Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to solve this kind of equation? I have an equation (in my homework) of the form
$a=\sqrt{x^2 + b^2} + \sqrt{x^2 + c^2}$
which I would like to solve for $x$. I am not sure how best to proceed. My thought is to square both sides of the equation, which gives me
$a^2=b^2+c^2+2x^2+\sqrt{(b^2+x^2)\times(c^2+x^2)}$
and ... | Suppose that our equation
$$\sqrt{x^2+b^2}+\sqrt{x^2+c^2}=a.\tag{1}$$
holds. Assume $a\ne 0$ and $b\ne c$. Flip both sides over. We get
$$\frac{1}{\sqrt{x^2+b^2}+\sqrt{x^2+c^2}}=\frac{1}{a}.$$
Multiply top and bottom on the left by $\sqrt{x^2+b^2}-\sqrt{x^2+c^2}$. We get
$$\frac{\sqrt{x^2+b^2}-\sqrt{x^2+c^2}}{b^2-c^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/744426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Maximum likelihood to throw exactly two 6s One throws a dice $n$ times. For which value of $n$ is maximum the probability to obtain exactly two 6s?
I get $$n=11 \text{ or } n=12.$$
My solution:
the probability to obtain exactly two 6s in $n$ throws is (Bernoulli distribution)
$$ \binom{n}{2} \left( \dfrac{1}{6} \right... | You do not need calculus, since you can consider
$$\dfrac{f(n+1)}{f(n)} = \dfrac{P(\text{sixes}=2|\text{dice}=n+1)}{P(\text{sixes}=2|\text{dice}=n)} = \dfrac{n+1}{n-1}\times \dfrac56$$
which is greater than $1$ when $2 \le n \lt 11$, is less than $1$ when $n \gt 11$, and is equal to $1$ when $n=11$.
So $11$ and $12$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/745859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proving that a function is bijective I have trouble figuring out this problem:
Prove that the function $f: [0,\infty)\rightarrow[0,\infty)$ defined by $f(x)=\frac{x^2}{2x+1}$ is a bijection.
Work: First, I tried to show that $f$ is injective. $\frac{a^2}{2a+1}=\frac{b^2}{2b+1}$
I got $a^2(2b+1)=b^2(2a+1)$. However, I ... | Notice $f(x) = \frac{x^2}{2x+1} = \frac{x}{2 + \frac{1}{x}}$. Suppose $f(a) = f(b) $, then
$$ \frac{a}{2 + \frac{1}{a} } = \frac{b}{2 + \frac{1}{b}} \iff \frac{a}{b} = \frac{ 2 + \frac{1}{a}}{2 + \frac{1}{b}}$$
We want to show that $a = b$. Suppose not. then either $a > b $ or $b >a $.
Say $a > b $. Then $\frac{1}{b} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/747294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Convergence of $ \sum_{n=1}^{\infty} (\frac{n^2+1}{n^2+n+1})^{n^2}$
Find if the following series converge: $$\displaystyle \sum_{n=1}^{\infty} \left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}$$
What I did:
$$a_n=\left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}$$
$$b_n=\frac {1}{(n+1)^{n^2}}$$
$$\forall n\ge 1 : a_n < b_n \ \Rightar... | I would take the following approach:
$$\begin{align}\left (\frac{n^2+1}{n^2+n+1}\right )^{n^2} &= \left (1-\frac{n}{n^2+n+1}\right )^{n^2}\\ &=e^{n^2 \log{\left (1-\frac{n}{n^2+n+1}\right )}} \\ &\sim e^{-n^2 (1/n - 1/(2 n^2))}\\ &\sim e^{-n+1/2} \end{align}$$
Thus the series converges by comparison with a geometric se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/748110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
An Inequality Problem with not nice conditions How to show that $\dfrac{a^3}{a^2+b^2} + \dfrac{b^3}{b^2+c^2} + \dfrac{c^3}{c^2+a^2} \ge \dfrac32$, where $a^2+b^2+c^2=3$, and $a,b,c > 0$ ?
| How $a^2+b^2+c^2=3$, then is equivalent to proof that
$$\frac{a^3}{3-a^2}+\frac{b^3}{3-b^2}+\frac{c^3}{3-c^2}\geq\frac{3}{2}$$
Now the function $f(x)=\frac{x^3}{3-x^2}$ have $f''(x)>0$ in $(0,\sqrt{3})$ then $f(x)$ is convex in this interval, then:
$$f(a)+f(b)+f(c)\geq3f\left(\frac{a+b+c}{3}\right)$$
how $(a+b+c)^2=a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/752038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
solving the derivative of a function with cos my question is
y=cos^4(2x^2-1)
here is my work
`Dy/dx=4cos^3(2x^2-1) d/dx cos(2x^2-1)
Dy/dx=4cos^3(2x^2-1) (-d/dx(2x^2-1)sin(2x^2-1))
Dy/dx=-4cos^3(2x^2-1)sin(2x^2-1)(4x)
$$dy/dx=-16x\cos^3(2x^2-1)\sin(2x^2-1)$$
any idea on where i went wrong
| Using Chain rule,
$$\frac{d\{\cos^4(2x^2-1)\}}{dx}$$
$$=\frac{d\{\cos^4(2x^2-1)\}}{d\{\cos(2x^2-1)\}}\cdot\frac{d\{\cos(2x^2-1)\}}{dx}$$
$$=4\cos^3(2x^2-1)\cdot\frac{d\{\cos(2x^2-1)\}}{d(2x^2-1)}\cdot\frac{d(2x^2-1)}{dx}$$
$$=4\cos^3(2x^2-1)\cdot\{-\sin(2x^2-1)\}\cdot(2\cdot2x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/754343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I solve this definite integral: $\int_0^{2\pi} \frac{dx}{\sin^{4}x + \cos^{4}x}$? $$\int_0^{2\pi} \frac{dx}{\sin^{4}x + \cos^{4}x}$$
I have already solved the indefinite integral by transforming $\sin^{4}x + \cos^{4}x$ as follows:
$\sin^{4}x + \cos^{4}x = (\sin^{2}x + \cos^{2}x)^{2} - 2\cdot\sin^{2}x\cdot\cos^{2... | Note that $\tan 2x$ is discontinuous at $\frac{\pi}4$ and some other values which can be easily found.
You have to break the integral from $0$ to $\frac{\pi}4$ and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/754750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Integer solutions of the equation $a^{n+1}-(a+1)^n = 2001 $ I am doing number theory and I came across that question $a^{n+1}-(a+1)^n = 2001 $. Find the integer solutions and show that they are the only solution.
I really tried hard but i am nowhere near solution.
| I'm assuming $a,n$ are positive integers.
First of all, $a^{n+1}-(a+1)^n \equiv -1 \equiv 2001 \;\; \text{mod} \; a ,$ so $a$ divides $2002=2 \times 7 \times 11 \times 13. $ Because $3|2001$ then $a^{n+1} \equiv (a+1)^n \;\; \text{mod} \; 3.$ You can easily eliminate $a \equiv 0,-1 \;\; \text{mod} \; 3,$ therefore, $a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/755265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Show that $(x + 1)^{(2n + 1)} + x^{(n + 2)}$ can be divided by $x^2 + x + 1$ without remainder I am in my pre-academic year. We recently studied the Remainder sentence (at least that's what I think it translates) which states that any polynomial can be written as $P = Q\cdot L + R$
I am unable to solve the following:
... | If $n=0$, then it is trivial.
For $n>0$, we have $$(x+1)^{2n+1}+x^{n+2}\\=(x^2+2x+1)(x+1)^{2n-1}+x^{n+2}\\=(x^2+x+1)(x+1)^{2n-1}+x(x+1)^{2n-1}+x^{n+2}\\=(x^2+x+1)(x+1)^{2n-1}+x((x+1)^{2n-1}+x^{n+1})$$.
By using induction, suppose $(x+1)^{2n-1}+x^{n+1}$ can be divided by $x^2+x+1$, then, $(x+1)^{2n+1}+x^{n+2}$ also can... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/757702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Stars and bars (combinatorics) with multiple bounds Count the number of solutions to the following:
$$x_1+x_2+\cdots+x_5=45$$
when:
$1$. $x_1+x_2>0$, $x_2+x_3>0$, $x_3+x_4>0$
$2$. $x_1+x_2>0$, $x_2+x_3>0$, $x_4+x_5>1$
($x_1,\ldots,x_5$ are non-negative integers)
Thanks!
| Using the principle of inclusion-exclusion, we can derive the required generating function and extract the coefficient for the formula.
1)
Let $\mathbb{N}$ indicate the number of solutions with the only condition being that any $x_i\ge 0$.
Let the given constraints be:
$A: x_1+x_2>0\\
B: x_2+x_3>0\\
C: x_3+x_4>0
$
We r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/760950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
integer solutions to $x^2+y^2+z^2+t^2 = w^2$ Is there a way to find all integer primitive solutions to the equation $x^2+y^2+z^2+t^2 = w^2$? i.e., is there a parametrization which covers all the possible solutions?
| For the equation:
$X^2+Y^2+Z^2+Q^2=R^2$
We can write the solution:
$X=2p^2+2(a+b+t)ps+(ab+at+bt)s^2$
$Y=2p^2+2(b+t-a)ps+(bt-a^2)s^2$
$Z=2p^2+2(a+t-b)ps+(at-b^2)s^2$
$Q=2p^2+2(a+b-t)ps+(ab-t^2)s^2$
$R=4p^2+2(a+b+t)ps+(a^2+b^2+t^2)s^2$
And more:
$X=-2p^2+2(a+b+t)ps+(4a^2+4b^2+4t^2-ab-at-bt)s^2$
$Y=-2p^2+2(b+t-5a)ps+(a^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/762769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding the limit of $\lim_{x\to 1} (x^2-\sqrt x)/(1-\sqrt x)$ How do I evaluate
$$\lim_{x\to 1} \frac{(x^2-\sqrt x)}{(1-\sqrt x)}$$
Can someone explain the steps by steps solution to this problem?
| Notice
$$ \frac{ x^2 - \sqrt{x}}{1 - \sqrt{x} } = \frac{ x^2 - \sqrt{x}}{1 - \sqrt{x} } \cdot \frac{1 + \sqrt{x}}{1+ \sqrt{x}} = \frac{x^2 + x^2 \sqrt{x} - \sqrt{x} - x}{1-x} = \frac{x(x-1) + \sqrt{x}((x-1)(x+1)}{-(x-1)}= \frac{x + \sqrt{x}(x+1)}{-1} \to_{x \to 1} -3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/763292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proving a trigonometric relation $$\frac{1+\cos{2\alpha}}{\sin^2{2\alpha}}=\frac{1}{2}\csc^2{\alpha}$$
Here the question is that I can prove that the $LHS=RHS$ if I use the variable $x$ but if we take
($2x=\pi$ on the left side) and ($x=\dfrac{\pi}{2}$ on the other side)
then the equation is not equal,why?
Then prove:
... | $$
\frac {1+\cos 2x}{\sin^2 2x} = \frac {2\cos^2 x}{4 \sin^2 x \cos^2 x} = \frac 1{2\sin^2 x} = \frac 12 \csc^2 x
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/764757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the indefinite integral $\int \frac{\sqrt{9-4x^{2}}}{x}dx$ $$\int \frac{\sqrt{9-4x^{2}}}{x}dx$$
How Can I attack this kind of problem?
| Let $x=\cfrac{3}{2}\sin\theta$, then $dx=\cfrac{3}{2}\cos\theta\,d\theta$.
\begin{align}
\require{cancel}
\int\frac{\sqrt{9-4x^2}}{x}\, dx&=\int\frac{\sqrt{9-4\left(\frac{3}{2}\sin\theta\right)^2}}{\cancel{\frac{3}{2}}\sin\theta}\, \cancel{\frac{3}{2}}\cos\theta\,d\theta\\
&=\int\frac{3\sqrt{1-\sin^2\theta}}{\sin\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/767463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Focus of parabola with two tangents
A parabola touches x-axis at $(1,0)$ and $y=x$ at $(1,1)$. Find its focus.
My attempt : All I can say is that as angle subtended by this chord at focus is $90^\circ$ as angle between tangents is $45^\circ$. I can find equation of directrix by taking mirror image of focus in tangent... | I have a different approach to this problem. Using the general conic equation
$$ A x^2 + B x y + C y^2 + D x + E y + F =0 $$
and its derivative (for tangents)
$$ {\rm d} x ( 2 A x + B y + D ) + {\rm d} y ( B x + 2 C y + E) =0 $$
With the parabola constraint $B^2 = 4 A C$, and the following 4 constraints, all coefficien... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/767631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Use mathematical induction to prove that $(3^n+7^n)-2$ is divisible by 8 for all non-negative integers. Base step: $3^0 + 7^0 - 2 = 0$ and $8|0$
Suppose that $8|f(n)$, let's say $f(n)= (3^n+7^n)-2= 8k$
Then $f(n+1) = (3^{n+1}+7^{n+1})-2$
$(3*3^{n}+7*7^{n})-2$
This is the part I get stuck. Any help would be really appr... | If for $n=k, f(k)=3^k+7^k-2$
for $n=k+1,$
$\displaystyle f(k+1)-3f(k)= 3^{k+1}+7^{k+1}-2-3(3^k+7^k-2)$
$\displaystyle=7^k(7-3)+6-2=4(7^k+1)$ which is divisible by $8$ as $7^k$ is odd
So, $f(k+1)$ will be divisible by $8\iff f(k)$ is
We can start with $f(k+1)-7f(k)$ as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/770344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Real and Imaginary $$Re\Big(({\frac{1+i\sqrt{3}}{1-i})^4\Big)} = 2$$
$$Im\Big(({\frac{1+i}{1-i})^5\Big)} = 1$$
I got that $Re\Big(({\frac{1+i\sqrt{3}}{1-i})^4\Big)} = 1 \ne 2$
And, that $\Big(({\frac{1+i}{1-i})^5\Big)} = i $ , which means that $Im\Big(({\frac{1+i}{1-i})^5\Big)} = 1$
Can you guys confirm that it's true?... | You can write
$$
1+i\sqrt{3}=2\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)=
2(\cos(\pi/3)+i\sin(\pi/3))
$$
so
$$
(1+i\sqrt{3})^4=16\cos(4\pi/3)+i\sin(4\pi/3)=
16\left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right).
$$
On the other hand
$$
1-i=\sqrt{2}\left(\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}\right)=
\sqrt{2}(\cos(-\pi/4)+i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/772779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Which is the minimum number of days that are required,so that all the exams are taken? In a university, the secretariat plans the examination period. There are $6$ subjects, $A,B,C,D,E,Z$ and $9$ students($1, \dots , 9$). At the subject $A$ the students $1,2,3$ are subscribed, at the subject $B$ the students $1,2,9$, a... | I added labels to the graph's edges, for anyone having trouble seeing what's going on.
In addition, here is a table of the classes and students.
$$
\newcommand{\x}{\blacksquare}
\begin{array}{|c|cccccc|}
\hline
& A & B & C & D & E & Z \\
\hline
1 & \x & \x & \x & & & \\
2 & \x & \x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/772917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove that the following series do not converge absolutely Prove that the following series do not converge absolutely:
$$\sum_{x=1}^\infty \frac{\sin x}{x}$$
| The idea behind the proof below is that for any two consecutive integers $x$ and $x+1$, at least one of $|\sin(x)|$ and $|\sin(x+1)|$ is "big."
We use the fact that $\sin(x+1)=\sin x\cos 1 +\cos x\sin 1$. Suppose that $|\sin x|\le \frac{1}{3}$. Then $|\cos x|\ge \sqrt{1-\frac{1}{9}}$.
Thus if $|\sin x\le \frac{1}{3}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/773577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
radicals and rational exponents. simplifying $4^{5/3}$ This question is not for homework.
The book says:
$$4^{5/3} = (\sqrt4)^5 = (\sqrt{2^2})^5 = 2^5 = 32$$
when a bunch of us tried it:
$$4^{5/3} = \sqrt[3]{4^5} = \sqrt[3]{2^3 \cdot 2^2} = 2 \cdot \sqrt[3]{2^2} = 2 \cdot \sqrt[3]{4}$$
I get the sense we missed somethi... | There is probably a typo in your book, since
$$(\sqrt 4)^5 = 4^{5/2}\neq 4^{5/3}.$$
However, you still made a mistake in your calculation. You replaced $4^5$ with $2^3\cdot 2^2$, when in reality, $4^5 = 4^3\cdot 4^2 = 2^6 \cdot 2^4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/773916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$P=\left\{\theta:\sin \theta-\cos \theta = \sqrt{2}\cos \theta\right\}$ Let $P=\left\{\theta:\sin \theta-\cos \theta = \sqrt{2}\cos \theta\right\}$ and $Q=\left\{\theta:\sin \theta+\cos \theta=\sqrt{2}\sin \theta\right\}$ be two sets , Then
which one is Right.
$(a)\;\;\; P\subset Q$ and $Q-P\neq \phi\;\;\;\;\;\; (b)\;\... | Unfortunately, you cannot get rid of the $\pm$ sign. When you squared the set expression for $P$, the resulting set was not equivalent to $P$; in general, you cannot assume $\{x \mid f(x) = g(x)\} = \{x \mid f(x)^2 = g(x)^2\}$. Usually the second set will be larger.
Here is an approach that avoids squaring by dividin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/775256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A closed form for the infinite series $\sum_{n=1}^\infty (-1)^{n+1}\arctan \left( \frac 1 n \right)$ It is known that $$\sum_{n=1}^{\infty} \arctan \left(\frac{1}{n^{2}} \right) = \frac{\pi}{4}-\tan^{-1}\left(\frac{\tanh(\frac{\pi}{\sqrt{2}})}{\tan(\frac{\pi}{\sqrt{2}})}\right). $$
Can we also find a closed form for th... | In the same spirit as this answer, note that
$$
\log\left(\frac{n+i}n\right)=\frac12\log\left(1+\frac1{n^2}\right)+i\arctan\left(\frac1n\right)
$$
Furthermore, using Gautschi's Inequality
$$
\begin{align}
\prod_{k=1}^{n-1}\frac{k+x}{k}
&=\frac{\Gamma(n+x)}{\Gamma(1+x)\Gamma(n)}\\
&\sim\frac{n^x}{\Gamma(1+x)}
\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/776182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 6,
"answer_id": 2
} |
Elegant proof of $\int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2} =\frac {\pi} {b \sqrt{2b+a}}$? Let $a, b > 0$ satisfy $a^2-4b^2 \geq 0$. Then:
$$\int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2} =\frac {\pi} {b \sqrt{2b+a}}$$
One way to calculate this is by computing the residues at the poles in the upper h... | Note
\begin{align} \int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2}
&= 2\int_{0}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2} \overset{x\to\frac b x}=
2\int_{0}^{\infty} \frac{\frac{x^2}b\ dx}{x^4 + a x^2 + b ^2}\\
&= \int_{0}^{\infty} \frac{1+ \frac{x^2}b}{x^4 + a x^2 + b ^2}dx=\frac1b \int_{0}^{\infty} \frac{d(x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/776812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 4
} |
Roots of Polynomial Equation?
$y=1/x$ so I plugged in $x=1/y$ into the equation above and got $y^{4}+y^{3}+y^{2}/c+y/4-1/2$, but apparently it's wrong, when I looked up the answer below.
What am I missing?
| $$\begin{align}
x^4 + x^3 + cx^2 + 4x - 2 & = 0 \\
\left(\frac{1}{y}\right)^4 + \left(\frac{1}{y}\right)^3 + c\left(\frac{1}{y}\right)^2 + 4\left(\frac{1}{y}\right) - 2 & = 0 \\
\frac{1}{y^4} + \frac{1}{y^3} + c\frac{1}{y^2} + 4\frac{1}{y} - 2 & = 0 \\
\text{multiply by }y^4\text{ since y=0 isn't a solution} \\
1 + y +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/777699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Equation of tangent perpendicular to line I've got the homework question which I cannot solve.
Find the equation of the tangents to $4x^2+y^2=72$ that are
perpendicular to the line $2y+x+3=0$.
What I have done so far:
I have found the gradient of the line which is $m_1 = -\frac12$.
Which means that the equation per... | HINT :
You found $m_1=-\dfrac{1}{2}$ and $m_2=2$. This is correct. Then
$$
\begin{align}
4x^2+y^2&=72\\
y^2&=72-4x^2\\
y&=\sqrt{72-4x^2}\\
\frac{dy}{dx}&=\frac{d}{dx}(72-4x^2)^{\Large\frac12}\\
m_2&=\frac12(72-4x^2)^{-\Large\frac12}(-8x)\\
2&=-\frac{4x}{\sqrt{72-4x^2}}\\
\sqrt{72-4x^2}&=-2x\\
72-4x^2&=(-2x)^2\\
72-4x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/778779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding the power-series of $\frac{1}{(2-x)^2}$ I am going through some old Calculus-tasks in preparation for an upcoming exam, and a seemingly simple task is being stubborn with me. We are simply to find the power-series of the function $$\frac{1}{(2-x)^2}$$
Now, we have our basic power-series:
$$\frac{1}{1-x} = 1 + x... | Using Newton's generalised binomial theorem or the Binomial Series and assuming the convergence ,
$$\frac1{(2-x)^2} =\frac14\left(1-\frac x2\right)^{-2}$$
$$=\frac14\left(1+\frac x2\cdot2+\frac{-2(-2-1)}{2!}\left(\frac x2\right)^2+\cdots\right)=\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/779379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to calculate $\sum_{k=1}^n \left(k \sum_{i=0}^{k-1} {n \choose i}\right)$ How do I calculate the following summation?
$$\sum_{k=1}^n \left[k \sum_{i=0}^{k-1} {n \choose i}\right]$$
| Here is a slightly different take on this that emphasises the use of
formal power series.
Suppose we seek to evaluate
$$Q_n = \sum_{k=1}^n k \sum_{q=0}^{k-1} {n\choose q}.$$
We have $$\sum_{q=0}^n {n\choose q} z^q = (1+z)^n$$
and therefore
$$\sum_{q=0}^{k-1} {n\choose q}
= [z^{k-1}] \frac{1}{1-z} (1+z)^n$$
because mu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/781645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Prove that $|f(x)| \le \frac{3}{2}$ when $f(x)=ax^2+bx+c$ Suppose $f(x) = ax^2+bx+c$ where $x \in [-1,1]$. If $f(-1),f(0),f(1)\in [-1,1]$ show that $|f(x)| \le \frac{3}{2}$ $\forall x \in [-1,1]$.
This is how I tried:
$f(0)=c$
$f(1)=a+b+c$
$f(-1)=a-b+c$
Putting $f(0)=c$ we get $f(1)-f(0)=a+b$, $f(-1)-f(0)=a-b$. Sol... | Though this has been answered before, to answer the question, I am posting it here. The answer credit goes to Tom Collinge who answered my question.
The proof for 3/2 follows, though intuitively, I think that you can limit it to 5/4: it's a quadratic and if you draw an extreme graph then it should go through (-1, 1), (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/783410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Given perimeter of triangle and one side, find other two sides In triangle ABC, all three sides have integer lengths. If AB = 21, the perimeter is 54, and the area is a positive integer, what are the lengths of BC and AC?
I tried using Heron's Formula, but I couldn't really get anywhere. Any suggestions? This question ... | Just to show that Heron's formula isn't so bad ...
In
$$\text{area}^2 = \frac{1}{16}(a+b+c)(-a+b+c)(a-b+c)(a+b-c)$$
you know $c=21$ and $a+b+c=54$, so that $b=33-a$. So, the above becomes
$$\begin{align}
\text{area}^2 &= \frac{1}{16}(54)(54-2a)(2a-12)(12) \\[4pt]
&= 2\cdot 3^4 \cdot (27-a)(a-6)
\end{align}$$
Thus, $a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/785764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Infinite Sum of products What is the infinite sum
$$S = {1 + \frac{1}{3} + \frac{1\cdot 3}{3\cdot 6} + \frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}+ ....}?$$
I attempted messing around with the $n$ th term in the series but didnt see any solution.
How should I proceed?
| The $n+1$ th term is $$\frac{1\cdot 3\cdot 5\cdots(2n-1) }{3^{n}(n)!}=\frac{(2n)!}{6^{n}(n!)^2}$$ Hence the series is $$\lim_{x\to 1/6}\sum_{n\ge 0}\binom{2n}{n}x^n=\frac{1}{\sqrt{1-4/6}}=\sqrt{{3}}$$ This type of sum appears while calculating probability of return to origin in $1D$ random walks.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/786162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Maximium value of $(b-a)\Big(\dfrac 34-\dfrac{a+b}2-\dfrac{a^2+ab+b^2}3\Big)$ For $b>a$ what is the maximum possible value of $(b-a)\Big(\dfrac 34-\dfrac{a+b}2-\dfrac{a^2+ab+b^2}3\Big)$ ?
| By partial differentiation:
Assume $b$ fixed, differentiate with respect to $a$, we get the condition
$$ a^2 + a - \frac {3}{4} = 0 $$
Assume $a$ fixed, differentiate with respect to $b$, we get the condition
$$ b^2 -b - \frac {3}{4} = 0 $$
This gives us $(a, b) = ( - \frac{3}{2}, \frac{1}{2} ) $. It remains to verify ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/789062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Let $a=\dfrac{3+\sqrt{5}}{2}.$ Show that $\lfloor a \lfloor an \rfloor \rfloor+n$ is divisible by $3$.
Let $a=\dfrac{3+\sqrt{5}}{2}.$
Show that for all $n\in\mathbb N$, $\lfloor a \lfloor an \rfloor \rfloor+n$ is divisible by $3$.
My teacher solve this problem with induction, I am just curious if we can do this exerc... | It is in fact the case that,
for $a = \frac{3 + \sqrt{5}}{2}$,
$$
\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}
\boldsymbol{\floor{a \floor{ a n}} + n = 3 \floor{a n}.}
$$
Proof. Let $r = a n - \floor{a n}$, $0 < r < 1$.
Then $\floor{a n} = (a n - r)$, so
we need to show that
\begin{align*}
3(a n - r) - n &\le ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/789343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Find the value of this combinatorial sum. $\sum_{k=4}^{100}\binom{k-1}{3}$ How to compute this sum without laboring?
$$\sum_{k=4}^{100}\dbinom{k-1}{3}.$$
Is it possible to reduce this to a single combinatorial term?
| $$\sum_{k=4}^{100}\dbinom{k-1}{3} = \sum_{k=4}^{100} \frac{(k-1)(k-2)(k-3)}{3!} = \sum_{k=4}^{100}\frac{k^3 - 6k^2 + 11k - 6}{6} \\ = \frac{1}{6} \sum_{k=4}^{100} \left(k^3 - 6k^2 + 11 k - 6\right) = \frac{1}{6} \sum_{k=1}^{100} \left(k^3 - 6k^2 + 11 k - 6\right)- \frac 1 6 \sum_{k=1}^{3} \left(k^3 - 6k^2 + 11 k - 6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/790023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
simplify $\sqrt[3]{11+\sqrt{57}}$ I read in a book (A Synopsis of Elementary Results in Pure and Applied Mathematics) that the condition to simplify the expression $\sqrt[3]{a+\sqrt{b}}$ is that $a^2-b$ must be a perfect cube.
For example $\sqrt[3]{10+6\sqrt{3}}$ where $a^2-b
=(10)^2-(6 \sqrt{3})^2=100-108=-8$ and $\sq... | I imagine you want to write, with integer $u$ and $v$,
$$11+\sqrt{57}=(u+\sqrt{v})^3$$
So that the cube root would simplify.
Hence
$$u^3+3u^2\sqrt{v}+3uv+v\sqrt{v}=11+\sqrt{57}$$
So you would need
$$\begin{eqnarray}
u(u^2+3v)&=&11\\
v(3u^2+v)^2&=&57
\end{eqnarray}$$
But $57=3 \cdot 19$ has no square factor, so $3u^2+v=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/790738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
If $\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\frac{3}{2}\sqrt{\frac{x}{x+\sqrt{x}}}$, how can it imply $x\ge1$? I encountered this problem in a book. Solve the equation $\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\frac{3}{2}\sqrt{\frac{x}{x+\sqrt{x}}}$.
According to the book, $x\ge1$ from observation. And after some algebra, they ... | By itself, the expression on the left is also defined at $x=0$, but then the expression on the right is not.
If $0\lt x\lt 1$, then $\sqrt{x}\gt x$, so $x-\sqrt{x}$ is negative. Presumably we are working in the reals, so there is no such thing as $\sqrt{x-\sqrt{x}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/791013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Limit of a sequence defined by a sum: $\lim_{n\to \infty} \frac1{2^n}\sum_{k=1}^n \frac1{\sqrt k}\binom nk$ $\lim_{n\to \infty} \frac{1}{2^n}\sum_{k=1}^n \frac{1}{\sqrt k}\binom{n}{k}$
Could it have a connection to Euler summability?
| $$
\frac{1}{2^n}\sum_{k=1}^n \frac{1}{\sqrt k}\binom{n}{k} = \frac{1}{2^n}\sum_{k=1}^{\log n-1} \frac{1}{\sqrt k}\binom{n}{k} + \frac{1}{2^n}\sum_{k=\log n}^{n} \frac{1}{\sqrt k}\binom{n}{k}
$$
The second term is upper bounded by
$$
\frac{1}{\sqrt{\log n}}\cdot\frac{1}{2^n}\sum_{k=\log n}^{n} \binom{n}{k} \leq \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/793093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Dirichlet series Suppose that the series $\sum \limits_{n=1}^{\infty}\dfrac{a_n}{n^{\sigma}} \quad(\sigma>0)$ converges. Prove that $\lim \limits_{n\to \infty}\dfrac{a_1+a_2+\dots+a_n}{n^{\sigma}}=0$.
I applied the Abel transformation, but unsuccessfully.
| For each $n$, let $R_n = \sum\limits_{k=n}^\infty \frac{a_k}{k^\sigma}$. Then $a_k = k^\sigma(R_k-R_{k+1})$. Since by assumption the series converges, we have $R_n \to 0$, thus $S_n = \sup\limits_{k\geqslant n} \lvert R_k\rvert$ is finite for all $n$ and $S_n \to 0$. Then we have
$$\begin{align}
\left\lvert \frac{a_1 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/793540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Arc Length polar curve $$r=a\sin^3\left(\frac{\theta}{3}\right) $$
I tried solving it using the equation for arc length with $dr/d\theta$ and $r^2$. Comes out messy and complicated.
| This solution borders on trivial when solved in the complex plane. First, the range of $\theta$ must be determined. It was pointed out in an earlier that $\theta\in[0,3\pi]$. I found the same thing by plotting the equation. Now, the equation for the arc length in the complex plane is given by
$$s=\int|\dot z|du$$
Thus,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/799708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show how to compute $2^{343}$ using the least multiplication. Show how to compute $2^{343}$ using the least multiplication.
| One way to do this (and probably the way the prof wants) is exponentiation by squaring. Note that $x^{2n} = (x^n)^2$ and $x^{2n + 1} = (x^n)^2 \cdot x$. We can either break down $343$ into smaller pieces this way, but it's easier to think about starting at the bottom:
$$
\begin{array}{ccc}
Value & Computation & Multipl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/802057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Evaluate the integral of primitive $\frac{1}{x(x+2)}$ I am doing this integral by example:
$$\int_1^\infty\frac{1}{x(x+2)}\ dx.$$
The example in the book, starts with $\dfrac{1}{x(x+2)} \leq \dfrac{1}{x^2}.$
Why is this important? Why does the example in the book start like that?
Thanks.
| If $\dfrac{1}{x(x+2)}\leq\dfrac{1}{x^2}$, then
$$\int^\infty_1\dfrac{1}{x(x+2)}dx\leq\int^\infty_1\dfrac{1}{x^2}dx=\left.-\dfrac{1}{x}\right|^\infty_1=1$$
i.e. $\displaystyle\int^\infty_1\dfrac{1}{x(x+2)}dx\leq1$. This shows convergence of the integral.
Now, split $\dfrac{1}{x(x+2)}$ as a partial fraction:
$$\dfrac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/802827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integral $I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}$ Hi how can we prove this integral below?
$$
I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}
$$
I tried to use
$$
I=\int_0^1 \frac{\log^2x}{1-x(1-x)}\mathrm dx
$$
and now tried changing variables to $y=x(1-x)$ ... | Real Part
Substituting $x\mapsto1/x$ says
$$
\int_0^1\frac{\log(x)^2}{x^2-x+1}\mathrm{d}x
=\int_1^\infty\frac{\log(x)^2}{x^2-x+1}\mathrm{d}x\tag{1}
$$
Therefore,
$$
\int_0^1\frac{\log(x)^2}{x^2-x+1}\mathrm{d}x
=\frac12\int_0^\infty\frac{\log(x)^2}{x^2-x+1}\mathrm{d}x\tag{2}
$$
Contour Integration Part
Putting the bran... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/803389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 8,
"answer_id": 0
} |
Surely You're Joking, Mr. Feynman! $\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}\,dx$
Prove the following
\begin{equation}\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}\,dx=\frac{\pi}{4}+\frac{\pi}{4e^2}\end{equation}
I would love to see how Mathematics SE users prove the integral preferably with the Feynman way (other methods a... | Here is my attempt:
\begin{align}
\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}dx&=\int_0^\infty\left[\frac{\sin^2x}{x^2}-\frac{\sin^2x}{1+x^2}\right]dx\\
&=\int_0^\infty\frac{\sin^2x}{x^2}dx-\frac{1}{2}\int_0^\infty\frac{1-\cos2x}{1+x^2}dx\\
&=\frac{\pi}{2}-\frac{1}{2}\int_0^\infty\frac{1}{1+x^2}dx+\frac{1}{2}\int_0^\infty\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/803954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
"answer_count": 3,
"answer_id": 0
} |
Solving the Integral using $\ln|u|$ Can Someone help me solve this
$$
\int\frac{19\tan^{-1}x}{x^{2}}\,dx
$$
We have been told to use $\ln|u|$ and $C$.
Thanks!
| Let $y=\arctan x\;\Rightarrow\;\tan y=x\;\Rightarrow\;\sec^2y\ dy=dx$, then the integral turns out to be
\begin{align}
\int\frac{\arctan x}{x^2}\ dx&=\int\frac{y}{\tan^2y}\cdot\sec^2y\ dy\\
&=\int\frac{y}{\sin^2y}\ dy.
\end{align}
The last integral can be solved by using IBP. Taking $u=y$ and $dv=\dfrac1{\sin^2x}\ dx\;... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/813209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find sum of $\frac{1}{\sin\theta\cdot \sin2\theta} + \frac{1}{\sin2\theta\cdot \sin3\theta} + \cdots + \frac{1}{\sin n \theta \sin (n+1)\theta}$ $$\sum_{k=1}^n \frac{1}{\sin k\theta \sin (k+1)\theta} = \dfrac{1}{\sin\theta\cdot \sin2\theta} + \dfrac{1}{\sin2\theta\cdot \sin3\theta} + \cdots + \frac{1}{\sin n \theta \si... | Note
$$\dfrac{\sin{\theta}}{\sin{(k+1)\theta}\sin{(k\theta)}}=\dfrac{\sin{((k+1)\theta-k\theta)}}{\sin{(k+1)\theta}\sin{(k\theta)}}=\cot{(k\theta)}-\cot{(k+1)\theta}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/813413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How do I find the sum of the infinite geometric series? $$2/3-2/9+2/27-2/81+\cdots$$
The formula is $$\mathrm{sum}= \frac{A_g}{1-r}\,.$$
To find the ratio, I did the following:
$$r=\frac29\Big/\frac23$$
Then got:
$$\frac29 \cdot \frac32= \frac13=r$$
and $$A_g= \frac23$$
Then I plug it all in and get:
$$\begin{align*}
\... | Hint: the terms are alternating in sign. $\displaystyle r = \frac{\frac{2}{9}}{-\frac{2}{3}} = -\frac{1}{3}$.
Note the minus sign.
Hence the sum is $\displaystyle \frac{\frac{2}{3}}{1 - (-\frac{1}{3})} = \frac{\frac{2}{3}}{\frac{4}{3}} = \frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/816306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Why does $\frac{(x^2 + x-6)}{x-2} = \frac{(x+3)(x-2)}{x-2}$? I'm not the best at algebra and would be grateful if someone could explain how you can get from,
$$\frac{x^2 + x-6}{x-2}$$
to,
$$\frac{(x+3)(x-2)}{x-2}$$
| All that was done was factoring the numerator. Notice that
$$
x^2+x-6=(x+3)(x-2)
$$
because $(x+3)(x-2)=x^2-2x+3x-6=x^2+x-6$. Then it follows that
$$
\frac{x^2+x+6}{x-2}=\frac{(x+3)(x-2)}{x-2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/817424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
How to find the following sum? $\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} $ I want to calculate the sum with complex analysis (residue)
$$
1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \frac{1}{{17}} - ...
$$ $$
1 + \sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} -... | (Posted as an answer in case my earlier comment is removed)
It is relatively easy to prove (either through elementary means or via complex analysis) the well-known identity $$1 + 2 \sum_{n=1}^\infty \frac{z^2}{z^2 - (n\pi)^2} = z \cot z.$$ Then the given series (not the one written in summation notation, but the one t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/817911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
limits of function without using L'Hopital's Rule $\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 - \ln x}}{{x\ln x+ 1 - x}} = 1$ Good morning.
I want to show that without L'Hopital's rule :
$\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 - \ln x}}{{x\ln x + 1 - x}} = 1$
I did the steps
$
\begin{array}{l}
\mathop {\lim... | Let $y= \dfrac{x-1-\ln x}{x\ln x - 1+ x} \to yx\ln x - y + xy = x - 1 - \ln x \to (xy + 1)\ln x = (y-1)(1-x) \to (xy+1)\cdot \dfrac{\ln x}{x-1} = 1 - y$.
Now using a well-known fact that: $\dfrac{\ln x}{x-1} \to 1$ when $x \to 1$. Taking limit as $x \to 1$ both sides of the above equation we have:
$y + 1 = 1 - y \to y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/818908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Factoring in the derivative of a rational function Given that
$$
f(x) = \frac{x}{1+x^2}
$$
I have to find
$$\frac{f(x) - f(a)}{x-a}$$
So some progressing shows that:
$$
\frac{\left(\frac{x}{1+x^2}\right) - \left(\frac{a}{1+a^2}\right)}{x-a} =
\frac{(x)(1+a^2)-(a)(1+x^2)}{(1+x^2)(1+a^2)}\cdot\frac{1}{x-a} =
\frac{x+xa^2... | Hint: you can guess that something interesting is going to happen near $x = a$, which suggests looking to factor $x-a$.
Indeed, $x + xa^2 - a - ax^2 = (x-a) + xa(x-a)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/819527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
A difficult integral evaluation problem How do I compute the integration for $a>0$,
$$
\int_0^\pi \frac{x\sin x}{1-2a\cos x+a^2}dx?
$$
I want to find a complex function and integrate by the residue theorem.
| I'll first consider the case $0<a<1$.
Let $\displaystyle f(z) = \frac{z}{a-e^{-iz}}$ and integrate around a rectangle with vertices at $\pm \pi$ and $\pm \pi + iR$.
The function $f(z)$ has poles where $a-e^{-iz}=e^{\ln a + 2 \pi i n}-e^{-iz}= 0$.
That is, when $z= i \ln a - 2 \pi n$.
If $ 0<a< 1$, all of those points... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/822484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Solving a simple matrix polynomial Does there exist a $2\times 2$ Matrix $A$ such that
$A-A^2=\begin{bmatrix} 3 & 1\\1 & 4\end{bmatrix}$ ?
| Let $A = \begin{bmatrix} a & b\\ c & d\end{bmatrix}$, then $A^2 = \begin{bmatrix} a & b\\ c & d\end{bmatrix}\cdot \begin{bmatrix} a & b\\ c & d\end{bmatrix} = \begin{bmatrix} a^2 + bc & ab + bd\\ ac + cd & bc + d^2\end{bmatrix}$. Thus:
$A - A^2 = \begin{bmatrix} a-a^2-bc & b-ab-bd\\ c-ac-cd & d-bc-d^2\end{bmatrix} = \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/827375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
On combining $n$ and $n^2$ into one number Consider the sequence $T_n$ formed by combining $n$ and $n^2$ into one number.
ie. (A053061)
$$T_n=\{11,24,39,416,525,636,749 \cdots\}$$
It is easy to see
$$T_n= 10^{\lceil 2 \log_{10}(n) \rceil } n+ n^2$$
I looked at the sequence closely trying to find if there are any perfec... | There are no squares in this sequence.
Elements of this sequence are of the form $b^2+10^n b$ for $10^{n-1} \leq b^2 < 10^n$. So we are looking for a special solution to the equation $a^2 =b^2+ 10^n b$. This is an easy Diophantine equation to solve: We have $a^2 - (b+10^n/2)^2 = - 10^{2n}/4$, or
$(a+b+10^n/2)(a-b-10^n/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/828086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that a given ideal is not maximal in $\mathbb C[x,y,z]$ I'm trying to prove this ideal:
$$(x^2+y^2+z^2+x+y+z,\ x^5+y^5+z^5+2(x+y+z),\ x^7+y^7+z^7+3(x+y+z))\subset
\mathbb C[x,y,z]$$
can't be maximal.
In order to do so, I'm using the Nullstellensatz theorem and showing this ideal is not of this form: $(x-a_1,y-... | To show your ideal is not maximal, I'll show that the set of common zeros of your ideal contains more than one point. Taking advantage of the symmetry of your ideal, we will look for solutions that even satisfy the extra condition $x+y+z=0.$ Then (using Newton's Identities and remember $e_1=x+y+z=0$) the first equation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/829820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Massive expression mod $2^{2013}$ Find $$1^1+3^3+5^5 + \cdots +(2^{2012}-1)^{(2^{2012}-1)}$$ modulo $2^{2013}$. By checking small cases I am pretty sure that it is $2^{2012}$. I tried applying induction but you "lose" powers of two at each step. So instead I used binomial theorem to reduce it to $$\sum_{i=1,i \text{ o... | I think induction does work to show that for $k\ge2$,
$$
\sum_{i=1}^{2^{k-1}} (2i-1)^{2i-1} \equiv 2^k \pmod{2^{k+1}}.
$$
One fact we'll use is that for $n\ge3$, we actually have $a^{2^{n-2}}\equiv1\pmod {2^n}$ for any odd $a$; this is one power of $2$ stronger than Euler's theorem.
Checking the $k=2$ case is easy. Sup... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/833503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Stronger version of AM-GM with condition If $a,b,c>0$ and $a^2+b^2+c^2=3$, prove that
$$a+b+c\ge 3(abc)^{13/76}.$$
I was thinking of using AM-GM :P but clearly some clever trick is needed. I was thinking about expanding but that is not feasible too. Note I would like a non-computational proof which can be done by hand.... | First we check that $abc \leq 1$, it's easy by AM-GM:
$1=\frac{a^2+b^2+c^2}{3} \leq (abc)^{\frac{2}{3}}$
Next we prove that $(a+b+c)^2 \geq 9(abc)^{\frac{13*2}{76}}$. If we prove this the inequality $a+b+c \geq 3(abc)^{\frac{13}{76}}$ will be proven, because $x \leq y$ implies $\sqrt{x} \leq \sqrt{y}$ for $x,y \geq 0$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/834740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proving $n^2(n^2+16)$ is divisible by 720 Given that $n+1$ and $n-1$ are prime, we need to show that $n^2(n^2+16)$ is divisible by 720 for $n>6$.
My attempt:
We know that neither $n-1$ nor $n+1$ is divisible by $2$ or by $3$, therefore $n$ must be divisible by both $2$ and $3$ which means it must be divisible by $6$.
... | $n=6k$ implies $n\equiv k \bmod 5$. If $k\equiv 0 \bmod 5$, then you're done.
Othwerwise, $n-1$ and $n+1$ prime implies $k\equiv n\equiv \pm 2 \bmod 5$ and so $9k^2+4\equiv -k^2+4 \equiv 0 \bmod 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/836482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Evaluating the limit of $\sqrt[2]{2+\sqrt[3]{2+\sqrt[4]{2+\cdots+\sqrt[n]{2}}}}$ when $n\to\infty$ The following nested radical $$\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$$ is known to converge to $2$.
We can consider a similar nested radical where the degree of the radicals increases:
$$\sqrt[2]{2+\sqrt[3]{2+\sqrt[4]{2+\cd... | One of the well known formula by Indian mathematician Ramanujan is this one:
$$f(x)=x+1=\sqrt { \left( x+1 \right) ^{2}}=\sqrt {1+{x}^{2}+2\,x}=\sqrt {1+x\sqrt {(x+2)^2}}$$
$$=\sqrt {1+x\sqrt {1+{x}^{2}+4\,x+3}}=\sqrt {1+x\sqrt {1+ \left( x+1 \right) \sqrt {(x+3)^2}}}$$
$$
=\sqrt {1+x\sqrt {1+ \left( x+1 \right) \sqrt ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/837189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
"answer_count": 4,
"answer_id": 3
} |
Find a closed expression for a formula including summation Let:
$$\sum\limits_{k = 0}^n {k\left( {\matrix{
n \cr
k \cr
} } \right)} \cdot {4^{k - 1}} \cdot {3^{n - k}}$$
Find a closed formula (without summation). I think I should define this as a "series" which generated by $F(x)$. I don't really have a lea... | The binomial expansion is given by
\begin{align}
(1+x)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^{k}.
\end{align}
Differentiating both sides with respect to $x$ yields
\begin{align}
n (1+x)^{n-1} = \sum_{k=0}^{n} k \ \binom{n}{k} x^{k-1}.
\end{align}
Multiplying this last expression by $3^{n} 4^{-1}$ leads to
\begin{align}
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/837594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
A formula for $\pi$ and an inequality For any $n\in \mathbb{N}$ prove the identity :
$$\pi =\sum_{k=1}^{n}\frac{2^{k+1}}{k\dbinom{2k}{k}}+\frac{4^{n+1}}{\dbinom{2n}{n}}\int_{1}^{\infty}\frac{\mathrm{d}x}{(1+x^2)^{n+1}}\tag{1}$$
and thus derive the inequality :
$$\frac{2}{2^n\sqrt{n}}<\pi -\sum_{k=1}^{n}\frac{2^{k+1}}{k... | Let
$$I_n = \int_1^\infty \frac{dx}{(1+x^2)^{n+1}}.$$
Then $I_0 = \frac{\pi}{4}$ is well-known, and $(1)$ holds for $n = 0$. Integration by parts yields
$$\begin{align}
I_n &= \int_1^\infty \frac{dx}{(1+x^2)^{n+1}}\\
&= \left[\frac{x}{(1+x^2)^{n+1}}\right]_1^\infty + 2(n+1) \int_1^\infty \frac{x^2\,dx}{(1+x^2)^{n+2}}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/840658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Probability Puzzle: Mutating Loaded Die Take an (initially) fair six-sided die (i.e. $P(x)=\frac{1}{6}$ for $x=1,…,6$) and roll it repeatedly.
After each roll, the die becomes loaded for the next roll depending on the number $y$ that was just rolled according to the following system:
$$P(y)=\frac{1}{y}$$
$$P(x)=\frac{1... | The transition matrix is given by $$\mathcal P = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ \tfrac{1}{10} & \tfrac{1}{2} & \tfrac{1}{10} & \tfrac{1}{10} & \tfrac{1}{10} & \tfrac{1}{10} \\ \tfrac{2}{15} & \tfrac{2}{15} & \tfrac{1}{3} & \tfrac{2}{15} & \tfrac{2}{15} & \tfrac{2}{15} \\ \tfrac{3}{20} & \tfrac{3}{20} & \tf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/841114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
$ 7\mid x \text{ and } 7\mid y \Longleftrightarrow 7\mid x^2+y^2 $
Show that
$$ 7\mid x \text{ and } 7\mid y \Longleftrightarrow 7\mid x^2+y^2 $$
Indeed,
First let's show
$7\mid x \text{ and } 7\mid y \Longrightarrow 7\mid x^2+y^2 $
we've $7\mid x \implies 7\mid x^2$ the same for $7\mid y \implies 7\mid ... | Let $x,y \in \mathbb{F}_p$ be with $x^2+y^2=0$. If $x=0$, then $y=0$. Now assume $x \neq 0$. Let $z:=y/x$, then $z^2=-1$. If $p=2$, this means $z = 1$. If $p > 2$, this means that $z$ has order $4$ in $\mathbb{F}_p^*$, which happens iff $4|p-1$ i.e. $p \equiv 1 \bmod 4$. Hence, for every odd prime $p$ with $p \not\equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/842406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Find an unknown coefficient in a line equation... So, I have to find the unknown coefficient in this line: $$x+y+C=0$$ so that it is a tangent to this circle: $$x^2+y^2-5x-7y+6=0$$
I've transformed the circle equation to this form:
$$(x-\frac{5}{2})^2+(y-\frac{7}{2})^2=\frac{25}{2}$$
But I can't see how it can help me ... | Substitute $y = C-x$ into the equation of the circle to get:
$x^2+(C-x)^2-5x-7(C-x)+6 = 0$
$2x^2+(2-2C)x+(C^2-7C+6) = 0$
If the line is tangent to the circle, this equation has only one solution for $x$. So the discriminant of the quadratic should be $0$, i.e.
$(2-2C)^2 - 4 \cdot 2 \cdot (C^2-7C+6) = 0$
This will giv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/843050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What does this log notation mean? Can someone please explain what $^2\log x$ means? Is it the same as saying $\log x^2$ or is it something completely different? Here is an image of it as an example:
| If $\log_2 \dfrac{x - y}{3} = 0$, then $\dfrac{x - y}{3} = 1 \Rightarrow x - y = 3$.
Now, $\log_4 y = \dfrac{\log_2 y}{\log_2 4} = \dfrac{1}{2}\log_2{y}$.
Then $\log_2 x + 2 \log_4 y = \log_2 x + \log_2 y = \log_2 xy$.
Thus, $\log_2 x + 2 \log_4 y = 2 \Rightarrow \log_2 xy = 2 \Rightarrow xy = 2^2 = 4$.
Now we have, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/844201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Double Integral I need help calculating these double integrals (in order to show they are not equal):
$$\int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} dy\,dx \ne \int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} dx\,dy$$
| Consider the double integral over the rectangle $[0,1]\times[0,1]$
$$\iint\limits_{[0,1]\times[0,1]} \! \frac{x^2-y^2}{(x^2+y^2)^2} \, \mathrm{d}y \, \mathrm{d}x.$$
Partial fraction decomposition yields
$$=\iint\limits_{[0,1]\times[0,1]} \! \frac{2x^2}{(x^2+y^2)^2} - \frac{1}{x^2+y^2} \, \mathrm{d}y \, \mathrm{d}x,$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/846278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Meaningful lower-bound of $\sqrt{a^2+b}-a$ when $a \gg b > 0$. I know that, for $|x|\leq 1$, $e^x$ can be bounded as follows:
\begin{equation*}
1+x \leq e^x \leq 1+x+x^2
\end{equation*}
Likewise, I want some meaningful lower-bound of $\sqrt{a^2+b}-a$ when $a \gg b > 0$.
The first thing that comes to my mind is $\sqrt{a... | Factor out an $a^2$ from the radical to get $a\sqrt{1+\frac{b}{a^2}}-a=a\left(\sqrt{1+\frac{b}{a^2}}-1\right)$
Which can then be expanded for $\left|\frac{b}{a^2}\right|<1$, which is true for $a \gg b > 0$.
This expansion, to first order, is $a\left(1+\frac{b}{2a^2}-1\right)=\frac{b}{2a}$.
EDIT: Forgot that you were lo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/847153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Simplifying polynomial fraction Working through an old book I got and am at this problem: Simplify:
$$\frac{3x^2 + 3x -6}{2x^2 + 6x + 4}.$$
The answer is supposed to be $\frac{3(x - 1)}{2(x - 1)}$. I thought I had all this polynomial stuff figured out well enough, but I'm having trouble seeing how they got to that answ... | $$\frac{3x^2 + 3x -6}{2x^2 + 6x + 4}=\frac{3(x^2 + x -2)}{2(x^2 + 3x + 2)}=$$
$$=\frac{3(x^2 -1+ x -1)}{2(x^2 + 2x+x + 2)}=\frac{3((x-1)(x+1)+(x-1))}{2(x(x+2)+(x+ 2)}=$$
$$=\frac{3(x-1)(x+1+1)}{2(x+2)(x+ 1)}=\frac{3(x-1)(x+2)}{2(x+2)(x+1)}=\frac{3(x-1)}{2(x+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/850148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Powers and differences of positive integers Assume that $a$, $b$, $c$, and $d$ are positive integers such that $a^5=b^4$, $c^3=d^2$, and $c-a=19$. Determine $d-b$.
I know this question isn't particularly hard, but I've been having trouble getting any substantial ground. I've tried substituting that $c=a+19$ and $a=c-19... | From $a^5 = b^4 \implies a = \left(\frac{b}{a}\right)^4$, we can deduce that $\frac{b}{a}$ is an integer. I will let $m = \frac{b}{a}$.
Similarly, from $c^3 = d^2 \implies c = \left(\frac{d}{c}\right)^2$, we can deduce that $\frac{d}{c}$ is an integer. I will let $n = \frac{d}{c}$.
Since $a = m^4$ and $c = n^2$, we ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/850450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
If $f(x) = \frac{x}{x+1}$ and $g(x) = 2x-1$, find $(g\circ f) (x)$. If $f(x) = \frac{x}{x+1}$ and $g(x) = 2x-1$, find $(g\circ f)(x)$.
My answer is $\frac{x-1}{x+1}$. However, the answer key in the book states $\frac{2x}{x+1}$. How is that? Is the book wrong?
| Wolframalpha and me agree with you, if
$$f(x)=\frac{x}{x+1}, g(x)=2x-1 $$ then
$$g \circ f(x)=g(f(x))= 2f(x)-1 = \frac{2x}{x+1}-1= \frac{2x-x-1}{x+1}= \frac{x-1}{x+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/851504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $ \lim_{x\rightarrow \infty}\left\{2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right\}$
Evaluate the limit
$$
\lim_{x\rightarrow \infty}\left(2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right)
$$
My Attempt:
To simplify notation, let $A = \left(\sqrt[3]{x^3+x^2+1}\right)$ and $B ... | It is easier to break into two limits, $x-\sqrt[3]{x^3+x^2+1}$ and $x-\sqrt[3]{x^3-x^2+1}$ then we shall see the first limits to $-\frac{1}{3}$ and the second $\frac{1}{3}$ so the sum limits to zero.
Note the more general result,
$$x-\sqrt[n]{x^n +ax^{n-1}+\cdots}\rightarrow -\frac{a}{n}$$
If we write $A=\sqrt[n]{x^n +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/851849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
Proving an expression is composite I am trying to prove that $ n^4 + 4^n $ is composite if $n$ is an integer greater than 1. This is trivial for even $n$ since the expression will be even if $n$ is even.
This problem is given in a section where induction is introduced, but I am not quite sure how induction could be us... | I am very impressive with Adam's solution. There is very neat. So, I beg for a chance to write the full description about the proof step-by-step.
*
*We can transform $n^4+4^n$ to $(n^2+2^n)^2-2^{n+1}n^2$ as Adam's suggestion by
*
*$n^{(2^2)}+(2^2)^n = (n^2)^2+(2^n)^2$ associative law
*Now, we mention $(a+b)^2 = (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/853615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Find the sum of the series $\sum \limits_{n=3}^{\infty} \dfrac{1}{n^5-5n^3+4n}$ Feel free to skip obvious steps, or use a calculator when required.
I just want to understand the theme of the solution.
Any help is appreciated
EDIT :
We can write$$ \dfrac{1}{n^5-5n^3+4n} = -\dfrac{1}{6 (n-1)}+\dfrac{1}{4 n}-\dfrac{1}{ ... | $$\sum_{n=3}^{\infty}\dfrac{1}{n^5-5n^3+4n} = \sum_{n=3}^{\infty}\dfrac{1}{(n-2)(n-1)n(n+1)(n+2)}$$
$ $
Rewrite this as $$ \dfrac{1}{4}\sum_{n=3}^{\infty}\dfrac{(n+2)-(n-2)}{(n-2)(n-1)n(n+1)(n+2)}$$
$ $
$$ \dfrac{1}{4}\sum_{n=3}^{\infty}\left[\dfrac{(n+2)}{(n-2)(n-1)n(n+1)(n+2)}-\dfrac{(n-2)}{(n-2)(n-1)n(n+1)(n+2)} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/854205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
what is a smart way to find $\int \frac{\arctan\left(x\right)}{x^{2}}\,{\rm d}x$ I tried integration by parts, which gets very lengthy due to partial fractions.
Is there an alternative
| It is not so lenghty. If $$I =\int \frac{\tan^{-1}x}{x^2}\,dx$$ let start with $u=\tan^{-1}x$ and $v'=\frac{\tan^{-1}x}{x^2}\,dx$. So $u'=\frac{\text{dx}}{x^2+1}$ and $v=-\frac{1}{x}$ and then $$I =\int \frac{\tan^{-1}x}{x^2}\,dx=-\frac{\tan ^{-1}(x)}{x}+\int \frac{dx}{x \left(x^2+1\right)}$$ Now, using partial fractio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/854997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Is this procedure for $5^{300} \bmod 11$ correct? I'm new to modular exponentiation. Is this procecdure correct?
$$5^{300} \bmod 11$$
$$5^{1} \bmod 11 = 5\\
5^{2} \bmod 11 = 3\\
5^{4} \bmod 11 = 3^2 \bmod 11 = 9\\
5^{8} \bmod 11 = 9^2\bmod 11 = 4\\
5^{16} \bmod 11 = 4^2 \bmod 11 = 5\\
5^{32} \bmod 11 = 5^2 \bmod 11 = ... | I see what you were trying to do with adding the numbers at the end.
Multiply them instead, and take that product $\mod 11$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/855657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
If $K = \frac{2}{1}\times \frac{4}{3}\times \cdots \times \frac{100}{99}.$ Then value of $\lfloor K \rfloor$
Let $K = \frac{2}{1}\times \frac{4}{3}\times \frac{6}{5}\times \frac{8}{7}\times \cdots \times \frac{100}{99}.$ Then what is the value of $\lfloor K \rfloor$, where $\lfloor x \rfloor$ is the floor function?
M... | Note that this answer is not completely rigorous, but it was too fun to pass up.
$$K^2 = 101 \cdot \frac{2 \cdot 2 \cdot 4 \cdot 4 \cdot \dots \cdot 100 \cdot 100}{1 \cdot 3 \cdot 3 \cdot 5 \cdot \dots \cdot 99 \cdot 101}$$
Now, $$\frac{2 \cdot 2 \cdot 4 \cdot 4 \cdot \dots}{1 \cdot 3 \cdot 3 \cdot 5 \cdot \dots} = \p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/855990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
} |
Proving $ \frac{x^3y^2}{x^4+y^4}$ is continuous. The problem asks to show that $$f(x,y) = \left\{ \begin{align} \frac{x^3y^2}{x^4+y^4}, & (x,y) \neq (0,0), \\ 0, & (x,y) = (0,0), \end{align} \right.$$ is continuous at the origin, however it has resisted my bravest efforts. I have attempted using $x^4+y^4 \geq y^4$ and ... | To apply the squeeze theorem, notice that if $x\neq 0$ and $y\neq 0$ then
$$\begin{align*}
\left|\frac{x^3y^2}{x^4+y^4}\right|&=\left|x\frac{x^2}{\sqrt{x^4+y^4}}\frac{y^2}{\sqrt{x^4+y^4}}\right|\\
&=\left|x\frac{x^{-2}}{x^{-2}}\frac{x^2}{\sqrt{x^4+y^4}}\frac{y^{-2}}{y^{-2}}\frac{y^2}{\sqrt{x^4+y^4}}\right|\\
& =\left|x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/856098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve $a$ and $b$ for centre of mass in $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ Given ellipse:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
What length do $a$ and $b$ have to be so the centre of mass is $S(4;2)$?
I've tried steps to solve the equation to $$y=b\sqrt{1-\frac{x^2}{a^2}}$$ and integrate
$$A=b\int_0^a{\sqrt{1-\frac{x... | Combining the results in the previous discussion and answers, we set $x=a t$, then $y=b\sqrt{1-t^2}$ and $dx=adt$. So:
$$\bar{x}=\frac{\int_0^a xy(x)\,\mathrm{d}x}{\int_0^a y(x)\,\mathrm{d}x}=\frac{a^2b}{ab}\frac{\int_0^1 t \sqrt{1-t^2}\,\mathrm{d}t}{\int_0^1 \sqrt{1-t^2}\,\mathrm{d}t}=a\frac{1/3}{\pi/4}=\frac{4a}{3\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/858213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Chinese Remainder Theorem $\ x\equiv m_k-2\pmod{m_k}$ I am trying to find all integers that give remainders 1,2,3 when divided by 3,4,5 respectively. So I start defining
$$a_1=1, a_2=2, a_3=3,$$$$ m_1=3, m_2=4, m_3=5,$$$$ m_1m_2=12, m_1m_3=15, m_2m_3=20,$$$$ m=60$$
All the moduli are pairwise coprime, so by the Eucli... | Another, more direct way of solving this runs as follows. You are looking at
$$x \equiv 1 \text{ mod } 3$$
$$x \equiv 2 \text{ mod } 4$$
$$x \equiv 3 \text{ mod } 5.$$
Now put $x=1+3k$ and mod this by $4$. Then $2\equiv 1 +3k \text{ mod } 4$, so $k\equiv -1 \text{ mod } 4$, say $k=4l-1$, hence $x=-2+12l$. Now take the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/860125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How find this P(x) if $ (x^3 - mx^2 +1 ) P(x+1) + (x^3+mx^2+1) P(x-1) =2(x^3 - mx +1 ) P(x) $ Let $m \neq 0 $ be an integer. Find all polynomials $P(x) $ with real coefficients such that
$$ (x^3 - mx^2 +1 ) P(x+1) + (x^3+mx^2+1) P(x-1) =2(x^3 - mx +1 ) P(x) $$
This problem is IMO Shortlist 2013
let $$P(x)=\sum_{i=0}... | EDIT (by abiessu): the original is wrong, but as it has been accepted, it seems that it would be good to make the accepted answer correct. The original approach is generally used but corrected where in error.
(original) I suspect that there is a typo in your OP, because as it is stated now it is rather uninteresting... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/863142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
An inequality about a sequence Let $(a_n)$ be a sequence such that $a_0=1 , a_1=2 , a_{n+1}=a_n+\dfrac {a_{n-1}}{1+ a_{n-1}^2} , \forall n \ge1 $ , then is it true that $52 < a_{1371} < 65$ ?
$ EDIT:-$ I am posing another question , so I'm not able to accept Oleg567 ' s very correct answer :
Let $(a_n)$ be a seque... | First, note that next recurrent relation is true for sequence $(a_n)$:
$$
a_0=1,\\
a_{n+1} = a_n+\frac{1}{a_n}, \qquad n\ge 0.\tag{1}
$$
Yes, $a_0=1$, $a_1 = 1+\frac{1}{1}=2$ for both definitions; if $a_{n+1}$ is defined by $(1)$, then
$$
a_{n+1} = a_n+\dfrac{1}{a_n} = a_n + \dfrac{1}{a_{n-1}+\frac{1}{a_{n+1}}} =
a_n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/864135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
$ \int \frac{1}{(x-a)(x+b)} dx $ Could you please explain how to integrate this integral:
$$ \int \frac{1}{(x-a)(x+b)} dx $$
| Since there is linear factor in denominator,
$$\int \frac{1}{(x-a)(x+b)} dx=\int \frac{1}{(x-a)(a+b)}+\frac{1}{(-b-a)(x+b)} dx$$
Factor out $\frac{1}{a+b}$,
$$=\frac{1}{a+b}\int \frac{1}{x-a}-\frac{1}{x+b} dx$$
The anti-derivatives of $\ln{(x+k)}$ is $\frac{1}{x+k}$
$$=\frac{1}{a+b}\bigg[\ln{(x-a)}-\ln{(x+b)}\bigg]+C$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/866727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Simplify [1/(x-1) + 1/(x²-1)] / [x-2/(x+1)] Simplify: $$\frac{\frac{1}{x-1} + \frac{1}{x^2-1}}{x-\frac 2{x + 1}}$$
This is what I did.
Step 1: I expanded $x^2-1$ into: $(x-1)(x+1)$. And got: $\frac{x+1}{(x-1)(x+1)} + \frac{1}{(x-1)(x+1)}$
Step 2: I calculated it into: $\frac{x+2}{(x-1)(x+1)}$
Step 3: I multiplied $x-\... | You need to multiply the +2 by (x-1) before you add it to $x^3-x$ because
$$\frac{2}{x+1}=\frac{2(x-1)}{(x-1)(x+1)}$$
Also, in step 6, you can't remove (x+2) from top and bottom. Firstly, $x^3-x+2=x^3-(x-2)$, so there is no $x+2$ in the denominator. Secondly, you need (x+2) to be a factor of the whole denominator, no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/866935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
What is the count of the strict partitions of n in k parts not exceeding m? Lets say we had a $k,m,n \in \mathbb{N}$ where $k < m \le n$. How many different sets $X_1,..,X_m$ with $|X_i|=k$ for $i=1,..,m$, where the sets do not include duplicates, for which the sum of their elements is equal to n?
$$n = X_{sum}^i := \s... | I'm assuming from your list of examples that what you are looking for is the number of partitions of $n$ into $k$ distinct parts with largest part equal to $m$ (this is different from the question in the title, but seems to fit your examples best).
The number of partitions of $n$ with at most $k$ parts, each of length ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/867760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Seeking concise proof: $\frac18(a^2+b^2)(b^2+c^2)(c^2+a^2)\ge\frac1{27}(ab+bc+ca)^3$, where $a$, $b$, $c$ are positive numbers I was just encountered an inequality in AoPs, Here it is:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=4569&view=next,
that is:
If $a$, $b$ and $c$ are positive numbers, th... | We'll prove that your inequality is true for all reals $a$, $b$ and $c$.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.
Since $\prod\limits_{cyc}(a^2+b^2)=\prod\limits_{cyc}(a^2+b^2+c^2-c^2)=-w^6+A(u,v^2)w^3+B(u,v^2)$,
we see that our inequality is equivalent to $f(w^3)\geq0$, whe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/868952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Trouble evaluating the sum involving logarithm I was trying to solve this problem: Closed form for $\int_0^1\log\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\mathrm dx$
In the procedure I followed, I came across the following sum:
$$\sum_{k=1}^{\infty} (-1)^{k-1}k\left(\frac{\ln(2k+1)}{2k+1}-\frac{\ln(2k-1)}{2k-1... | On the limit
Experimenting a bit I find
$$\lim_{m\to\infty} S_m = \ln\left(\frac{\Gamma(\frac14)}{2\Gamma(\frac34)}\right) = 0.391594392706836...$$
To be the exact limit
On convergence
Let $a_k := \frac{\ln(2k-1)}{2k-1}$ then
$$\begin{align*}
S_m & = \sum_{k=1}^m (-1)^{k-1} k (a_{k+1} - a_k) \\
& = \sum_{k=2}^{m+1} (-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/869772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How do I reduce radian fractions? For example, I need to know $\sin (19π/12)$.
I need to use the subtraction formula. How do I get $(\text{what}) - (\text{what}) = 19π/12$? I am stuck at what are the radians
Do I divde it by something? What is the process?
| $$\frac{19\pi}{12}=\pi+\frac{3\pi}4-\frac{\pi}6$$
Since $\sin(\pi+x)=-\sin x$, $\sin\left(\frac{19\pi}{12}\right)=-\sin\left(\frac{3\pi}4-\frac{\pi}6\right)$.
And since $\sin(A-B)=\sin A\cos B-\cos A\sin B$:
$$\begin{align}
-\sin\left(\frac{3\pi}4-\frac{\pi}6\right)&=-\left(\sin\left(\frac{3\pi}4\right)\cos\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/871177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
A problem on nested radicals Find the value of $x$ for all $a>b^2$ if:
$$\large x=\sqrt{a-b\sqrt{a+b\sqrt{a-b{\sqrt{a+b.......}}}}}$$
My attempt
$$\large x=\sqrt{a-b\sqrt{(a+b)x}}$$
$$\large x^4=(a-b)^2(a+b)x$$
$$\large x=((a-b)^2(a+b))^{1/3}$$ (real root)
Question: Is my solution correct??
| Here's another one:
$$\text{Let} x=\sqrt{a-b\sqrt{a+b\sqrt{a-b\sqrt{a+b\cdots}}}}$$
$$\text{Let}y=\sqrt{a+b\sqrt{a-b\sqrt{a+b\sqrt{a-b\cdots}}}}$$
Now $$x^2=a-by---1$$ and $$y^2=a+bx----2$$ $$\implies x^2-y^2=-b(x+y)$$ $$\implies(x+y)(x-y+b)=0$$(Considering only positive solutions of $x$) $$y=x+b$$, putting it in eq $2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/874672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
if $\frac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$,find $a_{n}$ Let
$$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$$
Find the closed form $$a_{n}$$
since
$$(1-x^4)(1-x^3)(1-x^2)=(1-x)^3(1+x+x^2+x^3)(1+x+x^2)(1+x)$$
so
$$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\dfrac{1}{(1-x)^3(1+x+x^2+x^3)(1+x+... | The $n$-th term is obtained from $(1+x^2+x^4+\cdots)(1+x^3+x^6+\cdots)(1+x^4+x^8+\cdots)$ which is the number of triples $(i,jk)$ that are solutions to $2i+3j+4k=n$. Hence, $$a_n=\sum_{2i+3j+4k=n}1=\sum_{j=0}^{\lfloor n/3\rfloor }\sum_{2i+4k=n-3j}1$$
Now we want to obtain $$b_m=\sum_{2i+4k=m}1$$
It is clear $b_{2m+1}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/875792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Intuitive ways to get formula of cubic sum Is there an intuitive way to get cubic sum? From this post: combination of quadratic and cubic series and Wikipedia: Faulhaber formula, I get $$1^3 + 2^3 + \dots + n^3 = \frac{n^2(n+1)^2}{4}$$
I think the cubic sum is squaring the arithmetic sum $$1^3 + 2^3 + \dots + n^3 = (1 ... | We have
$$
\sum_{k=1}^{n}k^3 = 1 + 8 + 27 + \ldots + n^3 = \\
\underbrace{1}_{1^3} + \underbrace{3+5}_{2^3} + \underbrace{7 + 9 + 11}_{3^3} + \underbrace{13 + 15 + 17 + 19}_{4^3} + \ldots = \\
\underbrace{\underbrace{\underbrace{1}_{1^2} + 3}_{2^2} + 5}_{3^2} + \ldots
$$
which is
$$
\left( \sum_{k=1}^{n}k \right)^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/876922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 4,
"answer_id": 0
} |
Ordered partitions of an integer (with a twist) I would like to know how to prove (preferably algebraically) that $P_1(2,n)=F_{2n+1}$, where $P_1(2,n)$ is what I define to be the number of ordered partitions of an integer, where the number $1$ has 2 possible colours. For example,
\begin{align}
2
&=2\\
&={\color\red{1}... | The derivation of the generating function can also done from first principles, without using the recurrence. We have straightforwardly that these partitions have the generating function
$$\sum_{q\ge 0} \left(\frac{z}{1-z} + z\right)^q$$
where we have added in the quantity $z$ to account for the fact that there are two ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/877816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int_{1}^{e}\frac{u}{u^3+2u^2-1}du.$ I'm trying to solve
$$\int_{1}^{e}\frac{u}{u^3+2u^2-1}du.$$
My first approach was to factorise and then do a partial integration. However the factorisation $(u+1)\left(u+\frac{1}{2}-\frac{\sqrt{5}}{2} \right)\left(u+\frac{1}{2}+\frac{\sqrt{5}}{2} \right)$ leads me to heavy... | Hint: $$u^3+2u^2-1=(u+1)(u^2+u-1).$$ Completing the square $$ u^2+u-1= u^2+u\frac{1}{4}-\frac{1}{4}-1=(u+1/2)^2-5/4.$$
Now use partial fraction
\begin{align}
\frac{1}{(u+1)((u+1/2)^2-5/4)}=\frac{A}{u+1}+\frac{Bu+C}{( u+1/2)^2-5/4}
\end{align}
We find that $ A=-1, B=1, C=0$, so that
\begin{align}
I&=\int { \frac{-1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/879722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving recurrence relation: Product form Please help in finding the solution of this recursion.
$$f(n)=\frac{f(n-1) \cdot f(n-2)}{n},$$
where $ f(1)=1$ and $f(2)=2$.
| Take some large n,
$$
\begin{align}
f(n) &= \frac{f(n-1) f(n-2)}{n} \\
&= {f(n-2)^2 f(n-3) \over n (n-1)} \\
&= {f(n-3)^3 f(n-4)^2 \over n (n-1) (n-2)^2} \\
&= \frac{f(n-4)^5 f(n-5)^3}{n (n-1) (n-2)^2 (n-3)^3} \\
&= \dots \\
&=\frac{f\left(n - (n-2) \right)^{F_{n-1}} f\left(n - (n-1) \right)^{F_{n-2}}}{\prod^{n-3}_{i=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/881121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$
Given $a$, $b$ and $c$ are positive real numbers. Prove that:$$\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$$
Additional info: We can't use induction. We should mostly use Cauchy inequality. Other inequalities can be used rare... | Since the inequality is homogeneous, WLOG assume that $a+b+c = 1$.
Then, the inequality becomes $\dfrac{a}{(1-a)^2}+\dfrac{b}{(1-b)^2}+\dfrac{c}{(1-c)^2} \ge \dfrac{9}{4}$.
Since the function $f(x) = \dfrac{x}{(1-x)^2}$ is concave up for $x > 0$, by Jensen's Inequality, we have:
$f(a)+f(b)+f(c) \ge 3f\left(\dfrac{a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/882272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
Proof for complex numbers and square root Use the polar form of complex numbers to show that every complex number $z\neq0$ has two square roots.
I know the polar form is $z=r(\cos(\alpha)+i \sin(\alpha))$. I'm just not sure how to do this one.
| De Moivre's formula is $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$ for any integer $n$. Therefore, if $w = s(\cos\theta + i\sin\theta)$ is a non-zero complex number such that $w^2 = z$, then
$$s^2(\cos(2\theta) + i\sin(2\theta)) = r(\cos\alpha + i\sin\alpha).$$
Therefore $s^2 = r$, so $s = \pm\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/883030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Recognizing the sequence 1/16, 1/8, 3/16, 1/4, 5/16, ...
What is the missing number?
$$\frac{1}{16}, \frac{1}{8}, \frac{3}{16}, \frac{1}{4}, \frac{5}{16}, \ \ \ [?]$$
$$A. \frac{5}{4}\quad B. \frac{3}{4}\quad C. \frac{5}{8}\quad D. \frac{3}{8}$$
Spoiler: Answer is $D$, but I don't know why.
Thanks
| $$\frac{1}{16}, \frac{1}{8}=\frac{2}{16}, \frac{3}{16}, \frac{1}{4}=\frac{4}{16}, \frac{5}{16}$$
So the $i$th term is of the form $$\frac{i}{16}$$ Therefore, the next term is $$\frac{6}{16}=\frac{3}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/885137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
What are some good questions for this trick, if $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\dots=\alpha$ then $\alpha=\frac{a+c+e+...}{b+d+f+...}$? I need some good algebra questions that are applications of this trick, often in a non obvious and elegant way: $$\text{If } \frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\dots=\alpha \t... | *
*If $\displaystyle\frac a{b+c}=\frac b{c+a}=\frac c{a+b};$ prove that each ratio $\displaystyle=\frac12$ if $\displaystyle a+b+c\ne0$
*If $\displaystyle\frac{a-b}{x^2}=\frac{b-c}{y^2}=\frac{c-a}{z^2}$ prove that $\displaystyle a=b=c$
*If $\displaystyle\frac{a-b}{a^2+ab+b^2}=\frac{b-c}{b^2+bc+c^2}=\frac{c-a}{c^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/885465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $f\left(x-\frac{2}{x}\right) = \sqrt{x-1}$, then what is the value of $f'(1)$
Find $f'(1)$ if $$f\left(x-\frac{2}{x}\right) = \sqrt{x-1}$$
My attempt at the question:
Let $(x-\dfrac{2}{x})$ be $g(x)$
Then $$f(g(x)) = \sqrt{x-1} $$
Differentiating with respect to x:
$$f'(g(x))\cdot g'(x) = \frac{1}{2\sqrt{x-1}} $... | Let us assume ${\left(x-\frac{2}{x}\right) = 1, \text{the part inside}}$
It is found that $x = 2$ or $x = -1$
Also $f(g(x))=f'(g(x)).g'(x)$ by the chain rule.
for $ g(x)=x-\frac {2}{x}$
$\Rightarrow g'(x) = 1+\frac {2}{x^2}$
for $f'(x)=\frac{1}{2} \frac {1}{\sqrt{(x-1)}}$
$f(g(x))=\left(1+\frac {2}{x^2}\right) \left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/886300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Evaluating $ \int x \sqrt{\frac{1-x^2}{1+x^2}} \, dx$ I am trying to evaluate the indefinite integral of
$$\int x \sqrt{\frac{1-x^2}{1+x^2}} \, dx.$$
The first thing I did was the substitution rule:
$u=1+x^2$, so that $\displaystyle x \, dx=\frac{du}2$ and $1-x^2=2-u$. The integral then transforms to
$$\int \sqrt{\fr... | Another way to solve the integral is as follows. Let $u=\sqrt{\frac{1-x^{2}}{1+x^{2}}}$ so that
$$u^{2}(1+x^{2})=1-x^{2}$$
$$u^{2}-1=-x^{2}(1+u^{2})$$
$$x^{2}=\frac{1-u^{2}}{1+u^{2}}$$
so
$$2xdx=\frac{(-2u)(1+u^{2})-(2u)(1-u^{2}))}{(1+u^{2})^{2}}du=\frac{-4u}{(1+u^{2})^{2}}du$$
Going back to the integral we have:
$$\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/887784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
} |
Solving the logarithimic inequality $\log_2\frac{x}{2} + \frac{\log_2x^2}{\log_2\frac{2}{x} } \leq 1$ I tried solving the logarithmic inequality:
$$\log_2\frac{x}{2} + \frac{\log_2x^2}{\log_2\frac{2}{x} } \leq 1$$
several times but keeping getting wrong answers.
| Let $\log_2 x=A$, then $\log_2 x^2=2\log_2 x=2A$ and $\log_2\frac{2}{x}=\log_22- \log_2x=1-A$. So the given inequality becomes:
$$(A-1)+\frac{2A}{1-A} \leq 1.$$
Consequently we get
$$\frac{4A-A^2-1}{1-A} \leq 1.$$
Furthermore you get
$$\frac{5A-A^2-2}{1-A} \leq 0.$$
Hopefully you can solve from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/888174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How many $3$ digit numbers with digits $a$,$b$ and $c$ have $a=b+c$ My question is simple to state but (seemingly) hard to answer. How many $3$ digit numbers exist such that $1$ digit is the sum of the other $2$. I have no idea how to calculate this number, but I hope there is a simple way to calculate it. Thank you in... | My answer is similar to that of @shooting-squirrel. Edit: Now it gives the same answer, silly mistakes :D
Consider the number $\overline{abc}$, we have three cases:
Case 1: $a=b+c$, then $a$ can be $1,\dots,9$. Now, for each value of $a$, there are $a+1$ values of $b$: $0,1,\dots, a$. Each of which gives one number. To... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/889687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Error in proving of the formula the sum of squares Given formula
$$
\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}
$$
And I tried to prove it in that way:
$$
\sum_{k=1}^n (k^2)'=2\sum_{k=1}^n k=2(\frac{n(n+1)}{2})=n^2+n
$$
$$
\int (n^2+n)\ \text d n=\frac{n^3}{3}+\frac{n^2}{2}+C
$$
But
$$
\frac{n^3}{3}+\frac{n^2}{2}+C $$ i... | This is from An Introduction to the Calculus of Finite Differences and Difference Equations by Kenneth S. Miller, 1960.
We have the difference operator $$ \Delta f(x) = f(x+1) - f(x) $$
Then we have the factorial polynomials,
$$ x^{(0)} =1, $$
$$ x^{(1)} =x, $$
$$ x^{(2)} =x(x-1) = x^2 - x, $$
$$ x^{(3)} = x(x-1)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"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.