int_0^{2\pi} u(t)\;dt$$
I have attempted so solve this and have found the residue to be $-\frac{1}{5}$ and therefore the integral to b... | It looks like your work is correct.
I get,
$$ \int_0^{2\pi} u(t)\; dt = \int_0^{2\pi} \frac{ie^{it}}{ie^{it} \left(13-12\frac{e^{it} + e^{-it}}{2}\right)}\; dt$$
$$ = \frac{1}{i} \oint_C \frac{dz}{-6z^2+13z-6},$$
where $C$ is the unit circle. From the quadratic equation, I find that $-6z^2+13z-6 = 0$ for $z = \frac{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1241213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_0^{1/\sqrt{3}}\sqrt{x+\sqrt{x^2+1}}\,dx$
I want to find a quick way of evaluating $$\int_0^{1/\sqrt{3}}\sqrt{x+\sqrt{x^2+1}}\,dx$$
This problem appeared on the qualifying round of MIT's 2014 Integration Bee, which leads me to think there should be a shortish way (no more than three minutes by hand) to ... | Using formula:
$$\sqrt{a+\sqrt{b}}=\sqrt{\frac{1}{2} \left(a+\sqrt{a^2-b}\right)}+\sqrt{\frac{1}{2} \left(a-\sqrt{a^2-b}\right)}$$
so:
$$\sqrt{x+\sqrt{x^2+1}}=\frac{\sqrt{-i+x}}{\sqrt{2}}+\frac{\sqrt{i+x}}{\sqrt{2}}$$
and then:
$$\int \left(\frac{\sqrt{-i+x}}{\sqrt{2}}+\frac{\sqrt{i+x}}{\sqrt{2}}\right) \, dx=\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1241346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 7,
"answer_id": 6
} |
Find the values of $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$
Calculate $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$. accurate upto two decimal places or in surds .
$\begin{align}\sin 69^{\circ}&=\sin (60+9)^{\circ}\\~\\
&=\sin (60^{\circ})\cos (9^{\circ})+\cos (60^{\circ})\sin (9^{\circ})\\~\\
&=\dfr... | Please refer to : derivation of sin 18 on this page
Once you know sin 18, you can find sin 9, cos 9 etc. by using half angle formulas.
In brief:
sin 72° = 2 sin 36° cos 36° by the double angle relationship.
sin 72° = 4 sin 18° cos 18° (1 - 2sin^2 18°) by the double angle relationship, again.
cos 18° = 4 sin 18° c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Computing $\int_3^5 \frac{x^2\,dx}{\sqrt{(x-3)(5-x)}}$
$$
\int_3^5 \frac{x^2\,dx}{\sqrt{(x-3)(5-x)}}
$$
how? $x^2/\sqrt{8x-x^2-15}$ and what to do then?
| Set $x=3+2\sin^2(t)$. We then obtain
$$\int_3^5 \dfrac{x^2dx}{\sqrt{(x-3)(5-x)}} = \int_0^{\pi/2} \dfrac{2(3+2\sin^2(t))\sin(2t)dt}{\sqrt{(2\sin^2(t))\cdot(2\cos^2(t))}} = \int_0^{\pi/2}2(3+2\sin^2(t))dt$$
I trust you can finish from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1245223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve for $x:1 + \tan^2(x) = 8\sin^2(x)$ I have a tricky problem , I tried several methods and I can't seem to get a definite answer.
$1 + \tan^2(x) = 8\sin^2(x), x \in [\frac{\pi}{6} , \frac{\pi}{2}]$
I got to $8\cos^4(x)-8\cos^2(x)+1=0$ and found that $\cos^2(x) = \frac{1}{4}[2-\sqrt{2}]$ but that is not too useful... | Here's another approach. Use the facts that $\tan^2(x) = \frac{\sin^2(x)}{\cos^2(x)}$ and $\cos^2(x) = 1 - \sin^2(x)$ to rewrite the equation so that the only trigonometric expression in it is $\sin^2(x)$. Then let $u = \sin^2(x)$. Now your equation should read:
$$1 + \frac{u}{1-u} = 8u$$
Now solve for $u$, then bac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1248525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Evaluation of $\int\frac{1}{x^2.(x^4+1)^{\frac{3}{4}}}dx$
Evaluate the integral
$$\int\frac{1}{x^2\left(x^4+1\right)^{3/4}}\,dx$$
My Attempt:
Let $x = \frac{1}{t}$. Then $dx = -\frac{1}{t^2}\,dt$. Then the integral converts to
$$
-\int \frac{t^3}{(1+t^4)^{3/4}}\,dt
$$
Now Let $(1+t^4) = u$. Then $t^3\,dt = \frac{1... | Let $$\displaystyle I = \int\frac{1}{x^2\cdot (x^4+1)^{\frac{3}{4}}}dx = \int \frac{(x^4+1)-x^4}{x^2\cdot (x^4+1)^{\frac{3}{4}}}dx$$
So we get $$\displaystyle I = \int\left[\frac{(x^4+1)^{\frac{1}{4}}-x^4\cdot (x^4+1)^{-\frac{3}{4}}}{x^2}\right]dx = \int \left[\frac{x(-x^3)\cdot (1+x^4)^{-\frac{3}{4}}+(x^4+1)^{\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1250935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Finding triplets $(a,b,c)$ such that $\sqrt{abc}\in\mathbb N$ divides $(a-1)(b-1)(c-1)$ When I was playing with numbers, I found that there are many triplets of three positive integers $(a,b,c)$ such that
*
*$\color{red}{2\le} a\le b\le c$
*$\sqrt{abc}\in\mathbb N$
*$\sqrt{abc}$ divides $(a-1)(b-1)(c-1)$
Exampl... | Too long for a comment:
In addition to the rather lengthy
\begin{align}
&(m^2,\\
&((-1)^{2 k} \left(2 (-1)^k k m+(-1)^{k+1} (m+2)+m-6\right)^2)/16,\\
&\left((-1)^k \left(2 (-1)^k k m+(-1)^{k+1} (m+2)+m-6\right)+1\right)^2/4)\\
\end{align}
we also have $(a,b,c):$
\begin{align}
&\left(k^3+k^2+k+1,k^3+k^2+k+1,k^4\right)\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1251576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "56",
"answer_count": 3,
"answer_id": 1
} |
Does $\sum 3^{-\sqrt{n}}$ converge or diverge? I need to find out whether this series converges or diverges:
$$\sum_{n=1}^\infty \frac 1{3^{\sqrt{n}}}$$
The $n$th term, ratio, and root tests are inconclusive, Abel's test doesn't apply (or I can't think of how to separate out part of the sequence), and I can't figure ou... | Here's something different, that doesn't require comparing $n^2$ with $3^{\sqrt{n}}$, or any similar comparison:
$$\frac{1}{3^{\sqrt{n}}}\leq\frac{1}{3^{\left\lfloor\sqrt{n}\right\rfloor}}$$
And the sequence $\{\left\lfloor\sqrt{1}\right\rfloor,\left\lfloor\sqrt{2}\right\rfloor,\left\lfloor\sqrt{3}\right\rfloor,\ldots\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
} |
Finding a matrix from equation we've got the following 4x4 Matrix
$$\begin{pmatrix}
4 & -2 & 3 & 2\\
3 & 5 & 1 & -4\\
-1 & 6 & -4 & -7\\
-2 & 0 & -2 & 4
\end{pmatrix}$$
and I need to find $B$ from the equation: $(A-3I)B=0$.
i started to solve it by finding first $A-3I$. and I got:
$$\begin{pmatrix}
1 & -2 & 3 & 2\\
... | by row reducing $A-3I,$ using my ti-83, i get $$\pmatrix{1&-2&3&2\\3&2&1&-4\\-1&6&-7&-7\\-2&0&-2&1}\to\pmatrix{1&0&1&-0.5\\0&1&-1&-1.25\\0&0&0&0\\0&0&0&0} $$ we can see that two linearly independent vectors $a, b$ such that $(A-3I)a = 0, (A-3I)b = 0$ where $$a=\pmatrix{0.5\\1.25\\0\\1}, b = \pmatrix{-1\\1\\1\\0}.$$ y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Let $x$ and $y$ be real numbers such that $x^2 + y^2 = 1$. Find the maximum value of $2x - 5y$. Let $x$ and $y$ be real numbers such that $x^2 + y^2 = 1$. Find the maximum value of $2x - 5y$.
I do know how to solve this problem using trigonometry, however I need to solve it by using vectors. Except for defining vector... | The maximum value $k$ occurs when the line $2x-5y = k$ is tangent to the circle $x^2+y^2=1$.
Solving for $x$, we get $x = \frac{1}{2} (k+5y)$, so $\frac{1}{4}(k^2+10ky+25y^2) + y^2 = 1$. Now set up the quadratic and set $\Delta = 0$, as we only want there to be one value of $y$ for the intersection of the circle and li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1254585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Show that $\sin\left(\frac\pi3(x-2)\right)$ is equal to $\cos\left(\frac\pi3(x-7/2)\right)$ Show that
$$\sin\left(\frac\pi3(x-2)\right)$$
is equal to
$$\cos\left(\frac\pi3(x-7/2)\right)$$
I know that $\cos(x + \frac\pi2) = −\sin(x)$ but i'm not sure how i can apply it to this question.
| Maybe you can try
\begin{align*}\sin\left(\frac{\pi}{3}(x-2)\right) &= \sin\left(\frac{\pi}{3}x-\frac{2\pi}{3}\right)\\
&=\sin\left( \pi - \frac{\pi}{3}x+\frac{2\pi}{3}\right) \\
&=\sin\left(\frac{5\pi}{3}-\frac{\pi}{3}x\right) \\
&=\sin\left(\frac{\pi}{2}+\frac{7}{2}\frac{\pi}{3} -\frac{\pi}{3}x\right)\\
&=\cos\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1255629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Diophantine equation $x^2 + xy + y^2 = \left({{x+y}\over{3}} + 1\right)^3$. Solve in integers the equation$$x^2 + xy + y^2 = \left({{x+y}\over3} + 1\right)^3.$$
| Hint $1$ :
Let $t=\frac{x+y}{3}\in \mathbb Z$ the equation becomes:
$$y^2-3ty+(3t)^2-(t+1)^3=0 $$
a quadratic equation on $y$ which is soluble up to the condition $4t+1$ is a square.
Hint $2$: $\Delta_y=(t-2)^2(4t+1)$
Solutions $(t,y)=(a^2+a,-a^3+3a+1),(a^2+a,a^3+3a^2-1)$ for any parameter $a$
And it's your turn to do ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1256663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solving differential equation I want to solve the following differential equation with initial conditions:
$$\frac{\mathrm{d}^2 y}{\mathrm{d} x^2}=\frac{x \, y(x)}{\sqrt{1-x}}$$
But do not know how to actually solve it. Any suggestion?
| Hint:
Let $u=\sqrt{1-x}$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=-\dfrac{1}{2\sqrt{1-x}}\dfrac{dy}{du}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(-\dfrac{1}{2\sqrt{1-x}}\dfrac{dy}{du}\right)=-\dfrac{1}{2\sqrt{1-x}}\dfrac{d}{dx}\left(\dfrac{dy}{du}\right)+\dfrac{1}{4(1-x)^\frac{3}{2}}\dfrac{dy}{du}=-\dfrac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1259324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\sum_{n=1}^\infty \frac{n^2}{3^n} $ The following series converges to 3/2 but I do not know why.
$$ \sum_{n=1}^\infty \frac{n^2}{3^n} = \frac{3}{2} $$
Searching via Google did not yield anything useful. I'm wondering if there's some sort of algorithm that can be used to solve this, or better yet, a closed-f... | A solution using only the fact that $\sum_{n=0}^\infty q^n = \frac{1}{1-q}$ when $|q|<1$:
Let $S_1 = \sum_{n=1}^\infty\dfrac{n^2}{3^n}$, then $3S_1 = \sum_{n=0}^\infty\dfrac{(n+1)^2}{3^n}$, so we have $$2S_1 = 3S_1 - S_1 = 1 + \sum_{n=1}^\infty\dfrac{2n+1}{3^n} = 1 + 2S_2 + \sum_{n=1}^\infty\dfrac{1}{3^n} = 1 + 2S_2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1261114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Deriving a contradiction How can I derive a contradiction from the following nasty statement:
Assume $\sqrt{5} = a + b\sqrt[4]{2} + c\sqrt[4]{4} + d\sqrt[4]{8},$ with $a,b,c,d \in \mathbb{Q}$?
This is the last piece of an effort to prove that the polynomial $x^4-2$ is irreducible over $\mathbb{Q}(\sqrt{5}).$
| From your assumption
$$(\sqrt5-a)^2 =\sqrt2b^2+2c^2+2d^2\sqrt2 +2bc\sqrt[4]{8}+2cd\sqrt[4]{32}+4bd$$ which after rearrangement is $2c^2+4bd +(b^2+2d^2+4cd)\sqrt2+2\sqrt2bc\sqrt[4]2$.
So multiplying by $\sqrt2$ we get $$\sqrt2(\sqrt5-a)^2 = 2(b^2+2d^2+4cd)+(2c^2+4bd)\sqrt2 + 4bc\sqrt[4]2$$ Denoting the integers $2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1261478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limit and convergence of $\frac{1}{n^{n-1}}\sum_{p=2}^{n-1} \left[ {n \choose p} (n-p)^{n-2} p (-1)^p \right]$ This is a part of larger question, in which I was asked to show that a certain ratio has a limit of $e^{-1}$.
After much of algebraic manipulation, I've found this ratio to be
$$
\frac{1}{n^{n-1}}\sum_{p=2}^{... | Suppose we seek to investigate
$$\frac{1}{n^{n-1}}
\sum_{p=2}^{n-1} {n\choose p} (-1)^p p (n-p)^{n-2}.$$
Taking $n\gt 2$ this becomes
$$\frac{1}{n^{n-1}}
\left(n\times (n-1)^{n-2}
+ \sum_{p=0}^{n} {n\choose p} (-1)^p p (n-p)^{n-2}\right).$$
The first term here is
$$\frac{(n-1)^{n-2}}{n^{n-2}}
= \left(1-\frac{1}{n}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1261689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Choosing n objects from k types of objects, each of which is in limited supply Suppose I wanted to light my Christmas tree. In my basement, I find a cord that has $5$ sockets in which I can screw bulbs. I also locate $5$ red bulbs, $4$ green bulbs, and $3$ blue bulbs. How many unique strings of bulbs can I form? Let's ... | You can use dynamic programming. The first step is to find the number of unique strings without modding out by reversion. Consider the expression
$$ (r+g+b)^5 = b^5+5 b^4 g+5 b^4 r+10 b^3 g^2+20 b^3 g r+10 b^3 r^2+10 b^2 g^3+30 b^2 g^2 r+30 b^2 g r^2+10 b^2 r^3+5 b g^4+20 b g^3 r+30 b g^2 r^2+20 b g r^3+5 b r^4+g^5+5 g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1264078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the change of basis matrix P from S to S'. Consider the following bases of $\mathbb{R}^2$:
$$S=\left\{\begin{pmatrix}
1\\
-2
\end{pmatrix},\begin{pmatrix}
3 \\
-4
\end{pmatrix}\right\}$$
$$S'=\left\{\begin{pmatrix}
1\\
3
\end{pmatrix},\begin{pmatrix}
3 \\
8
\end{pmatrix}\right\}$$
Find the change of basis m... | With your notation, I think your final matrix should be
$$\begin{pmatrix} a&c\\b&d \end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1266823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solutions of the Congruence If $x^{10}\equiv 1\pmod{\!55^2}$, how do I know one must have $x^{10}\equiv 1\pmod{\!5^2}$ and $x^{10}\equiv 1\pmod{\!11^2}$?
| If $55$ divides a number $n$, then any divisor of $55$ must also divides the number $n$. $5$ and $11$ are divisors of $55$. Hence, $5$ and $11$ must both divide $x^{10}-1$. This mean we want $x$ such that
\begin{align}
x^{10} & \equiv 1\pmod5\\
x^{10} & \equiv 1\pmod{11}
\end{align}
From Fermat's little theorem, we hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1267097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Weights in the Dynkin Basis and Eigenvalues of the Cartan Generators for SU(3)? The Cartan Generators of $SU(3)$ in the three dimensional rep have eigenvalues $(1,-1,0)$ and $\frac{1}{\sqrt{3}} (1,1,-2)$. Therefore we have the weights:
$$ (1,\frac{1}{\sqrt{3}}) \quad (-1,\frac{1}{\sqrt{3}}) \quad (1,\frac{-2}{\sqrt{3}... | The thing is, by multiplying the weights in den Dynkin basis with the metric tensor, we get the weights in the simple root basis. If we want to know the weights in terms of the Cartan generator $H_1,H_2,...$ eigenvalues, we need to use the simple roots in this basis.
For $SU(3)$ the simple roots are $\alpha_1= (\frac{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Why do we invert then multiply when dividing fractions? I have looked up various sites online. They only explain it in a very basic arithmetic way. Would someone explain to me why is $\displaystyle\frac{x}{\frac{1}{2}}$ is $\displaystyle\frac{2x}{1}$?
| $\frac{x}{\frac{1}{2}}=\frac{x}{\frac{1}{2}}\cdot1=\frac{x}{\frac{1}{2}}\cdot\frac{2}{2}=\frac{x\cdot 2}{\frac{1}{2}\cdot2}=\frac{2x}{1}$
Generally,
$\frac{a}{\frac{b}{c}}=\frac{a}{\frac{b}{c}}\cdot\frac{c}{c}=\frac{ac}{b}=a\cdot\frac{c}{b}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1267788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Finding $\int_{0}^{\infty }\frac{1}{1+x^4}dx$ finding $$\int_{0}^{\infty }\frac{1}{1+x^4}dx$$
My attempt is:
let $x=\sqrt{u}$
$dx=\frac{1}{2\sqrt{u}}$
$$\int_{0}^{\infty }\frac{1}{2\sqrt{u}(1+u^2)}du$$
here I stopped because I don't know how to complete this solution. Any help please.
| One way to deal with this is to write
$$
1+x^4 = (1 + 2x^2+x^4) - 2x^2 = (1+x^2)^2 - (x\sqrt 2)^2 = (1+x^2+x\sqrt2)(1+x^2-x\sqrt 2)
$$
Then use partial fractions:
$$
\frac {Ax+B} {1+x^2+x\sqrt2} + \frac {Cx+D} {1+x^2-x\sqrt2}
$$
Notice that you can tell that these quadratic polynomials are irreducible because their dis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1268550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Proving a little tough trigonometric identity Show that $$\frac{1+\sin A}{\cos A}+\frac{\cos B}{1-\sin B}=\frac{2\sin A-2\sin B}{\sin(A-B)+\cos A-\cos B}$$ How do I get the $A-B$ term in the denominator? Is RHS to LHS easier? Thanks.
| First, notice that
$$
\frac{1 + \sin A}{\cos A} = \frac{\cos A}{1 - \sin A}
$$
Next, regroup the denomitator as
$$
\sin(A - B) + \cos A - \cos B = \cos A(1 - \sin B) - \cos B(1 - \sin A)
$$
The right hand side becomes
$$
2\frac{\sin A - \sin B}{\cos A(1 - \sin B) - \cos B(1 - \sin A)} =
2\frac{\frac{\tan A}{\cos B} - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Solve $\int_{0}^{1} \log(x)\log(1-x) dx$ without convolution Maybe it's too much to ask for, but is there a way to solve $\int \limits_{0}^{1} \log(x)\log(1-x) dx$ without convolution?
Note that $\log x =\log_e x$.
| Step 1: We prove that for each $n\in \mathbb{N}$, $$\int_{0}^{1} x^n \log(x) \, dx = -\frac{1}{(n+1)^2}$$ Starting with the identity $$\int_{0}^{1} x^n \, dx = \frac{1}{n+1}$$ differentiate both sides with respect to $n$ to get the desired result.
Step 2: Expanding $\log(1-x)$ as its Taylor series, we have $$\log(x)\lo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1273925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate $\int_{-\infty}^{\infty} \frac{1}{x^6+1}\,dx$? I am trying to understand how to evaluate the following integral.
$$\int_{-\infty}^{\infty} \dfrac{1}{x^6+1}dx$$
We start by considering $f(z) = \dfrac{1}{z^6+1}dz.$ Then we find that $z_k = \exp \left (i \left (\frac{\pi}{6}+\frac{\pi k}{3}\right)\right )$... | We first resolve the integrand into partial fractions.
$$
\frac{1}{x^{6}+1}=\frac{1}{3\left(x^{2}+1\right)}-\frac{x^{2}-2}{3\left(x^{4}-x^{2}+1\right)}
$$
Integrating both sides yields
$$
3 \int_{0}^{\infty} \frac{d x}{x^{6}+1}=\int_{0}^{\infty} \frac{d x}{x^{2}+1}-\int_{0}^{\infty} \frac{x^{2}-2}{x^{4}-x^{2}+1} d x
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1277704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Prove/disprove an inequality $\sqrt{1 + x} < 1 + \frac{x}{2} - \frac{x^2}{8}$ Is $\sqrt{1 + x} < 1 + \frac{x}{2} - \frac{x^2}{8}$ TRUE in $(0, \frac{π}{2})$ ?
I proceed by taking two functions $f(x) = \sqrt{1 + x}-1 $ and $g(x) = \frac{x}{2} - \frac{x^2}{8} $.
Then $f(0) = 0 = g(0),$ while $f(\frac{π}{2}) > g (\frac{... | $$
\begin{align}
\sqrt{1+x}-(1+\frac x2-\frac{x^2}8)
&=\dfrac{1+x-(1+\frac x2-\frac{x^2}8)^2}{\sqrt{1+x}+(1+\frac x2-\frac{x^2}8)}\\
&=\dfrac{\frac{x^3}{64}(8-x)}{\sqrt{1+x}+(1+\frac x2-\frac{x^2}8)}\\
&=\dfrac{x^3}{8x+32-\frac{64}{3+\sqrt{1+x}}}
\end{align}
$$
The denominator is monotonically increasing, and at $x=0$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1278565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Proving $9^n - 8n - 1$ is divisible by $8^2$ for $n\ge 0$? My textbook provided the following proof:
*Base case: When $n=0, 9^n-8n-1=0=64\cdot0$, so $64\mid\left(9^n-8n-1\right)$.
Induction step: Suppose that $n\in\mathbb N$ and $64\mid\left(9^n-8n-1\right)$. Then there is some integer $k$ such that $9^n-8n-1=0=64... | Express $8n$ as $72n-64n$. We hence have
$$9^{n+1} - {\color{red}{8n}} - 9 = 9^{n+1} - \color{red}{(72n-64n)} - 9 = 9^{n+1} \color{red}{- 72n + 64n} - 9 = 9^{n+1} - \color{red}{72n} - 9 + \color{red}{64n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1279028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Proof of inequality problem from Spivak, ch 1, 16 b) If $$4x^2+8xy+4y^2 \ge 0$$ it follows that $$4x^2+6xy+4y^2 \gt 0 $$ unless $x=0$ and $y=0$.
How can I prove that?
| The way that I think of it is as follows. First, we need to recognise that
$$ 4x^2+6xy+4y^2=3(x+y)^2 + x^2 + y^2$$
Then, based on the stated assumption we know that:
$$ 4x^2+8xy+y^2=4(x+y)^2 \ge 0 $$
$$ (x+y)^2\ge0$$
$(x+y)^2 = 0$ when $x=-y$ or when $x=y=0$
For all other values $(x+y)^2\gt0$
Similarly,
$$ x^2 + y^2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1282848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Simplify the expression (combination and factorial) Simplify the following expression:
$\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!}$
My attempt:
$\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n+1-3)!} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n-2)!} * \frac{(n-1)! + (n-2)!}{(n+1)!}$
a... | $$n/6$$
as ${n+1\choose 3}$ can be reduced to $(n+1)(n)(n-1)/3!$, now $(n+1)(n)(n-1)$ is canceled from the denominator and the $(n-2)!$ is canceled too so we are left with $(n-1) + 1$ numerator and $3!$ denominator. So the answer is $n/6 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1283618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Prove by induction that $1+4+7+...+(3n-2) = 2n(3n-1)$ I have an exercise where I, using induction, have to prove the following:
\begin{equation*}
1 + 4 + 7 + \ldots + (3n-2) = 2n(3n-1).
\end{equation*}
I immediately got stuck on the base case with $n=1$ because the following should be true: $1 = 2 \cdot 1 \cdot (3 \cdo... | First of all, the claim is wrong. A direct proof shows that.
The sequence $\{1,4,7,\ldots,3n-2\}$ is an arithmetic progression with common difference $3$ and there are $n$ number of terms in it. So, the sum of all the terms is:
$$\begin{align}S =1+4+7+\ldots+(3n-2)&=1+(1+3)+(1+6)+\ldots+(1+3(n-1))&\\ &=\sum_{i=1}^n(1)+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1287809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find the eigenvectors corresponding to an eigenvalue I know how to find the eigenvectors corresponding to an eigenvalue of a matrix $A$: we basically need to find the vectors of the nullspace of $\lambda I - A$, but in my case, I have a matrix $A$ like this:
$$A = \left(\begin{matrix} \cos x & -\sin x \\ \sin x & \cos ... | Eigenvector for eigenvalue 3
Given that
$$
\left( \begin{array}{cc} 3 & 0 \\ 0 & 7 \end{array} \right)
\left( \begin{array}{c} 1 \\ 0 \end{array} \right)
=
3
\left( \begin{array}{c} 1 \\ 0 \end{array} \right),
$$
we can also write
$$
\left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x)
\end{array} \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
find an expression for $A^n$ for any positive integer N I have posted part of this question in another post which has gotten to long. I need help with the second part of the question which is finding an expression for $A^n$ for any positive integer $n$.
The previous question gave eigenvalues of $λ_1 =-\sqrt{2}, λ_2 =\... | I see you have the eigenvectors and the eigenvalues of matrix $A$... So build a unitary matrix $U$ (in your case the matrix $U$ is orthogonal), with columns the eigenvectors of $A$. That is, $U=\begin{pmatrix}\ \frac{1}{2} &\frac{1}{2} &\frac{\sqrt{2}}{2}\\ \frac{1}{2} &\frac{1}{2} &\frac{-\sqrt{2}}{2}\\ \frac{\sqrt{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1289490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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A limit with the harmonic series How can we prove the following (similar) limits?
$$\sum_{k=1}^n \frac{1}{k} (\ln 2 - \frac{1}{n+2} - \frac{1}{n+3} - \cdots -\frac{1}{2n + 2}) \to 0. $$
$$\sum_{k=1}^n \frac{1}{k} (\ln 3 - \frac{1}{n+2} - \frac{1}{n+3} - \cdots -\frac{1}{3n + 3}) \to 0. $$
| We have $$\sum_{k\leq n}\frac{1}{k}\left(\log\left(2\right)-\frac{1}{n+2}-\dots-\frac{1}{2n+2}\right)=H_{n}\left(\log\left(2\right)-\sum_{j=n+2}^{2n+2}\frac{1}{j}\right)=H_{n}\left(\log\left(2\right)-H_{2n+2}+H_{n+1}\right)
$$ where $H_{n}
$ is the $n
$-th armonic number. Recalling that $$H_{n}=\log\left(n\right)+\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Prove that $\alpha$ lies between $0$ and $4$.
Let $a,b,c$ be the length of the sides of the triangle $ABC$ .
Given $(a+b+c)(b+c-a)=\alpha bc$.Then Prove that the value of $\alpha$ lies in between $0$ and $4$.
$\begin{align}(a+b+c)(b+c-a)&=\alpha bc\\
\implies \alpha&=\dfrac{b^2+c^2-a^2}{bc}+2\\
\alpha&=2\cos A+2\... | Note that in the last line you should get $2\cos A+2$, not $\frac{\cos A}{2}+2$. Now you can easily use the trivial bound $-1\leq\cos A\leq 1$ (indeed, a strict inequalities hold, as $A$ is an angle of a triangle).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1291780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Need help with taylor series.
Evaluate the limit
$$\lim\limits_{x \to 1} \frac{1-x + \ln x}{1+ \cos πx}$$
The limit im trying to get is $-\frac{1}{π^2}$ as I've solved from l'Hopitals rule.
Now I need to solve the limit by using Taylor Series and this is what i did so far
$$\begin{align*}
f(x) &= 1-x + \ln x = 1 -x... | Hint. You have, near $x=1$,
$$1-x + \ln x = 1 -x + (x-1) -(x-1)^2/2+ O(x-1)^3$$
$$1-x + \ln x = -(x-1)^2/2+ O(x-1)^3$$
and
$$1+\cos πx = 1 -1+\frac{\pi^2}2 (x-1)^2 + O(x-1)^3$$
$$1+\cos πx = \frac{\pi^2}2 (x-1)^2 + O(x-1)^3$$ thus
$$\frac{1-x + \ln x}{1+ \cos πx} =\frac{-(x-1)^2/2+ O(x-1)^3}{\frac{\pi^2}2 (x-1)^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Mean Value Theorem to show inequalities about numbers
Show $\sqrt{65}-8=\frac{1}{2\sqrt{c}}$ for some $c \in (64,65)$, and hence show:
$$8+\frac{1}{18}<\sqrt{65}<8+\frac{1}{16}$$
I managed to do the first part easily, but I don't know how to do the "hence show" part.
Working For the First Part:
Consider the funct... | The first part looks good.
From the first part we know, that there exists a $c \in (64, 65)$, such that
$$\sqrt{65} = 8 + \frac{1}{2\sqrt{c}} \; .$$
Since $64 < c < 81$ and the square root is a monotone function, we see that $$8 < \sqrt{c} < 9 \; .$$
From this it follows that
$$ \frac{1}{18} = \frac{1}{2\cdot 9} < \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Finding $\int\frac{1}{x^{11}+4x^6}dx$ I wanted to find out if there is an easy way to evaluate $$\displaystyle\int\frac{1}{x^{11}+4x^6}dx.$$
I substituted $u=x^5$ and then used partial fractions, but maybe there is a simpler way to find this.
| Let $t=\frac{1}{x}$, so then $x=\frac{1}{t}$ and $dx=-\frac{1}{t^2}dt$.
Then $\displaystyle\int\frac{1}{x^{11}+4x^6}dx=\int\frac{1}{\frac{1}{t^{11}}+\frac{4}{t^6}}\left(-\frac{1}{t^2}\right)dt=-\int\frac{t^9}{1+4t^5}dt$
$=-\int\big(\frac{1}{4}t^4-\frac{\frac{1}{4}t^4}{1+4t^5}\big)dt=-\frac{1}{20}t^5+\frac{1}{80}\ln\ver... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
A not so hard basic calculus problem? But it appears to be very lengthy Find the coordinates of the two points on the curve $y=4-x^2$ whose tangents pass through the point $(-1,7)$.
My work:
Let the two points be $(a,b)$ and $(c,d)$. And $\frac{dy}{dx}=-2x$, so the gradients of the tangents are $-2a$ and $-2c$. We get ... | Your work is correct. Maybe you can shorten it by assuming that the points of tangency is $P=(t,4-t^2)$. Then slope at $P$ is $m_P=-2t$. Slope of line $LP=\dfrac{7-(4-t^2)}{-1-t}=\dfrac{3+t^2}{-(1+t)}$. Now this must be equal to $m_P$.
$$\dfrac{3+t^2}{-(1+t)}=-2t.$$
Solving this gives
$$t^2+2t-3=0.$$
Now solve for the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Integral $\int \frac{\tan^4 \theta \,d \theta}{1-\tan^2 \theta}$? I have to evaluate this indefinite integral $$\int \frac{\tan^4 \theta\, d \theta}{1-\tan^2 \theta}$$
I tried it as follows $$I=\int\frac{(\sec^2 \theta-1)\tan^2 \theta\, d \theta}{1-\tan^2 \theta}=\int\frac{\sec^2 \theta \tan^2 \theta\, d \theta}{1-\tan... | Here is another way to proceed:
$\displaystyle\int\frac{\tan^4\theta}{1-\tan^2\theta}d\theta=\int\frac{\tan^4\theta-1}{1-\tan^2\theta}d\theta+\int\frac{1}{1-\tan^2\theta}d\theta=-\int(\tan^2\theta+1)d\theta+\int\frac{\cos^2\theta}{\cos^2\theta-\sin^2\theta}d\theta$
$=\displaystyle-\int\sec^2\theta \;d\theta+\int\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
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Finding an angle which satisfies two equations I'd like to prove the following: Given any two real numbers $a$ and $b$, not both zero, there exists $c \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ such that $\sin c = \frac{a}{\sqrt{a^2 + b^2}}$ and $\cos c = \frac{b}{\sqrt{a^2 + b^2}}$. I'm trying to use as little geometry as p... | Your question is a trick question.
It is impossible to prove such a claim because it is not true. If you really want to prove the truthness of such a claim I'd start with the claim that for any real numbers $a$ and $b$ there exists an angle $\theta$ such that $\cos{\theta} = \frac{a}{\sqrt{a^2 + b^2}}$, $\sin{\theta} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1296235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to proceed with Euclidean algorithm for finding greatest common divisor of two polynomials. I am trying to find
\begin{equation*}
gcd(x^4-x^3-4x^2-x+5,x^2+x-2).
\end{equation*}
I have done the first step of long division and found.
\begin{equation*}
x^4-x^3-4x^2-x+5=(x^2-2x)(x^2+x-2)-5x+5
\end{equation*}
so we ha... | There is a theorem which states that if $f $ is a polynomial and $ f(a) =0 $ then $x-a$ is a factor of $f$, another theorem states that if $f$ is a polynomial the possible zeros of $ f$ are factors of $c/l$ where $l $is the leading coefficient and $ c$ is the constant. The two polynomials suppose $f(x)= x^4 - x^3 -4x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1296574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Product of two integers of the form $x^2+my^2$ is of the same form. Let $x,y,a,b\in \mathbb Z$. Prove that there are integers $c$ and $d$
so that
\begin{equation*}
(x^2+y^2m)(a^2+b^2m)=c^2+d^2m.
\end{equation*}
I'm stuck, I took the product and got $x^2a^2+b^2y^2m^2+(a^2y^2+x^2b^2)m$ but the numbers in the parenthesis ... | $$\begin{align}
(x^2+y^2m)(a^2+b^2m) &= x^2a^2 +mx^2b^2 +m^2y^2b^2+my^2a^2 \\
&= x^2a^2 +m^2y^2b^2 +m(x^2b^2 +y^2a^2) \\
&= x^2a^2 \color{red}{ +2mxyab } +m^2y^2b^2 +m(x^2b^2 \color{red}{ -2xyab }+y^2a^2)\\
&= (xa + myb)^2 + m(xb-ya)^2
\end{align}$$
Giving $c=(xa + myb)$ and $d=(xb-ya)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1298461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
taking the limit $\lim\limits_{n\rightarrow \infty} {\frac{(3^{n+1} + 4)(7^n-47)}{(7^{n+1}-47)(3^n +4)} }$ I need help with a guide on how i will deal with this kind of problem.. This a part of my solution in series convergence. I find it hard taking the limit of this: $$\lim_{n\rightarrow \infty} {\frac{(3^{n+1} + 4)(... | There are a couple of properties of limits you can abuse here. The one that should pop out first is $\lim f(x)g(x) = \lim f(x) \cdot \lim g(x)$.
$\lim_{n\rightarrow \infty} {\frac{(3^{n+1} + 4)(7^n-47)}{(7^{n+1}-47)(3^n +4)}} = \lim_{n\rightarrow \infty} {\frac{3^{n+1}+4}{3^n+4}} \cdot \lim_{n\rightarrow \infty} {\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
if $4^{\alpha} \equiv k+1 \pmod{2k+1}$ prove there is no $\beta$ where $4^{\beta} \equiv k\pmod{2k+1}$. Suppose that $3 \nmid 2k+1$ and there is $\alpha$ with $4^{\alpha} \equiv k+1 \pmod{2k+1}$ where $0 \leq \alpha \leq k$. I want to prove that there is no $\beta$, $0\leq \beta \leq k$ such that $4^{\beta} \equiv k \p... | Let $m$ be the order of $2$ modulo $2k+1$, $m = \min \bigl\{ n\in \mathbb{N}\setminus \{0\} : 2^n \equiv 1 \pmod{2k+1}\bigr\}$.
If $2^{2\alpha} = 4^\alpha \equiv k+1 \pmod{2k+1}$, then $2^{2\alpha+1} \equiv 1 \pmod{2k+1}$, so $m\mid (2\alpha + 1)$, hence $m$ is odd.
If on the other hand $2^{2\beta} = 4^\beta \equiv k \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1301649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show equality of a given function with a series in $ℝ$ Show that: $$2x\cos x-\sin x=4\sum_{n=2}^\infty \frac{(-1)^n}{n^2-1}\sin(nx)$$
Supposedly, this can be proved by using Fourier series, by choosing the right function but I have been thus far unable to do so. Any hints will be welcome, thanks in advance!
| By using the Fourier sine series
\begin{align}
f(x) &= \sum_{n=1}^{\infty} B_{n} \, \sin(nx) \hspace{15mm} x \in [0,\pi] \\
B_{n} &= \frac{2}{\pi} \, \int_{0}^{\pi} f(x) \, \sin(nx) \, dx
\end{align}
the function $f(x) = a x \cos(x) + b \sin(x)$ can be seen to be represented by
\begin{align}
a x \cos(x) + b \sin(x) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim_{x\to 0}\frac{\sin x-x\cos x}{x^3}$ $$\lim_{x\to 0}\frac{\sin x-x\cos x}{x^3}$$
How do I go about doing this? I can see no simple way of doing this. Application of l'Hopital's rule would be very laborious. A Taylor expansion seems feasible but is that the best way? It seems like it may be also very laborious... | I may use classic limits (which can easily be computed by hospital's rule
(twice)
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}} &=&-\frac{1}{6} \\
\lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}} &=&\frac{1}{2}
\end{eqnarray*}
The idea is to write the original expression using those above,
\begin{equation*}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Simple argument for $\frac{(x+y)^2}{x^2+xy+y^2}\le 4/3$ I would like to show that $\forall x,y\in\mathbb R^+:\frac{(x+y)^2}{x^2+xy+y^2}\le 4/3$.
The inequality is indeed true as the maximum of $\frac{(x+y)^2}{x^2+xy+y^2}$ is reached for $x=y$ and its value is $4/3$.
Except for the standard way of computing partial deri... | $$\frac{(x+y)^2}{x^2+xy+y^2}=1+\frac{xy}{x^2+xy+y^2}\le 1+\frac{1}{3}$$
$$\iff 3xy\le x^2+xy+y^2\iff (x-y)^2\ge 0,$$
which is true, with equality iff $x=y$.
Note we could multiply by $x^2+xy+y^2$ without flipping ineq sign because it equals $\left(x+\frac{1}{2}y\right)^2+\frac{3}{4}y^2>0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
integer as sum of three binomials Prove that for any nonnegative integer $n$ $\exists x,y,z \in \mathbb{N}$ and $0\leq x<y<z$ so
$$n=\binom{x}{1}+\binom{y}{2}+\binom{z}{3}$$
Please give me a hint, I don't have any idea.
| Easy to see that
$$
1 = {x_1 \choose 1}+{y_1 \choose 2}+{z_1 \choose 3}
$$
where $x_1 = 0$, $y_1 = 1$, $z_1 = 3$.
Now suppose
$$
i = {x_i \choose 1}+{y_i \choose 2}+{z_i \choose 3}
$$
where $0 \leq x_i < y_i < z_i$. We want to decide $0 \leq x_{i+1} < y_{i+1}<z_{i+1}$ such that
$$
i + 1 = {x_{i+1} \choose 1}+{y_{i+1} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"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.