Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to get the value of $a + b + c$? $(0 \leq a < b < c) \in Z$,
$a + b + c + ab + ac + bc + abc = 1622$
$a + b + c = ?$
I've assumed that $a = 0$, by that we can get rid of this part $ab + ac + abc$
Now $bc + b + c = 1622$.
But I found that was useless and got stuck.
| Hint:
$$a+b+c+ab+ac+bc+abc=(1+a)(1+b)(1+c)-1$$
Therefore $(1+a)(1+b)(1+c)=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How do I solve $\lim \limits_{x \to \frac{π}{3}} \frac{2 \sin x - \sqrt{3}}{\cos \frac{3x}{2}}$ Alright, I know, there are easier ways to solve this, like L'hopitals Rule etc.
But I'm not solving it for the answer, just doing it for the fun so I tried using substitution method.
Put $t= x- \dfrac{π}{3}$
$\lim \limits_{x... | So we already have, after putting $\;t:=x-\frac\pi3\;$ :
$$\dfrac{2 \sin \left(t+\frac{π}{3} \right) - \sqrt{3}}{\cos \left( \frac{3t}{2} + \frac{π}{2}\right)}=\frac{2\left(\sin t\cdot\frac12+\frac{\sqrt3}2\cos t\right)-\sqrt3}{-\sin\frac{3t}2}=-\frac{\sin t}{\sin\frac{3t}2}-\sqrt3\frac{\cos t -1}{\sin\frac{3t}2}=$$$${... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Proof that $x^{(n+4)} \bmod 10 = x^n \bmod 10\,$ for $\,n\ge 1$ While solving a programming challenge in which one should efficiently compute the last digit of $a^b$, I noticed that apparently the following holds (for $n > 0$)
$x^{(n+4)} \mod 10 = x^n \mod 10$
How can this be proven?
| We want to show that $x^{n+4}-x^n=x^n(x^4-1)$ is a multiple of $10$.
We first notice that if $x^n$ is odd then $x^4-1$ will be even and vice versa. Consequently, the product will always be even.
If $x$ is divisible by $5$ then the product is divisible by $10$ because we have an even product which is divisible by $5$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 3
} |
Solving $ \left(\frac{ ( x^{3}+1 )^{3}+8 }{16}\right) ^{3}+1=2x$
Solving :
$$\left(\frac{ ( x^{3}+1 )^{3}+8 }{16}\right) ^{3}+1=2x$$
My Try :
$$\left(\frac{ ( x^{3}+1 )^{3}+8 }{16}\right) ^{3}=\left(\dfrac{(x+1)^3+2^3}{2^4}\right)^3$$
$$ x^3+a^3=(x+a)^3-3ax(x+a)$$
$$\left(\frac{ (x^3+3)^3-6(x^3+1)(x^3+3)}{16}\rig... | Let $\frac{x^3+1}{2}=y$ and $\sqrt[3]{2x-1}=z.$
Thus, from the given we obtain $\frac{y^3+1}{2}=z.$
Now, let $x>y$.
Thus, $$x>y=\frac{x^3+1}{2}>\frac{y^3+1}{2}=z,$$ which says that
$$x>y>z.$$
But from $y^3+1=2z$ and $z^3+1=2x$ we obtain
$$0=y^3-z^3+2(x-z)>0,$$ which is a contradiction.
By the same way we'll get a contr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3051964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sin a + \sin b + \sin(a+b) = 4 \sin\frac12(a+b) \cos \frac12a \cos\frac12b$ Helping my son with his trigonometry review. We know that
$$\sin(a+b) = \sin a \cos b + \cos a \sin b$$
We also know that
$$\sin a \cos b = \frac12 \left(\sin(a-b) + \sin(a+b)\right)$$
And we have
$$\sin a + \sin b = 2 \sin\fra... | It may help setting $A=a/2$ and $B=b/2$, so the left-hand side becomes
$$
\sin2A+\sin2B+\sin2(A+B)
$$
The right-hand side screams “sum-to-product”! OK, let's apply the formula
$$
\sin2A+\sin2B=2\sin(A+B)\cos(A-B)
$$
so the left-hand side becomes
\begin{align}
\sin2A+\sin2B+\sin2(A+B)
&=2\sin(A+B)\cos(A-B)+2\sin(A+B)\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3053978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
A Difficult Definite Integral $\int_0^{2\pi}(t-\sin t)(1-\cos t)^2{\rm d}t.$ Problem
Evaluate $$\int_0^{2\pi}(t-\sin t)(1-\cos t)^2{\rm d}t.$$
Comment
It's very complicated to compute the integral applying normal method. I obtain the result resorting to the skillful formula
$$\int_0^{2\pi}xf(\cos x){\rm d}x=\pi\int_0^... | \begin{align}
&\int_0^{2\pi}(t-\sin t)(1-\cos t)^2{\rm d}t=\int_0^{2\pi}t+t{\cos }^2 t-2t\cos t-\sin t -\sin t{\cos }^2 t+\sin 2t{\rm d}t\\
&=\left[\frac{1}{2}t^2+\frac{1}{2}t^2+\frac{1}{4}t\sin 2t-\frac{1}{4}t^2+\frac{1}{8}\cos 2t-2t\sin t-2\cos t+\cos t+\frac{1}{3}{\cos }^3t-\frac{1}{2}\cos 2t\right]_0^{2\pi} \\
&=\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3054898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
} |
If $a$ has order 3 $\pmod p$, $p$ prime, then $1+a+a^2 \equiv 0 \pmod p$ and $1+a$ has order $6$ There is a little proposition left to the readers in my number theory book, and I'm not sure how to answer this in a proper way:
If $a$ has order $3 \pmod p$, $p$ prime, then $1+a+a^2 \equiv 0 \pmod p$ and $1+a$ has order... | How $a$ has order $3$, then $a\not\equiv 1$.
Now, note that $(a-1)(1+a+a^2) = a^3-1 \equiv 0$ (mod $p$). As $\mathbb{Z}_p$ is a field and $a-1\neq 0$, so $1+a+a^2=0$ (mod $p$).
To see why $1+a$ has order $6$, note that
$$(1+a)^6=(1+2a+a^2)^3=a^3=1$$
So, the order of $1+a$ divide $6$.
If $1+a$ has order $1$, then $1+a=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3055073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
If $-1\leq x, y \leq 1$ and $x\sqrt{1-y^2} + y\sqrt{1-x^2} = 1$, find $x^2+y^2$ Let $x,\, y\in\mathbb
R,\ -1\leq x,\, y\leq 1$ such that $x\sqrt{1-y^2} + y\sqrt{1-x^2} = 1.$
Find the sum $S = x^2+y^2.$
| I think Michael's solution is much easier, but it's not something most people would think of. Instead, let's try doing this algebraically. First, I am going to isolate one of the terms on the left, so subtract by $y\sqrt{1-x^2}$:
$$x\sqrt{1-y^2}=1-y\sqrt{1-x^2}$$
Now, when we square both sides, the square root on the l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the interval of convergence of the series $ \sum^{\infty}\limits_{k=0} ((-1)^k+3)^k(x-1)^k $ I wish to find the interval of convergence of the following series
\begin{align} \sum^{\infty}_{k=0} ((-1)^k+3)^k(x-1)^k \end{align}
PROOF
Wittingly,
\begin{align} \left[(-1)^k+3\right]^k= \begin{cases} 0,&\text{if}\;j=0;\... | Hint
$$
\sum^{n}_{k=0} ((-1)^k+3)^k(x-1)^k =\lim_{n\to \infty}\left(\sum_{k=0}^n 4^{2k}(x-1)^{2k}+2(x-1)\sum_{k=0}^n 2^{2k}(x-1)^{2k}\right)
$$
and now
$$
\sum_{k=0}^n 4^{2k}(x-1)^{2k} = \frac{1-4^{2(n+1)}(x-1)^{2(n+1)} }{1-4^{2}(x-1)^{2}}
$$
which converges if $|4(x-1)| < 1$ etc. so the result is
$$
\sum^{n}_{k=0} ((-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3058414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
$(x*(\log_{2}(x))^2)/2 = x^{3/2}$ how to solve it? Is there a manual solution for this equation? Or I should use Wolfram:
result from Wolfram.
| There is no algebraic solution, but you can use numeric methods like Newton-Raphson method or the Lambert $W$ function:
The Lambert $W$ function is defined by $z=W(ze^z)$ with the domain as $\mathbb{C}$ and there are two possible functions $W_0$ with $\textrm{Re}(W_0(x))\geq -1$ and $W_{-1}$ with $\textrm{Re}(W_{-1}) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Why is $\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$ $15\sqrt{5}$ and not $15\sqrt[4]{25}$? I have an expression I am to simplify:
$$\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$$
I arrived at $15\sqrt[4]{25}$. My textbook tells me that the answer is in fact $15\sqrt{5}$. Here is my thought process:
$\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}... | Write $$\sqrt[4]{\frac{125}{5}}=\sqrt[4]{25}=\sqrt{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluating an improper integral $\int_{0}^{\infty}\frac{x^2}{(x^4+1)^2}dx$ I tried to solve the integral: $$\int_{0}^{\infty}\frac{x^2}{(x^4+1)^2}dx$$
using $ x = \sqrt{\tan(u)}$ and $dx = \frac{ \sec^2(u)}{2\sqrt{\tan(u)}} du,$
but I ended up with an even worse looking integral $$ \int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\... | Write $$\frac{x^2}{(1+x^4)^2} = \frac{4x^3}{(1+x^4)^2} \cdot \frac{1}{4x}.$$ Then integration by parts with the choice $$u = \frac{1}{4x}, \quad du = -\frac{1}{4x^2} \, dx, \\ dv = \frac{4x^3}{(1+x^4)^2} \, dx, \quad v = -\frac{1}{1+x^4},$$ yields $$I_1(x) = \int \frac{x^2}{(1+x^4)^2} \, dx = -\frac{1}{4x(1+x^4)} - \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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prove $a,b,c$ in A.P if $\tan\dfrac{A}{2}=\dfrac{5}{6}$ and $\tan\dfrac{C}{2}=\dfrac{2}{5}$
In $\Delta ABC$, if $\tan\dfrac{A}{2}=\dfrac{5}{6}$ and $\tan\dfrac{C}{2}=\dfrac{2}{5}$, then prove that the sides $a,b,c$ are in A.P.
My Attempt
$$
\sin A=\frac{2.5}{6}.\frac{36}{61}=\frac{60}{61}\\
\sin C=\frac{2.2}{5}.\frac... | Hint:
Like In $\Delta ABC$, find $\cot\dfrac{B}{2}.\cot\dfrac{C}{2}$ if $b+c=3a$,
$$2b=a+c$$ will hold true if $\tan\dfrac A2\tan\dfrac B2=\dfrac13$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Super hard system of equations
Solve the system of equation for real numbers
\begin{split}
(a+b) &(c+d) &= 1 & \qquad (1)\\
(a^2+b^2)&(c^2+d^2) &= 9 & \qquad (2)\\
(a^3+b^3)&(c^3+d^3) &= 7 & \qquad (3)\\
(a^4+b^4)&(c^4+d^4) &=25 & \qquad (4)\\
\end{split}
First I used the identity
$$(a^2+b^2)(c^2+d^2)=(ac-bd)... | This is not an answer but it is too long for a comment.
Looking at this system of equations, I had a very strange feeling (which I cannot explain).
Using a CAS, I solved equations $(1)$, $(2)$, $(3)$ for $a,b,c$ as functions of $d$ and obtained $8$ solutions which are listed below
$$\left\{a= \frac{2}{d},b= \frac{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 4,
"answer_id": 2
} |
Global extremas of $f(x,y)=2x^2+2y^2+2xy+4x-y$ on the region $x\leq 0, y\geq 0, y\leq x+3$ I have the following two variable function $f(x,y)=2x^2+2y^2+2xy+4x-y$ and I need to find its global extremas on the region $x\leq 0, y\geq 0, y\leq x+3$.
As in the other questions I asked in the last days, I have no solution to... | A solution without calculus.
Adopt new variables $x = u - v$, $y = u + v$. The objective function:
\begin{align*}
2x^2 + 2y^2 + 2xy + 4x - y &= 2(u^2 - 2uv + v^2) + 2(u^2 + 2uv + v^2) + 2(u^2 - v^2) + 4(u-v) - (u+v) \\
&= 6u^2 + 2v^2 + 3u - 5v \\
&= 6 \left( u + \frac{1}{4} \right)^2 + 2 \left(v - \frac{5}{4}\right)^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Maximizing $f$ in $\mathbb{R}^3$
Find the domain and the maximum value that the function
$$f(x,y,z)=\frac{x+2y+3z}{\sqrt{x^2+y^2+z^2}}$$
may attain in its domain.
I have found the domain of the function to be $\mathbb{R^3\backslash\mathbf{0}}$. To maximize I differentiated in terms of $x,y,z$ having
$$f_x=\frac{-... | Consider the line
$x = t, y = 2t, z = 3t$
Along this line.
$f(t,2t,3t) = \frac {14 t}{\sqrt{14 t^2}} = \sqrt {14}$
let's find some orthogonal vectors.
$\mathbf u = (1,2,3)\\
\mathbf v = (2,-1,0)\\
\mathbf w = (0,3,-2)$
Any point in $\mathbb R^3$ is some linear combination
$c_1\mathbf u + c_2\mathbf v+ c_3\mathbf w$
$f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Prove the identity for $\tan3\theta$
Prove the identity for $$\tan3\theta= \frac{3\tan\theta - \tan^3 \theta}{1-3\tan^2 \theta}$$
Using de Moivre's theorem I have found that:
$$\cos3\theta = 4\cos^3\theta - 3\cos \theta$$
$$\sin 3\theta = 3\sin \theta-4\sin^3\theta$$
therefore:
$$\tan 3\theta = \frac{\sin 3\theta}{\c... | We have
$$
\tan3\theta =\frac{\frac{3\sin \theta}{4\cos^3\theta} - \tan^3\theta}{1-\frac{3\cos \theta}{4\cos^3 \theta}} =
\frac{\frac{3(\cos^2 \theta + \sin^2 \theta)\sin \theta}{4\cos^3\theta} - \tan^3\theta}{1-\frac{3(\cos^2 \theta + \sin^2 \theta)\cos \theta}{4\cos^3 \theta}} = \\
\frac{\frac{3\cos^2 \theta\sin \th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3067903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Prove all elements $x_n \geq \sqrt{2}.$ Given $x_1 = 2,$ and $$x_{n+1} = \frac{1}{2}\bigg(x_n + \frac{2}{x_n} \bigg),$$ show that for all $n \in \mathbb{N},$ $x_n \geq \sqrt{2}.$
I tried the following. Suppose, for contradiction, that $x_{n+1} < \sqrt{2}.$ Then, $$\frac{1}{2}\bigg(x_n + \frac{2}{x_n} \bigg) < \sqrt{2},... | $x_1\geq\sqrt{2}$, $x_2\geq\sqrt{2}$. Now suppose $x_{n-1}\geq\sqrt{2}$, then by AM-GM we have $x_n=\frac{1}{2}(x_{n-1}+\frac{2}{x_{n-1}})\geq\frac{1}{2}(2\sqrt{x_{n-1}\frac{2}{x_{n-1}}})=\sqrt{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3068291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that triangle $XYZ$ is equilateral Let $ABC$ be an acute angled triangle whose inscribed circle touches $AB$ and $AC$ at $D$ and $E$ respectively. Let $X$ and $Y$ be the points of intersection of the bisectors of the angles $ACB$ and $ABC$ with the line $DE$ and let $Z$ be the midpoint of the side $BC$. Prove tha... | Let
$$A=(0,\ a)\\
B=(-b,\ 0)\\
C=(b,\ 0)$$
and thus
$$Z=(0,\ 0)$$
then we get
$$\tan(\angle ABC)=\frac ab$$
and thus
$$\tan(\frac 12\ \angle ABC)=\frac{a/b}{1+\sqrt{1+a^2/b^2}}=\frac a{b+\sqrt{a^2+b^2}}$$
So we can deduce the center $M$ of the incircle to be
$$M=(0,\ \frac {ab}{b+\sqrt{a^2+b^2}})$$
Now define lines $g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3071076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Explicit calculation of $\int_0^{\infty} \frac{\sin(2\arctan x)}{(1+x^2)(\exp(2\pi x)-1)}\mathrm{d}x$ Is it possible to confirm the value of this integral using the methods of complex analysis or similar?
$$
\int_0^{\infty} \frac{\sin(2\arctan x)}{(1+x^2)(\exp(2\pi x)-1)}\mathrm{d}x=\frac{\pi^2-9}{12}
$$
Of course, on... | This approach is unnecessarily complicated, but quite fun:
As explained in the comments and the other answers, we may write
$$ I \equiv \int \limits_0^\infty \frac{\sin(2\arctan x)}{(1+x^2)(\exp(2\pi x)-1)} \, \mathrm{d}x = \int \limits_0^\infty \frac{2 x}{(1+x^2)(\mathrm{e}^{2\pi x}-1)} \, \mathrm{d} x \, . $$
We can ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3072149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
$\lim \limits _{x \rightarrow 0} \frac{x^3 e^{\frac {x^4}{4}} - {\sin^{\frac 3 2} {x^2}} }{x^7}$ evaluation using expansion series $$\lim \limits _{x \rightarrow 0} \frac{x^3 e^{\frac {x^4}{4}} - {\sin^{\frac 3 2} {x^2}} }{x^7}$$
This is the limit which I got from the book of Joseph Edwards' Differential Calculus for B... | We have $$\lim \limits _{x \rightarrow 0} \frac{x^3 e^{\frac {x^4}{4}} - {\sin^{\frac 3 2} {x^2}} }{x^7}{=\lim \limits _{x \rightarrow 0} \frac{e^{\frac {x^4}{4}} - \left({\sin x^2\over x^2}\right)^{3\over 2} }{x^4}\\=\lim \limits _{x \rightarrow 0} \frac{e^{\frac {x^4}{4}} -1+1- \left({\sin x^2\over x^2}\right)^{3\ove... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3073052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\lim\limits_{x \to 1}\dfrac{x-x^x}{1-x+\ln x}$. Problem
Evaluate $$\lim\limits_{x \to 1}\frac{x-x^x}{1-x+\ln x}$$.
Solution
Consider using Taylor's formula. Expand $x^x $ and $\ln x$ at $x=1$. We obtain
$$x^x=1+(x-1)+(x-1)^2+o((x-1)^2);$$
$$\ln x=(x-1)-\frac{1}{2}(x-1)^2+o((x-1)^2).$$
Therefore
\begin{align*}... | Another solution by L'Hospital's rule
Let $x=:1+h$. Then
\begin{align*}
\lim_{x \to 1}
\frac{x-x^x}{1-x+\ln x}&=\lim_{h \to 0}\frac{(1+h)[1-(1+h)^{h}]}{\ln(1+h)-h}\\
&=\lim_{h \to 0}\frac{e^{h\ln(1+h)}-1}{h-\ln(1+h)}\\
&=\lim_{h \to 0}\frac{h\ln(1+h)}{h-\ln(1+h)}\\
&=\lim_{h \to 0}\frac{h^2}{h-\ln(1+h)}\\
&=\lim_{h \to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3075787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Prove that $\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$ is an integer. Prove that $$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$$ is an integer using mathematical induction.
I tried using mathematical induction but using binomial formula also it becomes little bit complicated.
Please show me ... | Base case for $k=1$: $$\frac{1^7}{7}+\frac{1^5}{5}+\frac{2*1^3}{3}-\frac{1}{105}=1$$
Now, assume for some k that $\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$ is indeed in integer.
Then $$\frac{(k+1)^7}{7}+\frac{(k+1)^5}{5}+\frac{2(k+1)^3}{3}-\frac{k+1}{105}=\\\frac{\sum_{i=0}^7\binom{7}{i}k^i}{7}+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3075979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 8,
"answer_id": 5
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How do I solve for $a$ and $b$ in $\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2$? Given $\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2$
I need to solve for $a$ and $b$, so here we go,
$\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \d... | Your mistake lies in the stp where you replace the difference between logs with the ratio $(a-b) /x$. You can not replace an expression $A$ by another expression $B$ unless $A=B$. While evaluating limit you may find replacements of $A$ by $B$ even when $A\neq B$ but such replacements always represent some hidden steps ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3079016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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Evaluating $\lim_{x\to0^+}\frac{2x(\sin x)^2+\frac{2x^7+x^8}{3x^2+x^4}-\arctan(2x^3)}{\ln(\frac{1+x^2}{1-x^2})-2x^2+xe^{-{1\over x}}}$ To start with, the term $xe^{-{1\over x}}$ can be ignored.Then splitting the term $\ln(\frac{1+x^2}{1-x^2})=\ln(1+x^2)-\ln(1-x^2)$ and expanding both of them up to order 3 we have in th... | You made a mistake by considering only the first asymptotic term of the algebraic fraction. Be careful:
$$\frac{2x^5+x^6}{3+x^2}=\frac{2}{3}x^5 + \frac{1}{3}x^6 + o(x^6).$$
Then your limit gives $\frac{1}{2}$, as results from the comment by JimB.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$a+b=c+d$ , $a^3+b^3=c^3+d^3$ , prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$
Let $a, b, c, d$ be four numbers such that $a + b = c + d$ and $a^3 + b^3 = c^3 + d^3$. Prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$.
I've got $a^2+b^2-ab=c^2+d^2-cd$.
I tried squaring or cubing it repeatedly but I didn't get what I ... | $$(a+b)^3=(c+d)^3$$
$$\iff a^3+b^3+3ab(a+b)=c^3+d^3+3cd(c+d)$$
If $a+b\ne0, ab=cd\ \ \ \ (1)$
$\iff\dfrac ad=\dfrac cb=k$(say) $\ \ \ \ (2)$
$a+b=c+d\implies dk+b=bk+d\iff d(k-1)=b(k-1)$
Either $d=b\iff a=c$
or $k=1$ use this in $(2)$
Or Using $(1), ab=cda+b=c+d$ So if $a,b$ are roots of $t^2+bt+c=0,$
$c,d$ will ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
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Understanding some proofs-without-words for sums of consecutive numbers, consecutive squares, consecutive odd numbers, and consecutive cubes
I understand how to derive the formulas for sum of squares, consecutive squares, consecutive cubes, and sum of consecutive odd numbers but I don't understand the visual proofs ... | The second picture gives a visual proof for the formula
$$3(1^2+2^2+3^2+\dots +n^2)=\frac{n(n+1)}{2}\cdot (2n+1)$$
for $n=5$. The sum of the areas of the $3\cdot 5$ squares on the right
$$3(1^2+2^2+3^2+4^2+5^2)$$
is equal to the area of the rectangle on the left with height $1+2+3+4+5=\frac{6\cdot 5}{2}$ (see the firs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 3,
"answer_id": 0
} |
$\int\limits_0^\infty {x^4 \over (x^4-x^2+1)^4}\ dx$ I want to calculate
$$\int\limits_0^\infty \frac{x^4}{(x^4-x^2+1)^4}dx$$
I have searched with keywords "\frac{x^4}{(x^4-x^2+1)^4}" and "x^4/(x^4-x^2+1)^4". But there are no results
| For $b > 0$, define $I_b(a)$ by
$$ I_b(a)
= \int_{0}^{\infty} \frac{x^{a}}{((x - x^{-1})^2 + 1)^b} \, \mathrm{d}x
= \int_{0}^{\infty} \frac{x^{a+2b}}{(x^4 - x^2 + 1)^b} \, \mathrm{d}x. $$
This integral converges if $|a+1| < 2b$. We can also prove that $I_b(a) = I_b(-a-2)$ holds, by using the substitution $x \mapsto 1/x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How can I find the real and imaginary parts of $\sqrt{i+\sqrt{i+\sqrt{i+\dots}}}$? By using $x = \sqrt{i+\sqrt{i+\sqrt{i+\dots}}}$ , I can find a quadratic equation, but how can I separate the real and imaginary parts?
Edit:
My question was how to find the real and imaginary parts of the complex number $\sqrt{i+\sqrt{i... | x = $\sqrt{(i+\sqrt{(i+\dots)})}$ = $\sqrt{(i+x)}$
$x^2 - x - i = 0$
Solving for x we get,
$x_{1,2} = \frac{1}{2}\pm \frac{\sqrt{1+4i}}{2}$
Now,
$a + ib =\sqrt{1+4i}$
$\implies a^2 - b^2 + 2iab = 1+4i$
So we have, $a^2-b^2 = 1$ and 2ab = 4.
Solving for a and b we can get
$a = \sqrt(\frac{\sqrt{17}+1}{2})$ and $b = \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3085608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
$\sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n^2+n+4}}$ $$\sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n^2+n+4}}$$
We know that : $\arctan{x} - \arctan{y} = \arctan{\frac{x-y}{1+xy}}$ for every $ xy > 1 $
I need to find two numbers which satisfy: $ab = n^2+2n+3 $ and $ a-b =2$ in order to telescope.
Edit: I am very sorry ... | Hint:
$$\dfrac2{n^2+n+4}=\dfrac{\dfrac12}{1+\dfrac{n(n+1)}4}=\dfrac{\dfrac{n+1}2-\dfrac n2}{1+\dfrac n2\cdot\dfrac{n+1}2}$$
For the second, $$\dfrac{8n}{n^4-2n^2+5}=\dfrac{\dfrac{(n+1)^2}2-\dfrac{(n-1)^2}2}{\dfrac{(n+1)^2}2\cdot\dfrac{(n-1)^2}2+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
The greatest value of $|z|$ if $\Big|z+\frac{1}{z}\Big|=3$ where $z\in\mathbb{C}$
$\bigg|z+\dfrac{1}{z}\bigg|=3$ then the greatest value of $|z|$ is ___________
My Attempt
$$
\bigg|z+\frac{1}{z}\bigg|=\bigg|\dfrac{z^2+1}{z}\bigg|=\frac{|z^2+1|}{|z|}=3\\
\bigg|z+\frac{1}{z}\bigg|=3\leq|z|+\frac{1}{|z|}\implies |z|^2-3... | A possible approach is using polar representation
*
*$z = |z|e^{i\phi}$
*$\Big|z+\frac{1}{z}\Big|=3 \Leftrightarrow (z+\frac{1}{z})(\bar z + \frac{1}{\bar z}) = 9 \Leftrightarrow \boxed{\left(|z|e^{i\phi} + \frac{1}{|z|} e^{-i\phi}\right)\left(|z|e^{-i\phi} + \frac{1}{|z|} e^{i\phi}\right) = 9}$
Expanding gives
\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3091795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How can I simplify this fraction problem? I have the problem $\frac{x^2}{x^2-4} - \frac{x+1}{x+2}$ which should simplify to $\frac{1}{x-2}$
I have simplified $x^2-4$, which becomes:
$\frac{x^2}{(x-2)(x+2)} - \frac{x+1}{x+2}$
However, if I combine the fractions I get, $x^2-x-1$ for the numerator, which can't be factored... | \begin{align}
\frac{x^2}{(x-2)(x+2)} - \frac{x+1}{x+2}&=\frac1{x+2}\left(\frac{x^2}{x-2} - (x+1)\right)\\
&=\frac1{x+2}\left(\frac{x^2-(x-2)(x+1)}{x-2}\right)\\
&=\frac1{x+2}\left(\frac{x^2-(x^2-x-2)}{x-2}\right)\\
&=\frac1{x+2}\left(\frac{x+2}{x-2}\right)\\
&=\frac1{x-2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3093945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Show $\frac{n+1}{2n^2+5} \in \Theta (\frac{1}{n})$ using definition Show $\frac{n+1}{2n^2+5} \in \Theta (\frac{1}{n})$ using definition
First showing the upper bound
$\frac{n+1}{2n^2+5} \in O (\frac{1}{n})$
$\frac{n+1}{2n^2+5} \leq \frac{2n}{2n^2} = \frac{1}{n}$
for $n \geq 1$
now showing the lowerbound
$\frac{n+1}{2n... | Yes, you have shown found constant $c_1$ and $c_2$ such that
$$ \frac{c_1}{n}\le\frac{n+1}{2n^2+5} \le \frac{c_2}{n}.$$
You have proven what you want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3094491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
For the equation $y = 4x^2 + 8x + 5$, what are the integer values of x such that y/13 is an integer? For the equation $y = 4x^2 + 8x + 5$, what are the integer values of x such that y/13 is an integer?
For example, if x = 3, $y = 4(3^2) + 8(3) + 5$ = 65 which is divisible by 13
if x = 8, y = 325 which is divisible by 1... | You are looking for solutions to $4x^2+8x+5\equiv 0 \pmod {13}$. You can do the usual quadratic formula and find $x=\frac {-8 \pm \sqrt{64-80}}8$. Don't let the negative number under the square root bother you because $\pmod {13}$ we have $-16 \equiv 10$. You need to find whether $10$ is a square $\bmod 13$. One w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Simplify $\frac{x^3-x}{x^2+xy+x+y}$ $$\frac{x^3-x}{x^2+xy+x+y}$$
What I did:
$$\frac{x}{xy+x+y}$$
through simplifying the $x$'s.
But it's not right.
What did I do wrong?
| $$
\frac{x^3 - x}{x^2 +xy + x +y }
= \frac{x(x^2 - 1)}{x( x+y) + 1( x +y )}
= \frac{x(x - 1)(x+1)}{(x+1)( x+y)}
= \frac{x(x - 1)}{( x+y)}
$$
Here you had to remember that $x^2 -1 = (x+1)(x-1)$. The last step assumes $x\neq -1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3096139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to prove $\frac{(x+y)(y+z)(z+x)}{4xyz}≥\frac{x+z}{y+z}+\frac{y+z}{x+z}$ for $x,y,z>0$?
Prove that for $x,y,z$ positive numbers:
$$
\frac{(x+y)(y+z)(z+x)}{4xyz}≥\frac{x+z}{y+z}+\frac{y+z}{x+z}
$$
I tried to apply MA-MG inequality:
$x+y≥2\sqrt{xy}$ and the others and multiply them but it becomes
$\frac{(x+y)(y+z)... | We have
$$
\frac1 {4yz}+\frac1 {4xz}\ge \frac1{(y+z)^2}+\frac1{(x+z)^2}.
$$ Therefore
$$
\frac{(x+y)}{4xyz}(y+z)(z+x)\ge\left(\frac1{(y+z)^2}+\frac1{(x+z)^2}\right)(y+z)(z+x),
$$ which leads to the desired inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3096406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove that $1+b+(1+c)+1/c+1+a \ge 3$ if $a, b,$ and $c$ are positive real numbers. Let $a, b, c$ be positive real numbers. prove that
$$
\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\ge\frac{3}{1+abc},
$$
and that equality occurs if and only if $a = b = c = 1$.
What I tried:
$1$st part: I tried a brute force appro... | AM-GM helps:
$$\sum_{cyc}\frac{1}{a(1+b)}=\frac{1}{1+abc}\sum_{cyc}\frac{1+abc}{a(1+b)}=$$
$$=\frac{1}{1+abc}\sum_{cyc}\left(\frac{1+abc}{a(1+b)}+1\right)-\frac{3}{1+abc}=$$
$$=\frac{1}{1+abc}\sum_{cyc}\frac{1+a+ab(1+c)}{a(1+b)}-\frac{3}{1+abc}=$$
$$=\frac{1}{1+abc}\sum_{cyc}\left(\frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3100653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Show that $y^3=4x^2-1$ has no positive integer solutions Problem: Show that $y^3=4x^2-1$ has no positive integer solutions.
I tried to consider maybe that cubes are never equal to $3 \bmod 4$, but $3^3 \equiv 3\bmod 4$.
Next I tried letting $x$ be even , so that $y$ is odd, (and if $x$ is odd, then $y$ is still odd), b... | This equation is satisfied if and only if some $y^3$ exists that can be expressed as a product of two consecutive odd numbers, $2x - 1$ and $2x + 1$.
Consider $p$ a prime divisor of $y$. Note that $p > 2$. One of the two factors must be divisible by $p^2$. If it's not divisible by $p^3$, then other factor must be divis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3104378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Proving $\left(1+\frac{1}{m}\right)^m < \left(1+\frac{1}{n}\right)^n$ Let $m,n\in \mathbb{N}$. If $m > n$ show that
$$\left(1+\frac{1}{m}\right)^m > \left(1+\frac{1}{n}\right)^n$$
My works: I tried to show if $g(x)=\left(1+\frac{1}{x}\right)^x$ then $g'(x) > 0$.
\begin{align}
g'(x) &= \frac{d e^ { x \ln(1+\frac{1}{x}) ... | If $m<n$, then for all $j=0...m$ we have $1-\frac jm \le 1-\frac jn$ and so $\frac{m-j}{m} \le \frac{n-j}{n}$ (for $j=0$ we have equality) and therefore:
$$
\begin{align}
\left(1+\frac1m\right)^m & = \sum_{k=0}^m \binom{m}{k}\frac{1}{m^k} = \sum_{k=0}^m \frac {1}{k!}\prod_{j=0}^{k-1}\frac{m-j}{m}\\
& \le \sum_{k=0}^m \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How do we factor out $x^2 - x -2$ in this expression? Suppose we're given that
$$x^4 - 2x^3 +x-2$$
How do we factor out $x^2 - x -2$ in this expression?
$$(x^4 -x^3- 2x^2)-(x^3-x^2-2x)+(x^2-x-2) = x^2(x^2 -x-2)-x(x^2-x-2)+(x^2-x-2) = (x^2-x-2)(x^2-x+1)$$
This satisfies with what we want to get. However, I do not seem... | $$x^4-2x^3+x-2=x^4-x^3-2x^2-x^3+x^2+2x+x^2-x-2=$$
$$=x^2(x^2-x-2)-x(x^2-x-2)+(x^2-x-2)=(x^2-x-2)(x^2-x+1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3106729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
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Studying the convergence of the series $\sqrt{2}+\sqrt{2+\sqrt{2}}+\sqrt{2+\sqrt{2+\sqrt{2}}}+ \cdots$ Studying the convergence of the series $$\sqrt{2}+\sqrt{2+\sqrt{2}}+\sqrt{2+\sqrt{2+\sqrt{2}}}+ \cdots$$
I saw this problem and I tried to do it my own way but I don't know what I'm doing wrong because I'm getting a d... | Your conclusion that $\sum_{n=1}^\infty a_n$ diverges is correct. You don't need the explicit formula, it would suffice to observe that
$$
a_n = \underbrace{\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}_{n\text{ times}} \ge \sqrt 2
$$
or that
$$
0 < a_n < a_{n+1}
$$
for all $n \in \Bbb N$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3109128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Is $8n^3+12n^2-2n-3$ divisible by 5, with $n$ congruent with 1,2,3 mod 5? I make some test using python and I find that this is the case for $n = 5k +p$ with $k$ an integer and $p =$1,2,or 3.
I was able to prove for $p = 1$ and $p=2$ but not for $p = 3$.
What I'm doing wrong? This is my work:
Let's assume 5 is a diviso... | Alternative solution. Note that
$$8n^3+12n^2-2n-3=(2n+3)(2n+1)(2n-1)$$
which is divisible by $5$ iff $(2n+3)$ or $(2n+1)$ or $(2n-1)$ is divisible by $5$. Now
$$\begin{align}
&2n+3\equiv 0\leftrightarrow 2n\equiv -3\equiv 2\leftrightarrow n\equiv 1 \pmod{5}\\
&2n+1\equiv 0\leftrightarrow 2n\equiv -1\equiv 4\leftrighta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3112382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Proving that $\lim_{v\rightarrow \infty}\left[\frac{v^2}{v^3 + 1} + \frac{v^2}{v^3 + 2} +\cdots + \frac{v^2}{v^3+v}\right]= 1 $ I wonder if my solution that $\lim_{v\to\infty}\left[\frac{v^2}{v^3 + 1} + \frac{v^2}{v^3 + 2} + \cdots + \frac{v^2}{v^3+v}\right]= 1 $ is correct.
$$\frac{v^2}{v^3 + 1} + \frac{v^2}{v^3 + 2} ... | Consider the sequence$$1,\frac12+\frac12,\frac13+\frac13+\frac13,\ldots,\overbrace{\frac1n+\cdots+\frac1n}^{n\text{ times}},\ldots$$Its limit is $1$, right?! However, by your argument, the limit should be $0$.
Yes, $\lim_{n\to\infty}\frac1n=0$, but, on the other hand, you have $n$ terms. So, the fact that the limit of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find $K=a^2b+b^2c+c^2a$ for roots $a>b>c$ of a cubic. If $a>b>c$ are the roots of the polynomial $P(x)=x^3-2x^2-x+1$ find the value of $K=a^2b+b^2c+c^2a$.
Using Vièta's formulas:
$$a+b+c=2$$
$$ab+bc+ca=-1$$
$$abc=-1$$
Using those I found that
$$a^2+b^2+c^2=6$$
$$a^3+b^3+c^3=11$$
$$a^2b+b^2c+c^2a+ab^2+bc^2+ca^2=1$$
but... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, $u=\frac{2}{3},$ $v^2=-\frac{1}{3},$ $w^3=-1$ and
$$\sum_{cyc}a^2b=\frac{1}{2}\sum_{cyc}(a^2b+a^2c+a^2b-a^2c)=$$
$$=\frac{1}{2}(9uv^2-3w^3)+\frac{1}{2}(a-b)(a-c)(b-c)=$$
$$=\frac{1}{2}(9uv^2-3w^3)+\frac{1}{2}\sqrt{(a-b)^2(a-c)^2(b-c)^2}=$$
$$=\frac{1}{2}(9uv^2-3w^3)+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3117967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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What is the number of pairs of $(a,b)$ of positive real numbers satisfying $a^4+b^4<1$ and $a^2+b^2>1$?
The number of pairs of $(a,b)$ of positive real numbers satisfying
$a^4+b^4<1$ and $a^2+b^2>1$ is -
$(i)$0
$(ii)$1
$(iii)$2
$(iv)$ more than 2
Solution:We have $a,b>0$,
According to the given situation,$0<a^4+b^... | There are infinitely many solutions:
In fact for every $-1<a<1$ , $a\not = 0$ there exist infinitely many $b$'s such that the pair $(a,b)$ satisfies the condition.
Indeed, the conditions can be written as $b^4<1-a^4$ and $b^2>1-a^2$.
Therefore if $b$ satisfies $b<\sqrt[4]{1-a^4}, b>\sqrt[2]{1-a^2}$ then the pair $(a,b)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3122623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Evaluate $\lim\limits_{x\to+\infty} \sqrt{x}\left ( \sqrt[3]{x+1}-\sqrt[3]{x-1}\right )$?
How do you evaluate $$\lim\limits_{x\to+\infty} \sqrt{x}\left (\sqrt[3]{x+1}-\sqrt[3]{x-1}\right ) ?$$
Thanks in advance for your support.
| If it is ok for you to use only the fact that $\color{blue}{f(t)} =\sqrt[3]{1+t}$ and $\color{blue}{g(t)}=\sqrt[3]{1-t}$ are differentiable at $t= 0$, then without calculating any derivative you may proceed as follows:
*
*Set $x = \frac{1}{t}$ and consider $t \rightarrow 0^+$
\begin{eqnarray*} \sqrt{x}\left ( \sqrt[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3123640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find $A, B \in M_2(\mathbb{Q})$ Find $A, B \in M_2(\mathbb {Q}) $ so that $A^2+2B^2=\begin{pmatrix} 4& - 4\\
-2 & 2
\end{pmatrix}$ and $AB+BA=\begin{pmatrix}3 & - 3\\
-1 & 1
\end{pmatrix}$.
My work so far : $(A +\sqrt 2 B) ^2=A^2+2B^2+\sqrt 2 (AB+BA) $.
After taking determinants we get that $\det (A +\sqrt 2 B)=0$ and ... | If we denote $v=(1,1)$, then the given conditions imply that
$$
\ker (A+\sqrt 2B)^2 =\ker(A-\sqrt 2B)^2=\text{span}\{v\}.
$$ Thus $A\pm \sqrt{2}B$ are non-invertible matrices, which are not $O$, and it follows that
$$
\ker(A\pm \sqrt{2}B)=\text{span}\{v\}
$$ since
$$
(0)< \ker(A\pm \sqrt{2}B) \le \ker (A\pm \sqrt 2B)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3125288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Solving a tough equation involving integer functions I am stuck on solving the equation, given $k\lt\frac{n}{2},\ n,k\ge3$:
$$ m=\lceil 2k-\frac{2}{n}\displaystyle\left(\lfloor\frac{n-\lfloor\frac{n}{k+1}\rfloor}{2}\rfloor\right)(k+1)\rceil$$.
I think the value of $m$ would be $k$. But, the detailed solution is beyond ... | We calculate the first expression for integer values $k\geq 0, n\geq 1$.
Let $a,b$ be non-negative integer with
\begin{align*}
n=a(k+1)+b\qquad\quad a,b\geq 0,\ \ 0\leq b<k+1\tag{1}
\end{align*}
we obtain
\begin{align*}
\color{blue}{ a(n,k)}
&\color{blue}{=\left\lceil 2k-\frac{2}{n}\left(\left\lfloor\frac{n-\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3125924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Stuck on a Geometry Problem
$ABCD$ is a square, $E$ is a midpoint of side $BC$, points $F$ and $G$ are on the diagonal $AC$ so that $|AF|=3\ \text{cm}$, $|GC|=4\ \text{cm}$ and $\angle{FEG}=45 ^{\circ}$. Determine the length of the segment $FG$.
How can I approach this problem, preferably without trigonometry?
| Drop perpendiculars $FH, EI$ and $FJ$ and label $BH=y$, $EH=z$ and $\angle GEI=\alpha$:
$\hspace{2cm}$
Use the similarity of $\triangle FEH$ and $\triangle EIG$:
$$\frac{GI}{HE}=\frac{EI}{FH} \Rightarrow \\
\frac{4-CI}{z}=\frac{CI}{y+2z}\Rightarrow \\
\frac{4-\frac{y+z}{\sqrt{2}}}{z}=
\frac{\frac{y+z}{\sqrt{2}}}{y+2z}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 12,
"answer_id": 3
} |
integrate sin(x)cos(x) using trig identity. Book tells me the answer is:
$$ \int \sin(x)\cos(x) dx = \frac{1}{2} \sin^{2}(x) + C $$
however, I get the result:
$$
\sin(A)\cos(B) = \frac{1}{2} \sin(A-B)+\frac{1}{2}\sin(A+B)
$$
$$
\begin{split}
\int \sin(x)\cos(x) dx
&= \int \left(\frac{1}{2}\sin(x-x) + \frac{1}{2}\sin... | HINT
The answers are equivalent. Recall that
$$
\cos (2x) = \cos^2 x - \sin^2 x = \left(1-\sin^2 x\right) - \sin^2 x = 1 - 2 \sin^2x
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Find $z$ such that $(2+i)(1+i)=2+zi $ I tried multiplying both of the complex numbers, but the answer is $1+3i $.
Sorry for the poor English; I'm not a native speaker.
| Let $z = a + bi$ where $a$ and $b$ are real.
$(2+i)(1+i) = 2 +zi$
$1 + 3i = 2+ (a+bi)i$
$-1 + 3i = ai -b = -b + ai$
So $a =3$ and $-b = -1$ so $b = 1$.
So $z = 3 + i$.
=====
You should get used to the fact that Complex numbers are just like real numbers.
If you $m=a + dz$ then $z = \frac {m-a}d$ and this is no differen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to deduce the following complex number problem I am stuck with the following problem that says:
Using the result $$x^n-1=(x^2-1)\prod_{k=1}^{(n-2)/2}[x^2-2x\cos \frac{2k\pi}{n}+1],$$ if $n$ be an even positive integer, deduce that $$\sin \frac{\pi}{32}\sin \frac{2\pi}{32}\sin \frac{3\pi}{32}.........\sin \frac{15... | This doesn't use the given information, but it does give some intuition of what you are trying to show.
Consider $x^8 - 1$
The roots of unity are equally spaced around the unit circle.
$x^8-1 = (x-1)(x-r_1)(x-r_2)\cdots(x-r_7) = (x-1)(x^7 + x^6 + \cdots + 1)$
with $r_k = (\cos \frac {2k\pi}{8}, i\sin \frac{2k\pi}{8})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3129039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How many vertices so that the number of pentagons equals the number of triangles? Question: A set of points is chosen on the circumference of a circle so that the
number of different triangles with all three vertices among the points is equal to the
number of pentagons with all five vertices in the set. How many points... | Yes, this reasoning is correct. (Of course, $(n - 5)!$ is not defined for integers $\leq 4$, so we can discard those solutions outright for that reason, too.)
Alternatively, if there are $3 + 5 = 8$ vertices, then we can define a natural bijection $$\{\textrm{triangles}\} \leftrightarrow \{\textrm{pentagons}\}$$ taking... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3130991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
solve $\frac{\sqrt{5}(\cos \theta - \sin \theta)}{3\sqrt{2}}=\tan \theta$ I came across a question from another forum -
find the $x$ in the following diagram:
I managed to deduce an equation from the following diagram:
which is:
$\dfrac{\sqrt{5}(\cos \theta - \sin \theta)}{3\sqrt{2}}=\tan \theta$
and I know the answe... | If we start with this diagram:
we can deduce that:
$$
\overline{CE}=\sqrt{(k+1)^2+1}
\\
\overline{DE}=\frac{1}{k+1}\sqrt{(k+1)^2+1}
\\
\frac{\overline{DE}}{\overline{BC}}
=\frac{\frac{1}{k+1}\sqrt{(k+1)^2+1}}{k}
=\frac{\sqrt{(k+1)^2+1}}{k(k+1)}
=\frac{\sqrt{5}}{3\sqrt{2}}
\\
5k^2(k+1)^2=18(k+1)^2+18
\\
5k^4+10k^3-13k^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3132010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Calculate the sum of fractionals
Let $n \gt 1$ an integer. Calculate the sum: $$\sum_{1 \le p \lt q \le
n} \frac 1 {pq} $$ where $p, q$ are co-prime such that $p + q > n$.
Calculating the sum for several small $n$ value I found out that the sum is always $\frac 1 2$.
Now, I'm trying to prove the sum is $\frac 1 2$ ... | Let $$
s_n=\sum_{1\le p<q\le n\\(p,q)=1}\frac1{pq}-\sum_{p+q\le n\\p<q,(p,q)=1}\frac1{pq}=a_n-b_n.
$$ Then we have that
$$
a_{n+1}-a_n=\sum_{1\le p<q=n+1\\(p,n+1)=1}\frac{1}p\cdot\frac1{n+1}=\frac1{n+1}\sum_{1\le p<n+1\\(p,n+1)=1}\frac1{p}
$$ and $$\begin{align*}
b_{n+1}-b_n&=\sum_{p+q=n+1\\p<q, (p,q)=1}\frac1{pq}
\\&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Find the number of ordered triplets satisfying $5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right)$ Find the number of ordered triplets $(x,y,z)$ of real numbers satisfying $$5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right)$$
and
$$xy+yz+zx=1$$
My ... | Let $\left(x;y;z\right)\rightarrow \left(\tan \alpha ;\tan \beta ;\tan \gamma \right)\left(0<\alpha ,\beta ,\gamma <90^o\right)$
Then we have
\begin{align}
\begin{cases}
5\left(\tan \alpha +\frac{1}{\tan \alpha }\right)=12\left(\tan \beta +\frac{1}{\tan \beta }\right)=13\left(\tan \gamma +\frac{1}{\tan \gam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Confusion in this algebraic limit approaching infinity
Question : Evaluate $\lim_{x \to \infty}{\frac{1}{1+x^2} + \frac{2}{2+x^2} + \frac{3}{3+x^2}+...+\frac{x}{x + x^2}}$.
My working:
$$\lim_{x \to \infty}{\frac{1}{1+x^2} + \frac{2}{2+x^2} + \frac{3}{3+x^2}+...+\frac{x}{x + x^2}}$$
Dividing and transforming each fra... | I may be totally wrong but looking at the top formula, it seems that you are rather considering
$$S_n=\sum_{i=1}^n \frac i{i+n^2}=n \left(n H_{n^2}-n H_{n^2+n}+1\right)$$ where appear harmonic numbers.
Using the asymptotics
$$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12
p^2}+O\left(\frac{1}{p^4}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3135748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Help finding the derivative of $f(x) = \cos{\left(\sqrt{e^{x^5} \sin{x}}\right)}$ I am trying to find the derivative of $f(x) = \cos(\sqrt{(e^{x^5} \sin(x)})$.
I keep getting the wrong answer, and I'm not sure what I'm doing wrong.
$$\frac{d}{dx} e^{x^5} = e^{x^5} \cdot 5x^4$$
$$\frac{d}{dx} \sin(x) = \cos(x)$$
$$\fra... | This is one case where logarithmic differentiation can make life easier.
$$
\frac{d}{dx}\left[\cos\left(\sqrt{e^{x^5}\,\sin(x)}\right)\right]
=-\sin{\left(\sqrt{e^{x^5}\,\sin(x)}\right)}\,\,\frac{d}{dx}\left[\sqrt{e^{x^5}\,\sin(x)}\right]$$
Let
$$f=\sqrt{e^{x^5}\,\sin(x)}\implies \log(f)=\frac 12 x^5 +\frac 12 \log(\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3138417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Inverse of a function 2 What is the inverse of the function $f(x)=x+3[2x]+2[3x]$?
The function is one by one and has an inverse.
| Given
$$ f(x)=x+3\lfloor2x\rfloor+2\lfloor3x\rfloor $$
Since
$$ x=\left(x-\lfloor x \rfloor\right)+ \lfloor x \rfloor$$
we have
\begin{eqnarray}
f(x)&=&x-\lfloor x \rfloor + \lfloor x \rfloor\\
&&+3\lfloor2\left(x-\lfloor x \rfloor+ \lfloor x \rfloor\right)\rfloor+2\lfloor3\left(x-\lfloor x \rfloor+ \lfloor x \rfloor\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3141288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Developing a general expression for powers of matrices Is there a relationship for matrices that are squared?
I am trying to determine two possible matrices $P$ where
$$P^2=\begin{bmatrix}0.6&0.4\\0.4&0.6\end{bmatrix}$$
I know that $P$ has to be a $2×2$ matrix and that $P$ will also be symmetric.
Thanks
| If you assume that $$P=\begin{pmatrix}a&b\\b&c\end{pmatrix}$$
then you compute $$P^2=\begin{pmatrix}a^2+b^2&ab+bc\\ab+bc&b^2+c^2\end{pmatrix}=\begin{pmatrix}0.6&0.4\\0.4&0.6\end{pmatrix}$$
From this, you get the system of equations $$a^2+b^2=0.6,~ b(a+c)=0.4,~ b^2+c^2=0.6$$
Subtracting the third from the first, we get ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3142015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
minimum value of $2\cos \alpha\sin \beta+3\sin \alpha\sin \beta+4\cos \beta$ Let $\alpha,\beta$ be real numbers ; find the minimum value of
$2\cos \alpha\sin \beta+3\sin \alpha\sin \beta+4\cos \beta$
What I tried :
$\bigg|4\cos \beta+(2\cos \alpha+3\sin \alpha)\sin \beta\bigg|^2 \leq 4^2+(2\cos \alpha+3\sin \alpha)^2$... | By C-S twice we obtain:
$$2\cos\alpha\sin\beta+3\sin\alpha\sin\beta+4\cos\beta=$$
$$=\sin\beta(2\cos\alpha+3\sin\alpha)+4\cos\beta\geq$$
$$\geq-\sqrt{(\sin^2\beta+\cos^2\beta)((2\cos\alpha+3\sin\alpha)^2+16)}\geq$$
$$\geq-\sqrt{\left(\sqrt{2^2+3^2)(\cos^2\alpha+\sin^2\alpha}\right)^2+16}=-\sqrt{29}.$$
The equality occ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
A simpler method to prove $\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c} = 1?$ This is the way I think about it:
$
1 = \log_aa = \log_bb = \log_cc \\~\\
\textbf{Using the ‘change of base rule':} \\
\log_{a}{b} = \frac{\log_{b}{b}}{\log_{b}{a}}, \hspace{0.3cm} \log_{b}{c} = \frac{\log_{c}{c}}{\log_{c}{b}}, \hspace{0... | You may show it directly: let
$$x=\log_a(c)\log_c(b)\log_b(a),$$
hence
\begin{align}a^{\color{red}{x}}&=a^{\log_a(c)\log_c(b)\log_b(a)}\\
&=\left(\left(a^{\log_a(c)}\right)^{\log_c(b)}\right)^{\log_b(a)}\\
&=(c^{\log_c(b)})^{\log_b(a)}\\
&=b^{\log_b(a)}\\
&=a\\
&=a^{\color{red}{1}}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How is $\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 } $ equivalent to $\frac{ y dx + xdy - zdz}{0}=\frac{ xdx - ydy -zdz}{0}$? In the book of PDE by Kumar, it is given that
However, I couldn't figure out how is
$$\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 } $$
is equivalent ... | $$\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 } $$
Use the well known basic property of the fractions :
$$\text{if}\quad \frac{A}{B}=\frac{C}{D} \quad \text{then}\quad \frac{A}{B}=\frac{C}{D}=\frac{c_1A+c_2C}{c_1B+c_2D}$$
$c_1,c_2$are arbitrary constants (not both nul}.
This property is valid for m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3144760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is wrong with this proof of $3\arcsin x$? We know that
\begin{align}
2\arcsin x&= \arcsin \left(2x\sqrt{1-x^2}\right) \tag{1}\\
\arcsin x + \arcsin y &= \arcsin \left[x\sqrt{1-y^2}+y\sqrt{1-x^2}\right] \tag{2}\\
3\arcsin x &= \arcsin x + 2\arcsin x \tag{3}
\end{align}
Thus $x=x, y=2x\sqrt{1-x^2}$
using ($1$)... | For all positive $x$ (the case of negative $x$ is symmetric), $$\sin3x=3\sin x-4\sin^3x.$$
So with $x=\arcsin t$ we have
$$\sin(3\arcsin t)=3t-4t^3.$$
This allows us to write
$$3\arcsin t=\arcsin(3t-4t^3)\lor3\arcsin t=\pi-\arcsin(3t-4t^3).$$
As the range of the arc sine is $\left[0,\dfrac\pi2\right]$, the first ident... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the polynomials $f(x)$ and $g(x)$ with integer coefficients such that the following equation is true.
(a) Find the polynomials $f(x)$ and $g(x)$ with integer coefficients such that
$$
\dfrac{f(\sqrt 3 + \sqrt 5)}{g(\sqrt 3 + \sqrt 5)} = \sqrt 3
$$
(b) Find $f$ and $g$ so that $$
\dfrac{f(\sqrt 3 + \sqrt 5... | Just take $f(x)=x^2-2$, $g(x)=2x$ for part (a) and $f(x)=x^2+2$, $g(x)=2x$ for part (b). It's not hard to check that $f, g$ are satysfying all conditions.
However, how we can find such polynomials? Firstly, you can try to find $f, g$ in form $f(x)=a_1x^2+b_1x+c_1$ and $g(x)=a_2x^2+b_2x+c_2$ by solving corresponding sys... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3152807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Finding $\lim_{n \to \infty} \int_0^n \frac{dx}{1+n^2\cos^2x}$ Find $$\lim_{n \to \infty} \int_0^n \frac{dx}{1+n^2\cos^2x}$$
I tried:
*
*mean value theorem.
*variable change with $ \tan x = t $ but I need to avoid the points which are not in the domain of $\tan$ and it's complicated.
| Result
Defining
$$f(n) = \int_0^n \frac{1}{1+n^2 \cos(x)^2} \,dx\tag{1}$$
we find
$$\lim_{n\to\infty}f(n)=1$$
Derivation
We split the integration region $(0,n)$ into the intervals $(0,\pi/2) ,( k \pi -\pi/2, k\pi +\pi/2)$ with $k = 1, 2, ...,k_{max}= n/\pi$.
Hence letting
$x=k \pi + \xi$ so that $\cos(x) = (-1)^k \cos(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3152960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to prove that function $f(p)= \frac{1}{(p^2 - 1)} \sum_{q=3}^p \frac{q^2-3}{q}[q\text{ is prime}]$ is less than 0.3 for all $p$? How to prove that function $f(p)= \frac{1}{(p^2 - 1)} \sum_{q=3}^p \frac{q^2-3}{q}[q\text{ is prime}]$ is less than 0.3 for all $p$?
Let's suppose that we have a function defined as follo... | The result is true for $p = 3$, $5$, and $7$, so assume that $p = 2n+1$ for $n \ge 4$.
Note that all the primes $q$ occurring in the sum are odd. Thus
$$
\begin{aligned}
f(p) = f(2n+1) = & \ \frac{1}{4n^2 + 4n} \sum_{q=3}^{p, \text{ with $q$ prime}} \frac{q^2 - 3}{q}\\
< & \ \frac{1}{4(n^2 + n)} \sum_{q=3}^{p, \text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3153287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Finding a basis of the annihilator of a subspace Let $V\subset\Bbb R^4$ be the subspace spanned by $e_1+e_2+e_3+e_4$ and $e_1+2e_2+3e_3+4e_4$. Find a basis of the subspace $V^\circ$ in the dual space $(\Bbb R)^*$.
My attempt: Let $a = e_1+e_2+e_3+e_4$ and $b = e_1+2e_2+3e_3+4e_4$, then $a$ and $b$ are lin. independent ... | Did it the long way here. To find the basis of $V^{\circ}$, we want linear functionals $f\in V^{*}$ s.t. $f(v)=0$, where $v = a_1v_1+a_2v_2$ ; $a_1,a_2\in \mathbb{R}$.
$$v_1 = e_1+e_2+e_3+e_4 = \begin{pmatrix}
1 \\
1 \\
1 \\
1 \\
\end{pmatrix} \& \text{ }v_2 = e_1+2e_2+3e_3+4e_4 = \begin{pmatrix}
1 \\
2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3153452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Does this method always produce the minimal polynomial over $\Bbb{Q}$? I have used it a few times and so far I have not seen an instance in which it did not give me the minimal polynomial over $\Bbb{Q}$.
The method is the following: For an element $a \notin \Bbb{Q}$ set $a=X$ and use arithmetic operations until we get ... | For example, suppose you set $X = \sqrt{3 + 2\sqrt{2}}$, and you try to find the minimal polynomial using $X^2 = 3 + 2\sqrt{2}$; $X^2 - 3 = 2\sqrt{2}$; $(X^2 - 3)^2 = 8 = X^4 - 6X^2 + 9$; $X^4 - 6X^2 + 1 = 0$. Your method would thus find $X^4 - 6X^2 + 1$ as the minimal polynomial.
However, secretly, $X$ is actually eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Using Cardano's method to find an algebraic equation whose root is $\sqrt{2} +\sqrt[3]{3}$ $\sqrt{2} +\sqrt[3]{3}$ solution of an algebraic equation $ a_{0}x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+\cdots+a_{0}=0 $
How to find this equation?
I tried Cardano's method, noting that
$$\sqrt{2} +\sqrt[3]{3} = \sqrt[3]{\sqrt{8}} +\sq... | In the same spirit as Awe Kumar Jha, let $x=\sqrt 2 + \sqrt[3]{3}$, that is to say
$$x-\sqrt 2 =\sqrt[3]{3}\implies (x-\sqrt 2) ^3=3$$ Expand and group terms to get
$$x^3-3 \sqrt{2} x^2+6 x-(2 \sqrt{2}+3)=0$$ Use Cardano method; it will show that this is the only real root of the equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3159254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Sum of all values that satisfy $\frac{x^{x-3}}{x}=\frac{x}{x^\frac{4}{x}}$.
What is the sum of all values of $x$ that satisfy the equation
$\frac{x^{x-3}}{x}=\frac{x}{x^\frac{4}{x}}$?
I start off by cross multiplying.
$$x^2=x^{x-3+\frac{4}{x}}$$
Taking the square root of both sides gives me:
$$x=\pm x^{\frac{1}{2}x... | The reasoning here is faulty:
I can't log both sides because of the $-1$, so I raise both sides by a power of $2x$ to get rid of the fraction. Because $2x$ is even, the RHS becomes a $1$.
For one thing, what if $x$ is not an integer, and $2x$ is not even?
Starting from $x^\frac{x^2-5x+4}{2x}=-1$, you could absolute v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3160152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Help on $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{H_n^3}{n}$ From this post:
The sum: $$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{H_n^3}{n+1}=-\frac{9}{8}\zeta(3)\ln(2)+\frac{\pi^4}{288}-\frac{\ln^4(2)}{4}+\frac{\pi^2}{8}\ln^2(2)$$
by moving the $n+1$ back to $n$
What is the sum of: $$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{H_n^3}{n}?$... | Here might be a start. Observe that$$\begin{align*}\sum\limits_{n\geq1}(-1)^{n+1}\frac {H_n^3}{n+1} & =\frac {H_1^3}2-\frac {H_2^3}3+\frac {H_3^3}4-\cdots\\ & =\sum\limits_{n\geq2}(-1)^n\frac {H_{n-1}^3}n\end{align*}$$
Now use the fact that $H_n=H_{n-1}+\tfrac 1n$. Therefore, the original sum becomes$$\begin{align*}\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3161046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Compute $I = \int_{0}^{\infty} \frac{8}{\gamma^{4}} x^{3} e^{-\frac{2}{\gamma^{2}}x^{2}} e^{jxt} dx$ I want to evaluate the following integral:
$$I = \int_{0}^{\infty} \frac{8}{\gamma^{4}} x^{3} e^{-\frac{2}{\gamma^{2}}x^{2}} e^{jxt} dx $$
Where $\gamma \in \mathbb{R}$ and $j = \sqrt{-1}$.
The first thing I do is let... | Hint: find $$\int_0^\infty e^{-2 x^2/\gamma^2} e^{jxt}\; dx $$
and differentiate $3$ times with respect to $t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3161867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Pólya Enumeration Theorem and Chemical Compounds How many different chemical compounds can be made by attaching H, CH3, or OH radicals to each of the carbon atoms in the benzene ring of Figure 1? (Assume that all of the C–C bonds in the ring are equivalent.)
Figure 1:
I understand the overall method of the Pólya-Burns... | Here is a solution via the Polya Enumeration Theorem. The problem is the same as determining the number of ways of coloring the vertices of a hexagon with three colors.
The symmetry group of the hexagon is the dihedral group $D_6$. The cycle index of the group is
$$Z=\frac{1}{12} (x_1^6 + 4 x_2^3 +2 x_3^2 +2 x_6 + 3x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3162759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Linearizing the trigonometric functions or: Squaring the circle by Fourier transformation It's an easy exercise to approximate the cosine and the sine function by a piecewise linear function on the unit interval $[0,1]$. Let $\tau = 2\pi$.
Let
$$\boxed{\cos_\bigcirc(x) = \cos(\tau x)\\\sin_\bigcirc(x) = \sin(\tau x)}$$... | The work has been done by user J.M. (who seems to be a mathematician of sorts), so let me just put it in order.
First let me rewrite the definition of $\cos_\square(x)$ a bit:
$$\cos_\square(x) = \begin{cases}
+1 & \text{ for } -1 \leq 8x \leq 1 \\
+2 - 8x & \text{ for }\ \ \ \ \ 1 \leq 8x \leq 3 \\
-1 & \text{ for }\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3163034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
induction proof: number of compositions of a positive number $n$ is $2^{n-1}$ So my problem was to find the formula for the number of composition for a positive number $n$. Then prove its validity. I found the formula which was $2^{n-1}$. Having trouble with the induction.
$$Claim:2^0+2^1+2^2+...+(2^{n-2}+1)=2^{n-1}$$
... | You actually don't need induction to prove this. There's a telescoping effect.
$$Claim:2^0+2^1+2^2+...+(2^{n-2}+1)=2^{n-1}$$
Carry the one.
$$(1 + 2^0)+2^1+2^2 + ... + 2^{n-2}=2^{n-1}$$
$1 = 2^0$
$$(2^0 + 2^0)+2^1+2^2 + ... + 2^{n-2}=2^{n-1}$$
Collapse the parentheses
$$(2^1 + 2^1) + 2^2 + ... + 2^{n-2}=2^{n-1}$$
Colla... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3163924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
show this inequality to $\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $ Let $a,b$ and $c$ be positive real numbers. Prove that
$$\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $$
This problem is from Iran 3rd round-2017-Algebra final exam-P3,Now I can't find... | The proof of my inequality.
let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\tfrac{a^3b}{(2a+3b)^3}+\tfrac{b^3c}{(2b+2c)^3}+\tfrac{c^3a}{(2c+3a)^3}\geq\tfrac{a^2bc}{(2a+2b+c)^3}+\tfrac{b^2ca}{(2b+2c+a)^3}+\tfrac{c^2ab}{(2c+2a+b)^3}.$$
Indeed, by Holder and AM-GM we obtain:
$$\sum_{cyc}\tfrac{a^3b}{(2a+3b)^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3172778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why does the Lagrange multiplier $\lambda$ change when the equality constraint is scaled? Consider the problem
$$\begin{array}{ll} \text{maximize} & x^2+y^2 \\ \text{subject to} & \dfrac{x^2}{25} + \dfrac{y^2}{9} = 1\end{array}$$
Solving this using the Lagrange multiplier method, I get
$$x = \pm5, \qquad y = 0, \qquad... | Let $f(x,y)=x^2+y^2+\lambda(9x^2+25y^2-225).$
Thus, from $$\frac{\partial f}{\partial x}=2x+18\lambda x=0$$ and
$$\frac{\partial f}{\partial y}=2y+50\lambda y=0$$ we obtain two possibilities: $\lambda=-\frac{1}{9}$ or $\lambda=-\frac{1}{25}.$
The second gives a minimal value, wile the first gives a maximal value:
$$f(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3177834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
prove this inequality with $x^3+y^3+z^3$
Let $x,y,z>0$. Show that
$$x^3+y^3+z^3+2(xy^2+yz^2+zx^2)\ge 3(x^2y+y^2z+z^2x).$$
I have a solution by using $y=x+a,z=x+a+b$; my question is: can this be solved using simple methods (such as AM-GM, C-S, and so on)?
| We need to prove that $$\sum_{cyc}(x^3-3x^2y+2x^2z)\geq0$$ or
$$\sum_{cyc}(2x^3-6x^2y+4x^2z)\geq0$$ or
$$\sum_{cyc}(2x^3-x^2y-x^2z)\geq5\sum_{cyc}(x^2y-x^2z)$$ or
$$\sum_{cyc}(x+y)(x-y)^2\geq5(x-y)(x-z)(y-z).$$
Since our inequality is cyclic, we can assume that $x=\max\{x,y,z\}$ and since for $x\geq z\geq y$ we have $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to solve this with mathematical induction? $$\sum_{i=0}^n \frac{i}{2^i} = 2 - \frac{n+2}{2^n} $$
Let's skip the check, since when n = 1, I have $\frac{1}{2} = \frac{1}{2}$
What i will next do ? What for expression i may receive ?
$$\sum_{i=0}^{n+1}\frac{i}{2^i} = 2 - \frac{n+2}{2^n} $$
My trying :
$$ \biggl(\sum_... | For the inductive step you have to show, that $\sum_{i=0}^{n+1} \frac{i}{2^i}=2-\frac{(n+1)+2}{2^{n+1}}$.
We have to use the inductive claim and go like this:
$\begin{align}\sum_{i=0}^{n+1} \frac{i}{2^i}\\&=\color{red}{\sum_{i=0}^{n} \frac{i}{2^i}}+\frac{n+1}{2^{n+1}}\\&=\color{red}{2-\frac{n+2}{2^n}}+\frac{n+1}{2^{n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Discontinuity of Hypergeometric function along the branch cut I am trying to evaluate an expression involving the hypergeometric function evaluated near its (principal) branch cut discontinuity, which is placed on the real line from $1$ to infinity.
For $x>1$, DLMF gives the following expression for the difference (or ... | It turns out that to answer the question it is sufficient to employ in a suitable way some other formulas already present in the DLMF.
A possibility is to use the ''inversion formula''
$$
\frac{\sin(\pi(b-a))}{\pi}\,
\mathbf F\left(\begin{array}{} a,b\\\ \ c\end{array};z\right)
=
\frac{(-z)^{-a}}{\Gamma(b)\Gamma(c-a)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $t_n=\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}-\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}-\frac{1}{4n}$. Find $\lim_{n \to \infty} nt_n$ If $t_n=\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}-\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}-\frac{1}{4n}$
Find $\lim_{n \to \infty} nt_n$
First attempt: $t_n$ is positive(grouping two... | Here is a
completely elementary proof
by algebraic manipulation that
$\dfrac{1}{8n}-\dfrac{1}{16n^2}
\lt t_n
\lt \dfrac1{8n}
$.
$\begin{array}\\
t_n
&=\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}-\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}-\frac{1}{4n}\\
&=\sum_{k=2n+1}^{4n} \dfrac{(-1)^{k+1}}{k}\\
&=\sum_{k=n}^{2n-1} (\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Quickest way to find $a^5+b^5+c^5$ given that $a+b+c=1$, $a^2+b^2+c^2=2$ and $a^3+b^3+c^3=3$
$$\text{If}\ \cases{a+b+c=1 \\ a^2+b^2+c^2=2 \\a^3+b^3+c^3=3}
\text{then}\ a^5+b^5+c^5= \ ?$$
A YouTuber solved this problem recently and, though he spent some time explaining it, took over 40 minutes to solve it.
Like the v... | I think to use an identity must be the shortest way:-
${ (a^5+b^5+c^5)=(a^4+b^4+c^4)(a+b+c)-(a^3+b^3+c^3)(ab+bc+ca)+abc(a^2+b^2+c^2)} $
If we square the 1st equation we get
$ {ab+bc+ca=\frac {-1}2} $
On squaring the above equation we get
$ {(ab)^2+(bc)^2+(ca)^2+abc(a+b+c)=\frac 14} $
We also know
${ (a^3+b^3+c^3)-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
} |
Distance Formula Problem If two vertices of an equilateral triangle are $(1, -1)$ and $(-\sqrt{3}, - \sqrt{3})$, find the coordinates of the third vertex.
Step by Step procedure to get the answer.
Take $A=(1, -1)$, $B=(-\sqrt{3}, - \sqrt{3})$.
Let the third vertex be $C=(x, y)$.
The distance d between the two points $A... | Equilateral triangle
$A=(1,-1)$
$B=(-\sqrt{3},-\sqrt{3})$
$C=(x,y)$
$|AB|=2\sqrt{2}$
height of an equilateral triangle:
$h=\frac{1}{2}\sqrt{3}a$
$a=|AB|$
$h=\frac{1}{2}\sqrt{3}.2\sqrt{2}=\sqrt{6}$
noting that
$|BO|=\sqrt{(0-(-\sqrt{3}))^2+(0-(-\sqrt{3}))^2}=\sqrt{6}$
O = (0,0) origin Cartesian Cordinate System
BO is fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3183526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
$f$ dividing by $x + 1$ have remainder 4, when dividing with $x^2 + 1$ have remainder 2x+3. Find remainder dividing polynomial with($x+1$)($x^2+1$) Problem: The polynomial $f$ dividing by ($x + 1$) gives the remainder 4, and when dividing with ($x^2 + 1$) gives the remainder (2x+3). Determine the remainder when dividin... | The hint.
Since $\deg((x+1)(x^2+1))=3,$ the degree of $r$ must be less than $3$.
Let $r(x)=ax^2+bx+c$ and take $x=-1$, $x=i$ and $x=-i$.
Now, solve the following system.
$$a-b+c=4,$$
$$-a+bi+c=2i+3$$ and
$$-a-bi+c=-2i+3.$$
I got $$r(x)=1.5x^2+2x+4.5.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3185530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What am I doing wrong when testing for concavity with $f(x)=5x^{2/3}-2x^{5/3}$? I can't figure out what I'm doing wrong when testing for concavity with $f(x)=5x^{2/3}-2x^{5/3}$. I can't find the right intervals for where the graph is concave-down and concave-up.
My Steps:
find $f'(x)$:
$$f'(x)=\frac{10}{3}x^{-1/3}-\fr... | Keep going with fractional exponents, it's much easier:
$$
f'(x)=\frac{10}{3}x^{-1/3}-\frac{10}{3}x^{2/3}
$$
Therefore
$$
f''(x)=-\frac{10}{9}x^{-4/3}-\frac{20}{9}x^{-1/3}=-\frac{10}{9}x^{-1/3}(x^{-1}-2)
$$
This is undefined for $x=0$, and it should be easy study the sign elsewhere.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3186799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Computing $K$-theory elements in a $C^*$ algebra $A$ Let $A$ be a unital $C^*$ algebra. Let $p,q$ be projections in $M_n(A)$. Then $[p]-[q]$ defines an element in $K_0(A)$.
Now consider the matrices, the projections,
$$
\left[ \begin{pmatrix}
1-p & 0 \\
0 & q
\end{pmatrix} \right] - \left[\begin{pmatrix} 1 & 0 \\ 0 ... | I will assume you mean
$$\left[\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \right]-\left[ \begin{pmatrix}
1-p & 0 \\
0 & q
\end{pmatrix} \right] ,
$$
since your first matrix is not a projection. Recall that
$$
\left[\begin{pmatrix} p & 0 \\ 0 & 0 \end{pmatrix} \right]=\left[\begin{pmatrix} 0 & 0 \\ 0 & p \end{pmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3188041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Number of real roots of $p_n(x)=1+2x+3x^2+....+(n+1)x^n$ if $n$ is an odd integer
If $n$ be an odd integer. Then find the number of real roots of the polynomial equation $p_n(x)=1+2x+3x^2+....+(n+1)x^n$
$$
p_n(x)=1+2x+3x^2+....+(n+1)x^n\\
x.p_n(x)=x+2x^2+....nx^n+(n+1)x^{n+1}\\
p(x)[1-x]=1+x+x^2+....+x^n-(n+1)x^{n+1}... | You have found that $p(x)$ is a rational function. It is zero if and only if its numerator is zero. Its numerator is a polynomial of the form $Ax^{n+2}+Bx^{n+1}+C$, where $A,B,C$ are constants. The derivative of that numerator is of the form $Dx^{n+1}+Ex^n=x^n(Dx+E)$ for some constants $D,E$. It's easy to see how many ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3188364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Definite Integral of $\int_0^1\frac{dx}{\sqrt {x(1-x)}}$ We have to calculate value of the following integral :
$$\int_0^1\cfrac{dx}{\sqrt {x(1-x)}} \qquad \qquad (2)$$
What i've done for (2) :
\begin{align}
& = \int_0^1\cfrac{dx}{\sqrt {x(1-x)}} \\
& = \int_0^1\cfrac{dx}{\sqrt {x-x^2}} \\
& = \int_0^1\cfrac{dx}{\s... | Are you sure that you got the correct antiderivative for the integral of the secant function? The correct indefinite integral of secant is $\int\sec{x}=\ln{|\tan{x}+\sec{x}|}+C$. Thus:
$$
\int_{0}^{1}\sec{\theta}\,d\theta=\ln{|\tan{\theta}+\sec{\theta}|}\bigg|_{0}^{1}.
$$
EDIT:
Also note that:
$$
x-x^2=-(x^2-x)=-\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3191401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Solution to $\sqrt{\sqrt{x + 5} + 5} = x$ There are natural numbers $a$, $b$, and $c$ such that the solution to the equation
\begin{equation*}
\sqrt{\sqrt{x + 5} + 5} = x
\end{equation*}
is $\displaystyle{\frac{a + \sqrt{b}}{c}}$. Evaluate $a + b + c$.
I am not sure where I saw this problem. My guess is that it was fro... | use the following way
$$x=\sqrt{5+\sqrt{5 + x} }$$
$$x=\sqrt{5+\sqrt{5 + \sqrt{5+\sqrt{5 + \sqrt{5+\sqrt{5 + ....} }} }} } $$
or
$$x=\sqrt{5+x }$$
$$x^2-x-5=0$$
$$x=\frac{1}{2}\pm\frac{\sqrt{21}}{2}$$
now use long division to get the other roots and then check which which one satisfies the original equation
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3191768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Extracting coefficients from two-dimensional generating function We have the two-dimensional recurrent series $F(r+1,s+2) = F(r,s) + F(r,s+1) + F(r,s+2)$ and the boundary conditions $F(r,0)=1$, $F(0,s)=0$ for all $s>0$ and $F(0,0)=1$ and $F(r,1)=r$. This series is for all $r\geq0$ and $s\geq0$. How do we find the gener... | We use the coefficient of operator $[z^n]$ do denote the coefficient of $z^n$ in a series.
We obtain for $m,n\geq 1$:
\begin{align*}
\color{blue}{[x^my^n]}&\color{blue}{\frac{xy}{1-x(1+y+y^2)}}\\
&=[x^{m-1}y^{n-1}]\frac{1}{1-x(1+y+y^2)}\tag{1}\\
&=[x^{m-1}y^{n-1}]\sum_{j=0}^\infty x^j\left(1+y+y^2\right)^j\tag{2}\\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3193310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove polynomial map between two affine curves is bijection except origin I wanna to solve this problem dealing with a polynomial map between two affine curves.
First curve is $A \subset \mathbb{C}^2$ defined by equation $s(1+t^2)-1=0$
and the second curve is $B \subset \mathbb{C}^2$ defined by equation $(x^2+y^2)^2+3... | We can approach this by solving the polynomial system
$$
\left\{\begin{matrix}
x = ts^2(1-3t^2)\\
y = s^2(1-3t^2)
\end{matrix} \right.
$$
with $(x,y) \neq (0,0)$ satisfying $(x^2+y^2)^2+3x^2y-y^3=0$ and $(s,t)$ satisfying $s(1+t^2)=1$. First assume that $y\neq 0$, then $t = \frac{x}{y}$. From the equation of $A$ we see... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3193805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $a$ and $b$ for which $\int_{0}^{1}( ax+b+\frac{1}{1+x^{2}} )^{2}\,dx$ takes its minimum possible value.
Calculate for which values $a$ and $b$ the integral
$$\int_{0}^{1} \left( ax+b+\frac{1}{1+x^{2}} \right)^{2}\,dx$$
takes its minimum possible value?
For being honest I'm not sure how to try this, but my idea... | Let \begin{align}f(a,b)=\int_{0}^{1} \left( ax+b+\frac{1}{1+x^{2}} \right)^{2}dx\implies \frac{\partial f(a,b)}{\partial a}&=\int_0^12x\left( ax+b+\frac{1}{1+x^{2}} \right)\,dx\\&=\left[\frac{2ax^3}3+bx^2+\ln(1+x^2)\right]_0^1\end{align} so $$\frac{\partial f(a,b)}{\partial a}=\frac23a+b+\ln2=0\tag1$$ for critical poin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3195030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
For game where first player to N points wins, find the distribution of win probability and total number of points between players
Two players, A and B, play a series of points in a game with player A winning each point with probability p and player B winning each point with probability
q = 1 - p. The first player to... | Your calculation looks correct. As for why $\mathbb{P}(Y\in\{3,4,5\})=1,$ think in the following way.
$Y$ is the number of points at the end of the game.
Here $N=3,$ so $Y$ to be at least $3$, since a player can win at most $k$ points in $k$ trials. Now suppose we have $5$ points at the end, but neither player $A$ nor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3195166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
to prove $x^2 + y^2+1\ge xy + y + x$ $$x^2 + y^2+1\ge xy + y + x$$
$x$ and $ y$ belong to all real numbers
my attempt
$(u-2)^2\ge0\Rightarrow \frac{u^2}{4}+1\ge u $
let $u=x+y\Rightarrow \frac{(x+y)^2}{4}+1\ge x+y$
$\Rightarrow (x+y)^2+1\ge \frac{3}{4}(x+y)^2+(x+y)$
$but \frac{(x+y)^2}{4} \ge xy $ by AM-GM inequality
$... | Let $c=x^2+y^2+1-(xy+x+y)$
$\iff x^2-x(1+y)+1-y+y^2-c=0$
As $x$ is real, the discriminant must be $\ge0$
i.e., $$(1+y)^2\ge4(1-y+y^2-c)\iff4c\ge3(1-y)^2$$ which is $\ge0$ for real $y$
Alternatively,
$$ x^2-x(1+y)+1-y+y^2=\left(x-\dfrac{1+y}2\right)^2+\dfrac{3(1-y)^2}4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3197998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Advanced Complex numbers Let $\omega$ be a nonreal root of $z^3 = 1.$ Let $a_1,$ $a_2,$ $\dots,$ $a_n$ be real numbers such that
$$\frac{1}{a_1 + \omega} + \frac{1}{a_2 + \omega} + \dots + \frac{1}{a_n + \omega} = 2 + 5i.$$Compute
$$\frac{2a_1 - 1}{a_1^2 - a_1 + 1} + \frac{2a_2 - 1}{a_2^2 - a_2 + 1} + \dots + \frac{2a_... | Rationalize the denominator for each term of the left hand side & the right hand side
For integer $n$ not divisible by$3$
Let $p=\dfrac{2n\pi}3,2\cos p=-1\ \ \ \ (1)$
$$\dfrac1{a_j+w}=\dfrac1{a_j+\cos p+i\sin p}=\dfrac{(a_j+\cos p)-i\sin p}{a_j^2+2a_j\cos p+1}$$
Use $(1)$
Equate the real parts
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3198831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $A =\left[ \begin{smallmatrix} 3 & -4 \\ 1 & -1\end{smallmatrix}\right]$, then find $A^n$
If $$ A = \begin{bmatrix}
3 & -4 \\
1 & -1
\end{bmatrix}$$
Then find $A^n$
I have tried solving it using diagonalization $PDP^{-1}$ but I am getting only one independent eigenvector i.e $$ \begin{bmatrix}
2\\
1
\end{bmatrix... | Indeed, $A$ has one and only one eigenvalue: $1$. And $(2,1)$ is an eigenvector corresponding to that eigenvalue. Now, consider the vector $(1,0)$. Then$$A.(1,0)=(3,1)=(2,1)+(1,0).$$So, if $M=\left[\begin{smallmatrix}2&1\\1&0\end{smallmatrix}\right]$, then$$M^{-1}.A.M=\begin{bmatrix}1&1\\0&1\end{bmatrix},$$or$$A=M.\beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3201878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Jordan normal as transformation with respect to the basis of eigenvectors I have the following matrix
$$A = \begin{pmatrix}
2 & 0 & 1 & -3 \\
0 & 2 & 10 & 4 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 3 \\
\end{pmatrix}$$
and its Jordan normal form is
$$J = \begin{pmatrix}
2 & 0 & 0 & 0 \\
... | One way of seeing where equations like $\ A\mathbf{v_{3}} = \mathbf{v_{2}} + 2\mathbf{v_{3}}\ $ come from, which I found helpful when first introduced to Jordan forms, is the definition of the invariant subspace corresponding to an eigenvalue $\ \lambda\ $ as that spanned by the non-zero vectors $\ \mathbf{v} $ for wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3202610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integer solutions to $x^2 - 15 y^2 = 15$ How would one show that $x^2 - 15 y^2 = 15$ has no integer solutions ?
I got that $x = \pm \sqrt{15 (y^2 +1) }$ and then WLOG I assume that $x \in \mathbb{N}$ and $x = \sqrt{15 (y^2 +1) }$. From there, I want to show that for every $y \in \mathbb{Z}$, $15 (y^2 +1)$ is not a squ... | If $15(y^2+1)$ is a square then $y^2+1$ is divisible by $15$, so in particular it is divisible by $3$, a contradiction.
Note that your remark
Clearly, $y^2+1$ is not a square...
is false because $0^2+1=1^2=(-1)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3206087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to prove an inequality using mathematical induction? I have to prove the following:
$$ \sqrt{x_1} + \sqrt{x_2} +...+\sqrt{x_n} \ge \sqrt{x_1 + x_2 + ... + x_n}$$
For every $n \ge 2$ and $x_1, x_2, ..., x_n \in \Bbb N$
Here's my attempt:
Consider $P(n): \sqrt{x_1} + \sqrt{x_2} +...+\sqrt{x_n} \ge \sqrt{x_1 + x_2 +... | I'm will just show it my way, since I find it simpler.
First, we will start off with the basis step where $n = 1$.
Basis step
$\sqrt{x_1} \geq \sqrt{x_1}$, which is of course true.
Inductive step
We let the inequality be true for any $k$, where $0 \leq k \leq n$:
$\sqrt{x_1} + \sqrt{x_2} + ... + \sqrt{x_k} \geq \sqrt{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3208412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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Fractions in Questions and Answers
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