Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Positive integer $x$ such that $\sqrt{144+x^2}$is an integer How many positive integers $x$ are there so that $\sqrt{144+x^2}$ is an integer?
Attempt a solution:
I used trial and error and only got $x=5$
| Well then since $144 = 12^2$ then you want to find values of $x$ to satisfy the equation, $$12^2 + x^2 = y^2.$$ Now since $(a + b)^2 = a^2 + b^2 + 2ab$ and $(a - b)^2 = a^2 + b^2 - 2ab$ then when we multiply them together, we get that $$(a^2 - b^2)^2 + (2ab)^2 = (a^2 + b^2)^2.\tag1$$ What you have done to solve for $x$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2658837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to extract $(x+y+z)$ or $xyz$ from the determinant Prove
$$\color{blue}{
\Delta=\begin{vmatrix}
(y+z)^2&xy&zx\\
xy&(x+z)^2&yz\\
xz&yz&(x+y)^2
\end{vmatrix}=2xyz(x+y+z)^3}
$$
using elementary operations and the properties of the determinants without expanding.
My Attempt
$$
\Delta\stackrel{C_1\rightarrow C_1+C_2+C_3... | It is easy to calculate
\begin{vmatrix}
1&\frac{x+z}{2}&\frac{x+y}{2}\\
0&\frac{(x+z)^2}{2}&\frac{yz-x^2-xy-xz}{2}\\
0&\frac{yz-x^2-xy-xz}{2}&\frac{(x+y)^2}{2}
\end{vmatrix}
If you still want to use extract $x+y+z$, you can use
\begin{eqnarray}
\begin{vmatrix}
\frac{(x+z)^2}{2}&\frac{yz-x^2-xy-xz}{2}\\
\frac{yz-x^2-xy-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2659331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Laurent series expansion about z=0 I am to obtain the first 3 terms of the Laurent series about $z = 0$ for:
$$f(z)=1/(e^z-1-z)$$
I know that there is 1 singularity at $z=0$. My denominator is not in a polynomial form so I can convert it into such by taking a Taylor series of the function around z=0 which yields the de... | Borrowing from that answer I commented, we have
\begin{align}
\frac{1}{e^z - 1 - z} &= \frac{1}{(1 + z + z^2/2! + \dots) - 1 - z}\\
&= \frac{1}{z^2/2! + z^3/3! + z^4/4! + \dots}\\
&= \frac{2}{z^2} \cdot \frac{1}{1 + 2z/3! + 2z^2/4! + \dots}
\end{align}
Now letting $P(z) = 2z/3! + 2z^2/4! + \dots$, we have
\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2660585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the kernel of a 4x4 matrix $$
\begin{pmatrix}
1 & 2 & 3 & 4\\
5 & 6 & 7 & 8\\
9 & 10 & 11 & 12\\
13 & 14 & 15 & 16\\
\end{pmatrix}
$$
I am asked to find the kernel of the matrix $M$. After doing some row operation I get to
$$
\begin{pmatrix}
1 & 2 & 3 & 4\\
0 & -4 & -8 & -12\\
... | Your computations are correct and so is your answer. But you should really improve your MathJax skills.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Stuck on two phase simplex where RHS $= 0$ I'm stuck on the following linear program, it's a minimal example of the problem I'm having:
Minimize $Z = y$ where $x \ge 1$ and $-2x+y \ge 0$
We start by rewriting this as a system of linear equalities, introducing slack and artificial variables:
$\left\{
\begin{array}{c}... |
How did $s_2$ suddenly become negative? What did I do wrong?
You've chosen the wrong entering variable. This give rise to an unfeasible solution.
To set up the initial tableau, multiply the second equation of the system
$$
\left\{
\begin{array}{c}
x - s_1 + a_1=1\\
-2x +y - s_2=0,
\end{array}
\right.
$$
by $-1$, so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2664942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
real value of $k$ inirrational equation Find value of $k$ for which the equation $\sqrt{3z+3}-\sqrt{3z-9}=\sqrt{2z+k}$ has no solution.
solution i try $\sqrt{3z+3}-\sqrt{3z-9}=\sqrt{2z+k}......(1)$
$\displaystyle \sqrt{3z+3}+\sqrt{3z-9}=\frac{12}{\sqrt{2z+k}}........(2)$
$\displaystyle 2\sqrt{3z+3}=\frac{12}{\sqrt{2z+k... | Hint : From the start : $\sqrt{3z+3}-\sqrt{3z-9}=\sqrt{2z+k}$ leads to
$$ 3z+3 +3z-9-2\sqrt{3z+3}\sqrt{3z-9}=2z+k $$
so
$$2\sqrt{3z+3}\sqrt{3z-9}=4z-6-k $$
and squaring again to
$$ 4(3z+3)(3z-9)=(4z-6-k)^2 $$
so
$$20z^2 +4(48+2k)z-27\times 4 -(6+k)^2=0 \tag{3} $$
You won't have solutions if
$$ \Delta_k= (4(48+2k))... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Roots of $f(x)=x^3-x^2-2x+1$ We can prove using a monotony study that the function $f(x)=x^3-x^2-2x+1$ has three real roots. However, when I solve the equation $f(x)=0$ using Mathematica, I get
$$x_1=\frac{1}{3}+\frac{7^{2/3}}{3 \left(\frac{1}{2} \left(-1+3 i \sqrt{3}\right)\right)^{1/3}}+\frac{1}{3} \left(\frac{7}{2} ... | You need the complex numbers even when there are three real roots. That was a real puzzle when the cubic was first solved.
Read about the casus irreducibilis:
https://en.wikipedia.org/wiki/Casus_irreducibilis
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2666823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$Pr(A+B+C=X+Y+Z)$ with $A, B, C, X, Y, Z$ being iid random variables $A, B, C, X, Y, Z$ are iid random variables and they are uniformly distributed on the set of integers $0$ to $9$ (inclusive). What is $\Pr(A+B+C=X+Y+Z)$?
I am kind of lost and my attempt is to use generating functions such that $$G_{A+B+C}(t)\cdot G_{... | Let $A,B,C$ and $X,Y,Z$ are iid uniformly distributed on $\{0,1,\dots,9\}$, split the below probability by the value of the sums,
$\mathbb{P}(A+B+C=X+Y+Z)=\displaystyle \sum_{k=0}^{27}\mathbb{P}(A+B+C=k,\ X+Y+Z=k).$
Since $A+B+C$ and $X+Y+Z$ are independents, for all $k\in \mathbb{Z}$:
$\mathbb{P}(A+B+C=k,X+Y+Z=k)=\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show that $\frac{1}{a^n+b^n+c^n} = \frac{1}{a^n} + \frac{1}{b^n} + \frac{1}{c^n}$ Let $n \in N, n=2k+1, and \text{ } \frac{1}{a+b+c} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$.
Show that $$\frac{1}{a^n+b^n+c^n} = \frac{1}{a^n} + \frac{1}{b^n} + \frac{1}{c^n}$$
I have tried, but I don't get anything. Can you please gi... | The condition gives $$(a+b+c)(ab+ac+bc)=abc$$ or
$$\sum_{cyc}(a^2b+a^2c+abc)=abc$$ or
$$\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right)=0$$ or
$$\prod_{cyc}(a+b)=0.$$
We need to prove that
$$\prod_{cyc}(a^n+b^n)=0$$ or $$\prod_{cyc}(a+b)\prod_{cyc}\left(a^{2k}-a^{2k-1}b+...+b^{2k}\right)=0,$$ which is obvious.
We can u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Simplify $\frac14\log(3x+1)+\frac12\log(x-3)-\frac12\log(x^4-x^2-10)$ $$\frac14\log(3x+1)+\frac12\log(x-3)-\frac12\log(x^4-x^2-10)$$
I was able to find
$$\log(3x+1)^{1/4}(x-3)^{1/2}-\log(x^4-x^2-10)^{1/2}$$
through distributing the $^{1/2}$ but this answer is incorrect.
| I do not know how you would get a "simplified answer", but:
$$\dfrac 14\log(3x+1)+\dfrac 12 \log(x-3)-\dfrac 12\log(x^4-x^2-10)$$
Using the property $\log a^b=b\log a$:
$$\log{\sqrt[4]{3x+1}}+ \log{\sqrt{x-3}}-\log\sqrt{{x^4-x^2-10}}$$
Using the property $\log a+\log b=\log ab$:
$$\log{\sqrt[4]{3x+1}}+ \log{\sqrt{x-3}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2669932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Summation of binomial coefficients multiplied by their index
$\sum_{r=0}^{300}a_rx^r = (1+x+x^2+x^3)^{100}.$ If $a= \sum^{300}_{r=0}a_r$ then $\sum_{r=0}^{300}ra_r =? $
a) $300a$
b) $100a$
c) $150a$
d) $75 a$
Attempt:
First of all if we substitute $x=1$, we get $a= 4^{100}$
Then, using formula of summation of geome... | You have been given
$$\sum_{r=0}^{300}a_rx^r = (1+x+x^2+x^3)^{100}$$
substitute $x = 1 $ to get $a = \sum a_r = 4^{100}$
Differentiate once w.r.t $x$ to get
$$\sum_{r=0}^{300}ra_rx^{r-1} = 100(1+x+x^2+x^3)^{99} (1+2x+3x^2)$$
again substitute $x = 1$ to get $\sum ra_r = 4^{99} \cdot 100 \cdot 6 = 150a$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2671868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to calculate $ \int \frac{\sin^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx? $ How to calculate
$$ \int \frac{\sin^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx? $$
I already know one possible way, that is by :
$$ \int \frac{\sin^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx = \int 1 - \frac{\cos^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx $... | Let us take:
$$I=\int \frac{\sin^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx$$
then
$$I=\int \frac{-\cos^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx+x$$
giving$$2I=\int \frac{\sin^{6}(x)-\cos^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx+x$$
This can be written as (using identities like $a^3-b^3$ and $a^3+b^3$)
$$2I= \int\frac{(\sin^2(x)-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2671984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Greatest Common Divisor of two Polynomials. Find the $\gcd(x^3-6x^2+14x-15, x^3-8x^2+21x-18)$ over $\mathbb{Q}[x]$. Then find two polynomials $a(x),b(x) \in \mathbb{Q}[x]$ such that, $$a(x)(x^3-6x^2+14x-15) + b(x)(x^3-8x^2+21x-18)=\gcd(x^3-6x^2+14x-15, x^3-8x^2+21x-18)$$
I have managed to find,
$$x^3-6x^2+14x-15=(x-... | If I interpreted your question correctly and you want to do something like this, where in:
$a(x)(p)+b(x)(q)=gcd(p,q)$, $a=1$, and $b=1$
You can restate your expression as:
$$a(x)(x-3)(x^2-3x+5)+b(x)(x-3)(x-3)(x-2)$$
$$(x-3)\left(a(x)(x^2-3x+5)+b(x)(x-3)(x-2)\right)$$
Since this has to equal the $gcd$, which is $x-3$:
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2679525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Rattles with beads or necklace with beads? I came across this problem in a book called limits, sequences combinations great book for intro to combinatorics .
A rattle consists of a ring with $3$ white beads and $7$ red ones strung on it. Some rattles seemingly different can be made identical by arranging the rings and... | Let me just observe that with $(3,10) = 1$ and $(7,10) = 1$ Polya will
be very simple in this case. The OEIS uses necklace for cyclic
symmetry and bracelet for dihedral, so we have a bracelet here. The
cycle index is from first principles
$$Z(D_{10}) = \frac{1}{20} \sum_{d|10} \varphi(d) a_d^{10/d}
+ \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2680878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Area of intersection using integration
Suppose, for the sake of illustration, that $AF=2$, where $F\mathbb{'s}$ coordinates are $(2,0)$ and $A$ is the point of origin; $E=(0,1)$ and circle $A$ has $R=1.5$
How do I solve the area of the intersection of the circle and rectangle by integration?
If we let $f(x)=1$, then t... |
Consider $x=f(y)$ instead of $y=f(x)$.
Then the area of intersection is just
\begin{align}
\int_0^1 \sqrt{R^2-y^2}\,dy
&=\left.\left(
\tfrac12\,y\sqrt{R^2-y^2}
+\tfrac12\,R^2\arctan\frac{y}{\sqrt{R^2-y^2}}
\right)\right|_0^1
\\
&=
\left(
\tfrac12\,\sqrt{R^2-1}
+\tfrac12\,R^2\arctan\frac{1}{\sqrt{R^2-1}}
\right)
\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2683356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find $\int_0^{\sqrt{R^2-1}}g(x)-f(x)\,\mathrm dx$
How do you evaluate this integral?
$$\int_0^{\sqrt{R^2-1}}g(x)-f(x)\,\mathrm dx$$
where $f(x)=1$ and $g(x)=\sqrt{R^2-x^2}$.
Wolfram tells me I exceeded my computational limit. Mathematica gives me a long answer which is very difficult to read (for me).
EDIT: I was... | Trick: Split the integral into three parts.
$$\begin{align}\int_0^{\sqrt{R^2-1}}g(x)-f(x)\,\mathrm dx&=\int_0^{\sqrt{R^2-1}}\sqrt{R^2-x^2}-1\,\mathrm dx\\&=\left(\int_0^{\sqrt{R^2-1}}\frac{\sqrt{R^2-x^2}}2-\frac{x^2}{\sqrt{R^2-x^2}}+\frac{a^2}{\sqrt{R^2-x^2}}\,\mathrm dx\right)-\sqrt{R^2-1}\\&=\left(\int_0^{\sqrt{R^2-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2683423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Area of largest inscribed rectangle in an ellipse. Can I take the square of the area to simplify calculations? So say I have an ellipse defined like this:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
I have to find the largest possible area of an inscribed rectangle.
So the area ($A$) of a rectangle is $2x2y=4xy$. Also we can... | $$
\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1 \quad \Rightarrow \quad \dfrac{2x}{9}dx + \dfrac{y}{2}dy = 0 \quad (1)
$$
$$
A = 4xy \quad \Rightarrow \quad 0 = dA = 4ydx + 4xdy \quad (2)
$$
Therefore, from (1) and (2)
$$
\dfrac{dy}{dx} = -\dfrac{4x}{9y} \quad \text{and} \quad \dfrac{dy}{dx} = -\dfrac{y}{x} \quad \Rightarrow \q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2685225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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Why does $\int_0^1\binom{2n}{n}x^n(1-x)^ndF(x)=\frac{1}{2n+1}$? Why does $$\int_0^1\binom{2n}{n}x^n(1-x)^ndF(x)=\frac{1}{2n+1},$$
Where $F(x)$ is uniform over the interval $[0,1]$?
It's not clear from simple factoring:
$$\binom{2n}{n}\int_0^1x^n(1-x)^ndF(x)$$
$$=\frac{(2n)!}{n!^2}\int_0^1x^n(1-x)^ndF(x).$$
| By the binomial theorem,
\begin{align*}
\sum_{k=0}^{n} \left( \int_{0}^{1} \binom{n}{k} x^k(1-x)^{n-k} \, dx \right) z^k
&= \int_{0}^{1} (1 + (z-1)x)^n \, dx \\
&= \frac{1}{n+1} \cdot \frac{z^{n+1} - 1}{z-1} \\
&= \sum_{k=0}^{n} \frac{z^k}{n+1}.
\end{align*}
So we have
$$ \int_{0}^{1} \binom{n}{k} x^k(1-x)^{n-k} \, dx ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2685330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
If $\sin(x)=3\cos(x)$, compute $\sin(x)*\cos(x)$ I drew a triangle and was instructed to use a property unknown to me. Apparently the answer is $\frac{3}{\sqrt{10}}*\frac{1}{\sqrt{10}}=\frac{3}{10}$. Is this answer correct, and what theorem/formula is this?
| $$\begin{align} \\
\sin (2x)&=2\sin (x)\cos (x) \\
&=\frac {2\tan (x)}{1+\tan^2 (x)} \\
&=\frac {2\cdot3}{1+(3)^2} \\
&=\frac {6}{10} \\
\end{align}$$
thus
$$\sin (x)\cos (x)=\frac {3}{10}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2685713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
On the series $\sum \limits_{n=1}^{\infty} \frac{1}{n^2-3n+3}$ and $\sum\limits_{n=-\infty}^{\infty} \frac{1}{n^2-3n+3}$ Wolfram Alpha says that
$$\sum_{n=1}^{\infty} \frac{1}{n^2-3n+3} = 1 + \frac{\pi \tanh \left ( \frac{\sqrt{3}\pi}{2} \right )}{\sqrt{3}}$$
However I am unable to get it. It is fairly routine to prove... | The First Sum
$$
\begin{align}
\sum_{n=1}^\infty\frac1{n^2-3n+3}
&=1+\sum_{n=2}^\infty\frac1{n^2-3n+3}\tag1\\
&=1+\sum_{n=2}^\infty\frac1{\left(n-\frac32-i\frac{\sqrt3}2\right)\left(n-\frac32+i\frac{\sqrt3}2\right)}\tag2\\
&=1+\frac1{i\sqrt3}\sum_{n=2}^\infty\left(\frac1{n-\frac32-i\frac{\sqrt3}2}-\frac1{n-\frac32+i\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2686931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Simplifying $\frac{y}{y^2+b^2}$ where $y=b \cot\theta$ Could appreciate some help with this question.
I want to simplify the following trigonometric equation.
$$\frac{y}{y^2+b^2}$$
where $y=b \cot\theta$.
The solution I got was
$$\frac{1}b \cos\theta$$
Can someone verify and try and guide me through the solution... | $$1+\cot^2\theta=1+\frac{\cos^2\theta}{\sin^2\theta}=\frac1{\sin^2\theta}=\csc^2\theta$$ or
$$1+\cot^2\theta=\frac1{\sin^2\theta}$$
Now take reciprocal of both side:
$$\frac{1}{1+\cot^2\theta}=\sin^2\theta$$
If $y=b\cot\theta$, then:
$$\frac{y}{y^2+b^2}=\frac{b\cot\theta}{b^2\cot^2\theta+b^2}=\frac{\cot\theta}{b\left(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2690581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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$a+b+c=?$ by having $a^2+160=b^2+5$ and $a^2+320=c^2+5$ My question:
$a+b+c=?$ by having $a^2+160=b^2+5$ and $a^2+320=c^2+5$.
My work so far:
$a^2+160=b^2+5\Rightarrow (b-a)(a+b)=155=31\times 5$
$a^2+320=c^2+5\Rightarrow (c-a)(c+a)=315=5\times3^2\times 7$
And now, I'm stuck.
($a,b,c$ are a members of $\mathbb Z$ and ar... | We know that: $$ (b-a)(a+b)=31\times 5=155$$ as both $31$ and $5$ are prime numbers we can say either $b-a=5, a+b=31$ OR $b-a=1, a+b=155$. (becasue $a+b>b-a$)
*
*$b-a=5, a+b=31$, thus: $$a=13 , b=18$$
On the other hand, we have:
$$a^2+320=c^2+5$$
$$\to c^2=13^2+320-5 \to c=22$$
*$b-a=1, a+b=155$, thus: $$a=77 , b=7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2693530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the rational solution of the equation Find the rational solution of the equation:
$\frac{2x - 1}{2016} + \frac{2x - 3}{2014} + \frac{2x - 5}{2012}+ ...+ \frac{2x - 2011}{6} +\frac{2x - 2013}{4} + \frac{2x - 2015}{2} =\\ \frac{2x - 2016}{1} + \frac{2x - 2014}{3} + \frac{2x - 2012}{5}+ ...+ \frac{2x - 6}{2011} + \fr... | Every term is $\dfrac{2x-2k}{2015-2k+2}-\dfrac{2x-(2k-1)}{2016-2k+2}=\dfrac{2x-2017}{(2016-2k+2)(2015-2k+2)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2698770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show that $\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2} \, dx <\frac{\pi}{4}$ Show that $$\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2}\,dx <\frac{\pi}{4}$$
I want to use if $f<g<h$ then $\int f<\int g<\int h$ formula for Riemann integration.
$1+x^2<1+x+x^2$ and it will give RHS as $$\frac{1}{1+x+x^2}<\frac{1}{1+x^2}$$
How to choose... | You want to show $$\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2}dx <\frac{\pi}{4}$$
Note that for $0<x<1$ we have $$ 1+x^2 < 1+x+x^2 <3 $$ Thus $$ \frac {1}{3} < \frac {1}{1+x+x^2}< \frac {1}{1+x^2}$$
Upon integration we get $$\frac{1}{3}=\int^{1}_0\frac{1}{3}\,dx<\int^1_0\frac{1}{1+x+x^2}\,dx<\int^1_0\frac{1}{1+x^2}\,dx=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2699005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
What kind of matrix is this and why does this happen? So I was studying Markov chains and I came across this matrix \begin{align*}P=\left( \begin{array}{ccccc}
0 & \frac{1}{4} & \frac{3}{4} & 0 & 0\\
\frac{1}{4} & 0 & 0 & \frac{1}{4} & \frac{1}{2}\\
\frac{1}{2} & 0 & 0 & \frac{1}{4}& \frac{1}{4... | This is a periodic Markov chain (with period $2$). Otherwise, there's not much that's terribly unusual about it.
User "Iwillnotexist Idonotexist" raised an important point in the comments:
Well, something that can be noted for periodic Markov chains is that
by definition they cannot be ergodic, another very import... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2699410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 5,
"answer_id": 2
} |
Prove that the area of $\triangle DEF$ is twice the area of $\triangle ABC$ Let $\triangle ABC$ be an equilateral triangle and let $P$ be a point on its circumcircle. Let lines $PA$ and $BC$ intersect at $D$ , let lines $PB$ and $CA$ intersect at $E$, and let lines $PC$ and $AB$ intersect at $F$. Prove that the area o... | It is very interesting to solve this problem by only using our greek friends Pythagoras and Menelaus, together with some simple algebra.
Initial Observations
Without loss of generality assume that $P$ lies on the half-plane determined by $BC$ and not containing $A$. Note that there is no need to use the circle, once w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2701381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Find an upper bound for the modulus of a complex function For real $x$ let $r(x)$ be defined (implicitly) through
$$
e^{i\,x} = \left(1+i\,x\right)\,e^{\left(-(1/2)\,x^2+r(x)\right)},
$$
where $i$ is the complex unit. I need to prove that
$$
\left|r(x)\right|\leq \left|x\right|^3,\quad \text{for} ~~\left|x\right|\leq 1... | I think that the inequality does not hold. Let's find $r(x)$ explicitly. Let $p(x)$ and $q(x)$ be real functions such that $r(x)=p(x)+iq(x)$. To find $p(x)$, we take modulus on given equation:
$$
1=\sqrt{1+x^2}e^{-\frac{1}{2}x^2+p(x)}
$$
Then we get
$$
p(x)=\frac{1}{2}x^2 - \frac{1}{2}\ln(1+x^2).
$$
Substitute $p(x)$ t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2701547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving a linear congruence system How can this linear congruence system be solved by using Chinese remainder theorem?
\begin{align}
12x&\equiv -7 \pmod {13}\\
4x&\equiv 7 \pmod {9}\\
2x&\equiv -3 \pmod {11}
\end{align}
As far as I understand, we have to create some M numbers as it follows:
\begin{align}
M = 13 * 9 * 1... | Note that CRT guarantees that an unique solution exist mod $9\cdot 11 \cdot 13$ since those are relatively primes but doesn’t give any particular method to solve the problem others that those used for the proof that you can find here.
To find the solution here we can proceed as follow
*
*$12x \equiv -7 \iff x\equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2702272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Limit simplification I've been practicing limits today, and I've came across this exercise which confused me a bit:
$\lim_{x\to -3} \frac{(x^2-9)^2}{(x+3)^2}$
My approach was to do this:
$\lim_{x\to -3} \frac{(x^2-9)^2}{x^2+3x+9}$
$\lim_{x\to -3} \frac{0}{9}$
Can anyone explain why this is not a valid approach? I reali... | $$\begin{align} x^2 - 9 &= x^2 - 3^2 \\ \\ &= (x+3)(x-3) \\ \Rightarrow \dfrac{(x^2-9)^2}{(x+3)^2} &= \dfrac{\require{cancel}\cancel{(x+3)^2}(x-3)^2}{\cancel{(x+3)^2}} \\ &= (x-3)^2. \\ \\ \therefore \lim_{x\to -3}\frac{(x^2-9)^2}{(x+3)^2} &= \lim_{x\to -3} (x-3)^2 \\ \\ &= (-3 - 3)^2 \\ \\ &= 36.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Let C be the unit square with diagonal corners at $−1 − i$ and $1 + i$. Evaluate $\oint_C\frac{1}{2z+1}\mathrm{d}z$ I'm having trouble trying to understand what I'm doing wrong in solving this complex integral.
I partitioned the square into 4 different parameterized curves (each curve being 1 of the sides of the squar... | $$ I:=\oint_C\frac{1}{2z+1}\mathrm{d}z $$
The idea of your integration path is correct. Only in $c_3$ and $c_4$ you fail to implement it correctly, when you use $-t+i$ you actually have to go from $-1$ to $+1$ in positive direction $dt$ not in negative $-dt$ to get the right orientation. You do not have to transform an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Complex numbers $\left(\frac{1+i}{1-i}\right)^k = 1$ what is $k$? The smallest possible integer $k$ for which $\left(\frac{1+i}{1-i}\right)^k = 1$ is?
I tried solving this, but my answer doesn't match the given answer. Correct me if I'm wrong at some place
My solution:
\begin{align}
\left(\frac{1+i}{1-i}.\frac{1+i}{1+i... | $(\frac {1+i}{1-i})^k=1$
$(1+i)^k = (1-i)^k$
$\sum_{j=0}^k {k\choose j}i^j = \sum_{j=0}^k {k\choose j}i^j*(-1)^j$
As $(-1)^{even} = 1$ and $(-1)^{odd} = -1$
$\sum_{j=0;j odd}{k\choose j}i^j =0$
As $i^{4k + 1} = i$ and $i^{4k - 1} = -i$
we need to find the smallest $k$ where $\sum_{h=0}^{4h+1 \le k}{k\choose 4h+1} =\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
solving $x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$ I have to solve $x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$
Dividing by $dx$ we have
$x + xy^2 + yy' + yy'x^2=0$
From where,
$$\frac{yy'}{1+y^2}+\frac{x}{1+x^2}=\frac{y\,dy}{1+y^2}+\frac{x\,dx}{1+x^2}=\\ =\frac{d(y^2+1)}{1+y^2}+\frac{d(x^2+1)}{1+x^2}= \frac{1}{2}d\ln(1+y^2)+\... | $$d( \text{something})=0 \implies \text{something = constant}$$
So you get the solution $$\ln(1+y^2)(1+x^2) = C$$ (Where $C$ is arbitrary constant)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2710438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Vector spaces and basis I came across this problem on vector space basis:
Verify that $2-x^2, x^3-x, 2-3x^2$ and $3-x^3$ form a basis for $P^4$ and express each of the polynomial $x^2$ as a linear combination of this basis.
I have tried solving the problem with my knowledge of vector space but still don't know how to g... | 1) Let's verify that the system of vectors $(2−x^2,x^3−x,2−3x^2, 3−x^3)$(1) is a basis for $P^4$. Write the canonical basis for $P^4$: $(1 ,x ,x^2 ,x^3)$(2) and decompose vectors of the system (1) by basis (2):
\begin{pmatrix}
2 & 0 & 2 & 3\\
0 & -1 & 0 & 0\\
-1 & 0 & -3 & 0\\
0 & 1 & 0 & -1\\
\end{pmatrix}
Now we'll ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2711608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Evaluating $\lim _{t\to \infty}\frac{1-\frac{\sqrt{t}}{\sqrt{t+1}}}{2-\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}$ I am having difficulties evaluating this limit:
$$\lim _{t\to \infty \:}\left(\frac{1-\frac{\sqrt{t}}{\sqrt{t+1}}}{2-\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}\right)$$
I have tried to divide out by $\frac{\sqrt{t}}{\sq... | The Hospital Rule :
We separate the numerator and the denominator :
$$ f(x) = \frac{\sqrt{x}}{\sqrt{x+1}}$$
$$ g(x) = \frac{\sqrt{4x+1}}{\sqrt{x+2}}$$
we have that :
$$ f'(x) = \frac{1}{2 \sqrt{x} \cdot (x+1)^{3/2}} $$
$$g'(x) = \frac{2}{\sqrt {x+2} \sqrt{4x+1} } - \frac{4x+1}{2(x+2)^{3/2}}$$
then you do the quotient,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2712131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Help with a trigonometry problem. I have encountered some problems to solve the left side to the right.
$$ \cos^ 2x \sin x = \frac{\sin 3x + \sin x}{4}$$
I trying to solve a differential equation on the form $$ y'' + y = \cos^ 2x * \sin x$$ and need to rewrite it to $$ y'' + y = \frac{\sin 3x + \sin x}{4}$$
Have t... | Note that by product to sum formula
$$2\cos \theta \sin \varphi = {{\sin(\theta + \varphi) - \sin(\theta - \varphi)} }\implies \cos 2x \cdot \sin x = \frac{\sin 3x - \sin x}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $a_k = c^k-1$, where $c > 1$, what can be said about $s(n) =\sum_{k=1}^n \frac{a_k}{a_{k+1}} $? If
$a_k = c^k-1$,
where $c > 1$,
what can be said about
$s(n)
=\sum_{k=1}^n \dfrac{a_k}{a_{k+1}}
$?
This is a generalization of
$a_n=3^n-1$, prove that $\frac{a_1}{a_2}+\frac{a_2}{a_3}+\dots+\frac{a_n}{a_{n+1}}>\frac{n}{3... | Given the identity
$$ s(n)=\frac{n}{c}-\frac{c-1}{c}\left[\frac{1}{c^2-1}+\sum_{k=2}^{n}\frac{1}{c^{k+1}-1}\right] $$
we have $s(n)\leq \frac{n}{c}-\frac{1}{c(c+1)}$ for any $n\geq 2$. Under the same assumption
$$ \sum_{k=2}^{n}\frac{1}{c^{k+1}-1}\leq \sum_{k=2}^{+\infty}\frac{1}{c^{k+1}-1}\leq \sum_{k=2}^{+\infty}\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Variation on Harmonic Series I'm trying to establish the convergence or divergence of the following variant of the harmonic series:
$$\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}-\frac{1}{10}+\frac{1}{11}\cdots$$
Where the sign pattern has period 5, ie, it ... | The sum of the first two is
$\dfrac1{5n+1}+\dfrac1{5n+2}
\lt \dfrac{2}{5n}
$
and the sum of the last 3 is
$\dfrac1{5n+3}+\dfrac1{5n+4}+\dfrac1{5n+5}
\gt \dfrac{3}{5n+5}
$.
Therefore
$\begin{array}\\
\dfrac1{5n+1}+\dfrac1{5n+2}
-(\dfrac1{5n+3}+\dfrac1{5n+4}+\dfrac1{5n+5})
&\lt \dfrac{2}{5n}-\dfrac{3}{5n+5}\\
&=\dfrac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Finding $\int\frac{1}{(1+\sin^2\theta)^{\frac{3}{2}}}d\theta$
Finding $$\int\frac{1}{(1+\sin^2\theta)^{\frac{3}{2}}}d\theta$$
Try: $$I=\int\frac{1}{(1+\sin^2\theta)^{\frac{3}{2}}}d\theta =2\sqrt{2}\int \frac{1}{(3-\cos 2\theta)^{\frac{3}{2}}}d\theta$$
How can I write this in terms of the elliptic integral? Could some... | As stated by SmarthBansal in the comments, this can be described as an Incomplete Elliptic Integral of the Third Kind, defined as
$$\Pi(n;\phi,m)=\int_0^\phi \frac{1}{1-n\sin^2(\theta)}\frac{1}{\sqrt{1-m\sin^2(\theta)}} d\theta$$
So we can write this integral as $\Pi(-1,x|-1)$. However, this is not a very satisfying a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2715129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluating trigonometric limit $\csc^2(2x) - \frac{1}{4x^2}$
$$\lim_{x\rightarrow 0} \left[ \csc^2(2x) - \frac{1}{4x^2} \right]$$
I've tried to use l'Hôpital's rules but still can't find the answer. Here's my approach:
$$
\begin{aligned}
&\lim_{x\rightarrow 0} \left[ \csc^2(2x) - \frac{1}{4x^2} \right] \\
=& \lim_{x\... | Almost.
I'll edit your answer.
$\begin{aligned}
&\lim_{x\rightarrow 0} \left[ \csc^2(2x) - \frac{1}{4x^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{1}{\sin^2(2x)} - \frac{1}{4x^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{1}{\sin(2x)} - \frac{1}{2x} \right]\left[ \frac{1}{\sin(2x)} + \frac{1}{2x} \right] ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2718299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find all integer solutions of: $\;\frac{1}{m}+\frac{1}{n}-\frac{1}{mn^2}=\frac{3}{4}$ I found the following problem from the 10th Iranian Mathematical Olympiad in Crux Magazine.
Find all integer solutions of $$\frac{1}{m}+\frac{1}{n}-\frac{1}{mn^2}=\frac{3}{4}$$
Initially it looked like a typical quadratic problem, h... | As an alternative to Dr. Sonnhard Graubner's answer on how to find the complete set of solutions from
$m=\frac{4(n^{2}−1)}{3n^{2}−4n}$
consider the following:
We can factor that expression further to
$m=\frac{4(n+1)(n-1)}{n(3n−4)}$
That tells us a lot about $n$. First off, if $n$ is 1, then $m$ must be 0, but that i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2718551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 3
} |
Is this proof of the irrationality of the square root of $3$ valid? Assume that $\sqrt{3}=\frac{a}{b}$ where $\frac{a}{b}$ is in its simplest possible form. So $3=\frac{a^2}{b^2}$, hence $3b^2=a^2$. If $b$ is even then $a^2$ and $a$ are also even. That means they have a common factor of $2$. But this is impossible sinc... | It looks fine to me, except for a tiny detail: after saying that if $b$ is even, then $a$ must be even too, you should have added that the reverse is true: if $a$ is even, then $b$ must be even too.
And there's no need no introduce the numbers $c$ and $d$, though. You could just say that we can't have $4(3m^2+3m-k^2-k)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2718884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find all rational points on $x^2 + y^2 = 17$ So yes, this was asked before but I'm stuck on a specific step.
So to solve this, I tried to find the the intersection between the line $y=m(x-1)+4$ (we have $(1, 4)$ as a rational point on the circle) and $x^2 + y^2 = 17$. So simply substitution, $x^2 + (m(x-1)+ 4)^2 = 17 ... | It comes out better if you do
$$ (x,y) = (1,4) + t(p,q) $$
with integers $p,q$ and $\gcd(p,q) = 1.$
$$ x = 1 + tp, \; \; y = 4 + t q \; . $$
$$ 17 = x^2 + y^2 = 1 + 2 p t + p^2 t^2 + 16 + 8 q t + q^2 t^2 \; , $$
$$ 17 = 17 + (2p + 8 q) t + (p^2 + q^2 ) t^2 \; , $$
$$ 0 = (2p + 8 q) t + (p^2 + q^2 ) t^2 \; . $$
Ne... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2719566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, prove that $\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, prove that $\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$
My Attempt
$$
\frac{-2x}{2\sqrt{1-x^2}}-\frac{2y}{2\sqrt{1-y^2}}.\frac{dy}{dx}=a-a\frac{dy}{dx}\\
\implies \frac{dy}{dx}\bigg[a-\fra... | Let us remark that the looked for differential equation can be written under the form
$$\frac{dx}{\sqrt{1-x^2}}=\frac{dy}{\sqrt{1-y^2}}\tag{1}$$
involving solutions of the form :
$$-\arccos(x)=-\arccos(y)+a\tag{2}$$
otherwise said with cartesian equation :
$$y=\cos(\arccos(x)-a)\tag{3}$$
where $a$ is any real.
Settin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Elementary geometry problem about two squares.
Consider the picture above, where $ABCD$ and $EFGC$ are squares with areas respectively $A$ and $B$. Find the shaded area.
Well, I observerd the following: If the intersection between $BG$ and $EF$ is the midpoint of $EF$, then the area will be $\dfrac{A}{2} + \dfrac{3B}{... | Let $BC=p$ and $EC=q.$ The area of square $ABCD$ is $p^2,$ which we will also call $a.$ The area of $EFGC$ is $q^2,$ which we will also call $b.$ (The letters $A,B$ are already in use, for vertices.)
The area of $\triangle BCD$ is $\frac {p^2}{2}.$ The area of $\triangle BCG$ is $\frac{ pq}{2}.$
Let $H$ be the point o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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How to factor this polynomial over $\mathbb{F}_2$ On page 587 in Dummit and Foote, the authors say the polynomial $\frac{x^{16}-x}{x(x-1)(x^2+x+1)}$ can be factored into quartics over $\mathbb{F}_2$ as $(x^4+x^3+x^2+x+1)(x^4+x^3+1)(x^4+x+1)$.
I am having trouble seeing this. When I divide the polynomial $x^{16}-x$ by $... | The Berlekamp algorithm gives
$$
x^{16}-x=(x^4 + x^3 + x^2 + x + 1)(x^4 + x^3 + 1)(x^4 + x + 1)(x^2 + x + 1)(x + 1)x.
$$
So the statement in Dummit and Foote is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the horizontal asymptotes of $f(x) = \frac{x^3-2}{\lvert x\rvert^3+1}$ Taken from Thomas' Calculus 12e
Find the horizontal asymptote of the graph of:
$$f(x) = \frac{x^3-2}{\lvert x\rvert^3+1}$$
Solution: We calculate the limits as ${x \to \pm \infty}$
For $x\ge0$: $$\lim\limits_{x\to\infty}\frac{x^3-2}{\lvert x\rv... | As you find out, the limit as $x\to \infty$ is $1$ then $y=1$ is an horizontal asymptote.
For $x\to -\infty$ we have
$$\lim\limits_{x\to-\infty}\frac{x^3-2}{\lvert x\rvert ^3+1} =\lim\limits_{x\to-\infty}\frac{x^3-2}{-x^3+1}=\lim\limits_{x\to-\infty}\frac{1-(2/x^3)}{-1+(1/x^3)}=-1$$
then $y=-1$ is an horizontal asympto... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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In any triangle, if $\frac {\cos A+2\cos C}{\cos A+2\cos B}=\frac {\sin B}{\sin C}$, then the triangle is either isosceles or right-angled In any triangle, if $\dfrac {\cos A+2\cos C}{\cos A+2\cos B}=\dfrac {\sin B}{\sin C}$, prove that the triangle is either isosceles or right angled.
My Attempt:
Given:
$$\dfrac {\cos... | In your work try to factor out a term like $(b-c)$ from the expression:
$$c(ab^2+ac^2-a^3+2a^2c+2b^2c-2c^3) = b(ab^2+ac^2-a^3+2a^2b+2bc^2-2b^3)\\
ab^2c+ac^3-a^3c + 2a^2c^2 +2b^2c^2 - 2 c^4 = ab^3 + abc^2 -a^3b + 2a^2 b^2 + 2b^2 c^2 -2b^4\\
abc(b-c) + a(c^3 - b^3 ) +a^3 (b-c)+2a^2 (c^2 - b^2) +2(b^4-c^4)$$
From here eit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2726098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Integrate $\int\frac{1}{x^3+1}dx$ The problem is, as stated:
$$\int\frac{1}{x^3+1}dx$$
I tried using substitution: $t^3 = x^3 + 1$ but didn't get far with that. I also tried setting: $t = x^3 + 1$, with no luck again.
I tried partial decomposition but I didn't know how to integrate $$\int\frac{1}{x^2-x+1}$$ and I kept ... | $$\int\frac {1}{x^2-x+1} dx=\int\frac {1}{(x-\frac{1}{2})^2 +\frac{3}{4}}dx$$
And that take $x-\frac{1}{2}=t$.
Always use $ax^2+bx+c=a(x+\frac{b}{2a}) ^2 - \frac{b^2-4ac}{4a}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2726221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
In any $\triangle ABC$, prove that: $\frac {\cos B-\cos C}{\cos A +1}=\frac {c-b}{a}$ In any $\triangle ABC$, prove that: $\dfrac {\cos B-\cos C}{\cos A +1}=\dfrac {c-b}{a}$
My Attempt:
$$\begin{align}
\text{R.H.S.}&=\dfrac {c-b}{a} \\[4pt]
&=\frac {a\cos B+b\cos A-a\cos C-c\cos A}{b\cos C+c\cos B} \\[4pt]
&=\dfrac {a(... | We have$$a\cos B+b\cos A=c,$$$$a\cos C+c\cos A=b.$$Subtracting gives$$a(\cos B-\cos C)-(c-b)\cos A=c-b,$$which on rearranging yields the required result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
proof verification $\frac{3+2\sqrt{6}}{1-\sqrt{6}}$ is an algebraic integer
Is $$\frac{3+2\sqrt{6}}{1-\sqrt{6}}$$
an algebraic integer?
An algebraic integer means an algebraic number in some algebraic number field $K\supset \Bbb Q$ that is the root of a monic polynomial $f\in \Bbb Z[x]$. Here I guess we are in $\Bb... | Your approach - finding the minimal polynomial and observing that it is a monic polynomial over $\Bbb Z$ - is correct, but the following is probably easier:
$$
\frac{3 + \sqrt6}{1 - \sqrt6} =
\frac{3 + \sqrt6}{1 - \sqrt6}
\frac{1 + \sqrt6}{1 + \sqrt6} =
\frac{15 + 5 \sqrt6}{-5} =
-3 + \sqrt6.
$$
Since $\sqrt 6$ is an ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Determine the value of $ax^5 + by^5$ given that $a$, $b$, $x$, and $y$ satisfy the following system: Given
$\begin{cases} ax + by = 4 \\ ax^2 + by^2 = 8 \\ ax^3 + by^3 = 17 \\ ax^4 + by^4 = 42 \end{cases} $
Determine the value of $ax^5 + by^5$
Do you have any brilliant idea to solve this problem? By looking at the ri... | You may exploit the idea behind the Berlekamp-Massey algorithm. By denoting as $U_n = a x^n +b y^n $ we have that $\{U_n\}_{n\geq 0}$ is a linear recurrent sequence with a quadratic characteristic polynomial and such that
$$ U_0=a+b,\quad U_1=4,\quad U_2=8,\quad U_3=17,\quad U_4=42 $$
These informations are enough to r... | {
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"url": "https://math.stackexchange.com/questions/2728980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Real Analysis problem concerning recursive sequence limit. Define $$a_1=1$$, and define
$$
a_{n+1} = \left\{\begin{aligned}
&a_n+1/n &&: a_n^2\leq 2\\
&a_n-1/n &&: a_n^2>2
\end{aligned}
\right.$$
I am then asked to show that $$|a_n-\sqrt{2}|<2/n$$ for all indices n in order to show convergence by comparison.
My initial... | Induction works. Suppose $|a_n - \sqrt 2| < 2/n$, i.e.
\begin{equation*}
(1) \quad \sqrt 2 - 2/n < a_n < \sqrt 2 + 2/n.
\end{equation*}
There are two cases:
Case 1. If $a_n^2 \le 2$, i.e. $a_n \le \sqrt 2$, we can refine (1) to
\begin{equation*}
\sqrt 2 - 2/n < a_n \le \sqrt 2.
\end{equation*}
And we have $a_{n + 1} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2730296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate the following limit: $\lim_{n\to \infty} a_n = \frac{1}{n^5}(1^4 +2^4+....+n^4)$ Calculate the following sequence: $\lim_{n\to \infty} a_n = \frac{1}{n^5}(1^4 +2^4+....+n^4)$.
I know how to solve this using Riemann sums of the function $x^4$.
I was wondering if there's another way using other methods. Any hel... | The Binomial Theorem gives
$$
(k+1)^5-k^5=5k^4+10k^3+10k^2+5k+1\tag1
$$
Summing both sides yields
$$
\begin{align}
(n+1)^5-1^5
&=\sum_{k=1}^n\left(5k^4+10k^3+10k^2+5k+1\right)\\
&=5\sum_{k=1}^nk^4+O\!\left(n^4\right)\tag2
\end{align}
$$
since
$$
\begin{align}
\sum_{k=1}^n\left(10k^3+10k^2+5k+1\right)
&\le\sum_{k=1}^n\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2732137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Solving differential equation using Frobenius Method I've been given the problem to solve the following differential equation
\begin{equation}
x^2y''+(2x+3x^2)y'-2y=0
\end{equation}
using Frobenius Method around the regular singular point $x=0$. From the indicialequation I get $r=-2$ or $r=1$, which differ by an intege... | Let $y=\sum_{n=r}^{\infty}{a_nx^n}$. Then,
$x^2y''+(2x+3x^2)y'-2y=0$ gives us
$\sum_{n=r}^{\infty}{a_nx^n(n(n-1)+2n-2}+a_nx^{n+1}(3n)=0$
$\sum_{n=r}^{\infty}a_nx^n{(n^2+n-2)}=\sum_{n=r}^{\infty}{a_nx^{n+1}(3n)}$
Looking at the $x^r$ term gives us that $r^2+r-2=0$ and $r=-2,1$ as you mentioned.
Comparing the subsequent... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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$y^2+5xy+6x^2-9x-4y=0$ where $x$ and $y$ are integers Need some help solving this for integers $x$ and $y$:
$$
y^2+5xy+6x^2-9x-4y=0
$$
I managed to make something like this:
$$
(y+3x-4)(y+2x)=x\\
(y+3x)(y+2x-3)=y
$$
Find integers for $x$ and $y$ that satisfy the equations above.
But, what do I do next, or is this a bad... | $$ (y+3x -3)(y+2x - 1)= y^2 + 5 yx + 6 x^2 -9x-4y +3 $$
$$ (y+3x -3)(y+2x - 1)= 3 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2738734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find $\frac{\tan 54-\tan 2}{\frac{\sin 2}{\cos 6}+\frac{\sin 6}{\cos 18}+\frac{\sin 18}{\cos 54}}$ Find $$S=\frac{\tan 54-\tan 2}{\frac{\sin 2}{\cos 6}+\frac{\sin 6}{\cos 18}+\frac{\sin 18}{\cos 54}}$$
All the angles are in degrees.
My Try: I tried converting everything to $\sin$ and $\cos$ we get
$$S=\frac{\sin 52}{\... | Have a look at @labbhattacharjee's comment, we will use that identity in the answer.
\begin{align}
\frac{\sin x}{\cos3x} & = \frac{\sin(3x-x)}{2 \cos x \cos 3x} \\ \tag{Angle Difference}
& = \frac{\sin 3x \cos x - \cos 3x \sin x}{2\cos x \cos 3x} \\
& = \frac{\cos x \cos 3x(\tan 3x - \tan x)}{2\cos x \cos 3x} \\
&= \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2739597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Probability two withdrawn balls are the same color Suppose we have $n$ white and $m$ black balls in a urn. First, randomly withdraw two balls, what is the probability (Call it $P_1$) that they are the same color? Now, suppose a ball is randomly withdrawn and then replaced before second one is drawn, what is the probabi... | When working without replacement, we can either select $2$ white balls with probability
$$\frac{n}{n+m}\cdot\frac{n-1}{n+m-1}$$
or $2$ black balls with probability
$$\frac{m}{n+m}\cdot\frac{m-1}{n+m-1}$$
giving $$P_1=\frac{n}{n+m}\cdot\frac{n-1}{n+m-1}+\frac{m}{n+m}\cdot\frac{m-1}{n+m-1}$$
which is equivalent to what y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2740855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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How to solve this equation for c? I want to solve the equation
$$
-\frac{a}{2}\left(c+\sqrt{c^2+4}\right)=-\frac{a-1}{2}\left(c-\sqrt{c^2+4}\right)
$$
for $c$, where $a$ is just a constant.
What I get is
$$
\frac{c-\sqrt{c^2+4}}{c+\sqrt{c^2+4}}=\frac{a}{a-1}.
$$
I think there now is some "trick" to solve this for $c$.
| Rationalise the denominator to get $$\frac{2c^2+4-2c\sqrt{c^2+4}}{2c^2+4}=\frac a{a-1}.$$ Then after simplifying, you get $$\frac 1{1-a}=\frac c{c^2+2}\sqrt{c^2+4}.$$ If you then square both sides and make the substitution $B=1/(1-a)^2$, you should get a quadratic equation in $c^2$, which ultimately leads where you're ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2741138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Using the $N - \varepsilon$ definition to find the limit of a sequence Yesterday I had a post about this, and it cleared a lot up; however, even though I feel like I understand how to go about solving problems like this, I don't seem to get the right answers. For example:
$u_n = \frac{2n+3}{2n+1}$. We know that this se... | It might be useful to rewrite the expression in question before doing "epsilontic":
$$u_n = \frac{2n+3}{2n+1} = \frac{2n+1+2}{2n+1}= 1+\frac{2}{2n+1}$$
Now,
$$u_n - 1 = \frac{2}{2n+1} \lt \frac{2}{2n}= \frac{1}{n}$$
From here. you can surely find your $N_{\varepsilon}$ easily.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2741249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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If $\frac{b+c}{2k-1}=\frac{c+a}{2k}=\frac{a+b}{2k+1}$, then $\frac{\sin A}{k+1}=\frac{\sin B}{k}=\frac{\sin C}{k-1}$
If $$\frac{b+c}{2k-1}=\frac{c+a}{2k}=\frac{a+b}{2k+1}$$
show that
$$\frac{\sin A}{k+1}=\frac{\sin B}{k}=\frac{\sin C}{k-1}$$
where $k$ is an integer such that $2 < k \neq 4$, and where $a$, $b$, $c... | Hint:
$$\frac{b+c}{2k-1}=\frac{c+a}{2k}=\frac{a+b}{2k+1}=\frac{(b+c)+(c+a)+(a+b)}{(2k-1)+(2k)+(2k+1)}=\frac{a+b+c}{3k}$$
$$\frac{b+c}{2k-1}=\frac{a+b+c}{3k}=\frac{(a+b+c)-(b+c)}{(3k)-(2k-1)}$$
You can easily find $a:b:c$ and apply the sine formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2742306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factoring and expanding $1-x^8$ It's been a while since I've studied factoring, and I need it for a question. How did this go from $1-x^8$ to $(1-x^4)(1+x^4)$ and then to $(1-x)(1+x+...+x^7)$? I remember studying this a few years go but unfortunately I don't remember.
$$1-x^8=(1-x^4)(1+x^4)=(1-x)(1+x+\dots + x^7)$$
| Taking your question, “How did this go from $1-x^8$ to $(1-x^4)(1+x^4)$ and then to...” literally, it involves repeated applications of the identity $1-z^2=(1-z)(1+z)$: $$\begin{align} 1+x^8 &= (1-x^4)(1+x^4) \\ &= (1-x^2)(1+x^2)(1+x^4) \\ &= (1-x)(1+x)(1+x^2)(1+x^4) \\ &= (1-x)(1+x^2+x^3+x^4+x^5+x^6+x^7). \end{align}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2742424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find $\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}$
Find $$\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}$$
without L'Hôpital's rule.
My work:
1) I know that $$\lim_{x\rightarrow0}\frac{\sin x}{x}=1$$
2) Let $x=t-\frac{\pi}{6}$. Then
$$\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}=... | Hint: Multiply numerator and denominator by $$1+2\sin(x)$$
you can write your term in the form $$\frac{\csc \left(\frac{\pi }{4}-\frac{x}{2}\right) \csc
\left(\frac{x}{2}+\frac{\pi }{4}\right)}{2 (2 \sin
(x)+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2744745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Kummer solution to second order ODE I need to solve a second order linear ODE with non-constant coefficients of the form
$$
\frac{d^2Z}{dt^2}+(a+be^{-ct})\frac{dZ}{dt}+dZ=0
$$
where $a,b,c,f$ are positive real constants.
Mathematica provides me with this solution
$$
i^{(a - \sqrt{a^2 - 4 d})/c}\, b^{(a - \sqrt{a^2 - 4 ... | Making
$$
\xi = a+b e^{-ct}\Rightarrow \dot \xi = -c(\xi-a)
$$
Now changing variables
$$
\ddot Z + (a+b e^{-ct})\dot Z + d Z \equiv c^2\ddot Z(t-a)^2+c(c-1)\dot Z (t-a)+dZ = 0
$$
This last equation has an easy solution as
$$
Z(\xi) = C_1 (a-\xi)^{-\frac{\left(-\frac{\sqrt{1-4 d}}{\sqrt{d}}-\frac{1}{\sqrt{d}}\right) \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2745254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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attempt to solve a Bernoulli equation I tried solving the Bernoulli equation $y'-y\tan x = y^4 \cos x $ by equating the left hand side to 0 and finding the homogeneous solution, however the equation turned out to be too complex and without the ability to isolate x. Any hints on how to solve this?
| Another trick
$$y'-y\tan x = y^4 \cos x$$
$$\cos(x)y'-y\sin x = y^4 \cos^2 x$$
$$(\cos(x)y)' = y^4 \cos^2 x$$
Substitute $z=\cos(x)y$
$$z' = \frac {z^4} {\cos^2 x}$$
It's separable
$$\frac 1{z^3} =-3\int \frac {dx} {\cos^2 x}$$
$$\frac 1{z^3} =-3\tan x +K$$
$$z^3 =\frac 1 {-3\tan x +K}$$
$$\boxed{y^3(x) =\frac 1{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2745841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Rational function in integral form
If $f$ is quadratic function such that $f(0)=1$ and $\int\frac{f(x)}{x^2(x+1)^3}dx $ is a rational function, find the value of $f'(0)$.
I already tried solving this question by using general quadratic equation $ax^2+bx+c$ and then using partial fraction method but it became very co... | We have that $b=f'(0)$ and $c=f(0)=1$. Moreover, by using partial fraction method we get
$$\frac{ax^2+bx+1}{x^2(x+1)^3}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}+\frac{D}{(x+1)^2}+\frac{E}{(x+1)^3}.$$
If the integral is a rational function, i.e. a ratio of polynomials, then $A=0$ and $C=0$ because the integrals of those ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Isolating $a_n$ in a recursive formula I have three equations with three sequences, $a_n, b_n, c_n$.
I tried to isolate $a_n$ with no luck.
$$a_n = 2b_{n-1}+c_{n-1}$$
$$b_n=2a_{n-1}+2b_{n-1}+c_{n-1}$$
$$c_n = 4a_{n-1}+4b_{n-1}$$
Is it even possible to get an expression based only on $a_n$ terms here?
| Your coefficient matrix is
$$
M =
\left(
\begin{array}{ccc}
0 & 2 & 1 \\
2 & 2 & 1 \\
4 & 4 & 0 \\
\end{array}
\right)
$$
which satisfies (Cayley-Hamilton)
$$ M^3 - 2 M^2 - 12 M - 8I = 0 $$
$$ a_{n+3} = 2 a_{n+2} + 12 a_{n+1} + 8 a_n \; . $$
You also get
$$ b_{n+3} = 2 b_{n+2} + 12 b_{n+1} + 8 b_n \; , $$
$$ c_{n+3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2749300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Map From the Cantor Set to the Unit Interval Let $f: C \rightarrow I$ map each point of the middle third Cantor set $C$, expressed as a ternary number which contains only digits $0$ and $2$, to the set of real numbers in $I=[0,1]$ expressed in base $2$ according the the rule:
$0.a_1a_2a_3...\rightarrow 0.b_1b_2b_3...$
... | Fix a ternary expansion
$$x = \frac{a_1}{3} + \frac{a_2}{3^2} + \frac{a_3}{3^3} + \ldots,$$
where $a_n = 0, 2$ for all $n$. Further, fix $\varepsilon > 0$. Choose an $m \in \mathbb{N}$ such that $2^{-m} < \varepsilon$. Note that two binary expansions are within $2^{-m}$ if they agree up to their $m$th bit. That is, nec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove the following determinant Prove the following:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|=4abc$$
My Attempt:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
Using $R_1\to R_1+R_2+R_3$
$$\left |
\begin{matrix}
2(b+c)&2(a+c)&2(a+b) \\
b&(c... | You can use the rule of Sarrus in this case:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
\begin{align}
&=(b+c)(c+a)(a+b) +abc +abc
- c(c+a)a - cb(b+c) -ab(a+b)\\
&=a^2 b + a^2 c + a b^2 + 2 a b c + a c^2 + b^2 c + b c^2 +2abc -c^2a-a^2c-c^2b-b^2c -a^2b-b^2a\\
&=4abc.
\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Find $f_{10}$ of the sequence defined by $f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$ Find $f_{10}$ of the sequence defined by $$f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$$ given $f_0=0$
My Approach:
Letting $f_n=2^n b_n$ we get
$$b_{n+1}=\frac{4b_n}{5}+\frac{3 \sqrt{1-b_n^2}}{5}$$
Now letting $b_n=\... | $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
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"language": "en",
"url": "https://math.stackexchange.com/questions/2753141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
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Three factor pairs summing to consecutive numbers The number $144$ can be factored in three different ways such that $$12\times12=9\times16=8\times18$$ with$$12+12=24;\quad 9+16=25; \quad8+18=26$$ the sum of the factor pairs being consecutive numbers, Similarly $180$ also has three such factor pairs, $(12,15), (10,18)... | I have a partial algorithm to give you a start.
You are looking for numbers with factors $a\cdot b = k$ such that $(1-a-b)^2-4ab = a^2+b^2-2(a+ab+b)+1$ is a perfect square. Additionally, you need $a+b-1+\sqrt{a^2+b^2-2(a+ab+b)+1}$ be divisible by 2. Finally, you want $k$ to be divisible by $\dfrac{a+b-1+\sqrt{a^2+b^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2754959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Where have I gone wrong in evaluating $\lim_{x \to \infty}\sqrt x(\sqrt{x+c}- \sqrt x )$?
Evaluate
$$\lim_{x \to \infty}\sqrt x(\sqrt{x+c}- \sqrt x )$$
Attempt:
$$\begin{align}
\lim_{x \to \infty}\sqrt x(\sqrt{x+c}- \sqrt x ) &= \lim_{x\to \infty }(\sqrt{x^2+cx}- x) \\
&= \lim _{x\to \infty}x\left(\sqrt{\left(1+\... | Write like this (using $a^2-b^2=(a-b)(a+b)$)
$$ \sqrt{x^2+cx} - x = \frac{ x^2+cx - x^2}{\sqrt{x^2+cx}+x} = \frac{cx}{\sqrt{x^2+cx}+x}$$
Now, factor $x$ numerator and denominator cancel and we obtain
$$ \frac{ c }{ \sqrt{ 1 + \frac{c}{x} } + 1 } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2755909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
$\sqrt[5]{2^4\sqrt[3]{16}} = ? $ $$\sqrt[5]{2^4\sqrt[3]{16}} = ? $$
Rewriting this and we have
$$\sqrt[5\cdot 3]{2^{4\cdot 3}4}$$
$$\sqrt[15]{2^{12}2^2}$$
Finally we get
$$\sqrt[15]{2^{12}2^2} = \sqrt[15]{2^{14}} = 2^{\frac{15}{14}}$$
Am I right?
| $$\sqrt[5]{2^4\sqrt[3]{16}} = (2^4)^{1/5}(16^{1/3})^{1/5}=2^{4/5}(2^4)^{1/15}=2^{4/5}2^{4/15}=2^{16/15}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2756282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Show that $\int_0^{\pi/2}\frac{\sin x\cdot \cos x}{x+1}dx=\frac12(\frac12+\frac1{\pi +2}-A)$
Show that $\int_{0}^{\pi/2}\frac{\sin x\cdot \cos x}{x+1}dx=\frac{1}{2}(\frac{1}{2}+\frac{1}{\pi +2}-A)$
where $A=\int_0^\pi\frac{\cos x}{(x+2)^2}dx$.
I tried using partial integration on the integral $\int_0^{\pi/2}\frac{\... | Just a couple of tiny missteps by the OP.
\begin{align*}
\mathcal{I} \equiv\int_0^{\pi/2} \frac{ \sin x \cos x}{x+1}\,\mathrm{d}x &= \int_0^{\pi/2} \frac{ \sin 2x }{2x+2}\,\mathrm{d}x \tag*{, then denote $u = 2x$} \\
&= \frac12\int_0^{\pi} \frac{ \color{blue}{\sin u} }{u+2}\,\color{blue}{\mathrm{d}u} \tag*{, then by-pa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2756608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Function that sends $\sum_{i=0}^n x^i$ to $\sum_{i=0}^n x^{-i}$? Is there a known function $f(S,x)$ that takes
$$S=1+x+x^2+x^3+x^4+\dots + x^n$$
and $x$ as its input and returns
$$S'=1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^4}+\dots+\frac{1}{x^n} \ $$
for $x \in$ R, $x>0$ and $n>0$?
Edit: Simply evaluatin... | The following formula inverts the terms of the geometric series $\displaystyle{S=\sum_{i=0}^nx^i}$ independent of n:
$$S'=\frac{x}{\frac{1}{S}+x-1}.$$
As should be clear,
$$S'=\sum_{i=0}^nx^{-i}=\frac{S}{x^n}.$$
Since $S$ is a geometric series, $S=\frac{1-x^n}{1-x}$ and
$$
\begin{align}
1-x^n &= (1-x)S \\
x^n &= 1-S+xS... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2756730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
For what interval of $k$ does the equation have one postitive and one negative root? I was given an equation that looked like this
$(k-x)(1-x) + 4$
and was told to find the interval of k when the equation would have one positive root and one negative root.
So far I have found the solutions in terms of k
$x_1 = \frac{... | I made some silly errors
in my original answer.
The corrections now seem to show that
k < -4 is the condition.
Following dxiv's advice,
the roots are
$x_1 = \frac{1}{2}(-\sqrt(k^2-2k-15)+k+1)
$
and
$x_2 = \frac{1}{2}(\sqrt(k^2-2k-15)+k+1)
$.
For there to be
two real roots,
we must have
$k^2-2k-15 > 0$
or
$k^2-2k+1 > 16... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the integral $\int\frac{dx}{\sqrt{x^2-a^2}}$
Evaluate the integral of $\frac{1}{\sqrt{x^2-a^2}}$
Put $x=a\sec\theta\implies dx=a\sec\theta\tan\theta d\theta$
$$
\begin{align}
& \ \ \ \int \frac{dx}{\sqrt{x^2-a^2}} \\
&=\int \frac{a\sec\theta\tan\theta d\theta}{\sqrt{a^2\sec^2\theta-a^2}} \\
&=\int \frac{a\sec\... | Suppose that $a>0$.
The work is just for the case when $x>a$. The case for $x<-a$ is different, but the finals result is the same.
Let $x=a\sec\theta$, where $\theta\in[0,\frac{\pi}{2})\cup(\frac{\pi}{2},\pi]$. This is the domain of $\textrm{arcsec}$.
For $x< -a$, $\theta\in(\frac{\pi}{2},\pi]$ and so $\tan\theta\le0$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find all integral ordered pairs $(n,k)$ such that $\left\lfloor\frac{n^2+18n+10}{2}\right\rfloor = k^2$. I had two problems that I want to solve. The first one was easy, but the second one... not so much:
First Problem:
Find all values of $n$ such that$$\frac{(n+1)(n+9)+8n+1}{2} = n^2.$$
My Attempt:
$$(n+1)(n+9) = ... | A different approach to obtaining the Pell equations:
Note that $(n+1)(n+9)+8n+1=n^2+18n+10=(n+9)^2-71$ Let's write $N=n+9$. Then
$$\begin{align}
\left\lfloor(n+1)(n+9)+8n+1\over2 \right\lfloor=k^2
&\iff\left\lfloor N^2-71\over2\right\rfloor=k^2\\
&\iff k^2\le{N^2-71\over2}\lt k^2+1\\
&\iff71\le N^2-2k^2\lt73\\
&\iff 7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
how many positive dividers that aren't multiple of 2 are there in the number 52920? i need to know how many positive dividers that aren't multiple of 2 are there in the number 52920.
How do i eliminate multiples of 2?
| Just keep dividing by $2$ until it is no longer even:
$52920=2 \cdot 26460$
$2646- = 2 \cdot 13230$
$13230 = 2 \cdot 6615$
OK, so find out the dividers of $6615$.
Now, since it ends in $5$ it's divisible by $5$, so divide by $5$:
$6615 = 5 \cdot 1323$
Also, any number is divisible by $3$ iff the sum of its digits is di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2766723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find a corresponding eigenvector for each eigenvalue Let A =$\begin{pmatrix} 6 & 1 \\ 2 & 7 \end{pmatrix}$
(a) Find the eigenvalues of A.
(b) Find a corresponding eigenvector for each eigenvalue in part (a).
My attempt
a) Eigenvalues:
$$\begin{vmatrix} 6-\lambda& 1 \\ 2 & 7-\lambda \end{vmatrix}=0 \Rightarrow \lambda^2... | Both answers are correct. Eigenvectors that are multiples of each other share the same eigenvalue for a particular matrix.
$$Ae=\lambda e\\
A(ke)=k(Ae)=k(\lambda e)=\lambda (ke)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2767115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve differential equation $(1+x^2) \frac{dy}{dx} - 2xy = x$ Solve differential equation $$(1+x^2) \frac{dy}{dx} - 2xy = x$$
I simplified it to $$\frac{1}{1+2y} dy = \frac{x}{1+x^2} dx$$
$$ \int \frac{1}{1+2y} dy =\int \frac{x}{1+x^2} dx $$
$$\frac{1}{2} \int \frac{2}{1+2y} dy = \frac{1}{2} \int \frac{2x}{1+x^2} dx$$
... | solving the equation $$\ln(1+2y)=ln(1+x^2)+C$$ for $y$ we get
$$y=\frac{e^c}{2}(x^2+1)-\frac{1}{2}$$ now substitute $$k=\frac{e^C}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Show that $\cos(x) \le e^{-x^2/2}$ for $0 \le x \lt \pi/2$. Show that $\cos(x) \le e^{-x^2/2}$ for $0 \le x \lt \pi/2$.
This inequality came up
in my solution to
Show that the sequence $\sum\limits_{k=1}^n\cos\left(\frac kn\right)^{2n^2/k}$ converges.
This is in the category of
"There should be
a number of ways
to prov... | The function $\tan x$ has a Maclaurin series with positive coefficients, convergent for $|x|<\pi/2$, see series . Its integral $\log(\sec x)$ also has a series with positive coefficients
$$\log(\sec x)=\frac{x^2}{2} + \frac{x^4}{12} + \frac{x^6}{45} + \frac{17 x^8}{2520} + \frac{31 x^{10}}{14175}+\mathcal{O}(x^{12})$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2769766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
What order of operation should I use calculating $2^3 \times 4 - 6 \times 3 ÷ (4 + 3) - 16 ÷ 4$? I am struggling if I should use PEMDAS or BODMAS in this equation. What is the right method to get around these type of equations?
$$2^3 \times 4 - 6 \times 3 ÷ (4 + 3) - 16 ÷ 4 = \ ?$$
| In this case both PEMDAS and BODMAS give the same result:
$2^3 \times 4 - 6 \times 3 ÷ (4 + 3) - 16 ÷ 4 =$
$8\times 4 - 6\times 3\div 7 - 16\div 4 = $
$32 - 18\div 7 - 16\div 4 = $(PEMDAS) or $8\times 4 - 6\times \frac 37 - 4$ (BODMAS)
$32 - \frac {18}7 - 4=$
$\frac {206}7 - 4 = \frac {178}7$.
The ambiguity lies when ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2773501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
When does $\frac{a+b}{2}$ and $\sqrt{ab}$ have inversed tens digits and ones digits? Let $a$ and $b$ be natural numbers, and
$$A = \frac{a+b}{2}$$
$$B = \sqrt{ab}$$
It's given that $A$ and $B$ are two-digit numbers such that the tens digit of $A$ is the same as the ones digit of $B$, and the tens digit of $B$ is the ... | Note that $$A+B = 11(x+y)$$
and $$A-B = 9(x-y)$$
where $$1\le y \le x \le 9$$
Considering all our options, we see $x+y=9$ and $x-y=1$ provides the given solution of A=54 and B=45.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2775061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove $\sin(\alpha -\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\frac{\alpha-\beta}{2}\sin\frac{\alpha-\gamma}{2}\cos\frac{\beta-\gamma}{2}$ Here is a problem from Gelfand's Trigonometry:
Let $\alpha, \beta, \gamma$ be any angle, show that $$\sin(\alpha -\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\left(... | Use
\begin{eqnarray*}
\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)
\end{eqnarray*}
to give
\begin{eqnarray*}
\sin (\alpha-\beta) + \sin (\alpha-\gamma) = 2 \sin \left( \frac{2 \alpha-\beta-\gamma}{2} \right) \cos \left( \frac{\beta-\gamma}{2} \right).
\end{eqnarray*}
Now use ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2775981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Exact value of a convergent series: $\sum_{n=1}^\infty \frac{1}{n^3+6n^2+8n}$ I have a series
$$\sum_{n=1}^\infty \frac{1}{n^3+6n^2+8n}.$$
I know the series converges because
$$\frac{1}{n^3+6n^2+8n}\le \frac{1}{n^3}, $$
since $p=3>1$, I know that $\sum 1/n^3$ converges. But I am not sure how to figure out what it conv... | Note
$$ \frac{1}{n^3+6n^2+8n}=\frac1{n(n+2)(n+4)}=\frac14\bigg[\frac{1}{n(n+2)}-\frac{1}{(n+2)(n+4)}\bigg]. $$
Let
$$ f(x)=\sum_{n=1}^\infty\frac{1}{n(n+2)}x^n, g(x)=\sum_{n=1}^\infty\frac{1}{(n+2)(n+4)}x^{n+2}. $$
Then
\begin{eqnarray}
f(1)&=&\int_0^1\frac1{x^3}\int_0^x\frac{t^2}{1-t}dtdx\\
&=&\int_0^1\int_0^x\frac{t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2776091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Simplifying $\operatorname{tanh}(\operatorname{arsinh}(x))$ So I am trying to simplify $\tanh(\operatorname{arsinh}(x))$ to $\frac{x}{\sqrt{1+x^2}}$
In general,
$$\tanh(x) = \frac{e^x-e^{-x}}{e^x+e^{-x}}$$
and $$\operatorname{arsinh}(x)= \ln(x+\sqrt{x^2-1})$$
therefore
\begin{align}\tanh(\operatorname{arsinh}(x)) & =\... | Let $$ arsinh(x)=y$$
$$ x= sinh (y)$$
$$ \sqrt {1+x^2} = cosh (y)$$
$$tanh ^2 (y) = 1- sech ^2 (y) = 1-\frac {1}{1+x^2}=\frac {x^2}{1+x^2} $$
$$tanh (y)=\frac {x}{\sqrt {1+x^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2777721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding the residue of $\frac{1}{z^3 \sin z}$ at z = 0 Given the function $f(z) = \frac{1}{z^3 \sin{(z)}}$, what is the residue of $f(z)$ at $z = 0$?
I want to find $b_1$ from the Laurent expansion. So I did the following:
\begin{align*}
\frac{1}{z^3 \sin{(z)}}
&= ( \dots \frac{b_5}{z^5} + \frac{b_4}{z^4} + \frac{b_... | $\displaystyle z \mapsto -z$ does not change
$\displaystyle {1 \over z^{3}\sin\left(z\right)}.\quad$ So,
$\displaystyle {1 \over z^{3}\sin\left(z\right)}
\,\,\,\stackrel{\mathrm{as}\ z\ \to\ 0}{\sim}\,\,\, {b_{4} \over z^{4}}
+ {b_{2} \over z^{2}} + {\color{red}{\large 0} \over z} + a_{0}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2778613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Prove that $x^{3m}+x^{3n+1}+x^{3p+2}$ is divisible by $x^2+x+1$ Prove that $x^{3m}+x^{3n+1}+x^{3p+2}$ is divisible by $x^2+x+1$ in ring $ {\displaystyle \mathbb {R}}$$[x]$ where $m, n, p \in {\displaystyle \mathbb {N}}$
I have tried to use factor theroem to write down $x^{3m}+x^{3n+1}+x^{3p+2}$ as a product of $x^2... | Note:
$$\frac{x^{3m}+x^{3n+1}+x^{3p+2}}{x^2+x+1}=\frac{x^{3m}-1+x^{3n+1}-x+x^{3p+2}-x^2+(x^2+x+1)}{x^2+x+1}=\\
\frac{(x^{3m}-1)+x(x^{3n}-1)+x^2(x^{3p}-1)+(x^2+x+1)}{x^2+x+1}.$$
And:
$$x^{3k}-1 \equiv 0 \pmod{x^2+x+1}, k\in \mathbb N.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2779155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Power series of $f(x)=\ln (x^2+4)$ I am supposed to find a power series representation of
$$f(x)=\ln\left(x^{2}+4\right).$$
Then, I am to graph it and observe what happens as $n$ increases. My attempt at a solution:
$$\ln\left(x^2+4\right) = \int \frac{1}{x^2+4}\,dx = \frac{1}{4}\int \frac{1}{\frac{x^2}{4}+1}\,dx = \fr... | At first: the cubic member missed in OP, and the signs of the other ones inverted. So the series is wrong at all.
Corrrect realizaion of the OP idea is:
$$\ln(4+x^2) = \dfrac14x - \dfrac1{48}x^3+ \dfrac1{320}x^5- \dfrac1{1792}x^7+\dots.$$
Then, there are known the series
$$\ln(1+t) = t-\dfrac12t^2+\dfrac13t^3+\dots,\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Integrate $\sin^{-1}\frac{2x}{1+x^2}$
Integrate $\sin^{-1}\frac{2x}{1+x^2}$
The solution is given in my reference as: $2x\tan^{-1}x-\log(1+x^2)+C$.
But, is it a complete solution ?
My Attempt
$$
\int 2\tan^{-1}x \, dx=\int \tan^{-1}x \cdot 2\,dx=\tan^{-1}x\int2\,dx-\int\frac{1}{1+x^2}\int2\,dx\cdot dx\\
=\tan^{-1}x \... | According to the standard tangent half-angle substitution
\begin{align}
& x=\tan\frac\theta 2 \\[8pt]
& 2\arctan x = \theta \\[8pt]
& \frac {2\,dx}{1+x^2} = d\theta \\[8pt]
& dx = \sec^2\frac\theta 2 \,\,\frac{d\theta} 2 \\[8pt]
& \frac{2x}{1+x^2} = \sin\theta \\[8pt]
& \frac{1-x^2}{1+x^2} = \cos\theta
\end{align}
Thus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
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Prove the polynomial $x^2+2x+1$ divides the polynomial $x^{2(2n+1)}+2x^{2n+1}+1$ for every $n\in \mathbb{N}$ I've been trying to prove that the polynomial $x^2+2x+1$ divides the polynomial $x^{2(2n+1)}+2x^{2n+1}+1$ for all $n\in \mathbb{N}$ as follows and reached a "roadblock":
I wrote $x^2+2x+1$ as $(x+1)^2$ and showe... | Take the equation
$$(x^{2n}-x^{2n-1}+x^{2n-2}-\ ...\ +x^2-x+1)(x+1) = x^{2n+1}+1$$
This holds true, since
$$(x^{2n}-x^{2n-1}+x^{2n-2}-\ ...\ +x^2-x+1)(x+1) = \\ =x^{2n+1}-x^{2n}+x^{2n-1}-\ ...\ +x^3-x^2+x + \\ +\ x^{2n}-x^{2n-1}+x^{2n-2}-\ ...\ +x^2-x+1 = \\ = x^{2n+1} + 1$$
You can see that everything in the center c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2786213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Given that $a \gt b\ge 1$,show that $f(b)\gt f(a)$ Given that:
$$f(x)=\frac{x}{x^{2}+1}$$
If $a \gt b \ge 1$ and both are real numbers,show that $f(b)\gt f(a)$.
My attempt:
\begin{equation}\begin{aligned} f(b)-f(a)
&=\dfrac{b(a^2+1)-a(b^2+1)}{(b^2+1)(a^2+1)} \\
&=\dfrac{a^2b-ab^2+b-a}{(b^2+1)(a^2+1)} \\
&=\dfrac{ab(a-... | Your way is completely correct.
Another way to see it is just to calculate the derivative
*
*$f'(x) = \frac{1-x^2}{(x^2+1)^2} < 0$ for $x>1 \Rightarrow f$ is strictly decreasing on $[1,\infty)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve $2c x + x\dot x + (c^2-1)t = cd$ After computation on a certain problem, for a specific case I came across the differential equation $$2c x + x \dot x + (c^2-1) t = cd$$
with initial conditions $0 < x(0) = d $ and $1 < - \dot x(0) = c $. Does this have a closed form?
This differential equation doesn't app... | $$2c x + \frac{dx}{dt} x + (c^2-1) t =cD$$
The symbol $D$ is used instead of $d$ to avoid confusion with the differential symbol.
Let $\quad cD=a$
$$2c x + \frac{dx}{dt} x + (c^2-1) t - a=0$$
$y(t)=(c^2-1)t-a \quad;\quad \frac{dx}{dt}= \frac{dx}{dy}\frac{dy}{dt}= (c^2-1)\frac{dx}{dy}$
$$2c x + (c^2-1)x\frac{dx}{dy} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Testing $\sum\limits_{k=1}^∞(\frac{k+1}k)^{k^2}3^{-k}$ for convergence and absolute convergence
Test $$\sum_{k=1}^{\infty}\left(\frac{k+1}{k}\right)^{k^2}3^{-k}$$ for convergence and absolute convergence.
We apply the ratio test for $\displaystyle \sum_{k=1}^{\infty}\left|\left(\frac{k+1}{k}\right)^{k^2}3^{-k}\right|... | FYI, estimation of upper bound:
$\sum\limits_{k=1}^{\infty}\left(\frac{k+1}{k}\right)^{k^2}3^{-k}=\sum\limits_{k=1}^{\infty}\left(\frac{1}{k}+1\right)^{k^2}3^{-k}$
Using binominal theorem and the following unequality: $\begin{pmatrix}n \\ {k} \end{pmatrix}\lt \frac{n^k}{k!}$ we get:
$\sum\limits_{k=1}^{\infty}\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Inequality in number theory Prove that:
(i)$5<5^{\frac{1}{2}}+5^{\frac{1}{3}}+5^{\frac{1}{4}}$
(ii)$8>8^{\frac{1}{2}}+8^{\frac{1}{3}}+8^{\frac{1}{4}}$
(iii)$n>n^{\frac{1}{2}}+n^{\frac{1}{3}}+n^{\frac{1}{4}}$ for all integer $n\geq9$
Can raising both sides to exponent $12$ help
| The last one is equivalent to showing
$$1 > \frac{1}{n^{\frac{1}{2}}}+\frac{1}{n^{\frac{2}{3}}}+\frac{1}{n^{\frac{3}{4}}}$$
and
$$n\geq 9 \Rightarrow \frac{1}{3}\geq \frac{1}{n^{\frac{1}{2}}}$$
$$n^2\geq 9^2 > 3^3 \Rightarrow \frac{1}{3}> \frac{1}{n^{\frac{2}{3}}}$$
$$n^3\geq 9^3 > 3^4 \Rightarrow \frac{1}{3}> \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding factors of second order complex polynomial. It concerns with finding roots of complex polynomial:
$x^2+(i-1)x+(2+i)$. One way is to find roots by factoring, as:$(x-i)(x -(1-2i))=(x-i)(x-1+2i)=(x^2 -x +ix +i +2)=x^2 +x(i-1)+(i+2)$
But, this is a guess game for me, with no formal process to get in this case.
The... | $$x^2+(i-1)x+2+i=0$$
So
$$x=\frac{-(i-1)\pm\sqrt{-8-6i}}{2}=\frac{-(i-1)\pm i\sqrt{8+6i}}{2}$$
$$\sqrt{8+6i}=\sqrt{10}\left(\cos\left(\frac{\tan^{-1}\left(\frac{3}{4}\right)+2\pi k}{2}\right)+i\sin\left(\frac{\tan^{-1}\left(\frac{3}{4}\right)+2\pi k}{2}\right)\right)\space\space\space\text{k=0, 1}$$
Take $k=0$ since if... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2797468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove that $\frac{1}{\sqrt {a^2 - b^2}} = \frac{1/a}{\sqrt {1 - \frac{b^2}{a^2}}}$ Recently I have been reading physics book and saw interesting equation, like this:
$$\frac{1}{\sqrt {a^2 - b^2}} = \frac{1/a}{\sqrt {1 - \frac{b^2}{a^2}}}$$
But I still don't understand how to get the right part of the equation from left... | For nonnegative $u$ and $v$, you can write $\sqrt{uv}=\sqrt{u}\cdot\sqrt{v}$. You also have, for any $u$, $\sqrt{u^2}=|u|$. Use this here:
Assume $a^2\ge b^2$ and $a\ne0$, then
$$\sqrt{a^2-b^2}=\sqrt{a^2\left(1-\frac{b^2}{a^2}\right)}=\sqrt{a^2}\cdot\sqrt{1-\frac{b^2}{a^2}}=|a|\sqrt{1-\frac{b^2}{a^2}}$$
If you take the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Help finding the limit of the following function as x tends to 0 $$
\lim_{x\to0}\left(\sqrt{\frac{1}{x} + \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x}}}} - \sqrt{\frac{1}{x} - \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x}}}}\,\right)
$$
From Demidovich 5000 problems in mathematical analysis
| $\lim \limits_{x\rightarrow0} \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x}}}} - \sqrt{\frac{1}{x} - \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x}}}}=?$
Let $u=\sqrt{\frac{1}{x} + \sqrt{\frac{1}{x}}}$ then the function is
$\sqrt{\frac{1}{x}+ u} -\sqrt{\frac{1}{x}-u}=\frac{1}{\sqrt{x}}(1+ux)^{\frac{1}{2}}-\frac{1}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2801428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Cool way of finding $\cos\left(\frac{\pi}{5}\right)$ while trying to solve a problem, I stumbled upon a way of finding $\cos\left(\frac{\pi}{5}\right)$ using identities and the cubic formula. Is it possible to find other values of sine or cosine in a similar way ?
Consider
$$\cos\left(\frac{\pi}{5}\right) - \cos\left(\... | There is a slightly shorter way:
$$\sin\frac{\pi}{5}=\sin\frac{4\pi}{5}=2\sin\frac{2\pi}{5}\cos\frac{2\pi}{5}=4\sin\frac{\pi}{5}\cos\frac{\pi}{5}\cos\frac{2\pi}{5}$$
Hence
$$\cos\frac{\pi}{5}\cos\frac{2\pi}{5}=\frac14$$
You can also write this
$$\cos\frac{\pi}{5}\cos\frac{3\pi}{5}=-\frac14$$
Then, using the formula $2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
} |
Splitting up an infinite sum I am playing with the following example trying to determine if $\frac{1}{n}-\frac{1}{\sqrt{n}}$ is in $\ell^2$.
Here is what I have,
\begin{align}
\sum\limits_{n=1}^\infty \left( \frac{1}{n}-\frac{1}{\sqrt{n}}\right)^2 &= \sum\limits_{n=1}^\infty \left(\frac{\sqrt{n}-n}{n\sqrt{n}}\right)^2\... | You could note that$$\sum_{n=1}^\infty\frac{n-2n\sqrt n+n^2}{n^3}=\sum_{n=1}^\infty\frac{1-2\sqrt n+n}{n^2}$$and that$$\lim_{n\to\infty}\frac{\frac{1-2\sqrt n+n}{n^2}}{\frac1n}=\lim_{n\to\infty}\frac{1-2\sqrt n+n}n=1.$$Therefore, your series diveres, by the comparison test.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
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.