Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
The perimeter of an isosceles triangle $\triangle ABC$
An isosceles triangle $\triangle ABC$ is given with $\angle ACB=30^\circ$ and leg $BC=16$ $cm$. Find the perimeter of $\triangle ABC$.
We have two cases, right? When 1) $AC=BC=16$ and 2) $AB=BC=16$.
For the first case: let $CH$ be the altitude through $C$. Since ... | The first case.
Let $BK$ be an altitude of $\Delta ABC$.
Thus, $$BK=8,$$
$$CK=\sqrt{BC^2-BK^2}=\sqrt{16^2-8^2}=8\sqrt3$$ and
$$AB=\sqrt{AK^2+BK^2}=\sqrt{(16-8\sqrt3)^2+8^2}=$$
$$=\sqrt{8^2(2-\sqrt3)^2+8^2}=8\sqrt{(2-\sqrt3)^2+1}=16\sqrt{2-\sqrt3},$$
which gives the answer: $32+16\sqrt{2-\sqrt3}.$
We can solve the secon... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3673213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability problem on umbrellas
Two absent-minded roommates, mathematicians, forget their umbrellas in some way or another. $A$ always takes his umbrella when he goes out, while $B$ forgets to take his umbrella with probability $1/2$. Each of them forgets his umbrella at a shop with probability $1/4$. After visiting ... | I think your answer is correct. Each of them can forget the umbrella independently.
So:
Pr{A forgets}= $\frac{1}{4}+\frac{3}{4}.\frac{1}{4}+\frac{3}{4}\frac{3}{4}\frac{1}{4}= \frac{74}{128}$
Pr{B forgets}=$\frac{101}{128}$
Consequently: Pr{one umbrella} = Pr{A forgets}.(1-Pr{B forgets})+(1-Pr{A forgets}).Pr{B forgets}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3673481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$a+b+c+d=1, a,b,c,d ≠ 0$, then prove that $(a + \frac{1}{a})^2 + (b + \frac{1}{b})^2 + (c + \frac{1}{c})^2 + (d + \frac{1}{d})^2 \ge \frac{289}{4} $ If $a+b+c+d=1$, $a,b,c,d ≠ 0$, prove that $$\left(a + \frac{1}{a}\right)^2 + \left(b + \frac{1}{b}\right)^2 + \left(c + \frac{1}{c}\right)^2 + \left(d + \frac{1}{d}\right)... | Let $$f(x)=x^2+\frac{1}{x^2}.$$
Thus, $f$ is a convex function on $(0,+\infty)$ and $f$ is a convex function on $(-\infty,0)$.
We need to consider the following cases.
*
*$a\geq b\geq c\geq d>0$.
The case $0>a\geq b\geq c\geq d$ is impossible because $a+b+c+d=1.$
Now, our inequality follows from the Jensen's ine... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $z^2+z+2=0$, then $z^2 + \frac{4}{z^2} = -3$
If $z^2+z+2=0$, for $z\in\mathbb C$, demonstrate:
$$z^2 + \frac{4}{z^2} = -3$$
$z^2 = - 2 - z$, but it didn't help me.
Is there any other elegant solution?
| Since $z \neq 0$, you have
$$\begin{equation}\begin{aligned}
& z^2 + z + 2 = 0 \\
& z + 1 + \frac{2}{z} = 0 \\
& z + \frac{2}{z} = -1 \\
& \left(z + \frac{2}{z}\right)^2 = (-1)^2 \\
& z^2 + 4 + \frac{4}{z^2} = 1 \\
& z^2 + \frac{4}{z^2} = -3
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The polynomial $x^{2n} - 2x^{2n-1} + 3x^{2n-2}- \cdots -2nx + 2n + 1$ has no real roots.
Prove that the polynomial $$x^{2n} - 2x^{2n-1} + 3x^{2n-2}- \cdots -2nx + 2n + 1$$ has no real roots.
Here is the solution
Transcribed from this image
*For $x \leq 0$ we have obviously $p(x)>0$. Let $x>0$. We transform the p... | Since other users already explained the last part of the given proof, I propose another way where there is no need to distinguish between the positive and the negative case, and it does not involve any geometric sum.
By splitting the terms of even degree in a convenient way, the polynomial can be written as
$$(0+1)x^{2... | {
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"timestamp": "2023-03-29T00:00:00",
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Inequality $\sum_{cyc}\frac{a}{2a^2+a+1}\leq \frac{3}{4}$ Let $a,b,c\in\mathbb{R^+}$ such that $a+b+c=3$. Then prove that $$\sum_{cyc}\frac{a}{2a^2+a+1}\leq \frac{3}{4}$$
I tried to use tangent line method. Let $$f(x)=\frac{x}{2x^2+x+1}$$
Then $$f'(x)=\frac{1-2x^2}{2x^2+x+1}$$ and since we know that equality occurs if ... | This inequality is true for any reals $a$, $b$ and $c$ such that $a+b+c=3$.
Indeed, if $abc=0$ so the inequality is obvious.
But for $abc\neq0$, by AM-GM and your work (after assuming that $a\geq\frac{5}{2}$ and $x\neq0$ ) we obtain:
$$\sum_{cyc}\frac{a}{2a^2+a+1}\leq\frac{\frac{5}{2}}{2\left(\frac{5}{2}\right)^2+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3677136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Double Counting(combinatorial proof) for $1^2\binom{n}{1} + 2^2\binom{n}{2}+3^2\binom{n}{3}+...+n^2\binom{n}{n}$ = $n(n+1)2^{n-2}$ can anyone help me with double-counting proof for this equation:
$1^2\binom{n}{1} + 2^2\binom{n}{2}+3^2\binom{n}{3}+...+n^2\binom{n}{n}$ = $n(n+1)2^{n-2}$
I tried this example: we have n+1 ... | I only know how to prove it analytically. Let $f(x)=(1+x)^{n}=\sum_{k=0}^{n}{n \choose k}x^{k}.$
Differentiating, we have
$$
f'(x)=\sum_{k=1}^{n}kx^{k-1}{n \choose k}.
$$
Multiply both sides by $x$ and differentiate them again, then we have
\begin{eqnarray*}
\frac{d}{dx}\{xf'(x)\} & = & \frac{d}{dx}\sum_{k=1}^{n}kx^{k... | {
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"timestamp": "2023-03-29T00:00:00",
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Is there a way to find an upper bound for $n^2+an+b$? I was solving the Project Euler: Problem 27.
Considering quadratics of the form $n^2 + an + b$, where $|a| \lt 1000$ and $|b| \le 1000$
Find the product of the coefficients, $a$ and $b$, for the quadratic expression that produces the maximum number of primes for co... | I would like to extend the answer by @Ingix.
The given quadratic is $f(n)=n^2+an+b$. Let $n=ax+by$ where $x,y \in \mathbb{Z}$.
\begin{equation}
f(ax+by)=a^2(x^2+x)+aby(2x+1)+b^2y^2+b
\end{equation}
If $x^2+x=0$, then $b|f(n)$. This gives $x=0,-1$. For $x=0$, we get $n=by$ and $y=1$ gives the best answer ($y$ cannot be ... | {
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"timestamp": "2023-03-29T00:00:00",
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Calculate $\int_0^1 \Big( \int _0^x \sqrt{y^2+\frac{y^2}{x^2}+\frac{y^4}{x^4}} dy \Big) dx$
Calculate $\int_0^1 \Big( \int _0^x \sqrt{y^2+\frac{y^2}{x^2}+\frac{y^4}{x^4}} dy \Big) dx$
I have a problem with this task because I have no idea what parametrization should be used to calculate it. I tried $x=r\sin \alpha, y... | What do you mean by too complicated? Upon substitution $y=x\sqrt{z}$ the integral becomes $$\int_0^1 {\rm d}x \frac{x}{2} \int_0^1 \sqrt{x^2+z+1} \, {\rm d}z = \int_0^1 {\rm d}x \frac{x}{3} \left[ \left( x^2 + 2 \right)^{3/2} - \left(x^2+1\right)^{3/2} \right] \\
\stackrel{t=x^2}{=} \frac{1}{6} \int_0^1 {\rm d}t \left[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3684912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all ordered pairs $(x,y)$ of positive integers such that the expression $x^2+y^2+xy$ is a perfect square. My approach so far:
Let $x^2+y^2+xy=n^2,$ where $n\in\mathbb Z$.
$\implies (x+y) ^2-xy=n^2$
$\implies (x+y) ^2-n^2=xy$
$\implies (x+y+n) (x+y-n) =xy$
Only one case is possible:
When $x+y+n=xy$ and $x+y-n=1$. O... | Firstly, we can ssume Multiplying by $4$, we get $(2x+y)^2+3y^2$ is a perfect square. Thus we want to find $a,b$ such that $a^2+3b^2$ is a perfect square with $gcd(a,b)=1$
What follows is a well-known theory of finding out rational points on conic sections.
Now this amounts to finding rational points on the ellipse $E... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3688762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of a tricky binomial sum The Question:
$$
\mbox{To prove that:}\quad
\frac{3!}{2(n+3)} = \sum_{r=0}^{n}{(-1)^r\frac{\binom{n}{r}}{\binom{r+3}{r}}}
$$
My Attempt:
I start off by writing $\sum_{r=0}^{n}{(-1)^r\frac{\binom{n}{r}}{\binom{r+3}{r}}}$ as $\sum_{r=0}^{n}{(-1)^r\frac{n!3!}{(n-r)!(r+3)!}}$.
Now, since... | Start from
$$
\frac{3!}{(n+1)(n+2)(n+3)}\sum_{r=0}^{n}(-1)^r \binom{n+3}{r+3}
$$Reindex:
$$
=\frac{3!}{(n+1)(n+2)(n+3)}\sum_{r=3}^{n+3}(-1)^{r-3} \binom{n+3}{r}
$$
$$
=\frac{-3!}{(n+1)(n+2)(n+3)}\sum_{r=3}^{n+3}(-1)^{r} \binom{n+3}{r}
$$Now add and subtract the terms $0\leq r\leq 2$:
$$
=\frac{-3!}{(n+1)(n+2)(n+3)}\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3693032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding all real functions satisfying $f(x) + f \left( \frac{x - 1}{x} \right) = \frac{5x^2 - x - 5}{x}$
Find all functions $f:\mathbb R \to \mathbb R$ that satisfy
$$f(x) + f \left( \frac{x - 1}{x} \right) = \frac{5x^2 - x - 5}{x}$$
for all nonzero $x$.
| We set up a system of $3$ functional equations and one of the variables will be $f(x)$ and we will solve for it. First replace $x$ by $\frac{1}{1-x}$ and we get: $f(\frac{1}{1-x}) + f(x) = P(x)$, with $P(x)$ is the right hand side evaluated at $\frac{1}{1-x}$. Next replace $x$ by $\dfrac{x-1}{x}\implies f(\frac{x-1}{x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3697590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving a nonlinear periodic ODE Is it possible to solve the below ODE for arbitrary real values of $c$?
$$2 (\frac{d \theta}{dx})^2 [\sin(2\theta) -c \cos(2 \theta)]-[\cos(2\theta) +c \sin(2 \theta)]\frac{d^2}{dx^2}\theta=0,$$
where $\theta=\theta(x)$. Also, $\theta$ itself does not have meaning, but ($\cos(2\theta),\... | The given equation can be easily integrated:
$$-\dfrac12\left((\cos 2\theta+c\sin2\theta)\theta'\right)'=0\Rightarrow
(\cos 2\theta+c\sin2\theta)\theta' = \dfrac12C_1,\\
(\cos 2\theta+c\sin2\theta)\,\mathrm d(2\theta) = C_1\,\mathrm dx\Rightarrow
\sin 2\theta-c\cos2\theta = C_1 x+C_2 = y,$$
wherein the constants $C_1,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3697907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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I don’t understand how to reduce this fraction to the stated solution: The fraction is as follows:
$$
\frac{9 \cdot 11 + 18 \cdot 22 + 27 \cdot 33 + 36 \cdot 44 }{
22 \cdot 27 + 44 \cdot 54 + 66 \cdot 81 + 88 \cdot 108}
$$
That’s all fine. What I don’t get is that my textbook says this reduces to the following:
$$\frac... | First correct the error in your expression. The first addition sign on top should be a multiplication sign.
Observe for instance that on the numerator, you have the first term $9\cdot11= (9\cdot 1)(11\cdot 1) = 9\cdot11(1^2) $ and the second term is $18\cdot 22 = (9\cdot 2)(11\cdot 2) = 9\cdot11 (2^2)$.
Using exactly ... | {
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"url": "https://math.stackexchange.com/questions/3702238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Using the Maclaurin series for $\frac{1}{1-x}$ to find $\frac{x}{1+x^2}$ Suppose I know the Maclaurin series for $$\frac{1}{1-x}=1+x+x^2+x^3+...= \sum_{n=0}^{\infty}x^n \tag{1}$$
then I can find the Maclaurin series for $\frac{1}{(1-x)^2}$ by the substitution $x\to x(2-x)$, which is obtained by solving the following eq... | Hint:
Replace $x$ with $-y^2$ to find
$$\dfrac1{1-(-y^2)}=\sum_{r=0}^\infty(-y^2)^r$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limits: Epsilon and Delta questions on proving limits Suppose you have f(x)= $ \frac1{x^2+1}$ . We want to show that $\lim_{x\to {-1}} \frac1{x^2+1} = \frac12 $.
This is how I approached this issue
Suppose $ \lvert x+1 \rvert \lt \delta $ and $x \ne -1 $.
Then, $ \lvert f(x)-\frac12 \rvert = \lvert \frac1{x^2+1} -\f... | Looks good to me. But one can simplify it further by noticing that
$\displaystyle\left|\frac{1}{x^2+1}\right|\leq 1$ for any $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704781",
"timestamp": "2023-03-29T00:00:00",
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Proving That $\sum^{n}_{k=0} \bigl(\frac{4}{5}\bigr)^k < 5$
Using induction, prove that $$\sum_{k=0}^n \biggl(\frac 4 5 \biggr)^k = 1+\frac{4}{5}+\bigg(\frac{4}{5}\bigg)^2+\bigg(\frac{4}{5}\bigg)^3+\cdots +\bigg(\frac{4}{5}\bigg)^n<5$$ for all natural numbers $n.$
What I have tried is as follows.
Consider the stateme... | Assume
$$
1+\frac{4}{5}+\left(\frac{4}{5}\right)^2+\cdots +\left(\frac{4}{5}\right)^k < 5
$$
for some positive integer $k$.$\;$Then
\begin{align*}
&
\frac{4}{5}
{\,\cdot}
\left(
1+\frac{4}{5}+\left(\frac{4}{5}\right)^2+\cdots +\left(\frac{4}{5}\right)^k
\right) < \frac{4}{5}{\,\cdot\,}5
\\[4pt]
\implies\;&
\frac{4}{5}+... | {
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"timestamp": "2023-03-29T00:00:00",
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Smallest positive integer that has $x_{1},x_{2}, \dots, x_{t}$ with $x_1^3 + x_2^3 + \dots + x_t^3 = 2002^{2002}?$ What is the smallest positive integer $t$ such that there exist integers $x_{1},x_{2}, \dots, x_{t}$ with $x_1^3 + x_2^3 + \dots + x_t^3 = 2002^{2002}?$
I don't really get what the question means by "such ... | Evaluating $2002^{2002}$ mod $9$, we get
\begin{align*}
2002^{2002}
\!&\equiv 4^{2002}\;(\text{mod}\;9)\\[4pt]
\!&\equiv 4{\,\cdot\,}4^{2001}\;(\text{mod}\;9)\\[4pt]
\!&\equiv 4{\,\cdot\,}(4^3)^{667}\;(\text{mod}\;9)\\[4pt]
\!&\equiv 4{\,\cdot\,}1^{667}\;(\text{mod}\;9)\\[4pt]
\!&\equiv 4\;(\text{mod}\;9)\\[4pt]
\end{a... | {
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"timestamp": "2023-03-29T00:00:00",
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$\frac{3x+1}{x+1}+\frac{3y+1}{y+1}+\frac{3z+1}{z+1} \le \frac{9}{2}$ I'm having trouble proving that for any $x,y,z>0$ such that $x+y+z=1$ the following inequality is true:
$\frac{3x+1}{x+1}+\frac{3y+1}{y+1}+\frac{3z+1}{z+1} \le \frac{9}{2}$
It seems to me that Jensen's inequality could do the trick, but I'm having tro... | Hint:
$$\left(\frac1{x+1}+\frac1{y+1}+\frac1{z+1}\right)(x+1+y+1+z+1) \ge 9$$
And
$$\frac{3x+1}{x+1} = 3 - \frac{2}{x+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3717128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Convergence of $s_{n+1}=\sqrt{1+s_n}$ Does the sequence $s_{n+1}=\sqrt{1+s_n}$ always converge, no matter what the initial value of $s_1$ is?
Is this sequence always increasing and bounded? I think so, but what's throwing me off is that to find what the sequence converges to, we just solve $s^2=1+s$ to get $s=\frac{1+\... | For the case $s_1 = 3$, then $s_2 = 2 < 3 = s_1$ and $f(x) = \sqrt{1+x}\implies f'(x)=\dfrac{1}{2\sqrt{1+x}}> 0\implies f$ is an increasing function $\implies s_n$ is strictly decreasing sequence and is bounded below by $0$ as $s_n > 0, \forall n\ge 1$, hence is convergent to $L$ which is the solution of $L = \sqrt{1+L... | {
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"timestamp": "2023-03-29T00:00:00",
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Given a tetrahedron, whose sides are $AB=3,AC=4,BC=5,AD=6,BD=7,CD=8$ . Find the volume of the tetrahedral $ABCD$ .
Given a tetrahedron, whose sides are $AB= 3, AC= 4, BC= 5, AD= 6, BD= 7, CD= 8$ . Find the volume of the tetrahedral $ABCD$ .
Assume that the tetrahedral $ABCD$ has its height $DH$ , whose length I will ... | Let the coordinate of $D$ be ($x,y,z)$ and $\triangle ABC$ be on $xy$- plane, where $A$ is the origin, and $B$ and $C$ are on the coordinate axes.
Distance of $D$ from $B,A,C$ are
$$\begin{align}
(x-3)^2+y^2+z^2 &=49\\
x^2+y^2+z^2 &=36\\
x^2+(y-4)^2+z^2&=64
\end{align}$$ respectively.
Solving these equations we get th... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find maximum $k \in \mathbb{R}^{+}$ such that $ \frac{a^3}{(b-c)^2} + \frac{b^3}{(c-a)^2} + \frac{c^3}{(a-b)^2} \geq k (a+b+c) $ Find maximum $k \in \mathbb{R}^{+}$ such that $$ \frac{a^3}{(b-c)^2} + \frac{b^3}{(c-a)^2} + \frac{c^3}{(a-b)^2} \geq k (a+b+c) $$
for all $a, b, c$ that are distinct positive real numbers (... | Let $c\rightarrow0^+$.
Thus, $$\frac{a^3}{b^2}+\frac{b^3}{a^2}\geq k(a+b),$$ which gives that $k\leq1.$
We'll prove that $1$ is a maximal value.
Indeed, we need to prove that:
$$\sum_{cyc}\frac{a^3}{(b-c)^2}\geq a+b+c.$$
Now, let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Thus, we need to prove that:
$$(u^2-uv+v^2)^2a^3+3... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A problem of definite integral inequality? $$
\text { Minimum odd value of $a$ such that }\left.\left|\int_{10}^{19} \frac{\sin x d x}{\left(1+x^{a}\right)}\right|<\frac{1}{9} \text { is (where } a \in N\right)
$$
I proceed this way
As $|\sin x|<1$
Integration $|\frac{\sin x}{1+x^a}|<|\frac{1}{1+x^a}|$
And using trial ... | $$\left|\int_{10}^{19} \frac{\sin x \,dx}{1+x^{a}}\right|\le \int_{10}^{19}\left|\frac{\sin x }{1+x^{a}}\right|\,dx\le \int_{10}^{19}\frac{1 }{1+x^{a}}\,dx<\int_{10}^{19}\frac{1}{x^a}\,dx=
\frac{19^{1-a}-10^{1-a}}{1-a}.$$
Since the denominator contains $1-a$, we will assume that $a >1$. For $a=2$,
$$ \frac{19^{-1}-10^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719650",
"timestamp": "2023-03-29T00:00:00",
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} |
Solving a second-order recurrence relation with complex characteristic roots in polar form. I am self-studying this topic from a textbook and am stuck with trying work through one example.
Suppose we are solving the recurrence equation, $u_n = 2u_{n-1} - 2u_{n-2}$.
This has the characteristic equation $r^2 - 2r + 2 = 0... | The sequence you have found is a generalization of the Fibonacci sequence. Referring to a previous answer I have posted here, the general solution can be expressed as
$$f_n=\left(f_1-\frac{af_0}{2}\right) \frac{\alpha^n-\beta^n}{\alpha-\beta}+\frac{af_0}{2} \frac{\alpha^n+\beta^n}{\alpha+\beta}$$
where $\alpha,\beta=(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3720674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
An interesting limit
Let $x\in\mathbb{R}.$ For all $i,j\in\mathbb{N},$ define $a_{i0} = \frac{x}{2^i}, a_{ij} = a_{i,j-1}^2 + 2a_{i,j-1}.$ Find, with proof, $\lim\limits_{n\to\infty} a_{nn}.$
Below is my attempt.
Let for each $n, p_n(x) = a_{nn}$. Then observe that $a_{n+1,n} = p_n(\frac{x}2).$ As well, $p_{n+1}(x)+1... | For $n\in \mathbb{N}$ ,
$a_{n,0}=\frac{x}{2^n}$
$a_{n,1}+1=(a_{n,0}+1)^2=(\frac{x}{2^n}+1)^2$
$a_{n,2}+1=(a_{n,1}+1)^2=(\frac{x}{2^n}+1)^4$
Continuing this way,
For $1\le j\le n$,
$a_{n,j}+1=(\frac x{2^n}+1)^{2^j}$
Thus for $n=j$,
$a_{n,n}+1=(\frac x{2^n}+1)^{2^n}$
Taking limit as $n\to \infty$, we get
$lim_{n\to \inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3720806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Group of order $1+2+3+ \cdots + n$ with class equation $1+2+3+ \cdots + n$ It is known that the symmetric group $S_3$ of order $1+2+3$ has the class equation $1+2+3$. But a non-abelian group of order $10 = 1+2+3+4$(which is just $D_5$) cannot have the class equation $1+2+3+4$. My question is the following: "are there f... | Note that one necessary condition for this to occur is that $\binom{n+1}{2} = \frac{(n+1)n}{2}$ must be divisible by $1, 2, \ldots, n-1, n$, since all the terms in the class equation are indices of centralizers, hence are divisors of the order of the group. Since $n$ and $n-1$ are relatively prime, this means that $n(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3720982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find the minimum of $x^3+\frac{1}{x^2}$ for $x>0$ Finding this minimum must be only done using ineaqualities.
$x^3+\frac{1}{x^2}=\frac{1}{2}x^3+\frac{1}{2}x^3+\frac{1}{3x^2}+\frac{1}{3x^2}+\frac{1}{3x^2}$
Using inequalities of arithemtic and geometric means:
$\frac{\frac{1}{2}x^3+\frac{1}{2}x^3+\frac{1}{3x^2}+\frac{1}... | A bit late answer but I thought it might be worth noting.
I was wondering whether we could squeeze out the minimum directly from Young's inequality:
$$ab\leq \frac{a^p}{p} + \frac{b^q}{q};\: a,b \geq 0;\: p,q>1 \text{ and } \frac 1p + \frac 1q = 1$$
And, yes indeed, this works well, too. Since we need the powers of $x$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3727632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Find the pedal equation of the ellipse $\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$ Find the pedal equation of the ellipse $\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$
My Attempt:
Given equation of ellipse is
$$\frac {x^2}{a^2} + \frac {y^2}{b^2}=1$$
Differentiating both sides wrt $x$
$$\frac {2x}{a^2} + \frac {2y}{b^2} \cdo... | The equation of the tangent to given ellipse at the point $~(X,Y),~$$~\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2} = 1~,\tag1$ is$~\dfrac {xX}{a^2} + \dfrac {yX}{b^2} = 1~.\tag2$
Compared the equation of the tangent of the ellipse with $~AX + BY + C = 0 ~,$ we have $~A=\frac {x}{a^2},~B=\frac {y}{b^2}~$ and $~C=-1~.$
Now perp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3729777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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For how many natural numbers(<=100) is $1111^n +2222^n+3333^n+4444^n$ divisible by 10? For how many natural numbers (0 not included) $n \leq 100$ is $1111^n +2222^n+3333^n+4444^n$ divisible by 10?
I factored out $1111^n$ and got $1111^n(1+2^n+3^n+4^n)$. So $1+2^n+3^n+4^n$ must be divisible by 10. I figured out that thi... | This can be simplified using the patterns for exponents. $1^n$ always ends in $1$. $2^n$ repeats a pattern where its last digit ends in $2, 4, 8, 6$. $3^n$ repeats a pattern where its last digit ends in $3, 9, 7, 1$. $4^n$ repeats a pattern where its last digit ends in $4, 6$. So adding up our final digits for the four... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Integrate $\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$ Evaluate $$\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$$
I tried $u=\sqrt{2x^4-2x^2+1}$, $u=\dfrac{1}{x}$ and $u=\sqrt{x}$ but none of these worked.
A friend gave this and said it's from IIT JEE and answer is $$\dfrac{\sqrt{2x^4-2x^2+1}}{2x^2}+... | Following @AdityaDwivedi's suggestion, $dt=4x^{-5}(x^2-1)dx$ so the integral is $\int\frac{dt}{4\sqrt{2+t}}=\tfrac12\sqrt{2+t}+C$. As to how you'd come up with this idea, note the original integral is $\int\frac{x^{-3}-x^{-5}}{\sqrt{2-2x^{-2}+x^{-4}}}dx$, which suggests the proposed substitution, or better still $u=2-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
$f(xy + x +y) = f(xy) + f(x) + f(y)$ and $f(x)(y - x) + f(y)(x - y) \geq 0$.
Let $f : \mathbb R \to \mathbb R$ that satisfies both 2 conditions ,
$f(xy + x + y) = f(xy) + f(x) + f(y)$ and $f(x)(y - x) + f(y)(x - y) \geq 0$ $\forall x,y \in \mathbb R$.
Determine all such $f$.
My solution.
Let $P(x,y) $ be $f(x)(y - x)... | Putting $y = 1$,
$f(x + x + 1) = f(x) + f(x) + f(1)$
$f(2x + 1) = 2f(x) + f(1)$
Under the assumption that $f$ is continuously differntiable, you can say the following:
$2f'(2x+1) = 2f'(x) \implies f'(2x+1) = f'(x)$
This gives us,
$f'(x) = f'(2(\frac{x-1}{2}) + 1) = f'(\frac{x-1}{2})$
$f'(\frac{x-1}{2}) = f'(2(\frac{x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3732370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int_0^{\frac{\pi}{4}} \left( \frac{\sin^2{(5x)}}{\sin^2{x}} -\frac{\cos^2{(5x)}}{\cos^2{x}} \right)\mathop{dx}$ Evaluate $$\int_0^{\frac{\pi}{4}} \left( \frac{\sin^2{(5x)}}{\sin^2{x}} -\frac{\cos^2{(5x)}}{\cos^2{x}} \right)\mathop{dx}$$
I tried substitutions like $u=\frac{\pi}{4}-x$, and trig identities like... | De Moivre's law says that:
$(\cos x + i\sin x)^5 = \cos 5x + i\sin 5x$
To find $\cos 5x, \sin 5x$ we just need to separate the real and imaginary parts of the left hand side.
$\cos^5 x + 5i\cos^4x\sin x - 10\cos^3x \sin^2x - 10i\cos^2x\sin^3 x+ 5\cos x\sin^4x + i\sin^5 x$
$\cos 5x = \cos^5x - 10\cos^3x\sin^2x + 5\cos x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3733349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Evaluate integral: $\int_0^{\frac{\pi}{2}}\ln(a^2\cos^2 x +b^2\sin^2x)dx$? For $a,b>0$, show that
$$\int_0^{\frac{\pi}{2}}\ln(a^2\cos^2 x+b^2\sin^2 x) dx=\pi\ln\left(\frac{a+b}{2}\right)$$
I came across to this problem in the book Table of integrals, series and product. So here is my try to prove the closed form.
For... | I thought it might be instructive to present an approach that relies on contour integration in the complex plane. To that end, we proceed.
MODIFYING THE INTEGRAL
Let $I(a,b)$ be defined by the integral
$$I(a,b)=\int_0^{\pi/2}\log(a^2\cos^2(x)+b^2\sin^2(x))\,dx\tag1$$
Using the double angle formulas, $\sin^2(x)=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3733804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
Calculating $\binom {2016}0 - \binom {2016}3 + \binom {2016}6 - \binom {2016}9 + ... +\binom {2016}{2016}$ Hello everyone how can I calculate this expression
$\binom {2016}0 - \binom {2016}3 + \binom {2016}6 - \binom {2016}9 + ... +\binom {2016}{2016}$?
I tried to mark $\omega = \frac{\sqrt{3}i-1}{2}$ and $\omega^3 = 1... | Hint: Instead, let $\omega$ be such that $\omega^6 = 1$ (so $\omega = \frac 12 + i\frac{\sqrt{3}}2$). Let $f(x) = (1 + x)^{2016}$. Consider the sum
$$
f(\omega) + f(\omega^3) + f(\omega^5).
$$
We can calculate the above quantity as
$$
(1 + \omega)^{2016} + (1 + \omega^{3})^{2016} + (1 + \omega^5)^{2016} = \\
(1 + \om... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve for $y$ in $\frac{dy}{dx}-\frac{3y}{2x+1}=3x^2$ I saw a challenge problem on social media by a friend, solve for $y$ in $$\frac{dy}{dx}-\frac{3y}{2x+1}=3x^2$$
I think this is an integration factor ODE
$$\frac{1}{{(2x+1)}^{\frac{3}{2}}} \cdot \frac{dy}{dx}-\frac{3y}{{(2x+1)}^{\frac{5}{2}}}=\frac{3x^2}{{(2x+1)}^{\f... | hint
For the last integral, put
$$2x+1=u^2$$
$$x=\frac{u^2-1}{2}$$
$$dx=udu$$
$$3x^2=\frac 34(u^4+1-2u^2)$$
it becomes
$$\frac 34\int \frac{u^4-2u^2+1}{u^3}udu$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
$\int \frac{x^3+3x+2}{(x^2+1)^2 (x+1)} \ dx$ without using partial fraction decomposition One way to evaluate the integral $$\int \frac{x^3+3x+2}{(x^2+1)^2(x+1)} \ dx $$ is to rewrite it as $$ \int \frac{x^3+x+2x+2}{(x^2+1)^2(x+1)} dx \\=\int\frac{x(x^2+1) +2(x+1)}{(x^2+1)^2(x+1)}dx\\=\int\frac{x}{(x^2+1)(x+1)}dx+\int\... | To avoid partially fractionizing the integrand, proceed with the substitution below instead
\begin{align}
\int\frac{x^3+3x+2}{(x^2+1)^2(x+1)}\overset{x=\frac{1-y}{1+y}}{dx}
=&\ \frac12\int \frac{y^3-3y^2-3y-3}{(1+ y^2)^2}dy\\
=& \ \frac12\int \frac{(y^2-3)y}{(1+ y^2)^2}-\frac3{1+ y^2}\ dy\\
=& \ \frac1{1+ y^2}+\frac14\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Simplify $\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}$
Simplify:
$$\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}$$
After the substitution as $\cos(x)=a$ and $\s... | Let us express everything in term of $s:=\sin x$:
$$\frac{4(1-2s^2)^2-4(1-s^2)+3s^2}{4s^2-4s^2(1-s^2)}=\frac{16s^4-9s^2}{4s^4}=4-\frac9{4s^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3736352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3 +\dots+ n^3$ is divided by $n+5$ the remainder is $17.$
Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3 +\dots+ n^3$ is divided by $n+5$ the remainder is $17.$
Letting $k= n+5$ we get that $1^3+2^3+3^3 +\dots+ (k-5)^3 \equiv 1... | As $\sum\limits_{i=1}^ni^3=\frac14 n^2(n+1)^2$ we consider cases $n=2k$ and $n=2k-1$.
*
*$n=2k$, $\ \frac14 n^2(n+1)^2=(n+5)\left(2 k^3 - 3 k^2 + 8 k + \frac{100}{2 k + 5} - 20\right)$ so $(2k+5)|(100-17)$, but $100-17=83$ is a prime, thus $2k+5=83$ and $n=78$.
2a. $n=4k+1$, $\ \frac14 n^2(n+1)^2-17=(n+5)\left(16 k^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3737447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
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prove $\sum\cos^3{A}+64\prod\cos^3{A}\ge\frac{1}{2}$ In every acute-angled triangle $ABC$,show that
$$(\cos{A})^3+(\cos{B})^3+(\cos{C})^3+64(\cos{A})^3(\cos{B})^3(\cos{C})^3\ge\dfrac{1}{2}$$
I want use Schur inequality
$$x^3+y^3+z^3+3xyz\ge xy(y+z)+yz(y+z)+zx(z+x)$$
then we have
$$x^3+y^3+z^3+6xyz\ge (x+y+z)(xy+yz+zx)$... | Problem: Let $a, b, c > 0$ with $a^2+b^2+c^2+abc=4$. Prove that $a^3+b^3+c^3+(abc)^3\ge 4$.
Solution: It suffices to prove that, for $a, b, c > 0$,
$$a^3+b^3+c^3+(abc)^3 - 4 - 2(a^2+b^2+c^2+abc - 4) \ge 0. \tag{1}$$
It is verified by Mathematica.
By Vasc's Equal Variable Theorem [1, Corollary 1.9], we only need to prov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3737759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
A square is cut into three equal area regions by two parallel lines find area of square. A square is cut into three equal area regions by two parallel lines that are 1 cm apart, each one passing through exactly one of two diagonally opposed vertices. What is the area of the square ?
|
Let $AE=x$. By symmetry (equal area etc.) $FC=x$. Let the side of square be $a$. Then $DE=BF=\sqrt{a^2+x^2}$.
Area of the parallelogram $FBED$ = $\sqrt{a^2+x^2} \cdot 1=\sqrt{a^2+x^2}$.
Area of $\triangle DAE$ = Area of $\triangle FCB$ = $\frac{ax}{2}$.
$$\frac{ax}{2}=\sqrt{a^2+x^2} \implies \color{red}{x^2=\frac{4a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$2^x\leq x+1$ for $x\in [0,1]$ I tried using mean value theorem but couldn't show $2^{2^c} < e$ for $c ∈ (0,x)$. Writing taylor expansion of $2^x$ also don't work because we need to show $2^x$ is smaller than something, not greater
| Since other people have explained the typical way to do this, I'll mention that there's a very tricky way to show $2^x \leq 1 + x$ for $0 \leq x \leq 1$ with an infinite series expansion. But not a Taylor series - an infinite series of binomial coefficients.
If we define the binomial coefficients $\binom{x}{k}$ for non... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3748080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 6
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$\lim_{x \to 0}\left(\frac{\sin^2(x)}{1-\cos(x)}\right)$ without L'Hopital's rule I'm trying to calculate $\lim_{x \to 0}\left(\frac{\sin^2(x)}{1-\cos(x)}\right)$ without L'Hopital's rule.
The trigonometrical identity $\sin^2(x) = \frac{1-\cos(2x)}{2}$ doesn't seem to lead anywhere. I also attempted to calculate using ... | The way you tried to solve:
$(i) \ \displaystyle \sin^2(x) = \displaystyle \dfrac{1-\cos(2x)}{2}$
$$\displaystyle \lim _{x \rightarrow 0} \dfrac{1}{2}\left(\dfrac{1-\cos (2 x)}{1-\cos (x)}\right)$$
$$=\dfrac{1}{2} \displaystyle \lim _{x \rightarrow 0} \frac{\cos (0)-\cos (2 x)}{\cos(0)-\cos (x)}=\frac{1}{2} \displaysty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Evaluate: $\int_0^{\infty}\frac{\ln x}{x^2+bx+c^2}\,dx.$
Prove that:$$\int_0^{\infty}\frac{\ln x}{x^2+bx+c^2}\,dx=\frac{2\ln c}{\sqrt{4c^2-b^2}}\cot^{-1}\left(\frac{b}{\sqrt{4c^2-b^2}}\right),$$ where $4c^2-b^2>0, c>0.$
We have:
\begin{align}
4(x^2+bx+c^2)&=(2x+b)^2-\left(\underbrace{\sqrt{4c^2-b^2}}_{=k\text{ (say)}... | I will just show how to solve it in the case $b=0$, since the idea is the same, and the calculation becomes easier. Substituting $x=c\tan\theta$, we have:
\begin{equation}
\begin{split}
\int_0^\infty\frac{\ln x}{x^2+c^2}dx&=c^{-1}\int_0^{\pi/2}\ln(c\tan\theta)d\theta\\
&=c^{-1}\int_0^{\pi/2}\ln(c)d\theta+\int_0^{\pi/2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the minimum of the set $A=\left\{\int_0^1(t^2 - at-b)^2 dt\, : \,a,b \in \mathbb{R}\right\}$.
Let $$A=\left\{\int_0^1(t^2 - at-b)^2 dt\, : \,a,b \in \mathbb{R}\right\}\,.$$ Find the minimum of $A$.
$\textbf{My attempt:}$
Well, we have
$ 0 \leq\int_0^1(t^2 - at-b)^2 dt = \frac{1}{5} - \frac{a}{2} + \frac{a^2-b}... | Your calculation is incorrect. For $a,b\in\mathbb{R}$, if $f(a,b):=\displaystyle\int_0^1\,(t^2-at-b)^2\,\text{d}t$, then
$$f(a,b)=\frac{a^2}{3}+ab+b^2-\frac{a}{2}-\frac{{\color{red}2}b}{3}+\frac{1}{5}\,.$$
Thus,
$$f(a,b)=\frac{1}{3}\,\left(a+\frac{3(2b-1)}{4}\right)^2+\frac{1}{4}\left(b+\frac{1}{6}\right)^2+\frac{1}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Let $a_2,a_3,\cdots,a_n$ be positive real numbers and $s=a_2+a_3+\cdots+a_n$. Show that $\sum\limits_{k=2}^na_k^{1-\frac{1}{k}}I tried this problem in this way:-
Let the real numbers are $b_2,b_3,\cdots,b_n$. such that $(b_2,b_3,\cdots,b_n)$is a permutation of the numbers given $(a_2,a_3,\cdots,a_n)$. Hence $s=b_2+b_3+... | We can prove that $a_k^{1-1/k} < a_k + \frac{2}{k} \sqrt{a_k}$.
Indeed, if $a_k \ge 1$, it is obvious;
and if $0 < a_k < 1$, by Bernoulli inequality $(1+x)^r \le 1 + rx$ for $0 < r \le 1$ and $x > -1$, we have
$a_k^{1-1/k} = a_k (a_k^{-1/2})^{2/k} = a_k(1 + a_k^{-1/2} - 1)^{2/k}
\le a_k [1 + (a_k^{-1/2} - 1)\frac{2}{k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Question about finding roots of a polynomial and studying the nature The number of real roots of the equation $1+\frac{x}{1}+\frac{x^{2}}{2}+\frac{x^{2}}{3}+\cdots+\frac{x^{7}}{7}=0$
(without factorial) is
My work
Let,$\mathrm{f}(\mathrm{x})=1+\frac{x}{1}+\frac{x^{2}}{2}+\frac{x^{3}}{3}+\cdots+\frac{x^{6}}{6}$
[Let, f... | Define
$$f(x)=1+\sum_{n=1}^7 \frac{x^n}{n}$$
to be the given function. Then we have
\begin{align}
f'(x)
&=\sum_{n=1}^7x^{n-1}\\
&=\cases{\frac{x^7-1}{x-1}&$x\ne1$\\7&$x=1$}\\
\end{align}
But $x^7=1$ has a single real root namely $x=1$ because it's derivative is $7x^6\ge0$. Thus $f'(x)$ has no real roots. So $f(x)$ is s... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 0
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Other way to evaluate $\int \frac{1}{\cos 2x+3}\ dx$? I am evaluating
$$\int \frac{1}{\cos 2x+3} dx \quad (1)$$
Using Weierstrass substitution:
$$ (1)=\int \frac{1}{\frac{1-v^2}{1+v^2}+3}\cdot \frac{2}{1+v^2}dv =\int \frac{1}{v^2+2}dv \quad (2) $$
And then $\:v=\sqrt{2}w$
$$ (2) = \int \frac{1}{\left(\sqrt{2}w\right)... | HINT:
Using Euler's formula, we have $\cos(2x)=\frac12(e^{i2x}+e^{-i2x})$.
Now make the substitution $z=e^{i2x}$ with $dx=\frac1{i2z}\,dz$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Simplify $\sum^{20}_{k=10} k\binom{k-1}{9}$.
Simplify $$\sum^{20}_{k=10} k\binom{k-1}{9}$$ as much as possible.
I feel like I could utilize the hockey stick identity, but have not found a way to do so with that extra $k$.
Any help would be appreciated.
| If you couldn't think about how to get rid of the '$k$':
What we need to evaluate is
\begin{align*}
\sum^{20}_{k=10} k\binom{k-1}{9}&=20{19\choose 10}+19{18\choose 9}+\cdots+10{9\choose 9}\\
\end{align*}
Using Hockey-stick identity, we can evaluate
\begin{align*}
\sum^{20}_{k=10} k\binom{k-1}{9}&=10{9\choose 9}+\cdots+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Use linearisation of a certain function to approximate $\sqrt[3]{30}$ Background
Find the linearisation of the function
$$f(x)=\sqrt[3]{{{x^2}}}$$
at
$$a = 27.$$
Then, use the linearisation to find
$$\sqrt[3]{30}$$
My work so far
Applying the formula
$${f\left( x \right) \approx L\left( x \right) }={ f\left( a \right) ... | There is something wrong : your definition of $f$ is not consistant : sometimes you use $f(x)=\sqrt[3]{x^2}$ and other times $f(x)=\sqrt[3]{x}$... You should only use the second one I think. Get that right and you got the good idea !
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to solve this ODE: $x^3dx+(y+2)^2dy=0$? I am trying to solve $$ x^3dx+(y+2)^2dy=0 \quad( 1)$$
Dividing by $dx$, we can reduce the ODE to seperate variable form, i.e
$$ (1) \to (y+2)^2y'=-x^3 $$
Hence,
$$ \int (y(x)+2)^2y'(x) dy = \int -x^3dx = - \frac{x^4}{4} + c_1$$
This LHE seems to be easy to solve using integ... | Substitute $Y=y+2$ initially. Solving $x^3dx+Y^2dY=0$ yields $Y=\left(C^\prime-\frac{3x^4}4\right)^{\frac13}=y+2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3765155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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What is the smallest integer $n>1$ for which the mean of the square numbers $1^2,2^2 \dots,n^2 $ is a perfect square?
What is the smallest integer $n>1$ for which the mean of the square numbers $1^2,2^2 \dots,n^2 $ is a perfect square?
Initially, this seemed like one could work it out with $AM-GM$, but it doesn't see... | Following a hint by Mostafa Ayaz, we have $(n+1)(2n+1)=6k^2$ for some integer $k$. That is,
$$(4n+3)^2-3(4k)^2=1\,.$$
Hence, $(4n+3)+(4k)\sqrt{3}=(2+\sqrt{3})^m$ for some nonnegative integer $m$. Therefore,
$$4n+3=\sum_{r=0}^{\left\lfloor\frac{m}{2}\right\rfloor}\,\binom{m}{2r}\,2^{m-2r}\,3^r\,.$$
If $m$ is odd, then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3766779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Let $\Phi$ be standard Gaussian CDF and $u > 0$. What is good u-bound for $\int_0^1\Phi(u/r - ur)dr$ as a function of $u$? Let the function $\Phi(x) := (1/\sqrt{2\pi})\int_{-\infty}^x e^{-t^2/2}dt$ be the standard Gaussian CDF. For $u > 0$, define $I(u) := \int_0^1 \Phi(u/r-ur)dr$.
Question. What are good upper-bounds... | A good simple upper bound for $u > 1$:
Using integration by parts and then the substitution $w = \frac{u}{r} - ur$, we have
\begin{align}
I(u) &= \frac{1}{2} + \frac{1}{2u\sqrt{2\pi}} \int_0^\infty \mathrm{e}^{-\frac{1}{2}w^2}
(\sqrt{w^2+4u^2} - w)\mathrm{d} w\\
&\le \frac{1}{2} + \frac{1}{2u\sqrt{2\pi}} \int_0^\infty... | {
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"source": "stackexchange",
"question_score": "3",
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Show that the elements of the sequence are divisible by $2^n$ I am trying to prove the following:
Consider the sequence defined by $A_{n+2}=6A_{n+1}+2A_n, A_0=2, A_1=6$. Show that $2^n|A_{2n-1}$ but $2^{n+1}\nmid A_{2n-1}$.
The first terms of this sequence are 2, 6, 40, 252, 1592, 10056, 63520.
In fact, the maximal e... | We can't do induction on just the odd terms without generalizing something about the even terms.
Let's try a few terms and see why this seems to occur...and to see what else we will need to assure this will always be true:
$2^1|a_{2*1-1} =a_1 = 6$ because ... it does and $2^2\not \mid a_1 = 6$ because ... it doesn't.
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Find the value of $k$ in the equation $\sin {1^\circ}\sin {3^\circ}.......\sin {179^\circ} = \frac{1}{{{2^k}}}$ Find the value of 'k' in the equation $\sin {1^\circ}\sin {3^\circ}.......\sin {179^\circ} = \frac{1}{{{2^k}}}$
My approach is as follow
$\sin {1^\circ} = \sin {179^\circ}$
$T = {\sin ^2}{1^\circ}{\sin ^2}{3^... | The connection of these kind of products with Chybshev polynomials in my opinion is the most important. Because, these polynomials show up everywhere like a unification of many things. Some nice properties of them are listed here. I know so few of them!: https://en.wikipedia.org/wiki/Chebyshev_polynomials
$T_n(x)=2^{n-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How can this integral be convergent? According to ${\tt Mathematica}$, the following integral converges if
$\beta < 1$.
$$
\int_{0}^{1 - \beta}\mathrm{d}x_{1}
\int_{1 -x_{\large 1}}^{1 - \beta}\mathrm{d}x_{2}\, \frac{x_{1}^{2} + x_{2}^{2}}{\left(1 - x_{1}\right)\left(1 - x_{2}\right)}
$$
How is this possible ?. For $x... | Changing a little notations (and trying to show the steps), considering
$$I=\int^{1-\beta}_{0}dx \int^{1-\beta}_{1-x} \frac{x^2+y^2}{(1-x)(1-y)}\,dy$$
$$\int \frac{x^2+y^2}{(1-x)(1-y)}\,dy=\frac{\left(x^2+1\right) \log (y-1)+\frac{1}{2} (y-1)^2+2 (y-1)}{x-1}$$
$$J(x)=\int^{1-\beta}_{1-x} \frac{x^2+y^2}{(1-x)(1-y)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3770391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to integrate $\int {2\over (x^2+2)\sqrt{x^2+4}}dx$?
Solve the following indefinite integral:
$$\int \frac{2}{(x^2+2)\sqrt{x^2+4}} dx$$
My approach:
I used the substitution: $x=2\tan t$, $dx=2\sec^2t dt$
$$\int \frac{2}{(x^2+2)\sqrt{x^2+4}} dx=\int \frac{2}{(4\tan^2t+2)\sqrt{4\tan^2t+4}}\cdot 2\sec^2t\ dt$$
$$=\in... | The problem's already solved but just another method.
Substitute $x = \frac{1}{t}$
$$=\int \frac{2}{\big(\frac{1}{t^2} + 2\big) \sqrt{\frac{1}{t^2} + 4}} \frac{-dt}{t^2}$$
$$=\int \frac{2t^3}{(1+2t^2) \sqrt{1+4t^2}} \frac{-dt}{t^2}$$
Now substitute $1+4t^2 = u^2$.
Integral will simplify to
$$\int - \frac{du}{u^2 + 1}$$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why is the convergence point of $ \sum _{n=1}^{\infty }\frac{1}{2^n}-\frac{1}{2^{n+1}} $ negative? I am trying to evaluate $\frac1{2^1} - \frac1{2^2} + \frac1{2^3} - \frac1{2^4} + \cdots$
I re-wrote the sum using sigma notation as:
$$ \sum _{n=1}^{\infty } \left( \frac{1}{2^n}-\frac{1}{2^{n+1}} \right) \quad (1) $$
Hen... | I reckon your original sum is
$$\sum _{n=1}^{\infty }\left(\frac{1}{2^{2n-1}}-\frac{1}{2^{2n}}\right).$$
Of course, this is a GP.
| {
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"source": "stackexchange",
"question_score": "1",
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Definite integration $\int _{-\infty}^\infty \frac{\tan^{-1}(2x-2)}{\cosh(\pi x)}dx$ How do I integrate $$\int _{-\infty}^\infty \frac{\tan^{-1}(2x-2)}{\cosh(\pi x)}dx\quad ?$$
The actual integral that I encountered is:
$$\int_{-\infty}^\infty dx \left(\frac{N}{\cosh(\frac{\pi }{c}(x-1))}+\frac{1}{\cosh(\frac{\pi}{c}x)... | For $a>0$ and $b\in \mathbb{R}$,
$$\tag{*}\color{blue}{\int_{ - \infty }^\infty {\frac{{\arctan (ax + b)}}{{\cosh \pi x}}dx} = 2\Im\left[ \log\Gamma(\frac{3}{4}+\frac{i (b-i)}{2 a})- \log\Gamma(\frac{1}{4}+\frac{i (b-i)}{2 a})\right]}$$
Here, $\log\Gamma$ is the log gamma function.
To begin, assume $\Im(c)>0, \Re(s)<... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 2
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Finding the limit of $f(f(x))$ type problem Let $f$ be a differentiable function and equation
of normal to the graph of $y = f(x)$ at $x = 3$ is
$3y = x + 18$. If $L = \mathop {\lim }\limits_{x \to 1} \frac{{f\left( {3 + {{\left( {4{{\tan }^{ - 1}}x - \pi } \right)}^2}} \right) - f\left( {{{\left( {3 + f\left( 3 \right... | Hint:
Normal $3y=x+18$ gives us $y=7$ for $x=3$ so it intersects the curve at $(3,7)$ so $f(3)=7$ and the slope of the normal is $\frac{1}{3}$, so $f'(3)=-3$ (normal being perpendicular to the tangent etc.).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrating $\int \sqrt{a^2+x^2} \ \mathrm{d} x$ with trig. substitution I am trying to come up with all the formulas I have myself and I stumbled upon a roadblock again.
Integrating $\int \sqrt{a^2+x^2} \ \mathrm{d} x$ with Trig Substitution.
So I imagined a triangle where the hypotenuse is $1$, $\sin(y) = x$ opposite... | Let $x=a\tan\theta\implies dx=a\sec^2\theta\ d\theta$
$$\int \sqrt{a^2+x^2}dx=\int \sqrt{a^2+a^2\tan^2\theta}\ a\sec^2\theta \ d\theta$$
$$=\int a\sec\theta\ a \sec^2\theta \ d\theta$$
$$=a^2\int \sec^3\theta \ d\theta$$
Using integration by parts
$$=a^2\left(\frac{1}{2}\sec\theta \tan\theta+\frac12\ln|\sec\theta+\tan\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .
Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .
What I tried: In some step I messed up with this problem and so I think I am getting my answer wrong, so please correct me.
We have $x^2 - 3x + ... | $P(x)=(x-1)^{100}+(x-2)^{200}=Q(x)×(x^2-3x+2)+ax+b$
$P(1)=(-1)^{200}=a+b, a+b=1$
$P(2)=(1)^{100}=2a+b, 2a+b=1$
$a=0 , b=1$
Hence remainder :- $R(x)=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3779596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to make a perfect square from a number given in a surd form $a+b\sqrt{c}$? Is there a way of checking that a number can be written as a perfect square and hence finding it if the number is given in the surd form?
For example, if I expand and simplify $$(1+\sqrt{2})^{2}=3+2\sqrt{2}$$.
Is there a way of finding that ... | $$(a + b \sqrt{c})^2 = a^2 + b^2 c + 2 a b \sqrt{c}$$
So if you are given $s + t \sqrt{c}$ (with $s, t, c$ rational, $t \ne 0$ and $c$ not a square)
and want to write it in this form, you want to find rationals $a$ and $b$ to solve the equations
$$ \eqalign{a^2 + b^2 c &= s\cr
2 a b &= t\cr}$$
Since $a=0$ ... | {
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"timestamp": "2023-03-29T00:00:00",
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If $xyz=32$, find the minimal value of If $xyz=32;x,y,z>0$, find the minimal value of $f(x,y,z)=x^2+4xy+4y^2+2z^2$
I tried to do by $A.M.\geq M.G.$:
$\frac{x^2+4y^2+2z^2}{2}\geq\sqrt{8x^2y^2z^2}\to x^2+4y^2+2z^2\geq32$
But how can I maximaze 4xy?
| The Alexey Burdin's hint gives the following substitution.
Let $x=4a$, $y=2b$ and $z=4c$.
Thus, $abc=1$ and by AM-GM we obtain:
$$x^2+4xy+4y^2+2z^2=16a^2+32ab+16b^2+32c^2=$$
$$=16(a^2+2ab+b^2+2c^2)\geq16\cdot6\sqrt[6]{a^4b^4c^4}=96.$$
The equality occurs for $a=b=c=1,$ which says that we got a minimal value.
| {
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"url": "https://math.stackexchange.com/questions/3782030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to compute $\int_{0}^{\frac{\pi}{2}}\frac{\sin(x)}{\sin^3(x)+ \cos^3(x)}\,\mathrm{d}x$?
How to compute the following integral?
$$\int_{0}^{\frac{\pi}{2}}\frac{\sin(x)}{\sin^3(x)+ \cos^3(x)}\,\mathrm{d}x$$
So what I did is to change $\sin(x)$ to $\cos(x)$ with cofunction identity, which is $\sin(\frac{\pi}{2} -x) ... | Note
\begin{align}
\int_{0}^{\infty} \frac{\mathrm{d}u}{1+u^3}
&=\frac13
\int_{0}^{\infty} \left(\frac{1}{1+u}+ \frac{2-u}{u^2-u+1}\right)du\\
&= \frac13\int_{0}^{\infty} \left(\frac{1}{1+u}-\frac12 \frac{2u-1}{u^2-u+1}+ \frac32\frac1{(u-\frac12)^2+\frac34} \right)du\\
&=\frac13 \left( \ln\frac{u+1}{\sqrt{u^2-u+1}}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3784579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding all possible values of $z$ given three conditions. Suppose that $x,y,z$ are positive integers satisfying $x \le y \le z$, and such that the product of all three numbers is twice their sum. What is the sum of all possible values of $z$?
I have only found $2$ sets of positive integers solutions satisfying all con... | $\frac{1}{2} = \frac{1}{yz} + \frac{1}{xz} + \frac{1}{xy} \leq \frac{3}{x^2}$.
Therefore, $x^2 \leq 6$ which only leaves $x=1$ and $x=2$
For $x=1$, you get $(x,y,z) = (1,3,8)$ and $(1,4,5)$. I'll leave it to you to prove that these are the only possibilities.
Hint: Use the same technique as above to get an upper bound ... | {
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If $f(x)$ is differentiable for all real numbers, then what is the value of $\frac{a+b+c}{2}$?
If $f(x)=\begin {cases} a^2 + e^x & -\infty <x<0 \\ x+2 & 0\le x \le 3 \\ c -\frac{b^2}{x} & 3<x<\infty \end{cases}$, where $a,b,c$ are positive quantities. If $f(x)$ is differentiable for all real numbers, then value of $\f... | First see my comment, and Peter Foreman's comments.
As $x \to 3^-$ (that is, as $x$ approaches 3 from below)
$f'(x) = 1.$
Also, $f(3) = 5.$
As $x \to 3^+$:
$f(x) \to c - \frac{b^2}{3}.$
$f'(x) \to \frac{b^2}{3^2}.$
Thus, $\frac{b^2}{3^2} = 1 \Rightarrow b^2 = 9.$
Further, $c - \frac{b^2}{3} = 5 \Rightarrow c - 3 = 5.$... | {
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How can I show the quotient of the $k$th partial sums of $\sum\limits_{n=1}^{k} n$ and $\sum\limits_{n=1}^{k} n^2$ is $\frac{3}{2k+1}$? I've found using pen and paper that any trivial case of the sum of a sequence of integers from $1$ to $k$ divided by the sum of the squares of these integers is equal to $\frac{3}{2k+... | As we know or we can easily prove, $\sum_{n=1}^kn = \dfrac{k(k+1)}{2}$
Now expanding, $(n-1)^3 = n^3 - 3n^2 + 3n - 1$
$n^3 - (n-1)^3 = 3n^2 - 3n + 1$
$ \sum_{n=1}^k n^3 - (n-1)^3 = 3\sum_{n=1}^kn^2 - 3\sum_{n=1}^kn + \sum_{n=1}^k1$
$ k^3 = 3\sum_{n=1}^kn^2 - 3\dfrac{k(k+1)}{2} + k$
$ 3\sum_{n=1}^kn^2 = k^3 + 3\dfrac{k(... | {
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Problem with factoring $x^4-x^3+x^2-x+1$ I want to calculate following integral Using partial fraction: $$\int{1\over x^5+1}dx$$So I decompose the denominator:
$$x^5+1=(x+1)(x^4-x^3+x^2-x+1)$$
For the next step I searched on internet and find out I should decompose$x^4-x^3+x^2-x+1$ like this:
$$x^4-x^3+x^2-x+1=(x^2-ax+... | In general, the two polynomials are given up to multiplication of a constant (you can multiply one by $k$ and other by $1/k$), so you can arrange it in a way that $a=d=1$ is guaranteed. For example $x^2+4x+4$ can be factored as $(x+2)(x+2)$ but also as $(2x+4)(\frac{1}{2}x+1)$. So we are free to fix one of the coeffici... | {
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"source": "stackexchange",
"question_score": "2",
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Complex Matrix is Orthogonal if and only if... Let D be a 2x2 matrix with entries in the complex numbers. Prove that D is orthogonal if and only if, it is of the form:
\begin{pmatrix}
a & -b\\
b & a
\end{pmatrix}
or
\begin{pmatrix}
a & b\\
b & -a
\end{pmatrix}
Proof. I've already proved that if D is equal to those fo... | To Prove: If A is a 2x2 orthogonal matrix, A is of the form given in the question.
Proof:
For any general non-singular matrix, $A= \begin{pmatrix}a & b \\
c & d\end{pmatrix}$,
$\begin{equation}\det(A) = ad-bc \end{equation}$ and
$A^T= \begin{pmatrix}a & c \\
b & d\end{pmatrix}$.
Using $ \begin{align}A A^T = I \end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Evaluate $\int e^{2x}(7+e^x)^{1/2}\,dx$ $\int e^{2x}(7+e^x)^{1/2}\,dx$
Let $u=7+e^x \rightarrow du=e^xdx$
So the integral becomes:
$\int u^{\frac{3}{2}}-7u^{\frac{1}{2}}du$ and so our answer is
$\frac{2}{5}(7+e^x)^{\frac{5}{2}}-\frac{14}{3}(7+e^x)^{\frac{3}{2}}+C$
But this is not what wolfram says. Did I make a mistake... | The integral becomes
\begin{align}
\int (u-7)u^{\frac{1}{2}}du
&=\frac{2}{5}(7+e^{x})^{\frac{5}{2}}-\frac{14}{3}(7+e^{x})^{\frac{3}{2}}+C\\
&=\frac{2}{15}(7+e^{x})^{\frac{1}{2}}(3(7+e^{x})^{2}-35(7+e^{x}))+C\\
&=\frac{2}{15}\sqrt{7+e^{x}}(7e^{x}+3e^{2x}-98)+C
\end{align}
as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3796828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can $\int_{0}^{2\pi} \frac{d\theta}{\sqrt{R^2+x^2-2Rx\cos\theta}},$ where $R$ and $x$ are positive constants, be solved using substitution? While I was finding the potential at a point in the plane of a uniformly charged ring, I got the following integral as the solution.
$$\int_{0}^{2\pi} \frac{d\theta}{\sqrt{R^2+x^2-... | Once even I tried to derive this result and got stuck on the same step. After searching many articles and abstracts on internet, I concluded that there is no simple analytical solution of this integral rather a lot of heavy computation is required.
The potential equation was like $$V=\int_0^{2\pi}k\cdot\frac{Q\ d\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3799421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Pointwise and uniform convergence of ${\sum_{n=1}^{+\infty}\left (1+\frac{1}{2n}\right )^{4n^2}x^n}$ I want to check the pointwise and uniform convergence of the below power series.
*
*$\displaystyle{\sum_{n=1}^{+\infty}\left (1+\frac{1}{2n}\right )^{4n^2}x^n}$
We have that \begin{equation*}\sqrt[n]{\left (1+\frac{1}... | You have
$$\ln \left[\left( 1 + \frac{1}{2n}\right)^{4n^2} \left( \frac{1}{e^2}\right)^n \right] = 4n^2 \ln \left( 1 + \frac{1}{2n}\right) - 2n = 4n^2 \left( \frac{1}{2n} - \frac{1}{8n^2} + o\left( \frac{1}{n^2}\right)\right) - 2n$$
$$= -\frac{1}{2} + o \left( 1\right)$$
So
$$\lim_{n \rightarrow +\infty} \left( 1 + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does $3\cos x + 4\sin x$ become $5\cos(x - \arctan\frac{4}{3})$? I'm not sure how any rule is being applied to manipulate the $4\sin x$
None of the double angle/compound angle formulas have the trig functions in this layout
| First observe that $\sqrt{3^2 + 4^2} = 5,$ and so $\sqrt{\left( \frac 3 5 \right)^2 + \left( \frac 4 5 \right)^2} = 1.$ So there is some angle $\varphi$ for which $\cos\varphi=\frac 3 5$ and $\sin\varphi=\frac 4 5.$ Since the sine and cosine of that angle are both positive, it is in the first quadrant. Its tangent is i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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A summation of a series based on the Fibonacci sequence. A sequence is defined as follows: $$a_n=a_{n-1}+a_{n-2}\ \forall \ n\geq3\ ,\ n\in Z. $$
If $a_1=a$ and $a_2=b$, find $$S=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{1}{a_{2i-1}a_{2i+1}}$$ in terms of a and b.
My Approach:
I first found the n-th term as $a_n=... | Telescopic summation works
Let a=b=1, for simplicity. Use the property of Fibonacci numbers $F_m$:
$$F_{m+1} F_{m+2} -F_m F_{m+3}=(-1)^m.$$ Here $A_m=F_m.$
Let $m=2(n-1)$, we get
$$A_{2n-1}A_{2n} - A_{2n-2} A_{2n+1}=1.$$
Hence $$\frac{1}{A_{2n-1} A_{2n+1}}= \frac{A_{2n-1}A_{2n} - A_{2n-2} A_{2n+1}}
{A_{2n-1} A_{2n+1}}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve $I^{\prime}(a)=\int_{0}^{\pi} \frac{2 a-2 \cos x}{1-2 a \cos x+a^{2}} d x$ I tried to compute $I^{\prime}(a)=\int_{0}^{\pi} \frac{2 a-2 \cos x}{1-2 a \cos x+a^{2}} d x$
\begin{equation}\begin{aligned}
&\int_{0}^{\pi} \frac{2 a-2 \cos x}{1-2 a \cos x+a^{2}} d x&\\
&=\int_{0}^{\pi} \frac{2(a-1)+4 \sin ^{2} \frac{x... | Well,your mistake is in the last step, I am assuming that you made the substitution $\frac{a+1 }{a-1}\tan x=u$ and when $-1<a<1$ the limits would be $-\infty$ and $0$.
And your integral would vanish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3804655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the inequality with the best possible $k= constant$ (with the condition $x^{2}+ y^{2}\leq k$).
Find the inequality with the best possible $constant$
*
*Given two non-negative numbers $x, y$ so that $x^{2}+ y^{2}\leq \frac{2}{7}$. Prove that
$$\frac{1}{1+ x^{2}}+ \frac{1}{1+ y^{2}}+ \frac{1}{1+ xy}\leq \frac{3}... | Hint.
After rotating CCW the inequality
$$
\frac{3}{\frac{1}{4} (a+b)^2+1}-\frac{1}{a^2+1}-\frac{1}{a b+1}-\frac{1}{b^2+1}\ge 0
$$
we have
$$
\frac{6}{y^2+2}-\frac{2}{-x^2+y^2+2}-\frac{2}{(x+y)^2+2}-\frac{2}{(x-y)^2+2}\ge 0\ \ \ \ \ \ (1)
$$
this has at the equality, the trace in blue shown in the figure below.
Thus to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3804775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Morrie's law with sines While it's trivial to prove $\prod_{k=0}^{n-1}\cos(2^kx)=\frac{\sin(2^nx)}{2^n\sin x}$, Wikipedia refers to a "similar" identity $\sin\tfrac{\pi}{9}\sin\tfrac{2\pi}{9}\sin\tfrac{4\pi}{9}=\frac{\sqrt{3}}{8}$. How does this generalize to a result for $\prod_{k=0}^{n-1}\sin(2^kx)$? Failing that, ho... | I can show you a proof of the similar identity, but I don't know a general formula for $\prod_{k=0}^{n-1}\sin(2^kx)$.
According to Showing $\sin{\frac{\pi}{13}} \cdot \sin{\frac{2\pi}{13}} \cdot \sin{\frac{3\pi}{13}} \cdots \sin{\frac{6\pi}{13}} = \frac{\sqrt{13}}{64}$,
we have that
$$\prod _{k=1}^{n-1}\,\sin \left({\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Have I followed through on this proof by induction correctly? We are asked to prove that $1^3+2^3+\cdots+n^3 = (1+2+\cdots + n)^2$ by induction.
Basis: $n=1 \\ 1^3 = 1^2 \quad\checkmark \\ n=2 \\ 1^3+2^3 = (1+2)^2 = 9 \quad\checkmark
$
We rewrite both series using the following summation notation:
$$\sum_{i=1}^{n}i^3 =... | Your last step is not justified, unless you just want to say that the result is trivial and be a bada%# mathematician. If not, then we need to recall that $$\sum_{i=1}^{k}i=\frac{k(k+1)}{2} $$
Continuing from the last line, $$\begin{align}\bigg(\sum_{i=0}^{k} i \bigg)^2 +(k+1)^3&= \frac{k^2(k+1)^2}{4} +(k+1)^3\\&=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Fast way to solve $4 = \sqrt[3] {x+10}-\sqrt[3] {x-10}$ The question is this:
$4 = \sqrt[3] {x+10}-\sqrt[3] {x-10}$
For some reason, I keep on getting 289/3, even though it is the wrong answer. This is from a timed test, and my way is wrong and extremely slow.
| Use this formula $$\boxed{(a-b)^3 = a^3-3ab(a-b)-b^3}$$
$$4 = \sqrt[3] {x+10}-\sqrt[3] {x-10}\;\;\;/^3$$
$$64 = x+10 -3\sqrt [3]{(x+10)(x-10)}(\underbrace{\sqrt[3] {x+10}-\sqrt[3] {x-10}}_{4})-x+10$$
So $$11 = -3\sqrt [3]{x^2-100}\implies x = \pm\sqrt{{-11^3\over 27}+100}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3812014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the power series of $\frac{3x+4}{x+1}$ around $x=1$. I'm trying to find the power series of
$$
\frac{3x+4}{x+1}
$$
around $x=1$.
My idea was to use the equation
$$
\left(\sum_{n\ge0}a_n (x-x_0)^n\right)\left(\sum_{k\ge0}b_k (x-x_0)^k\right) = \sum_{n\ge0}\left(\sum_{k=0}^n a_k b_{n-k} \right)(x-x_0)^n \tag{1}
$$
t... | To make life simpler, let $x=y+1$ $$A=\frac{3x+4}{x+1}=\frac{3y+7}{y+2}=3+\frac 1 {y+2}$$ Let $y=2t$
$$A=3+\frac 12 \frac 1 {1+t}$$
Make the expansion around $t=0$ and go back to $y$ and then $x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3812400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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$3x^2+y^2 \ge -ax(x+y)$, solve for $a$ so this inequality is true for all $x,y$ I started trying to reorganize the equation
$$3x^2+y^2\ge -ax(x+y)$$
$$3x^2+y^2+ax^2+axy\ge0$$
notice:$$\left(\frac {ax}2+y\right)^2=axy+\frac14x^2a^2+y^2$$
now i should be able to simply add $\frac14x^2a^2$ as this number is always bigger ... | We have that
$$3x^2+y^2\ge-ax(x+y) \iff(3+a)x^2+axy+y^2 \ge 0$$
which requires $a\ge -3$ for $y=0$ and for $y\neq 0$ by $\frac x y =t$ we have
$$\iff y^2\left((3+a)t^2+at+1\right) \ge 0 \iff a^2-4(3+a) \le 0 \iff a^2-4a-12 \le 0$$
which requires $-2\le a \le 6$.
Note that adding the factor $\frac14x^2a^2$ affects the i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3822701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Let $r, s, t$ be integers. Suppose $\gcd(a, b) = 1$. Let $d = \gcd(a − b, ra^2 + sab + tb^2 )$. Show that $d|(r + s + t)$. Let $r, s, t$ be integers. Suppose $\gcd(a, b) = 1$. Let $d = \gcd(a − b, ra^2 + sab + tb^2)$. Show that $d|(r + s + t)$.
Any hints for this question? I am stuck with it for days.
| Since $d$ divides $a - b$, there exists an integer $c$ such that $a = b +cd$. Since we also have $ra^2 + sab + tb^2 \equiv 0 \bmod d$, one gets
$$
ra^2 + sab + tb^2 = r(b +cd)^2 + s(b +cd)b + tb^2 \equiv 0 \bmod d
$$
whence
$$
(r+s+t)b^2 = rb^2 + sb^2 + tb^2 \equiv 0 \bmod d.
$$
Since $\gcd(a,b) = 1$, there exist by Bé... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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A tricky inequality: $n(n+1)^{\frac{1}{n}}+(n-2)n^{\frac{1}{n-2}}>2n,\ n\geq3.$ Let $H_n=1+\frac{1}{2}+\cdots+\frac{1}{n}$ be the $n-$th harmonic number,
it is not difficult to prove that:
(1) $n(n+1)^{\frac{1}{n}}<n+H_n,$ for $n>1$; (use AM-GM inequality)
(2) $(n-2)n^{\frac{1}{n-2}}>n-H_n$, for $n>2$. (Hint: $n=3$ is ... | By Bernoulli
$$n(n+1)^{\frac{1}{n}}+(n-2)n^{\frac{1}{n-2}}=\frac{n}{\left(1+\frac{1}{n+1}-1\right)^{\frac{1}{n}}}+\frac{n-2}{\left(1+\frac{1}{n}-1\right)^{\frac{1}{n-2}}}\geq$$
$$\geq\frac{n}{1-\frac{1}{n+1}}+\frac{n-2}{1+\frac{1}{n-2}\cdot\frac{1-n}{n}}=n+1+\frac{n(n-2)^2}{n^2-3n+1}=$$
$$=2n+\frac{1}{n^2-3n+1}>2n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3829630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding $x$ when $\sqrt{3x - 4} + \sqrt[3]{5 - 3x} = 1$ Find $x$ if $\sqrt{3x - 4} + \sqrt[3]{5 - 3x} = 1.$
I was thinking of trying to substitute some number $y$ written in terms of $x$ than solving for $y$ to solve for $x.$ However, I'm not sure what $y$ to input, so can someone give me a hint?
| Let $z:=\sqrt{3x-4}\ge0$. We have
$$z+\sqrt[3]{1-z^2}=1$$
or by cubing,
$$(z-1)^3+(1-z^2)=z^3-4z^2+3z=z(z-1)(z-3)=0.$$
This yields three solutions in $x$,
$$\frac43,\frac53,\frac{13}3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3830329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $\frac{1+\cos\theta}{\sec\theta-\tan\ \theta}+\frac{\cos\theta-1}{\sec\theta+\tan\ \theta}=2+2\tan\ \theta$
Prove this trigonometric identity:
$$\frac{1+\cos\theta}{\sec\theta-\tan\ \theta}+\frac{\cos\theta-1}{\sec\theta+\tan\ \theta}=2+2\tan\ \theta$$
I've simplified it until
$$\frac{2\cos^2\theta}{1-\sin\th... | We have that
$$\frac{1+\cos\theta}{\sec\theta-\tan\theta}+\frac{\cos\theta-1}{\sec\theta+\tan\theta}=\frac{\cos\theta(1+\cos\theta)}{1-\sin\theta}+\frac{\cos\theta(\cos\theta-1)}{1+\sin\theta}=$$
$$=\frac{\cos\theta(1+\cos\theta)(1+\sin\theta)+\cos\theta(\cos\theta-1)(1-\sin\theta)}{1-\sin^2\theta}=$$
$$=\frac{(1+\cos\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3830933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Minimum of a function without calculus. $a=\frac{{(1+t^2)}^3}{t^4}$
Find the the minimum value of $a$. .$$a=\frac{{(1+t^2)}^3}{t^4}$$.
Instead of calculus i tried using the AM-GM inequality.,as follows:we have $$3+\frac{1}{t^4}+\frac{3}{t^2}+3\left( \frac{t^2}{3}\right)\ge 3+{\left(\frac{1}{9}\right)}^{1/5}$$ which ... | A very short proof that the minimal value is $\frac{27}4$ can be given by noting that $$(1+t^2)^3-\frac{27}4 t^4=\frac{(\sqrt 2 - t)^2 (\sqrt 2 + t)^2 (1 + 4 t^2)}4\geq0$$
with equality if and only if $t=\sqrt 2$ or $t=-\sqrt 2$.
EDIT: Systematically we can proceed as follows: For any $a>0$, we have by weighted AM-GM,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding cube roots of a unity - proper explanation is needed SQUARE ROOT
Let's say that I have:
\begin{align*}
x^2 &= 9
\end{align*}
And I $\sqrt{\phantom{x}}$ an entire equation to get:
\begin{align*}
x_{1,2} &= \sqrt{9}\\
x_{1} &= +3\\
x_{2} &= -3
\end{align*}
Experience and "fundamental theorem of algebra" thought m... | There is really no "first" root, "second" root, etc.
In the case of $x^2=1$, there are two roots, $-1$ and $1$. It is your choice to give these two roots whatever names you like.
In the case of $x^3=1$, there are three distinct roots. Again, it does not matter which one you call $x_1$.
In general, (complex) solutions t... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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There are triads of perfect squares that are consecutive terms of arithmetic progression? Prove that there exist infinitely many triples of positive integers $ x , y , z $ for which the numbers $ x(x+1) , y(y+1) , z(z+1) $ form an increasing arithmetic progression.
$ \bigg( $ It is equivalent to find all triples of $ 4... | Looking at your examples,
$$ 2(5^2)=1^2+7^2=(4-3)^2+(4+3)^2 \rightarrow (3,4,5) $$
$$ 2(13^2)=7^2+17^2=(12-5)^2+(12+5)^2 \rightarrow (5,12,13) $$
$$ 2(17^2)=7^2+23^2=(15-8)^2+(15+8)^2 \rightarrow (8,15,17) $$
$$ 2(25^2)=17^2+31^2=(24-7)^2+(24+7)^2 \rightarrow (7,24,25) $$
Use any Pythagorean triplet such that $... | {
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"timestamp": "2023-03-29T00:00:00",
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To Prove $\frac{1}{b}+\frac{1}{c}+\frac{1}{a} > \sqrt{a}+\sqrt{b}+\sqrt{c}$ The three sides of a triangle are $a,b,c$, the area of the triangle is $0.25$, the radius of the circumcircle is $1$.
Prove that $1/b+1/c+1/a > \sqrt{a}+\sqrt{b}+\sqrt{c}$.
what I've tried:
$$\frac{1}{4} = \frac{1}{2}ab\sin C \Rightarrow ab=\fr... | $ \sum_{cyc}\frac{1}{a} = \frac{1}{2} \sum_{cyc}\frac{1}{a} + \frac{1}{b} \ge \sum_{cyc}\frac{1}{\sqrt {ab}} = \sum_{cyc}\sqrt {a} \,$ (using AM-GM and $abc = 1)$
EDIT: I think I missed an important point earlier that when $a = b = c = 1$, R = $\frac{1}{2\sin 60^0} = \frac{1}{\sqrt3}$ which does not meet the condition... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3835903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How prove this limit via epsilon delta? $$\lim_{(x,y) \to (4,1)}{\frac{y}{2x-y}}=\frac{1}{7}$$
I know so far that $|x-4|<\delta\quad \&\quad |y-1|<\delta$, and that I can use
$$\bigg|\frac{y-1}{2x-y} + \frac{1}{2x-y} - \frac{1}{7}\bigg|$$ and I get one delta, but how to continue from here, or is it even correct?
| To simplify we can change coordinates $u=x-4$ and $v=y-1$ such that
$$\lim_{(x,y) \to (4,1)}{\frac{y}{2x-y}}=\lim_{(u,v) \to (0,0)}{\frac{v+1}{2u-v+7}}=\frac{1}{7}$$
and assuming wlog $|u|,|v|\le 1$
$$\left|\frac{y-1}{2x-y} - \frac{1}{7}\right|=\left|\frac{v+1}{2u-v+7} - \frac{1}{7}\right|=\left|\frac{8v-2u}{7(2u-v+7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $\{a_n\}$ be a sequence of real numbers such that $a_1=2$, $a_{n+1} = a_n^2 -a_n+1$, for $n=1,2,3..$. [Cont]
Let $\{a_n\}$ be a sequence of real numbers such that $a_1=2$, $a_{n+1} = a_n^2 -a_n+1$, for $n=1,2,3..$. Let $S=\frac{1}{a_1}+\frac{1}{a_2} ....+\frac{1}{a_{2018}}$, then prove that
*
*$S>1-\frac{1}{20... | You have already shown that
$S = 1-\frac{1}{a_{2019}-1}$
So, $S \gt 1-\frac{1}{2018^{2018}}$ if $\, a_{2019} \gt {2018}^{2018} + 1$
By induction -
As $a_{2} = {a_1}^2 - a_1 + 1 = 3 \gt 1^1 + 1$
and $a_{3} = {a_2}^2 - a_2 + 1 = 7 \gt 2^2 + 1$
For a value of $n \ge 3$,
If $a_{n+1} \gt n^n + 1$, we need to show that $a_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3839954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Rewriting this equation without the square roots? Given the following system of equations:
$\sqrt{(x-x_1)^2+(y-y_1)^2}+s(t_2-t_1) = \sqrt{(x-x_2)^2 + (y-y_2)^2}$
$\sqrt{(x-x_2)^2+(y-y_2)^2}+s(t_3-t_2) = \sqrt{(x-x_3)^2 + (y-y_3)^2}$
$\sqrt{(x-x_3)^2+(y-y_3)^2}+s(t_3-t_1) = \sqrt{(x-x_1)^2 + (y-y_1)^2}$
How could I writ... | $$
\sqrt{a^2}+b = \sqrt{c^2} \\
\implies a^2 + b^2 + 2b\sqrt{a^2} = c^2 \\
\implies a^2 + b^2 + 2b(\sqrt{c^2}-b) = c^2 \\
\implies \sqrt{c^2} = \frac{c^2 -a^2 - b^2}{2b} + b \\
\implies c^2 = \left(\frac{c^2 -a^2 - b^2}{2b} + b \right)^2
$$
$$
\iff c = \pm \left(\frac{c^2 -a^2 - b^2}{2b} + b \right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Troubles finding the Fourier series of a sawtooth function plus a cosine function I have a function $y(x) = y_1(x) + y_2(x)$ consisting of two other waveforms, where
$ y_1(x) = \cos{\left(\dfrac{16 \pi}{5} x \right)}; \, y_2(x) = \displaystyle \sum_{k=-\infty}^{\infty} y_3(x - k); \, y_3(x) = \begin{cases} x & \text{if... | I find it suspicious that $a_n=0$ since the function $y-\dfrac{1}{2}$ is not odd. According to your calculations $y-\dfrac{1}{2}$ is a pure sinus wave.
You should check the calculations of $a_n$ when $n=8$, the integral now involves
$$\cos^2\left(\dfrac{16\pi t}{5}\right)$$
and is unlikely to vanish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3845129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Find the values of $x$ for which $x^{12}-x^9+x^4-x+1>0$
Find the values of $x$ for which $x^{12}-x^9+x^4-x+1>0$.
I tried to substitute some basic values like $-1,0,1$ and try to find the roots of the function but couldn't.
Then I graphed the function on desmos and this is the graph.
So from this, we can say that $x^{... | Another way.
For $x\leq0$ it's obvious.
But for $x>0$ by AM-GM we obtain:
$$x^{12}-x^9+x^4-x+1=x^{12}-x^9-x^4+x+2x^4-2x+1=$$
$$=x(x^8-1)(x^3-1)+2x^4+3\cdot\frac{1}{3}-2x\geq$$
$$\geq4\sqrt[4]{2x^4\cdot\left(\frac{1}{3}\right)^3}-2x=2\left(\sqrt[4]{\frac{32}{27}}-1\right)x>0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3845812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
use mathematical induction to show that $n^3 + 5n$ is divisible by $3$ for all $n\ge1$ What I have so far
Base: $n^3 + 5n$
Let $n=1$
$$
1^3 + 5(1) = 6
$$
$6$ is divisible by $3$
Induction step: $(k+1)^3 + 5(k+1)$
$(k^3 + 3k^2 + 8k + 6)$ is divisible by $3$
I kind of get lost after this point. For starters, how do I pro... | Just note that
$$
f(n+1)-f(n)=(n+1)^3 + 5(n+1)-(n^3 + 5n)=3 (n^2 + n + 2)
$$
I suspect this is the intended solution.
Actually, no induction is needed:
$$
n^3 + 5n = 6 \binom{n}{1} + 6 \binom{n}{2} + 6 \binom{n}{3}
$$
and so is always a multiple of $6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3846655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
How to solve such a system of quadratic equations: $x^2+y^2-xy=a^2, x^2+z^2-xz=b^2, y^2+z^2-yz=c^2$ I don't know how to solve this this system:
$$x^2+y^2-xy=a^2\\x^2+z^2-xz=b^2\\y^2+z^2-yz=c^2$$ The system of quadratic equation in symmetry form has many geometric meaning,
this seems to be a triangular pyramid, three ad... | Solve the first and second equations as quadratic in $y$ and $z$ to get four pairs of solutions
$$y=\frac{1}{2} \left(x-\sqrt{4 a^2-3 x^2}\right)\qquad z=\frac{1}{2}
\left(x-\sqrt{4 b^2-3 x^2}\right)$$
$$y=\frac{1}{2} \left(x-\sqrt{4 a^2-3 x^2}\right)\qquad z=\frac{1}{2}
\left(x+\sqrt{4 b^2-3 x^2}\right)$$
$$y=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3849602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Where is the mistake when finding $\int x \sqrt{4+5 x} \ dx$? We have to find the integral of $\frac{dy}{dx}=x \sqrt{4+5x}$. This is from Morris Kline's book, Chapter $7$, exercise $5$, question $1$m.
I try to solve like following:
Let $u=4+5x$. Then $x=\frac{u-4}{5}$. Hence $\frac{d y}{d x}=\left(\frac{u-4}{5}\right) ... | Remember change of variables requires substituting the differentiable:
$x = \frac{u - 4}{5} \rightarrow dx = \frac{1}{5} du$
So really after substitution you should have:
$\int{\frac{u-4}{5}(\sqrt{u})(\frac{1}{5}) du}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3850797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Comparing $(2+\frac{1}{2})(3+\frac{1}{3})(4+\frac{1}{4})(5+\frac{1}{5})$ with $(2+\frac{1}{5})(3+\frac{1}{4})(4+\frac{1}{3})(5+\frac{1}{2})$ This question appeared in one of the national exams (MCQs) in Saudi Arabia.
In this exam;
*
*Using calculators is not allowed,
*The student have $72$ seconds on average to answ... | My intuition is to notice that
$$\frac{2+\frac{1}{2}}{2+\frac{1}{5}}\ ?\ \frac{5+\frac{1}{2}}{5+\frac{1}{5}}$$
(I have used $?$ since I do not know how these expressions relate)
$$\frac{5}{2}\frac{5}{11}\ ?\ \frac{11}{2}\frac{5}{25}$$
$$\frac{25}{22}\ ?\ \frac{55}{50}$$
$$\frac{50}{44}\ ?\ \frac{55}{50}$$
Since $50-44=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3852464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
The tangent at $(1,7)$ to the curve $x^2=y-6$ touches the circle $x^2+y^2+16x+12y+c=0$ at...
The tangent at $(1,7)$ to the curve $x^2=y-6$ touches the circle $x^2+y^2+16x+12y+c=0$ at...
What I tried...
The equation $x^2=y-6$ is of a parabola. To find the slope of the tangent to the parabola at the point $(1,7)$,
$$\... | Because the line "touches" the circle, there is only one point of intersection with the circle. Therefore, the equation,
$$
x^2+(2x+5)^2+16x+12(2x+5)+c=0 \implies 5 x^2 + 60 x + 85 + c= 0
$$
has one solution.
So the determinant $\Delta = 3600 - 4\cdot5\cdot(85+c)=0$. Ergo, $c=95$.
And solving the equation $5x^2 + 60x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3853199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}+\frac{c}{1+a^2+b^2+c^2}+\frac{d}{1+a^2+b^2+c^2+d^2}\leq\frac{3}{2}$
For any reals $a$, $b$, $c$ and $d$ prove that:
$$\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}+\frac{c}{1+a^2+b^2+c^2}+\frac{d}{1+a^2+b^2+c^2+d^2}\leq\frac{3}{2}$$
C-S in the IMO 2001 stile does not help here:
\... | A proof for $n=2$.
We need to prove that:
$$\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}\leq\sqrt{\frac{207+33\sqrt{33}}{512}}.$$
Since $x\leq|x|$ and for $ab=0$ it's obvious, it's enough to prove this inequality for positive variables.
Now, by AM-GM
$$\frac{b}{1+a^2+b^2}=\frac{1}{\frac{1+a^2}{b}+b}\leq\frac{1}{2\sqrt{1+a^2}}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3853444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
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Fractions in Questions and Answers
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