Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Matrices operations My professors asked to find the inverse of M:
$$ M=
\begin{bmatrix}
I_p&0\\
A&I_q\\
\end{bmatrix}
$$
Therefore, to find $M^{-1}$, we calculate:
$$
\begin{bmatrix}
I_p&0\\
A&I_q\\
\end{bmatrix}
\begin{bmatrix}
C&D\\
E&F\\
\end{bmatrix} =
\begin{bmatrix}
I_p&0\\
-A&I_q\\
\end{bmatrix}
$$
How do I go f... | Assuming the matrix equation
$\begin{bmatrix} C&D\\AC+E&AD+F\\\end{bmatrix} =
\begin{bmatrix} I_p&0\\ 0&I_q\\\end{bmatrix} \tag 1$
is to be interpreted in terms of matrix blocks, we immediately have
$C = I_p, \tag 2$
$D = 0, \tag 3$
$AC + E = 0, \tag 4$
$AD + F = I_q; \tag 5$
then
$A + E = AI_p + E = AC + E = 0, \tag ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3407230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Taylor series for $\sqrt{x^0+\sqrt{x^1+\sqrt{x^2+\cdots}}}$ Let
$$f(x) =\sqrt{x^0+\sqrt{x^1+\sqrt{x^2+\cdots}}} = A_0 + A_1(x-1)+A_2(x-1)^2+\cdots$$
I got the following:
*
*$A_0 = (1+\sqrt{5})/2$
*$A_1 = 1/5$
*$A_2=-1/25$
*$A_3 = -1/168$
The values seem correct for $A_0, A_1, A_2$ but only a good approximation ... | Let $f_n(z) = \sqrt{z^n + \sqrt{z^{n+1} + \sqrt{\ldots}}}$.
Thus $f_n(z)^2 = z^n + f_{n+1}(z)$. Since we're expanding
around $z=1$, it may help to write $z = 1+t$. Letting
$f_n(z) = a_0(n) + a_1(n) t + a_2(n) t^2 + \ldots$ and working formally, we have
$$ (a_0(n) + a_1(n) t + a_2 (n) t^2 + \ldots)^2 = (1+t)^n +
a_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3407391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
$\lim_{x \to 0} \frac{x^{11}-3x^2+\sin x}{e^x - \cos x}$ Is this solution correct? Just want to double check whether all my operations are legal. The result seems to be correct but I want to make sure I didn't make any mistakes.
$\require{cancel}$
$$\lim_{x \to 0} \frac{x^{11}-3x^2+\sin x}{e^x - \cos x}=$$
$$=\lim_{x \... | The result is correct but we can simplify some step as follows
$$\frac{x^{11}-3x^2+\sin x}{e^x - \cos x}=\frac{x^{11}-3x^2}{e^x - \cos x}+\frac{\sin x}{e^x - \cos x}$$
and since the first term tends to $0$
$$\frac{x^{11}-3x^2}{e^x - \cos x}=\frac{x^{10}-3x}{\frac{e^x - 1}x+x\frac{1-\cos x}{x^2}}\to 0$$
we obtain
$$\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Recursive function proof using induction We get a $D_0, D_1, D_2, ...$ sequence, for the recursive function $D_n = (n-1)(D_{n-1} + D_{n-2})$ for every n ≥ 2, starting at $D_0 = 1, D_1 = 0$.
Using induction, prove that for every n≥0, $D_n = n! \sum_{k=0}^n = \frac{(-1)^k}{k!}$.
Now, taking for example n=2, the recursive... | We have
$$D_{n+1}=(n+1)!\sum_{k=0}^{n+1}\frac{-1^k}{k!}=(n+1)*n!*(\sum_{k=0}^{n}\frac{-1^k}{k!}+\frac{-1^{n+1}}{(n+1)!})=nD_n+n*n!\frac{-1^{n+1}}{(n+1)!}+n!(\sum_{k=0}^{n-1}\frac{-1^k}{k!}+\frac{-1^n}{n!}+\frac{-1^{n+1}}{(n+1)!})=nD_n+nD_{n-1}+n!(n\frac{-1^{n+1}}{(n+1)!}+\frac{-1^{n+1}}{(n+1)!}+\frac{-1^{n}}{(n)!})=n(D... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals. I need to prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals.
I started by using Bernoulli's inequality:
$(1+\frac{2}{\sqrt{n}}) < (1 + \frac{1}{\sqrt{n}})^2$
I can say that:
$(1+\frac{2}{\sqrt{n}}) = (1+\frac... | It is equivalent to prove
$$ n < \left(1 + \frac{1}{\sqrt{n}}\right)^{2n}.$$
By the binomial identity, the right hand side is at least
\begin{align*}
& 1 + \binom{2n}{1}\frac{1}{\sqrt{n}} + \binom{2n}{2}\frac{1}{n} \\
= & 1 + 2\sqrt{n} + 2n - 1 \\
= & 2n + 2\sqrt{n} \\
> & n.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
} |
$\int \frac{1}{(x^2-4)^2}dx$ Calculate:
$$\int \frac{1}{(x^2-4)^2}dx.$$
I tried Partial Fractions method first I write:
$$\frac{1}{(x^2-4)^2}=\frac{A}{X-2}+\frac{Bx+C}{(x-2)^2}+\frac{D}{x+2}+\frac{Ex+F}{(x+2)^2}.$$
We have:
$$A(x-2)(x+2)^2+(Bx+C)(x+2)^2+D(x+2)(x-2)^2+(Ex+F)(x-2)^2=1.$$
$$(A+B+D+E)x^3+(4A-2A+4B+C-4D+2D... | Hint:
Another way
Write numerator as $$\left(\dfrac{x+2-(x-2)}4\right)^2$$
Now $\dfrac1{(x+2)(x-2)}=\dfrac{x+2-(x-2)}{4(x+2)(x-2)}=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Problem in the properties of limit: $\lim\limits_{x \to\frac{\pi}{3}}\frac{\sin\left(x-\frac{\pi}{3}\right)}{1-2\cos\left(x\right)}$
$$\lim\limits_{x \to\frac{\pi}{3}}\frac{\sin\left(x-\frac{\pi}{3}\right)}{1-2\cos\left(x\right)}$$
I used the following property:
if $$\lim\limits_{\large x \to\frac{\pi}{3}}f(x)=L$$
th... | Your derivation is absolutely fine and right, but we have that
$$1-2\cos\left(u+\frac{\pi}{3}\right)=1-2\frac12\cos u+2\frac {\sqrt 3} 2\sin u=1-\cos u+\color{red}{\sqrt 3}\sin u$$
therefore
$$\lim\limits_{x \to\frac{\pi}{3}}\frac{\sin\left(x-\frac{\pi}{3}\right)}{1-2\cos\left(x\right)}=\frac1{\sqrt 3}$$
Note also that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3410852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Square-root equation
Solve square-root equation: $\left (\sqrt{5+2\sqrt{6}} \right )^x+\left (\sqrt{5-2\sqrt{6}} \right )^x=10$
$\left (\sqrt{5+2\sqrt{6}} \right )^x+\left (\sqrt{5-2\sqrt{6}} \right )^x=10\\
\left (\sqrt{\left (\sqrt{3}+\sqrt{2} \right )^2} \right )^x+\left (\sqrt{\left (\sqrt{3}-\sqrt{2} \righ... | Expanding on @DanielFischer's comment, let $y=(\sqrt{3}+\sqrt{2})^x$ so $y+1/y=10$ and $y=5\pm2\sqrt{6}$, so $x=\pm2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Last step in evaluating: $\int {dx\over a\cos x + b\sin x}$ Seems like I've made a mistake somewhere in my calculation and am now struggling to find it without any success. Here is the problem statement:
Evaluate:
$$
\int {dx\over a\cos x + b\sin x}
$$
I've decided to use a $t = \tan{x\over 2}$ substitution:
$$
dx... | $\tan(\frac x2+\frac\phi 2) = \dfrac{\tan\frac x2+\tan\frac\phi 2}{1 - \tan\frac x2\tan\frac\phi 2}$
but
$$\tan\frac\phi 2 = \dfrac{\sin\phi}{1+\cos\phi} = \dfrac{a}{b+\sqrt{a^2+b^2}} = \dfrac{\sqrt{a^2+b^2} - b}{a} = -R_2.$$
Thus,
$$\ln\tan\left(\frac x2+\frac\phi 2\right) = \ln\dfrac{\tan\frac x2 - R_2}{1+R_2\tan\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3421982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Determine the value of $k$ such that the line determined by the points $(3,2)$ and $(1,-2)$ is tangent to the graph of $y=\frac{k}{x+1}$. I have done the following work but I am stuck on solving for k. Am I doing this right? If so what do I do next?
| We have
$1.$ $k/(x+1)=2x-4$
$2.$ $-k/(x+1)^2=2$
First equation
Multiplying it by $(x+1)$
$k=(x+1)\cdot (2x-4)=2x^2-4x+2x-4=2x^2-2x-4 \quad (1)$
Second equation
Multiply it by $(x+1)^2$
$-k=2(x+1)^2=2x^2+4x+2 \Rightarrow k=-2x^2-4x-2 \quad (2)$
Set $(1)$ equal to $(2)$
$2x^2-2x-4=-2x^2-4x-2$
$4x^2+2x-2=0$
$2x^2+1x-1=0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3423556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all $n≥1$ natural numbers such that : $n^{2}=1+(n-1)!$ Problem :
Find all $n≥1$ natural numbers such that : $n^{2}=1+(n-1)!$
My try :
$n=1$ we find : $1=1+1$ $×$
$n=2$ we find : $4=1+1$ $×$
$n=3$ we find : $9=1+2$ $×$
$n=4$ we find : $16=1+6$ $×$
$n=5$ we find : $25=1+24$ $√$
Now how I prove $n=5$ only the ... | Hint If $n \geq 6$ then
$$1+(n-1)! \geq 1+2 \cdot 3 \cdot (n-2) \cdot (n-1)$$
Show that
$$2(n-1) \geq n \\
3(n-2) \geq n$$
for $n \geq 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3427795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Find the units digit of $572^{42}$ The idea of this exercise is that you use the modulus to get the right answer.
What I did was:
$$572\equiv 2\pmod {10} \\
572^2 \equiv 2^2 \equiv 4\pmod{10} \\
572^3 \equiv 2^3 \equiv 8\pmod{10} \\
572^4 \equiv 2^4 \equiv 6\pmod{10} \\
572^5 \equiv 2^5 \equiv 2\pmod{10} \\
572^6 \... | Another approach is:
The reminder is repeating between:
$$R=\{2,4,8,6\}$$
Solve the following, since $|R|=4$
$$42\equiv x\pmod{4}$$
Have $x=2$, which means the reminder is the second term in $R$ that is $4$
Therefore $$572^{42}\equiv4\pmod{10}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3428727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Find all triples of non-negative real numbers $(a,b,c)$ Find all triples of non-negative real numbers $(a,b,c)$ such that:
$a^2+ab=c$
$b^2+bc=a$
$c^2+ca=b$.
This question was problem number of 3 in the RMO(India) Olympiad in 2019 held on $10^{th}$ November. link .
My attempt-
Assume $a \geq b\geq c$,
$\therefore a^2 ... | Eliminating $b,c$ from our system we get for $a$ the following equation:
$$2 a^8+a^7+a^6+3 a^5+6 a^4+2 a^3-a^2-a=0$$
and this is $$a (2 a-1) \left(a^6+a^5+a^4+2 a^3+4 a^2+3 a+1\right)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3431214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Simplifying the result of integration of $\int\frac{e^x + e^{3x}}{1-e^{2x}+e^{4x}}\mathop{dx}$
Evaluate:
$$
\int\frac{e^x + e^{3x}}{1-e^{2x}+e^{4x}}\mathop{dx}
$$
I'm trying to simplify my answer so that it matches the keys section, no success so far. The integral itself is pretty simple. Factor $e^x$ in the denomi... | Let's start off from
$$\sinh(x) = \frac{e^x-e^{-x}}{2} = -\frac{e^{-x}-e^x}{2} = -\frac{\frac{1}{e^x}-e^x}{2} = -\frac{1-e^{2x}}{2e^x} = -\frac{4-4e^{2x}}{8e^{2x}} = \\ = -\frac{1-(4e^{2x}-3)}{8e^x} = -\frac{1-(2e^x+\sqrt{3})(2e^x-\sqrt{3})}{8e^x}.$$
For conveniency, I'll rewrite this as:
$$2\sinh(x)=-\frac{1-(2e^x+\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3431568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Proof $a_{n}$ by induction This topic is a continuation of Prove that $a_{n} = a_{n-1} + a_{n-2}$.
I need to prove that $a_{n} = (\frac{5+3\sqrt{5}}{10})(\frac{1+\sqrt{5}}{2})^{n} + (\frac{5-3\sqrt{5}}{10})(\frac{1-\sqrt{5}}{2})^{n}$ for all $n \geq 1$.
This is how I tried to prove this.
Basic step: $a_{1} = (\frac{5+3... | Rather than solve the problem through the standard methods for more general linear recurrences, I'll address your questions about induction:
*
*Your choice of induction hypothesis is right. We already know that $a_{n+2}=a_{n+1}+a_n$ for all $n\geq 1$ from your previous question. Here you want to assume that $a_{n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3432924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find $I=\int_0^1\frac{\arctan^2x}{1+x}\left(\frac{\ln x}{1-x}+\ln(1+x)\right)dx$ $$I=\int_0^1\frac{\arctan^2x}{1+x}\left(\frac{\ln x}{1-x}+\ln(1+x)\right)dx=-\frac{\pi^4}{512}+\frac{3\pi^2}{128}\ln^22+\frac{\pi}{8}G\ln2-\frac{21}{64}\zeta(3)\ln2$$
This integral was proposed to me by a friend, but without the so... | Continuing on @AliShather's argument, let's focus on the integral
\begin{align*}
I_1&=\int_0^1\frac{\arctan^2 x\ln x}{1-x^2}dx\\
&\overset{\rm IBP}=-\int_0^1\frac{\arctan^2 x\mathop{\rm artanh} x}{x}+2\underbrace{\frac{\arctan x\mathop{\rm artanh} x \ln x}{1+x^2}}_{x\mapsto \frac{1-x}{1+x}}dx\\
&=-\int_0^1\frac{\arcta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3440789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Proving the inequality $\sum_{\text{cyc}} \frac{a}{a^2+b^3+c^3}\le\frac1{5abc}$ Let $a,b,c> 0$ be three real numbers such that $a+b+c=1$. I want to prove that
$$\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+a^3+c^3}+\frac{c}{c^2+a^3+b^3}\le\frac{1}{5abc}.$$
My attempt: Using AM-GM on each denominator gives (here, LHS denotes the... | Also, by AM-GM and C-S we obtain:
$$\sum_{cyc}\frac{a}{a^2+b^3+c^3}=\sum_{cyc}\frac{a}{a^3+b^3+c^3+a^2b+a^2c}\leq\sum_{cyc}\frac{a}{3abc+a^2b+a^2c}=$$
$$=\sum_{cyc}\frac{1}{3bc+ab+ac}\leq\frac{1}{(3+1+1)^2}\sum_{cyc}\left(\frac{3^2}{3bc}+\frac{1^2}{ab}+\frac{1^2}{ac}\right)=\frac{1}{5}\sum_{cyc}\frac{1}{ab}=\frac{1}{5a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3442048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $3^{2n-1} + 2^{n+1}$ is divisible by $7$ for all values of $n$ I have tried to prove this through mathematical induction but I can't seem to prove that the proposition works for $k+1$.
| $3^{2n-1} \begin{matrix} n& 1 & 2 & 3 & 4 & 5 & \cdots \\ mod 7 &3 & 6 & 5 & 3 & 6 & \cdots \end{matrix}$
$2^{n+1} \begin{matrix} n& 1 & 2 & 3 & 4 & 5 & \cdots \\ mod 7 &4 & 1 & 2 & 4 & 1 & \cdots \end{matrix}$
Since $3+4 = 7$, $6+1=7$, $5+2 = 7$ we can see that $3^{2n-1}+2^{n+1} \equiv 0 \pmod{7}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3443306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Probability of taking out coins by order
The question goes as following:
There is a jar with 65 coins. 30 are green, 15 are yellow and 20 are red.
When taking out 20 coins from the jar, one by one, what is the probability that the first three coins and last three coins selected will be green?
My logic was that if... | Let's see for $7$ balls:
$$P(\color{green}{G_1G_2G_3}(\color{green}G \text{ or } G')_4\color{green}{G_5G_6G_7})=\\
P(\color{green}{G_1G_2G_3G_4G_5G_6G_7})+P(\color{green}{G_1G_2G_3}G'_4\color{green}{G_5G_6G_7})=\\
\color{green}{\frac{30}{65}\cdot \frac{29}{64}\cdot \frac{28}{63}}\cdot \color{green}{\frac{27}{62}\cdot \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3447903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Obtain the maximum and minimium of the function $f$ I have to calculate the max and min of the function
\begin{equation}
f(x_1,x_2 \ldots ,x_n)=x_1^3 + x_2^3+\cdots + x_n^3
\end{equation}
subject to
\begin{equation}
x_1 + x_2+\cdots + x_n =0
\end{equation}
\begin{equation}
x_1^2 + x_2^2+\cdots + x_n^2 =1
\end{equation... | The Lagrange multipliers tell us that there exist $\lambda$ and $\mu$ such that for all $i=1,\ldots,n$
\begin{equation}
3 x_i^2 = 2 \lambda x_i + \mu
\end{equation}
by summing these equalities one gets $3 = n\mu$. We still don't know $\lambda$ but as all the $x_i$ satisfy the same 2nd degree equation, it means that the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Help evaluating a multivariable limit I tried approaching from y=0, from x=0, from x=y, I also tried using polar coordinates, I always get 0/0. What do I need to do to?
$$\lim_{(x,y)\to(1,1)} \dfrac{(x-y)(x+y-2)}{\sqrt{(x-1)^2+(y-1)^2}}$$
Mulivariable limit problem
| Set $x=1+r\cos\theta$, $y=1+\sin\theta$. You obtain
\begin{align}
\frac{(x-y)(x+y-2)}{\sqrt{(x-1)^2+(y-1)^2}}&=\frac{(r\cos\theta-r\sin\theta)(r\cos\theta+r\sin\theta)}{\sqrt{r^2}}\\&=r(\cos^2\theta-\sin^2\theta)=r\cos2\theta,
\end{align}
which tends to $0$ as $r$ tends to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3449790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Computation of a 2D limit I came across this limit and I used Mathematica to solve it, (it shows that it is equal to $0$).
Any thoughts on how to prove this ? The limit is:
$$\lim \limits_{(x,y)\to (4,0)} \frac{y^2(x^2+y^2)}{(x-4)^2+y^2}$$
Thanks.
| Let $u=x-4\to 0$ then
$$\lim \limits_{(x,y)\to (4,0)} \frac{y^2(x^2+y^2)}{(x-4)^2+y^2}=\lim \limits_{(u,y)\to (0,0)} \frac{y^2((u+4)^2+y^2)}{u^2+y^2}$$
and for
*
*$y=0$
$$\frac{y^2((u+4)^2+y^2)}{u^2+y^2}=0$$
but for
*
*$u=y$
$$\frac{u^2((u+4)^2+u^2)}{2u^2}=\frac{2u^2+8u+16}{2}=u^2+4u+8\to 8$$
therefore the given ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\sum_\text{cyc} \frac{ab}{ab+b^2+ca}\le 1$ Hello ladies and gentlemen, here I have another inequality that I am struggling with:
Let $a,b,c>0$ Then
$$\sum_\text{cyc} \frac{ab}{ab+b^2+ca}\le 1.$$
I try to show $$\frac{ab}{ab+b^2+ca}\le\frac{a}{a+b+c}$$ but this is doesn't work because $$\frac{ab}{ab+b^2+ca}... | Multiply both sides with $\prod_{\text{cyc}} ab+b^2+ac$, subtract $ab\cdot(bc+c^2+ab)\cdot(ac+a^2+bc)$ from both sides and you get by expanding that the original inequality is equivalent to:
\begin{equation}\tag 1\label 1\sum_{\text{cyc}} a^4bc+\sum_{\text{cyc}} a^3b^3\geq 2\sum_{\text{cyc}} a^3b^2c.\end{equation}
By A... | {
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Logical statements: $A+B+C=A+B+C+AB+BC+AC$
Let $A,B, C$ be logical statements.
Then:
$A+B+C=A+B+C+AB+BC+AC$
Prove it without a table.
My attempt:
$$A=A\cdot(1+B+C)$$
$$B=B\cdot(1+A+C)$$
$$C=C\cdot(1+A+B)$$
$$\implies A+B+C= A+AB+AC+B+AB+BC+C+BC+AC$$
$$\implies A+B+C=A+B+C+AB+AB+BC+BC+AC+AC$$
$$\implies A+B+C=A+... | Yes. it is completely correct.
You could have shortened it a bit by writing
$$A+B+C+AB+BC+AC = A(1+B+C) + B(1+C) + C $$ $$
\stackrel{1+x=1,1x=x}{=}A+B+C$$
| {
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Convergence of $\int_0^{\infty}\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $
Find out if the following integral diverges or converges:
$$
\int_0^\infty \frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx
$$
First I split the integral as $\displaystyle \int_0^1 \frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx + \int_1^\infty... | Just for the fun of it.
The convergence issues having been clearly explained and one answer for the value showing that a CAS is able to find a (nasty) antiderivative, the result is in fact
$$\int_0^{\infty}\frac{\log (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx=\frac{1}{3} \log \left(\frac{1}{4} \left(-1+i+\sqrt{-1-2 i}\right)^{... | {
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Prove that the minimum values of $x^2+y^2+z^2$ is $27$ with given condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$. Question: Prove that the minimum values of $x^2+y^2+z^2$ is $27$, where $x,y,z$ are positive real variables satisfying the condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$.
From AM$\ge$ GM, we h... | You can still finish it off if you notice that $( xy+yz+zx )^2 = (xyz)^2 = xy\cdot yz \cdot zx \le \dfrac{(xy+yz+zx)^3}{27} \implies xy+yz+zx \ge 27 \implies \dfrac{x^2+y^2+z^2}{3} \ge \sqrt[3]{27^2} = 9 \implies x^2+y^2+z^2 \ge 9\cdot 3 = 27 $.
| {
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Show that following determinant is divisible by $\lambda^2$ and find the other factor. Show that $\begin{vmatrix}
a^2+\lambda &ab &ac \\
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}=0$ is divisible by $\lambda^2$ and find the other factor.
My attempt is as follows:-
$$R_1\rightarrow R_1+R_2+R_3$$
$$\be... | Alternatively:
$$\begin{vmatrix}
a^2+\lambda &ab &ac \\
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}=
\begin{vmatrix}
a^2 &ab &ac \\
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}+
\begin{vmatrix}
\lambda&ab&ac \\
0 & b^2+\lambda & bc \\
0 & bc & c^2+\lambda
\end{vmatrix}=\\
a^2\begin{v... | {
"language": "en",
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"answer_id": 4
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Infinite series from Euler I am reading Euler's Introduction to analysis of the infinite" and is stumped at this section:
How does he derive the part that I circled in red?
I thought:
$\dfrac{a+bz+cz^2+...}{z(1-\alpha\cdot z-\beta\cdot z^2...)}=A+Bz+Cz^2+Dz^3+Ez^4...$
is gonna be:
$a+bz+cz^2+...=z(1-\alpha\cdot z-\be... | You are happy with geometrically expanding ?
\begin{eqnarray*}
\frac{1}{1- \alpha z - \beta z^2 -\cdots} &=& 1+(\alpha z + \beta z^2 -\cdots)+( \alpha z + \beta z^2 +\cdots)^2+ \cdots \\
&=& 1+pz+qz^2+ \cdots.
\end{eqnarray*}
Now let multiply in the numerator
\begin{eqnarray*}
\frac{a+bz+cz^2+\cdots }{1- \alpha z -... | {
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Find the closed form for this series I found this interesting series from the , it is from an old math books. It is as followed:
$\dfrac{1}{2}-\dfrac{x^2}{6}+\dfrac{x^4}{12}-\dfrac{x^6}{20}+\dfrac{x^8}{30}-...$
I notice that one can rewrite this series as followed:
$\dfrac{1}{2}-\dfrac{x^2}{2\cdot 3}+\dfrac{x^4}{3\cdot... | $$\sum_{n=0}^{\infty} \dfrac{(-1)^{n}x^{2n}}{(n+1)(n+2)}=\sum_{n=0}^\infty(-1)^nx^{2n}\int_0^1\int_0^1 y^nz^{n+1}dydz$$
$$=\int_0^1\int_0^1z\sum_{n=0}^\infty(-x^2yz)^ndydz$$
$$=\int_0^1\int_0^1\frac{z}{1+x^2yz}dydz$$
$$=\int_0^1z\left(\int_0^1\frac{dy}{1+x^2yz}\right)dz$$
$$=\int_0^1z\left(\frac{\ln(1+x^2z}{x^2z}\right... | {
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"answer_id": 3
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Number of empty subway cars I have task:
There are $9$ passengers and they get into empty $5$-car subway train. What is the probability of the fact, that exactly two cars will be empty?
I know that the power of all possibilities is $5^9$. Then I thought to pick two empty cars that is ${5\choose 2}$ possibilities, th... | The three cars must have at least $1$ passenger each. So the cases can be easily calculated:
$$\{1,1,7\} \Rightarrow {9\choose 1}{8\choose 1}3=216\\
\{1,2,6\} \Rightarrow {9\choose 1}{8\choose 2}6=1512\\
\{1,3,5\} \Rightarrow {9\choose 1}{8\choose 3}6=3024\\
\{1,4,4\} \Rightarrow {9\choose 1}{8\choose 4}3=1890\\
\{2,2,... | {
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"answer_count": 2,
"answer_id": 1
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Solve degree-four polynomial equation $\left(x^2-3x-5\right)^2-3\cdot \left(x^2-3x-5\right)-5=x$ How to solve
$$\left(x^2-3x-5\right)^2-3\cdot \left(x^2-3x-5\right)-5=x?$$
I think there is a trick, because, if $X=x^2-3x-5$, I get
$$ X^2-3X-5=x.$$
I don't know how to continue.
I've also tried to factorize but it does... | Factorize as follows,
$$(x^2-3x-5)^2-3(x^2-3x-5)^2-4=x+1$$
$$(x^2-3x-9)(x^2-3x-4)=x+1$$
$$(x^2-3x-9)(x-4)(x+1)-(x+1)=0$$
$$(x+1)(x^3-7x^2+3x+35)=0$$
$$(x+1)[(x^3-5x^2)-(2x^2-3x-35)]=0$$
$$(x+1)[x^2(x-5)-(2x+7)(x-5)]=0$$
$$(x+1)(x-5)(x^2-2x-7)=0$$
$$(x+1)(x-5)(x-1+2\sqrt2)(x-1-2\sqrt2)=0$$
Thus, the roots are $x=-1, 5, ... | {
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"answer_id": 2
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Let $A$ be a real $2×2$ matrix such that $A^6=I$. The total number of possibilities for the characteristic polynomial of $A$ is: Let $A$ be a real $2×2$ matrix such that $A^6 = I$ (where $I$ denote the identity $2×2$ matrix). The total number
of possibilities for the characteristic polynomial of $A$ is:
Annihilating po... | To find the possibilities for the characteristic polynomial of an $n\times n$ matrix $A$, it helps to know some facts about the minimal polynomial of $A$. Let $c(x)$ be the characteristic polynomial of $A$, and let $m(x)$ be the minimal polynomial. Here are some facts about $m(x)$:
*
*The degree of $m(x)$ is at leas... | {
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"answer_id": 1
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Why is $\frac{1}{\sqrt 5}\left[\left(\frac{1+\sqrt 5}{2}\right)^n-\left(\frac{1-\sqrt 5}{2}\right)^n\right]$ an integer?
I am looking for a proof on why
$$\frac{1}{\sqrt 5}\left[\left(\frac{1+\sqrt 5}{2}\right)^n-\left(\frac{1-\sqrt 5}{2}\right)^n\right]$$
an integer.
I have seen many proofs on this, but they al... | One of the easiest proofs goes by induction.
*
*Set $a= \frac{1+\sqrt{5}}{2}$ and $b= \frac{1-\sqrt{5}}{2}$
Then you have
$$ab = -1 \mbox{ and } a+b = 1$$
Induction start $n= 1$: $\frac{1}{\sqrt 5}\left(a-b\right) = 1$ (Note, that for $n=0$ it is trivially true.)
Induction hypothesis: $\frac{1}{\sqrt 5}\left(a^k-b^k... | {
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Does exist $a,b,c \in \Bbb N$ such that $(a+b)(b+c)(c+a)=340$? Does exist $a,b,c \in \Bbb N$ such that $(a+b)(b+c)(c+a)=340$?
$340=2\cdot2\cdot5\cdot17$
I just noticed that $(a+b)+(b+c)+(c+a)=2(a+b+c)$ can it be useful to prove first equation?
Also I tried to construct numbers $a,b,c$ such that $a+b$ would be even, $b... | We may assume that $a \le b \le c$. Then
$ 2a \le a+b \le 2b $
$ 2b \le b+c \le 2c $
$ 2a \le c+a \le 2c $
and so
$ 8a^3 \le 8a^2b \le 340 \le 8 bc^2 \le 8 c^3$. Therefore, $a \le 3$ and so $a \in \{0,1,2,3\}$. Consider all four cases.
*
*$a=0$. Then $b(b+c)c=340$ and so $2b^3 \le 340$. Therefore, $b \le 5$ and so $... | {
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"source": "stackexchange",
"question_score": "1",
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Inequality Proof $(x^2+1)(y^2+1)(z^2+1)\leq...$ I want to show, that for positive numbers $x,y,z$ with $xy,yz,zx\geq1$, $(x^2+1)(y^2+1)(z^2+1)\leq\left(\left(\frac{x+y+z}{3}\right)^2+1\right)^3$.
Using the AM-QM inequality for $(x^2+1),(y^2+1)$ and $(z^2+1)$, I already got $(x^2+1)(y^2+1)(z^2+1)\leq\left(\frac{x^2+y^2+... | Hint: First show that, if $ab\geq 1$,
$$(a^2+1)(b^2+1)\leq \left(\left(\frac{a+b}{2}\right)^2+1\right)^2$$
(hint for this: expand out and use that $(a-b)^2\geq 0$ and $ab\geq 1$). Then apply this to get that
$$(a^2+1)(b^2+1)(c^2+1)(d^2+1)\leq \left(\left(\frac{a+b+c+d}{4}\right)^2+1\right)^4$$
if $ab,bc,cd,da,ac,bd\geq... | {
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"answer_id": 0
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On the biconditional $I(n^2) = 2 - \frac{5}{3q} \iff (k = 1 \land q = 5)$, where $q^k n^2$ is an odd perfect number MOTIVATION
Let $N$ be an odd perfect number given in the so-called Eulerian form
$$N = q^k n^2,$$
i.e., $q$ is the special / Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
In what... | Let $q^k n^2$ be an odd perfect number given in Eulerian form.
Since the equation
$$I(n^2) = 2 - \frac{5}{3q}$$
is true if and only if the conjunction
$$k=1 \land q=5$$
holds, then $I(n^2) = 2 - {5/(3q)}$ is true if and only if
$$I(n^2) = 2 - \frac{5}{3\cdot{5}} = 2 - \frac{1}{3} = \frac{5}{3}.$$
It follows that
$$I(n^... | {
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"source": "stackexchange",
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How to solve $x^{\log(x)-1 } = \frac{1}{\sqrt[4]{10}}$ Question:
$x^{\log(x)-1 } = \frac{1}{\sqrt[4]{10}}$
My attempts to solve:
$x^{\log(x)-1 } = \frac{1}{\sqrt[4]{10}}$
$x^{\log(x)-\log(10) } = 10^\frac{-1}{4}$
$x^{\log(\frac{x}{10}) } = 10^\frac{-1}{4}$
$-\frac{1}{4}\log{_x 10} = \log(\frac{x}{10})$
$-\frac{1}{4}\l... | There is a mistake later as well, but here's the first, from the fourth to fifth line:
$$
\boxed{
\log \left(\frac x{10}\right) \neq \frac{\log_x x}{\log_x {10}}}
$$
Because the left hand side is $\log_{10} x - 1 = \frac{1}{\log_{x}10} - 1$ while the RHS is $\frac 1{\log_x 10}$ , and these two are clearly not equal, in... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Finding $\int \sqrt{3-2x-x^2}dx$ $\int \sqrt{3-2x-x^2}\,dx$
First I did:
$$\begin{align}\int \sqrt{3-2x-x^2}\,dx &= \int \sqrt{-((x+1)^2-4)}\,dx \\
&=\int \sqrt{4-(x+1)^2}\,dx \\
\end{align}$$
Then I set $(x+1)=2\sin(t)$, $dx = 2\cos(t)\,dt$
$$\begin{align}
\int \sqrt{4-(2\sin(t))^2}2\cos(t)\,dt &= \int\sqrt{4-4\sin^2... | Hint:
$$\sin\left(2\arcsin\left(\dfrac u2\right)\right) = u\sqrt{1 - \dfrac{u^2}4}$$
In your case, $u = x + 1$. This may be helpful.
| {
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"answer_count": 2,
"answer_id": 0
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$\int \frac{dx}{a^2-x^2}=\frac{1}{2a}\ln\left|\frac{x+a}{x-a}\right|+C$
$\int\frac{dx}{a^2-x^2}=\frac{1}{2a}\ln\left|\frac{x+a}{x-a}\right|+C$
how to derive this formula?
$\int \frac{dx}{a^2-x^2}=\int\frac{dx}{(a-x)(a+x)}\boxed{=}\\\frac{1}{(a-x)(a+x)}=\frac{A}{a-x}+\frac{B}{a+x}\\1=A(x+a)+B(a-x)\\
x=a:1=A(a+a)\Right... | Beginning from your partial fractions:
\begin{align} \int{\frac{dx}{a^2-x^2}} &= \frac{1}{2a} \int{\frac{1}{a-x} + \frac{1}{a+x} \; dx} \\ &= \frac{1}{2a} \left(-\ln|a-x|+\ln|a+x|\right) + C \\ &= \frac{1}{2a} \; \ln\left|\frac{x+a}{x-a}\right| + C
\end{align}
| {
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"question_score": "1",
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How to prove $\frac{a}{7a+b}+\frac{b}{7b+c}+\frac{c}{7c+a}\le\frac38$ Suppose that $a,b,c>0$. How to prove $$\frac{a}{7a+b}+\frac{b}{7b+c}+\frac{c}{7c+a}\le\frac38$$
?
My first idea: By AM-GM, $$7a+b\geq \sqrt{7ab}$$ so $$\sum_{cyc} \frac{a}{7a+b}\le\sum_{cyc}\sqrt{\frac{a}{7b}}$$ but I am not sure if we can continue ... | Can be much detail, please
$$\frac{3}{7}-\frac{1}{7}\sum_{cyc}\frac{b^2}{7ab+b^2}\leq\frac{3}{7}-\frac{1}{7}\cdot\frac{(a+b+c)^2}{\sum\limits_{cyc}(7ab+b^2)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Find $\lim_{x\to0}\frac{1-\cos x\cos2x\cos3x}{x^2}$
Find $$\lim_{x\to0}\dfrac{1-\cos x\cos2x\cos3x}{x^2}$$
My attempt is as follows:-
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(2\cos x\cos2x\right)\cos3x}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(\cos3x+\cos x\right)\cos3x}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}... | Consider that
$$
1-\cos x\cos2x=1-\cos x+\cos x-\cos x\cos2x=(1-\cos x)+\cos x(1-\cos 2x)
$$
Going a further step on,
$$
1-\cos x\cos2x\cos3x=(1-\cos x)+\cos x(1-\cos2x)+\cos2x(1-\cos3x)
$$
and, by induction,
$$
1-\prod_{k=1}^n\cos(kx)=\sum_{k=1}^{n}\cos((k-1)x)(1-\cos(kx))
$$
Now
$$
\lim_{x\to0}\frac{1-\cos(kx)}{x^2}=... | {
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"question_score": "4",
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Finding solutions if existed in this number theory - prime numbers problem. I was working on a problem involving prime number and I reached a point where I could not solve an equation, and the equation is.
Given that $a,b$ and $c$ are positive integers which satisfies that both $a$ and $b $ are less than $c$.
And satis... | Changing your letters to become $$ y^2 - z x = -1. $$
The automorphisms (with positive determinant) of the quadratic form $y^2 - zx$ are known, and parametrized by the modular group. So, given $ps-qr = 1,$ so
$$
\det
\left(
\begin{array}{cc}
p & q \\
r & s \\
\end{array}
\right) = 1
$$
we construct
$$
W =
\left(
\begi... | {
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If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$, find the value of $p^3+q^3+r^3$. If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$, find the value of $p^3+q^3+r^3$.
Here's what I have got,
By Vieta's rule
$p+q+r=1\text{. ...........}(1)$
$pq+qr+pr=1\text{. ...........}(2)$
$p... | You can use the strict $p^3+q^3+r^3-3pqr=(p+q+r)(p^2+q^2+r^2-pq-qr-rp)$. It's easy to compute $p^2+q^2+r^2$ and use Viete's rule.
| {
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"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Finding $n$ if $|\sum_{r=0}^{3n-1}\beta^{2^r}|=4\sqrt{2}$ where $\beta=\exp(i2\pi/7)$
Find $n$ if $\left|\displaystyle\sum_{r=0}^{3n-1}\beta^{2^r}\right|=4\sqrt{2}$ where $\beta=\exp(i2\pi/7)$.
My Attempt
$$\begin{aligned}\left|\sum_{r=0}^{3n-1}\beta^{2^r}\right|&=\left|\sum_{r=0}^{n}(\beta+\beta^2+\beta^4)\right|\\... | Let $z=\beta+\beta^2+\beta^4$, then $\bar{z}=\beta^3+\beta^5+\beta^6$.
Notice that $\beta^7=1$ and $\beta+\beta^2+\beta^3+\beta^4+\beta^5+\beta^6=-1$.
So $\left|z\right|^2=z\bar{z}=(\beta+\beta^2+\beta^4)(\beta^3+\beta^5+\beta^6)$
$=\beta^4+\beta^6+\beta^7+\beta^5+\beta^7+\beta^8+\beta^7+\beta^9+\beta^{10}$
$=3+\beta+\... | {
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"url": "https://math.stackexchange.com/questions/3489309",
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"answer_id": 0
} |
Find the remainder when $f(x) = x^{2016}+2x^{2015}-3x+4$ is divided by $g(x)=x^2+3x+2$ Find the remainder when $f(x) = x^{2016}+2x^{2015}-3x+4$ is divided by $g(x)=x^2+3x+2$
I try to factor them, but I failed. I know that it's impossible to divide it
| First note $x^2+3x+2=(x+1)(x+2)$.
Modulo $x+2$, $x\equiv-2$, so $f(x) = x^{2016}+2x^{2015}-3x+4\equiv2^{2016}-2^{2016}+6+4=\color{blue}{10}.$
Modulo $x+1$, $x\equiv-1$, so $f(x) = x^{2016}+2x^{2015}-3x+4\equiv1-2+3+4=\color{blue}{6}.$
$\color{red}1(x+2)+\color{red}{-1}(x+1)=1$,
so by the Chinese remainder theorem $f(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
I am stuck on problem involving absolute values and inequalities The problem states to solve for $x$, and then to write the answer with absolute value notation.
The problem is: $(x-2)/(x-4) > (x+2)/(x)$
The correct answer is abs$(x-2) > 2$
(abs stands for absolute value).
Can someone explain to me how to arrive at th... | First, the given inequality is invalid when $x = 0$ or $x = 4$, since it contains undefined expressions in either of these cases.
Suppose $x \neq 0$ and $x \neq 4$. Then we may multiply both sides of the inequality by $x(x-4)$ to clear denominators. We should be careful: for some values of $x$, in particular $0 < x <... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\sum_{k=0}^{\lfloor (n-1)/2 \rfloor} (-1)^k {n+1 \choose k} {2n-2k-1 \choose n} =\frac{ n(n+1)}2 $ $$\sum_{k=0}^{\Big\lfloor \frac{(n-1)}{2} \Big\rfloor} (-1)^k {n+1 \choose k} {2n-2k-1 \choose n} = \frac{n(n+1)}{2} $$
So I feel like $(-1)^k$ is almost designed for the inclusion-exclusion principle. And the... | We seek to show that
$$\sum_{k=0}^{\lfloor (n-1)/2 \rfloor}
(-1)^k {n+1\choose k} {2n-2k-1\choose n}
= \frac{1}{2} n (n+1).$$
The LHS is
$$\sum_{k=0}^{\lfloor (n-1)/2 \rfloor}
(-1)^k {n+1\choose k} {2n-2k-1\choose n-1-2k}
\\ = [z^{n-1}] (1+z)^{2n-1}
\sum_{k=0}^{\lfloor (n-1)/2 \rfloor}
(-1)^k {n+1\choose k}
z^{2k} (1+z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3497919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
If $z = \frac{(\sqrt{3} + i)^n}{(\sqrt{3}-i)^m}$, find the relation between $m$ and $n$ such that $z$ is a real number. I am given the following number $z$:
$$z = \dfrac{(\sqrt{3} + i)^n}{(\sqrt{3} - i)^m}$$
with $n, m \in \mathbb{N}$. I have to find a relation between the natural numbers $n$ and $m$ such that the numb... | $\sqrt {3} - i$ is the conjugate to $\sqrt {3} + i$
$\frac {(\sqrt 3 + i)^n}{(\sqrt 3 - i)^m} = \frac {(\sqrt 3 + i)^{n+m}}{(\sqrt 3 - i)^m(\sqrt 3 + i)^m}$
This will rationalize the denominator
$\frac {(\sqrt 3 + i)^n}{(\sqrt 3 - i)^m} = \frac {(\sqrt 3 + i)^{n+m}}{4^{m}}$
We must find where $(\sqrt 3 + i)^{n+m}$ is r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3498784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
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Calculate the limit of function I was trying to find it for some time, but couldn't, so please help me.
$ \lim_{x\to\infty} ( \sqrt[100]{(x + 3*1)(x + 3*2)...(x +3*100)} - x)$
| \begin{gather*}
\lim _{x\leadsto \infty }(\sqrt[100]{}( x+3\times 1)( x+3\times 2) ...( x+3\times 100) \ -\ x\\
Taking\ x\ common\ \ from\ all\ brackets\ ,\ \\
\lim _{x\leadsto \infty }( x\sqrt[100]{}( 1+3\times 1/x)( 1+3\times 2/x) ...( 1+3\times 100/x) \ -\ x\\
Apply\ the\ binomial\ series\ expansion\ because\ \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3499311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\lim_{x \to 1}\frac{\sin(x^2-x)}{x^2-1}$ and $\lim_{x \to 1}\frac{\ln(x^2+x-1)}{\ln(x^3+x-1)}$ I found these limits and I was unable to solve them due to the occurring indeterminations
$$\lim_{x \to 1}\frac{\sin(x^2-x)}{x^2-1}$$
$$\lim_{x \to 1}\frac{\ln(x^2+x-1)}{\ln(x^3+x-1)}$$
Can someone help me, please... | Since $\lim_{t \to 0}\frac{\log_{a}\left(t+1\right)}{t}=\log_{a}\left(e\right)$,
we have:
$$\lim_{x \to 1}\frac{\ln(x^2+x-1)}{\ln(x^3+x-1)}$$$$=\lim_{x \to 1}\frac{\ln\left(1+\left(\color{blue}{x^{2}+x-2}\right)\right)}{\color{blue}{x^{2}+x-2}}\cdot\frac{x^{2}+x-2}{x^{3}+x-2}\cdot\frac{\color{red}{x^{3}+x-2}}{\ln\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
A Diophantine equation: solve $(3x^2+y^2-4y-17)^3-(2x^2+2y^2-4y-6)^3=(x^2-y^2-11)^3$ (without using Fermat's last theorem) Solve this Diophantine equation: $(3x^2+y^2-4y-17)^3-(2x^2+2y^2-4y-6)^3=(x^2-y^2-11)^3$
My attempt (use Fermat's last theorem)
$$(3x^2+y^2-4y-17)^3-(2x^2+2y^2-4y-6)^3=(x^2-y^2-11)^3$$
$$\Leftrighta... | Let
$$3x^2 + y^2 - 4y -17 = a\text{ ,}$$
$$-(2x^2 +2y^2 +4y -6)= b$$
and observe
$$x^2-y^2 -11= a+b$$
Your problem may be reexpressed as $$a^3 + b^3=(a+b)^3$$ implying
$$3ab(a+b)=0.$$ This gives your proposed solution set, plus the case where $a=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3502327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How many solutions does the expression $3^x(3^x+1)+2=|3^x-1|+|3^x+2|$ have? I can solve this by opening every single modulus, doing alternate combinations of $\pm$ and checking whether every $3^x$ is $>0$ or not. But that consumes a lot of time. Is there is a shorter way to solve such questions.
The answer is 1
| Simplify the equation by substituting $a=3^x$ and using the fact that $a>0$.
$a(a+1)+2=|a-1|+a+2$ since $a+2>0$
There are only two cases to be checked:
Case 1: $a^2 + a+2=a-1+a+2$
$\Longleftrightarrow a^2 -a+1=0$
Here $D=b^2-4ac=1-4=-3<0$
Therefore no real solutions.
Case 2: $a^2 + a+2=-a+1+a+2$
$\Longleftrightarrow a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3503998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Calculating limit approaches inifnity having infinite terms On my sample calculus mid-term exam, there is a weird question that asks us to calculate the limit that has an infinite term:
$\displaystyle\lim_{n \to \infty} {\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}}$
I am not sure what kind of technique should be used... | Note that
$$\displaystyle\lim_{n \to \infty} {\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}}=\displaystyle\lim_{n \to \infty} \frac{1}{n}{\sum_{k=1}^n\frac{1}{1+\frac{k}{n}}}$$
This is a Riemann Sum for $f(x)=\frac{1}{1+x}$ over the interval $[0,1]$, with the satndard partition $x_i=\frac{k}{n}$ for $1 \leq k \leq n$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3504583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Can we deduce the characteristic polynomial for this matrix? Given a square $n \times n$ matrix $A$ that satisfies $$\sum\limits_{k=0}^n a_k A^k = 0$$ for some coefficients $a_0, a_1, \dots, a_n,$ can we deduce that its characteristic polynomial is $\sum\limits_{k=0}^n a_k x^k$?
| The answer is no.
You are given a degree $n$ polynomial $p(x)$ that an $n\times n$ matrix $A$ satisfies. This is not enough information to find the characteristic polynomial $c_A(x)$, although you will be able to narrow it down to finitely many possibilities.
Let's look at an example to see why. Suppose your friend pic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3506928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 3
} |
Solving $3|2-x| + |2x-4| = 8$ I'm not sure how to solve $3|2-x| + |2x-4| = 8$. I don't exactly need an answer, the method is what I'm interested in. I've done equations with one absolute value element. >_<
| Well, it THIS case, $|2x-4| = 2\cdot |x-2|= 2\cdot |2-x|$ so $3|2-x| + |2x-4| = 3|2-x|+2|2-x| = 5|2-x| =8$ so
$|2-x| =|x-2|= \frac 85$ and so $x-2 =\pm \frac 85$ and $x = \frac 85 + 2=\frac {18}5=3\frac 35$ or $x = 2-\frac 85 = \frac 25$.
But in general do cases.
Do cases:
$2-x \ge 0$ if $x \le 2$ and $2-x < 0$ if $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find summation definition of a certain function Consider a tuple $T$ to be an object (list of numbers) similar to a row vector of $n$ elements which are real numbers: $$T=(x_{1}, x_{2},\ldots, x_{n}); n\in \Bbb N, x_{i}\in \Bbb R$$
A function $F$ can be defined from the tuples $T$ to $\Bbb R$ which takes the alternatin... | We build multiplication tables of $x_j\ast x_k$ for small $n$ and check for a pattern of the signs. Since the pattern differs slightly for $n$ even and $n$ odd we consider $n=6$ and $n=7$.
Case $n=6$:
We have in this case
\begin{align*}
&x_1x_2-x_1x_3+x_1x_4-x_1x_5+x_1x_6\\
&\qquad-x_2x_3+x_2x_4+x_2x_5-x_2x_6\\
&\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3512161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve inverse trigonometric equation $\frac{\pi}{6}=\tan^{-1} \frac{11}{x} -\tan^{-1} \frac{1}{x}$ How do I go about solving for $x$ when I have:
$\frac{\pi}{6}=\tan^{-1} \left( \frac{11}{x} \right)-\tan^{-1}\left( \frac{1}{x} \right)$.
| Make a right triangle ABC with C=90, AC=x, BC=11. Try to represent $\sin B$ in two ways: the straightforward way:
$$
\sin B=\frac{x}{\sqrt{121+x^2}},
$$
and - let $D$ be on $BC$ with $CD=1$, apply law of sine to triangle ADB and get
$$
\frac{\sqrt{1+x^2}}{\sin B}=\frac{10}{\sin 30}=20,
$$
thus get
$$
\frac{x}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3513047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
A contestant participates in a game show where three important prizes are offered. A contestant participates in a game show where three important prizes are offered. His chances of winning the three prizes are $\frac{1}{6}$, $\frac{1}{3}$ and $\frac{1}{2}$, respectively.
*
*How many prizes can the contestant expect... | $X$ does not have a Poisson distribution because it only takes finitely many values. Observe that $X=X_1+X_2+X_3$ where $X_1\sim\mathrm{Ber}\left(\frac16\right)$, $X_2\sim\mathrm{Ber}\left(\frac13\right)$, and $X_3\sim\mathrm{Ber}\left(\frac12\right)$ are independent. Therefore
\begin{align}
\mathbb P(X=0) &= \mathbb P... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Flux through the Surface Find the flux through the surface $\iint_S F\cdot NdS$ where N is the normal vector to S.
i) $F=3z\hat i-4\hat j+y\hat k$
$~~~S:z=1-x-y$ (first octant)
ii) $F=x\hat i+y\hat j-2z\hat k$
$~~~~S:\sqrt{a^2-x^2-y^2}$
I have evaluated $N$ vector as : $-\hat i-\hat j\over\sqrt2$ for i) and -$\frac... | Take the example: $F=x\hat i+y\hat j-2z\hat k$; $~S: z=\sqrt{a^2-x^2-y^2}$. Rewrite the surface as $f(x,y,z)=x^2+y^2+z^2 = a^2$ and calculate its unit normal vector
$$N=\frac{(f_x, f_y, f_z)}{\sqrt{f_x^2+ f_y^2+ f_z^2}}=\frac1a(x,y,z)$$
Then,
$$F\cdot N = (x,y,-2z)\cdot \frac1a(x,y,z)=\frac1a (x^2+y^2-2z^2)=a-\frac3a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3515937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Derivative of $\dfrac{\sqrt{3-x^2}}{3+x}$ I am trying to find the derivative of this function
$f(x)=\dfrac{\sqrt{3-x^2}}{3+x}$
$f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}}\frac{d}{dx}(3-x)(3+x)-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-2x(3+x)}{2\sqrt{3-x^2}}-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-x(3+x)}{\sqrt{3-... | your first line is not entirely clear
\begin{eqnarray*}
f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}} \color{red}{ \left( \frac{d}{dx}(3-x^{\color{blue}2}) \right)}(3+x)-\sqrt{3-x^2}}{(3+x)^2}
\end{eqnarray*}
But everything is right after this and you just need to do a common denominator at the end & you will get th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3525571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
} |
Minimum value of $\frac{ax^2+by^2}{\sqrt{a^2x^2+b^2y^2}}$ If $$x^2+y^2=1$$
Prove that Minimum value of $$f(x,y)=\frac{ax^2+by^2}{\sqrt{a^2x^2+b^2y^2}}$$ is
$$\frac{2\sqrt{ab}}{a+b}$$
My try:
I used basic Trigonometry:
Let $x=\cos t$ and $y=\sin t$
Then we get $$f(x,y)=g(t)=\frac{a\cos^2 t+b\sin^2 t}{\sqrt{a^2\cos^2 t+b... | Let us show
$$
\Big(\ a^2x^2+b^2y^2\ \Big)(x^2+y^2)\cdot\frac{4ab}{(a+b)^2}\le \Big(\ ax^2+by^2\ \Big)^2\ .
$$
It is useful to substitute $u=x^2$, $v=y^2$, so let us show:
$$
4ab\Big(\ a^2u+b^2v\ \Big)(u+v)\le \Big(\ au+bv\ \Big)^2\cdot (a+b)^2 \ .
$$
It turns out that after moving all terms to the R.H.S. we can factor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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How does this fraction $\frac{1}{y} = \frac{1}{a}$ become $y = a$? I was learning SET. To solve a problem for Y at some point I stopped right here $\frac{1}{y} = \frac{1}{a}$, and I couldn't go further. Then I opened the solution book and there I found that everything I did was right up-to this $\frac{1}{y} = \frac{1}... | There’s plenty of ways. You can take reciprocals on both sides, since $r=s$ implies $\frac{1}{r} = \frac{1}{s}$ when they are not zero. So you get
$$\frac{1}{\quad\frac{1}{y}\quad} = \frac{1}{\quad\frac{1}{a}\quad}$$
and this is the same as $y=a$.
Or you can multiply through by $ay$ and then cancel. You get
$$\begin{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3529082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Torsion group of $y^2 = x^3 \pm nx$ Let $n\in \mathbb{Z}$ be not divisible by $15$ and suppose $n$ and $-n$ are both not perfect squares. Prove that at least one of the elliptic curves $Y^2 = X^3 + nX$ and $Y^2 = X^3 - nX$ has torsion group given by $\{{\bf o},(0,0)\}$ (where ${\bf o}$ is the point at infinity).
Here i... | Since $n$ and $-n$ are not perfect squares, the point $(0, 0)$ is the only point of order $2$ on both curves.
Now if $P = (x, y)$ is a rational point of order $4$ on the curve $Y^2 = X^3 - nX$, then by the doubling formula, we have $$X(2P) = \frac{x^4 + 2nx^2 + n^2}{4y^2}.$$ However, $2P$ must be the point $(0, 0)$, he... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3530422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove that $x + \frac{2x^3}{3} + \cdots + \frac{2\cdot 4 \cdot \cdots 2nx^{2n+1}}{3\cdot 5 \cdot (2n+1)}+\cdots = \frac{\arcsin(x)}{\sqrt{1-x^2}}$
Prove that $x + \dfrac{2x^3}{3} + \cdots + \dfrac{2\cdot 4 \cdot \cdots 2nx^{2n+1}}{3\cdot 5 \cdot (2n+1)}+\cdots = \dfrac{\arcsin(x)}{\sqrt{1-x^2}}.$
Let $f(x) = x + \dfr... | Like this
$$ \sum_{n=0}^{\infty} \frac{x^{2n+1}(2n)!!}{(2n+1)!!}=\sum_{n=0}^{\infty} \int_{0}^{\frac{\pi}{2}}x^{2n+1}\sin^{2n+1} t dt=\int_{0}^{\frac{\pi}{2}} \frac{x \sin t}{1-x^{2}\sin^{2} t} dt=-\int_{0}^{\frac{\pi}{2}}\frac{1}{1-x^{2}+x^{2}\cos^{2} t} d(x\cos t)=\frac{1}{\sqrt{1-x^{2}}} \arctan{\frac{x}{\sqrt{1-x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3534871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Find whether $\sum_{n=1}^\infty \frac{2^{3n-3}3^{n+2}}{5^{2n-2}} $ is geometric and convergent and if so, state the value of the sum $$\sum_{n=1}^\infty \frac{2^{3n-3}3^{n+2}}{5^{2n-2}} $$
First I tried to find the ratio by doing $\frac{a_{n+1}}{a_n}$. Many calculations were made, and the result was $24/25$, so it is ... | HINT
You have
$$
\frac{2^{3n-3}3^{n+2}}{5^{2n-2}}
= \frac{2^{-3} 3^2}{5^{-2}} \frac{2^{3n} 3^n}{5^{2n}}
= \frac{25 \cdot 9}{8} \frac{8^n 3^n}{25^n}
= \frac{25 \cdot 9}{8} \left(\frac{24}{25}\right)^n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Calculate $\int_{0}^{3}\sqrt{4-x^2}\,dx$ using a Riemann sum What is the Riemann intergral of $\sqrt{4-x^2}$ from $0$ to $3$?
I tried to write down the Riemann sum $\sum_{i=1}^n\sqrt{4-\frac{(3i)^2}{n^2}}\frac{3}{n}$. Then I tried to take limit as $n \to \infty$, but I don't know how to evaluate it. I don't know what t... | Building on (Unfortunately, it does not fit in the comment section) from Integrand and Dark Malthrop). Hence posting as a full version.
We can start with the series expansion,
\begin{eqnarray*}
\sqrt{a^{2}-x^{2}} &=& \sum_{m=0}^{\infty}{(-1)^{m} a^{1-2m} \binom{\frac{1}{2}}{m} x^{2 m}}
\end{eqnarray*}
\begin{eqnarray... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Limit of $\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$ $$\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$$
I tried to used $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ but it did not worked out so I tried to use the squeeze theorem.
$$0=\sqrt[3]{x^3}-\sqrt{x^2}\leq \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}\leq\sqrt[3]{8x^3}-\sqrt{4x^2}=... | $\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}
=\lim_{x\to \infty} \frac{(\sqrt[3]{x^3+2x}^6-\sqrt{x^2-2x}^6)}{(\sqrt[3]{x^3+2x}^5+\sqrt[3]{x^3+2x}^4\sqrt{x^2-2x}+\sqrt[3]{x^3+2x}^3\sqrt{x^2-2x}^2+\sqrt[3]{x^3+2x}^2\sqrt{x^2-2x}^3+\sqrt[3]{x^3+2x}\sqrt{x^2-2x}^4+\sqrt{x^2-2x}^5)}=
=\lim_{x\to \infty} \frac{((x^3+2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
$a_n=\sqrt{2-a_{n-1}},a_1=\sqrt{2}$ Given
$$a_n=\sqrt{2-a_{n-1}},a_1=\sqrt{2}$$
I calculated $a_1$ to $a_5$
$$\sqrt{2},
\sqrt{2-\sqrt{2}},
\sqrt{2-\sqrt{2-\sqrt{2}}}, \\
\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}, \\
\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}}$$
which made me think of $\sin/\cos$. So I divided each by $2$, c... | If the question is about finding the limit, let's consider $a_{n+1}=f(a_n)$, where $f(x)=\sqrt{2-x}$. Then we have
If $0\leq x \leq \sqrt{2}$ then $0\leq f(x)\leq\sqrt{2}$
Indeed $0\leq x\leq\sqrt{2}
\Rightarrow 0\geq-x \geq -\sqrt{2}
\Rightarrow 2\geq 2-x \geq 2-\sqrt{2}>0
\Rightarrow \sqrt{2}\geq \sqrt{2-x}=f(x)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3538786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is $x^3 + x + 1$ for $x=\frac{\sqrt 5 - 1}{2}$
What is $x^3 + x + 1$ for $x=\frac{\sqrt 5 - 1}{2}$
Of course we can subtitute and expand it
${(\frac{\sqrt 5 -1}{2})}^3 + \frac{\sqrt 5 -1}{2} + 1$
Is there a better way to do it?
| Yes. $\dfrac{\sqrt{5}-1}{2}$ is a root of $x^2+x-1=0$. So $x^2=1-x$.
Multiply by $x$ to get
$$x^3=x-x^2=x-1+x=2x-1$$
Therefore
$$x^3+x+1=3x=\frac{3}{2}(\sqrt{5}-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Calculating Inverse modulo 101 A lot of problems are calculating $n\mod m$ with $n<m$. But I have the following problem: Find the inverse of $71 \mod 101$. Now using Euclid, I get the $\gcd(71,101)=1$ and
$$
1=3-(1\cdot(8-2\cdot(11-1\cdot(30-2\cdot(71-2\cdot(101-71)))))
$$
I know the solution is $37$ but I am not able ... | $1=3-1\times2=3-1\times(8-2\times3)=3\times3-1\times8=3\times(11-1\times8)-1\times8$
$=3\times11-4\times8=3\times11-4\times(30-2\times11)=11\times11-4\times30$
$=11\times(71-2\times30)-4\times30=11\times71-26\times30=11\times71-26\times(101-71)$
$=\color{red}{37}\times71-26\times101$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3541509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating Diverging Integral? I'm looking at:
$$\int_{-\infty}^\infty \sin(2cx)\frac{1-2x^2-\cos(2x)}{2x^3}dx$$
How come this converges for $c \in \mathbb{R}$? And how does mathematica compute it as:
$$\frac{\pi \left(-2 \left(c^2+1\right) \left| c\right| +(c-1) c \left| c-1\right| +c (c+1)
\left| c+1\right| \righ... | First, the convergence analysis:
*
*Observe that for $x$ small we have $1 - 2x^2 - \cos(2x) = O(x^4)$, and so the integral is not singular near the origin.
*When $|x|$ is large, the integral $\displaystyle \int\sin(2cx) \frac{1 - \cos(2x)}{2x^3} dx$ is absolutely integrable.
*The remaining term is $\displaystyle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3542216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to calculate $\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}$ $$\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}$$
someone please help i’m not sure how to compute this. i’ve tried to do it this way:$$\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{... | $$\begin{align}
\frac13 \frac{7^{2x} + 7^{-2x}}{7^{2x} - 7^{-2x}}
&= \frac13 \frac{7^{2x}(7^{2x} + 7^{-2x})}{7^{2x}(7^{2x} - 7^{-2x})} \\
&= \frac13 \frac{7^{4x} + 1}{7^{4x} - 1} \\
&= \frac13 \frac{(7^{4x} -1) + 2}{7^{4x} - 1} \\
&= \frac13 \Big(1 + \frac{2}{7^{4x} - 1}\Big) \\
&= \frac13 + \frac{2}{3(7^{4x} - 1)}
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3544652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\lim_{n\to\infty} \sin^2 (\pi\sqrt{n^{2014} + n^{2012} + 1})$ I tried to use $\sin^2(\pi x) = \sin^2 (\pi x - \pi)$
So,
$$\lim_{n\to\infty}\sin^2 (\pi\sqrt{n^{2014} + n^{2012} + 1}) \\
=\lim_{n\to\infty}\sin^2 (\pi(\sqrt{n^{2014} + n^{2012} + 1} - 1)) \\
\lim_{n\to\infty}\sin^2 (\frac{\pi (n^{2014} + n^{2012}... | NOTE: this is not a complete answer but a small collection of some partial results.
First remark
$$
\begin{gathered}
\pi \sqrt {n^2 + n + 1} = \pi n\left( {1 + \frac{{\text{1}}}
{n} + \frac{1}
{{n^2 }}} \right)^{1/2} = \hfill \\
\hfill \\
= \pi n\left( {1 + \frac{1}
{{2n}} + o\left( {\frac{1}
{n}} \right)} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3545447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Whats maximum and minimum of $4x + y^2$ without Lagrange
Whats maximum and minimum of $4x + y^2$ without Lagrange
constraints = $2x^2 + y^2 =4$
Substitute $y^2= 4-2x^2$ into $4x+y^2$ we get $4x+4-2x^2$ manipulate it into $-2(x-1)^2 + 6$ the maximum is 6. Or using differentiation we get $x=1$ and only produces maximum... | $f(x)=4x+4-2x^2=$
$-2(x-1)^2+6$;
1)$2x^2+y^2=4$;
An ellipse: $x^2/(√2)^2+y^2/2^2=1$.
Semi minor axis : $√2$ ;
$y^2=0: x_{1,2}=\pm √2$;
Minimum:
$x=-√2$: $f_{min}(-√2)=-2(-√2-1)^2 +6=-2(1+√2)^2 +6$.
See also comments.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3546579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A question involving inequality $x=cy+bz\ ,\ y=az+cx$ and $z=bx+ay$ has a non zero solution and atleast one of a,b,c is a proper fraction(a rational number whose absolute value is less than 1) , prove that $a^3+b^3+c^3<3$ and $abc>-1$.
For non zero solution implies , the determinant of following marix be zero because ... | Let $|c|<1.$
Thus, $$1=a^2+b^2+c^2+2abc=(a+bc)^2+b^2+c^2-b^2c^2\geq b^2+c^2-b^2c^2,$$
which gives
$$(1-b^2)(1-c^2)\geq0$$ and $$b^2\leq1.$$
Similarly, $$a^2\leq1.$$
Thus, $$-1<abc<1$$ and $$a^3+b^3+c^3<3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3547569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\frac{a}{a+2b+c}+\frac{b}{b+2c+a}+\frac{c}{c+2a+b}\geq\frac{3}{4}$ I found the following exercise:
Prove that
$$\frac{a}{a+2b+c}+\frac{b}{b+2c+a}+\frac{c}{c+2a+b}\geq\frac{3}{4}$$
for any positive $a$, $b$, $c$.
I tried substituting the denominators but it led me nowhere. I don really have other ideas.
| Using Cauchy-Schwarz:
$$ \left(\sum_{cyc}\frac{a^2}{a^2+2ab+ca}\right)\left[\sum_{cyc}(a^2+2ab+ca)\right] \geq \left(\sum_{cyc}\frac{a}{\sqrt{a^2+2ab+ca}}\cdot \sqrt{a^2+2ab+ca}\right)^2$$
$$=(a+b+c)^2$$
Since $\displaystyle\sum_{cyc}(a^2+2ab+ca) =a^2+b^2+c^2+3(ab+bc+ca)$, we get:
$$\sum_{cyc}\frac{a^2}{a^2+2ab+ca}\geq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3548517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\int_a^b \frac{1}{x}\sqrt{-(x-a)(x-b)}dx = (\frac{a+b}{2}-\sqrt{ab})\pi$ Does anyone know how to solve $$\int_a^b \frac{1}{x}\sqrt{-(x-a)(x-b)}dx$$
After trying Wolfram Alpha, I conclude that it possibly equals $(\frac{a+b}{2}-\sqrt{ab})\pi = \frac{(\sqrt{a}-\sqrt{b})^2}{2}\pi$.
The book I am reading says one ca... | Noticing that $(x-a)(b-x)=\left(\frac{b-a}{2}\right)^2-\left(x-\frac{a+b}{2}\right)^2$, we let $x-\frac{a+b}{2}=\frac{b-a}{2} \sin \theta$ and transforms the integral into $$
\begin{aligned}I&=\left(\frac{b-a}{2}\right)^2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos ^2 \theta}{\frac{b-a}{2} \sin \theta+\frac{a+b}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3549527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
} |
How to prove that $\tan \left( \frac{\pi}{2} - \theta \right) = \cot \theta$ I am asked to simplify the following expression:
$$\tan \left( \frac{\pi}{2} - \theta \right)$$
The book gives me the answer $\cot \theta$ but, when I try to derive that formula using
$$\tan \left( A \pm B \right) = \frac{\tan A \pm \tan B}{1... | $\tan\left(\frac{\pi}{2}-\theta\right)=\frac{\sin\left(\frac{\pi}{2}-\theta\right)}{\cos\left(\frac{\pi}{2}-\theta\right)}=\frac{\sin\frac{\pi}{2} \cos\theta-\cos\frac{\pi}{2} \sin\theta}{\\cos\frac{\pi}{2}\cos\theta+\sin\frac{\pi}{2}\sin\theta}=\frac{\cos\theta}{\sin\theta}=\cot\theta.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3550289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Find $\left\lfloor \sum_{k=1}^n\sqrt[2k+1]{\frac{2k+1}{2k-1}}\right\rfloor$
If $n$ is a positive integer, find:
$$\left\lfloor \sum_{k=1}^n\sqrt[2k+1]{\frac{2k+1}{2k-1}}\right\rfloor$$
I computed a few values for small $n$ and it seems it is always $n$, so I think I should prove:
$$n\le \sum_{k=1}^n\sqrt[2k+1]{\frac{... | We can also use AM-GM:
$$\sqrt[2k+1]{\frac{2k+1}{2k-1}}=\sqrt[2k+1]{1+\frac{2}{2k-1}} < \frac{1+\frac{2}{2k-1}+1+\ldots+1}{2k+1}$$
$$=\frac{2k+1+\frac{2}{2k-1}}{2k+1}=1+\frac{2}{(2k-1)(2k+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3553159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Is this algebra allowed Given
$$\frac{\frac{a}{a+b}}{\frac{c}{c+d}} = 1.8 \text{ (1)}$$
and
$$\frac{a}{a+b}-\frac{c}{c+d} = \frac{2}{15} \text{ (2)}$$
and
$a = 30$
and
$a+b+c+d=220 \text{ (3)}$
Solve the remaining
This is what I did
According to $(1)$ $$\frac{a}{a+b} = 1.8 \frac{c}{c+d}$$
$$\Leftrightarrow \frac{5... | Your answer is perfectly fine. Giving fractional algebraic values to another variable fraction is as good as substituting non-fractional variables such as $p=q$. You can always do this unless the denominator equals $0$.
| {
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"source": "stackexchange",
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How to simplify the trigonometric functions sine and cosine in a derivative? I am learning calculus and came across the following problem:
Find $\frac{dx}{d\theta}$ when $x=\sin^2\theta \cos^3\theta$
I solved this using the product rule.
$$u= \sin^2\theta \qquad v=\cos^3\theta$$
$$u'= \sin2\theta \qquad v'=-3\cos^2\t... | With
$u = \sin^2\theta, \; v = \cos^3 \theta, \tag 1$
the chain rule yields
$u'(\theta) = \dfrac{d\sin^2 \theta}{d\theta} = \dfrac{d\sin^2 \theta}{d\sin \theta}\dfrac{d\sin \theta}{d\theta} = 2(\sin \theta)(\cos\theta), \tag 2$
and
$v'(\theta) = \dfrac{d\cos^3\theta}{d\cos \theta} \dfrac{d\cos \theta}{d\theta} = 3(\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3556475",
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"source": "stackexchange",
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Find all the relative extrema of the function $f(x)=2x-24x^{1/3}$ Find all the relative extrema of the function $f(x)=2x-24x^{1/3}$
Solution:
Step 1: Find the values of $x$ where $f'(x)=0$ and $f'(x)$ DNE.
$f'(x)=2-8x^{\frac{-2}{3}}=2-\frac{8}{x^{\frac{2}{3}}}$
We can see that $f'(x)$ DNE when $x=0$. Let's figure out ... | Let $t=x^{1/3}$ so $x=t^3$.
Then we want extrema of $2t^3-24t$.
These occur when $6t^2-24=0$; i.e., $t=\pm2$.
Then $x=\pm8$, and $f(x)=\mp32$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3558930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Finding series from its sum, then finding its reciprocal's sum If $$\sum_{r=1}^nt_r=\frac{n(n+1)(n+2)}{12}$$, then value of $$\sum_{r=1}^n\frac{1}{t_r}$$ is
Now, how can we find the series (i.e. formulae of $n^{th}$ term) from the sum?
The answer is $\frac{4n}{n+1}$
| Observe that
$$
t_n=\frac{n(n+1)(n+2)}{12}-\frac{(n-1)n(n+1)}{12}=\frac{n(n+1)}{4}
$$
and
$$
\frac{1}{t_n}=\frac{4}{n(n+1)}=\frac{4}{n}-\frac{4}{n+1},
$$
and hence
$$
\sum_{k=1}^n\frac{1}{t_k}=\left(\frac{4}{1}-\frac{4}{2}\right)+\left(\frac{4}{2}-\frac{4}{3}\right)+\cdots+
\left(\frac{4}{n-1}-\frac{4}{n}\right)+\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3559514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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proving the inequality $n \cdot a^{n-1} \cdot(b-a) < b^n - a^n < n \cdot b^{n-1} \cdot (b-a)$ let $b>a>0$ and $n>1$ I need to prove that:
$n \cdot a^{n-1} \cdot(b-a) < b^n - a^n < n \cdot b^{n-1} \cdot (b-a)$
I though of proving it using induction, or to build a function on where n is the variable but failed both times... | There are two ways to prove this.
Method 1:
Using the fact that $\dfrac{b^n-a^n}{b-a}=b^{n-1}+a^{n-2}b+\cdots+a^{n-1}$, then
$$b^{n-1}+a^{n-2}b+\cdots+a^{n-1}>a^{n-1}+a^{n-1}+\cdots+a^{n-1}=na^{n-1}\\b^{n-1}+a^{n-2}b+\cdots+a^{n-1}<b^{n-1}+b^{n-1}+\cdots+b^{n-1}=nb^{n-1} \\na^{n-1}<\dfrac{b^n-a^n}{b-a}<nb^{n-1} \\ na^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3559796",
"timestamp": "2023-03-29T00:00:00",
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Find convolution of $f=x+2$ and $g=\frac{\sin\frac{7x}{2}}{\sin\frac{x}{2}}$.
I have to find the convolution of the functions $f$ and $g$ on the interval $(-\pi, \pi)$.
$$
f(x) = x+2;\ \ \ g(x) =\frac{\sin\frac{7x}{2}}{\sin\frac{x}{2}}
$$
Here is my attempt:
$$
\begin{aligned}
&(f*g)(x)=\int\limits_{-\pi}^{\pi}f(t)... | You can first evaluate Fourier series for $f(x)$ and then use its relation to the Dirichlet kernel:$$S_n(x)=\int_{-\pi}^{\pi}f(x)D_n(x-t)dt$$ where $D_n(t)=\frac{1}{2\pi}\frac{\sin((n+\frac{1}{2})t)}{\sin(\frac{t}{2})}$ and $S_n(x)$ is n-th partial sum of Fourier series.
In your case, $g(t)=\frac{\sin\frac{7t}{2}}{\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3560140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Compute $\lim_{n\to\infty}(1+a)(1+a^2)(1+a^4)...(1+a^{2^n})$? Compute
$$\lim_{n\to\infty}(1+a)(1+a^2)(1+a^4)...(1+a^{2^n})$$
Where $|a|<1$.
| $$(1-a)f=(1-a)(1+a)(1+a^2)(1+a^4)(1+a^8)...(1+a^{2^n})=(1-a^{2^{n+1}})$$
Use $(x-y)(x+y)=x^2-y^2$ sucsessively to get
$$f_n=\frac{1-a^{2^{n
+1}}}{1-a}.$$
As $|a|<1$
$$\lim_{n \rightarrow \infty} f_n= \frac{1}{1-a}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3560850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find where $f(x)=\frac{x^2+4}{x^2-4}$ is concave upwards and concave downwards Find where $f(x)=\frac{x^2+4}{x^2-4}$ is concave upwards and concave downwards
Solution: We are going to take the second derivative, find the critical points, and then test each region. Lets get to the second derivative by doing the quotien... | It's good to keep in mind that the critical points of $f''(x)$ are places where either $f''(x)=0$ or $f''(x)$ is undefined. Hence concavity can change at places where $f''(x)=0$ and at places where $f''(x)$ is undefined.
We were given that
$$f(x)=\frac{x^2+4}{x^2-4},$$
and you correctly calculated that
$$f''(x)=\frac{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3561246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Can a sum arithmetic square ever equal to power of two? Does there exist any $2^t,\ t\in\mathbb{Z}_+$ which can be express as Sum of arithmetic square number.
Or
Can it be shown that
$$\begin{split}2^t&\ne \sum_{q=0}^u (n+qd)^2=n^2+(n+d)^2+(n+2d)^2+\cdots+(n+ud)^2\end{split}$$
Where $n,u,d\in\mathbb{Z}_+$
Example: le... | Let's focus on the case $n=d=1$.
$$\sum_{q=0}^u (n+qd)^2=\sum_{q=0}^u (1+q)^2=(u+1)+u(u+1)+\frac{u(u+1)(2u+1)}6$$
$$=(u+1)\frac{6+6u+u(2u+1)}6$$
Therefore, we have two cases: $\,u=2^k-1\,$ or $\,u=3\cdot 2^k-1$, for a certain $\,k$.
First case ($\,u=2^k-1\,$):
$$\sum_{q=0}^u (1+q)^2=2^{k-1}\frac{6+6(2^k-1)+(2^k-1)(2(2^... | {
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"url": "https://math.stackexchange.com/questions/3564360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the MacLaurin series of $\frac{x+3}{2-x}$ I did:
$$\frac{x+3}{2-x} = \frac{x}{2-x}+\frac{3}{2-x} = x(\frac{1}{2-x})+3(\frac{1}{2-x}) = \\
= \frac{x}{2}(\frac{1}{1-\frac{x}{2}})+\frac{3}{2}(\frac{1}{1-\frac{x}{2}}) = \\
= \frac{x+3}{2}\sum(\frac{x}{2})^n = ...?$$
The answer my professor got was $\frac{(3 + x)}{(... | \begin{align}
\frac{x+3}2\sum_{n=0}^\infty\left(\frac{x}2\right)^n
&=\frac{x}2\sum_{n=0}^\infty\frac{x^n}{2^n}+\frac32\sum_{n=0}^\infty\frac{x^n}{2^n}\\
&=\sum_{n=0}^\infty\frac{x^{n+1}}{2^{n+1}}+\sum_{n=0}^\infty\frac{3x^n}{2^{n+1}}\\
&=\overbrace{\sum_{n=1}^\infty\frac{x^n}{2^n}}^{n+1\,\mapsto\, n}+\overbrace{\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3570936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Given positives $a, b, c$ such that $a + b + c = 3$, prove that $\sum_{cyc}\frac{1}{a^2 + 4b^2 + c^2} \le \frac{1}{2}$.
Given positives $a, b, c$ such that $a + b + c = 3$, prove that $$\frac{1}{c^2 + 4a^2 + b^2} + \frac{1}{a^2 + 4b^2 + c^2} + \frac{1}{b^2 + 4c^2 + a^2} \le \frac{1}{2}$$
We have that $$a^2 + 4b^2 + c... | Another way.
We'll prove that our inequality is true for any real $a$, $b$ and $c$ such that $a+b+c=3.$
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v^2$ can be negative, and $abc=w^3$.
Thus, by my first proof we need to prove that:
$$\sum_{sym}(a^6-4a^5b+13a^4b^2-2a^4bc-6a^3b^3-12a^3b^2c+10a^2b^2c^2)\geq0$$ or
$$27... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $x^4+12x-5$ has roots $x_1,x_2,x_3,x_4$ find polynomial with roots $x_1+x_2,x_1+x_3,x_1+x_4,x_2+x_3,x_2+x_4,x_3+x_4$ I have the polynomial $x^4+12x-5$ with the roots $x_1,x_2,x_3,x_4$ and I want to find the polynomial whose roots are $x_1+x_2,x_1+x_3,x_1+x_4,x_2+x_3,x_2+x_4,x_3+x_4$.
I found the roots $x_1=-1+\sqrt{... | Let $s_1,p_1$ the sum and product of any two roots and $s_2,p_2$ the sum and product of the other two roots. From Vieta's:
$$
\begin{cases}
s_1+s_2=0 \\
s_1s_2+p_1+p_2=0\\
p_1s_2+p_2s_1=-12\\
p_1p_2=5
\end{cases}
$$
Substitute $s_2=-s_1$
$$
\begin{cases}
p_1+p_2=s_1^2\\
-p_1+p_2=\frac{12}{s_1}\\
p_1p_2=5
\end{cases}
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3574530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Prove $\cos x +\cos y = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$ Prove that $$\cos(x) + \cos(y) = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$$ holds true for any $x, y \in \mathbb{R}$.
Even though I managed to prove its brother $\sin(x) + \sin(y)$, I haven't been able to tackle this one.
Important identities needed for th... | Add these equations:
$\cos x=\cos\dfrac{x+y}2\cos\dfrac{x-y}2-\sin\dfrac{x+y}2\sin\dfrac{x-y}2$
$\cos y=\cos\dfrac{x+y}2\cos\dfrac{x-y}2+\sin\dfrac{x+y}2\sin\dfrac{x-y}2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3576912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to find the first four terms of the maclaurin series of $f(x) = e^{e^x}$ I tried substituting the maclaurin series for $e^x$ into the equation, but that doesn’t seem to give the right answer.
I’m not sure what do besides just taking the derivatives.
| We know
$$
e^x = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots
$$
so then
$$\begin{align}
\exp(e^x) &= \exp\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots\right)
\\ &= e^1 e^x e^{x^2/2} e^{x^3/6}\dots
\\
&= e \left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots\right)
\left(1+\frac{x^2}{2}+\frac{x^4}{8}+\dots\right)
\left(1+\frac{x^3}{6}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3577668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Use epsilon-delta definition of limit to establish the following: $\displaystyle\lim_{x\to 1}\frac{1}{2+\sqrt{x}}=\frac{1}{3}$ I understand that my solution here is probably not the most efficient (My professor's solution is "cleaner") but it is how my mind attacked the problem. I have been losing lots of points for m... | I think the following salvages the initial strategy while still holding true to the mathematical principals of epsilon-delta definition of limits. Feedback welcome. Please let me know if I have made another error. I posted this as an answer so that the original question could remain for reference.
\begin{align*}
\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3578018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Confused about positive and negative signs: Find the value of $\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$. Without tables or a calculator, find the value of $\displaystyle\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$.
I do not understand how the positive/negative signs are obtained as shown in the book; is there... | It might be interesting that you may avoid tedious calculations with roots if you use recurrence relations:
*
*Set $t_1 =2+\sqrt 5$ and $t_2 = 2-\sqrt 5$.
So, the searched for value is
$$\frac{t_1^6-t_2^6}{8\sqrt{5}}$$
This is $a_6$ in the recurrence relation
$$a_{n+2} - (t_1+t_2)a_{n+1} + t_1t_2 = a_{n+2} - 4a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3578191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
$\left| \frac{1}{z}+ \sum_{n=1}^\infty \frac{2z}{z^2-n^2}\right| \le C + C \sum_{n=1}^\infty \frac{y}{y^2+n^2}$ for $y>1$ and $|x|\le 1/2$. Suppose $|x| \le \frac{1}{2}$, and $y>1$ for $z=x+iy$.
Then we have
$$\frac{1}{z} + \sum_{n=1}^\infty \frac{2z}{z^2-n^2} = \frac{1}{x+iy} + \sum_{n=1}^\infty \frac{2(x+iy)}{x^2-y... | The idea is that under these restrictions, $z^2 \approx -y^2$ lies near the negative real axis, so that $|z^2 -n^2| \approx y^2 + n^2$.
Now the precise estimate: You have $x^2 \le \frac 14 \le \frac{y^2}{4}$, so that
$$
|z^2-n^2| \ge |\operatorname{Re}(z^2-n^2)| = y^2+n^2-x^2 \ge \frac 34 (y^2+n^2)
$$
and
$$
|2z| = 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3578505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show $\sum_{k=0}^{n}2^{n-k}\binom{a+k}{k}\frac{a-k}{a+k}=\binom{a+n}{n}$ How it can be shown that:
$$\sum_{k=0}^{n}2^{n-k}\binom{a+k}{k}\frac{a-k}{a+k}=\binom{a+n}{n}$$
Where $a \ne 0$
My try:
$$\sum_{k=0}^{n}2^{n-k}\binom{a+k}{k}\frac{a-k}{a+k}=\sum_{k=0}^{n}2^{n-k}\binom{a+k}{a}\frac{a-k}{a+k}$$
$$=\frac{1}{a}\sum... | Here is an algebric proof :
Let $ a $ be a real, let's define the function $ f_{a} $ as follows : $$ \left(\forall x\in\left]-\frac{1}{2},\frac{1}{2}\right[\right),\ f_{a}\left(x\right)=\displaystyle\frac{1-2x}{\left(1-x\right)^{a+1}} $$
know : \begin{aligned} \left(\forall x\in\left]-\frac{1}{2},\frac{1}{2}\right[\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3579670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Interpreting an olympiad inequality as a convex function We need to prove the following inequality:
$$\frac{y^3xz}{x^3(xy+z^2)}+\frac{z^3xy}{y^3(x^2+yz)}+\frac{x^3yz}{z^3(xz+y^2)}\geq \frac{3}{2}$$
This equation is convex in each of the variables $x,y,z$. Moreover, its minimum seems to be at the origin (the equation ... | By C-S and AM-GM we obtain: $$\sum_{cyc}\frac{x^3yz}{z^3(y^2+xz)}=\sum_{cyc}\frac{x^4y^2}{z^2xy(y^2+xz)}\geq\frac{\left(\sum\limits_{cyc}x^2y\right)^2}{\sum\limits_{cyc}z^2xy(y^2+xz)}=$$
$$=\frac{\left(\sum\limits_{cyc}x^2y\right)^2}{xyz\sum\limits_{cyc}(y^2z+z^2x)}=\frac{\left(\sum\limits_{cyc}x^2y\right)^2}{2xyz\sum\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate: $S=\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}$ Evaluate of this sum:
$$S=\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}$$
Expand out the sum:
$$S=\prod_{k=1}^{1}\frac{2k}{k+2}+\prod_{k=1}^{2}\frac{2k}{k+3}+\prod_{k=1}^{3}\frac{2k}{k+4}+\cdots$$
$$S=\frac{2}{3}+\frac{2}{4}\cdot\frac{4}{5}+\frac{2}{... | Correcting $(2n)!$ to $(2n+1)!$ in OP's nth term, we have
$$S=\sum_{n=1}^{\infty} \frac{2^n n! (n+1)!}{(2n+1)!} =\sum_{n=1}^{\infty}\frac{2^n n ~\Gamma(n) \Gamma(n+2)}{\Gamma(2n+2)}=\sum_{n=1}^{\infty}2^n n~B(n,n+2)= \sum_{n=1}^{\infty} 2n\int_{0}^{\pi/2} 2^n \sin^{2n-1} x ~\cos^{2n+3} xdx.$$
$$S=\int_{0}^{\pi/2} \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3585115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\lim\limits_{n\to \infty}\frac{1^p+3^p+\dots+(2n-1)^p}{n^{p+1}}$
Evaluate
$\lim\limits_{n\to \infty}\frac{1^p+3^p+\dots+(2n-1)^p}{n^{p+1}}$ using Stolz-Cesaro theorem .
Now this is my attempt :$\lim\limits_{n\to \infty}\frac{1^p+3^p+\dots+(2n-1)^p+(2n)^p-1^p-3^p-\dots-(2n-1)^p}{(n+1)^{p+1}-n^{p+1}}$=$\l... | No, not correct.
\begin{align*}
&\quad \ \ \frac {(2n)^p} {(n+1)^{p+1} - n^{p+1}} \\
&=\frac 1 n \cdot \frac {2^p} {(1 + 1/n)^{p+1} - 1} \\
&\sim \frac 1 n \cdot \frac {2^p} {(p+1)/n} \\
&= \frac {2^p} {p+1}. \tag {$n \to \infty$}
\end{align*}
$p$ is a real positive number, so generally the binomial theorem fails. In... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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.