Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
is $y = \sqrt{x^2 + 1}− x$ a injective (one-to-one) function? I have a function $y = \sqrt{x^2 + 1}− x$ and I need to prove if it's a Injective function (one-to-one). The function f is injective if and only if for all a and b in A, if f(a) = f(b), then a = b
$\sqrt{a^2 + 1} − a = \sqrt{b^2 + 1} − b$
$\sqrt{a^2 + 1} +... | When you wrote $$2b\sqrt{a^2 + 1} = 2a\sqrt{b^2 + 1}$$ this gave you a new condition : $a$ and $b$ must have same sign. (The square roots are positive). Hence $a^2=b^2$ implies $a=b$ in this case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Finding stagnation points and stream function
Sorry for the lack of latex. The question I want to ask would need all this info and it would take very long to write it.
(a) Irrotational flow means $\nabla \times \textbf u =0$ so we can define potential to be $\textbf u =\nabla \phi$ and since it is irrotational and inv... | For (b), we can take partial derivatives of $\phi$ to obtain
$$u_r = \frac{\partial \phi}{\partial r} = U \cos \theta - U\frac{a^3}{r^3} \cos \theta, \\ u_\theta = \frac{1}{r} \frac{\partial \phi}{\partial \theta}= -U \sin \theta - U\frac{a^3}{2r^3} \sin \theta.$$
For (c), on the surface of the sphere $r = a$ and the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\sqrt{4+\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\cdots}}}}$
Find the value of $$\sqrt{4+\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\cdots}}}}$$
I know how to solve when all surds are of the same order, but what if they are different?
Technically, (as some users wanted to know exactly what is to be found), find:
$$\lim_{n\to\inf... | Put $$y=\sqrt{4+\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+...}}}}\qquad (1)$$ stopping successively at the $n$-th root we have a sequence strictly increasing and bounded so the limit $y$ is well defined.
From $(1)$ we can easily get a sequence $\{P_n\}$ of polynomials such that its largest real roots $$\alpha_n=\sqrt{4+(4+(4+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1782032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
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Calculating $\sqrt[3]{\sqrt 5 +2}-\sqrt[3]{\sqrt 5 -2}$ We want only the real 3rd root.
By calculation, $[\sqrt[3]{\sqrt 5 +2}-\sqrt[3]{\sqrt 5 -2}]^3= 4-3[\sqrt[3]{\sqrt 5 +2}-\sqrt[3]{\sqrt 5 -2}]$
Therefore, the answer is a root of $t^3=4-3t$ , which obviously has the real solution $t =1$.
But I want another way o... | By observation,
$$(\sqrt5+1)^3=16+8\sqrt5=8(\sqrt5+2)$$
$$\implies\left(\dfrac{\sqrt5+1}2\right)^3=\sqrt5+2$$
Similarly,$$\left(\dfrac{\sqrt5-1}2\right)^3=\sqrt5-2$$
Motivation:
$$(\sqrt5+a)^3=a^3+15a+\sqrt5(3a^2+5)$$
Let us find $a$ such that $$\dfrac{a^3+15a}{3a^2+5}=\dfrac21\iff0=a^3-6a^2+15a-10=(a-1)(a^2-5a+10)$$
O... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the limit of $\lim_{n\to \infty}\frac{2^n n^2}{n!}$ I am trying to find the limit of the following, $$\lim_{n\to \infty}\frac{2^n n^2}{n!}$$
L'Hospital is not going to work. Hoping for a squeeze, by the observation that $2^n<n!$ for $n\geq 4$ does not help either as one side of the limit goes to $\infty$. How can ... | $n!>2\cdot 3^{n-2}$ for $n\ge2$ is apparent from the product definition, so:
$$\frac{2^nn^2}{n!}\le\frac{2^nn^2}{2\cdot 3^{n-2}}=\frac{9n^2}{4\left(\frac{3}{2}\right)^n}$$
Note that $\left(\frac{3}{2}\right)^n=\left(1+\frac{1}{2}\right)^n=\sum_{k=0}^n\binom{n}{k}\left(\frac{1}{2}\right)^k\ge\binom{n}{3}\left(\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1782490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find the integral of a quotient of rational functions? How do I compute the following integral: $$\int \dfrac{x^4+1}{x^3+x^2}\,dx$$
My attempt:
We can write $$\dfrac{x^4+1}{x^3+x^2} = \dfrac{A}{x^2} + \dfrac{B}{x} + \dfrac{C}{x+1}$$
It is easy to find that
$A=1$,
$B=2$, and
$C=-1$.
Therefore
$$\frac{x^4+1}{x^3+... | Observe that
$$\frac{x^4+1}{x^3+x^2}=\frac{x^4+x^3-x^3-x^2+x^2+1}{x^3+x^2}=\frac{(x-1)(x^3+x^2)+x^2+1}{x^3+x^2}=x-1+\frac{x^2+1}{x^2(x+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
Closed form of $\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{\lfloor|\tan x|\rfloor}{|\tan x|}dx$? I am trying to find a closed form for the integral $$I=\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{\lfloor|\tan x|\rfloor}{|\tan x|}dx$$ So far, my reasoning is thus: write, by symmetry through $x=\pi/2$, $$I=2\sum_{n=1}^{\i... | $$\begin{align}
I
&=\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{\lfloor\left|\tan{\left(x\right)}\right|\rfloor}{\left|\tan{\left(x\right)}\right|}\,\mathrm{d}x\\
&=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\lfloor\left|\cot{\left(y\right)}\right|\rfloor}{\left|\cot{\left(y\right)}\right|}\,\mathrm{d}y;~~~\small{\left[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1785414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to calculate the nullity for a matrix with variable? In order to find the number of (x,y) satisfying
$$f(x,y) =\begin{cases}
xy^2-y+x^2+1=0\\
x^2y^2+y-1=0\\
\end{cases}$$
I use the Jacobi-Darboux Theorem by fix 1.x; 2.y and get the Bezoutian matrix, but I don't know how to find the nullity of the both m... | We can solve the two polynomial equations using $S$-polynomials (from Buchberger's algorithm). Apart from the trivial solution $(x,y)=(0,1)$ this yields
$$
x=y^4 + y^3 - 2y^2 + y - 1,
$$
so that we obtain a polynomial in $y$, namely
$$
y^6 + y^5 - y^4 + 2y^3 - 2y^2 + y - 1=0.
$$
This has $6$ solutions over the complex ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving an identity involving floor function
Prove that :
$$\left \lfloor \dfrac{2 a^2}{b} \right \rfloor - 2 \left \lfloor \dfrac{a^2}{b} \right \rfloor = \left \lfloor \dfrac{2 (a^2 \bmod b)}{b} \right \rfloor $$
Where $a$ and $b$ are positive integers.
Please provide some hints/solutions. Thanks.
| Let us assume that $b>0$. By division algorithm $a^2=bq+r$ with $0 \leq r <b$. Then
\begin{align*}
a^2&=bq+r\\
\frac{2a^2}{b} & =2q+\frac{2r}{b}\\
\left\lfloor \frac{2a^2}{b} \right\rfloor & = 2q+\left\lfloor \frac{2r}{b} \right\rfloor\\
\left\lfloor \frac{2a^2}{b} \right\rfloor -2q& =\left\lfloor \frac{2r}{b} \right\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can composite numbers be uniquely written as a sum of two squares?
Let $X = a^2 +b^2$ where all the terms are positive integers and $X$ is a composite number and $\gcd(a,b)=1$ . Do there exist positive integers $c$ and $d$ other than $a$ and $b$ such that $X = c^2+d^2$ ?
By Fermat's Two Square Theorem, since $\gcd(a... | If $N$ is a product of two numbers of the form $4k+1$, then it usually has at least two representations as a sum of two squares, because:
$$\begin{align}
(a^2+b^2)(c^2+d^2) & = (ac-bd)^2 + (ad+bc)^2 \\
& = (ac+bd)^2 + (ad-bc)^2
\end{align}$$
This is known as
Brahmagupta's identity.
By applying Brahma... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $f(z)=\dfrac{U(z)}{V(z)}=\dfrac{2z^3-3z^2+7z-8}{z^4-5z^3+4z^2-6z+1}$ find $f(1-\sqrt{2}i)$ without lots of complex arithmetic. Such a problem is usually done either by direct substitution (ugh!) or synthetic division.
Synthetic division after several complex products and additions gives $U(1-\sqrt{2}i)=-8-3\sqrt{... | Since it has been 12 hours and no more answers have been suggested, I am posting a solution based upon a suggestion by @almagest.
\begin{equation}
[z-(1-\sqrt{2}i)]\cdot[z-(1+\sqrt{2}i)]=z^2-2z+3
\end{equation}
Using long division (or better yet, quadratic synthetic division) no complex arithmetic is required to obta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1787748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Inclusion–Exclusion Identical Computers Problem
Find the number of ways to distribute 19 identical computers to four schools, if School A must get at least three, School B must get
at least two and at most five, School C get at most four, and School D gets the rest.
a) Solve using inclusion-exclusion
b) Solve using... | Addendum to the generating function part of the answer of @callculus. Assuming WA is not available, it's not too hard to calculate the coefficient by hand. In order to do so it's convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series. This way we can write e.g.
\begin{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1788439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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How to factor $a^3 - b^3$? I know the answer is $(a - b)(a^2 + ab + b^2)$, but how do I arrive there? The example in the book I'm following somehow broke down $a^3 - b^3$ into $a^3 - (a^2)b + (a^2)b - a(b^2) + a(b^2) - b^3$ and factored that into $(a-b)(a^2 + ab + b^2)$ from there, but I don't quite understand how it w... | Long division makes it as easy as $1$, $2$, $3$:
$$\begin{align}
\frac{a^3-b^3}{a-b}&=a^2+\frac{ba^2-b^3}{a-b} \tag 1\\\\
&=a^2+ab+\frac{b^2a-b^3}{a-b} \tag 2\\\\
&=a^2+ab+b^2 \tag 3
\end{align}$$
as was to be shown!
Note that in $(1)$, the term $\frac{ba^2-b^3}{a-b}$ is the remainder of $a^2$ in $\frac{a^3-b^3}{a-b}$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Please help me compute this$ \sum_m\binom{n}{m}\sum_k\frac{\binom{a+bk}{m}\binom{k-n-1}{k}}{a+bk+1}$ Compute following:
$$
\sum_m\binom{n}{m}\sum_k\frac{\binom{a+bk}{m}\binom{k-n-1}{k}}{a+bk+1}
$$
Only consider real numbers a, b such that the denominators are never 0.
Now I simplify it into
$$
-\frac{1}{n}\sum_k\binom{... | Here is a slightly different variation of the theme. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write e.g.
\begin{align*}
[z^k](1+z)^n=\binom{n}{k}
\end{align*}
We obtain
\begin{align*}
\sum_{m=0}^{n}&\binom{n}{m}\sum_{k=0}^{n}\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1789981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Sketch the heart and indicate its orientation with arrows $ r = 1 - \cos(\theta)$. Find the area enclosed by the heart Hi all I am trying to figure out how to sketch the heart. Here is what I have tried so far:
$$r = 1 - \cos(\theta) \\
r(r = 1 - \cos(\theta)) \\
r^2 = r - r\cos(\theta) \\
$$
Use the fact that
$$r =\sq... | Drawing the figure is the hard part by far. For example, Matlab doesn't have a method for directly adding arrows to a curve like this. What I tried was to draw a little right-pointing arrowhead with its tip at $(0,0)$ and some reasonable size. Then at a selection of points along the path, I rotated that arrow countercl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1792115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove $\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3$ $a,b,c >0$ and $a+b+c=3$, prove
$$\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3$$
I try to app... | Only a partial answer.
Assuming $a\le b\le c$, then we have $0<a\le1$ and $1\le c<3$. Now we have two cases, $b\le1$ and $b>1$. The case $b\le1$ is easy to deal with.
Assuming $0<a\le b\le1\le c<3$, we have
\begin{align}
\left(\frac{a+1}{a+b} \right)\left(\frac{b+1}{b+c}\right)\left(\frac{c+1}{c+a} \right) \ge 1
&\iff... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1793442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 2,
"answer_id": 0
} |
Prove $\sum\limits_{cyc}\left(\frac{a^4}{a^3+b^3}\right)^{\frac34} \geqslant \frac{a^{\frac34}+b^{\frac34}+c^{\frac34}}{2^{\frac34}}$ When $a,b,c > 0$, prove
$$\left(\frac{a^4}{a^3+b^3}\right)^{\frac34}+\left(\frac{b^4}{b^3+c^3}\right)^ {\frac34}+\left(\frac{c^4}{c^3+a^3}\right)^{\frac34} \geqslant \frac{a^{\frac34}+b^... | Partial answer :
Hint :
Using WRCF theorem see https://journalofinequalitiesandapplications.springeropen.com/articles/10.1186/1029-242X-2011-101
We have $c=\min{(a,b,c)}$ and let $f(x)$ such that :
$$f\left(x\right)=\left(\frac{1}{1+x^{3}}\right)^{\frac{3}{4}}$$
It seems we have $f''(x)>0$ for $x\geq 1$
For and $a,b,c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1793680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 4
} |
Is $x^4 + 4$ irreducible in $\mathbb{Z}_5$? Well, I'm having doubts, isnt that $\mathbb{Z}_5$ has no zero divisors, and now you cant factor $x^4 + 4$ ?
| \begin{align}
x^4+4&=x^4-1\\
&=(x^2-1)(x^2+1)\\
&=(x-1)(x+1)(x^2-4)\\
&=(x-1)(x+1)(x+2)(x-2)\\
&=(x-1)(x-2)(x-3)(x-4).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Solve summation expression For a probability problem, I ended up with the following expression
$$\sum_{k=0}^nk\ \binom{n}{k}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k$$
Using Mathematica I've found that the result should be $\frac{n}{3}$. However, I have no idea how to get there. Any ideas?
| \begin{align*}
\sum_{k=0}^n k \binom{n}{k}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k &=\sum_{k=0}^n k\frac{n!}{k!(n-k)!}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k\\
&=\sum_{k=0}^n \frac{n!}{(k-1)!(n-k)!}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k\\
&=\frac{n}{3}\sum_{k=0}^n \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1798085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
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Verify the correctness of $\sum_{n=1}^{\infty}\left(\frac{1}{x^n}-\frac{1}{1+x^n}+\frac{1}{2+x^n}-\frac{1}{3+x^n}+\cdots\right)=\frac{\gamma}{x-1}$ $x\ge2$
$\gamma=0.57725166...$
(1)
$$\sum_{n=1}^{\infty}\left(\frac{1}{x^n}-\frac{1}{1+x^n}+\frac{1}{2+x^n}-\frac{1}{3+x^n}+\cdots\right)=\frac{\gamma}{x-1}$$
Series (1) co... | For $x =2$ the identity is true.
Claim. $\displaystyle \sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^k}{2^n+k} = \gamma. $
Proof. We first rearrange the given sum:
$$ S
:= \sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^k}{2^n+k}
= \sum_{l = 2}^{\infty} \bigg( \sum_{\substack{(n,k) \ : \ 2^n + k = l \\ n \geq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1798296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Evaluating the arc length integral $\int\sqrt{1+\frac{x^4-8x^2+16}{16x^2}} dx$
Find length of the arc from $2$ to $8$ of $$y = \frac18(x^2-8 \ln x)$$
First I find the derivative, which is equal to $$\frac{x^2-4}{4x} .$$
Plug it into the arc length formula $$\int\sqrt{1+\left(\frac{dy}{dx}\right)^2} dx$$ and get
$$\in... | Hint
$$1 + \frac{x^4 - 8 x^2 + 16}{16 x^2} = \frac{16 x^2}{16 x^2} + \frac{x^4 - 8 x^2 + 16}{16 x^2} =\frac{x^4 + 8 x^2 + 16}{16 x^2}$$
Can you write the rightmost expression as a square of another expression?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1800505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Use $\epsilon$, $\delta$ to prove that $\lim\limits_{x\to\ b} \frac{1}{a+x}$ = $\frac{1}{a+b}$. I've been working on this epsilon delta proof for the longest time now, and I can't quite get it.
Let $a>0$ and $b>0$. Use $\epsilon$, $\delta$ to prove that $\lim\limits_{x\to\ b} \frac{1}{a+x}$ = $\frac{1}{a+b}$.
I've fo... | Let $\varepsilon > 0$. Want some $\delta > 0$ such that $0 < |x-b| < \delta$ implies $\left| \frac{1}{a + x} - \frac{1}{a + b} \right| < \varepsilon$. We start with the inequality $\left| \frac{1}{a + x} - \frac{1}{a + b} \right| < \varepsilon$ and try to manipulate it to give us insight into what $\delta$ should be. W... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1802735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solve $ord_x(2) = 20$ Given that the (multiplicative) order of $2$ mod $x$ is $20$, how can I work out what $x$ is?
| Method 1:-
From Euler Totient Function(or theorem) we know-
$$a^{\phi(n)}\equiv1\pmod n$$ for $\gcd(a,n)=1$
Here you have,$2^{20}\equiv1\pmod n$.
Comparing the two equations we have,$\phi(n)=20$.Try to find $n$ from it.
Method 2:-
We know that for all $n$,$(a-b)|(a^n-b^n)$ and $a^n-b^n$ is also divisible by $a+b$ if ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1809770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Evaluation of $\sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
Prove that $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{2n}xdx = \frac{\pi}{2}\frac{1}{4^{n}}\binom{2n}{n}$ and also find value of $\displaystyle \sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
$\bf{My\; Try::}$ Let $$\displaystyle I_{n} = \int_{0}^{\fr... | Here's another probabilistic approach. We don't evaluate any (Riemann) integrals.
Consider a simple symmetric random walk on ${\mathbb Z}$ starting from $0$ at time $0$. The probability that at time $2n$ the walk is at zero is equal to $\binom{2n}{n}2^{-2n}$, and the probability that at time $2n+1$ the walk is at zero... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 0
} |
Why is this incorrect $\int_{0}^{1}{\ln(x)\over (1+x)^3}dx=-\sum_{n=0}^{\infty}{(-1)^n(n+2)\over 2(1+n)}$ $$I=\int_{0}^{1}{\ln(x)\over (1+x)^3}dx$$
Recall
$${1\over (1+x)^3}=\sum_{n=0}^{\infty}{(-1)^n(n+1)(n+2)\over 2}x^n$$
$$\int_{0}^{1}x^n\ln(x)dx=-{1\over (n+1)^2}$$
Substitute in
$$I=\sum_{0}^{\infty}{(-1)^n(n+1)(n+... | You are working on the assumption that
\begin{align*}
&\int_{0}^{1} \left( \sum_{n=0}^{\infty} \frac{(-1)^n(n+1)(n+2)}{2} x^n \right) \log x \, \mathrm{d}x \\
&\hspace{5em} = \sum_{n=0}^{\infty} \frac{(-1)^n(n+1)(n+2)}{2} \int_{0}^{1} x^n \log x \, \mathrm{d}x.
\tag{1}
\end{align*}
But this is not true because the latt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$\int\frac{\sin x}{\sqrt{1-\sin x}}dx=?$ Calculate this integral $\displaystyle\int\dfrac{\sin x}{\sqrt{1-\sin x}}dx=?$
Effort;
$1-\sin x=t^2\Rightarrow \sin x=1-t^2\Rightarrow \cos x=\sqrt{2t^2-t^4}$
$1-\sin x=t^2\Rightarrow-\cos x dx=2tdt\Rightarrow dx=\frac{2t}{\sqrt{t^4-2t^2}}dt$
$\displaystyle\int\frac{1-t^2}{t}... | By setting $x=\frac{\pi}{2}-t$ the problem boils down to finding:
$$ \int \frac{\cos t}{\sqrt{1-\cos t}}\,dt = \frac{1}{\sqrt{2}}\int\frac{1-2\sin^2\frac{t}{2}}{\sin\frac{t}{2}}\,dt $$
where:
$$ \int \frac{1}{\sin\frac{t}{2}}\,dt = C + 2\log\left(\tan\frac{t}{4}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How we can show this ;$\frac{x^2+y^2+z^2}{2}\times\frac{x^5+y^5+z^5}{5}=\frac{x^7+y^7+z^7}{7}$ Let be $\quad x+y+z=0$
show this:
$$\frac{x^2+y^2+z^2}{2}\times\frac{x^5+y^5+z^5}{5}=\frac{x^7+y^7+z^7}{7}$$
I solved ,but Im interesting what are you thinking about this,how can we arrive to solution quickly?
| Your polynomials are invariant by the action of $S_3$ on $(x,y,z)$.
A standard result says that the graded ring $\Bbb Q[x,y,z]^{S_3}$ is $\Bbb Q[x+y+z;xy+yz+xz;xyz]$.
Furthermore here, quotienting by $(x+y+z)$ we can work in the ring $(\Bbb Q[x,y,z]/(x+y+z))^{S_3} \cong \Bbb Q[xy+yz+xz;xyz]$.
Now in this graded ring, t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
derivative with square root I have been trying to figure this equation for some time now, but have come up empty. I have tried multiple ways on solving it. Whether by using the Quotient Rule or some other method, I can't seem to figure it out. Any help would be appreciated.
Find the derivative of the function
$
y = \f... | While using the quotient rule for $f(x)=\dfrac{h(x)}{g(x)}\rightarrow f'(x)=\dfrac{h'(x)g(x)-h(x)g'(x)}{[g(x)]^2}$ is straightforward and gives
$y'=\dfrac{(2x+8)\sqrt{x}-\frac{1}{2}(x^2+8x+3)x^{-\frac{1}{2}}}{x}=(2x+8)x^{-\frac{1}{2}}-\frac{1}{2}(x^2+8x+3)x^{-\frac{3}{2}}=x^{-\frac{3}{2}}\left[(2x+8)x-\frac{1}{2}(x^2+8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Problem solving a word problem using a generating function
How many ways are there to hand out 24 cookies to 3 children so that they each get an even number, and they each get at least 2 and no more than 10? Use generating functions.
So the first couple steps are easy.
The coefficient is $x^{24}$
$g(x) = x^6(1+x^2+x... | Here is variation of the theme which might be helpful when doing the calculation. It's convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.
We obtain
\begin{align*}
[x^{24}]&(x^2+x^4+x^6+x^8+x^{10})^3\tag{1}\\
&=[x^{24}]x^6(1+x^2+x^4+x^6+x^8)^3\\
&=[x^{18}]\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the rightmost digit of: $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n$ Find the rightmost digit of: $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n(n$ arbitrary positive integer)
First of all I checked a few cases for small $n$'s and in all cases the rightmost digit was $5$, so maybe this is the case for all values of $n$.
Then I tho... | Modulo $10$ easy inductions show that
$$\begin{align*}
&1^n+9^n\equiv\begin{cases}
0,&\text{if }n\text{ is odd}\\
2,&\text{if }n\text{ is even}\;,
\end{cases}\\
&2^n+8^n\equiv\begin{cases}
0,&\text{if }n\text{ is odd}\\
8,&\text{if }n\equiv 2\pmod4\\
2,&\text{if }n\equiv 4\pmod4\;,
\end{cases}\\
&3^n+7^n\equiv\begin{ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
A probability question that uses the binomial expansion The question is as follow:
(i) Find the binomial expansion of $(1-x)^{-3}$ up to and including $x^{4}$.
(ii)
A player throws a 6-sided fair die at random. If he gets an even number, he loses the game and the game ends. If he gets a "1", "3" or "5" he throws the d... | Binomial expansion of
$$(1 + x)^a = 1 + ax + \frac{a(a-1)}{2!} x^2 + \frac{a(a-1)(a-2)}{3!} x^3 + \frac{a(a-1)(a-2)(a-3)}{4!} x^4 + {\cal O}(x^5)$$
when $|x| < 1$.
Part 2) I have just written out the pattern of him winning in 3 steps, 4 steps, 5 steps, etc.
$$ \Pr(Win) = \frac{1}{3^3} \bigg[ 1 + \Big(\frac{1}{6}\Big) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
A curious algebraic fraction that converges to $\frac{\sqrt{2}}{2}$ I have noticed that the algebraic fraction
$\frac{3a+2b}{4a+3b} $
Gives better and better approximations to $\sin 45^\circ = \frac{\sqrt{2}}{2} $
For $ a = b = 1$ we get $5/7 \approx 0.714 $
Now, taking $ a = 5, b = 7$, we get $ 29/41 \approx 0.707... | I think we can split the ratio $(3a + 2b)/(4a + 3b)$ further. Let $a/b$ be a rational approximation to $1/\sqrt{2}$. Then I prove that $(a + b)/(2a + b)$ is a better approximation to $1/\sqrt{2}$ but in opposite direction.
Clearly we have $$\left(\frac{a + b}{2a + b}\right)^{2} - \frac{1}{2} = \frac{b^{2} - 2a^{2}}{2(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1817854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
} |
Limit of gamma and digamma function In my answer of the previous OP, I'm able to prove that
\begin{align}
I(a)&=\int_0^\infty e^{-(a-2)x}\cdot\frac{1-e^{-x}(1+x)}{x(1-e^{x})(e^{x}+e^{-x})}dx\tag1\\[10pt]
&=\int_0^1\frac{y^{a-1}}{1+y^2}\left(\frac{y}{1-y}+\frac{1}{\log y}\right)dy\tag2\\[10pt]
&=\log\Gamma\!\left(\frac{... | We have that $$\log\left(\Gamma\left(x\right)\right)\sim x\log\left(x\right)-x-\frac{1}{2}\log\left(\frac{x}{2\pi}\right)+O\left(\frac{1}{x}\right)
$$ and $$\psi\left(x\right)\sim\log\left(x\right)+O\left(\frac{1}{x}\right)
$$ as $x\rightarrow\infty$ (see here and here) hence we have to evaluate $$\begin{align}
&\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1818207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solve congruence Solve:$$ \underbrace{2 ^ {2 ^ { {...} ^ 2 }}}_\text{2016} \pmod {2016}$$
So $ 2016 = 2^5 \cdot 3^2 \cdot 7$
And $$ \underbrace{2 ^ {2 ^ { {...} ^ 2 }}}_\text{2016} \pmod{2^5} \rightarrow \underbrace{2 ^ {2 ^ { {...} ^ 2 }}}_\text{2016} \pmod{2016} $$
$$ \underbrace{2 ^ {2 ^ { {...} ^ 2 }}}_\text{2016} ... | The lambda-function : https://en.wikipedia.org/wiki/Carmichael_function
helps us to find a quick answer. Because of $\lambda(2016)=24$ and considering $2016=2^5\cdot 3^2\cdot 7$, we can conclude $a^k=a^{k+m\cdot 24}$ for all $a$ and $m$, when $k\ge 5$.
The given number can be written by $2^M$, where $M$ is a power towe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
On the composition of formal power series I an attempt to compute the coefficients of the composition $f(g(x))$ of two power series $f(x) = \frac{1}{1-x}$ and $g(x) = \frac{1}{1-x}-1$, I used the definition of composition to get to
$$f(g(x)) = \sum_{n=0}^{\infty} \left(\frac{1}{1-x} - 1\right)^n = \sum_{n=0}^{\infty} ... | We have $$f(g(x)) = \sum_{k = 0}^\infty \left( \dfrac x {1 - x} \right)^k .$$
Now, for $k \ge 1$, using $\dfrac 1 {(1 - x)^k} = \sum\limits_{m = 0}^\infty \binom{m + k- 1}{m} x^m$, we get
\begin{align*}
f(g(x)) & = 1 + \sum_{k=1}^\infty x^k \sum_m \binom{m + k - 1}{m} x^m\\
& = 1 + \sum_{k=1}^{\infty} \sum_m \binom{m +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove if $n \in \mathbb N$, then $\frac{1}{2!}+\cdots+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$
Prove if $n \in \mathbb N$, then $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{n}{\left(n+1\right)!} = 1-\frac{1}{\left(n+1\right)!}$
So I proved the base case where $n=1$ and got $\frac{1}{2}$
Then since $n=k$ implies $... | The equality you want to prove is
$$
\underbrace{\frac{1}{2!}+\dots+\frac{k}{(k+1)!}}_{*}+
\frac{k+1}{(k+2)!}=1-\frac{1}{(k+2)!}
$$
The term marked $*$ is equal, by the induction hypothesis, to
$$
1-\frac{1}{(k+1)!}
$$
and so you need to manipulate
$$
1-\frac{1}{(k+1)!}+\frac{k+1}{(k+2)!}
$$
Hint:
$$
1-\frac{1}{(k+1)!}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1821557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Solutions to a linear system of equations. I want to find the solution of this system when the parameter $a \in R$ varies.
\begin{cases} (a+2)x_2 + x_4 = 1 \\ -x_1 +x_3 = a+1 \\ (a+1)x_1 + 2x_2 -x_3 = 0 \\ x_1 -2x_2 -(a+1)x_3 = -2 \end{cases}
I notice that the matrix associated with the homogeneous system has rank $=... | Transforming the system of equations into an augmented matrix, we have:
$$\begin{bmatrix} 0 & (a+2) & 0 & 1 &|& 1 \\ -1 & 0 & 1 & 0 &|& (a+1) \\ (a+1) & 2 & -1 & 0 &|& 0 \\ 1 & -2 & -(a+1) & 0 &|& -2\end{bmatrix}$$
(note that the vertical lines between columns 4 and 5 seperate the sides of the equations). Row reduction... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1823877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that $\lim\limits_{x\to \infty} a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}=0$ if and only if $ a+b+c=0$ Prove that
$$ \displaystyle \lim_{x\to\infty } \left({a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}}\right)=0$$
$$\text{if and only if}$$
$$ a+b+c=0.$$. I tried to prove that if $a+b+c=0$, the limit is $0$ first, but after ge... | Lessee...
first prove $\lim_{x\rightarrow \infty}\frac{\sqrt {x + j}}{\sqrt{x+k}}= 1$ which should be easy (albeit tedious) enough. [$\frac{\sqrt{x + j}}{\sqrt{x + k}} =\frac{ \sqrt{1 + j/x}}{\sqrt{1 + k/x}}$ so limit is 1].
$\lim_{x\rightarrow \infty}a\sqrt{x + 1} + b \sqrt{x+2} + c\sqrt{x+3} =$
$\lim_{x\rightarrow \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Factoring the sequence ${10}^{2n}+10^{n}+1$ While I am waiting for the basketball NBA game between Cleveland Cavaliers and Golden State Warriors to begin I sort of played with the sequence $a_n={10}^{2n}+10^{n}+1$ in a way that I looked for the prime factors for small values of $n$.
For $n=1$ we have $111=3 \cdot 37$.
... | $a_n$ is divisible by $p^2$ if $x = 10^n$ is a solution of $x^2 + x + 1 \equiv 0 \mod p^2$. We may assume $p$ is not $2$ or $5$, so $x$ is coprime to $p$. Now $4 (x^2 + x + 1) = (2x + 1)^2 + 3$, so $2x + 1$ must be a square root of $-3$ mod $p^2$. If $p$ is an odd prime (other than $3$) for which $-3$ is a quadratic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\int_{0}^{\infty}{1\over x^4+x^2+1}dx=\int_{0}^{\infty}{1\over x^8+x^4+1}dx$ Let
$$I=\int_{0}^{\infty}{1\over x^4+x^2+1}dx\tag1$$
$$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx\tag2$$
Prove that $I=J={\pi \over 2\sqrt3}$
Sub: $x=\tan{u}\rightarrow dx=\sec^2{u}du$
$x=\infty \rightarrow u={\pi\over 2}$, $x=0\... | \begin{align}
\int_0^{\infty}\frac{\mathrm dx}{a^2x^4+bx^2+c^2}
&= \int_0^{\infty}\frac{\mathrm dx}{\left(ax-\frac{c}{x}\right)^2+b+2ac}\cdot \frac{1}{x^2}\\[9pt]
&=\frac{c}{a} \underbrace{\int_0^{\infty}\frac{\mathrm dy}{\left(ay-\frac{c}{y}\right)^2+b+2ac}}_{\large\color{blue}{x=\frac{c}{ay}}}\\[9pt]
&=\frac{c}{a} \u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 9,
"answer_id": 1
} |
The largest root of a recursively defined polynomial Suppose that for all $x \in \mathbb{R}$, $f_1(x)=x^2$ and for all $k \in \mathbb{N}$,
$$
f_{k+1}(x) = f_k(x) - f_k'(x) x (1-x).
$$
Let $\underline{x}_k$ denote the largest root of $f_k(x)=0$.
I want to prove the following conjectures:
*
*$0=\underline{x}_1 < \un... | From
$f_{k+1}(x) = f_k(x) - f_k'(x) x (1-x)
$,
let
$f_k(x)
=x^2 g_k(x)
$,
so
$f_k'(x)
=x^2 g_k'(x)+2xg_k(x)
=x(x g_k'(x)+2g_k(x))
$.
Then
$\begin{array}\\
x^2g_{k+1}(x)
&= x^2g_k(x) - x (1-x)x(x g_k'(x)+2g_k(x))\\
&= x^2(g_k(x) - (1-x)(x g_k'(x)+2g_k(x)))\\
\text{so}\\
g_{k+1}(x)
&= g_k(x) - (1-x)(x g_k'(x)+2g_k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Can't find minimum using Lagrange multipliers I want to find the minimum of the function $f(x,y) = x + y^2$ with the constraint $2x^2 +y^2 = 1$.
Here are my partial derivatives:
$$f_x = 1$$
$$f_y = 2y$$
$$g_x = 4x$$
$$g_y = 2y$$
I have the following system of equations:
\begin{align*}
1 = \lambda4x\\
2y = 2y\lambda\\
2... | There's no need to use a Lagrange multiplier. The ellipse
$$\{ (x,y) \in \mathbb{R}^2 \mid 2 x^2 + y^2 = 1 \}$$
is parametrized as follows
$$x (\theta) = \frac{\sqrt 2}{2} \, \cos (\theta) \qquad \qquad \qquad y (\theta) = \sin (\theta)$$
Hence,
$$g (\theta) := f (x (\theta), y (\theta)) = \frac{\sqrt 2}{2} \, \cos (\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Find the values of a and b so that $x^4+x^3+8x^2+ax+b$ is exactly divisible by $x^2+1$ I have been trying this question for a long time but I am not getting it. So please help me and try to make it as fast as possible
| You simply write out a skeleton $(x^4+x^3+8x^2+ax+b)=(x^2+1)(cx^2+dx+e)$.
(1) comparing the $x^4$ term we must have $c=1$.
(2) comparing the $x^3$ term we have $d=1$.
(3) comparing the constant term $e=b$.
So we now have $(x^4+x^3+8x^2+ax+b)=(x^2+1)(x^2+x+b)$.
(4) comparing the $x^2$ term $8=b+1$, so $b=7$.
(5) compari... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
If the sides of a triangle satisfy $(a-c)(a+c)^2+bc(a+c)=ab^2$, and if one angle is $48^\circ$, then find the other angles.
In triangle $ABC$ one angle of which is $48^{\circ}$, length of the sides satisfy the equality:
$$(a-c)(a+c)^2+bc(a+c)=ab^2$$
Find the value in degrees the other two angles of the triangle.
... | From $c^2-a^2=ab,$
using $(i)$Sine Law,
$(ii)$Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,
and $(iii)\sin(C+A)=\cdots=\sin B$
$$\sin B\sin(C-A)=\sin A\sin B$$
As $\sin B\ne0,\implies\sin(C-A)=\sin A$
$\implies C-A=n\pi+(-1)^nA$ where $n$ is any integer
If $n$ is odd, $n=2m+1$(say) $\implies C-A=(2m+1)\pi+(-1)A$
$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
For all positive integer $n$ prove the equality: $\sum_{k=0}^{n-1}\frac{\binom{n-1}{k}^2}{k+1}=\frac{\binom{2n}{n}}{2n}$
For all positive integer $n$ prove the equality:
$$\sum_{k=0}^{n-1}\frac{\binom{n-1}{k}^2}{k+1}=\frac{\binom{2n}{n}}{2n}$$
My work so far:
$$\frac{n\binom{n-1}{k}}{k+1}=\frac{n(n-1)!}{(k+1)k!(n-k-1... | Another approach.
$$ \sum_{k=0}^{n-1}\binom{n-1}{k}^2\frac{1}{k+1}=\int_{0}^{1}\color{blue}{\sum_{k=0}^{n-1}\binom{n-1}{k}\binom{n-1}{n-1-k}x^k}\,dx \tag{1}$$
and the blue term is the coefficient of $y^{n-1}$ in the following product:
$$\left(\sum_{k=0}^{n-1}\binom{n-1}{k}x^k y^k\right)\cdot\left(\sum_{k=0}^{n-1}\binom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
What is the fastest method to find which of $\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ and $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $ is bigger manually? What is the fastest method to find which number is bigger manually?
$\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ or $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $
| Claim:
$$\frac {3\sqrt {3}-4}{7-2\sqrt {3}}<\frac {3\sqrt {3}-8}{1-2\sqrt {3}}$$
Proof:
Let $u=\sqrt 3-1$. Notice $1>u>\frac12$.
If
$$\frac{3u-1}{5-2u}<\frac{3u-5}{-2u-1}$$
$$\Leftarrow-6u^2-u+1>-6u^2+25u-25$$
$$\Leftarrow26>26u$$
$$\Leftarrow1>u=\sqrt3-1$$
which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Solve definite integral: $\int_{-1}^{1}\arctan(\sqrt{x+2})\ dx$ I need to solve:
$$\int_{-1}^{1}\arctan(\sqrt{x+2})\ dx$$
Here is my steps, first of all consider just the indefinite integral:
$$\int \arctan(\sqrt{x+2})dx = \int \arctan(\sqrt{x+2}) \cdot 1\ dx$$
$$f(x) = \arctan(\sqrt{x+2})$$
$$f'(x) = \frac{1}{1+x+2} \... | It is sometimes better to take $x+a$ (for some constant $a$) as the antiderivative of $1$. In this case
$$
\int \arctan\sqrt{x+2}\,dx=(x+a)\arctan\sqrt{x+2}-\frac{1}{2}\int(x+a)\frac{1}{1+x+2}\cdot\frac{1}{\sqrt{x+2}}\,dx,
$$
so $a=3$ seems to be a good choice.
I think you can continue from here.
| {
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"url": "https://math.stackexchange.com/questions/1835533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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There are 10 boxes, 15 balls; 10 red, 5 blue. Each is randomly placed in a box in an independent manner. What's E[X=the number of empty boxes?] There are 10 boxes, 15 balls; 10 red, 5 blue.
Each is randomly placed in a box in an independent manner. The red balls are placed in boxes 1-10, blue balls are placed in 1-6. ... | Let $A_i=1$ if box $i$ is empty, 0 if not. Consider the four boxes numbered 7-10. They can only get red balls. For each one we have $p(A_i=1)=\left(\frac{9}{10}\right)^{10}$. So $E(A_i)=\left(\frac{9}{10}\right)^{10}$. But expectation is linear, so $E(\sum_7^{10}A_i)=4\left(\frac{9}{10}\right)^{10}\approx1.395 $.
The r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Closed form for $\prod\limits_{l=1}^\infty \cos\frac{x}{3^l}$ Is there any closed form for the infinite product $\prod_{l=1}^\infty \cos\dfrac{x}{3^l}$? I think it is convergent for any $x\in\mathbb{R}$.
I think there might be one because there is a closed form for $\prod_{l=1}^\infty\cos\dfrac{x}{2^l}$ if I'm not wron... | We follow the method of this paper, using the Fourier transform.
Following his normalizations,
$$
\hat f (\omega) = \frac{1}{2\pi} \int_{\mathbb R} f(x) e^{-i \omega x} \, dx, \qquad f(x) = \int_{\mathbb R} \hat f(\omega) e^{i \omega x} \, d\omega,
$$
the key facts to be used are:
*
*The identity $\cos(bx) = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Prove that $\sqrt{n+1}>\sqrt{n}+\frac{1}{2\sqrt{n}}-\frac{1}{8n\sqrt{n}}$
Prove that $\sqrt{n+1}>\sqrt{n}+\frac{1}{2\sqrt{n}}-\frac{1}{8n\sqrt{n}}$ if $n>0$.
I didn't see an easy way of proving this without doing a lot of algebra and rearranging. Is there an easier way?
| It does not look so terrible to me! We have:
$$ \sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}} \tag{1}$$
and:
$$ \frac{1}{2\sqrt{n}}-\frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n}\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{1}{2\sqrt{n}\left(\sqrt{n}+\sqrt{n+1}\right)^2}\tag{2}$$
so the inequality b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1839351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How to get the correct angle of the ellipse after approximation I need to get the correct angle of rotation of the ellipses. These ellipses are examples. I have a canonical coefficients of the equation of the five points.
$$Ax ^ 2 + Bxy + Cy ^ 2 + Dx + Ey + F = 0$$
Ellipses:
Points:
Zero ellipse: [16,46] [44,19] ... | There're two principal axes in general, so
\begin{align*}
\theta &=\frac{1}{2} \tan^{-1} \frac{B}{A-C}+\frac{n\pi}{2} \\
&= \tan^{-1}
\left(
\frac{C-A}{B} \color{red}{\pm} \frac{\sqrt{(A-C)^{2}+B^{2}}}{B} \:
\right) \\
\end{align*}
The centre is given by $$(h,k)=
\left(
\frac{2CD-BE}{B^2-4AC}, \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1839510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Is it really necessary to learn how to write a quadratic in standard to vertex form? How crucial is this skill or form of writing and polynomial function? Can't we just always use the $-b/2a$ trick for the $x$ intercept and just plug it back into the function to find the $y$?
| To answer your question, it isn't crucial in any serious way. However, it is as handy as, say, "point-slope form" or "slope-intercept form" for lines, it gives you a way to immediately recognize certain bits of information. Additionally, as Daniel M. mentions, it covers completing the square which is more often than no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1839822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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How can I solve this nice rational equation I am trying solve this equation
$$\dfrac{3x^2 + 4x + 5}{\sqrt{5x^2 + 4x +3}}+\dfrac{8x^2 + 9x + 10}{\sqrt{10x^2 + 9x +8}} = 5.$$ Where $x \in \mathbb{R}$. I knew that $x=-1$ is a given solution. But I can't solve it. I tried
We rewrite the given equation in the form
$$\dfrac{... | Hint : Square the equation , isolate the term with the square-root and square the equation again.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
Summation Notation Question in McMillan's Theorem Proof Let me preface by saying that this question does not pertain as much to coding theory, as it does to mathematical notation. Every symbol in this question is a natural number.
Anyhow, I am currently reading through a proof of McMillan's Theorem. In the proof, there... | Would this help?
\begin{align}
\left( \sum_{j=1}^2 \dfrac{\alpha_j}{r^j} \right)^3
&= \left( \dfrac{\alpha_1}{r}+\dfrac{\alpha_2}{r^2} \right)^3 \\
&= \left( \dfrac{\alpha_1}{r^1}+\dfrac{\alpha_2}{r^2} \right)
\left( \dfrac{\alpha_1}{r^1}+\dfrac{\alpha_2}{r^2} \right)
\left( \dfrac{\alpha_1}{r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Summing $3+7+14+24+37...$ up to $10$ terms
What is $3+7+14+24+37...$ up to $10$ terms?
I only see that difference between two consecutive terms is in AP ie $7-3=4,14-7=7,24-14=10$ and so on. Any ideas on how to do it?
| This answer seeks to expand on the "one can check that" comment in Foobaz's answer. First, let $A=\{3,7,14,24,37,\ldots\}$. The pattern for the difference, $d_n$, between terms $n$ and $n+1$ is clear enough:
$$
d_n=4+(n-1)3=1+3n.
$$
Hence,
$$
a_{n+1}=\frac{n}{2}(5+3n)+3
$$
or, more usefully,
\begin{align}
a_n&=\frac{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Questions regarding dot product (possible textbook mistake) I am given the following exercise:
Show that $\Vert \overrightarrow{a} + \overrightarrow{b} \Vert = \Vert
\overrightarrow{a} \Vert + \Vert \overrightarrow{b} \Vert $ if and
only if $\overrightarrow{a}$ and $\overrightarrow{b}$ are parallel and
point to t... | The first question is valid. For instance, if we let $0\neq b = -a$, then $||a+b|| = 0 \neq ||a||+||b||.$ So, $a$ and $b$ must point in the same direction.
As for a proof of the first statement: going backwards, assume $a$ and $b$ point in the same direction, i.e. (without loss of generality) $b = ca$ for some scalar ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to find the determinant of a 5 x 5 matrix? Finding a 3x3 matrix is easy, but how can I find the determinant of this 5x5 matrix?? I just need an example of the first couple steps to mimic
$A =$
$\begin{bmatrix} 7&1&9&-4&3\\0&-3&4&9&-6\\0&0&-6&-6&-9\\0&0&0&7&6\\0&0&0&0&2\end{bmatrix}$
then the $\det(A) = ?$
By th... | Developing with respect the last row, you get that the determinant equals
$$2\cdot\begin{vmatrix}7&1&9&\!-4\\0&\!-3&4&9\\0&0&\!-6&\!-6\\0&0&0&7\end{vmatrix}$$
Again deloping the last row we get
$$2\cdot7\cdot\begin{vmatrix}7&1&9\\0&\!-3&4\\0&0&\!-6\end{vmatrix}=2\cdot7\cdot(-6)\cdot\begin{vmatrix}7&1\\0&\!-3\end{vmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1848861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving differential equation of third degree If differential equation of the curves $$c(y+c)^2 = x^3$$ where 'c' is an arbitary constant is $$12y(y')^2 + ax = bx(y')^3$$
What is the value of a+b?
I tried differentiating the curve given and obtain a linear relation with the equation.
I got $$cyy'y'' + c(y')^3 + c^2y'y'... | $$c(y+c)^2=x^3$$
Differentiate :
$$2c(y+c)y'=3x^2$$
$2c(y+c)^2y'=3x^2(y+c)=2x^3y'$
$$y+c=\frac{2}{3}xy'\quad\to\quad c=\frac{2}{3}xy'-y$$
$c(y+c)^2=x^3=(\frac{2}{3}xy'-y)(\frac{2}{3}xy')^2$
$x=\frac{8}{27}xy'^3-\frac{4}{9}yy'^2$
$-\frac{8}{27}xy'^3+\frac{4}{9}yy'^2+x=0$
$$12yy'^2+27x-8xy'^3=0$$
Compared to
$$12yy'^2+a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\sin \frac{\theta}{2} + \cos \frac{\theta}{2} = \sqrt{2}$ Does anyone have some tips for me how to go about the problem in the image?
$$\sin \frac{\theta}{2} + \cos \frac{\theta}{2} = \sqrt{2}$$
I know it's supposed to be simple, but I can't figure out why the solution is:
$90^{\circ}+ 720^{\circ}k$ where $k... | Note that: $$\sin \frac{\theta}{2} + \cos \frac{\theta}{2} = \sqrt{2} \cos \left(\frac{\theta}{2} - \frac{\pi}{4} \right) \quad \quad \quad \quad (\star)$$
So your equation reduces down to $\cos \left(\frac{\theta}{2} - \frac{\pi}{4}\right) = 1 \iff \frac{\theta}{2} - \frac{\pi}{4} = 2\pi k$ where $k$ ranges over the i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1849934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Evaluation of $\int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$
$$\int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$$
$\bf{My\; Try::}$ Let $$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{... | your change of variable is wrong.
$$\int\frac{1}{t(1+t^{2})}dt=\int\frac{1}{t^{3}\left(1+\frac{1}{t^{2}}\right)}dt
$$
and now we use the change of variable $u=\frac{1}{t^{2}} $ => $du=-\frac{dt}{t^{3}}$,then the integral take the form:
$$-\int\frac{1}{1+u}du=-\ln(1+u)=\ln\left(\frac{1}{1+u}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1850568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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$\prod\limits_{k=0 \atop 2k+1\ne 0 \bmod n}^\infty \frac{2k+1+(-1)^{\lfloor (2k+1)/n\rfloor}}{2k+1}=\sqrt{n}$ for $n\ge 2$ I found tasks in an old script. For the following problem I have no idea how to solve.
How can one prove $$\prod\limits_{k=0 \atop 2k+1\ne 0 \bmod n}^\infty \frac{2k+1+(-1)^{\lfloor (2k+1)/n\rfloor... | Literature: Sierpinski Oeuvres Choisies I (page 222-226)
http://www.plouffe.fr/simon/math/Sierpinski%20Oeuvres%20Choisies%20I.pdf
Gamma function product: $\displaystyle \Gamma(1+x)=\lim\limits_{m\to\infty} m^x/\prod\limits_{k=1}^m \left(1+\frac{x}{k}\right)$
or written äquivalent $\displaystyle m^x/\prod\limits_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1850808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Concyclicity of $4$ points using algebraic geometry
Consider the line $L_1:ax+4y-1=0$ and a circle $S:x^2+y^2-10x+2y+10=0$. The line intersects the circle at $2$ distinct points $A$ and $B$. Another line $5x-12y-67=0$ intersects the circle $x^2+y^2+6x+14y-28=0$ at $2$ distinct points $C$ and $D$. Find the value of $a$... | HINT:
The equation of the circle passing through the intersection of $L_1, S$
$$x^2+y^2-10x+2y+10+A(ax+4y-1)=0\ \ \ \ (1)$$
Similarly for $5x-12y-67=0, x^2+y^2+6x+14y-28=0$
$$x^2+y^2+6x+14y-28+B(5x-12y-67)=0\ \ \ \ (2)$$
$A,B,C,D$ will be concyclic if $(1)$ & $(2)$ become identical.
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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number of non differentiable points in $g(x) = \tan \left(\frac{1}{2}\arcsin\left(\frac{2f(x)}{1+(f(x))^2}\right)\right)$
If $\displaystyle f(x) = \lim_{n\rightarrow \infty}\frac{x^2+2(x+1)^{2n}}{(x+1)^{2n+1}+x^2+1}\;,n\in \mathbb{N}$
and $\displaystyle g(x) = \tan \left(\frac{1}{2}\arcsin\left(\frac{2f(x)}{1+(f(x)... | if you plot function f, you will see $-2 \lt f \lt 2$ with 2 discontinuity points at -2, 0, which are not differentiable.
Next you can say the following x are not differentiable, $$\arcsin\left(\frac{2f(x)}{1+(f(x))^2}\right) = \pm n\pi$$
This leads to the points satifying the following equation.
$$\frac{2f(x)}{1+(f(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $x^2+\frac{1}{2x}=\cos \theta$, evaluate $x^6+\frac{1}{2x^3}$. If $x^2+\frac{1}{2x}=\cos \theta$, then find the value of $x^6+\frac{1}{2x^3}$.
If we cube both sides, then we get $x^6+\frac{1}{8x^3}+\frac{3x}{2} \cdot \cos \theta=\cos ^3 \theta$ but how can we use it to deduce required value?
| Let $c = \cos \theta$. You have $x^3 = c x - 1/2$, so $x^6 = (c x - 1/2)^2$ while $1/x = 2 c - 2 x^2$. Then I get
$$\eqalign{x^6 + \dfrac{1}{2x^3} &= - 3 c^2 x^2 - 3 c x + 4 c^3 - \frac{3}{4}\cr & =
-\dfrac{3}{2} (1 + \cos(2\theta)) x^2 - 3 \cos(\theta) x + \cos(3\theta) + 3 \cos(\theta) - \dfrac{3}{4}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
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Integrate $\int \frac{x^2-2}{(x^2+2)^3}dx$ $$\int \frac{x^2-2}{(x^2+2)^3}dx$$
What I did :
Method $(1) $ Re writing $x^2-2 = (x^2+2)-4 $ and partial fractions.
Method $(2) $ Substituting $x^2 = 2\tan^2 \theta $
Is there any other easy methods ?
Some substitution ?
| $$\int \frac { x^{ 2 }-2 }{ \left( x^{ 2 }+2 \right) ^{ 3 } } dx=\int { \frac { { x }^{ 2 }+2-4 }{ { \left( x^{ 2 }+2 \right) }^{ 3 } } } dx=\underbrace { \int { \frac { dx }{ { \left( x^{ 2 }+2 \right) }^{ 2 } } } }_{ { I }_{ 1 } } -4\underbrace { \int { \frac { dx }{ { \left( x^{ 2 }+2 \right) }^{ 3 } } } }_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Is there a difference between $\sqrt{x+2}+x=0$ and $x^2-x-2=0$ Is there a difference between $\sqrt{x+2}+x=0$ and $x^2-x-2=0$
Solutions are $x=2$ or $x=-1$.
But $x=2$ does not satisfy $\sqrt{x+2}+x=0$.Because $\sqrt{4}+2 \neq0$
So does it mean that they are different ? Why ?
| Yes. By convention $\sqrt{x+2}$ is non-negative.
You can multiply the first by $\sqrt{x+2}-x$ to get $x+2-x^2=0$. You introduced the solution $x=2$ when you assumed $\sqrt{x+2}\sqrt{x+2}=x+2$. This is not true, since $x+2$ can be negative.
Note that this does not violate the fundamental theorem of algebra since you did... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Prove: $x^3+y^3\geq \frac{1}{4}(x+y)^3$
Prove: $x^3+y^3\geq \frac{1}{4}(x+y)^3$ for all $x,y$ positive.
Let's look at
$$\begin{split} &(x-y)^2(x+y)\geq 0 \\
\iff &(x-y)(x+y)(x-y)\geq 0\\
\iff& (x-y)(x^2-y^2)\geq 0 \\
\iff &x^3-xy^2-yx^2+y^3\geq 0\\
\iff & 3x^3+3y^3\geq +3xy^2+3yx^2\\
\iff &3x^3+3y^3\geq (x+y)^3 -x^3... | Using AM-GM Inequality:
$$\frac{x+y}2\ge\sqrt{xy}\iff xy\le \left(\frac{x+y}2\right)^2$$
We have:
\begin{align}
(x+y)^3&=x^3+y^3+3xy(x+y)\\
&\le x^3+y^3+3\left(\frac{x+y}2\right)^2(x+y)\\
&=x^3+y^3+\frac 34 (x+y)^3
\end{align}
So
$$x^3+y^3\ge \frac14 (x+y)^3$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Convergence of the series $\sum \frac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}}$
To prove that nature of the following series : $$\sum \dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}}$$
they use in solution manual :
My questions:
*
*I don't know how to achieve ( * ) could someone complete my... | Let $f(n)=\frac{1}{n^{2/3}+n^{1/3}+A}$ and consider $\phi_m(n)=n^{-m/3}$ for $m=2,3,...$ Then you should be after an asymptotic expnasion for $f$ of the form $$f(n)=\sum a_m\phi_m(n).$$ We have $$a_2=\lim_{n\to\infty}\frac{f(n)}{\phi_2(n)}=1$$ and $$a_3=\lim_{n\to\infty}\frac{f(n)-a_2\phi_2(n)}{\phi_3(n)}=-1$$ and $$a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Integrate $\int\sin^2x\cos4x\,dx$ I'm having a difficult time solving this integral.
I tried integrating by parts:
$\int\sin^2x\cos4x\,dx$
$u=\sin^2x$, $dv=\cos4x\,dx$
I used the power reducing formula to evaluate $\sin^2x$
$du = 2\sin x\cos x\,dx$, $v=1/4\sin4x$
$uv - \int\ v\,du$
$\dfrac{1}{4}\sin^2x\sin4x - \dfrac{... | $$\begin{align*} \sin^2 x \cos 4x &= \sin x (\sin x \cos 4x) \\ &= \frac{1}{2} \sin x (\sin 5x - \sin 3x) \\ &= \frac{1}{4}(\cos 4x - \cos 6x + \cos 4x - \cos 2x) \\ &= \frac{1}{2} \cos 4x - \frac{1}{4} \cos 6x - \frac{1}{4} \cos 2x. \end{align*}$$
Consequently, we immediately and rather trivially obtain $$\int \sin^2 ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Can $x^2+y^2,y^2+z^2,z^2+x^2$ and $x^2+y^2+z^2$ all be square numbers? I know that if we want $x^2+y^2$ to be square number, we are looking for pythagorean triple; if we want $x^2+y^2+z^2$ to be a square number, we are looking for pythagorean quadruple. But have we ever found any positive integers $x,y,z$ such that $x^... | It's a famous open problem : the perfect cuboid problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$f(x)$ is a quadratic polynomial with $f(0)\neq 0$ and $f(f(x)+x)=f(x)(x^2+4x-7)$
$f(x)$ is a quadratic polynomial with $f(0) \neq 0$ and $$f(f(x)+x)=f(x)(x^2+4x-7)$$
It is given that the remainder when $f(x)$ is divided by $(x-1)$ is $3$.
Find the remainder when $f(x)$ is divided by$(x-3)$.
My Attempt:
Let $f(x)=ax... | $ac + b + 8=0$, $ac - a - c + 11 = 0$. If $a = 1$, then $10 = 0$. So $a = -1$, hence $c = 6$, $b = -2$, $f(x) = -x^2 -2x + 6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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How to solve this inequality problem? Given that $a^2 + b^2 = 1$, $c^2 + d^2 = 1$, $p^2 + q^2 = 1$, where $a$, $b$, $c$, $d$, $p$, $q$ are all real numbers, prove that $ab + cd + pq\le \frac{3}{2}$.
| Subtracting 2ab, 2cd and 2pq from the three equations gives us:
$$ a^2+b^2-2ab = 1 - 2ab$$$$
c^2 + d^2 - 2cd = 1 - 2cd$$$$p^2+q^2 - 2pq = 1-2pq$$
Noting that the LHS of each of these equations is a perfect square and all perfect squares are non-negative we have:$$1-2ab≥0$$$$1-2cd≥0$$$$1-2pq≥0$$
Adding the three equatio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Doubt in finding number of integral solutions Problem :
writing $5$ as a sum of at least $2$ positive integers.
Approach :
I am trying to find the coefficient of $x^5$ in the expansion of $(x+x^2+x^3\cdots)^2\cdot(1+x+x^2+x^3+\cdots)^3$ .
which reduces to coefficient of $x^3$ in expansion of $(1-x)^{-5}$ ,which is $$... | There are $2^{5-1} -2^{(2-1)-1}=15$ compositions of $5$ into at least $2$ positive integers while there are $6$ partitions of $5$ into at least $2$ positive integers. With compositions, $1+1+3$ is distinct from $3+1+1$, while with partitions they are the same.
But your first attempt counts even more than compositions.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
integrate $\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx$
$$\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx$$
$$\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx=\int \frac{\sin^2x \cos^2x \sin x }{1+\sin^2x}dx=\int \frac{(1-\cos^2x )\cos^2x \sin x }{1+1-\cos^2x}dx=\int \frac{(1-\cos^2x )\cos^2x \sin x }{2-\cos^2x}dx$$
$u=cosx$
$du=-... | The decomposition in partial fractions should be
$$\frac{1}{u^2-2}==\frac{A}{u +\sqrt{2}}+\frac{B}{u -\sqrt{2}}.$$
But every one should know that
$$\int\frac1{a^2-x^2}\,\mathrm d x=\frac1{2a}\ln\biggl\lvert\frac{x+a}{x-a}\biggr\rvert=\frac1a\,\operatorname{argtanh}\Bigl(\frac xa\Bigr).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
$3(x-\sqrt{xy}+y)^2\geq x^2+xy+y^2$ Prove that $3(x-\sqrt{xy}+y)^2\geq x^2+xy+y^2$ for all $x,y\geq 0$.
Expanding, the inequality becomes
$$3x^2+3xy+3y^2-6x\sqrt{xy}-6y\sqrt{xy}+6xy\geq x^2+xy+y^2$$
which is
$$x^2+4xy+y^2\geq3\sqrt{xy}(x+y)$$
We can try using AM-GM:
$$x^2+xy+xy\geq 3\sqrt[3]{x^4y^2}$$
This is close to ... | $\Leftrightarrow (\sqrt{x}-\sqrt{y})^2(x-\sqrt{xy}+y)\geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How does $\int_{\pi/3}^{\pi/2} \frac{1-\cos^2x}{\sqrt{\sin^2(x/2)}}dx$ simplify to $\int_{\pi/3}^{\pi/2} 4\sin(x/2)\cos^2(x/2)dx$? $$\int_{\large{\frac{\pi}{3}}}^{\large{\frac{\pi}{2}}} \frac{1-\cos^2x}{\sqrt{\sin^2\left(\frac x2\right)}}dx$$
How does the above simplify to the below?
$$\int_{\large{\frac{\pi}{3}}}^{\la... | $$\frac{1-\cos^2 x}{\sin\frac{x}{2}}=\frac{\sin^2 x}{\sin\frac{x}{2}}=\frac{\left(\color{red}{2}\sin\frac{x}{2}\cos\frac{x}{2}\right)^2}{\sin\frac{x}{2}}=\color{red}{4}\cos^2\frac{x}{2}\sin\frac{x}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Evaluate $\int \frac{x}{x^4+5}dx$
$$\int \frac{x}{x^4+5}dx$$
$u=x^2$
$du=2xdx\Rightarrow \frac{du}{2}=xdx$
$$\int \frac{x}{x^4+5}=\frac{1}{2}\int \frac{du}{u^2+5}$$
I want to get to the expression in the form of $\frac{da}{a^2+1}$ so I factor out $5$ to get to:
$$\frac{1}{2}\int \frac{du}{5[(\frac{u}{\sqrt{5}})^2+1]}... | The First is correct. Let $t=u/\sqrt5$ then $dt=du/\sqrt5$:
$$\frac 12 \frac 1{\sqrt5}\int\frac {dt}{1+t^2}=\frac 12 \frac 1{\sqrt5}\tan^{-1}(t)+c=\frac 12 \frac 1{\sqrt5}\tan^{-1}(\frac {x^2}{\sqrt5})+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Combinatorial identity's algebraic proof without induction. How would you prove this combinatorial idenetity algebraically without induction?
$$\sum_{k=0}^n { x+k \choose k} = { x+n+1\choose n }$$
Thanks.
| Suppose we seek to verify that
$$\sum_{k=0}^n {q+k\choose k} = {q+n+1\choose n}.$$
The difficulty here lies in the fact that the binomial coefficients on
the LHS do not have an upper bound for the sum wired into them. We use
an Iverson bracket to get around this:
$$[[0\le k\le n]]
= \frac{1}{2\pi i}
\int_{|w|=\gamma} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1868416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find all triples satisfying an equation Another question I saw recently:
Find all triples of positive integers $(a,b,c)$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$.
Can someone help me with it?
| As $\frac{1}{n}>0$ for all natural $n$, all of $a,b,c$ have to be at least two.
Let $a\leq b\leq c$. Unless $a=b=c=3$, $a=2$. Then we have $\frac{1}{b}+\frac{1}{c}=\frac{1}{2}$, so similarly either $b=c=4$ or $b=3$. In the latter case, we get $c=6$.
So, $(a,b,c)=(3,3,3),(2,4,4),(2,3,6)$ are all the solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Integral $\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$ $$\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$$
$$\tan x + \cot x + \csc x + \sec x=\frac{\sin x + 1}{\cos x} +\frac{\cos x + 1}{\sin x} $$
$$= \frac{\sin x +\cos x +1}{\sin x \cos x}$$
$$t= \tan {\frac{x}{2}}$$
On solving ,
$$\frac{1}{\tan x + \cot x +... | A trigonometric formula can be used：
\begin{align}
&(\sin x+\cos x+1)(\sin x+\cos x-1)=\sin 2x\\
I&=\int\frac{\sin x\cos x}{\sin x+\cos x+1}dx=\int\frac{1}{2}(\sin x+\cos x-1)dx=\frac{1}{2}(-\cos x+\sin x-x)+C\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Show that $a_n$ is decreasing
$a_1 = 2, a_{n+1} = \frac{1}{3 - a_n}$ for $n \ge 2$. Show $a_n$ is decreasing.
First we need to show $a_n > 0$ for all $n$.
$a_2 = 1/2$ and $a_3 = 2/5$ and $a_4 = 5/13$
One way we can do this is by showing $3- a_n > 0$. Thus suppose it holds for $n$ then we need to show $\frac{3(3 - a_... | First compute, $a_2=\frac{1}{3-2}=1\leq a_1<3$. Then, notice that
$$
a_{n+1}-a_{n}=\frac{1}{3-a_n}-\frac{1}{3-a_{n-1}}=\frac{a_n-a_{n-1}}{(3-a_n)(3-a_{n-1})}\leq 0
$$
which gives you the induction step for a strong induction argument for a dual hypothesis: $a_n< 3$ and $a_{n}\leq a_{n+1}$ for all $n\geq 1$. Note that i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to solve $y'' + y = -2\sin(x)$? I don't know how to find the particular solution of $$y'' + y = -2\sin(x)$$
I started with $$y'' + y = 0$$ to find the homogeneous form $$A\cos(x)+B\sin(x)$$
But now i am stuck.
| You need to find the particular solution.
Let $y_{p}=x(a\cos x+b\sin x)$.
\begin{align*}
y'_{p} &= a\cos x+b\sin x+x(-a\sin x+b\cos x) \\
y''_{p} &= 2(-a\sin x+b\cos x)-x(a\cos x+b\sin x) \\
y''_{p}+y_{p} &= 2(-a\sin x+b\cos x) \\
-2\sin x &= 2(-a\sin x+b\cos x) \\
(a,b) &= (1,0) \\
y(x) &= A\cos x+B\sin x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Quadratic Inequality in terms of variable $x$
Find the values of $a$ for which the inequality $x^2+ax+a^2+6a<0\;\forall x \in (1,2)$
$\bf{My\; Try::}$ We can Write Equation as $$x^2+ax+\frac{a^2}{4}+\frac{3a^2}{4}+6a<0$$
So $$\left(x+\frac{a}{2}\right)^2+\frac{3a^2+24a}{4}<0$$
Now how can i solve after that, Help req... | We are interested in $f(x)=x^2+ax+a^2+6a<0\;\forall x \in (1,2)$
$$f(1)=1+a+a^2+6a \leq 0$$
That is $$a^2+7a+1\leq 0$$
$$ \frac{-7-\sqrt{45}}{2}\leq a \leq \frac{-7+\sqrt{45}}{2}$$
Also, we want, $$f(2)=4+2a+a^2+6a \leq 0$$
$$a^2+8a+4 \leq 0$$
$$\frac{-8-\sqrt{48}}{2} \leq a \leq \frac{-8+\sqrt{48}}{2}$$
Hence, overall... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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Show that $3^n+4^n+\dots+(n+2)^n=(n+3)^n$ has no answers for $n\ge 6$. Considering
$$3^n+4^n+\dots+(n+2)^n=(n+3)^n$$
Clearly $n=2$ and $n=3$ are solutions of this equation and this equality does not hold for $n=4$ and $n=5$.
How can I show this equation has no solutions for $n>5$.
Thanks.
| For $n \geq 6$ we have
$$\frac{(n+3)^n}{(n+2)^n} = \left( 1+\frac{1}{n+2} \right)^n = \left( 1+\frac{1}{n+2} \right)^{n+2} \cdot \left(\frac{n+2}{n+3}\right)^2$$
Both factors are monotonely increasing, hence this is at least its value for $n=6$, which is $\left(\frac{9}{8}\right)^6 \approx 2.072\dotsc > 2$.
Since $k \m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Find the value $\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$ The value of $$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$$is equal to
*
*$\frac{\pi}{6}$
*$\frac{\pi}{4}$
*$\frac{\pi}{3}$
*$\frac... | Using the fact that $\tan { \left( \alpha -\beta \right) =\frac { \tan { \alpha -\tan { \beta } } }{ 1+\tan { \alpha \tan { \beta } } } } ,\tan { \left( \tan ^{ -1 }{ \alpha } \right) =\alpha } $$$\tan ^{ -1 } \left( \frac { 1 }{ \sqrt { 2 } } \right) -\tan ^{ -1 } \left( \frac { \sqrt { 5-2{ \sqrt { 6 } ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Integrate $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx$
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx$$
$$f(x) = (\tan \ x)^{\frac{2}{3}}, \ f'(x) = \frac{2}{3} \cdot (\tan \ x)^{-\frac{1}{3}} \cdot \sec^2x$$
$$\therefore \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\fra... | Put $u = \tan x\implies I = \displaystyle \int_{1}^{\sqrt{3}} u^{-\frac{1}{3}}du= \left[\frac{3}{2}u^{\frac{2}{3}}\right]|_{u=1}^{u = \sqrt{3}}= \dfrac{3}{2}\left(\sqrt[3]{3}-1\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\int_{3}^{\infty}\frac{dx}{x^2-x-2}$ $$\int_{3}^{\infty}\frac{dx}{x^2-x-2}=\int_{3}^{\infty}\frac{dx}{(x-2)(x+1)}$$
$$\frac{1}{(x-2)(x+1)}=\frac{A}{(x-2)}+\frac{B}{(x+1)}$$
$$1=Ax+A+Bx-2B$$
$$1=(A+B)x+A-2B$$
$A+B=0\iff A=-B$
$-3B=1$
$B=-\frac{1}{3}$, $A=\frac{1}{3}$
$$\int_{3}^{\infty}\frac{dx}{(x-2)(x+1)}=\f... | Observe $\infty-\infty$ is indeterminate form, so you need to get rid from it by writing:
$$\lim_{t\to \infty}(\frac{1}{3}\ln(t-2)-\frac{1}{3}\ln(t+1))=\lim_{t\to \infty}\frac{1}{3}\ln{\frac{(t-2)}{(t+1)}}=\frac{1}{3}\ln{1}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Laurent series of $f(z) = \frac{z^2+1}{\sin(z)}$ How does one find the first four terms of the Laurent series of $f(z) = \frac{z^2+1}{\sin(z)}$? My approach was this:
$(z^2+1) = f(z)\sin(z) = \left(\sum\limits_{n=-\infty}^\infty c_n z^n \right)\sin(z)=\left(\sum\limits_{n=-\infty}^\infty c_n z^n \right)\left(z-\frac{z^... | For example, very close to zero we have
$$\frac{z^2+1}{\sin z}=\frac{z^2+1}{z-\frac{z^3}{3!}+\frac{z^5}{5!}-\ldots}=\frac{z^2+1}{z\left(1-\frac{z^2}6+\frac{z^4}{120}-\ldots\right)}\stackrel{\text{Devel. of geom. series}}=$$
$$=\left(z+\frac1z\right)\left(1+\frac{z^2}6+\frac{z^4}{36}+\ldots\right)=\frac1z+\frac76z+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Overcounting when doing Counting Problems Problem:
How many ways are there to split a dozen people into 3 teams, where one team has 2 people, and the other two teams have 5 people each?
Attempted Solution:
$$N_\text{Teams} = {12 \choose 2}{10 \choose 5}{ 5 \choose 5} = {12 \choose 2}{10 \choose 5}$$
Arriving at this ... | There's one more (cool, in my opinion) way to look at such problems.
For each possible allocation of people into groups, let's take the same example as in the OP, there are a total of
$$
\binom{12}{2}\binom{10}{5}\binom{5}{5} =16632
$$
allocations, such as (letters are people, numbers are groups):
$$
(1)AB|(2)CDEFG|(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Question on the inequality Question.
prove that if ${ a }_{ 1 },{ a }_{ 2 },...{ a }_{ n }>0$ then $$ \frac { { a }_{ 1 }+{ a }_{ 2 }+...+{ a }_{ n } }{ n } \ge \frac { n }{ \frac { 1 }{ { a }_{ 1 } } +\frac { 1 }{ { a }_{ 2 } } +...+\frac { 1 }{ { a }_{ n } } } $$
Proof
$$\\ \left( { a }_{ 1 }+{ a }_{ 2 }+...+{... | $$(a_1+a_2+...+a_n)(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n})$$ gives clearly $n^2$ terms but each $a_i$ gives $$1+\sum_{i\ne j}\frac{a_i}{a_j}$$ so you have
$$(a_1+a_2+...+a_n)(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n})=(1+1+...+1)+\sum_{i=1}^{n-1}(\frac{a_i}{a_{i+1}}+\frac{a_{i+1}}{a_i})$$
$$(a_1+a_2+...+a_n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove a geometric inequality, if 3 numbers are satisfying the condition I just got this problem, but I have no idea on how to prove that.
Prove that if $x,y,z\in\mathbb{R},\ x,y,z\ge 0$ and $2\cdot(x\cdot z+x\cdot y+y\cdot z)+3\cdot x\cdot y\cdot z = 9$, then $(\sqrt x + \sqrt y + \sqrt z )^4 \ge 72$. This is a geometr... | We need to prove that $x+y+z\geq\sqrt[4]{72}$, where $x$, $y$ and $z$ are non-negatives such that
$$\sum\limits_{cyc}(2x^2y^2+x^2y^2z^2)=9$$
Let $x+y+z<\sqrt[4]{72}$, $x=ka$, $y=kb$ and $z=kc$ such that $k>0$ and $a+b+c=\sqrt[4]{72}$.
Hence, $k<1$ and $9=\sum\limits_{cyc}(2x^2y^2+x^2y^2z^2)=k^4\sum\limits_{cyc}(2a^2b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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A question about substitute equivalent form into limit: $\lim_{x\to \infty} \sqrt{x^2+x+1}-\sqrt{x^2+3x+1} = -1$ When I had the calculus class about the limit, one of my classmate felt confused about this limit:
$$\lim_{x\to \infty} \sqrt{x^2+x+1}-\sqrt{x^2+3x+1} = -1$$
What he thought that since $x^2 > x$ and $x^2 >... | Let our problem be of the form $\displaystyle\lim_{x \to \infty} \sqrt[n]{p(x)} - \sqrt[n]{q(x)}$, where $p$ and $q$ are $n$th degree monic(coefficient of $x^n$ is $1$) polynomials. First of all, without loss of generality, we will rewrite $p$ and $q$ in the forms:
$$
p = (x + a)^n + \sum_{i=0}^{n-1} a_ix^i, q = (x + b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Definite integral and limit $\lim_{n \rightarrow \infty} (n((n+1)I_{n}-\frac{\pi}{4})$ Given $I_{n} = \int_{0}^{1} x^{n} \arctan(x)dx $
Calculate:
$\lim_{n \rightarrow \infty} (n((n+1)I_{n}-\frac{\pi}{4})$
| Integrating by parts we have $$I_{n}=\int_{0}^{1}x^{n}\arctan\left(x\right)dx=\frac{\pi}{4\left(n+1\right)}-\frac{1}{\left(n+1\right)}\int_{0}^{1}\frac{x^{n+1}}{1+x^{2}}dx
$$ $$ \stackrel{x^{2}=u}{=}\frac{\pi}{4\left(n+1\right)}-\frac{1}{2\left(n+1\right)}\int_{0}^{1}\frac{u^{n/2}}{1+u}du
$$ and now since $$ \frac{u^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How can we prove this inequality? Let $a$ , $b$ and $c$ be positive real numbers and $a+b+c=1$ How can we show this inequality?
$$a^2+b^2+c^2+2\sqrt{3abc}\le 1$$
Thanks.
| We have that
$$1=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca),$$
so it suffices to show that
$$ab+bc+ca\geq \sqrt{3abc}.$$
Now $X^2+Y^2+Z^2\geq XY+YZ+XZ$ implies
$$ab+bc+ca=(\sqrt{ab})^2+(\sqrt{bc})^2+(\sqrt{ca})^2
\\\geq b\sqrt{ac}+c\sqrt{ab}+a\sqrt{bc}
=(\sqrt{a}+\sqrt{b}+\sqrt{c})\sqrt{abc}.$$
Hence, we still have to prove tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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An integral involving a the dilogarithmic function Consider a following definite integral:
\begin{equation}
\phi_a(x):=\int\limits_0^x \frac{Li_2(\xi)}{\xi+a} d \xi
\end{equation}
By using the integral representation of the dilogarithmic function and swapping the integration order we have shown that our integral satisf... | Here we are computing the antiderivative in closed form rather than solving the functional equation above. In our calculations we will be transforming the integrand by adding and subtracting terms to it in order to -- at every step-- extract terms which are full derivatives. In here we are using the following identity:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Given a matrix $B$, what is $det(B^4)$? My task is this:
Compute det$(B^4)$, where
$$B =\begin{pmatrix}
1 & 0 & 1\\
1 & 1 & 2\\
1 & 2 & 1
\end{pmatrix}$$
My work so far:
It can be shown that
$B =
\begin{pmatrix}
1 & 0 & 1\\
1 & 1 & 2\\
1 & 2 & 1
\end{pmatrix} \sim \begin{pmatrix}
1 & 0 & 1\\
0 & 1 & 2\\
0 & 0 & -4
\... | The issue here is that you are computing the determinant wrong. $\det(B) = -2$, therefore $\det(B^4) = 16$. To check it, you can use Sarrus rule:
$$\det(B) = b_{11}b_{22}b_{33}+b_{12}b_{23}b_{31}+b_{13}b_{21}b_{32}-b_{11}b_{23}b_{32}-b_{12}b_{21}b_{33}-b_{13}b_{22}b_{31}$$
$$= 1+0+2-4-0-1 = 3-5 = -2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
$5$ divides one of $x,y,z$ in $x^2+y^2=z^2$? 1) If $x$, $y$, and $z$ are positive integers for which $gcd(x,y,z) =1$ and $x^2+y^2=z^2$, show that $3$$∣$$xy$.
2) Now again if $x$, $y$, and $z$ are positive integers for which $gcd(x,y,z) =1$ and $x^2+y^2=z^2$, can a similar result to the one in (part 1) be said mod $5$? ... | Squares are $0,1$ or $-1\bmod 5$
If both $x$ and $y$ are not multiples of $5$ then $x^2\equiv \pm 1$ and $y^2\equiv \pm 1\bmod 5$
This implies $x^2+y^2\equiv -2,0$ or $2\bmod 5$.
Of course, since $x^2+y^2$ is a square it must be $0\bmod 5$ given the previous options.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $7$ Divides $\binom{2^n}{2}+1$ then $n =3k+2$ for some positive integer $k$ It is a straightforward question.
If $$ 7 \text{ }\Bigg | \binom{2^n}{2}+1$$ then $n=3k+2$ for some
positive integer $k$.
This is just curiosity no motivation just rummaging through some old question in a notebook. A simple counter exam... | Something slightly more general: we have ${k\choose 2} \equiv -1 \pmod{7}$ if and only if $k\equiv 4\pmod{7}$.
To see this, write ${k\choose 2} = \frac{k(k-1)}{2}$, so the equation is equivalent to $k^2 - k \equiv -2\pmod{7}$. But $k^2 - k + 2 \equiv (k-4)^2 \pmod{7}$, so this is satisfied exactly when $k\equiv 4\pmod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Prove that if $a+b+c=1$ then $\sum\limits_{cyc}\frac{1}{\sqrt{a^2+b^2}}\le\frac{9\sqrt{2}}{2}$ Let $a,b,c>0$,and such $a+b+c=1$,prove or disprove
$$\dfrac{1}{\sqrt{a^2+b^2}}+\dfrac{1}{\sqrt{b^2+c^2}}+\dfrac{1}{\sqrt{c^2+a^2}}\le\dfrac{9\sqrt{2}}{2}\tag{1}$$
My try:since
$$\sqrt{a^2+b^2}\ge\dfrac{\sqrt{2}}{2}(a+b)$$
it ... | It is false. Try $a=\frac{9}{10},\, b=c=\frac{1}{20}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to compute the Pythagorean triple by one of the numbers that belonged to it? I have a positive number $n>2$. How to compute the Pythagorean triple containing $n$? $n$ may be the hypotenuse and leg.
| So thanks to @Henning Makholm, I realized that it does hold for every integer $n >2$.
Case 1. Suppose $n$ is an odd integer, say $n=2k+1$. ($k\geq1$)
\begin{align*}
n&=2k+1\\
c&=2k^2+2k+1\\
b&=2k^2+2k\\
c^2&=b^2+n^2
\end{align*}
Case 2. Suppose $n$ is two times an odd integer, say $n=4k+2$.($k\geq1$)
\begin{align*}
n&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.