Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Need help proving an interval $\frac {1} {ek} \le \frac {1}{k} (1 - \frac {1}{k} )^{k-1} \le \frac {1}{2k}$ I am trying to proof
$$\frac {1} {ek} \le \frac {1}{k} (1 - \frac {1}{k} )^{k-1} \le \frac {1}{2k} $$
for k>=2
to prove this I first multiply by k getting
$$\frac {1} {e} \le \left(1 - \frac {1}{k} \right)... | Use the well-known inequality proved here: https://math.stackexchange.com/a/1161287/148510.
For $0 < x < 1$,
$$1-x \leqslant -\ln x \leqslant \frac{1-x}{x}.$$
Then with $x = 1-1/k$, we have
$$\ln(1 -1/k) \geqslant \frac{-\frac1{k}}{1-\frac1{k}}=\frac{-1}{k-1}.$$
Hence,
$$(k-1)\ln(1 -1/k) \geqslant -1 \implies \left(1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Find the maximum and minimum of $x^2+2y^2$ if $x^2-xy+2y^2=1$.
Find the maximum and minimum of $x^2+2y^2$ if $x,y\in\mathbb R$ and $$x^2-xy+2y^2=1$$
My attempt:
Clearly, since $x^2-x(y)+(2y^2-1)=0$ and $2y^2-y(x)+(x^2-1)=0$, we have that
$$\Delta_1=y^2-8y^2+4=4-7y^2\ge 0$$
and
$$\Delta_2=x^2-8x^2+8=8-7x^2\ge 0$$
so... | An inequality approach: using $a^2+b^2\geq 2ab$ inequality below, we have
$$
x^2-xy+2y^2=1\implies 1+xy=x^2+2y^2\geq2\sqrt{2}xy\implies xy\leq\frac{1}{2\sqrt{2}-1}\cdot
$$
It follows that
$$
x^2+2y^2=1+xy\leq 1+\frac{1}{2\sqrt{2}-1}=
\boxed{\frac{2\sqrt{2}}{2\sqrt{2}-1}}\cdot
$$
Similarly, using $a^2+b^2\geq -2ab$, we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Finding polynomial Coefficients
Let $f(x) = x^5 - x^4 + ax^3 + bx^2 + 8x + 4$
The root will make sure, $f(2) = 0$ Which shows:
$$2^5 - 2^4 + a2^3 + 4b + 16 + 4 = 0$$
$$16 + 8a + 4b + 20 = 0 \implies 8a + 4b = -36 \implies 2a + b = -9$$
It also follows that $f'(2) = 0$ but:
$$f'(x) = 5x^4 - 4x^3 + 3ax^2 + 2bx + 8$$
$$f... | Your idea is correct, but you made a tiny mistake in solving the system. $a=-5$ is correct, but $b$ should be equal to $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1169442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimizing the product $xy$ subject to a polynomial constraint on $x, y$
Given that $$16y(x^2+1)=25x(y^2+1),$$ where $x,y$ are positive integers, find the smallest possible value of $xy$.
I wrote my expression as a quadratic in $x$ and calculated it in form of $y$. Then $xy$ became in $y$ but it had square roots, so ... | Let's first note that $x > y > 1.$ Since $(y, y^2+1) = 1,$ we have $y\mid 25x,$ and similarly, $x\mid 16y.$ Suppose that $5\not\mid y,$ so that $y\mid x.$ Writing $x = ky$ yields $ky\mid 16y \Rightarrow k\mid 16,$ and clearly $k\not = 1.$ Hence we now have $16y(k^2y^2+1) = 25ky(y^2+1),$ which after rearranging becomes ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1171417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Parametrizing intersection of a plane and surface I'm working on…
Parametrize the curve which is the intersection of the plane $2x+4y+z=4$ with the surface $z=x^2+y^2$.
I tried eliminating $z$ by plugging it into the first equation and also tried parametrizing each equation before finding the intersection but was not... | Step 1: As you suggest, eliminate $z$ by plugging it into the first equation:
$$2x + 4y + x^2 + y^2 = 4$$
Now, complete the square. You get the equation of a circle:
$$(x + 1)^2 - 1 + (y + 2)^2 - 4 = 4$$
$$(x + 1)^2 + (y + 2)^2 = 9$$
This is a circle with center $(-1,-2)$ and radius $3$. So $x = -1 + 3\cos(t)$ and $y =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1172384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the slope of the tangent line to the graph of $f^{-1}$ Given function $f$, find the slope of the line tangent to the graph $f^{-1}$ at the point on the graph $f^{-1}$. $f(x)=\sqrt{5x}$; $(4,\frac{16}{5})$?
Here is what I have thus far:
$f'(x)= \frac{\sqrt{5}}{2\sqrt{x}}$
can it also be $f'(x)= \frac{2.5}{\sqrt{5x... | y= f(x) = $\sqrt{5x}$ $\rightarrow$ $y^2$ = $5x$ $\rightarrow$ $\frac{1}{5}y^2$ = x
So $f^{-1}(x)$ = $\frac{1}{5}x^2$ , which is easily verified by taking $f^{-1}(f(x))$= x.
Now:
$(f^{-1})'(x)$ = $\frac{2}{5}x$. The slope of the tangent line of $f^{-1}$ at (4,$\frac{16}{5}$) is $(f^{-1})'(4)$ = $\frac{2}{5}*4$ =$\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1174350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all solutions to $x^{10} = 1 \pmod {377}$ Find all solutions to $x^{10} \equiv 1 \pmod {377}$
I noticed that $x^{10} \equiv 1 \pmod {377}$ can be written as:
$(x^5+1)(x^5-1)\equiv 0 \pmod {377}$
also $377 = 13 \times 29$
Any help would be greatly appreciated
| Hint: by the Chinese remainder theorem, we can find all solutions to this equation by first finding all solutions to
$$
x^{10} \equiv 1 \pmod{13}\\
x^{10} \equiv 1 \pmod{29}
$$
To find the solutions mod $13$, it is slightly useful to note that $x^{10} = x^{12 - 2} = x^{-2}$.
Full solution: The first equation has solut... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1174606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Number of solutions of the equation $x_1 + x_2 + x_3 = 9$ How to calculate the number of solutions of the equation $x_1 + x_2 + x_3 = 9$ when $x_1$, $x_2$ and $x_3$ are integers which can only range from 1 to 6.
| We can find the number of solutions using binomial theorem.
The coefficient of $x^9$ in the following will be the required answer.
$$(x+x^2+\cdots+x^6)^3$$
This above, is a Geometric Progression. Therefore,
$$=\left (\frac{x-x^7}{1-x}\right )^3$$
$$=(x-x^7)^3(1-x)^{-3}$$
Now apply binomial theorem to get the coefficien... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1178080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Tough inequality in positive reals numbers. Let $a, b, c$ be positive real. prove that
$$\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\geq\left(1+\frac{a+b+c}{\sqrt[3]{abc}}\right)$$
Thanks
| The inequality is unchanged if you multiply $a$, $b$, and $c$ by a positive number $k$, so WLOG $abc = 1$. Then the left hand is greater than or equal to
$$1 + \frac{a}{b} + \frac{b}{c} + \frac{c}{a}$$
and $$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge a + b + c$$
since
\begin{align}&\frac{a}{b} + \frac{b}{c} + \frac{c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1178979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solve $a_{n+1} - a_n = n^2$ using generating functions The Full Question
Using the method of generating functions, solve $a_{n+1} - a_n = n^2$ where $a_0 = 1$
My Research
Scanned the website for similar answers, reviewed the following links:
Solve the following recurrences using generating functions.
can we use generat... | You made a mistake in solving
$$ x(1+x)= A(1-x)^3+B(1-x)^2+C(1-x)+D$$
A faster way to solve this is the following:
$$x = 1 \Rightarrow D=2$$
Next derivate
$$1+2x=-3A(1-x)^2-2B(1-x)-C$$
plug in $x=1$. Then derivate again, and plug in $x=1$. Continue...
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Minimise $x+y$ under the condition $2x+32y \leq 9xy$ where $x,y > 0$ Let $x$ and $y$ be positive real numbers such that $2x+32y\leq 9xy$. Then find the smallest possible value of $x+y$. I tried using AM-GM but it is difficult to find the case of equality. Is Cauchy-Schwarz also helpful? Thanks.
| Rearranging the inequality yields
\begin{align*}
0 &\le 9xy -2x -32y \\
\frac{64}9 &\le 9xy -2x -32y + \frac{64}9 = \bigg(3 x - \frac{32}3\bigg) \bigg(3 y - \frac23\bigg).
\end{align*}
The AM-GM inequality* now gives
$$
\frac83 \le \sqrt{\bigg(3 x - \frac{32}3\bigg) \bigg(3 y - \frac23\bigg)} \le \frac12 \bigg(3 x - \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The $ABCD$ paralelograms sides are $AB,BC,CD,DA$. On these line segments there are points in the same order: $X,Y,Z,V$. The $ABCD$ paralelograms sides are $AB,BC,CD,DA$. On these line segments there are points in the same order: $X,Y,Z,V$. We know, that: $$\frac{AX}{XB}=\frac{BY}{YC}=\frac{CZ}{ZD}=\frac{DV}{VA}=k$$
$k$... | Let $AB = CD = a$ and $BC = AD = b$. Then we have:
$$\frac{AX}{XB} = k \implies \frac{a}{XB} = k+1 \implies XB = \frac{a}{k+1}$$
From this we obtain that:
$$AX = \frac{ka}{k+1}$$
We get simular results for all other segments. Now check that $XYZV$ is inscribed in $ABCD$ and we have:
$$P_{ABCD} - P_{XBY} - P_{YCZ} - P_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1183946",
"timestamp": "2023-03-29T00:00:00",
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Prove that $ab \leq \frac14$ and $ (1+1/a)(1+1/b)\ge 9$ when $a+b=1, a \ge 0, b \ge 0$ Our teacher gave us some identities firstly and said we could use one of them to prove it. The identities are: $$\frac{(a^2+b^2)}{2}≥\left(\frac{(a+b)}{2}\right)^2$$ $$(x+y)^2≥2xy$$ and $$\frac{(x+y)}{2} \ge \sqrt{xy}$$
Here's what ... | Subtract the inequality from the equation
$$
1=(a+b)^2=a^2+2ab+b^2\\
0\le(a-b)^2=a^2-2ab+b^2
$$
to get
$$
1\ge4ab\implies\frac14\ge ab\tag{1}
$$
Apply $(1)$ to get
$$
\begin{align}
\left(1+\frac1a\right)\left(1+\frac1b\right)
&=1+\frac1a+\frac1b+\frac1{ab}\\
&=1+\frac{a+b}{ab}+\frac1{ab}\\[3pt]
&=1+\frac2{ab}\\[3pt]
&\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1185465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Prove $(1 + \frac{1}{n})^n$ is bounded above I've checked similar questions on the site but couldn't find satisfactory solutions or hints.
Also, is there a more general approach to proving whether a given sequence is bounded below or above?
| By the binomial formula:
\begin{eqnarray*}
\left(1+\frac{1}{n}\right)^n=1+1+\sum_{k=2}^n\binom{n}{k}\cdot\left(\frac{1}{n}\right)^k.
\end{eqnarray*}
Notice that
\begin{eqnarray*}
\binom{n}{k}\cdot\left(\frac{1}{n}\right)^k=\frac{n(n-1)\cdots(n-k+1)}{n^k}\cdot\frac{1}{k!}<\frac{1}{k(k-1)}=\frac{1}{k-1}-\frac{1}{k}.
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1185660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 0
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Find the derivative of the function. y = $\sqrt{7x+\sqrt{7x+\sqrt{7x}}} $ This question is really tricky. I am wondering if I am right?
| $$\dfrac{dy}{dx}=\dfrac{d\left( \sqrt{7x+\sqrt{7x+\sqrt{7x}}}\right) }{dx}\\
=\dfrac{1}{2\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}\dfrac{d\left( 7x+\sqrt{7x+\sqrt{7x}}\right) }{dx}\\
=\dfrac{1}{2\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}\left ( 7+\dfrac{d\left(\sqrt{7x+\sqrt{7x}}\right) }{dx}\right)\\
=\dfrac{1}{2\sqrt{7x+\sqrt{7x+\sqrt{7x}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1190891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Fourier Series Expansion, error in coefficients? After reworking the problem many times I keep getting the same (incorrect?) answer.
So the problem as stated is
Find the Fourier expansion of :
$$
f(x) = \begin{cases}
x &\text{ if }0 < x < \pi,\\
2(\pi - x) &\text{ if }\pi < x < 2\pi
\end{cases}
$$
I get t... | Consider the shifted variant $f(x)=\pi-|x|$ on $[-π,π]$ as fundamental period.
Then, since it is an even function, $b_n=0$ and
\begin{align}
a_0&=\frac1{\pi}\int_0^\pi(\pi-x)dx\\&=\frac1{2\pi}[-(\pi-x)^2|_0^\pi=\frac{\pi}2
\\
a_n&=\frac{2}{\pi}\int_0^\pi(\pi-x)\cos(nx)dx
\\&=\frac2{n\pi}[(\pi-x)\sin(nx)]_0^\pi+\frac2{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Which mixed numbers have the property $\sqrt{n + \frac{p}{q}}=n\sqrt{\frac{p}{q}}$? Could I please have help with describing mixed numbers (aka mixed fractions) that have this property:
Show that $\sqrt{9\frac{9}{80}}=9\sqrt{\frac{9}{80}}$ and $\sqrt{4\frac{4}{15}}=4\sqrt{\frac{4}{15}},\;$ where $\sqrt{9\frac{9}{80}} =... | HINT
For $x,y\in \mathbb{Z^{+}}$
$$\begin{align}\sqrt{x\dfrac{x}{y}} &= x\sqrt{\dfrac{x}{y}}\\ &\iff \\\sqrt{x+\dfrac{x}{y}}&=\sqrt{\dfrac{x^3}{y}}\\\sqrt{\dfrac{\color{blue}{x(y+1)}}{y}} &=\sqrt{\dfrac{\color{blue}{x^3}}{y}}\end{align}$$
Comparing you get $$\color{blue}{y = x^2-1}$$
That means it works for all $x,y$ t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why is $\sqrt{3}=[1;1,2,1,2,\dots]$?
Why is $\sqrt{3}=[1;1,2,1,2,\dots]$ ?
$\displaystyle[1;1,2,1,2,\dots]=1+\frac{1}{[1;2,1,2,\dots]}=1+\frac{1}{1+\frac{1}{2+\frac{1}{[1;2,1,2,\dots]}}}$
If I set $x=[1;2,1,2,\dots]$ then;
$\frac{x+1}{x}=\frac{5x+2}{3x+1}$
with solutions $x_{1,2}=\frac{1\pm\sqrt{3}}{2}$
what did go w... | Let's begin by computing $y=[1;2,1,2,\cdots]$. This is $$1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\ddots}}}.$$
Now, $\frac{1}{y-1}$ equals $$2+\frac{1}{1+\frac{1}{2+\ddots}}.$$
Then, $$\frac{1}{\frac{1}{y-1}-2}=y.$$
Simplifying the fractions results in $2y^2-2y-1$. Solving for $y$ results in $\frac{1\pm\sqrt{3}}{2}$.
Since... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Algebraic manipulation with $e^x$ I have these inequalities;
$ 1 + x \le e^x \le \frac{1}{1-x} $ for $|x| <1$
This is what i require;
$ 1 - \frac{|x|}{1-x} \le e^x \le 1 + \frac{|x|}{1-x} $ for $0 < x <1$
This is my attempt;
$ 1 + x \le e^x \le \frac{1}{1-x} $ for $|x| <1$
$\implies x \le e^x - 1 \le \frac{1}{1... | Note that for $x\in(0,1)$
$$
e^x\leq\frac{1}{1-x}=\frac{1-x+x}{1-x}=1+\frac{x}{1-x}=1+\frac{|x|}{1-x}\cdot
$$
This takes care of the the inequality on the right. For the one on the left, (still assuming $x\in(0,1)$)
$$
e^x\geq 1+x>1>1-\underbrace{\frac{|x|}{1-x}}_{+}\cdot
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove or disprove $n \geq 2 ~\rightarrow~ \prod \limits_{i=1}^{n} \left ( 1 - \frac{1}{i^2} \right ) ~=~ \frac{n+1}{2n}$ I am working on one of my HW assignments
$$
\forall n \in \mathbb{Z}, ~
n \geq 2
~\rightarrow~
\prod \limits_{i=1}^{n} \left (
1 - \frac{1}{i^2}
\right )
~=~
\frac{n+1}{2n}
$$
And i am not cl... | \begin{align}
\prod \limits_{i=1}^{n} \Big(1 - \frac{1}{i^2}\Big)&=\prod \limits_{i=1}^{1} \Big(1 - \frac{1}{i^2}\Big) \times \prod \limits_{i=2}^{n} \Big(1 - \frac{1}{i^2}\Big)\\
&=\prod \limits_{i=1}^{1} \Big(1 - \frac{1}{i^2}\Big) \times \frac{\prod_{i=2}^{n}(i^2-1)}{\prod_{i=2}^{n}i^2}\\
&=\prod \limits_{i=1}^{1} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1196627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show by induction: $ \frac{a_{2n+4}}{a_{n+2}}=\frac{a_{2n+2}}{a_{n+1}}+\frac{a_{2n}}{a_{n}}$ I would appreciate if somebody could help me with the following problem:
Q: Sequence $\{a_n\}$; satisfy $a_{n+2}=a_{n+1}+a_{n}, a_1=1,a_2=1$
Show by induction:
$$ \frac{a_{2n+4}}{a_{n+2}}=\frac{a_{2n+2}}{a_{n+1}}+\frac{a_{2n}}... | We are give that $a_{n+2}=a_{n+1}+a_{n}$, $a_1=1$, and $a_2=1$.
Thus, it is easy to see that $a_3=a_2+a_1=1+1=2$, $a_4=a_3+a_2=2+1=3$.
Step 1:
Let's use induction on $n$ to show that for $m\ge1$, $n\ge2$, we have
$$a_{n+m}=a_{n}a_{m+1}+a_{n-1}a_{m}$$
For the first benchmark, let $n=2$. Note that
$$a_{m+2}=a_{m+1}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1197120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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INMO Problem with even function proof.
Let $n$ be a natural number. Show that
$$\left[ \frac{n}{1} \right ] + \left[ \frac{n}{2} \right ] + \left[ \frac{n}{3} \right ] + \cdots + \left[ \frac{n}{n} \right ] + [\sqrt{n}]$$ is even. Here $[x]$ means the greatest integer less than or equal to $x$.
Let $$f(n) = [\sqrt... | Hint: Prove by induction.
When is $ \lfloor \frac{n}{k} \rfloor \neq \lfloor \frac{n+1}{k} \rfloor $?
When they are not equal, what is the difference of these 2 terms?
Ans: The terms are different if there exists some integer $i$ such that $ \frac{n}{k} < i \leq \frac{ n+1}{k}$. Multiplying throughout by $k$, we get $ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solving a radical equation for real roots I'm attempting to solve the derivative of my function $f(x)$ for real roots.
$$
\\ \begin{align*}
\\ f(x) &= 3x^2 + 3\arcsin{x}
\\ f^{\prime}(x) &= 6x + \dfrac{3}{\sqrt{1-x^2}}
\\
\\ 0 &= 6x + \dfrac{3}{\sqrt{1-x^2}}
\\
\\ 0 &= 6x \sqrt{1-x^2} + 3
\\ 0 &= \sqrt{36x^2 - 36x^4} ... | Your work is correct, $1/\sqrt 2$ is indeed a root of $-4x^4+4x^2-1$. The rational root test just tells you that the only possible rational roots of your polynomial are $\pm 1$, $\pm 1/2$ and $\pm 1/4$, it doesn't tell you anything about possible irrational roots, so there's no contradiction. For example, the rational ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1202949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
A quick way to prove the inequality $\frac{\sqrt x+\sqrt y}{2}\le \sqrt{\frac{x+y}{2}}$ Can anyone suggest a quick way to prove this inequality?
$$\frac{\sqrt x+\sqrt y}{2}\le \sqrt{\frac{x+y}{2}}$$
| Divide by $\sqrt y$ and let $t=\frac xy$ to obtain $\frac{\sqrt t+1}2\leq\sqrt\frac{t+1}2$. Comparing the derivatives of both sides, we see
$$\begin{array}{c}\left(\frac{\sqrt t+1}2\right)'&&\left(\sqrt\frac{t+1}2\right)'\\
\|&&\|\\
\frac1{2\sqrt{4t}}&\leq&\frac1{2\sqrt{2t+2}}&\quad\text{for }t\geq1\\
\frac1{2\sqrt{4t}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1203291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
Chinese Remainder Theorem for $x\equiv 0 \pmod{y}$ Can anyone solve the following system of congruences using CRT step-wise, without skipping any part?
$$\begin{cases} x\equiv 3 \pmod{7}\\ x\equiv 3 \pmod{13}\\ x\equiv 0 \pmod{12}\end{cases}$$
The answer to this problem $276$ but, I keep getting the answer $264$. What ... | $x$ is a multiple of $12$ so $x=12k$. We also have $12k\equiv 3\bmod 7$, $12$ is congruent to $5\bmod 7$ and the inverse of $5\bmod 7$ is $3$ by inspection. So $k\equiv 3\cdot5\cdot k\equiv 3\cdot3\equiv 2\bmod 7$. Therefore $k$ is $7l+2$.
So $x=12(7l+2)=84l+24$.
We now have $84l+24\equiv 3 \bmod 13$. This tells us $84... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1203678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How to solve $\int_0^{\infty} \frac{\log(x+\frac{1}{x})}{1+x^2}dx$? Here is my question
$$\int_0^{\infty} \frac{\log(x+\frac{1}{x})}{1+x^2}dx$$
I have tried it by substituting $x$ = $\frac{1}{t}$. I got the answer $0$ but the correct answer is $\pi log(2)$. Any suggestion would be appreciated.
| You may write
$$
\begin{align}
\int_0^{\infty} \frac{\log(x+\frac{1}{x})}{1+x^2}dx&=\int_0^{\infty} \frac{\log(1+x^2)-\log x}{1+x^2}dx\\\\
&=\int_0^{\infty} \frac{\log(1+x^2)}{1+x^2}dx-\int_0^{\infty} \frac{\log x}{1+x^2}dx.
\end{align}
$$ Clearly, by the change of variable $ x \to \dfrac 1 x$, we get
$$
\int_0^{\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1205543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Set A = {1,2,3,4,5} Pick randomly one digit and remove it. What is the prob. that we pick an odd digit the 2nd time. The probability that we pick any number for the first time is $\dfrac{1}{5}$
the sample space of sample spaces after the first event is then
{2,3,4,5}
{1,3,4,5}
{1,2,4,5}
{1,2,3,5}
{1,2,3,4}
prob. to pic... | There is a simpler way. No digit should have an inherent advantage of being picked second. So since 3 of the 5 digits are odd, there is a $\frac{3}{5}$ chance of the second number picked being odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1206670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Pascals Identity Let there be a group of n boxers and we want to select k people out of it, suppose one of the persons name is ‘Prem’ , so no of ways to choose k people = (combinations in which Prem is present + combinations in which Prem is not present) i.e $ {n-1} \choose {k-1} $ + $ {n-1} \choose k $, this is equal ... | Short answer: Your middle term should be $2\binom{n-2}{k-1}$.
Explanation: If, from $n$ people—Prem, Ram, and $n-2$ others—you must choose Prem but not Ram, and $k-1$ others, then the $k-1$ are drawn from the last group of $n-2$ others. Likewise if you choose Rem but not Pram, and $k-1$ others. Both these terms are $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1208011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding value of $\sin (15\, ^{\circ})$ with half angle identity The answer I got when trying to solve it was $\sqrt{\frac{1 - \sqrt3}{2} }$ but the book says it's $\sqrt{ \frac{2 - \sqrt3}{2}}$ and I don't know how the two on the top half gets there.
| By the half angle formulas $$\sin\dfrac{\pi}{6}=2\sin\dfrac{\pi}{12}\cos\dfrac{\pi}{12}.$$ Therefore if you substitute $x=\sin\dfrac{\pi}{12},$ you will get the equation $$\dfrac{1}{2}=2x\sqrt{1-x^2}.$$ One way to solve this equation is squaring both sides. But, if you square an equation number of roots will be double... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1208582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find All Solutions to System of Congruence $$
\begin{cases}
x\equiv 2 \pmod{3}\\
x\equiv 1 \pmod{4}\\
x\equiv 3 \pmod{5}
\end{cases}
$$
$
n_1=3\\
n_2=4\\
n_3=5\\
N=n_1 * n_2 * n_3 =60\\
m_1 = 60/3 = 20\\
m_2 = 60/4 = 15\\
m_3 = 60/5 = 12\\
gcd(20,3)=1=-20*1+3*7\\
gcd(15,4)=1=-15*1+4*4\\
gcd(12,5)=1=12*3-5*7\\
x=-20*2-1... | I think you're a bit confused about the recipe used to solve the system of congruences using the Chinese Remainder Theorem. Using your notation, the actual solution is given by
\begin{align*}
x = 2 \cdot m_1 \cdot ({m_1}^{-1} \text{ mod } 3) + 1 \cdot m_2 \cdot ({m_2}^{-1} \text{ mod } 4) + 3 \cdot m_3 \cdot ({m_3}^{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1209672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How do I know that $\ln(x^2+1)-x \arctan(x)$ is always negative or zero? I did google the function and I can clearly see that it is always negative or zero, but I have no idea how I would have found this on my own. Both the logarithm and the $x\cdot \arctan(x)$ are positive.
What I did do is derive the function but tha... | let $$y = \ln(1 + x^2 ) - x \tan^{-1} x .$$ the function is even and $y = 0$ at $x = 0 \text{ and } \lim_{x \to \infty} y = -\infty. $ taking the derivative, you find $$\frac{dy}{dx} = \frac{2x}{1+x^2} - \frac{x}{1+x^2} - \tan^{-1} x =\frac{x}{1+x^2} - \tan^{-1} x$$ we will have established that $y < 0 \text{ for a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1210071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
To Find $a$ such that $2^{1990} \equiv a\pmod {1990}$. To Find $a$ such that $2^{1990} \equiv a\pmod {1990}$.
$1990 = 2 \times 5 \times 199$. Now $a \equiv 0 \pmod {2}$, $a \equiv 4 \pmod{5}$ and $a \equiv 29 \pmod{199}$. Taking first two together we get $a \equiv 4 \pmod {10}$.
$199x \equiv 1\pmod {10}$ has a solutio... | Checking your answer of $a=1024=2^{10}$:
$$\begin{align}2^{1990}&\equiv 0\equiv 2^{10}&\pmod 2\\
2^{1990}&\equiv \left(2^{4}\right)^{495}2^{10}\equiv 2^{10}&\pmod{5}\\
2^{1990}&\equiv \left(2^{198}\right)^{10}2^{10}\equiv 2^{10}&\pmod{199}
\end{align}$$
After your edits, your answer is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1210866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Simplify $\left(\sqrt{\left(\sqrt{2} - \frac{3}{2}\right)^2} - \sqrt[3]{\left(1 - \sqrt{2}\right)^3}\right)^2$ I was trying to solve this square root problem, but I seem not to understand some basics.
Here is the problem.
$$\Bigg(\sqrt{\bigg(\sqrt{2} - \frac{3}{2}\bigg)^2} - \sqrt[3]{\bigg(1 - \sqrt{2}\bigg)^3}\Bigg)^... | This is a mistake that might have been avoided by being aware of the numerical values of some of the expressions involved. A very rough approximation for $\sqrt{2}$ is $1.41$. Clearly $\frac{3}{2} = 1.5$. Then $\sqrt{2} - \frac{3}{2} \approx -0.09$, and that squares to $0.0081$, and the square root of that is $0.09$, n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1212000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Hiding 3 coins in a pie, and slicing the pie in 8 equal pieces - probability Okay so I have this question on my maths sheet and it's in the permutations section but I can't get my head round it (probably just being dumb)
I know that $P(n,r) = \frac{n!}{(n-r)!}$
and that $C(n,r) = \frac{n!}{r!(n-r)!}$
And old lady bak... | If an event has probability $p$, then the probability that it will occur exactly $k$ times in $n$ trials is
$$\binom{n}{k}p^k(1 - p)^{n - k}$$
Since the pie has been cut into eight pieces of equal size, the probability that a particular coin is in that slice is $p = 1/8$. The number of trials is three since we must c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to solve this limit: $\lim_{x\to\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}$ Compute the limit $\lim\limits_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}$
my attempt:
I tried to multiply top and bottom by the conjugate
$$\begin{align}
\lim_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}&=\lim_{x\to+\infty}\left(\sqrt{x+\sqrt{x}}-\sq... | You may just observe that, as $x\to \infty$:
$$
\frac{1+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\sim \frac{\sqrt{x}\left(1+\frac1{\sqrt{x}}\right)}{\sqrt{x}\left(\sqrt{1+\frac{1}{\sqrt{x}}}\right)+\sqrt{x}\left(\sqrt{1-\frac{1}{x}}\right)}\sim \frac12.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
} |
Simplifying trig expression I have
$$\frac{\tan{15^\circ}}{1-\tan{15^\circ}^2}$$
and need to simplify it. The only equation I have that is even close to a match for it is $2\frac{\tan{15^\circ}}{1-\tan{15^\circ}^2}$. But the numerator isn't right with the $\times 2$. The answer should be $\frac{\sqrt{3}}{6}$. What do ... | $$
2 \frac{\tan(15^{\circ})}{1-\tan(15 ^{\circ})^2} = \tan(30) = \frac{1}{\sqrt{3}} \tag{1}
$$
because:
$$
\frac{\tan(\alpha) + \tan(\beta)}{1-\tan(\alpha)\tan(\beta)} = \tan(\alpha+\beta) \tag{2}
$$
Following equation (1):
$$
\frac{\tan(15^{\circ})}{1-\tan(15 ^{\circ})^2} = \frac{1}{2}\tan(30) = \frac{1}{2\sqrt{3}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1219565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluating a definite integral by definition: $A(x) = \int_{-1}^x (t^2 + 1)\space dt$ I have an area function $A(x)$ defined as
$$A(x) = \int_{-1}^{x} (t^2 + 1)\space dt$$
... and I would like to use the definition of definite integral to evaluate it.
I started this way
$$A(x) = \lim_{n \to \infty} \sum_{k=1}^n { (t^2... | Remember that
$$\int_a^b dt \, f(t) = \lim_{n \to \infty} \frac{b-a}{n} \sum_{k=0}^n f \left [ a + (b-a) \frac{k}{n} \right ] $$
Here, $a=-1$ and $b=x$, so that the sum we are interested in is
$$ \begin{align} \int_{-1}^x dt \, (1+t^2) &= \lim_{n \to \infty} \frac{x+1}{n} \sum_{k=0}^n \left [ 1+ \left ( -1 + \frac{x+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1222271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Converge or diverge $\sum^{\infty}_{n=2}\frac{4^n(n!)^2}{(2n-1)!}$ Show if the series converges or diverges.$$\sum^{\infty}_{n=2}\frac{4^n(n!)^2}{(2n-1)!}$$Can someone please help with proving this? (I think it converges)
| $a_n=\dfrac{4^n(n!)^2}{(2n-1)!} = \dfrac{(2^n\cdot n!)^2}{(2n-1)!} = \dfrac{2\cdot 4\cdots (2n-2)\cdot (2n)}{1\cdot 2\cdot 3\cdots (2n-2)\cdot (2n-1)}\cdot 2\cdot 4\cdots (2n) > \dfrac{2\cdot 4\cdot 6\cdots (2n-2)\cdot (2n)}{1\cdot 2\cdot 3\cdots (2n-2)\cdot (2n-1)}\cdot 1\cdot 3\cdot 5\cdots (2n-1) = \dfrac{(2n-1)!2n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1227868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Compute $\lim_{x\rightarrow0}\frac{e^{x^2} - \cos x}{\sin^2 x}$ Find the limit as $x$ approaches $0$ of
$\dfrac{e^{x^2} - \cos x}{\sin^2 x}$
What I tried is as $x$ approaches $0$, $e^{x^2}$ tends to $1$ and so the numerator tends to $1-\cos x$ and after doing some trigonometric simplifications I got the answer as $\fr... | Use Taylor expansion. As $x\rightarrow0$, you have
$$e^{x^2}=1+x^2+O(x^4)$$
$$\cos x=1-\frac12x^2+O(x^4)$$
$$\sin^2x=x^2+O(x^4)$$
Thus
$$\frac{e^{x^2}-\cos x}{\sin^2 x}=\frac{1+x^2-1+\frac12x^2+O(x^4)}{x^2+O(x^4)}=\frac{\frac32+O(x^2)}{1+O(x^2)}\underset{x\rightarrow 0}{\longrightarrow} \frac32$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1228251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
When $p=2$ or $p$ prime, with $p=1\pmod{4}$, $x^2\equiv -1\pmod{p}$ is soluble - trouble understanding proof Theorem: When $p=2$ or $p$ prime, with $p=1\pmod{4}$, $x^2\equiv -1\pmod{p}$ is soluble
Proof: When $p=2$, the statement is clear.
Assume $p\equiv 1\pmod{4}$, let $r=\frac{p-1}{2}$ and $x=r!$
Then since $r$ is e... | Suppose $\;p=4k+1\;$ , so that
$$\frac{p-1}2=2k\implies\;\; \text{since}\;\;j=p-j\pmod p\;\;\text{for}\;\;0\le j\le p \,, $$
we get that
$$\frac{p-1}2+1=\frac{p-1}2-1\pmod p\;,\;\;\frac{p-1}2+2=\frac{p-1}2-2\pmod p\,,\ldots$$
so that
$$1\cdot2\cdot\ldots\cdot k\cdot\ldots\cdot\left(\frac{p-1}2-1\right)\cdot\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1228931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrating $ \int^{\pi}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx $ I am trying to show that
$$ \int^{\pi}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx = \frac{\pi}{a(a+b)} $$ where $ a,b > 0$.
I have tried a few things, but none have worked.
For example, one approach was
$$ \int^{\pi}_{0} \frac{\... | If we set $x=\arctan t$ we simply have:
$$I(a,b)=\int_{0}^{\pi}\frac{\cos^2 x}{b^2\sin^2 x+a^2\cos^2 x}\,dx = 2\int_{0}^{+\infty}\frac{dt}{(1+t^2)(a^2+b^2 t^2)}$$
that is straightforward to compute through partial fraction decomposition or through the residue theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1230302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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System of equations with radicals: $2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}y^2}$ and $2\sqrt[4]{\frac{y^4}{3}+4} = 1+\sqrt{\frac{3}{2}x^2}$
Solve the system of equations (in $\mathbb R$):
$$\begin{matrix}
2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}y^2}
\\
2\sqrt[4]{\frac{y^4}{3}+4}
=
1+\sqrt{\frac{3}{2}x^2... | Without loss of generality, we can assume that $x,y\ge0$. (Notice that all exponents are even. We can then add signs to get the remaining solutions.)
So our equations are changed to
$$2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}}y\tag{1}$$
$$2\sqrt[4]{\frac{y^4}{3}+4}=1+\sqrt{\frac{3}{2}}x\tag{2}$$
If we subtract the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1232550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $dy/dx$ of $(xy^2)+5 = x + 2y^2$ For the solution I got $$\frac{y^2-1}{ 4y-2xy} = dy/dx$$
I just want to know if this is correct. Also it says to evaluate $dy/dx$ at $(1,2)$.
Would the solution to that be $3/4$?
| Here are the steps
$$ xy^2+5=x+2y^2 $$
$$ \frac{d}{dx}\left[xy^2+5\right]=\frac{d}{dx}\left[x+2y^2\right] $$
$$ y^2\frac{d}{dx}\left[x\right]+x\frac{d}{dx}\left[y^2\right]=1+2\frac{d}{dx}\left[y^2\right] $$
$$ y^2+2xy\frac{d}{dx}[y]=1+4y\frac{d}{dx}[y] $$
$$ y^2-1=\left(4y-2xy\right)\frac{d}{dx}[y] $$
$$ \frac{d}{dx}[y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
sum of the squares of the reciprocals of the two parts of the focal chord of a parabola
Find the sum of the squares of the reciprocals of the two parts of the focal chord of a parabola.
My attempt:
Let $y^2=4ax$ be a parabola. Let PQ be the focal chord through the focus S$(a, 0)$ of the parabola such that the co-ordi... | Loosely speaking, the "axis chord" has parts $a$ and $\infty$, so the sum of the squares of the reciprocals of the lengths is
$$\frac{1}{a^2}+\frac{1}{\infty^2} = \frac{1}{a^2} + 0 = \frac{1}{a^2}$$
However, the focal chord perpendicular to the axis has parts $2a$ and $2a$, so the corresponding sum is
$$\frac{1}{4a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1235318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the coefficient of $x^{30}$. Find the coefficient of $x^{30}$ in the given polynomial
$$
\left(1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^{10}+x^{11}+x^{12}\right)^5
$$
I don't know how to solve problems with such high degree.
| This turns out to be a detailed explanation of Jack's answer.
$$
\begin{align}
\left(\frac{1-x^{13}}{1-x}\right)^5
&=(1-x^{13})^5(1-x)^{-5}\\
&=\sum_{j=0}^5(-1)^j\binom{5}{j}x^{13j}\sum_{k=0}^\infty(-1)^k\binom{-5}{k}x^k\tag{1}\\
&=\sum_{j=0}^5(-1)^j\binom{5}{j}x^{13j}\sum_{k=0}^\infty\binom{k+4}{k}x^k\tag{2}\\
&=\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1239132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Proving $\left(a+\frac{2}{a}\right)^2+\left(b+\frac{2}{b}\right)^2\ge \frac{81}{2}$ for all positive real $a,b$ such that $a+b=1$ I approached this problem in two different ways, but only one was successful. I'll post the latter as an answer, while here follows the first approach:
I expanded the squares:
$$a^2+\frac{4}... | To begin with, we note that equality occurs when $a=b=1/2$. Thus, assuming WLOG $a>b$ and using $b=1-a$, let us differentiate the LHS to show it is increasing: $$2\left(a+\frac{2}{a}\right)\left(1-\frac{2}{a^2}\right)+2\left(1-a+\frac{2}{1-a}\right)\left(-1+\frac{2}{(1-a)^2}\right)>0 $$ $$ \left(a+\frac{2}{a}\right)\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1240173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int_{-\pi}^\pi \! \cos(kx)\cos^n(x) \, \mathrm{d}x$ My question is:
Evaluate $$\int_{-\pi}^\pi \! \cos(kx)\cos^n(x) \, \mathrm{d}x$$ for $k=0,1,...,(n-1)$ and $n \in \mathbb{N}$.
I've tried integration by parts but without much success. Any help/trick for this integral? Thanks!
| $$\begin{align} 2^n \cos^n x &= \left(e^{ix} + e^{-ix}\right)^n\\
&= e^{inx} + {n \choose 1}e^{i(n-2)x} + {n \choose 2}e^{i(n-4)x}+ \cdots+{n \choose 2}e^{-i(n-4)x}+{n \choose 1}e^{-i(n-2)x} + e^{-inx}\\
&=2\cos nx+ 2{n \choose 1}\cos(n-2)x+2{n \choose 2}\cos(n - 4)x + \cdots
\end{align}$$
multiplying by $\cos kx,$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1241163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Complex Numbers Question, IIT JEE [2006]. Please tell me whether I solved it properly? $Q.$The value of $\sum\limits_{k=1}^{10}(\sin{\frac{2k\pi}{11}-i\cos\frac{2k\pi}{11}})$ is-?
I solved it like this-
$\frac{\sum\limits_{k=1}^{10}(\cos{\frac{2k\pi}{11}+i\sin\frac{2k\pi}{11}})}{i}$
If we observe these are the roots o... | $$\frac1i\sum_{k=1}^{10}\left(\cos\frac{2k\pi}{11}+i\sin\frac{2k\pi}{11}\right)=-i\sum_{k=1}^{10} \left(e^{\frac{2\pi i}{11}}\right)^k=$$
$$=-ie^{\frac{2\pi i}{11}}\frac{e^{\frac{2\cdot10\pi i}{11}}-1}{e^{\frac{2\pi i}{11}}-1}=-ie^{\frac{2\pi i}{11}}\frac{e^{\frac{-2\pi i}{11}}-1}{e^{\frac{2\pi i}{11}}-1}=-i\frac{e^{\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1241529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Basis of a $2\times2$ matrix How would I find the basis for an arbitrary matrix W such that:
$$
W =\left\{
\begin{pmatrix} a & b \\ c & a +b +c\end{pmatrix} \ \big| \ \ a ,b ,c \in \mathbb{R}
\right\}
$$
| Hint
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\in W\iff d=a+b+c\iff\begin{pmatrix}a&b\\c&d\end{pmatrix}=a\begin{pmatrix}1&0\\0&1\end{pmatrix}+b\begin{pmatrix}0&1\\0&1\end{pmatrix}+c\begin{pmatrix}0&0\\1&1\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1241733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Show that if $a\neq 1$, then $\sum_{k=0}^{n-1}ka^k = \frac{1-na^{n-1}+(n-1)a^n}{(1-a)^2}$
Need to show that if $a\neq 1$, then
$$\sum_{k=0}^{n-1}ka^k = \frac{1-na^{n-1}+(n-1)a^n}{(1-a)^2}$$
Here is my attempt:
$$\begin{aligned}
S & =\sum_{k=0}^{n-1}ka^k \\
&= \sum_{k=0}^{n}(k-1)(a^{k-1}) \\
\end{aligned}$$
from he... | let $$\begin{align}1+S_n &= 1 + a + 2a^2 + 3a^3 + \cdots + (n-1)a^{n-1}\\
aS_n &= a + a^2 +2a^3+ \cdots + (n-1)a^n\end{align}$$ subtracting the second equation from the first, we get $$(1-a)S_n + 1 = 1+a^2 + a^3 + \cdots+a^{n-1} - (n-1)a^n $$ rearranging this, we get $$(1-a)S_n+1+a+(n-1)a^n = 1+ a + a^2 + \cdots + a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Generating function for recurrence in two variables Given characteristic polynomial for the recurrence in two variables (say $F(x,y)$)
$$
(y^2-1)^x
$$
and initial values can generating function for $F(x,y)$ be derived?
I know how to do it for a recurrence with one variable but have no idea how to do it in the case of t... | We want $F(z,w)$ a generating function such that
$$F(z,w) = \sum_{x=0}^{\infty} \sum_{y=0}^{\infty} (y^2-1)^x z^xw^y \tag{1}$$
$(y^2-1)^x z^x$ is a power series.
$$ F(z,w) = \sum_{y=0}^{\infty} \frac{w^y}{1 - (y^2-1)z} \tag{2}$$
$$ F(z,w) = -\frac1{z} \sum_{y=0}^{\infty} \frac{w^y}{y^2 - 1 - \frac1{z}} \tag{3}$$
$$ F(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solution verification: Prove by induction that $a_1 = \sqrt{2} , a_{n+1} = \sqrt{2 + a_n} $ is increasing and bounded by $2$ I have the following recursive relation (sequence):
\begin{align}
a_1 = \sqrt{2}, \quad a_{n+1} = \sqrt{2 + a_n}
\end{align}
My Try:
I'm a little skeptical of my manipulations near the end but it... | Your base case and inductive step are both overly complicated.
Base case: We have $a_1 = \sqrt{2}$, which is less than $2$.
Inductive step: Assume we have $a_k < 2$
Now $a_{k+1} = \sqrt{2 + a_k} < \sqrt{2 + \sqrt{2}} < \sqrt{2 + 2} = 2$.
It may also be useful to remark that the sequence is strictly positive, so it is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How find this minimum Help me!
Let $x,y,z\ge0$ such that: $xy+yz+zx=1$.
Find the minimum value of: $A=\dfrac{1}{x^2+y^2}+\dfrac{1}{y^2+z^2}+\dfrac{1}{z^2+x^2}+\dfrac{5}{2}(x+1)(y+1)(z+1)$
I found minimum value of $A$ is $\dfrac{25}{2}$ iff $(x,y,z)=(1,1,0)$ or any permutation. But I can't solve that.
| first to prove:
$\dfrac{1}{x^2+y^2}+\dfrac{1}{y^2+z^2}+\dfrac{1}{z^2+x^2} \ge \dfrac{5(1+xyz)}{2}$ ...(1)
$A \ge \dfrac{25}{2} \implies x+y+z+2xyz\ge 2 $...(2)
(2) is easy to prove. But (1) is hard. I have a ugly solution and I think to wait some days to post it if there is no better way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1244325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integral using trigonometric substitution I'd like to ask for feedback on my calculation for this integral:
$$\int{\frac{dx}{2-\cos{x}}}$$
Using half-angle substitution:
$$t = \tan{\frac{x}{2}}$$
$$\cos{x} = \frac{1-t^2}{1+t^2}$$
$$dx = \frac{2dt}{1+t^2}$$
So
$$\int{\frac{dx}{2-\cos{x}}} = \int{\frac{1}{2-\frac{(1-t^2)... | Note also that it is $\tan^{-1}(\sqrt 3t)$, not $\tan^{-1}\left(\frac{t}{\sqrt 3}\right)$.
Setting $t=\frac{1}{\sqrt 3}\tan\theta$ gives you
$$\begin{align}\frac 23\int\frac{1}{\left(\frac{1}{\sqrt 3}\right)^2+t^2}dt&=\frac 23\int\frac{1}{\frac 13+\frac 13\tan^2\theta}\cdot\frac{d\theta}{\sqrt 3\cos^2\theta}\\&=\frac 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1245493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the type of triangle from equation. In triangle $ABC$, the angle($BAC$) is a root of the equation
$$\sqrt{3}\cos x + \sin x = \frac{1}{2}.$$
Then the triangle $ABC$ is
a) obtuse angled
b) right angled
c) acute angled but not equilateral
d) equilateral.
Thanks in advance.
| $\frac{\sqrt{3}}{2}\cdot \cos x + \frac{1}{2}\sin x = \frac{1}{4}$
$\sin ({x + \frac{\pi}{3}}) = \frac{1}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1246484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Calculating the work integral of a vector field. Let F be the vector field
F = $<\dfrac{1}{x} + 2x + y, \ \ \dfrac{1}{y} + x + 1>$ .
Compute the work integral $\int_{c} F dr$ where C is the path
r(t) = (1 + cos t)i + (2 + sin t) j $ \ \ \ \ -\dfrac{\pi}{2} \leq t \leq \dfrac{\pi}{2}$.
I am very confused as to how to a... | Fundamental Theorem for Line Integrals
$$\eqalign{
& W = \int\limits_C {F \bullet dr} \cr
& - - - - Test\;For\;Gradient\;Field \cr
& F = \left( {{1 \over x} + 2x + y\,,\,{1 \over y} + x + 1} \right) = \left( {M\;,\;N} \right) \cr
& F = \nabla f\;\;if\;\;\;{\partial _y}M = {\partial _x}N \cr
& {\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1247682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How many planes are there with the desired property? Two points $A$ and $B$ are given, for example $A(2/5/7)$ and $B(4/11/16)$.
The object is to find all planes containing the points $A$ and $B$ with distance
$2$ to the origin.
I tried the hesse-normal-form, but since only one direction-vector is given,
the normal-vect... | Let plane $\alpha$ be $p(X)=Ax+By+Cz+D=0$, then $\rho(\alpha,X)=\frac{|p(X)|}{\sqrt{A^2+B^2+C^2}}$.
So we put $A^2+B^2+C^2=1$ and get equations
$D=\pm 2$;
$2A+5B+7C+D=0$;
$4A+11B+16C+D=0$.
$2A+5B+7C+D=0 \Leftrightarrow A=-\frac{5}{2}B-\frac{7}{2}C\mp1$
Plugging this into $4A+11B+16C\pm 2=0$ we get
$-10B-14C\mp 4 +11B+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1248664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Simplifying Square Roots Frustration Okay, I'm really frustrated with this.
So, when you have $3 \sqrt 5 + 5 \sqrt 5$,
you get $8\sqrt5$, right?
Okay, so what do I do for here:
$\sqrt{11} - 3 \sqrt{11}$
Is it just $-3 \sqrt{11}$ ?
What about for $6 \sqrt 2 + 4 \sqrt{50}$ ?
Do I multiply the $6 \sqrt{2}$ so that I can... | *
*$\sqrt{11}-3\sqrt{11}=\color{red}{1}\cdot\sqrt{11}-3\sqrt{11}=-2\sqrt{11}.$
*$6\sqrt 2+4\sqrt{50}=6\sqrt 2+4\color{blue}{\sqrt{2\cdot 5^2}}=6\sqrt{2}+4\cdot \color{blue}{5\sqrt 2}=6\sqrt 2+20\sqrt 2=26\sqrt 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1249138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Calculate a determinant. Let $a_{1}, \cdots, a_{n}$ and $b$ be real numbers. I like to know the determinant of the matrix
$$\det\begin{pmatrix}
a_{1}+b & b & \cdots & b \\
b & a_{2}+b & \cdots & b \\
\vdots & \vdots & \ddots & \vdots \\
b & b & \cdots & a_{n}+b
\end{pmatrix}=?$$
I guess the answer is $$a_{1... | Applying multilinearity and an inductive argument:
$$\begin{vmatrix}
a_{1}+b & b & \cdots & b \\
b & a_{2}+b & \cdots & b \\
\vdots & \vdots & \ddots & \vdots \\
b & b & \cdots & a_{n}+b
\end{vmatrix}=\begin{vmatrix}
a_{1} & b & \cdots & b \\
0 & a_{2}+b & \cdots & b \\
\vdots & \vdots & \ddots & \vdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1249516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
How prove $a + b +c + \frac{1}{abc} \geq 4\sqrt{3}$? Let $a, b, c >0$ and $a^2+b^2+c^2 =1$ How prove $a + b +c + \frac{1}{abc} \geq 4\sqrt{3}$?
| Let's consider the function
$$f(a,b,c) = a+b+c+\frac{1}{abc}.$$
We want to minimize it.
Then
$$f'_a = 1 - \frac{1}{(abc)a}, f'_a = 1 - \frac{1}{(abc)b}, f'_c = 1 - \frac{1}{(abc)c}$$
Imposing all of them to be $0$, we have:
$$\left\{\begin{array}{l}(abc)a = 1\\(abc)b = 1\\(abc)c = 1\end{array}\right.$$
This means that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1251680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving the existence of a square root of $ -1_{A} $ in a $ 2 $-dimensional unital algebra $ A $ over $ \Bbb{R} $. Suppose that $ A $ is a $ 2 $-dimensional unital algebra over $ \Bbb{R} $ with a basis $ \{ 1_{A},u \} $, and assume that $ A $ does not have any zero divisors. Show that $ A $ contains an element $ b $ su... | Not sure how old Frobenio handled this one, but here's how I do it:
Since $A = \text{span}(\{ 1_A, u \})$, we have
$u^2 = \alpha 1_A + \beta u, \;\; \alpha, \beta \in \Bbb R. \tag{1}$
Then
$u^2 - \beta u = \alpha 1_A, \tag{2}$
or, "completing the square",
$u^2 - \beta u + \dfrac{\beta^2}{4} 1_A = \alpha 1_A + \dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1261069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
The complex roots of a biquadratic polynom In my recent post I have a problem with the following function: $x^4-4x^2+16$, and what I need is to find the complex roots.
Here is my answer:
First step, I make the substitution $x^2=y$ which involving $y^2-4y^2+16$, with $x^2=2\pm i\sqrt{12}$. Therfore:
$x^4-4x^2+16=(x^2-... | Note that your answer gives that same result as author's answer, but the more important thing to pay attention with is when you use $\sqrt z$ with $z$ a complex numbers this is not defind so in I would say that your answer is incomplete because you did not find the complex square roots of $2\mp i\sqrt {12}$, and when y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1262094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Solving $\int dx {\sqrt{x^2+a}} e^{-A x^2} erf \left( c(x-b) \right)$ I got as far as:
$$\int dx {\sqrt{x^2+a}} e^{-A x^2} erf \left( c(x-b) \right) $$
$$=\frac{2}{\sqrt{\pi}} \int dx \int^{c(x-b)}_0 dy {\sqrt{x^2+a}} e^{-A x^2 - y^2}$$
$$=\frac{-2 c}{\sqrt{\pi}} \int db \int dx {\sqrt{x^2+a}} e^{-A x^2 - c^2 (x-b)... | Hint:
$\int\sqrt{x^2+a}e^{-Ax^2}\text{erf}(c(x-b))~dx$
$=\dfrac{2}{\sqrt\pi}\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nc^{2n+1}(x-b)^{2n+1}\sqrt{x^2+a}e^{-Ax^2}}{n!(2n+1)}~dx$
$=\dfrac{2}{\sqrt\pi}\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n+1}\dfrac{(-1)^{n-k+1}(2n)!b^{2n-k+1}c^{2n+1}x^k\sqrt{x^2+a}e^{-Ax^2}}{n!k!(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1264811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Residue $\frac{e^z}{z^3\sin(z)}$ I want to find the residue of $$\frac{e^z}{z^3\sin(z)}$$ and get $$ \frac 1 {3!} \lim_{z \to 0} \left( \frac{d^3}{dz^3} \left(\frac{ze^z}{\sin(z)} \right)\right) = \frac{1}{3}$$
Can anyone confirm this? I tried using the Laurent Series, but I didn't know how to compute it.
Oh, I was ab... | Near zero,
$$ e^z = 1 + z + \frac{z^2}{2} + \frac{z^3}{6} + O(z^4). $$
Also,
$$\sin{z}=z-\frac{1}{6}z^3 + O(z^5),$$
so the denominator is
$$ z^{-4} \left(1-\frac{1}{6}z^2+ O(z^4)\right)^{-1} = \frac{1}{z^4}\left( 1-\frac{z^2}{6} +O(z^4) \right), $$
using the binomial theorem.
Hence the Laurent series has principal part... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1266716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that $\dfrac{m - \sqrt[3]{2}}{\sqrt[3]{3}}$ is an algebraic integer. Let $m$ be an integer such that $m \equiv 2 \pmod 3$. Show that the number $$\dfrac{m - \sqrt[3]{2}}{\sqrt[3]{3}}$$ is an algebraic integer.
The usual technique, doing $x = \dfrac{m - \sqrt[3]{2}}{\sqrt[3]{3}}$ and trying to find an algebraic exp... | $x\sqrt[3]{2} = m - \sqrt[3]{2} \to \sqrt[3]{2} = \dfrac{m}{x+1} \to 2 = \dfrac{m^3}{x^3+3x^2+3x+1} \to 2x^3+6x^2+6x+2-m^3=0$. Thus $x$ is algebraic number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1267149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Find the standard matrix representation of the linear transformation T in M2,2 let $T: M_{2,2} \rightarrow M_{2,2}$ be a linear transformation defined by:
$$T \left(\begin{bmatrix}
a & b\\
c & d\\
\end{bmatrix}\right) = \begin{bmatrix}a + b& b + a \\ c - d&d+b\end{bmatrix}
$$
Find the standard ... | Let $v_1,v_2,v_3,v_4$ denote the standard basis. Note that
$$
T(v_1) = T \left(\begin{bmatrix}
1 & 0\\
0 & 0\\
\end{bmatrix}\right) = \begin{bmatrix}1& 1 \\ 0&0\end{bmatrix} =
1v_1 + 1v_2 +0v_3 + 0v_4\\
T(v_2) = T \left(\begin{bmatrix}
0 & 1\\
0 & 0\\
\end{bmatrix}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1268191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Find the value of : $\lim_{x\to\infty}\frac{\sqrt{x-1} - \sqrt{x-2}}{\sqrt{x-2} - \sqrt{x-3}}$ I'm trying to solve evaluate this limit
$$\lim_{x\to\infty}\frac{\sqrt{x-1} - \sqrt{x-2}}{\sqrt{x-2} - \sqrt{x-3}}.$$
I've tried to rationalize the denominator but this is what I've got
$$\lim_{x\to\infty}(\sqrt{x-1} - \sqrt{... | \begin{align}
\frac{\sqrt{x-1} - \sqrt{x-2}}{\sqrt{x-2} - \sqrt{x-3}}
&=\frac{\sqrt{x-1} - \sqrt{x-2}}{\sqrt{x-2} - \sqrt{x-3}}\,\,
\frac{\sqrt{x-1} + \sqrt{x-2}}{\sqrt{x-1} + \sqrt{x-2}}\,\,
\frac{\sqrt{x-2} + \sqrt{x-3}}{\sqrt{x-2} + \sqrt{x-3}}\\ \ \\
&=\frac{x-1-(x-2)}{x-2-(x-3)}\,\frac{\sqrt{x-2} + \sqrt{x-3}}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Solve this equation: $\log_3(3-2\cdot3^{x+1})=2+2x$ Solve this equation: $\log_3(3-2\cdot3^{x+1})=2+2x$. I put $(2+2x)^3=3-2\cdot3^{x+1}$. But I don't know how to go on.
| We have
$$3-2\cdot 3^{x+1} = 3^{2+2x}$$
Setting $3^{x+1}=t$, we obtain
$$3-2t = t^2 \implies t^2+2t-3 =0 \implies (t+3)(t-1) = 0 \implies t \in \{-3,1\}$$
Hence, if we want $x \in \mathbb{R}$, we have
$$3^{x+1} = 1 \implies x+1 =0 \implies x=-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1271656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to solve $\int_{-1}^{1} (x^{4/3} + 4x^{1/3}) dx$? I started by integrating it this way:
$$(\frac{3}{7} x^{7/3} + \frac{16}{4}x^{4/3})$$
What is wrong with it?
| The answer is
$\frac{6}{7}$
Integrating,
$ \int^1_{-1} x^{4/3} 4 x^{1/3} \,dx = (\frac{3}{7}x^{7/3} + 3 x^{4/3})^{1}_{-1} = \frac{3}{7} + 3 + \frac{3}{7} - 3 = \frac{6}{7}. $
In the OP as it reads now, it appears that you increased the power on the second term and multiplied by the new power instead of dividing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1271847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
If $a+b+c\le 1$ then $3(a+b+c)-(a^2+b^2+c^2-ab-bc-ac)\ge (\sqrt{a}+\sqrt{b}+\sqrt{c})^2$
Let $a,b,c\ge 0$ such that $a+b+c\le 1$, prove that
$$3(a+b+c)-(a^2+b^2+c^2-ab-bc-ac)\ge (\sqrt{a}+\sqrt{b}+\sqrt{c})^2\tag{1}$$
I conjecture:
Let $a_{i}\ge 0$, $i=1,2,\cdots$, $a_{1}+a_{2}+\cdots+a_{n}\le 1$, $n\ge 3$,then
$... | Since $a+b \leqslant 1 $ by Cauchy - Schwarz we get $$\sqrt{a} +\sqrt{b} \leq \sqrt{2}\cdot \sqrt{a+b} \leqslant \sqrt{2} $$
hence $$\frac{(\sqrt{a} +\sqrt{b})^2}{2} \leqslant 1$$
multiplying both sides of the above inequality by $(\sqrt{a} -\sqrt{b} )^2 $ we get $$\frac{(a-b)^2 }{2} \leqslant (\sqrt{a} -\sqrt{b})^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1272534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Finding pole order and calculating residue of: $\frac1{z-\tan z} - \frac1{z}$ Part of the the problem I'm trying to do involves finding the residue of
$\frac{1}{z-\tan z} - \frac{1}{z}$ at z=0
I am not sure of the order of the pole
Here's what I did to find the order of the pole
f(z) = $\frac{1}{z-\tan z} - \frac{1}{z... | You may write
$$
\begin{align}
\frac{1}{z-\tan z} - \frac{1}{z}&=\frac{1}{z-\left(z+\frac{1}{3}z^3 + \frac{2}{15}z^5 + \mathcal{O}(z^7)\right)} - \frac{1}{z}\\\\
&=\frac{1}{-\frac{1}{3}z^3 - \frac{2}{15}z^5 + \mathcal{O}(z^7)} - \frac{1}{z}\\\\
& =-\frac{3}{z^3}\frac{1}{1 + \frac{2}{5}z^2 +\mathcal{O}(z^4)} - \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1272770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving that $12^n + 2(5^{n-1})$ is a multiple of 7 for $n\geq 1$ by induction
Prove by induction that $12^n + 2(5^{n-1})$ is a multiple of $7$.
Here's where I am right now:
Assume $n= k $ is correct:
$$12^k+2(5^{k-1}) = 7k.$$
Let $n= k+1 $:
$$12^{k+1} + 2(5^k)$$
$$12^k(12) + 2(5^k)$$
Any ideas on how I can proceed f... | You should write $12^k+2\cdot 5^{k-1}=7r$ (using $k$ with two different meanings is wrong).
Then you have $12^k=7r-2\cdot 5^{k-1}$; now
\begin{align}
12^{k+1}+2\cdot 5^k
&=12\cdot 12^k+2\cdot 5^k\\
&=12\cdot(7r-2\cdot 5^{k-1})+2\cdot 5^k\\
&=7\cdot12r-24\cdot5^{k-1}+10\cdot5^{k-1}
\end{align}
and you should be able to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 4
} |
Boolean Algebra - proof without associativity? I would like to prove the following:
$(x\cdot y) + (\overline{x} + \overline{y}) = 1$
without the Associativity Property. I can't seem to do this algebraically (without truth tables).
| So I cannot use Associative:
$X + (Y + Z) = (X + Y) + Z = (X + Z) + Y = X + Y + Z $
$(x\cdot y) + (\overline{x} + \overline{y})$
Expand minimized terms using complement
$(x\cdot y) + (\overline{x} \cdot (\overline{y} + y) + \overline{y} \cdot (\overline{x} + x))$
$(x\cdot y) + (\overline{x}\cdot\overline{y} + \overlin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1277880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$f(x)=\sum_{i=0}^{\infty} (x^{2^n})/(1-x^{2^{n+1}})$. Find $f(99)$. $f(x)=\sum_{i=0}^{\infty} (x^{2^n})/(1-x^{2^{n+1}})$. Find $f(99)$.
ATTEMPT: The following series can be re-written as $f(x)=\sum_{i=0}^\infty \left(\frac{1}{1-x^{2^n}}\right) \cdot \left( \frac{1}{1+x^{2^n}}\right)$ and then expanded along.
| If this problem is $$f(x)=\sum_{n=0}^{\infty}\dfrac{x^{2^n}}{1-x^{2^{n+1}}}$$
Hint:Let $x^{2^n}=t$,then
$$\dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\dfrac{t}{1-t^2}=\dfrac{1}{1-t}-\dfrac{1}{1-t^2}=\dfrac{1}{1-x^{2^n}}-\dfrac{1}{1-x^{2^{n+1}}}$$
so
$$f(x)=\dfrac{1}{1-x}-\lim_{N\to\infty}\dfrac{1}{1-x^{2^{N+1}}}=\dfrac{1}{1-x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1278610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What's wrong?: Find the infinite sum $S = 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} + \ldots$ The answer I got by hand is not the same to the one I found using a spreadsheet.
$\displaystyle S = 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} + \ldots$
$\displaystyle \frac{1}{4}S =... | We can pattern match this to a difference of geometric series. The sum in question is $$\begin{align} 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} = 1+\sum_{k=1}^\infty \frac{2^{k+1}-1}{4^{k}} \\ = 1+\sum_{k=1}^\infty \frac{2^{k+1}}{4^{k}}-\sum_{k=1}^\infty \frac{1}{4^{k}}\end{align}$$ Now use geom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1280948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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Limit and Integral problem work verification-2 I have to calculate the following:
$$\large\lim_{x \to \infty}\left(\frac {\displaystyle\int\limits_{x^{2}}^{2x}t^{4}e^{t^{2}}dt}{e^{x}-1-x - \frac{x^2}{2}- \frac{x^3}{6}-\frac{x^4}{24}}\right)$$
My attempt:
Let $F(x)=\displaystyle\int\limits_0^xt^4e^{t^2}dt$. Then,
$$\l... | Hint: Notice that $$\left(\int_{x^2}^{2x}t^4 e^{t^2} dt\right)' = \color{red}2\dot\,(2x)^4 e^{(2x)^2} - \color{red}{2x}\dot\,(x^2)^4 e^{(x^2)^2}$$
and $$e^x = \sum_{k=0}^\infty \frac{x^k}{k!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1281044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Use generating function to find coefficient Use a generating function to find the coefficient of $x^{22}$ in:
$$\frac{1+3x}{(1-x)^8}$$
I know I need to use a binomial expansion on the lower term, but what about the upper term?
| Recall that $$\frac1{1-x}=\sum_{n=0}^\infty x^n.$$
So $$\frac{\mathsf d^7}{\mathsf dx^7}\left[\frac1{1-x}\right] = \frac{7!}{(1-x)^8}$$
and
$$\begin{align*}
\frac{\mathsf d^7}{\mathsf dx^7}\sum_{n=0}^\infty x^n =\sum_{n=7}^\infty \frac{n!}{(n-7)!}x^{n-7}=\sum_{n=0}^\infty\frac{(n+7)!}{n!}x^{n}
\end{align*}$$
Hence
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1283786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve the equation $\log_{2} x \log_{3} x = \log_{4} x$ Question:
Solve the equations
a)
$$\log_{2} x + \log_{3} x = \log_{4} x$$
b)
$$\log_{2} x \log_{3} x = \log_{4} x$$
Attempted solution:
The general idea I have been working on is to make them into the same base, combine them and take the inverse to get the soluti... | I think that you made two mistakes.
The first mistake is in a) when factorizing, instead you typed it should have been $\frac{\log_{2} x(2\log_{2} 3+ 2 -\log_{2} 3)}{2\log_{2} 3}=0$ though it doesn´t affect the answer.
And the other mistake in b) is the one that @Gregory Grant says, once you do that you get to answer $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Show that $ \frac{\cos5x + \cos4x} {1-2\cos3x} = -\cos2x -\cos x $ The Question reads -
$$ \frac{\cos5x + \cos4x} {1-2\cos3x} = -\cos2x -\cos x $$
I tried using the obvious approach by converting $5x , 4x $ and $ 3x$ to either $2x$ or $x$ but all that seemed to do was to further complicate the fraction. Any hints wou... | Let's apply the identity $\cos (x+y)+\cos(x-y)=2\cos x \cos y$ to the product
$$(1-2\cos 3x)(-\cos 2x-\cos x)=-\cos 2x -\cos x+2\cos 3x \cos 2x+2\cos 3x \cos x$$
Note that $2\cos 3x \cos 2x=\cos x+\cos 5x$ while $2\cos 3x \cos x=\cos 2x+\cos 4x$
Thus
$$\begin{align}
(1-2\cos 3x)(-\cos 2x-\cos x)&=-\cos 2x -\cos x+2\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Solving a Diophantine equation3 The Diophantine equation that I have to solve is:
$$343x^2-27y^2=1$$
This question has already been posted by other user but it has not received an answer.
I proved to solve it.
This is my attempt:
substituting for $x^2=u$ and $y^2=v$ the equation becomes a Diophantine linear equation:$$... | $343x^2-27y^2=1$
$⇒343x^2≡1 mod 27$
$343≡19 mod 27$
we can find that if $x=8$ we get:
$343 . 8^2 ≡ 1 mod 27$
We can see that x belong to a set its members make an arithmetic progression and have following specification:
$s . 19 ≡ 1 mod 27$; $s = [10, 37, 64, . . a+27k]$
where $a=10$; and $d=27$ is common ratio.
With $k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Prove that $7$ divides $1 + 2^{(2^n)} + 2^{(2^{n+1})}$ by induction Prove that $7$ divides $1 + 2^{(2^n)} + 2^{(2^{n+1})}$ by induction.
I ran into the above problem. The base case $n=1$ gives $21$ which is divisible by $7$.
Now assume it is true for $n$. Then for $n+1$, we have the expression
$$ 1 + 2^{(2^{n+1})} + 2^... | As in Winther's post let $x_n=2^{2^n}$. Then the sequence is
$I_n=1+x_n+x_n^2$, and $x_{n+1}=x_n^2$.
We suppose $I_n\equiv0\pmod 7 \implies x_n+x_n^2\equiv-1 \pmod 7$,
squaring both sides:
$x_n^2+2x_n^3+x_n^4\equiv1 \implies I_{n+1}=1+x_n^2+x_n^4\equiv 2(1-x_n^3)\equiv 0 \pmod 7$
since $x_n^3=2^{3\cdot2^n}\equiv1^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
A supposed to be easy calculus problem Find the values of $m$ if the line $y=mx+2$ is a tangent to the curve $x^2-2y^2=1$.
My working:
First we differentiate $x^2-2y^2=1$ with respect to $y$ to get the gradient. We get $y^2=\frac{1}{2}x^2-\frac{1}{2}\implies y=\pm\sqrt{\frac{1}{2}x^2-\frac{1}{2}}$.
We take the positive... | Let $(a,b)$ be a point of tangency. We have $2x-4y\frac{dy}{dx}=0$, so the slope of the tangent line at $(a,b)$ (if $b\ne 0$) is $\frac{a}{2b}$.
The tangent line has equation $y-b=(x-a)(a/2b)$. Simplifying , and comparing with $y=mx+2$, we find that $b-a^2/(2b)=2$. It follows that $2b^2-a^2=4b$. Since $a^2-2b^2=1$, we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Finding value of a quadratic The polynomial
\begin{equation*}
p(x)= ax^2+bx+c
\end{equation*}
has $1+\sqrt{3}$ as one of it's roots and also $p(2)=-2$. Is there any way to know the value of $a$, $b$ and $c$? I tried but I can form only $2$ equations how can I figure out value of $3$ unknown with $2$ equations so some... | Hint: If one of the roots is $1 + \sqrt{3}$ and $a, b, c$ are rational then what must the other root be?
Solution 1:
The other root must then be $1 - \sqrt{3}$.
You have two options now: You can make a system of three linear equations in three variables and simply solve.
Or, you can simply expand $\alpha[x - (1 - \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding value of $m$ such that such that the polynomial is factorized A polynomial $2x^2+mxy+3y^2-5y-2$ Find the value of $m$ much that $p(xy)$ can be factorized into two linear factors
| Putting $m = 7$ you obtain
\begin{align*}
(x+3y+1)(y+2x-2) &= xy+2x^2-2x+3y^2+6xy-6y+y+2x-2 \\
&= 2x^2 + 7xy +3y^2-5y-2
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1296810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving a Radical Equation $5(\sqrt{1-x} + \sqrt{1+x}) = 6x + 8\sqrt{1-x^2}$ (squaring doesn't help) How should I approach this problem:
$$ 5(\sqrt{1-x} + \sqrt{1+x}) = 6x + 8\sqrt{1-x^2} $$
I've tried squaring both sides but to get rid of all the radicals requires turning it into a quartic equation, which I don't know... | Here is a really nice method: when looking at terms such as $\sqrt{1-x}$, $\sqrt{1+x}$, and $\sqrt{1-x^2}$, think trigonometry!
Make the trig substitution:
$$x = \cos{\theta}$$
Now:
$$\sqrt{1-x}=\sqrt{2}\sin{\frac{\theta}{2}}$$
$$\sqrt{1+x}=\sqrt{2}\cos{\frac{\theta}{2}}$$
$$\sqrt{1-x^2}=\sin{\theta}$$
The equation wit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1298438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
What will be the equation of side $BC$.
The equation of two equal sides $AB$ and $AC$ of an isosceles triangle $ABC$ are $x+y=5$ and $7x-y=3$ respectively . What will be the equation of the side $BC$ if the area of the triangle $\triangle ABC$ is $5$ square units.
$a.)x+3y-1=0\\
b.)x-3y+1=0\\
c.)2x-y-5=0\\
\color{gre... | So, the coordinate of $A(1,4)$
We can set the coordinates of $B(a,5-a);C(b,7b-3)$
We have $|AB|^2=|AC|^2\implies(a-1)^2+(a-1)^2=(b-1)^2+\{7(b-1)\}^2$
$\implies a-1=\pm(5b-5)=\pm5(b-1)$
$$\triangle ABC=\dfrac12\begin{vmatrix}1 & 4 & 1 \\ a & 5-a & 1 \\ b & 7b-3 &1\end{vmatrix}=4|(a-1)(b-1)|$$
$\implies\triangle ABC=4\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1298619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Get rid of the square roots of the denominator: $\dfrac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}}$
How to get rid of the square roots of the denominator: $\dfrac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}}$?
*
*I squared the whole denominator, but that didn't help.
*Also I searched for a propriety or identity like $A^2-B^2$, but I did... | $$\begin{align}\frac{1}{\sqrt 7-2\sqrt 5+\sqrt 3}&=\frac{1}{\sqrt 7-2\sqrt 5+\sqrt 3}\cdot \frac{\sqrt 7-2\sqrt 5-\sqrt 3}{\sqrt 7-2\sqrt 5-\sqrt 3}\\&=\frac{\sqrt 7-2\sqrt 5-\sqrt 3}{(\sqrt 7-2\sqrt 5)^2-3}\\&=\frac{\sqrt 7-2\sqrt 5-\sqrt 3}{4(6-\sqrt{35})}\cdot\frac{6+\sqrt{35}}{6+\sqrt{35}}\\&=\frac{(\sqrt 7-2\sqrt ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1301362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Decomposition into partial fractions to compute an integral I'm having problems with:
$$\int_{-\infty}^{\infty}\frac{x^4+1}{x^6+1}dx$$
I was thinking: $\frac{x^4+1}{x^6+1}$ is an even function and the interval $(-\infty,\infty)$ is symmetric about 0, we could write the integral like:
$$2\int_{0}^{\infty}\frac{(x^4+1)}... | $$\begin{eqnarray*}\int_{\mathbb{R}}\frac{x^4+1}{x^6+1}\,dx&=&2\int_{\mathbb{R}^+}\frac{x^4+1}{x^6+1}\,dx=2\int_{0}^{1}\frac{x^4+1}{x^6+1}\,dx+2\int_{1}^{+\infty}\frac{x^4+1}{x^6+1}\,dx\\&=&4\int_{0}^{1}\frac{x^4+1}{x^6+1}\,dx=4\int_{0}^{1}\sum_{k\geq 0}(-1)^k\left(x^{6k}+x^{6k+4}\right)\,dx\\&=&4\sum_{k\geq 0}(-1)^k\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1302247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
How to find $\frac{0}{0}$ limit without L'Hôpital's rule I am having trouble solving this limit. I tried applying L'Hôpital's rule but I got $\frac{0}{0}$.
$$\lim_{x\to0} {\frac{\frac{1}{1+x^3} + \frac{1}{3}\log{\left(1+3x^3\right)}-1}{2\sin{\left(3x^2\right)}-3\arctan{\left(2x^2\right)}}}$$
I would appreciate any hint... | Recall some standard limits (which can be computed without using
l'Hospital's rule ):
\begin{eqnarray*}
\lim_{u\rightarrow 0}\frac{\frac{1}{1+u}-1+u}{u^{2}} &=&1 \\
\lim_{u\rightarrow 0}\frac{\log (1+u)-u}{u^{2}} &=&-\frac{1}{2} \\
\lim_{u\rightarrow 0}\frac{\sin u-u}{u^{3}} &=&-\frac{1}{6} \\
\lim_{u\rightarrow 0}\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1302336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Maximum of given expression? Suppose $a,b,c>0$ and further that $a^{2} + b^{2} + c^{2}=2abc + 1 $.
The problem is to find $\max \big(a-2bc\big) \big(b-2ca\big) \big(c-2ab\big) $.
Give me some help. I've tried $X=a-2bc$, $Y=b-2ca$, $Z=c-2ab$
which yields $X^2 + Y^2 + Z^2 = 1-2XYZ$, but $\frac12$ is not the maximum... | let $$x=a-2bc,y=b-2ca,z=c-2ab$$
since
$$a^2+b^2+c^2=2abc+1\Longrightarrow x^2+y^2+z^2+2xyz=1$$
so we only find $xyz$ maximu,since
$$1=x^2+y^2+z^2+2xyz\ge 3(x^2y^2z^2)^{\frac{1}{3}}+2xyz\Longrightarrow xyz\le\dfrac{1}{8}$$
so
$$(a-2bc)(b-2ac)(c-2ab)\le\dfrac{1}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find value of $xy\sqrt{y^2 - x^2}$ for the given differential equation.
If $(y^3 - 2x^2y)dx + (2xy^2 - x^3)dy = 0 $ , then prove that the value of $xy\sqrt{y^2 - x^2}$ is a constant.
This is what I've tried :
$$ y(y^2 - 2x^2) dx + x(2y^2 - x^2) dy = 0 \\
\cfrac{dy}{dx} = \cfrac{(2x^2 - y^2)y}{(2y^2 - x^2)x} \\
$$
I... | From the very beginning, use $y=x\,v$, $y'=x v'+v$ and replace in the original equation. You will get a common factor $x^3$; simplify to get $$x \left(2 v^2-1\right) v'+3 v \left(v^2-1\right)=0$$ which is separable becoming $$\frac 1x \frac{dx}{dv}=-\frac{2 v^2-1}{3 v \left(v^2-1\right)}$$ Use partial fraction decompos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proof of Different Polynomial Decompositions into Linear Factors From G. Polya "Mathematics and Plausible Reasoning" p. 18.
How do you prove that provided the roots of a polynomial are different from zero,
$$a_0 + a_1x+a_2x^2 + ... + a_nx^n$$
$$\,= a_0\left(1-\frac{x}{\alpha_1}\right)\left(1-\frac{x}{\alpha_2}\right).... | \begin{align}
&a_0+a_1x+\ldots+a_nx^n=\\
&\qquad=a_n(x-\alpha_1)\cdot\ldots\cdot(x-\alpha_n)=\\
&\qquad=a_n\alpha_1\frac{x-\alpha_1}{\alpha_1}\cdot\ldots\cdot\alpha_n\frac{x-\alpha_n}{\alpha_n}=\\
&\qquad=a_n\alpha_1\cdot\ldots\cdot\alpha_n\left(\frac{x}{\alpha_1}-1\right)\cdot\ldots\cdot\left(\frac{x}{\alpha_n}-1\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Finding kernel and range of a linear transformation We are given:
Find $\ker(T)$, and $\textrm{rng}(T)$, where $T$ is the linear transformation given by
$$T:\mathbb{R^3} \rightarrow \mathbb{R^3}$$
with standard matrix
$$ A = \left[\begin{array}{rrr}
1 & -1 & 3\\
5 & 6 & -4\\
7 & 4 & 2\\
\end{array}\r... | $$
A = \left[\begin{array}{rrr}
1 & -1 & 3\\
5 & 6 & -4\\
7 & 4 & 2\\
\end{array}\right]
$$
Consider a linear map represented as a $m × n$ matrix $A$ .
The kernel of this linear map is the set of solutions to the equation $Ax = 0$
$$
ker(A)=\{x \in R^n|Ax=0\}
$$
$$
det(A)=1(12+16)-(-1)(10+28)+3(20-42)=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
How i show this beautiful inequality :$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3} {2}(\frac{1}{\sqrt{3}})^{n-m}$? let $m,n$ be integers, show that if $ n>m\geq 0 $ :
$$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3}
{2}\left(\frac{1}{\sqrt{3}}\right)^{n-m}$$
where ... | This is a sketch with some details I didn't want to fill in but they'll check out. Let $n>m\geq 0$ be fixed and
$$f(x,y,z) = \frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}$$
we'll try to find the minimum under $xz+zy+yx=1$ using Lagrange multipliers. Let
$$ g(x,y,z) = xz + zy+ yx$$
Following the idea of th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
} |
Calculate the sum $\sum_{n=3}^{\infty}\frac{4n-3}{n^3-4n}$ $$\sum_{n=3}^{\infty}\frac{4n-3}{n^3-4n}$$
I think it is related to power series, because it is the topic, but I have no idea how to get there.
Could you give a hint?
| By partial fractions
$$\begin{align}\sum\limits_{n=3}^\infty\frac{4n-3}{n^3-4n}&=\sum\limits_{n=3}^\infty\frac{5}{8(n-2)}+\frac{3}{4n}-\frac{11}{8(n+2)}
\\&=\sum\limits_{n=1}^4\frac{5}{8n}+\sum\limits_{n=3}^4\frac{3}{4n}+\sum\limits_{n=5}^\infty\frac{5}{8n}+\frac{3}{4n}-\frac{11}{8n}
\\&=\sum\limits_{n=1}^4\frac{5}{8n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Is it possible to have an $n\times n$ real matrix $A$ such that $A^TA$ has an eigenvalue of $-1$?
Question: Is it possible to have an $n\times n$ real matrix $A$ such that $A^TA$ has an eigenvalue of $-1$?
I can prove that it is not possible for $n=1,2$, but I am not sure for the general case.
Case $n=1$: $a^2v=-v$ $... | According to wiki, the matrix $A^TA$ is positive definite for any non-singular $A$. The proof there is easily modified to show that singular $A$ give $A^TA$ positive semi-definite.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Calculus I: Find $y'$ equation: $x^4y+\sqrt{xy}=5$ I need to find $y'$ from the equation $x^4y+\sqrt(xy)=5$
Wolffram and Mathway(pro) are giving me way different answers and I can not follow what they are doing and where I am going wrong.
My attempt:
$$x^4y+\sqrt{xy}=5$$
$$x^4y+(xy)^{1/2} = 5$$
$$x^4y'+y(4x^3)+\frac... | Use multiplication rule:
$$(x^4y)'=4x^3y+x^4y'$$
$$(\sqrt{xy})'=\frac{y}{2\sqrt{xy}}+\frac{xy'}{2\sqrt{xy}}$$
$$5'=0$$
We get
$$4x^3y+x^4y'+\frac{y}{2\sqrt{xy}}+\frac{xy'}{2\sqrt{xy}}=0$$
$$x^4y'+\frac{xy'}{2\sqrt{xy}}=-4x^3y-\frac{y}{2\sqrt{xy}}$$
$$y'\bigg(x^4+\frac{x}{2\sqrt{xy}}\bigg)=-4x^3y-\frac{y}{2\sqrt{xy}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$(1+i)^6$ in polar form $re^{i\theta}$ I used De Moivre's formula and got
$$
\left(\frac{\sqrt{2}}{2}\right)^6 \times
\cos\left(6 \times \frac{1}{4\pi}\right) +
i\sin\left(6 \times \frac{1}{4\pi}\right) =
\frac{1}{8} e^{\frac{3}{2\pi}}.
$$
But the answer is $8e^{\frac{3}{2\pi}}$, can someone explain where t... | $$(1+i)^6=\left(\sqrt{2}e^{\frac{\pi}{4}i}\right)^6=\left(\sqrt{2}\right)^6e^{6\left(\frac{\pi}{4}i\right)}=8e^{\frac{3}{2}\pi i}=8e^{-\frac{\pi}{2}i}=-8i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Combinatorial identity of $\sum\limits_{k=1}^n \frac{{n \choose k}.{(-1)^{k}}}{k+1}$ I don't know how to deal with this example: Find a closed form of
$$
\sum\limits_{k=1}^n \frac{{n \choose k}.{(-1)^{k}}}{k+1}
$$
| $\sum_{k=0}^n {n \choose k}x^k=(1+x)^n$
Integrating both sides to get
$\sum_{k=0}^n {n \choose k}\frac{x^{k+1}}{k+1}=\frac{(1+x)^{n+1}}{n+1}+c$
For $x=0$ we have $0=\frac{1}{n+1}+c$, i.e. $c=-\frac{1}{n+1}$
$x\sum_{k=0}^n {n \choose k}\frac{x^{k}}{k+1}=\frac{(1+x)^{n+1}}{n+1}-\frac{1}{n+1}$
For $x=-1$ we have
$-\sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1312025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Minimal Polynomial of $\sqrt{2}+\sqrt{3}+\sqrt{5}$ To find the above minimal polynomial, let
$$x=\sqrt{2}+\sqrt{3}+\sqrt{5}$$
$$x^2=10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}$$
Subtracting 10 and squaring gives
$$x^4-20x^2+100=4(31+2\sqrt{60}+2\sqrt{90}+2\sqrt{150})$$
$$x^4-20x^2+100=4(31+4\sqrt{15}+6\sqrt{10}+10\sqrt{6})$$
$... | The 'easy' way to get at the minimal polynomial in this case is to take the product of all the terms $x\pm\sqrt2\pm\sqrt3\pm\sqrt5$; this is, in essence, because the Galois group over $\mathbb{Q}$ in this case is just $(\mathbb{Z}/2\mathbb{Z})^3$, with each of the three copies of $\mathbb{Z}/2\mathbb{Z}$ corresponding ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 0
} |
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.