Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove $x^{y+1}z+y^{z+1}x+z^{x+1}y\geq x^2y^2z^2$
Let $x>1$, $y>1$ and $z>1$ be such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$. Prove that:
$$x^{y+1}z+y^{z+1}x+z^{x+1}y\geq x^2y^2z^2.$$
When $x=y=z=3$, both sides are equal to $3^6$. The difficulty is how to deal with the variables appearing in the exponents, and ... | It's just AM-GM:
$$\sum_{cyc}x^{y+1}z=xyz\sum_{cyc}\frac{1}{y}x^{y}\geq xyz\prod_{cyc}\left(x^y\right)^{\frac{1}{y}}=x^2y^2z^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the equation $\sqrt{x^3+1}\left(4x-1\right)=2x^3+x^2+1$ Solve the equation $$\sqrt{x^3+1}\left(4x-1\right)=2x^3+x^2+1$$
$$\Leftrightarrow \sqrt{x^3+1}=\frac{2x^3+x^2+1}{4x-1}$$
$$\Leftrightarrow \sqrt{x^3+1}-\left(x+1\right)=\frac{2x^3+x^2+1}{4x-1}-\left(x+1\right)$$
$$\Leftrightarrow \frac{x^3+1-\left(x+1\right... | After squaring and rearranging we have:
$$4x^6-12x^5+9x^4+3x^3-14x^2+8x=0$$
Since there is no constant term, $x=0$ is a solution. Using the rational root theorem, $p$ and $q$ have to be coprime. Since the only factors of $4$ and $8$ are powers of $2$, then the only solutions are when either $1$ is in the numerator or $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find $ \lim\limits_{n \rightarrow \infty} \int_{0}^{1} \left(1+ \frac{x}{n}\right)^n dx$
Find the limit of
$$ \lim\limits_{n \rightarrow \infty} \int_{0}^{1} \left(1+ \frac{x}{n}\right)^n dx$$
Let $$u= 1 +\frac{x}{n} \implies du =\frac{1}{n} dx \implies n \cdot du = dx$$
at $x=0$ $u=1$ and at $x=1$ $u=1+\frac{1}{n}$ ... | Since
$\left(1+ \frac{x}{n}\right)^n
\to e^x
$,
$
\lim\limits_{n \rightarrow \infty} \int_{0}^{1} \left(1+ \frac{x}{n}\right)^n dx
= \int_{0}^{1} e^x dx
=e-1
$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that for all prime numbers $ \ a,b,c \ , a^{2} + b^{2} \neq c^{2}$ Question:
Prove that for all prime numbers $ \ a,b,c \ , a^{2} + b^{2} \neq c^{2}$
My attempt:
Proof by contradiction:
Assume $ \ \exists a,b,c $ prime numbers such that $ \ a^{2} + b^{2} = c^{2}$.
Then $ \ a^2 = c^2 - b^2 \implies a^2 = (c-b)(c+... | For any integer $a$, $a\equiv 0,1,2(\mod 3)\Rightarrow a^2\equiv 0,1,4(\mod 3)\Rightarrow a^2\equiv 0,1(\mod 3)$
Now $a,b,c$ are prime numbers and $a^2+b^2=c^2$. First, important thing to note that $a,b,c$ are distinct.
If $a=3$, then $(c+b)(c-b)=9$. Check that $b,c$ can not prime numbers.
If $b=3$, then $(c+a)(c-a)=9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum to infinity An infinite G.P. has a finite sum with initial term $u_1 = 2$. Find the sum to infinity of this G.P. If it is also a Fibonacci sequence
My approach-
*
*since it is a G.P., $ u_3 = u_2 * u_2/u_1$
*since it is a Fibonacci sequence, $u_3 = u_1+u_2$
*Therefore, $u_2 * u_2/u_1 = u_1+ u_2$
By substit... | Yes, your reasoning looks correct. I get the same answer you do in a slightly different form:
$$\begin{align*}
\sum_{i=1}^\infty u_i &= \frac{a}{1-r}\\
& = \frac{2}{1-\frac{1-\sqrt{5}}{2}}\\
&= \frac{4}{2-(1-\sqrt{5})}\\
&= \frac{4}{1+\sqrt{5}}
\end{align*}$$
(Since $1-\sqrt{5} < 0$, every other term is negative and th... | {
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Evaluating $\int_0^{\pi/4} (\cos 2x)^{3/2} \cos x \, dx$ The question is to evaluate $$\int_0^{\pi/4} (\cos 2x)^{3/2} \cos x \, dx$$ I tried replacing $x$ by $\pi/4 -x$ and solving but couldn't get the answer.please help me in this regard.thanks.
| $$\int_{0}^{\pi/4}\left(\cos(2x)\right)^{3/2}\,dx \stackrel{x\mapsto\frac{z}{2}}{=} \frac{1}{2}\int_{0}^{\pi/2}\left(\cos z\right)^{3/2}\,dz \stackrel{z\mapsto\frac{\pi}{2}-z}{=} \frac{1}{2}\int_{0}^{\pi/2}\left(\sin z\right)^{3/2}\,dz $$
equals:
$$ \frac{1}{2}\int_{0}^{1}u^{3/2}(1-u^2)^{-1/2}\,du \stackrel{u\mapsto v^... | {
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"timestamp": "2023-03-29T00:00:00",
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find $f(\frac{1}{2014})+f(\frac{2}{2014})+.....+f(\frac{2013}{2014})$ of $f(x)=\frac{2}{2+4^x}$ $f(x)=\frac{2}{2+4^x}$
find
$f(\frac{1}{2014})+f(\frac{2}{2014})+.....+f(\frac{2013}{2014})$
Please guide me through it, the only step I know is probably to eliminate the denominator
ps. Not a homework
| HINT:
$$f(1-x)=\dfrac2{2+4^{1-x}}=\dfrac{2\cdot4^x}{2\cdot4^x+4}=\dfrac{4^x}{4^x+2}=\dfrac{4^x+2-2}{4^x+2}=1-f(x)$$
Set $x=\dfrac r{2014}, 1\le r\le2013$ and add
| {
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How many positive integers less than $1000$ divisible by $3$ with sum of digits divisible by $7$?
How many positive integers less than $1000$ are divisible by $3$ with their sum of digits being divisible by $7$?
Well, I got Answer: $28$.
$a+b+c = 7k$ and $a,b,c$ are multiples of 3 so $a+b+c$ is a multiple of both $7... | The digit sum has to divisible by $3$ and by $7$, hence by $21$. This implies that we have to deduct a total of $6$ points from the maximal possible $9+9+9=27$. By stars and bars these $6$ points can be distributed in ${8\choose 2}=28$ ways onto the three decimal places.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402018",
"timestamp": "2023-03-29T00:00:00",
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If a, b, c>0 show that $\frac{a^2}{b^2+bc}+\frac{b^2}{c^2+ac}+\frac{c^2}{a^2+ab} \ge \frac{3}{2}$
For positive real numbers $a$, $b$, and $c$ prove that: $$\frac{a^2}{b^2+bc}+\frac{b^2}{c^2+ac}+\frac{c^2}{a^2+ab} \ge \frac{3}{2}.$$
I let $x=\frac{a}{b}$, $y=\frac{b}{c}$, and $z=\frac{c}{a}$. Then inequality becomes
$... | You can end your way.
Since $xyz=1$, by using your work and AM-GM we obtain:
$$2(x+y+z)^2=(x+y+z)^2+(x+y+z)^2\geq9+3(xy+xz+yz),$$
which you want to get.
Also, by C-S
$$\sum_{cyc}\frac{a^2}{b^2+bc}=\sum_{cyc}\frac{a^4}{a^2b^2+a^2bc}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^2b^2+a^2bc)}\geq\frac{3}{2}$$
because the... | {
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"url": "https://math.stackexchange.com/questions/2402336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $\frac{a(a^2+2)}{3}$ is an integer for all integer $a\geqslant 1$ If $\frac{a(a^2+2)}{3}$ then $3\mid{a(a^2+2)}$.
By induction:
Lets define the set, $S=\left\{a\in N:a\geqslant1, 3\mid a(a^2+2) \right\}$
If $a=1$ then, $1\in S$
So we have to prove that if $k(k^2+2)=3m$ then $(k+1)((k+1)^2+2)=3n$ with $m,n\in Z$
I... | I'd suggest writing the proof in a more linear fashion which is easier to read. Below is how I would approach the induction step.
Let $P(n)$ be the statement $3\mid n(n^2+2)$.
Proving $P(k)\implies P(k+1)$:
$$P(k)\implies 3\mid k(k^2+2)$$
$\implies$
$$3\mid k(k^2+2)+3(k^2+k+1)$$
$\implies$
$$3\mid k^3+3k^2+5k+3$$
$\im... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Find the value of $\frac{9}{5}(a+b)$ given that $a\sqrt{a}+b\sqrt{b}=183$ and $a\sqrt{b}+b\sqrt{a}=182$
Suppose $a,b$ are positive real numbers such that $a\sqrt{a}+b\sqrt{b}=183$, $a\sqrt{b}+b\sqrt{a}=182$. Find $\frac{9}{5}(a+b)$.
It is my equation. I subtracted the second equation from the first one and found $(a-... | Let $A^2=a$ and $B^2=b$ then we have
\begin{eqnarray*}
A^3+B^3 = 183 \\
A^2B+B^2 A =182
\end{eqnarray*}
Multiply the second equation by $3$ and add the first
\begin{eqnarray*}
A^3+B^3+ 3(A^2B+B^2 A) =(A+B)^3= 729 \\
A+B =9 \\
AB = \frac{182}{9} \\
(A-B)^2=(A+B)^2-4AB= \frac{1}{9} \\
A= \frac{14}{3} \\
B= \frac{13}{3} ... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$ Prove that $ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$ when $A,B,C$ are the angles in a triangle.
I need a help on this question, one can enter $A=B=C$ and get the maximum value $\frac{1}{8}$, but that is not a so... | Let $a, b, c$ be the sides of a triangle with given angles $A, B, C$.
Recall for any $3$ positive numbers $a, b, c$, it can form the sides of a non-degenerate triangle if and only if we can find $3$ positive numbers
$u,v,w$ such that
$$a = v + w,\quad b = u + w,\quad c = u + v$$
This is known as Ravi substitution for a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Asymptotic behaviour of sum I would like to evaluate the number $c$ given by
$$
c = \lim_{m\to\infty} \frac{1}{\log m}\sum_{n=1}^m \frac{1}{n^2 \sin^2(\pi n \tau)}
$$
where $\tau = (1+\sqrt{5})/2$.
My attempt: my guess was this sum would be dominated by the terms for which $n$ is a Fibonacci number. I considered the s... | $\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}\def\cI{\mathcal{I}}$Amazingly, your sum can actually be computed in closed form. The answer is
$$c=\frac{2 \pi^2}{15 \sqrt{5} \log \tau}.$$
Notation: Let $\tau = \tfrac{1+\sqrt{5}}{2}$ and $\bar{\tau} = \tfrac{1-\sqrt{5}}{2}$, so $\tau+\bar{\tau}=1$ and $\tau \bar{\tau} = -1$.
L... | {
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Determine all pairs $(a,b)$ of positive integers such that $ab^2+b+7$ divides $a^2b+a+b$
Determine all pairs $(a,b)$ of positive integers such that $ab^2 + b + 7$ divides $a^2b + a +b$.
I guess the trivial solution is $a=b=7$.
Also any $a$ and $b$ which are divisible by $7$ will seem to suit.
we can also write
$b(\fr... | First we will dealt with the cases when $b=1,2,3,4,5,6$ and then we can assume that $b \geq 7$,
For the case $b=1$ we get that $a+8 | a^2+a+1$ which means $a(a+8) +r = a^2+a+1$ and we want $r=0 \mod a+8$ so solving the above gives that $r = 1 - 7 a$ so we want that $a+8|1-7a$ which means that $-7(a+8)+r = 1-7a$ and we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can you find a Pythagorean triple with $a^2+b^2=c^2$ and $a/b$ close to $5/7$? How can you find a Pythagorean triple with $a^2+b^2=c^2$ and $a/b$ close to $5/7$?
I've been reading the Plimpton 322 news, and this fits in the gap in the Babylonian table between 0.6996 ($a=1679,b=2400$) and 0.75 ($a=3,b=4$). The Baby... | Pythagorean triplets are characterized by being of the form
$$a= m^2-n^2 \\ b= 2mn \\ c=m^2+n^2$$
So you could look for two integers $m>n$ such that
$$\frac{m^2-n^2}{2mn} \approx \frac{5}{7}$$
or, equivalently,
$$\frac{m}{n}- \frac{n}{m} \approx 2 \cdot \frac{5}{7}$$
Now, let $x$ be unique positive solution of
$$x-x^{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2409907",
"timestamp": "2023-03-29T00:00:00",
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Find the locus of $w=1/z$ when $z$ lies on $y=2x+1$ For the transformation $w=\frac{1}{z}$, find the locus of $w$ when $z$ lies on the line with equation $y=2x+1$
I'm not quite sure how or where to start on this question. I know for instance that the locus of $z$ such that
$$|z-2| = |z+2-2i|$$
is the line $y=2x+1$. I'm... | Suppose $z=x+iy$ then as $y=2x+1$ we have $w=\dfrac{1}{x +(2x+1)i}$
$\dfrac{1}{x +(2x+1)i}=\dfrac{x-(2x+1)i}{5 x^2+4 x+1}$
If $w=a+bi$ then we have $(a,b)=\left(\dfrac{x}{5 x^2+4 x+1},\dfrac{-2 x-1}{5 x^2+4 x+1}\right)$
So
$a=\dfrac{x}{5 x^2+4 x+1};\;b=\dfrac{-2 x-1}{5 x^2+4 x+1}$
$x=\dfrac{1-4a\pm\sqrt{-4 a^2-8 a+1... | {
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"timestamp": "2023-03-29T00:00:00",
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Recurence relation $D(n) \le D(\frac{2}{3}n) + 2$ We know that
$D(1) = 0$.
Now from the inequality,
$$D(n) \le D\left(\frac{2}{3}n\right) + 2$$
it should follow that
$$D(n) \le 2\cdot \log_{\frac{3}{2}}(n).$$
*Edit: the logarithm has a base $\frac{3}{2}$
How do we get that last bound?
| (Disclaimer: the following assumes this is one of those contexts where certain hand-waving is allowed/expected, otherwise see the posted comments for why the problem is not well posed.)
Let $\,n = (3/2)^k \iff k = \log_{3/2}(n)\,$, and define $T(k)=D\big((3/2)^k\big)=D(n)\,$, then:
$$
T(k)=D \left(\left(\frac{3}{2}\rig... | {
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"timestamp": "2023-03-29T00:00:00",
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Period of $f(x)=\sin{2x}-\sin{\frac{x}{2}}$ I know the periods of the two functions $f(x)=\sin{2x}$ and $g(x)=\sin{\frac{x}{2}}$, which are $\pi$ and $4\pi$ respectively. But what is their period when you subtract them?
More generally, if $f(x)=\sin{(ax+c)}$ and $g(x)=\sin{(bx+d)},$ what is the period of $f(x)-g(x)?$ ... | In short, period of $\sin 2x$ is $\pi$ and period of $\sin \dfrac{x}{2}$ is $4\pi$. Then answer will be $\text{lcm} (1,4)\cdot\pi=4\pi$
Now, elaborately:
$\sin 2x-\sin \dfrac{x}{2}=2\sin (x-\dfrac{x}{4})\cos(x+\dfrac{x}{4})=2\sin \dfrac{3x}{4}\cos\dfrac{5x}{4}$
If $x=4k\pi$ and $k\in \mathbb{Z}$, either $\cos\dfrac{5x... | {
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"timestamp": "2023-03-29T00:00:00",
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Convergence of series $(\frac{1}{3})^{2}+(\frac{1.2}{3.5})^{2}+(\frac{1.2.3}{3.5.7})^{2}+...$ $$\left(\frac{1}{3}\right)^{2}+\left(\frac{1\cdot2}{3\cdot5}\right)^{2}+\left(\frac{1\cdot2\cdot3}{3\cdot5\cdot7}\right)^{2}+...$$
I am not able to find a general equation and that's creating problem for me as I can't proceed ... | We have to deal with:
$$ \sum_{n\geq 1}\left(\frac{n!}{(2n+1)!!}\right)^2=\sum_{n\geq 1}\left(\frac{2^n n!^2}{(2n+1)!}\right)^2 = \sum_{n\geq 1}\frac{4^n}{(2n+1)^2\binom{2n}{n}^2} $$
and since $\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}$, the above series is clearly convergent. We may recall that
$$ \frac{\arcsin z}{\sq... | {
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"timestamp": "2023-03-29T00:00:00",
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Show $x=0$ and $x=1$ are the only integer solutions I'm trying to show that the only solutions to $x^2-x+1=y^2$ are when $x=0$ and $x=1$. All I can think of is completing the square gives $x^2-x+1=(x-\frac{1}{2})^2+\frac{3}{4}$, which is clearly not a perfect square. Does this suffice? Should I use induction?
| $$x^2-x+1=y^2\\
4x^2-4x+4=4y^2 \\
(2y)^2-(2x-1)^2=3 \\
\left( 2y-2x+1\right) \left(2y+2x-1 \right)=3
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that the expression $5^{2n+1} \cdot 2^{n+2} + 3^{n+2} \cdot 2^{2n+1}$ is divisible by 19.
Prove that the expression
$$5^{2n+1} * 2^{n+2} + 3^{n+2} * 2^{2n+1}$$
is divisible by $19$.
I'll skip the basis step (as I have done last time) but I can conclude that it's only divisible by 19 for integers n ≥ 0 (or ... | $$5^{2n+1}2^{n+2}+3^{n+2}2^{2n+1}\equiv 5^{2n}2^n-3^n2^{2n}\pmod{19}\qquad (1)$$
since $5^12^2=20\equiv 1\pmod{19}$ and $3^22^1=18\equiv -1\pmod{19}$. Hence, $(1)$ yields the following
$$2^n(5^{2n}-3^n2^n)=2^n(25^n-6^n)=50^n-12^n\equiv 12^n-12^n\pmod{19}\equiv 0\pmod{19}$$
since $50\equiv 12\pmod{19}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 4
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What will be the $n^{th}$ term of this given series? Given series is:
$$1+\frac{1\times x^2}{2\times 4}+\frac{1\times 3\times 5\times x^4}{2\times 4\times 6\times 8}+\frac{1\times 3\times 5\times 7\times 9\times x^6}{2\times 4\times 6\times 8\times 10\times 12}+.....\infty$$
I need to find it's $n^{th}$ term but am hav... | The correct power of $X$ for when $n$ is the $n^{th}$ term of the series is
$$ a_n=K \times x^{2(n-1)} $$
You can deduce the correct formula starting from that
then for $ n>0 $ :
$$ a_n = \frac { K \times x^{2(n-1)} }{ 2^{2(n-1)} (2(n-1))!} $$
For $ n > 1 $ :
$$ k = \frac {(4n-7)!}{2^{2(n-2)} (2(n-2))!}$$
Therefore,
... | {
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"timestamp": "2023-03-29T00:00:00",
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Solving complex equation $z^2 + (1+i) \overline{z} + 4i = 0$ Consider the following equation, where $z \in \mathbb{C}$, $i$ is the imaginary unit and $\overline{z}$ is the conjugate of $z$:
$$
z^2 + (1+i) \overline{z} + 4i = 0
$$
What is the method to deal with equations such as this?
I have tried various things: I tri... | $$a^2-b^2+2iab+(1+i)(a-ib)+4i=0\\ a^2-b^2+a+b+(2ab+a-b+4)i=0\\ \begin{cases}a^2-b^2+a+b=0\\ 2ab+a-b+4=0\end{cases}\\ \begin{cases}(a+b)(a-b+1)=0\\ 2ab+a-b+4=0\end{cases}\\ \begin{cases}a=-b\\ -2a^2+2a+4=0\end{cases}\vee\begin{cases}b=a+1\\ 2a^2+2a+3=0\end{cases}$$
An these can be solved for $a,b\in\Bbb R$ accordingly.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\frac{1}{1^2}+\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2+3^2}+...\to ?\;\;$ (Click here.) $$\frac{1}{1^2}+\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2+3^2}+\dots\to~?$$
I tried like below
$$\frac{1}{1^2}+\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2+3^2}+\dots=\\\sum_{n=1}^{\infty}\frac{1}{1^2+2^2+3^2+\dots+n^2}=\\\sum_{n=1}^{\infty}\frac{1}{\fr... | Trying to avoid Daniel Fischer's approach.
$$\frac2{(2n)(2n+1)(2n+2)}=\frac1{(2n)(2n+1)}-\frac1{(2n+1)(2n+2)}$$
Collect the terms by their signs:
$$S=12\sum_{n=2}^\infty\frac{(-1)^n}{n(n+1)}$$
Now apply partial fractions on this nicely and it becomes
$$S=12\sum_{n=2}^\infty\frac{(-1)^n}n+\frac{(-1)^{n+1}}{n+1}$$
You ca... | {
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"source": "stackexchange",
"question_score": "7",
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Proof verification: Determine whether $f(x)$ is one-to-one or not
$$f:\mathbb{R}\mapsto\mathbb{R}
\text{ defined as } f(x)=2x^3+3x-4 \\
\underline{\textit{proof(by contradiction)}}\\
\text{By definition a function is called one-to-one if}\\
f(x)=f(y),\text{ } x,y\in\mathbb{R} \text{ implies that x=y.}\\
\text{Suppose... | In your edit, Case 1 is fine, but the "proof" in Case 2 isn't. If the assumption of Case 2 is that precisely one of $x$ and $y$ is negative, then you have to add Case 3 to you proof: the case when both $x<0$ and $y<0$. If the assumption of Case 2 is that at least one of $x$ and $y$ is negative, then it also includes th... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Evaluating $\max(ab+bc+ac)$
Let a,b,c be real numbers such that $a+2b+c=4$. What is the value of $\max(ab+bc+ac)$
My attempt:
Squaring both the sides:
$a^2 +4b^2+c^2+2ac+4bc+4ab=16$
Then I tried factoring after bringing 16 to LHS but couldn't. It's not even a quadratic in one variable or else I could have directly f... | Note that $4xy \leq (x+y)^2$ for all $x$, $y$.
You have $$ab+bc+ca = b(a+c) + ac \leq b(a+c) + \frac{1}{4}(a+c)^2 = (a+c) \frac{a+c+4b}{4} = \frac{1}{4}(a+c)(8-(a+c)) \leq \frac{1}{4}\frac{1}{4}8^2 = 4.$$
Equality is happen when $a=c$ and $a+c = 8-(a+c)$, or $a+c=4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx})$ Evaluate $$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx})$$
I tried the following:
$$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx}) \cdot \frac{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}$$
But ende... | another method: asymptotics
For $x>0$, we have as $x \to \infty$:
$$
\sqrt{x^2+ax} = x\sqrt{1+\frac{a}{x}} = x\left(1+\frac{a}{2x}+o(x^{-1})\right) = x+\frac{a}{2}+o(1)
\\
\sqrt{x^2+bx}= x+\frac{b}{2}+o(1)
\\
\sqrt{x^2+ax}-\sqrt{x^2+bx}= \frac{a-b}{2}+o(1)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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The arithmetic sequence is 13, 21, 29, 37... Find the least number of terms required for the sum of the sequence terms to exceed 1000. All I know is that a=13 and common difference=8. I am unaware how to continue from here.
| Just to do it slightly differently.
If I have $a_0 = 13$ and $a_k = 13 + 8*k$ and I have a list of $n$ of these terms $a_0=13, a_1=13+8, a_2=13+2*8,......, a_n=13 + n*8$, what is the average value of all of them.
Well as they are all exactly $8$ apart from each other, the average value will be directly in the middle o... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve trigonometric equation with $\arcsin$ and $\arccos$
Solve the following equation
$$2\arccos(x)-\arcsin(2x\sqrt{1-x^2})=0.$$
I tried to solve this equation, but I don't know how can I change $\arcsin$ to $\arccos$ to solve the equation.
| Hmmm what about the following approach ( there is a fair bit of trig )
Take the cosine of both sides to get
$$ \cos\left(\underbrace{2 \cos^{-1}(x)}_{\text{=Part 1}} - \underbrace{\sin^{-1}\left(2x\sqrt{1-x^2}\right)}_{\text{=Part 2}} \right) = 1. \tag{1} \label{eqn:main} $$
Look at each part individually.
$$ \cos... | {
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"source": "stackexchange",
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same area and perimeter
The perimeter and the area of a given triangle have the same numerical value when measured via a certain unit of measure. If the lengths of the three altitudes of the triangle are p, q, and r, what is the numerical value of $1/q + 1/p + 1/r$?
So first I set up the three sides' lengths as $a, b... | $$A = \frac{ap}{2}= \frac{bq}{2}= \frac{cr}{2}=a+b+c.$$
$$\frac{ap}{2}=a+b+c$$
$$p=\frac{2(a+b+c)}{a}$$
Similarly,
$$q=\frac{2(a+b+c)}{b}$$
$$r=\frac{2(a+b+c)}{c}$$
Now
$$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{a}{2(a+b+c)}+\frac{b}{2(a+b+c)}+\frac{c}{2(a+b+c)}$$$$$$
$$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{a+b+c}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of terms $1^2-2^2+3^2-...$ Prove that: $$1^2-2^2+3^2-4^2+...+(-1)^{n-1}n^2=\frac{(-1)^{n-1}n(n+1)}{2}$$
I proved it by induction but is there any other way to solve it?
If it was not a proof but rather a question like find the term,how to solve it?
I realized that alternate terms were under same sign but can't unde... | Here's a nice trick. For each $k$,
$$k^2=\frac{(k-1)k}2+\frac{k(k+1)}2=a_{k-1}+a_k$$
say.
Then
$$1^2-2^2+3^2-4^2+\cdots\pm n^2
=a_0+a_1-a_1-a_2+a_2+a_3-\cdots\pm a_{n-1}\mp a_n$$
and almost all of these terms cancel.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 7
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graph of $y = a\cos(bx+c)+d$
Attached is the graph of $y = a\cos(bx+c)+d$, where $a>0$, $b>0$, and $c>0$, and $c$ is as small as possible. Find $a + b + c + d$.
I'm having a hard time trying to solve this problem. First of all, the period seems to be $3\pi$, which means that $b=2/3$, and the amplitude, or a, is 2 a... | Firstly we need to find the midline of the graph, which will be the variable $d$.
Since it's maximun is $3$ and it's minimun is $-1$, the midline is:
$$y=\frac{(3)+(-1)}{2} = \fbox{d = 1}$$
We may also notice that two of the maximuns of the curve occur at: $-\frac{3\pi}{2}$ and $\frac{3\pi}{2}$, So the period of the fu... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Verification of infinite square root equation proof
b) Show also that the infinite square root
$$\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\ldots}}}}}=\frac{1+\sqrt5}{2}$$
Now, I've devised two proofs for this statement (both of which include using substitution), and it entails assuming that the case is true, and the... | Let's show that the sequence $(a_n)_{n=1}^\infty$ given by $a_1 = 1$, $a_{n+1} = \sqrt{a_n + 1}$ is monotonically increasing and bounded from above.
We shall prove by induction that $a_n \leq \frac{1+\sqrt{5}}{2}$.
$$a_1 = 1 < \frac{1+\sqrt{5}}{2}$$
Assume $a_n \leq \frac{1+\sqrt{5}}{2}$, for some $n\in\mathbb{N}$.
$$a... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Prove this sequence using induction Prove the following formula for all positive intergers $n$ using induction:
$1^2+3^2+5^2+...+(2n-1)^2=(n(2n-1)(2n+1))/3$
I have pretty gotten through the whole induction, but something seems wrong with my algebra. Can someone provide an answer to correct my algebra?
Induction step:
$... | For $2n-1 = 1$:
$$1^2 = \frac{1(2*1-1)(2*1+1)}{3} = \frac{3}{3} = 1$$
Assuming the formula works for $n=2k-1$, where $k$ is some integer, for $n=2(k+1)-1$:
$$1^2+3^2+...+(2k-1)^2+(2(k+1)-1)^2 =$$
$$ \frac{k(2k-1)(2k+1)}{3} + (2(k+1)-1)^2 \stackrel{?}{=} \frac{(k+1)(2(k+1)+1)(2(k+1)-1)}{3} $$
By expanding the right hand... | {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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If $f(x+1) +f(x-1) =\sqrt{3}\,f(x)$ and $f(2) =2$, what is the value of $f(4)$? My Attempt
$f(2)=2$. So, $f(1) + f(3)=2\sqrt{3}$ and $f(2) + f(4)=\sqrt{3}\,f(3)$. After solving these equations I got the value of $f(3)=2\sqrt{3}$ and $f(4)=4$. But are there any other methods than this? Any suggestions are welcome. ... | I'll be more general.
$f(n+1)+f(n-1)
= cf(n)
$.
Suppose
$f(n)
= b^n
$.
Then
$b^{n+1}+b^{n-1}
= cb^n
$
or
$b^2-cb+1
= 0
$.
Then
$b
=\dfrac{c\pm\sqrt{c^2-4}}{2}
$.
If $c^2=4$,
$b = c/2 = \pm 1$,
so
$f(n)=1$
or
$(-1)^n$.
If
$c^2 > 4$,
then
$b$ has two possible values,
one with
$|b|>1$
and one with
$|b|<1$.
If
$c^2 < 4$,
t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2433417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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determinant of the $7\times7$ matrix How to find determinant of the following $7\times7$ matrix
\begin{bmatrix}
a&1&0&0&0&0&1&\\
1&a&1&0&0&0&0&\\
0&1&a&1&0&0&0&\\
0&0&1&a&1&0&0&\\
0&0&0&1&a&1&0&\\
0&0&0&0&1&a&1&\\
1&0&0&0&0&1&a&
\end{bmatrix}
| An elementary proof. Let $J$ be the permutation matrix $(1\rightarrow 7,2\rightarrow 1,3\rightarrow 2\cdots)$; clearly $J^7=I$. Then we consider the matrix $A=aI+J+J^{-1}$. Let $\Delta_p=J^p+J^{-p}$ and $X=J+J^{-1}$.
$X^7=2I+7\Delta_5+21\Delta_3+35\Delta_1$. We remove $\Delta_5$:
$X^5=\Delta_5+5\Delta_3+10\Delta_1$ imp... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Are these sums of squares identities new? The identities are
\begin{align}
u^2 + \left( (2u-1)(2u^2-u+1) \right)^2 + \left( u (2u-1)(2u^2-u+1) \right)^2 &= \left( 4u^4-4u^3+5u^2-3u+1 \right)^2 \\
u^2 + \left( (2u+1)(2u^2+u+1) \right)^2 + \left( u (2u+1)(2u^2+u+1) \right)^2 &= \left( 4u^4+4u^3+5u^2+3u+1 \right)^2
\end{... | The complete solution to $x^2 + y^2 + z^2 = w^2$ is known: $$x = (a^2-b^2-c^2)t,\; y = 2 a b t,\; z = 2 a c t,\; w = (a^2 + b^2 + c^2) t$$
Your first is the case
$$ \eqalign{a &= (2 u^2 - u + 1) \cr
b &= 2u-1\cr
c &= u (2u-1)\cr
t &= \frac{1}{2}\cr }$$
Similarly for the second, with... | {
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Find the limit to $\lim\limits_{(x,y)\to(0,0)}\frac{x^5+y^2}{x^4+|y|}$ My problem is evaluating the following limit:
$$\lim_{(x,y)\to(0,0)}\frac{x^5+y^2}{x^4+|y|}$$
The answer should be 0. I tried to convert the limit into polar form, but it didn't help because I couldn't isolate the $r$ and $\theta$-variables of the e... | A common tool is to homogenise the denominator by setting $\displaystyle u=\frac{y}{x^4}$, then discuss the limit in function of $u$.
$\displaystyle f(x,y)=\frac{x^5+y^2}{x^4+|y|}=\frac{x^5+u^2x^8}{x^4+|u|x^4}=\frac{x+u^2x^4}{1+|u|}=\frac{x+uy}{1+|u|}=x\left(\frac 1{1+|u|}\right)+y\left(\frac u{1+|u|}\right)$
Since bot... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Find $\lim\limits_{x\to 0} \frac{\sqrt{\ 1+x} - \sqrt{\ 1-x}}{\sqrt[3]{\ 1+x} - \sqrt[3]{\ 1-x}}$ This problem
$$\lim_{x\to 0} \frac{\sqrt{\ 1+x} - \sqrt{\ 1-x}}{\sqrt[3]{\ 1+x} - \sqrt[3]{\ 1-x}}$$
is from Silverman's "Modern Calculus and Analytical Geometry" Section 22, #16d. I've been struggling on it for a while a... | Added for your curiosity.
For the limit itself, you aleary received good answers. The problem can also be addressed using Taylor series or the generalized binomial theorem
$$(1+x)^a=1+a x+\frac{1}{2} a(a-1) x^2+\frac{1}{6} a(a-1) (a-2) x^3+O\left(x^4\right)$$ making
$$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}... | {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Arithmetic and Geometric progression in 3 numbers Suppose
$$a,b,c \textrm{ is an arithmetic progression}$$ and
$$a^2,b^2,c^2\textrm{ is a geometric progression}$$
$$a+b+c = \frac{3}{2}.$$
From these equations I get
$$2b=a+c \textrm{, from A.P.}$$
$$b^4=a^2c^2 \textrm{, from G.P.}$$
and finally $$a=b=c=\frac{1}{2}.$$... | $2b=a+c$ and $a+b+c=\frac{3}{2}$ gives $b=\frac{1}{2}$ and $a+c=1$.
Also we have $b^4=a^2c^2$, which gives $$a^2c^2=\frac{1}{16}.$$
Thus, $$a^2(1-a)^2=\frac{1}{16},$$
which gives $a(1-a)=\frac{1}{4}$ and $a=c=\frac{1}{2}$ or
$$a(1-a)=-\frac{1}{4},$$
which gives $$(a,c)\in\left\{\left(\frac{1+\sqrt2}{2},\frac{1-\sqrt2}{... | {
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"source": "stackexchange",
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EV of probability given outcome Michael has a crush on a girl. Every night, he texts her and asks to go on a date. There is a 1/7
chance that the girl
says yes, a 2/7
chance that the girl says no, and a 4/7 chance that the girl asks Michael to text her again tomorrow, which
Michael does. Given that she said no, what is... | I have assumed that the No the first day is assumed to be 1. If that were to be the case, then the expected no of days on we which she decided would be the following:
Probability of a decision made = Probability of a No + Probability of a Yes
$P(Decide) = \frac{2}{7}+\frac{1}{7} = \frac{3}{7}$
USing Bayes' theorem:
Pr... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Ellipse $4x^2+y^2+ax+by+c=0$ passes through the point $(-1,2)$ and is tangent to the x-axis
Question:
Find the values of the constants $a$,$b$ and $c$ such that the ellipse $4x^2+y^2+ax+by+c=0$ passes through the point $(-1,2)$ and is tangent to the x-axis
What I've done so far:
Substitute $(-1,2)$ in $4x^2+y^2+ax+by... | There's not one, but an infinite number of ellipses that satisfy the two conditions that you have listed. You have 3 unknowns ($a$,$b$ and $c$) to solve for, but only 2 conditions given by which you can determine them. So, no unique solution.
Note that the equation of the ellipse can be written as
$$\frac{(x+\frac{a}{... | {
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Number of three-term arithmetic progressions in [n]
Three numbers are chosen at random between 1 and $n$ (say $n=500$).What will be the probability of those numbers to be in arithmetic progression?
I don't know how to count the number of favorable events.
Sample space={(1,2,3),(4,5,6),(18,20,21)........}
favorable e... | (Assuming three distinct numbers are chosen.)
A three-element arithmetic progress $a,b=a+k,c=a+2k$ with $k>0$ is entirely determined by $a,c$, and thus the number of them is equal to the number of ways of picking $a,c$ with the same parity (either both odd or both even.)
The number of odd numbers from $1$ to $n$ is $\... | {
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Is this a correct way to prove by induction? I understand that most people use the inductive hypothesis, but I find that counterintuitive. Is the below proof correct? In particular, I am concerned with my use of $n$ in (I); is the reason people use another variable, e.g. $k$, for conceptual reasons, or does the use o... | The first proof is perfectly fine. The choice of letters is mostly/often just for aestetical/pedagogical reasons.
The second (and to some extent the first) uses the $\dots$ informal notation. This should probably be seen as a shorthand or more visual way of handling summations. Formally one should have used $\sum$-nota... | {
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Minimum value of an expression in two variables $$\sqrt{x^2-7\sqrt{2}x+49} + \sqrt{x^2+y^2-\sqrt{2}xy} + \sqrt{50+y^2-10y}$$
$x$ and $y$ are positive real numbers, what is the minimum value?
I have tried finding the minimum value of the first expression and the third expression, with the first expression being $\frac{7... | Let $OABCD$ be a pentagon such that $OA=5\sqrt2$, $OB=y$, $OC=x$, $OD=7$, $\measuredangle AOB=45^{\circ}$, $\measuredangle BOC=45^{\circ}$ and $\measuredangle COD=45^{\circ}$.
Thus, $$\sqrt{x^2-7\sqrt{2}x+49} + \sqrt{x^2+y^2-\sqrt{2}xy} + \sqrt{50+y^2-10y}=$$
$$=DC+CB+BA\geq AD=\sqrt{(5\sqrt2)^2+7^2-2\cdot5\sqrt2\cdot... | {
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"source": "stackexchange",
"question_score": "2",
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Confusing Integral Expression: $\int{\frac{1}{x+\sqrt{x^2-x+1}}dx}$ I have this integral expression to be evaluated:
$$\int{\frac{1}{x+\sqrt{x^2-x+1}}dx}$$
I've followed these steps:
*
*Completed the square in $x$ and then substituted $u=x-\frac{1}{2}.$
*Trig substituted $u = \frac{\sqrt{3}}{2}\tan t.$ and then s... | Hint
Start considering the integrand as $$\frac{1}{x+\sqrt{x^2-x+1}}=\frac{1}{x+\sqrt{x^2-x+1}}\times\frac{x-\sqrt{x^2-x+1}}{x-\sqrt{x^2-x+1}}$$
$$\frac{1}{x+\sqrt{x^2-x+1}}=\frac{x-\sqrt{x^2-x+1}}{x-1}=\frac x {x-1}-\frac{\sqrt{x^2-x+1}}{x-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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What are the harmonic conjugates of the following rational function? List all the harmonic conjugates of the following rational function. The integration is almost impossible, Symbolab and Microsoft Math fails to arrive at the answer:
\begin{equation}
\mu(x,y) =\frac{x^2+x+y^2}{x^2+y^2}
\end{equation}
| Look at
$\mu(x, y) = \dfrac{x^2 + x + y^2}{x^2 + y^2} \tag 1$
in polar coordinates:
$\mu(x, y) = \dfrac{x^2 + x + y^2}{x^2 + y^2} = \dfrac{r^2 + r\cos \theta}{r^2} = 1 + \dfrac{\cos \theta}{r}; \tag 2$
with
$z = x + iy = r\cos \theta + ir\sin \theta =re^{i\theta} \tag 3$
we have
$z^{-1} = r^{-1}e^{-i\theta} = \dfrac{\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $5+4\cdot 5+4\cdot5^2+4\cdot5^3+4\cdot5^4+4\cdot5^5$
Evaluate
$$5+4\cdot 5+4\cdot5^2+4\cdot5^3+4\cdot5^4+4\cdot5^5.$$
The options are $5^6$, $5^7$, $5^8$, $5^9$, $5^{10}$.
I'm new to this site. I came across this question in an Olympiad foundation site. I have no idea how to solve it. Can I get the answe... | Hint. Note that the given sum can be written as
$$5+(5-1)\cdot 5+(5-1)\cdot5^2+(5-1)\cdot5^3+(5-1)\cdot5^4+(5-1)\cdot5^5.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Minimum value of $S=\frac{3a}{b+c}+\frac{4b}{a+c}+\frac{5c}{a+b}$ Find Minimum value of Minimum value of $$S=\frac{3a}{b+c}+\frac{4b}{a+c}+\frac{5c}{a+b}$$
My Try:
we have $$S=3 \times \left (\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)+\frac{b}{a+c}+\frac{2c}{a+b}$$
By Standard $AM$ $GM$ inequality we know that
$$... | For any positive numbers $\alpha, \beta, \gamma$, we have
$$\frac{\alpha^2 a}{b+c} + \frac{\beta^2 b}{c+a} + \frac{\gamma^2 c}{a+b}
= (a+b+c)\left(\frac{\alpha^2}{b+c} + \frac{\beta^2}{c+a} + \frac{\gamma^2}{a+b}\right) - \left(\alpha^2+\beta^2+\gamma^2\right)$$
By Engel's form of Cauchy Schwarz, we have
$$\frac{\alpha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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An integral of a rational function with high degrees: Evaluate $\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx$.
Calculate: $$\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx.$$
What I have tried is to divide both numerator and denominator with $x^4 + 1$ and then get two following integrals, because o... | More a comment than an answer:
According to Wolfy,
the indefinite integrals
involve arctan and
complex values.
But the definite integrals
come out like this:
$\int_0^∞ \dfrac{dx}{x^4 - \sqrt{3} x^2 + 1}
= \frac12 \sqrt{2 + \sqrt{3}} π≈3.0345
$
and
$\int_0^∞ \dfrac{dx}{x^4 + \sqrt{3} x^2 + 1} dx = \dfrac{π}{2 \sqrt{2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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Existence of Limit for $x_{n+1} = 1+1/x_{n}$ I'm attempting to prove that if one defines a sequence with $x_{1} = 1$ and
$$x_{n+1} = 1+\frac{1}{x_{n}}$$
for all $n >1$ then the limit of $\{x_{n}\}_{n \ge 1}$ exists.
My Attempt:
Since $\{x_{n}\}_{n \ge 1}$ was defined recursively I immediately thought to try and prove... | This is a trick I have learned here a few years ago.
Let's say you have a sequence $x_n$ defined by some sort of iteration.
If the limit $\lambda$ of the sequence satisfies a quadratic equation with roots $\alpha, \beta$ and one construct an auxiliary sequence by
$$y_n = \frac{x_n - \alpha}{x_n - \beta}$$
Sometimes, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2455280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Proof that no non-trivial integer solution exists? The equation system
$$2ab+2ac+2bc=4a+4b+4c$$
$$2ab+2ac+2bc=abc$$
In short $$2ab+2ac+2bc=4a+4b+4c=abc$$
has, according to Wolfram Alpha, only the trivial integer solution $a=b=c=0$. I would like to prove this. My approach (not sure whether it is helpful) is using the eq... | One has $$(ab+bc+ca)^2 = a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a+b+c) = a^2b^2 + b^2c^2 + c^2a^2 + 2(2ab+2bc+2ca)(\frac{1}{2}(ab+bc+ca))$$
Then $$a^2b^2 + b^2c^2 + c^2a^2 = -(ab+bc+ca)^2$$
So, one gets $ab=bc=ca=ab+bc+ca=0$,or $a=b=c=0$.
EDIT: i make it clear at the last step as @peter's comment:
One gets from $ab=bc=ca=0$ t... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit Evaluation (Conjugate Method)–Further algebraic manipulation? $$\lim_\limits{x\to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
I have tried evaluating the above limit by multiplying either both the conjugate of numerator and the denominator with no avail in exiting the indeterminate form.
i.e. $$\frac{\sqrt{6-x}-2}{\... | The problem with $\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ is that, when you let $x=2$, you get $\dfrac 00$. So we have to assume that $x \ne 2$. This is not necessarily a problem because $\displaystyle \lim_{x \to x_0}f(x)$ does not care about what happens to $f(x)$ at $x=x_0$.
Notice below that a factor of $(2-x)$ appear... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Sylvester Equation over GF(2) I know that a Sylvester equation $$AX+XB=0$$ has nontrivial solutions exactly when there is a common eigenvalue of $A$ and $-B$. This is because if there is a common eigenvalue $k$, then there exists a column eigenvector $v$ of $A$ and a row eigenvector $w$ of $-B$. Then $vw$ is a nontrivi... | Let us assume that the characteristic polynomials $\chi_A$ and $\chi_{-B}$ have a common irreducible factor $p$. Let $C$ be the companion matrix of $p$. Then $A$ is similar to an upper triangular block matrix with $C$ in the upper left corner and $-B$ is similar to a lower triangular block matrix with $C$ in the upper ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2457715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How would one prove that for all positive real $a$ and $b$, $\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}$? I think that the best way to prove this would be to prove by contradiction. Am I right?
If so, can I just assume that $\sqrt{a+b} = \sqrt{a} + \sqrt{b}$ and say $a = 1$ and $b = 1$ and show that $\sqrt{1+1} \neq \sqrt{1} ... | a,b positive real numbers.
Let $x^2:= a$, $y^2:=b,$ $x,y \gt 0.$
Consider a right triangle with lengths of legs $x,y$,
length of hypotenuse is $(x^2+y^2)^{1/2}.$
Sum of $2$ sides in a triangle is greater than the third side.
(Euclid)
$\Rightarrow:$
$(x^2+y^2)^{1/2} \lt x + y$,
or reverting to $a,b:$
$(a+b)^{1/2} \lt ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
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Can the inequality $\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \geq 6$ be proved with differentiation? $$\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \geq 6,\quad \text{with}\quad a,b,c > 0$$
I could do it with letting $x=\frac{a}{b}$, $y=\frac{b}{c}$, $z=\frac{c}{a}$, but I wonder if it is solvable somehow with d... | This is a standard example: it's equivalent to
$$(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\ge9$$
which is AM/HM or Cauchy-Schwarz (provided all $a$, $b$, $c$ are non-negative).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2462791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Solve differential equation $yy'(yy'-2x)=x^2-2y^2$.
$$yy'(yy'-2x)=x^2-2y^2$$
I've tried to divide by $y^2$ and substitude $\frac{x}{y}=z$, but it led to:
$$y'^2-2zy'=z^2-2$$
$$(y'-z)^2=2(z^2-1)$$
$$|y'-z|=\sqrt{2}\sqrt{z^2-1}$$ hence
$y'=\sqrt{2}\sqrt{z^2-1}+z$ or $y'=-\sqrt{2}\sqrt{z^2-1}+z$ and I don't know any w... | $$yy'(yy'-2x)=x^2-2y^2$$
divide by $y^2$:
$$(y')^2-2\frac{x}{y}y'+(2-(\frac{x}{y})^2)=0$$
$$y'=\frac{x}{y}±\sqrt {2(\frac{x}{y})^2-2}$$
substitute $y=vx$
$$v+x\frac{dv}{dx}=\frac{1}{v}±\sqrt {2(\frac{1}{v})^2-2}$$
$$\frac{dx}{x}=\frac{dv}{\frac{1}{v}-v±\sqrt {2(\frac{1}{v})^2-2}}$$
$$\frac{dx}{x}=\frac{vdv}{1-v^2±\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2463150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Prove that $2^n +1$ is divisible by $3$ for all positive integers $n$. I just want to know if I went on the right direction. With induction
Let $n=1$, then $2^1+1= 3$, which is divisible by $3$. Then show proof for
$n+1.$
$2^n+1=3k$
So we get $2^{n+1}+1, \rightarrow 2^n+2+1, \rightarrow 3k+3= 3(k+1)$. Thus $2^n+1$ is ... | *
*$2^1+1=3$;
*$2^{2n+1}+1=3m\implies4(3m)-3=2^{2(n+1)+1}+1=3m'$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How to find $\lim_{x \to 1} \frac{x+x^2+\sqrt{x}-3}{\sqrt[3]{x}-1}$? How to find
$$
\lim_{x \to 1} \frac{x+x^2+\sqrt{x}-3}{\sqrt[3]{x}-1}?
$$
My try : $$(\sqrt[3]{x}-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)=(x-1)$$
So we have :
$$\lim_{x \to 1} \frac{x+x^2+\sqrt{x}-3}{\sqrt[3]{x}-1} \cdot\frac{(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}{(\... | Hint:
$$x+x^{ 2 }+\sqrt { x } -3=x-1+x^{ 2 }-1+\sqrt { x } -1=x-1+\left( x-1 \right) \left( x+1 \right) +\sqrt { x } -1=\left( x-1 \right) \left( 1+x-1 \right) +\sqrt { x } -1=x\left( x-1 \right) +\sqrt { x } -1=\\ =x\left( \sqrt { x } -1 \right) \left( \sqrt { x } +1 \right) +\sqrt { x } -1=\left( \sqrt { x } -1 \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find $\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{x+x^2-2}$? How to find $$\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{x+x^2-2}$$
My Try :
$$x^2+x-2=(x-1)(x+2)$$
$$\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{(x-1)(x+2)}\cdot \frac{(\sqrt{x}+\sqrt{x+3})+3}{(\sqrt{x}+\sqrt{x+3})+3}=\\ =\lim_{x \to 1} \frac{2x+2\s... | First factor $x^2+x^2-2=(x-1)(x+2)$. $$\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{(x-1)(x+2)}=$$
$$=\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{x-1}\lim_{x\to 1}\frac{1}{x+2}=$$
$$=\left(\lim_{x\to 1}\left(\frac{\sqrt{x}-1}{x-1}+\frac{\sqrt{x+3}-2}{x-1}\right)\right)\cdot \frac{1}{3}$$
Hint: multiply the numerator and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Lagrange multiplier to function $x^2+y^2+z^2$ Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given condition:
$$f(x,y,z)=x^2+y^2+z^2; \quad x^4+y^4+z^4=1$$
My solution: As we do in Lagrange multipliers I have considered $\nabla f=\lambda \nabla g$ where $g(x,y,z)=x^4+y^4+... | Yeah so you have $x^{2}=\frac{1}{2\lambda}$ which says $x^{4}=\frac{1}{4\lambda^2}$. And you want $x^{4}+y^{4}+z^{4}=1$. So we have $$\frac{1}{4\lambda^2}+\frac{1}{4\lambda^2}+\frac{1}{4\lambda^2}=1\implies \frac{3}{4\lambda^2}=1 \implies\lambda =\frac{\sqrt{3}}{2}$$ Now you have $$x^{2}=\frac{1}{2\lambda}=\frac{1}{2\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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How to compute the following limit$ \lim\limits_{x\to \infty }[( 1+x^{p+1})^{\frac1{p+1}}-(1+x^p)^{\frac1p}],$ I am trying to find
$$ \lim\limits_{x\to \infty }[( 1+x^{p+1})^{\frac1{p+1}}-(1+x^p)^{\frac1p}],$$
where $p>0$. I have tried to factor out as
$$(1+x^{p+1})^{\frac1{p+1}}- \left( 1+x^{p}\right)^{\frac{1}{p}} =... | Thanks this answer's here : Find $\lim_{n \to \infty } \sqrt[3]{n^3+1} - \sqrt{n^2+1}$
Consider
$$
f(x)=[( 1+x^{p+1})^{\frac{1}{p+1}}-(1+x^p)^{\frac1p}]
$$
Then
$$
f'(x)=x^p[( 1+x^{p+1})^{\frac{1}{p+1}-1}-x^{p-1}(1+x^p)^{\frac1p -1}]
$$
If $p>1$ therefore $f'(0)=0$. Therefore
$$
0=f'(0)=\lim_{x\to0^+}x^p[( 1+x^{p+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2482868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help proving Jacobi Symbol property $\left(\frac{-1}{n}\right) = (-1)^{\frac{n-1}{2}}$ Using the properties of Legendre symbols, I need to show $\left(\frac{-1}{n}\right) = (-1)^{\frac{n-1}{2}}$
Here is what I have so far
$\left(\frac{-1}{n}\right) = \left(\frac{-1}{p_1}\right)\left(\frac{-1}{p_2}\right) \cdots \left(\... | You are making hard work of this. The point is to prove that
$$\left(\frac{-1}n\right)=\chi(n)\tag1$$
where
$$\chi(n)=\begin{cases}1&\text{if }n\equiv1\pmod 4,\\
-1&\text{if }n\equiv-1\pmod 4\end{cases}$$
for odd $n$. Now $\chi(mn)=\chi(m)\chi(n)$ and
also $\left(\frac{-1}{mn}\right)=\left(\frac{-1}m\right)
\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2485368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Calculate $\lim\limits_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}$
Calculate
$$\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}.$$
My Attempt :
$$=\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}
=\lim_{x\to 3} \frac {1-(\pi x)/2! +1}{(5)^{1/3} + \dfrac {x}{3\cdot 5^{2/3}} -... | Your approach is not correct. You are using the expansions at $0$ of the functions $\cos(\pi x)$ and the $(x+5)^{1/3}$, whereas the limit is for $x$ that goes to $3$.
Note that by letting $x=t+3$, we have that $t\to 0$ and
$$\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{1/3} -2}=\lim_{t\to 0} \frac {\cos (\pi t +3\pi)+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2491005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why do I get two different indefinate integrals for the same integrand? (u-substitution) Using u-substitution I'm able to integrate the following two integrands:
$$\int \frac{1}{2x+3}dx = \frac{1}{2}\ln|2x+3|+C$$
$$\int \frac{1}{2x+5}dx = \frac{1}{2}\ln|2x+5|+C$$
Following this pattern I would assume:
$$\int \frac{1}{... | Because $$\frac{1}{2}\ln|2x+4|+C=\frac{1}{2}\ln|x+2|+\frac{1}{2}\ln2+C,$$
where $\frac{1}{2}\ln2+C$ is a constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2491090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Does $n+1$ divides $\binom{an}{bn}$? Suppose that $a>b>0$ be integers. Is it true that for an integer $n>2$ that
$$n+1|\binom{an}{bn}$$
or is there a counter example. Certainly i think the right hand side would reduce to
$$\frac{an(an-1)(an-2)...((a-1)n+1)}{n(n-1)(n-2)...2\cdot 1}$$
But I'm not seeing how this could ... | As counterexamples for every $n=p-1$ for any prime $p$ and considering $a=n+2=p+1$ and any $b$ with $0 \lt b \lt a$
Then $(an)! = (p^2-1)!$ is divisible by $p^{p-1}=p^n$ but not $p^{p}=p^{n+1}$, while $(bn)!(an-bn)!$ is divisible by $p^{b-1}p^{a-b-1}=p^{n}$, so ${an \choose bn}$ is not divisible by $p=n+1$
This gives f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2492260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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let $ \ \ 0let $ \ \ 0<a \leq b \leq c \in \mathbb{R}$
then prove that :
$$\frac{(c-a)^2}{6c}\leq \frac{a+b+c}{3}-\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$
I do not know where to start please help me !
| Your inequality is true!
We need to prove that
$$\frac{a+b+c}{3}-\frac{3abc}{ab+ac+bc}\geq\frac{(c-a)^2}{6c}$$ or
$$\frac{\sum\limits_{cyc}c(a-b)^2}{ab+ac+bc}\geq\frac{(c-a)^2}{2c}$$ or
$$2c(a-b)^2+2a(b-c)^2\geq\left(\frac{ab}{c}+a-b\right)(c-a)^2.$$
Now, by C-S
$$2c(a-b)^2+2a(b-c)^2=2ac\left(\frac{(b-a)^2}{a}+\frac{(c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2493515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Self dual functions My book says that $f(x,y,z) = xy + yz + zx$ is self dual function but $(x+y)(y+z)(z+x)$ is not.
I understood how $xy + yz + zx$ is self dual but I think $(x+y)(y+z)(z+x)$ is also self dual because in k-map if we represent $(x+y)(y+z)(z+x)$ then it's minimized SOP is $xy + yz + zx$ and if we take dua... | For all boolean expression $e(x_{1},\ldots,x_{n})$, let us represent the dual expression of $e(x_{1},\ldots,x_{n})$ by $e(x_{1},\ldots,x_{n})^{D}$.
Expression $xy+yz+zx$ is self dual. In effect, in virtue of elementals properties of boolean algebra, the following equalities hold:
\begin{align*}
(x+y)(y+z)(z+x)&=x(y+z)... | {
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"url": "https://math.stackexchange.com/questions/2493779",
"timestamp": "2023-03-29T00:00:00",
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} |
Prove $\lim_{(x,y)\rightarrow (0,3)}\frac{\tan(xy)}{x}=3$ exists Prove this limit exists:
$$\lim_{(x,y)\rightarrow (0,3)}\frac{\tan(xy)}{x}=3$$
My work:
Let $\epsilon >0$, $\delta=\cdots$
If $\sqrt{(x-0)^2+(y-3)^3}<\delta$
Then
\begin{align}
& \left|\frac{\tan(xy)}{x}-3\right|=\left|\frac{\tan(xy)-3x}{x}\right| = \left... | A not so good solution (which can be made better of course) is
\begin{align*}
\lim_{(x,y)\rightarrow(0,3)}\dfrac{\tan(xy)}{x}&=\lim_{(x,y)\rightarrow(0,3)}\dfrac{1}{x}\left(xy+\dfrac{1}{3}(xy)^{3}+\cdots\right)\\
&=\lim_{(x,y)\rightarrow(0,3)}\left(y+\dfrac{1}{3}x^{2}y^{3}+\cdots\right)\\
&=3.
\end{align*}
If we decid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2494310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\forall n \in \mathbb{N}: 4^n + 6n - 10$ is divisible by $18$
Prove that $\forall n \in \mathbb{N}: 4^n + 6n - 10$ is divisible by $18$
Base case: for $n = 1: 4^1 +6\cdot 1 - 10 = 0$ is divisible by 18.
Inductive Assumption: Assume that for for some $k \in \mathbb{N} :4^k +6k-10$
Proving that $4^{k+1}+6(k... | For $n=1$ we have $4^1+6(1)-10=0$, which is divisible by $18$. If there exists $k \in \mathbb{N}$ such that $4^n+6n-10=18k$, for some $n \in \mathbb{N}$, then $$4^{n+1}+6(n+1)-10=4\cdot 4^n+6n+6-10=(4\cdot 4^n+24n-40)-24n+40+6n-4=4\cdot 18k-18n+36=18(k-n+2),$$ and since $m-n+2 \in \mathbb{Z}$ you have that $4^{n+1}+6(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2495561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 5
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Sketch the contour plot and the graph of the function: $f(x,y) = \sqrt{36 - 9x^2 - 4y^2}$ I've done a bit:
range of function: $[0,6]$. So we need level curves for $k = 0,1,\ldots,6$
$$f(x,y) = k$$
$$\sqrt{36 - 9x^2 - 4y^2} = k$$
$$36 - 9x^2 - 4y^2 = k^2$$
$$36-k^2 = 9x^2 + 4y^2$$
Not really sure how to go further. I tr... | $$36-k^2 = 9x^2+4y^2$$
$$1=\frac{x^2}{\left(\frac{\sqrt{36-k^2}}{3}\right)^2}+\frac{y^2}{\left( \frac{\sqrt{36-k^2}}{2}\right)^2}$$
Notice that $1=\frac{x^2}{a^2}+\frac{y^2}{b^2}$ is an equation of an ellipse.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2497578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $1+3^{\frac{1}{5}}+3^{\frac{2}{5}}+3^{\frac{3}{5}}+3^{\frac{4}{5}}$ is irrational I tried to follow this example to resolve this exercise, this
example to resolve this exercise, however if:
$1+3^{\frac{1}{5}}+3^{\frac{2}{5}}+3^{\frac{3}{5}}+3^{\frac{4}{5}}=x$,then
$3^{\frac{1}{5}}+3^{\frac{2}{5}}+3^{\frac{3}... | Observe that
$$3^{1/5} x = 3^{1/5} + 3^{2/5} + 3^{3/5} + 3^{4/5} + 3 = x + 2,$$
so by rearranging $x = 2/(3^{1/5} - 1)$.
Now if $x \in \mathbb{Q}$, we would get $3^{1/5} \in \mathbb{Q}$, say $3^{1/5} = a/b$ with $a, b \in \mathbb{Z}^+$ coprime. Then $$3b^5 = a^5,$$ forcing $3 \mid a$, whence $3^5 \mid a^5$, so $3 \mid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2499076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Proof by Contradiction: let a,b $\in$Z . Prove that if 3 $\nmid$ a and 3 $\nmid$ b then $3 | a^2 - b^2$ Just wondering if this works for this question, book had a different answer and I couldn't find another answer for the question.
Assume, to the contrary, that 3 | a and 3 | b, then a = 3k, and b = 3x for x,k $\in$ Z,... | If you don't have to use contradiction, here is a direct proof.
*
*$a = 3A+1$ and $b = 3B+1$.
Then $a^2-b^2=(a+b)(a-b)=(a+b)(3A-3B)=3(a+b)(A-B)$.
*$a = 3A+1$ and $b = 3B+2$.
Then $a^2-b^2=(a+b)(a-b)=(3A+3B+3)(a-b)=3(A+B+1)(a-b)$.
The case $a = 3A+2$ and $b = 3B+1$ follows by symmetry.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2499349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Prove that area of ellipse $Ax^2+Bxy+Cy^2=1$ is equal to $\frac{2\pi}{\sqrt{4AC-B^2}}$ Question:
If $B^2-4AC \lt 0$, the equation
$$Ax^2+Bxy+Cy^2=1$$
represents an ellipse. Prove that the area of the ellipse is
$\frac{2\pi}{\sqrt{4AC-B^2}}$.
I know that, if $2a$ and $2b$ are the major and minor axis of the ellipse resp... | In polar coordinates, the ellipse is given by
$$r^2 (\theta)= \frac1{A\cos^2\theta +B\cos\theta\sin\theta +C\sin^2\theta}$$
Denote $\Delta = \sqrt{(A-C)^2+B^2}$ and integrate its area as follows
\begin{align}
A=\frac12 \int_0^{2\pi} r^2(\theta)\> d\theta
&= \frac12 \int_0^{2\pi} \frac{1}{A\cos^2\theta +B\cos\theta\sin\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2499695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Will the series $\sum\limits_{n=1}^\infty \frac{n^2-n-1}{n!}$ converge when it runs from 1 to infinity? $$\sum_{n=1}^\infty \frac{n^2-n-1}{n!}$$
I usually divide by $n$ or $n^2$ but I'm unable to do so because of the factorial term.
| The series evaluation:
\begin{align}
\sum_{n=1}^{\infty} \frac{n^{2} - n - 1}{n!} &= \sum_{n=1}^{\infty} \left(\frac{1}{(n-2)!} - \frac{1}{n!}\right) \\
&= \sum_{n=2}^{\infty} \frac{1}{(n-2)!} - (e^{1} -1) \\
&= \sum_{n=0}^{\infty} \frac{1}{n!} - (e^{1} - 1) = e^{1} - (e^{1} - 1) = 1.
\end{align}
Convergence by ratio t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Why does $3^{16} \times 7^{-6}$ become $\frac{3^{16}} {7^{6}}$? I was doing an exercise on exponents:
$$\begin{align}
\left(3^{-8} \times 7^3\right)^{-2} &= \left(3^{-8}\right)^{-2}\times \left(7^3\right)^{-2} \\
&= 3^{16} \times 7^{-6} \\
&= \frac{3^{16}} {7^{6}} \\
\end{align}$$
Why did $7^{-6}$ turn to $7^{6}$? Mor... | The OP asked for kindergarten language. At that grade level math rules are stressed. So if you are taking algebra-precalculus and this is your last math course, here are a couple of rules;
$\tag 1 \frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d} \;\text{ where } b \ne 0 \text{, } d \ne 0$
$\tag 2 x = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 0
} |
Determining the limit of a function containing $(x-1)^5$ without l'hopital`s rule. Find the limit of $$\begin{equation*}
\lim_{x \rightarrow 0}
\frac{(x-1)^5 + (1 + 5x)}{x^2 + x^5}
\end{equation*}$$
Shall I use the binomial theorem?
Any hint will be appreciated!
| Divide top and bottom by $x^2$ to get $$\lim_{x \to 0} \frac{\frac{1}{x^2}(1+5x) + \frac{1}{x^2}(x-1)^5}{1+x^3}$$
This is a quotient where the denominator has limit $1$, so the limit exists iff the following limit exists, and it has the same value as this limit: $$\lim_{x \to 0} \left[\frac{1}{x^2}(1+5x) + \frac{1}{x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502678",
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"source": "stackexchange",
"question_score": "1",
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Proof by induction that $\sum_{k=1}^n (-1)^{k-1}k^2 = (-1)^{n-1}\frac{(n)(n+1)}{2}$ Through Induction I tried to prove that:
$$\sum_{k=1}^n (-1)^{k-1}k^2 = (-1)^{n-1}\frac{(n)(n+1)}{2}$$
I first let $n=1$, so that on the left hand side and the right hand side you get 1.
Then I tried to prove that this also works when ... | If
$$\sum_{k=1}^{n} (-1)^{k-1} \, k^{2} = (-1)^{n-1} \, \frac{n \, (n+1)}{2} = (-1)^{n-1} \, \binom{n+1}{2}$$
then when $n \to n+1$ it is seen that
$$\sum_{k=1}^{n+1} (-1)^{k-1}k^2 = \sum_{k=1}^n (-1)^{k-1}k^2 +(-1)^{n}(n+1)^{2}$$
which leads to
\begin{align}
\sum_{k=1}^{n+1} (-1)^{k-1} \, k^{2} &= \sum_{k=1}^n (-1)^{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2506079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Checking that $2+\sqrt3$ is a cube root of $26 + 15 \sqrt3$ I am trying to show that $$\sqrt[3]{26 + 15 \sqrt{3}} = 2 + \sqrt{3}$$
My idea is to find the cube roots of $z=26 + 15\sqrt{3}$ via De Moivre's formula.
So $r=\sqrt{26^2 + (15\sqrt{3})^2} = \sqrt{1351}$, and $\theta = \tan^{-1} \big(\frac{15\sqrt{3}}{26} \big)... | I would try to find a cubic root of $26+15\sqrt3$ of the form $a+b\sqrt3$. Since$$(a+b\sqrt3)^3=a^3+9ab^2+(3ab+3b^3)\sqrt3,$$I would try to find numbers $a$ and $b$ such that$$\left\{\begin{array}{l}a^3+9ab^2=26\\3a^2b+3b^2=15\iff(a^2+b)b=5.\end{array}\right.$$Of course, I would try first to find integer solutions. But... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2506923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Show that $\lim_{n\to\infty}\left(\prod_{k=1}^n\frac{2k+1}{3k-2}\right)^{\frac{1}{n}}=\frac{2}{3}$ How can I show
$$\lim_{n\to\infty}\left(\prod\limits_{k=1}^n\frac{2k+1}{3k-2}\right)^{\frac{1}{n}}=\frac{2}{3}$$
with the aid of Gamma function？
| We have
$$ \left( \prod_{k=1}^{n} \frac{2k+1}{3k-2} \right)^{\frac{1}{n}}
= \frac{2}{3} \left( \prod_{k=1}^{n} \left( 1 + \frac{7}{6k-4} \right) \right)^{\frac{1}{n}}. $$
Therefore the claim follows from
$$ 1
\leq \left( \prod_{k=1}^{n} \left( 1 + \frac{7}{6k-4} \right) \right)^{\frac{1}{n}}
\leq \exp\left(\frac{1}{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2512083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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query on finding numeric function of a generating function For the given generating function $A(z)=1/(1-x^3)$ what will be the numeric function..
I can find out for $A(z)=1/(1-x^2)$ but in the first case we get two factor in the denominator $(1-x)$ and $(x^2+x+1)$, now how do we find the numeric function for $1/(x^2+x+... | We don't need to factorize the denominator. Let's recall the geometric series expansion.
\begin{align*}
\frac{1}{1-y}=1+y+y^2+y^3+\cdots\qquad\qquad |y|<1
\end{align*}
We obtain with $y=x^3$
\begin{align*}
\frac{1}{1-x^3}=1+x^3+x^6+\cdots\qquad\qquad |x|<1
\end{align*}
and the coefficients are $1$ if the powers ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2514000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $abc = (1-a)(1-b)(1-c)$ and $0 \le a,b,c \le 1$ then find minimum of expression If $abc = (1-a)(1-b)(1-c)$ and $0 \le a,b,c \le 1$ then find minimum of expression:
$$\min \{a(1-c)+b(1-a)+c(1-b)\}$$
The given expression can be rewritten as:
$$a+b+c-ab-ac-bc = S_1-S_2 \tag{1}$$
Then from the given condition:
$$\begin... | $$a(1-c)+b(1-a)+c(1-b)=1-2abc\ge \dfrac{3}{4}$$
$$(abc)^2=(a(1-a))\cdot (b(1-b))\cdot(c(1-c))\le \dfrac{1}{4^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2518051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Sum of a sequence of reciprocals of square of odd natural numbers If $$a_{n}=\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+....+\frac{1}{(2n+1)^2}$$where $n\in N$.
Then prove that
$a_{n}<\frac{1}{4}$
| Here's a sketch of a direct proof that only requires calculus. Note that for each $k$, we have $$\frac{1}{(2k + 1)^2} \leq \int_{k-1}^k \frac{1}{(2x + 1)^2}\,dx.$$
This implies that each $a_n$ is less than $$\frac{1}{9} + \frac{1}{25} + \frac{1}{49} + \int_{3}^\infty \frac{1}{(2x + 1)^2}.$$
Calculating this integral ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2518609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Expressing ${_3F_2}$ in terms of gamma functions I have a generalized hypergeometric funtion of the following form
$$\,_3F_2\left(\frac{1}{2},\frac{\beta }{2},\frac{1}{4} (\beta +2 p-2);\frac{\beta +1}{2},\frac{1}{4} (\beta +2 p+5);1\right)$$ where $\beta$ can take integer values $2,1,0,-1,\ldots$ and $p = 0$ or, $\fra... | Since $\beta$ is an integer, I will rename it $n$. So we have:
$$n=2,1,0,-1,-2,\dots$$
And two functions (for each value of $p$):
$$F_n=\,_3F_2\left(\frac{1}{2},\frac{n }{2},\frac{n-2}{4} ;\frac{n +1}{2},\frac{n+5}{4};1\right)$$
$$G_n=\,_3F_2\left(\frac{1}{2},\frac{n }{2},\frac{n-1}{4} ;\frac{n +1}{2},\frac{n+6}{4};1\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2519717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve a quartic equation using trigonometric power identitiy? According to this post, we can solve a cubic equation
$$t^3+pt+q=0$$
by the trigonometric identity
$$4\cos^3\theta-3\cos\theta-\cos3\theta=0$$
So I've tried to solve the quartic equation
$$t^4+pt^2+qt+r=0$$
using the identity
$$8\cos^4\theta-4\cos2\t... | In general, you cannot parameterize a two-dimensional set, like
$$
\left(-\frac{8q}{A^3},\frac{8r}{A^4}\right)
$$
with a one-edimensional curve, like
$$
(\cos3\theta,1+\sin3\theta\sin\theta)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2520277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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Value of $e$ given that $\frac{a+b+c+d+e}{5} = 2 = \sqrt{\frac{a^2 + b^2 + c^2 + d^2 + e^2}{5}}$ I have five real numbers $a,b,c,d,e$ and their arithmetic mean is $2$. I also know that the arithmetic mean of $a^2, b^2,c^2,d^2$, and $e^2$ is $4$. Is there a way by which I can prove that the range of $e$ (or any ONE of t... | If we have
$$
\frac{a+b+c+d+e}{5} = 2 = \sqrt{\frac{a^2 + b^2 + c^2 + d^2 + e^2}{5}}
$$
then all the five numbers are necessarily equal to $2$, as dictated by the AM-QM inequality.
PS. Technically, the power mean inequality is only valid for positive real numbers, but if any of the numbers were negative, then we could ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2529262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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To find formal products in given expressions For each of the following expressions, list the list of all formal products in which exponents sum to 4.
(a)$(1+x+x^{2})^{2} (1+x)^{2}$
(b) $(1+x+x^{2}+x^{3}+x^{4})^{3}$
(c)$(1+x^{2}+x^{4})^{2} (1+x+x^{2})^{2}$
(d)$(1+x+x^{2}+x^{3}+...)$
The answers are
. (a) 7 products—$xxx... | Assuming that a "formal product" is an unsimplified term in the expanded product, then here is how one should reason about (a).
The expression in (a) is $(1 + x + x^2)(1 + x + x^2)(1 + x)(1 + x)$. A formal product is formed by picking a single term from each factor and then multiplying these terms together. For example... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2530262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $p(x)$ be a real $7$ degree polynomial with $p(\pi)=\sqrt 3$ and $\int_{-\pi}^{\pi}x^k p(x)=0$ for $0\le k\le 6$. Find $p(0)$ and $p(-\pi)$.
Q.Let $p(x)$ be a real $7$ degree polynomial with $p(\pi)=\sqrt 3$ and $\int_{-\pi}^{\pi}x^k p(x)=0$ for $0\le k\le 6$. Find $p(0)$ and $p(-\pi)$.
Let $p(x)=a_0 + a_1 x+ a_2... | One way.
Consider the vector space $V$ of polynomials of degree $\le7$. We know from linear algebra that $\dim V=8$ and $\{1,x,x^2,\ldots,x^7\}$ is a basis.
We also know that the subspace $U$ spanned by $\{1,x,\ldots,x^6\}$ has dimension seven. We further know that
$$
(f,g)=\int_{-\pi}^\pi f(x)g(x)\,dx
$$
is an inner p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2530638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find the minimum of the value $f=(1+\sin^2{x})(1+\sin^2{y})$ Let $\tan{x}\tan{y}=-\frac{1}{2}$,find the minumin of the value $$f=(1+\sin^2{x})(1+\sin^2{y})$$
My ugly solution:
$$f=\dfrac{1+\sin^2{x}}{\sin^2{x}+\cos^2{x}}\cdot\dfrac{1+\sin^2{y}}{\sin^2{y}+\cos^2{y}}=\dfrac{2m^2+1}{m^2+1}\cdot\dfrac{2n^2+1}{n^2+1}$$where... | You have done the hard work.
On rearrangement
$$m^4(8-4f)+m^2(8-5f)+2-f=0$$ which is a Quadratic Equation in $m^2$ which is real
So, the discriminant must be $\ge0$
$$(8-5f)^2-4(8-4f)(2-f)=-16f+9f^2\ge0$$
$\iff f(9f-16)\ge0$
As $f>0, 9f-16\ge0\iff f\ge?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2530730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding Christoffel Symbol
Given a sphere, find the christoffel symbols
$X(\theta,\phi)=(r \sin\phi \cos\theta,r \sin\phi \sin\theta, r \cos \phi)$
So
$$(g_{ij})=\begin{pmatrix} r^2\sin^2\phi& 0\\ 0& r^2 \end{pmatrix}$$
$$(g^{ij})=\frac{1}{r^4 \sin^2 \phi}\begin{pmatrix} r^2& 0\\ 0& r^2\sin^2\phi\end{pmatrix}=\begin{... | I think the Christoffel symbol you want is $\Gamma_{\theta\theta}^{\phi}$, which is
$$\begin{align}\Gamma_{\theta\theta}^{\phi}&=\frac{1}{2}g^{\phi k}(\partial_{\theta}g_{\theta k}+\partial_{\theta}g_{\theta k}-\partial_{k}g_{\theta\theta})\\
&=\frac{1}{2}g^{\phi\phi}(\partial_{\theta}g_{\theta\phi}+\partial_{\theta}g_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2531430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Evaluating $\int_1^\infty \frac{1}{x(x^2+1)}\ dx$ $$I=\int_1^\infty \frac{1}{x(x^2+1)}\ dx$$
I tried to use partial fractions, but am unsure why I cann't evaluate it using partial fractions. Would appreciate any explanation about which step is incorrect.
$$\frac{1}{x(x^2+1)} = \frac{1}{2} \left(\frac{1}{x}-\frac{1}... | put $x = \tan \theta$ given integral is
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\sec^2 \theta}{\tan \theta \sec^2 \theta} = \frac{d(\sin \theta)}{\sin \theta} = \ln [|\sin \theta|]^{\frac{\pi}{2}}_{\frac{\pi}{4}} = 0-\ln \frac{1}{\sqrt{2}}=\ln \sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2533478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a,b,c$ are three complex numbers Find possible values of $\lvert a+b+c \rvert$ Given three complex numbers $a,b,c$ such that $\lvert a \rvert=\lvert b \rvert=\lvert c \rvert=1$ and $$\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=-1$$
Find which of the following are possible values of $\lvert a+b+c \rvert$
A)0
B)2
C)... | The four complex numbers$\frac{a^2}{bc},\frac{b^2}{ac},\frac{c^2}{ab},1$ all have length $1$ and sum to zero. Added as vectors,'head to tail', in the Argand diagram they form a closed shape and must consist of two pairs of equal but opposite vectors.
Without loss of generality we therefore have $\frac{a^2}{bc}=-\fra... | {
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"url": "https://math.stackexchange.com/questions/2533698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sketching $(x^2-2)^2+(y^2-2)^2=4$ I was recently asked to sketch $(x^2-2)^2+(y^2-2)^2=2$, which did not prove to be too problematic, for establishing the range and domain of the expression gives nearly all of it away.
I then asked myself what would happen as I change the constant on the R.H.S. I recommend you try this... | Well, you can go backwards, and multiply $x^2 - \sqrt2xy+y^2-2$ and $x^2 + \sqrt2xy+y^2-2$ to get the product $x^4-4x^2+y^4-4y^2 + 4$, equal to $(x^4-4x^2+4) + (y^4-4y^2+4)-4$, but I did not see the reverse path when I looked at your question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2533916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Range of root of cubic equation having least absolute value Suppose $a$ and $b$ are two positive real numbers such that the roots of the cubic equations $x^3-ax+b=0$ are all real. Let $\alpha$ is a root of this cubic equation with minimum absolute value. Then range of values of $\alpha$ is given by(choose the correct o... | Let $\alpha,\beta,\gamma$ be the solutions of $x^3-ax+b=0$ with $|\alpha|\le|\beta|$ and $|\alpha|\le|\gamma|$.
By Vieta's formula,
$$\alpha+\beta+\gamma=0,\quad \alpha\beta+\beta\gamma+\gamma\alpha=-a\lt 0,\quad \alpha\beta\gamma=-b\lt 0$$
Since $\alpha\beta\gamma\lt 0$, we have either that only one of them is negativ... | {
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"url": "https://math.stackexchange.com/questions/2536975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $\log x=\log 2+\log y$ and $2^x+8^y=4$ then find the $x$
Let $\log x=\log 2+\log y$ and $2^x+8^y=4$ then find the $x$
My Try :
$$\log x=\log 2+\log y \\ \log x=\log 2y \\x=2y $$
So we have :
$$2^x+8^y=4\\2^{2y}+2^{3y}=4 \\t=2^y \\t^2+t^3=4$$
now what ?
| from the first equation we have $$\log(x)=\log(2y)$$ and we get $$x=2x$$ plugging this in your second equation we have
$$2^{2y}+8^y=4$$
this can be written as $$(2^y)^2+(2^y)^3=4$$ with $t=2^y$ we have to solve
$$t^3+t^2-4=0$$
one solution is given by $$t=\frac{1}{3} \left(-1+\sqrt[3]{53-6
\sqrt{78}}+\sqrt[3]{53+6 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2539310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find a primitive root of $71$. In my Number Theory Class we found that $7$ was a primitive root of 41 by first finding two integers who have order $5$ and $8$ $modulo 41$ respectively, these being $16$ and $3$. Since $16(3)=7(mod41)$ 7 is a primitive root.
I'm trying to do this for $71$ and so far have that $5$ has o... | $G=\mathbb{Z}/(71\mathbb{Z})^*$ is a cyclic group with order $70=2\cdot 5\cdot 7$. It follows that any $g\in G$ such that $g^{10}\not\equiv 1\pmod{71}$, $g^{14}\not\equiv 1\pmod{71}$ and $g^{35}\not\equiv 1\pmod{71}$ is a generator for $G$. There are $\varphi(70)=1\cdot 4\cdot 6=24$ generators, so a random non-quadrati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2540224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Sum of series of fractions: $\frac{4}{1!}+ \frac{8}{2!}+ \frac{14}{3!} + \frac{22}{4!}+\cdots$
Find the sum of the series:
$$\frac{4}{1!}+ \frac{8}{2!}+ \frac{14}{3!} + \frac{22}{4!}+\cdots$$
I am not able to understand how to proceed. The numerator terms have nothing in common which might result in an AP or GP.
Pl... | Observe that the first order differences of the numerators, $4,6,8$ are increasing linearly (and hopefully continue doing so - there are few known terms), so that the general form of the numerators must be quadratic, $N_n:=an^2+bn+c$.
By extrapolation, the zero-th term is $2$, and as the second order difference is $2$,... | {
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"timestamp": "2023-03-29T00:00:00",
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Irreducible components of $I = (x^3 - y^3, x^2 - y^2 + 2x - 2y)$.
I am trying to find the irreducible components of $V(I),$ where $I = (x^3 - y^3, x^2 - y^2 + 2x - 2y)$.
So far I've factorized as follows:
$$x^2 - y^2 + 2x - 2y = (x-y)(x + y + 2)$$
$$x^3 - y^3 = (x -y)(x - \omega y)(x - \omega^2 y)$$
From this I've ... | We have
$$x^2 - y^2 + 2x - 2y = (x-y)(x + y + 2), \quad x^3 - y^3= (x -y)(x^2+xy+y^2)$$
So, $(x,y) \in V(I)$ iff $x=y$ or $x + y + 2=0=x^2+xy+y^2$.
The last equation has two solutions: $x=-1 \pm \sqrt3i, y=-1 \mp \sqrt3i$.
Therefore, $V(I)$ has three irreducible components: a line and two isolated points.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2544302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Series expansion of $\frac{x^n-1}{x-1}$ at $ x=1$. $$\lim_{x\to 1}\left(\frac{x^n-1}{x-1}\right)=n$$
I thought that at $x=1$ the expansion is:
$$n+n(n-1)(x-1)+n(n-1)(n-2)(x-1)^2+...$$
but the answer is:
$$n+\frac{1}{2}n(n-1)(x-1)+\frac{1}{6}n(n-1)(n-2)(x-1)^2+...$$
Can you explain the origin of $$\frac{1}{2},\frac{1}{6... | hint
Let $f (x)=\frac {x^n-1}{x-1}$.
we expand $f (x+1) $ near zero as
$$f (x+1)=((x+1)^n-1)/x=$$
$$(1+nx+\frac {n (n-1)}{2!}x^2+\frac{n (n-1)(n-2)}{3!}x^3+...-1)/x $$
$$=n+\frac {n (n-1)}{2!}x+.... $$
to get the expansion arround $x=1$, replace $x $ by $x-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2546536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Let $a^{4} + b^{4} + a^{2} b^{2} = 60 \ \ a,b \in \mathbb{R}$ Then prove that : Let $a^{4} + b^{4} + a^{2} b^{2} = 60 \ \ a,b \in \mathbb{R}$ Then prove that :
$$4a^{2} + 4b^{2} - ab \geq 30$$
My attempt: :
$$4a^{2} + 4b^{2} - ab \geq 30 \\ 4(a^2+b^2)-ab \geq30 \\4(60-a^2b^2)-ab\geq30\\ 240-30\geq4(ab)^2+ab\\ 4(ab)^2+... | Let $a^2+b^2=2kab$.
Hence, we need to prove that
$$(4a^2+4b^2-ab)^2\geq900\cdot\frac{a^4+b^4+a^2b^2}{60}$$ or
$$(8k-1)^2\geq15(4k^2-1)$$ or
$$(k-2)^2\geq0.$$
Done!
| {
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"url": "https://math.stackexchange.com/questions/2552804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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.