Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Proof by induction $\sum_{k=1}^{2^n} \frac1k \ge 1 + \frac{n}2 $ Show that for all $n \in N , n \ge 1$ :
$\sum_{k=1}^{2^n} \frac1k \ge 1 + \frac{n}2 $
Here is what I got:
I. $A(1)$: $\sum_{k=1}^{2} \frac1k = 1.5 \ge 1.5 \;\; \implies A(1) \text{ is correct} $
II. $A(n)$ is correct: $\sum_{k=1}^{2^n} \frac1k ... | Base case:
Let $n=1$. Then $\sum_{k=1}^{2^1} \frac{1}{k} \geq 1 + \frac{1}{2}$ by evaluation / inspection.
Induction step (examining the $n+1$ case against the $n$ case):
$(\sum_{k=1}^{2^{n+1}} \frac{1}{k}) - (\sum_{k=1}^{2^n} \frac{1}{k}) = \sum_{k=2^n+1}^{2^{n+1}} \frac{1}{k}$
$(1+\frac{n+1}{2}) - (1+\frac{n}{2}) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1520295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Knowing that m and n are two positive integers, solve the equation m/n = n.m (ex.: 5/2 = 2.5) Knowing that $m$ and $n$ are two positive integers, find all the solutions for the equation:
$\frac{m}{n} = n.m$
For example, the pair $(m=5, n=2)$ is a solution because $\frac{5}{2} = 2.5$ while $(m=294, n=17)$ comes close ... | $m/n = n.m= n + m/10^k$ so $10^kn^2 + mn - 10^km$
So by quadratic equation $n = \frac{-m + \sqrt{m^2 + 4m*10^{2k}}}{2*10^k} $
Notice:
m has k digits.
So n = $\frac{-m + \sqrt{m^2 + 4m*10^{2k}}}{2*10^k}$ has $\lceil k/2 \rceil$ digits. But m/n $\approx$ n then has ${\lceil k/2 \rceil}$ digits, so $m$ has between $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1523261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine the complex contour integral $\oint \limits_{C} \frac{2}{z^3+z}dz$ without using Residue Theorems
Without residue theory, determine $$\oint \limits_{C} \frac{2}{z^3+z}dz$$ if $C: \big|~z~-~\frac{i}{2}~\big|=1$ is positively oriented.
We first find that our integrand has three distinct singular points $z=\{-... | It's correct. Essentially what you've done is exactly how the proof of the residue theorem goes anyway - you surround each isolated pole with a small contour and apply Cauchy's theorem to each individual pole. The result is the sum of the residues, multiplied by $2\pi i$, which is the residue theorem. Now you know wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1527581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Calculate the limit $\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)}$ $$\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)}$$
My attempt
\begin{align}
\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)} &= \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)^{(-n+1)/(n+1)} \\... | For all $n$, we have $0 < \frac{n^2}{n^2 + 1} < 1.$
For $n > 1$, we have $\frac{n-1}{n+1} > 0.$
Therefore, for $n > 1$,
$$ \left(\frac{n^2}{n^2 + 1}\right)^{(n-1)/(n+1)} < 1.$$
On the other hand, $\frac{n^2 + 1}{n^2} > 1$ for all $n$, and for $n > -1$ we have $\frac{2}{n+1} > 0$, so
\begin{align}
\left(\frac{n^2}{n^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Trigonometry equation, odd-function. So I have the following equation:
$\sin\left(x-\frac{\pi}{6}\right) + \cos\left(x+\frac{\pi}{4}\right)=0$
It should be solved using the fact that Sin is an odd function, I can not really get the gripp of how and what I need to do? Any sugestions?
| Sorry that this approach does not use the oddness of the Sine function, after I finishing writing this it was too late for me to turn back and utilize that property.
Using trigonometric addition formulae as mentioned in the comments:
$$\sin(A \pm B)= \sin A\cos B\pm\cos A\sin B$$
$$\cos(A \pm B)= \cos A\cos B\mp\sin A\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Finding a sequence satisfying this recurrence relation? I just don't even know where to start with this,
Find a sequence $(x_n)$ satisfying the recurrence relation:
$2x_n$$_+$$_2$ = $3x_n$$_+$$_1$ + $8x_n$ + $3x_n$$_-$$_1$ Where n is a natural number and
$x_0$ = -1,
$x_1$ = 3 and
$x_2$ = 3
Thanks in advance!
| Let $2x_{n+2} = 3 \, x_{n+1} + 8 \, x_{n} + 3 \, x_{n-1}$, with $x_{0} = -1,
x_{1} = 3$ and $x_{2} = 3$ for which letting $x_{n} = p^{n}$ the equation
$$2 \, p^{3} - 3 \, p^{2} - 8 \, p - 3 = 0$$
is obtained. This equation can be factored to $(p-3)(p+1)(2p+1) = 0$ and leads to the roots $p \in \{ 3, -1, -1/2 \}$. Fro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1532747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Simultaneous equations How can I find the solution $(3, 3, \frac{-3}{2})$ from the following system of equations:
$x+(y-1) z = 0$
$(x-1) z+y = 0$
$x (y-1)-y+2 z = 0$
I have done eq1 - eq2 to find the other solutions. How would i get to the solution $(3, 3, \frac{-3}{2})$
| $x+(y-1) z = 0$ : (eq1)
$(x-1)z+y = 0$ : (eq2)
$x (y-1)-y+2 z = 0$ : (eq3)
Rearranging eq1 and we have: $x=-(y-1)z$. Subbing this into eq2 and eq3 gives:
$(-(y-1)z-1)z+y=0$ : (eq4)
$-(y-1)z(y-1)-y+2z=0$ : (eq5)
Rearranging eq5 we get: $z=\frac{y}{2-(y-1)^2}$
Subbing this into eq4 gives:
$$\left(-(y-1)\frac{y}{2-(y-1)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that T is a rotation through an angle, and find the angle $T\left[\begin{array}{c}x \\ y \end{array}\right] = \frac{1}{\sqrt{2}}\left[\begin{array}{c}x+y \\ -x+y \end{array}\right]$
This is my attempt, but I'm stuck. Can someone continue?
$T\left[\begin{array}{c} x \\ y \end{array}\right] = \left[\begin{array}{cc}... | You want to find a rotation matrix that has the same effect as the matrix you have. You can then easily find the angle of rotation. We can write it like this to make it clearer (I've just re-written your first equation as a
matrix product rather than the way you have it):
$$
\left( \begin{array}{ccc}
\cos\theta & -\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1534454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to rationalize denominator? Suppose $c$ is not a complete square integer, ${a_0},{a_1} \in \mathbb{Q}$, we have
$$
\frac{1}{{{a_0} + {a_1}\sqrt c }}
= \frac{{{a_0} - {a_1}\sqrt c }}{{a_0^2 - a_1^2c}}.
$$
We need to show ${{a_0} + {a_1}\sqrt c } = 0$ iff ${a_0} = {a_1} = 0$. I know it's not hard.
Suppose $c$ is not... | I found a way. Let ${p^3} = c$, we write
$\left\{ \begin{gathered}
{a_0} + {a_1}p + {a_2}{p^2} = S \hfill \\
{a_2}c + {a_0}p + {a_1}{p^2} = pS \hfill \\
{a_1}c + {a_2}cp + {a_0}{p^2} = {p^2}S \hfill \\
\end{gathered} \right.$
So
$\left| {\begin{array}{*{20}{c}}
{{a_0} - S}&{{a_1}}&{{a_2}} \\
{{a_2}c - pS}&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1536441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Find the length of $\sqrt x$. Let $f(x) = \sqrt x$. Find the length of the curve for $0\le x \le a$.
So I know the formula is:
$$\int_0^a \sqrt{(1+(f')^2(x)} \ dx = \ldots = \int_0^a \sqrt{1+\frac{1}{4x}} \ dx$$
Now, how do I evaulate this integral?
I tried the subtitution $t = \sqrt {1+\frac{1}{4x}}$ but it became so... | $$\int\sqrt{1+\frac{1}{4x}}\space\text{d}x=$$
$$\int\sqrt{1+\frac{x+\frac{1}{4}}{x}}\space\text{d}x=$$
Substitute $u=\frac{x+\frac{1}{4}}{x}$ and $\text{d}u=\left(\frac{1}{x}-\frac{x+\frac{1}{4}}{x^2}\right)\space\text{d}x$:
$$-\frac{1}{4}\int\frac{\sqrt{u}}{(1+u)^2}\space\text{d}u=$$
Substitute $s=\sqrt{u}$ and $\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1536832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Double integral with Polar coordinates - hard example Calculate using polar coordinates:
$$\iint_{D}^{} (x^2+y^2)^\frac{1}5 \ dx \ dy $$
where D is the region inside the circle with radius 1.
Working:
D: $ \ x^2+y^2=1 \\ $ so
$ 0 \leq r \leq 1 \ \ , $ $0 \leq \theta \leq \pi \ $
and $x=rcos\theta \ ,$ $y=rsin\theta $ s... | According to your calculations it should go on as:
$$\iint_{D}^{} (x^2+y^2)^\frac{1}5 \ dx \ dy $$
$$=\int_{0}^{1} \int_{0}^{2\pi} (r^2\cos^2\theta + r^2\sin^2\theta)^\frac{1}{5} \cdot r d\theta \ dr $$
$$=\int_{0}^{1} \int_{0}^{2\pi} r^\frac{2}{5} \cdot r d\theta \ dr$$
$$=\int_{0}^{1} r^\frac{7}{5} dr\int_{0}^{2\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find number of words of length n over alphabet {A, B, C} where any nonterminal A must be followed by B. Let $a_n$ be the number of words of length n over the alphabet {A, B, C} such that any nonterminal A has to be immediately followed by B. Find $a_n$.
Here's what I know:
If the word starts with $C$ or $B$, any valid ... | This can be done recursively. We have $a_1 = 3$ and $a_2 = 7$ by inspection.
Then for $n>2$, we see that there are three possible ways to start a sequence: $AB$, $B$, or $C$. Depending on which one we choose, we recurse into a smaller sequence of known size, where the counting process is repeated all over again.
This s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1538638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do you prove that $\arcsin( \frac{1}{2} \sqrt{2-\sqrt {2-2x}})=\frac{\pi}{8} + \frac{1}{4} \arcsin x$? I have the task to prove that
$$
\arcsin( \frac{1}{2} \sqrt{2-\sqrt {2-2x}})=\frac{\pi}{8} + \frac{1}{4} \arcsin x ,\left|x\right|\le 1
$$
I do not have any ideas from where I should start.
Can anyone help m... | Let $\arcsin( \frac{1}{2} \sqrt{2-\sqrt {2-x}})=y$
$\implies-\dfrac\pi2\le y\le\dfrac\pi2$
and $\dfrac12\sqrt{2-\sqrt {2-x}}=\sin y$
$\implies2-\sqrt {2-x}=(2\sin y)^2$
Using $\cos2A=2\cos^2A-1=1-2\sin^2A,$
$\sqrt {2-x}=2-4\sin^2y=2\cos2y$
$\implies x=2-(2\cos2y)^2=-2\cos4y$
$\implies\cos4y=-\dfrac x2$
$\arccos\left(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Solving the integral $\int_{-1}^1 2\sqrt{2-2x^2}\,dx$ I'm working on a triple integral and have managed to get it to a certain point:
$$\int_{-1}^1 2\sqrt{2-2x^2}dx $$
When I check this with WolframAlpha it gives the answer $\pi\sqrt{2}$, which is the right answer to the problem.
I know I should do a trig substitutio... | $$\int 2\sqrt{2-2x^2}\, dx=2\sqrt{2}\int \sqrt{1-x^2}\, dx$$
For the expression to make sense, we must have $1-x^2\ge 0$, i.e. $-1\le x\le 1$, so exists $t\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ such that $x=\sin t$. Then $dx=\cos t\, dt$.
$$=2\sqrt{2}\int\cos^2 t\, dt=2\sqrt{2}\int \frac{1+\cos(2t)}{2}\, dt$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of a fraction with a square root
Find $$\lim_{x \to 2} \frac{4-x^2}{3-\sqrt{x^2+5}}$$ (without L'Hopital)
Where is the following wrong? (The limit is 6.)
\begin{align}\lim_{x \to 2} \frac{4-x^2}{3-\sqrt{x^2+5}}& =\lim_{x \to 2} \frac{(2-x)(2+x)}{\sqrt{9-x^2-5}}= \\
& =\lim_{x \to 2} \frac{(2-x)(2+x)}{\sqrt{-x^2... | Let $f(x) = x^2, g(x) = \sqrt {x^2 + 5}.$ The expression equals
$$\frac{(f(x) - f(2))/(x-2)}{(g(x) - g(2))/(x-2)}.$$
By definition of the derivative, as $x\to 2$ this $\to f'(2)/g'(2),$ which is easy to compute.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Confused about multinomials. Can we write $\binom{n}{a,b,c}=\binom{n}{a}\binom{n-a}{b}\binom{n-a-b}{c}$ if $a+b+c \le n$? Can we write $\binom{n}{a,b,c}=\binom{n}{a}\binom{n-a}{b}\binom{n-a-b}{c}$ if $a+b+c \le n$?
The definition for multinomial says $a+b+c=n$ must hold or else $\binom{n}{a,b,c}=0$.
I found that if $a+... | No, for any $(a,b,c)\in\Bbb N^3$ and $a+b+c\leq n$ then,
$$\dbinom{n}{a}\dbinom{n-a}{b}\dbinom{n-a-b}{c}=\dbinom{n\qquad\qquad\qquad}{a,b,c, n-a-b-c}\;(n-a-b-c)!$$
Unlike the binomial coefficient, the usual convention is not to leave the last lower term of the multinomial coefficient implicit. The sum of the lower te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding $\binom n0+\binom n3+\binom n6+\cdots $ How to get
$$\binom n0 + \binom n3 + \binom n6 + \cdots$$
MY ATTEMPT
$$(1+\omega)^n = \binom n0 + \binom n1 \omega^1 + \binom n2 \omega^2 + \cdots$$
$$(1+\omega^2)^n = \binom n0 + \binom n1 \omega^2 + \binom n2 \omega^4 + \cdots $$
$$(1 + 1)^n = 2^n = \binom n0 + \binom... | $$(1+\omega)^n+(1+\omega^2)^n+2^n\\
=(-\omega^2)^n+(-\omega)^n+2^n\\
=(-1)^n(\omega^{2n}+\omega^n)+2^n\\
$$
i) $n=3m$:
$$(-1)^n(\omega^{2n}+\omega^n)+2^n=2\cdot(-1)^n+2^n$$
ii) $n=3m+1$ or $3m+2$:
$$(-1)^n(\omega^{2n}+\omega^n)+2^n=(-1)^{n+1}+2^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding $\sqrt{(14+6\sqrt 5)^3}+\sqrt{(14-6\sqrt 5)^3}$ Find $$\sqrt{(14+6\sqrt{5})^3}+ \sqrt{(14-6\sqrt{5})^3}$$
A.$72$
B.$144$
C.$64\sqrt{5}$
D.$32\sqrt{5}$
How to cancel out the square root?
| The square of that sum equals:
$$(14+6\sqrt{5})^3+(14-6\sqrt{5})^3+2\sqrt{(14^2-36\cdot 5)^3}$$
that is:
$$ 2\cdot 14^3 +2\cdot 3\cdot 14\cdot 36\cdot 5+2\sqrt{16^3} = 2^8\cdot 3^4$$
so the original sum equals $2^4\cdot 3^2 = \color{red}{144}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1548512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Indefinite integral with substitution For my engineering math course I got a couple of exercises about indefinite integrals. I ran trought all of them but stumbled upon the following problem.
$$\int \frac{1-x}{\sqrt{1+x-2x^2}}\,dx $$
We can write $1+x-2x^2$ as $(1-x)(2x+1)$
So I got:
$$
\int \frac{1-x}{\sqrt{1+x-2x^2... | Substitute $t = \sqrt{6} \sin(u)$, so $dt = \sqrt{6} \cos(u) du$ and the integrand becomes (up to losing the $-\frac{1}{\sqrt{2}}$ at the start)
$$\sqrt{6} \int \sqrt{6} \sin^2(u) du$$
I think you can do that!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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If $x_{n+1}= \frac{x_n^2+x_n+1}{x_n+1}$ find$ \sum_{n=1}^{p}\frac{1}{1+x_n}$ If $x_{n+1}= \frac{x_n^2+x_n+1}{x_n+1}$ find $ \sum_{n=1}^{p}\frac{1}{1+x_n}$,
given $x_1=1/2$
The original question was for p=2014, but I want to find the general solution.
I have tried using $x^2+x+1=(x- \omega)(x-\omega^2)$ and breaking $... | Observe that $x_{n + 1} = x_n + \frac {1} {1 + x_n}$ so $\frac {1} {1 + x_n} = x_{n + 1} - x_n$ so $$\sum_{n = 1}^{p} \frac {1} {1 + x_n} = \sum_{n = 1}^{p} (x_{n + 1} - x_n) = x_{p + 1} - x_1 = x_{p + 1} - \frac {1} {2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1551594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Question on indefinite integrals I have to integrate:
$$I_2 = \int \frac{e^{2x} - e^{x} + 1}{(e^x\cos(x) - \sin(x))\cdot \left(e^x\sin(x) + \cos(x)\right)} \text{d}x$$
I simply can't understand from where to begin with. Please help me in solving this problem.
| I consulted Moor and he suggested an approach involving differentiating the denominator.
Let $f(x) = (e^x \cos (x) - \sin(x))$ and $g(x) = (e^x \sin x + \cos x)$. Then, note that, by the product rule $(fg)'=f'g +g'f = (e^x\cos(x)-e^x\sin(x) - \cos(x))(e^x \sin (x) + \cos (x)) + (e^x\sin (x) + e^x \cos(x)-\sin(x))(e^x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Zero divided by zero must be equal to zero What is wrong with the following argument (if you don't involve ring theory)?
Proposition 1: $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}$ is not equal to $0$
$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac... | Using the definition of a Ring, it is pretty simple to show that
$$\forall a, 0\times a=a\times 0=0$$
Thus, If we were to define division (only in the case where $c=ab=ba$):
$$\forall a, a=\frac{0}{0}$$
Which can be true if and only if your ring is the Zero ring
$$\{0\}$$
Therefore, when $\frac{0}{0}$ exists, it is act... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
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"answer_id": 9
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2 problems related to the number 2015
*
*Let $p=\underbrace{11\cdots1}_\text{2015}\underbrace{22\cdots2}_\text{2015}$. Find $n$, where $n(n+1) = p$
*Prove that $\frac{1}{2^2} + \frac{1}{3^2} + \cdots + \frac{1}{2015^2} < \frac{2014}{2015}$
For 1, I tried dividing in various ways until I got a simpler expression, bu... | $1)$ $n=\frac{10^{2015}-1}{3}=3...3$ with $2015$ $3's$ :
$$\frac{10^{2015}-1}{3}\times \frac{10^{2015}+2}{3}=\frac{10^{4030}+10^{2015}-2}{9}=10^{2015}\times \frac{10^{2015}-1}{9}+2\times \frac{10^{2015}-1}{9}=10^{2015}R_{2015}+2\times R_{2015}$$
$2)$ follows from $\frac{1}{2^2}+...+\frac{1}{2015^2}<\sum_{j=2}^{\infty} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Summation of Greatest Integer $ \sum_{k=1}^{n} \left \lfloor{\sqrt{k}}\right \rfloor $ I want to know the concept behind summing up a Greatest Integer function like $\sum_{k=1}^{n} \left \lfloor{x}\right \rfloor $ can be evaluated by simply writing the series but something like $$ \sum_{k=1}^{n} \left \lfloor{\sqrt{k}}... | For any given $m\geq0$, we have $\lfloor\sqrt{k}\rfloor=m$ if and only if $m^2\leq k<(m+1)^2$. So the number of times $m$ occours in the sum is
$$(m+1)^2-m^2=2m+1.$$
Setting $\ell:=\lfloor\sqrt{n}\rfloor$ this indeed leads to the simplification you already found:
$$\sum_{k=0}^n\lfloor\sqrt{k}\rfloor=\ell\cdot(n+1-\ell^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1557966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate the integral $\int \sqrt{(x-a)(b-x)}$ I'm trying to figure out how to evaluate the following integral:
$$\int \sqrt{(x-a)(b-x)} \, dx $$
I've tried various trig substitutions, but can't seem to get anywhere. This is an exercise in Apostol's Calculus Volume 1 (Section 6.22, Exercise 46). The solution provided ... | $$\sqrt{(x-a)(b-x)} = \sqrt{\left(\frac{a+b}{2}\right)^2 - ab - \left(x^2 - (a+b)x + \left(\frac{a+b}{2}\right)^2 \right)}$$
$\left(\frac{a+b}{2}\right)^2 - ab = \left(\frac{a-b}{2}\right)^2 $, call it $k^2$ for some $k$. The remaining term is $\left(x - \frac{a+b}{2}\right)^2$, and you may make the substitution $ u = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
If $ c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}\;,$ prove that it is Isosceles Triangle
In a $\triangle ABC\;,$ If $\displaystyle c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}\;,$ Then how can we prove that $\triangle ABC$
is an Isoceles $\triangle.$
$\bf{My\; Try::}$ Using $\displaystyle \frac{a}{\sin A}=\frac{b... | using the sine rule in the form:
$$
b = 4R\sin\frac{B}2\cos\frac{B}2 \\
c = 4R\sin\frac{C}2\cos\frac{C}2
$$
we obtain:
$$
(a+c)\sin\frac{B}2 = (a+b)\sin\frac{C}2
$$
which, together with the original relation:
$$
c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}
$$
gives
$$
b \tan\frac{B}2 = c\tan\frac{C}2
$$
but the functi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1560934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Evaluation of $\int_{0}^{1} \frac{1}{x+\sqrt{1-x^2}} \space dx$ Evaluate
$$\int_{0}^{1} \frac{1}{x+\sqrt{1-x^2}} \space dx$$
My main concern is finding the indefinite integral as once i have that the rest is fairly straight forward. Please give a detailed answer with reference to why you made each substitution (what in... | BIG HINT:
$$\int\frac{1}{x+\sqrt{1-x^2}}\space\text{d}x=$$
Substitute $x=\sin(u)$ and $\text{d}x=\cos(u)\space\text{d}u$.
Then $\sqrt{1-x^2}=\sqrt{1-\sin^2(u)}=\cos(u)$ and $u=\arcsin(x)$:
$$\int\frac{\cos(u)}{\sin(u)+\cos(u)}\space\text{d}u=$$
$$\int\frac{\sec^3(u)}{\sec^3(u)}\cdot\frac{\cos(u)}{\sin(u)+\cos(u)}\spa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1561559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Diophantine equation $x^2 + y^2 = z^3$ I have found all solutions to the Diophantine equation $x^2 + y^2 = z^3$ when
$z$ is odd. I am having some difficulty finding the solutions when $z$ is even. I am asking for a proof that provides the solutions where $z$ is even. I want the proof to be elementary and use only Numb... | Unfortunately, there isn't (apparently) one complete polynomial parameterization to
$$x^2+y^2 = z^k\tag1$$
when $k>2$. For $k=2$, the complete solution is,
$$x,\,y,\,z = (a^2-b^2)s,\; (2ab)s,\; (a^2+b^2)s$$
where $s$ is a scaling factor. Using complex numbers $a+b i$, one can generalize the method. For $k=3$, it is
$$x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1561774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Limit of function $x^2-x\cot\left(\frac{1}{x}\right)$ How To compute $\lim_{x \to \infty} x^2 -x \cot(1/x)$? Wolfram says it is $\frac{1}{3}$ and I know it is supposed to converge to a number other than 0 but I keep getting infinity.
| WithOut L'Hospital:
i will guess that the limit exist so :
$$l=\lim_{y\to 0}\frac{1-y \cot y}{y^2}$$
$$l=\lim_{y\to 0}\frac{\sin y-y \cos y}{y^2\sin y}$$
$$l=\lim_{y\to 0}\frac{\sin y-y \cos y}{y^3}$$
$y\to 2y$
$$l=\frac{1}{4}\lim_{y\to 0}\frac{\sin y \cos y-y (1-2\sin^2y)}{y^3}$$
$$l=\frac{1}{4}\lim_{y\to 0}\frac{\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1561829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Length of a side of a triangle given the angles and the area We have $\triangle ABC$ with the following measures:
$A = 65^\circ$,
$B= 75^\circ$,
$\text{area}= 88\,m^2$.
How can I determine the longest side?
| We want to find $AC=b$.
We have
$$88=\frac 12ac\sin75^\circ\tag1$$
By the law of sines, we have
$$\frac{a}{\sin 65^\circ}=\frac{b}{\sin 75^\circ}=\frac{c}{\sin 40^\circ}$$
So,
$$a=\frac{\sin 65^\circ}{\sin 75^\circ}b,\quad c=\frac{\sin 40^\circ}{\sin 75^\circ}b\tag2$$
From $(1)(2)$,
$$88=\frac 12\cdot\frac{\sin 65^\cir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1564322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Finding $\int_{-\infty}^\infty \frac{x^2}{x^4+1}\;dx$ $$\int_{-\infty}^\infty \frac{x^2}{x^4+1}\;dx$$
I'm trying to understand trigonometric substitution better, because I never could get a good handle on it. All I know is that this integral is supposed to reduce to the integral of some power of cosine. I tried $x^2=... | Via contour integration $\oint_C \frac{z^{\alpha-1}}{{1+z}} dz$ it was shown that
$$\int_0^\infty \frac{y^{\alpha-1}}{1+y} dy = \frac{\pi}{\sin \pi \alpha}.$$
Let $y=x^4$ so that $dy = 4x^3 dx.$
$$\int_{-\infty}^\infty \frac{x^2}{1+x^4} dx =\frac{1}{4} \int_{-\infty}^\infty \frac{y^{-1/4}}{1+y} dy = \frac{1}{2} \int_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 13,
"answer_id": 12
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Find all incongruent solutions of $x^8\equiv3\pmod{13}$. Find all incongruent solutions of $x^8\equiv3\pmod{13}$.
I know that $2$ is a primitive root of $13$ and that $2^4\equiv3\pmod{13}$, so we want to solve $x^8\equiv2^4\pmod{13}$.
Now, $\gcd(8,\phi(13))=\gcd(8,12)=4$ divides the exponent of $2$, which is $4$, so $x... | $$x^8-3\equiv x^8-16\equiv \left(x^2-2\right)\left(x^2+2\right)\left(x^4+4\right)\pmod{13}$$
$x^2\equiv 2\pmod{13}$ and $x^2\equiv -2\pmod{13}$ are both unsolvable (by Quadratic Reciprocity), because $13\equiv 5\pmod{8}$.
$$x^4+4\equiv x^4-9\equiv \left(x^2+3\right)\left(x^2-3\right)\pmod{13}$$
$$x^2+3\equiv x^2-36\equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How do you calculate this limit without using L'Hopital or Taylor? I need to calculate this limit $$\lim_{x\rightarrow 0}\frac{xe^x-(e^x-1)}{x^2}$$ but only with elementary methods, so no L'Hopital/Taylor.
I've done quite a bit of manipulation but nothing seems to work, could you give me a hint?
| Clearly we have $$\frac{xe^{x} - (e^{x} - 1)}{x^{2}} = \frac{xe^{x} - x - (e^{x} - 1 - x)}{x^{2}} = \frac{e^{x} - 1}{x} - \frac{e^{x} - 1 - x}{x^{2}}\tag{1}$$ and we know that $(e^{x} - 1)/x \to 1$ as $x \to 0$ hence it follows from the equation $(1)$ that our job is done if we can calculate the limit $$\lim_{x \to 0}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find the product xy. Given
$$
\log (x) + \frac{\log (xy^8)}{(\log x)^2+(\log y)^2} = 2\\
\log (y) + \frac{\log \left(\frac{x^8}{y}\right)}{(\log x)^2+(\log y)^2} = 0
$$
Find the product $xy$ if both $x$ and $y$ are real.
After applying basic log identities, I tried equating value of $ \large\frac{1}{(\log x)^2}+\frac{1... | hint put $\log(x)=a\,\log(y)=b$ this leads to two equations with two unknowns.
They are:
(I) $a+\frac{a+8b}{a^2+b^2}=2$ and
(II) $b+\frac{(8a-b)}{a^2+b^2}=0.$
Mutiplying (I) by $b$ and (II) by $a$ we get $ab+\frac{ab+8b^2}{a^2+b^2}=2b$ and $ab+\frac{8a^2-ab} {a^2+b^2}=0$, adding them up, we get $2ab+8=2b$ and $ab+4=b$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Show $\sin(\frac{\pi}{3})=\frac{1}{2}\sqrt{3}$ I have to show that
$$\sin\left(\frac{\pi}{3}\right)=\frac{1}{2}\sqrt{3}$$
and
$$\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}$$
Should I use the exponential function?
| Without using triangles:
We use the following core formulas to prove this identity:
*
*Pythagorean identity: $cos^2(x) + sin^2(x) = 1$
*Angle transformation formula: $sin(x - y) = sin(x)cos(y) - cos(x)sin(y)$
*Double-angle formula: $sin(2x) = 2sin(x)cos(x)$
To show: $sin(\frac{\pi}{3}) = \frac{\sqrt3}{2}$
We start... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 2
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Find $\int _{0}^{2\pi}{1\over a\sin t+b \cos t +c}$ where $\sqrt{a^2+b^2}=1Find $\int _{0}^{2\pi}{1\over a\sin t+b \cos t +c}$ where $\sqrt{a^2+b^2}=1<c$. I am lead to believe I should be using curves but I really don't understand what curves to choose and how to properly use them. I would appreaciate some help on the ... | Suppose we seek to evaluate
$$\int_{0}^{2\pi} \frac{1}{a\sin x + b\cos x + c} \; dx$$
where $a^2+b^2 = 1 < c.$
Introduce $z=\exp(ix)$ so that $dz=iz \; dx$ to get
$$\int_{|z|=1}
\frac{1}{a(z-1/z)/2/i + b(z+1/z)/2 + c}
\frac{dz}{iz}
\\ = \int_{|z|=1}
\frac{1}{az^2/2-a/2 + ibz^2/2+ib/2 + ciz}
\; dz.$$
Call this function ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Rational solutions $(a,b)$ to the equation $a\sqrt{2}+b\sqrt{3} = 2\sqrt{a} + 3\sqrt{b}$ Find all rational solutions $(a,b)$ to the equation
$$a\sqrt{2}+b\sqrt{3} = 2\sqrt{a} + 3\sqrt{b}.$$
I can see that we have the solutions $(0,0), (2,0), (0,3), (3,2), (2,3)$, and I suspect that there are no more.
I tried to do the ... | We can use the interesting
Proposition. The square roots of the square-free naturals are $\Bbb Q$-linearly independant.
Write $a=c^2n$, $b=d^2m$ with $c,d\in\Bbb Q$ and $n,m$ square-free (note that $a,b$ cannot be negative anyway). Then we have
$$ a\sqrt 2+b\sqrt 3-2c\sqrt n-3d\sqrt m=0.$$
By the proposition, the follo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Calculating the convolution of piecewise constant functions Let $f(x) = \frac{1}{2}$ on $[-1,1]$. Find $f*f*f$
$(f*g)(x)=\int\limits_{-\infty}^\infty f(t)g(x-t)\,dt$.
So $(f*f)(x)=\begin{cases} \frac{1}{4}x+\frac{1}{2} & -2\le x \le 0 \\ \frac{-1}{4}x+\frac{1}{2} & 0 < x \le 2 \\ 0 & \text{otherwise} \end{cases}$
Thi... | The convolution is $f*f*f(w)=\int_{x}\int_{y} f(x)f(y)f(w-x-y) dydx$. We seek the integration limits, firstly for $x$ given $w$ (with $-3\lt w\lt 3$), then for $y$ given $w,x$.
Limits for $x$ given $w$:
*
*$-1\lt x\lt 1$
*$w-2\lt x\lt w+2\qquad \text{(lower bound comes from $x=w-y-z\;$ with $y=z=1$)}$
$\qquad\qquad... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1570990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve the given Differential Equation Solve the non-linear first order differential equation
$$\frac{dy}{dx}=\frac{x+2y-5}{2x+xy-4} $$
I tried substituting $x=X+h$ and $y=Y+k$ but the $xy$ term is creating problem.
How to solve it?
Any suggestion is appreciated.
| $\dfrac{dy}{dx}=\dfrac{x+2y-5}{2x+xy-4}$
$(x+2y-5)\dfrac{dx}{dy}=(y+2)x-4$
This belongs to an Abel equation of the second kind.
Let $u=x+2y-5$ ,
Then $x=u-2y+5$
$\dfrac{dx}{dy}=\dfrac{du}{dy}-2$
$\therefore u\left(\dfrac{du}{dy}-2\right)=(y+2)(u-2y+5)-4$
$u\dfrac{du}{dy}-2u=(y+2)u-2y^2+y+6$
$u\dfrac{du}{dy}=(y+4)u-2y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1571133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
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I want to show that $\sum_{n=1}^\infty \sin{\left(n\frac{\pi}{2}\right)}\cdot\frac{n^2+2}{n^3+n} $ converges or diverges. I want to show that $$\sum_{n=1}^\infty \sin{\left(n\frac{\pi}{2}\right)}\cdot\frac{n^2+2}{n^3+n} $$ converges (absolutly?) or diverges.
My idea was: $n=2k+1$ and then it becomes: $$\sum_{n=1}^\inft... | The sum converges by the alternating series test.
It turns out that the sum may be evaluated using complex analysis, i.e., the residue theorem. First simplify the sum
$$\sum_{n=0}^{\infty} \sin{\left (n \frac{\pi}{2} \right )} \frac{n^2+2}{n^3+n} $$
as
$$\sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} \left [1+\frac1{ (2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1572075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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show that the given series converges Prove that the series:
$$
\sum _{n=1} ^\infty \frac{(-1)^{\lfloor n/3 \rfloor}}{n}
$$
I tried to use Leibniz rule but I did not succeed. Any suggestions? Thanks for helping!
| \begin{align}
\sum _{n=1}^{\infty} \frac{(-1)^{\lfloor n/3\rfloor}}{n} &= \frac{3}{2}+\sum _{n=1}^{\infty/3} \left[ \frac{(-1)^{n}}{3n} + \frac{(-1)^{n}}{3n+1} +\frac{(-1)^{n}}{3n+2} \right]\\
&=\frac{3}{2}+\sum _{n=1}^{\infty} (-1)^n \left(\frac{1}{3 n}+\frac{1}{3 n+1} + \frac{1}{3 n+2}\right)\\
\end{align}
By the Al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1573687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Factoring Multivariable polynomials Determine a constant $k$ such that the polynomial
$$ P(x, y, z) = x^5 + y^5 + z^5 + k(x^3+y^3+z^3)(x^2+y^2+z^2) $$
is divisible by $x+y+z$.
Im having trouble utilizing the multivariable factor theorem for this problem>
| Write these symmetric expressions in terms of $s = x+y+z$, $q = x y + x z + y z$, $p=x y z$. We get
\begin{eqnarray}
x^5 + y^5 + z^5 &=& s^5 - 5 q s^3 + 5 q^2 s + 5 p s^2 - 5 p q\\
x^3 + y^3 + z^3 &=& s^3 - 3 q s + 3 p \\
x^2 + y^2 + z^2 &=& s^2 - 2 q
\end{eqnarray}
Therefore
$$x^5 + y^5 + z^5 + k (x^3 + y^3 + z^3)(x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1574284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calculus - limit of a function: $\lim\limits_{x \to {\pi \over 3}} {\sin (x-{\pi \over 3})\over {1 - 2\cos x}}$ How do you compute the following limit without using the l'Hopital rule?
If you were allowed to use it, it becomes easy and the result is $\sqrt{3}\over 3$ but without it, I am not sure how to proceed. $$\li... | Proceed as follows:
$$\lim_{x\to\frac{\pi}{3}} \frac{\sin(x-\pi/3)}{1-2\cos x}$$
$$=\lim_{t\to0} \frac{\sin t}{1-2\cos (t+\frac{\pi}{3})}$$
$$=\lim_{t\to0} \frac{\sin t}{1-2\cos t\cos \frac{\pi}{3}+2\sin t \sin\frac{\pi}{3}}$$
$$=\lim_{t\to0} \frac{\sin t}{(1-\cos t)+2\sin t \sin\frac{\pi}{3}}=\frac{1}{2\sin\frac{\pi}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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When does $nx^4+4x+3=0$ have real roots?
Find all positive integers $n$ such that the equation $nx^4+4x+3=0$ has real roots.
I think the answer must also include the cases with $2$ real roots. But my main question is, how do I start? Thanks.
| We have
\begin{align}
nx^4+4x+3 & = (n-1)x^4 + x^4+4x+3 = (n-1)x^4 + (x+1)^2((x-1)^2+2)\\
& = (n-1)x^4 + (x^2-1)^2 + 2(x+1)^2
\end{align}
Hence, for $n \geq 2$, the function is a sum of squares and is always positive, since all three of them cannot be simultaneously zero and therefore has no real roots.
For $n=1$, we s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1578553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Is it possible to find $g^k(x)$? Given $g(x)=\frac{x}{2}+\frac{1}{x}$, is it possible to find an expression for $g^k(x)=(g\underbrace{\circ \cdots \circ}_k g)(x)$, where $k$ is some positive whole number?
For example, given $h(x)=\frac{x}{2}$, the expression for $h^k(x)$ is $h^k(x)=\frac{x}{2^k}$.
| Looking at
$$
g(x)=\frac{x}{2}+\frac{1}{x} = \frac{x^2 + 2}{2x}
$$
and cranking some iterations through the Maxima CAS, I get
$$
g^2(x) =
{{x^4+12\,x^2+4}\over{4\,x^3+8\,x}} \\
g^3(x) =
{{x^8+56\,x^6+280\,x^4+224\,x^2+16}\over{8\,x^7+112\,x^5+224\,x^3+
64\,x}} \\
g^4(x) =
{{x^{16}+240\,x^{14}+7280\,x^{12}+64064\,x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1578882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Method of completing squares with 3 variables I want to use the method "completing squares" for this term:
$x^2-2xy +y^2+z^2*a+2xz-2yz$
The result should be $(x-y+z)^2 +(a-1)*z^3$
Is there a "recipe" behind how to do this? Hope someone could help
| Aras' answer is really good. But you might be wondering how one develops intuition and what happens if one doesn't immediately see $x^2 - 2xy + y^2$ is a square. (Part of developing intuition is that this does become obvious.)
This'll get messy. But
So you have:
$\color {blue}{x^2}-2xy +y^2+z^2*a+2xz-2yz$
And you hav... | {
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"timestamp": "2023-03-29T00:00:00",
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What is the probability that $x^4-y^4$ is divisible by 5? Two numbers $x$ and $y$ are chosen at random without replacement from the set $\{1,2,3...,5n\}$.
What is the probability that $x^4-y^4$ is divisible by $5$?
I divided the numbers into groups of 5 $(1,2,3,4,5),(6,7,8,9,10),...$.The probability in the first group ... | Hint: If $x \equiv 0\pmod{5}$, then $x^4 \equiv 0\pmod{5}$. If $x \not\equiv 0\pmod{5}$, then $x^4 \equiv 1\pmod{5}$ by Fermat's Little Theorem.
Alternatively, if you don't know modular arithmetic, you can do the following to get the same result.
If $x = 5q+r$ for some integers $q$ and $r$ with $0 \le r \le 4$, then we... | {
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How to compute the area of the portion of a paraboloid cut off by a plane? How to compute:
The area of that portion of the paraboloid $x^2+z^2=2ay$ which is cut off by the plane $y=a$ ?
I think I have to compute $\iint f(x,z) dx dz$ , where $f(x,z)=\sqrt{(x^2+z^2)/2a}$ ; but I can't figure out what should be the li... | So we have the paraboloid $x^2+z^2=2ay$ and the plane $y=a$. Let me just rotate the whole shebang: this is the same as $x^2+y^2=2az$ cut by $z=a$. Just to ease my 3D imagination :).
The picture shows the paraboloid for $a=\frac12$, so it is $x^2+y^2=z$. The section we must find the area of is evidently a circle. So al... | {
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How to handle indices with fractional degree? An algebra problem ate my head!!!
$x$ and $y$ are positive real numbers such that
$$\sqrt{x^2 + \sqrt[3]{x^4 y^2}} + \sqrt{y^2 + \sqrt[3]{x^2 y^4}} = 512.$$
Find $x^{2/3} + y^{2/3}$.
It would be a great help if anybody helps me in solving this problem. I tried takin... | Hint: Notice that $\sqrt{x^2+\sqrt[3]{x^4y^2}}+\sqrt{y^2+\sqrt[3]{x^2y^3}}$ $= \sqrt{x^2+x^{4/3}y^{2/3}}+\sqrt{y^2+x^{2/3}y^{4/3}}$
$= \sqrt{x^{4/3}(x^{2/3}+y^{2/3})}+\sqrt{y^{4/3}(y^{2/3}+x^{2/3})}$ $= x^{2/3}\sqrt{x^{2/3}+y^{2/3}}+y^{2/3}\sqrt{y^{2/3}+x^{2/3}}$
$= (x^{2/3}+y^{2/3})\sqrt{x^{2/3}+y^{2/3}}$ $= (x^{2/3... | {
"language": "en",
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Why does this method to solve a quadratic equation for $x$ omit $x=0$? Here is a simple quadratic equation:
$$9x^2 - 36x = 0$$
We proceed as following:
\begin{align*}
9x^2 & = 36x\\
9x & = 36\\
x & = 4
\end{align*}
So, we get $x=4$.
But, here's another way:
\begin{align*}
9x^2 - 36x & = 0\\
x(9x-36) & = 0\\
\end{align... | To solve an equation, you have to establish a sequence of logical equivalences.
In your first method, you only established a sequence of logical implications. This is the reason why you lost the solution $x=0$.
Remark: In order to solve an equation, an sequence of implications can fails for two reasons:
*
*You lost ... | {
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Nesbitt's Inequality $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$ I'm reading a book which focus in inequality.
I'm stuck in this question. Let $a,b,c$ be positive real numbers. (Nesbitt's inequality) Prove the inequality $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$$
So the first step of so... | The first step of the solution says that $X+\frac1X+Y+\frac1Y+Z+\frac1Z\geq2+2+2$.
where $X=\frac{a+b}{b+c}$ and so on.
Do you know that $X+\frac1X$ is always $2$ or more whenever $X>0$?
| {
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Solving functional equation $f(x+y)=f(x)+f(y)+xy$ We are given $f(0)=0$. Then when $x+y=0$:
$$0=f(-y)+f(-x)+xy$$
Can I now use $x=0$ and obtain:
$$0=f(-y)?$$ Is this correct? Is there a better way to solve this equation?
| I show that the only solution is $f = \frac{1}{2}x^2 + bx$ (assuming $f$ continuous).
My strategy was to transform $f(x+y) = f(x) + g(y) + xy$ into a linear equation.
Write $f(x+y) = f(x) + g(y) + xy$ as $f(x+y) - \frac{1}{2}(x+y)^2 = (f(x) - \frac{1}{2}x^2) + (g(y) - \frac{1}{2}y^2)$. Substitute $g(z) = f(z) - \frac{1... | {
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Better proof for $\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$ It's required to prove that
$$\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$$
I managed to go about out it two ways:
*
*Show it is equivalent to $\mathsf{true}$:
$$\frac{1+\cos x + \sin x}{1... | Since $1-\cos^2 x = \sin^2 x$, we have $f(x) := \dfrac{1+\cos x}{\sin x} = \dfrac{\sin x}{1-\cos x}$. Therefore,
\begin{align*}\dfrac{1+\cos x + \sin x}{1-\cos x + \sin x} &= \dfrac{f(x)\sin x + f(x)(1-\cos x)}{1-\cos x + \sin x} \\ &= \dfrac{f(x)[1-\cos x + \sin x]}{1-\cos x + \sin x} \\ &= f(x) \\ &= \dfrac{1+\cos x... | {
"language": "en",
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Prove that $xy+yz+zx+\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}\ge 2x^2+2y^2+2z^2$ for every $x,y,z$ strictly positive I checked the inequality for $y=z$. In this particular case, after some simplifications, the inequality becomes:
$$
2x^3+y^3\ge 3x^2y,
$$
which is true, according to the arithmetic-geomet... | Use Schur inequality
$$a^4+b^4+c^4+abc(a+b+c)\ge \sum_{cyc}(a^3b+ab^3)\ge \sum_{cyc}2a^2b^2$$
Let
$$a=xy,b=yz,c=xz$$
then is your inequality
| {
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Prove $(1+2+...+k)^2 = 1^3 + ... + k^3$ using induction I need to prove that
$$(1+2+{...}+k)^2 = 1^3 + {...} + k^3$$
using induction.
So the base case holds for $0$ because $0 = 0$ (and also for $1$: $1^2 = 1^3 = 1$)
I can't prove it for $k+1$ no matter what I try! Can you give me a hint?
| After checking this for k=1, k=2 you need to follow this:
Let's say $A_1=(1+2+...+k)^2$; $B_1=1^3+...k^3$ and $A_2=(1+2+...+ (k+1))^2$
$B_2=1^3+...(k+1)^3$
Then to use induction it is needed to prove that $A_2-A_1=B_2-B_1$ or:
$$(1+2+...+ (k+1))^2-(1+2+...+k)^2=1^3+...(k+1)^3-1^3+...k^3$$
And it is equal to:
$$(k+1)(1+... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the real ordered pairs $(x,y)$ satisfying $x^2+x=y^4+y^3+y^2+y.$ Find the real ordered pairs $(x,y)$ satisfying $x^2+x=y^4+y^3+y^2+y.$
$x^2+x=y^4+y^3+y^2+y$
$4x^2+4x=4y^4+4y^3+4y^2+4y$
$4x^2+4x+1=4y^4+4y^3+4y^2+4y+1$
$(2x+1)^2=4y^4+4y^3+4y^2+4y+1$
I am stuck here.Is there a general method to solve such type of eq... | Under the assumption that the problem asks for integer solutions, (which is probably the case)
We have $$(2y^2+y)^2<(2x+1)^2=4y^4+4y^3+4y^2+4y+4<(2y^2+y+1)^2$$ for $y > 3$ or $y <-1$, so we must have $-1 \le y \le 3$. Now bash.
| {
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$\lim \frac{\cos{x}}{x^2}$ as x goes to infinity $\displaystyle \lim_{x \to \infty} \frac{\cos{x}}{x^2} =\lim_{x \to \infty} \frac{\frac{d}{dx} \cos{x}}{\frac{d}{dx}x^2} = -\frac{1}{2}\lim_{x \to \infty}\frac{\sin{x}}{x}$. But $-\frac{1}{x} \le \frac{\sin{x}}{x} \le \frac{1}{x}$ so $\displaystyle -\lim_{x \to \infty}\f... | You don’t need to use the derivatives: For every $x > 0$ we have
$$
-\frac{1}{x^2} \leq \frac{\cos(x)}{x^2} \leq \frac{1}{x^2}.
$$
Because $\lim_{x \to \infty} \frac{1}{x^2} = 0$ it follows that $\lim_{x \to \infty} \frac{\cos(x)}{x^2} = 0$.
| {
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Find all real numbers $x,y > 1$ such that $\frac{x^2}{y-1}+\frac{y^2}{x-1} = 8$
Find all real numbers $x,y > 1$ such that $$\frac{x^2}{y-1}+\frac{y^2}{x-1} = 8$$
Attempt
We can rewrite this as $x^2(x-1)+y^2(y-1) = 8(x-1)(y-1)$. Then I get a multivariate cubic, which I find hard to find all solutions to.
| As suggested by matt, let $a=x-1$, $b=y-1$ and apply the inequality $r+s\ge 2\sqrt{rs}$ for $r,s>0$ with equality if and only if $r=s$. Then $a,b>0$ and
$$\frac{(a+1)^2}{b}+\frac{(b+1)^2}{a}=\left(\frac{a^2}{b}+\frac{1}{b}\right)+\left(\frac{b^2}{a}+\frac{1}{a}\right)+2\left(\frac{a}{b}+\frac{b}{a}\right)$$
$$\ge \left... | {
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If $x=2+i$, $gcd(a,b,c)=1$, and $ax^4+bx^3+cx^2+bx+a=0$, then what is $|c|$?
Suppose
$$a(2+i)^4 + b(2+i)^3 + c(2+i)^2 + b(2+i) + a = 0,$$
where $a,b,c$ are integers whose greatest common divisor is $1$.
Determine $|c|$.
So I first simplified the exponents and combined like terms.
I received $$a(-6+24i)+b(4+12i)... | We use your calculation. Note that as the other answers show there would have been nicer ways to proceed.
From the equation that you obtained we get $-6a+4b+3c=0$ and $24a+12b+4c=0$. Divide the second expression through by $4$, and add the first expression. We get $7b+4c=0$.
Thus $4c=-7b$ and therefore $c=7k$, $b=4k$ f... | {
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Strange Mean Inequality This problem was inspired by this question.
$\sqrt [ 3 ]{ a(\frac { a+b }{ 2 } )(\frac { a+b+c }{ 3 } ) } \ge \frac { a+\sqrt { ab } +\sqrt [ 3 ]{ abc } }{ 3 } $
The above can be proved using Hölder's inequality.
$\sqrt [ 3 ]{ a(\frac { a+b }{ 2 } )(\frac { a+b+c }{ 3 } ) } =\sqrt [ 3 ]{ (\fra... | This result was conjectured by professor Finbarr Holand, and then it was proved by
K. Kedlaya in an article in the American Mathematical Monthly that could be found here, and then it was generalized by Professor Holand in an article that could be found here.
| {
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Denoting the sum of $n$ odd numbers $1+3+5+\cdots+(2n-1) = 1+3+5+\cdots+(2n+1)$? Let's say we are denoting the sum of $n$ odd numbers.
Then in symbols $1+3+5+\cdots+(2n-1)$.
If we substitute $(k+1)$ for $n$. $2n-1=2k+1$
So $1+3+5+...+2k+1$
Then can we use $1+3+5+\cdots+(2n+1)$ instead of $1+3+5+\cdots+(2n-1)$?
Logic... | If you're writing $1+3+5+\cdots+45$, for instance, you could describe that as
$1 + 3 + 5 + \cdots + (2n-1)$ for $1 \le n \le 23$, and there are 23 terms.
You could also describe it as
$1 + 3 + 5 + \cdots + (2n+1)$, but this time we have $0 \le n \le 22$, and there are still only 23 terms: even though the last value of ... | {
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$\frac{2^{2n}}{2n} \le {2n \choose n}$? Show that $\frac{2^{2n}}{2n} \le {2n \choose n}$, $n\in\mathbb{N}$.
What I can see easily is that $\frac{2^{2n}}{2n + 1} \le {2n \choose n}$, since
$(1+1)^{2n} = \sum_{m=0}^{2n}{2n \choose m} = 2^{2n}$. This sum contains $2n+1$ terms with the largest being ${2n \choose n}$. So ${... | The first and last terms are $\left( \begin{array}{c} 2n \\ 0 \end{array} \right) = \left( \begin{array}{c} 2n \\ 2n \end{array} \right) = 1$.
So with $n \ge 1$, the sum of these two terms is $\le \left( \begin{array}{c} 2n \\ n \end{array} \right)$ and we only have $2n$ terms in the sum now.
| {
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Prove that if $a$ and $b$ are nonnegative real numbers, then $(a^7+b^7)(a^2+b^2) \ge (a^5+b^5)(a^4+b^4)$ Prove that if $a$ and $b$ are nonnegative real numbers, then $(a^7+b^7)(a^2+b^2) \ge (a^5+b^5)(a^4+b^4)$
My try
My book gives as a hint to move everything to the left hand side of the inequality and then factor an... | \begin{align*}
(a^7 + b^7)(a^2 + b^2) - (a^5 + b^5)(a^4 + b^4) &= a^7 b^2 + a^2 b^7 - a^5 b^4 - a^4 b^5 \\
&= a^2 b^2 \big(a^5 + b^5 - a^3 b^2 - a^2 b^3\big) \\
&= a^2 b^2 \big(a^2 (a^3 - b^3) + b^2 (b^3 - a^3)\big) \\
&= a^2 b^2 (a^2 - b^2)(a^3 - b^3)
\end{align*}
Now regardless of how $a$ and $b$ are related to each ... | {
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General values of $\theta$ for which $(\sin\theta)x^2+(2\cos\theta)x+\frac{\cos\theta+\sin\theta}{2}$ is the square of a linear function. Find the general values of $\theta$ for which the quadratic function $(\sin\theta)x^2+(2\cos\theta)x+\frac{\cos\theta+\sin\theta}{2}$ is the square of a linear function.
As the quad... | Since it is the square of a linear function, it has to be positive. So the first coefficient $\sin \theta$ has to be positive. This gives the criteria that $\theta$ is in the first or second quadrant. Hence the answer $2n\pi+\cdots$
| {
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Prove that $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$
Let $a,b,c$ be three nonnegative real numbers. Prove that $$a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$$
It seems that the inequality $a^2+b^2+c^2 \geq ab+bc+ca$ will be of use here. If I use that then I will get $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2}... | $x^3=a^2,y^3=b^2,z^3=c^2 \implies x^3+y^3+z^2 +3xyz \ge 2(\sqrt{(xy)^3}+\sqrt{(yz)^3}+\sqrt{(xz)^3})$
we have $x^3+y^3+z^3 +3xyz \ge xy(x+y)+yz(y+z)+xz(x+z)$
$xy(x+y)\ge 2xy\sqrt{xy}=2\sqrt{(xy)^3} \implies xy(x+y)+yz(y+z)+xz(x+z)\ge 2(\sqrt{(xy)^3}+\sqrt{(yz)^3}+\sqrt{(xz)^3})$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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prove that $(x-y)^3+(y-z)^3+(z-x)^3 > 0.$
If $x<y<z$ are real numbers, prove that $$(x-y)^3+(y-z)^3+(z-x)^3 > 0.$$
Attempt
We have that $(x-y)^3+(y-z)^3+(z-x)^3=-3 x^2 y+3 x^2 z+3 x y^2-3 x z^2-3 y^2 z+3 y z^2$. Grouping yields $3(x^2z+xy^2+yz^2)-3(x^2y+xz^2+y^2z)$. Then I was thinking of using rearrangement but then... | Because it is only the differences in the variables that matter, we can subtract the smallest value ($x$) from each term:
$-y^3 + (y-z)^3 + z^3 > 0$.
Let $y = d_1$ and $z-y = d_2$ and note $z = d_1 + d_2$. Then:
$-d_1^3 - d_2^3 + (d_1 + d_2)^3 > 0$.
Simplify:
$3d_1 d_2^2 + 3 d_1^2 d_2 >0$.
or
$3(d_1 d_2)(d_1 + d_2) > ... | {
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Value of $X^4 + 9x^3+35X^2-X+4$ for $X=-5+2\sqrt{-4}$
Find the value of $X^4 + 9x^3+35X^2-X+4$ for $X=-5+2\sqrt{-4}$
Now the trivial method is to put $X=5+2\sqrt{-4}$ in the polynomial and calculate but this is for $2$ marks only and that takes a hell lot of time for $2$! So I was thinking may be there is some tri... | $X+5=4i$ squaring you get $x^2+10x+41=0$ now just divide the given polynomial with this equation so solving $$\frac{x^4+9x^3+35x^2-x+4}{x^2+10x+41}$$ you get it equal to $(160)$
| {
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Prove that if $a,b,$ and $c$ are positive real numbers then $(a+b)(a+c) \geq 2 \sqrt{abc(a+b+c)}.$
Prove that if $a,b,$ and $c$ are positive real numbers then $$(a+b)(a+c) \geq 2 \sqrt{abc(a+b+c)}.$$
This looks like a simple question. We can apply AM-GM twice to get $(a+b)(a+c) \geq 4a\sqrt{bc}$. Then how do I use th... | we get after squaring and factorizing $$ \left( {a}^{2}+ab+ac-cb \right) ^{2}$$ and this is nonnegative
| {
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Evaluating integral $\int\frac{e^{\cos x}(x\sin^3x+\cos x)}{\sin^2x}dx $ $$\int\frac{e^{\cos x}(x\sin^3x+\cos x)}{\sin^2x}dx $$
The usual form $\int e^x(f(x)+f'(x))dx $ does not apply here. What substitution should I make ?
| The integral is of the form
\begin{equation*}
\int \left( x\sin x+\frac{\cos x}{\sin ^{2}x}\right) e^{\cos x}dx=\int
h(x)e^{g(x)}dx.
\end{equation*}
This form recalls the well-known formula
\begin{equation*}
\int \left( f^{\prime }(x)+g^{\prime }(x)f(x)\right)
e^{g(x)}dx=f(x)e^{g(x)}+C.
\end{equation*}
Its proof maybe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove the identity $\binom{2n+1}{0} + \binom{2n+1}{1} + \cdots + \binom{2n+1}{n} = 4^n$ I've worked out a proof, but I was wondering about alternate, possibly more elegant ways to prove the statement. This is my (hopefully correct) proof:
Starting from the identity $2^m = \sum_{k=0}^m \binom{m}{k}$ (easily derived from... | Let $A$ be the powerset (i.e., the set of subsets of) of $\{1,\ldots,2n\}$. Let $B$ the set of all subsets of $\{1,\ldots,2n+1\}$ at most $n$ elements. Then "clearly" $|A|=2^{2n}=4^n$ and $|B|=\sum_{k=0}^n{2n+1\choose k}$.
Define the following map $f\colon B\to A$:
$$f(S)=\begin{cases}S&\text{if }2n+1\notin S\\\{1,\ldo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1607936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Minimum value of $\frac{(1 + x + x^2)(1 + y + y^2)}{xy}$
What is the minimum value of $$\frac{(1 + x + x^2)(1 + y + y^2)}{xy},~~(x \neq 0)$$
Should we find the minimum value of each quadratic?
| $$\frac{(1+x+x^2)(1+y+y^2)}{xy} = \frac{1+x+x^2}{x} \cdot \frac{1+y+y^2}{y}$$
By AM–GM inequality $(1+x+\frac1x)(1+y+\frac1y)≥3*3=9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1609174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Given $a+b+c=3$ .Prove $ \sum \limits_{cyc} \frac {1}{a^2+b^2+2} \le \frac 34$ Yesterday I found this on the Internet:
Give 3 non-negative numbers $a,b,c$ that $a+b+c=3$. Prove
$$ \sum _{cyc} \frac {1}{a^2+b^2+2} \le \frac 34 $$
I have tried to solve this using AM-GM:
From AM-GM we got:
$$a^2+b^2+2\ge2(a+b)$$
$$\Rig... | It's suffice to show the following inequality $$ \sum\limits_{sic}{\frac{a^2+b^2}{a^2+b^2+2}} \ge \frac{3}{2} $$ By using Cauchy, we have $$ LHS \ge \frac{\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)^2}{2\left(a^2+b^2+c^2\right)+6} \ge \frac{\sqrt{3\left(a^2b^2+b^2c^2+c^2a^2\right)}+2\left(a^2+b^2+c^2\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1609457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Converting $(1+...+n)^2*(n+1)^3$ to $(2+...+2n)^2$ I'm currently going through Calculus by Spivak by myself, and came across a proof by induction requiring to prove $1^3+...+n^3 = (1+...+n)^2$
Naturally, to prove this, I need to somehow convert $(1+...+n)^2+(n+1)^3$ to $(2+...+2n)^2$.
After quite a bit of thinking, I'm... |
Naturally, to prove this, i need to somehow convert $(1+...+n)^2*(n+1)^3$ to $(2+...+2n)^2$.
Unfortunately, "$(1+...+n)^2*(n+1)^3 = (2+...+2n)^2$" is not true for all $n$. Once you simplify the summations, the left side will be a polynomial with degree $2 \cdot 2 + 3 = 7$ while the right side will be a polynomial wit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1611750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Why is $\lim_\limits{x\to 0}\frac{\sin(6x)}{\sin(2x)} = \frac{6}{2}=3$? Why is $\lim_\limits{x\to 0}\frac{\sin(6x)}{\sin(2x)} = \frac{6}{2}=3$?
The justification is that $\lim_\limits{x\to 0}\frac{\sin(x)}{x} = 1$
But, I am not seeing the connection.
L'Hospital's rule? Is there a double angle substitution happening?
| As user Daniel has written you can express the ratio $\frac{\sin 6x}{\sin 2x}$ in a way which makes it amenable to the use of standard limit $$\lim_{x\to 0} \frac{\sin x}{x} = 1\ .$$ Notice that $$\lim_{x\to 0} \frac{\sin x}{x} = \lim_{x\to 0} \frac{\sin 2x}{2x} = \lim_{x\to 0} \frac{\sin 6x}{6x}\ .$$ As long as the ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 1
} |
Find the value of $A/B$ for this definite integration. $$\int_{0}^{\infty}e^{-\sqrt 3 x^{\frac{2}{3}}}\sin(x^{\frac{2}{3}})\mathrm{d}x$$
Given that the integral above is equal to $\frac{3\pi^A}{B}$ for rational numbers $A$ and $B$, find the value of $A/B$.
I thought of taking $x^{2/3}$ as t but that didn't help. Any ot... | Letting $t=x^{\frac13}$ and noting
$$ (-\frac{1}{8} e^{-\sqrt{3} t^2} (\sqrt{3} \sin(t^2)+\cos(t^2)))'=te^{-\sqrt 3 t^2}\sin(t^2)$$
then we have
\begin{eqnarray}
&&\int_{0}^{\infty}e^{-\sqrt 3 x^{\frac{2}{3}}}\sin(x^{\frac{2}{3}})\mathrm{d}x\\
&=&3\int_{0}^{\infty}e^{-\sqrt 3 t^2}\sin(t^2)t^2\mathrm{d}t\\
&=&-\frac{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Second derivative of $x^3+y^3=1$ using implicit differentiation I need to find the $D_x^2y$ of $x^3+y^3=1$ using implicit differentiation
So,
$$
x^3 + y^3 =1 \\
3x^2+3y^2 \cdot D_xy = 0 \\
3y^2 \cdot D_xy= -3x^2 \\
D_xy = - {x^2 \over y^2}
$$
Now I need to find the $D_x^2y$.
I am pretty sure that means the second d... | If
$x^n + y^n =1
$
then
$nx^{n-1} +ny^{n-1}y' = 0$
or
$y'
=-\frac{x^{n-1}}{y^{n-1}}
$.
Differentiating again,
$\begin{array}\\
0
&=(n-1)x^{n-2}+y^{n-1}y''+(n-1)y^{n-2}(y')^2\\
&=(n-1)x^{n-2}+y^{n-1}y''+(n-1)y(\frac{x^{n-1}}{y^{n-1}})^2\\
&=y^{n-1}y''+(n-1)x^{n-2}(1+\frac{x^n}{y^{2n-3}})\\
&=y^{n-1}y''+(n-1)x^{n-2}\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\frac{1}{a+ab}+\frac{1}{b+bc}+\frac{1}{c+ca} \geq \frac{3}{2}.$
Let $a,b,$ and $c$ be positive real numbers such that $abc = 1$. Prove that $$\dfrac{1}{a+ab}+\dfrac{1}{b+bc}+\dfrac{1}{c+ca} \geq \dfrac{3}{2}.$$
I thought about substituting in $abc = 1$ to get $$\dfrac{1}{a+\dfrac{1}{c}}+\dfrac{1}{b+\dfra... | Notice that the inequality is equivalent to
$(abc+1)(\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)})+3\ge 6$
This is equivalent to proving that
$\frac{1+abc+a+ab}{a(1+b)}+\frac{1+abc+b+bc}{b(1+c)}+\frac{1+abc+c+ca}{c(1+a)}\ge6$.
However, $\frac{1+abc+a+ab}{a(1+b)}+\frac{1+abc+b+bc}{b(1+c)}+\frac{1+abc+c+ca}{c(1+a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $ax+by+cz+2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{a+b+c}$ Let $a,b,c,x,y,z$ be positive real numbers such that $x+y+z=1$. Prove that
$$ax+by+cz+2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{a+b+c}$$.
my try:
$2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{\frac{2(a+b+c)}{3}}$
But this is not the right choice because
$ax+by+cz\le{\frac{a+b+c}... | Let $d = \sqrt{2(ab+bc+ca)}$ and $t = \sqrt{2(xy+yz+xz)}$. It is easy to check
$$
\begin{cases}
a^2 + b^2 + c^2 + d^2 &= a^2+b^2+c^2 + 2(ab+bc+ca) = (a+b+c)^2\\
x^2 + y^2 + z^2 + t^2 &= x^2+y^2+z^2 + 2(xy+yz+xz) = (x+y+z)^2 = 1
\end{cases}
$$
Apply Cauchy Schwarz to the two 4-vectors $(a,b,c,d)$ and $(x,y,z,t)$,
we im... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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A line is drawn through the point $A(1,2)$ to cut the line $2y=3x-5$ in $P$ and the line $x+y=12$ in $Q$. If $AQ=2AP$, find $P$ and $Q$. A line is drawn through the point $A(1,2)$ to cut the line $2y=3x-5$ in $P$ and the line $x+y=12$ in $Q$. If $AQ=2AP$, find the coordinates of $P$ and $Q$.
I found the lengths of the ... | Here is a method you might prefer.
We can write $$P(2s+1, 3s-1)$$ where $s$ is a parameter. Clearly the parametrization is not unique, and I have chosen this for convenience.
Similarly we can write $$Q(t, 12-t)$$
Now $$\overrightarrow{AP}=\left(\begin{matrix}2s\\3s-3\end{matrix}\right)$$ and $$\overrightarrow{AQ}=\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find the last $4$ digits of $2016^{2016}$
Find the last $4$ digits of $2016^{2016}$.
Technically I was able to solve this question by I used Wolfram Alpha and modular arithmetic so the worst I had to do was raise a $4$ digit number to the $9$th power. I would do $2016^2 \equiv 4256^2 \equiv \cdots$
and then continue... | Allow me give a different answer from the three ones above and easy to achieve with a calculator, please. This direct procedure can be applied to similar calculations.
One can simply use one or both properties, $A^{abc}=(A^a)^{bc}\equiv(B)^{bc}\pmod{M}$ and $A^n=A^r\cdot A^s\equiv(B)\cdot (C)\pmod{M}$ where $r+s=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1614501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\frac{a^3}{(1+b)(1+c)}+\frac{b^3}{(1+a)(1+c)}+\frac{c^3}{(1+a)(1+b)} \geq \frac{3}{4}.$
Let $a,b,$ and $c$ be positive real numbers such that $abc = 1$. Prove that $$\dfrac{a^3}{(1+b)(1+c)}+\dfrac{b^3}{(1+a)(1+c)}+\dfrac{c^3}{(1+a)(1+b)} \geq \dfrac{3}{4}.$$
Attempt
We have $$\dfrac{a^3}{(1+b)(1+c)}+\dfr... | From the AM-GM inequality :
$$\frac{a^3}{(1+b)(1+c)}+\frac{1+b}{8}+\frac{1+c}{8} \geq 3\sqrt[3]{\frac{a^3}{(1+b)(1+c)} \cdot \frac{1+b}{8}\cdot\frac{1+c}{8}}=\frac{3a}{4}$$
Now do this for the others and add them all to get :
$$\frac{a^3}{(1+b)(1+c)}+\frac{b^3}{(1+a)(1+c)}+\frac{c^3}{(1+a)(1+b)}\geq \frac{a+b+c}{2} -\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the least value of $4\csc^{2} x+9\sin^{2} x$
Find the least value of $4\csc^{2} x+9\sin^{2} x$
$a.)\ 14 \ \ \ \ \ \ \ \ \ b.)\ 10 \\
c.)\ 11 \ \ \ \ \ \ \ \ \ \color{green}{d.)\ 12} $
$4\csc^{2} x+9\sin^{2} x \\
= \dfrac{4}{\sin^{2} x} +9\sin^{2} x \\
= \dfrac{4+9\sin^{4} x}{\sin^{2} x} \\
= 13 \ \ \ \ \ \ \ \... | Without nice tricks, you can use calculus. Consider
$$
f(x)=\frac{4}{\sin^2x}+9\sin^2x
$$
defined over $(0,\pi)$. Note that the general function is periodic of period $\pi$.
Then
$$
f'(x)=-\frac{8\cos x}{\sin^3x}+18\sin x\cos x
=2\frac{\cos x}{\sin^3x}(9\sin^4x-4)
=2\frac{\cos x}{\sin^3x}(\sqrt{3}\sin x-\sqrt{2})(\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Occurrence of 5 consecutive tails before occurrence of 2 consecutive heads In repeated tossing of a fair coin find the probability that $5$ consecutive tails occur before occurrence of $2$ consecutive heads.
My attempt:
I tried to find the probability of non-occurrence of two consecutive heads in $n$ throws.
Let $a_{n}... | Here is a slightly different proof which may be of interest although
it is not as elegant as the one by @leonbloy.
Suppose we treat the problem of $t$ tails before $h$ heads.
Encoding this in a generating function with $u$ marking sequences of
tails of length at least $t$ and $v$ sequences of heads of length at
l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
To evaluate the limits $\lim\limits_{n \to \infty} \{\frac{1}{1+n^3}+\frac{4}{8+n^3}+\ldots +\frac{n^2}{n^3+n^3}\}$ To me it seems like that we need to manipulate the given sum into the Riemann sum of some function. First writing in the standard summation form;
$$\{\frac{1}{1+n^3}+\frac{4}{8+n^3}+\ldots +\frac{n^2}{n^3... | You can proceed from here as $$\lim_{n\to \infty}\sum\limits_{k=0}^n \frac{k^2}{k^3+n^3}= \lim_{n\to\infty}\sum\limits_{k=0}^n \;\frac{1}{n}\frac{\frac{k^2}{n^2}}{\frac{k^3}{n^3}+1}= \lim_{h\to 0} \;h\sum_{k=1}^n \frac{{(kh)}^2}{1 + {(kh)}^3}= \int_0^1 \frac{x^2}{1+x^3} \mathrm{d}x\;.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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If $\omega$ is an imaginary fifth root of unity, then $\log_2 \begin{vmatrix} 1+\omega +\omega^2+\omega^3 -\frac{1}{\omega} \\ \end{vmatrix}$ =
If $\omega$ is an imaginary fifth root of unity, then $$\log_2 \begin{vmatrix}
1+\omega +\omega^2+\omega^3 -\frac{1}{\omega} \\
\end{vmatrix} =$$
My approach :
$$\omega^5 = ... | Since
$$
\begin{align}
1+\omega+\omega^2+\omega^3+\omega^4
&=\frac{1-\omega^5}{1-\omega}\\[3pt]
&=0
\end{align}
$$
we have
$$
\begin{align}
1+\omega+\omega^2+\omega^3-\frac1\omega
&=-\omega^4-\frac1\omega\\
&=-\frac{\omega^5+1}\omega\\
&=-\frac2\omega
\end{align}
$$
Therefore
$$
\begin{align}
\log_2\left|1+\omega+\omeg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Find the minimum value of $\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
Find the minimum value of
$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
$a.)\ 1 \ \ \ \ \ \ \ \ \ \ \ \ b.)\ 3 \\
c.)\ 5 \ \ \ \ \ \ \ \ \ \... | Using standard trigonometric identities, we see this is $1+2\tan^2\theta+1+2\cot^2\theta+1$.
Now we can use AM/GM to show that $2\tan^2\theta+2\cot^2\theta\ge 4$, and the value $4$ is attained at $\pi/4$.
Remark: Your AM/GM argument is enough to identify the right answer of this multiple choice question. For as you sa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1620239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
Compute $\lim_{n \to +\infty} n^{-\frac12 \left(1+\frac{1}{n}\right)} \left(1^1 \cdot 2^2 \cdot 3^3 \cdots n^n \right)^{\frac{1}{n^2}}$
How to compute
$$\displaystyle \lim_{n \to +\infty} n^{-\dfrac12 \left(1+\dfrac{1}{n}\right)} \left(1^1\cdot 2^2 \cdot 3^3 \cdots n^n \right)^{\dfrac{1}{n^2}}$$
I'm interested in ... | We have $$\exp \left (- \frac {1} {2} \left (1 + \frac {1} {n}\right) \log n + \frac {1} {n^2} (1 \log 1 + 2 \log 2 + \cdots + n \log n)\right) = \exp \left (- \frac {1} {2} \left (1 + \frac {1} {n}\right) \log n + \frac {1} {2} \log n - \frac {1} {4} + \frac {\log n} {2n} + o \left (\frac {1} {n}\right) \right) = \exp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Complex Numbers - Quartic Find two distinct real roots of the equation $z^4-3z^3+5z^2-z-10$, and hence solve this equation completely.
The problem is how do you find the two distinct real roots?
| $$z^4-3z^3+5z^2-z-10=0\Longleftrightarrow$$
$$(z-2)(z+1)((z-2)z+5)=0$$
So, we've got three equations:
*
*$$z-2=0\Longleftrightarrow z=2$$
*$$z+1=0\Longleftrightarrow z=-1$$
*$$(z-2)z+5=0\Longleftrightarrow$$
$$z^2-2z+5=0\Longleftrightarrow$$
$$z=\frac{-\left(-2\right)\pm\sqrt{\left(-2\right)^2-4\cdot(1)\cdot(5)}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the maximum value of $(12\sin x-9\sin^{2} x)$
The maximum value of $(12\sin x-9\sin^{2} x)$
is equal to
$a.)\ 3 \\
\color{green}{b.)\ 4} \\
c.)\ 5 \\
d.)\ \text{none of these}$
As
$-1\leq \sin x\leq 1 ,\\
12\sin x-9\sin^{2} x \\
=12-9=3 \\
$
But the answer given is $4.$
I am looking for a short and simple w... | As you noted $\sin(x)$ takes on all values from $-1 \dots 1$.
So consider $f(y) = 12y - 9y^2$, where $y$ is in the range $-1 \dots 1$.
What is the maximum of $f$? The $y^2$ coefficient is negative, so the maximum occurs at the vertex of the parabola.
The vertex of a parabola $f(x) = ax^2 + bx + c$ occurs at $x = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Bounding the remainder I need to find the 3rd order Taylor polynomials and bound the remainder term at $(0,0)$.
The function is $$f(x,y)=\cos(x)\sin(y)$$
Here is what I did: first, I found the taylor expansions of sin and cos functions and multiply them, and then I get the following remainder term
$$R_3(x)-\frac{x^2}{2... | Lets refer to a set of notes for Multivariable Taylor and MV Taylor.
Lets do it the long way first.
*
*$(a, b) = (0, 0)$
*$f(x, y) = \cos x \sin y$
*$(x + y)^1 = x + y$, so we need partials $f_x, f_y$
*$f_{x}(x, y) = -\sin x \cos y \rightarrow f_x(0, 0) = 0$
*$f_{y}(x, y) = -\cos x \cos y \rightarrow f_y(0, 0) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving Trig Identities (Complex Numbers)
Question: Prove that if $z = \cos (\theta) + i\sin(\theta)$, then
$$ z^n + {1\over z^n} = 2\cos(n\theta) $$
Hence prove that $\cos^6(\theta)$ $=$ $\frac 1{32}$$(\cos(6\theta)$ + $6\cos(4\theta)$ + $15\cos(2\theta)$ + $10$)
I learnt to prove the first part in another post l... | This is simply because for such a $z$, $\;\frac1{z^n}=\bar z^n=\overline{z^n}$, and
$$z^n+\overline{z^n}=2\operatorname{Re}(z^n)=2\cos n\theta\qquad\quad\text{by De Moivre's formula.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove that one of the elements can't be in the interval $(0,1)$ Let be $a,b,c\in\mathbb{R}$ so that the sum of two of them is never equal to $1$. Prove that atleast on of $\frac{ab}{a+b-1},\frac{bc}{b+c-1},\frac{ca}{c+a-1}$ can't be in the interval $(0,1)$.
I aproached it with contradiction, but can't get sth that is n... | Let $0<\frac{ab}{a+b-1}<1$, $0<\frac{ac}{a+c-1}<1$ and $0<\frac{bc}{b+c-1}<1$.
Hence, $0<\frac{a^2b^2c^2}{\prod\limits_{cyc}(a+b-1)}<1$.
In another hand, $0<1-\frac{ab}{a+b-1}<1$, $0<1-\frac{ac}{a+c-1}<1$ and $0<1-\frac{bc}{b+c-1}<1$, which gives
$0<\frac{(1-a)(b-1)}{a+b-1}<1$, $0<\frac{(1-c)(a-1)}{a+c-1}<1$ and $0<\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Summation up to $n$ terms : $\sum r\cdot (r+1)^2$ Summation up to $n$ terms : $$\sum_{r=1}^{n} r\cdot (r+1)^2$$
My attempt :
$$\sum_{r=1}^{n} r\cdot (r+1)^2=\sum_{r=1}^{n} r^3+2\sum_{r=1}^{n} r^2+\sum_{r=1}^{n} r$$
$$\sum_{r=1}^{n} r\cdot (r+1)^2=\sum_{r=1}^{n} r^3+2\sum_{r=1}^{n} r^2+\sum_{r=1}^{n} r$$
But for this me... | Let $f(r)=a+br+cr^2+dr^3+er^4$.
Then $$f(r+1)-f(r)$$
$$=a+b(r+1)+c(r^2+2r+1)+d(r^3+3r^2+3r+1)+e(r^4+4r^3+6e^2+4e+1)-(a+br+cr^2+dr^3+er^4)$$
$$=b+2cr+c+3dr^2+3dr+d+4er^3+6er^2+4er+e$$
$$=(b+c+d+e)+(2c+3b+4e)r+(3d+6e)r^2+4er^3$$
We need this to equal $r(r^2+2r+1)=r+2r^2+r^3$.
So $4e=1$ etc...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\sum_{n=1}^{\infty }\int_{0}^{\frac{1}{n}}\frac{\sqrt{x}}{1+x^{2}}\,\mathrm{d}x$ I have some trouble in evaluating this series
$$\sum_{n=1}^{\infty }\int_{0}^{\frac{1}{n}}\frac{\sqrt{x}}{1+x^{2}}\mathrm{d}x$$
I tried to calculate the integral first, but after that I found the series become so complicated.
Bes... | Let
$$I(n) = \int_{0}^{\frac{1}{n}}\frac{\sqrt{x}}{1+x^{2}}\mathrm{d}x.$$
Substitution:
$$x = t^2,\quad \mathrm{d}x = 2t\mathrm{d}t:$$
$$I(n) = \int\limits_{0}^{\frac{1}{\sqrt{n}}}\frac{2t^2}{1+t^{4}}\mathrm{d}t.$$
$$I(n) = \int\limits_{0}^{\frac{1}{\sqrt{n}}}\frac{2}{t^2 + \dfrac{1}{t^2}}\mathrm{d}t.$$
$$I(n) = \int\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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How to evaluate $\int_{0}^{\pi }\theta \ln\tan\frac{\theta }{2} \, \mathrm{d}\theta$ I have some trouble in how to evaluate this integral:
$$
\int_{0}^{\pi}\theta\ln\left(\tan\left(\theta \over 2\right)\right)
\,\mathrm{d}\theta
$$
I think it maybe has another form
$$
\int_{0}^{\pi}\theta\ln\left(\tan\left(\theta \over... | Obviously we have
$$\int_{0}^{\pi }\theta \ln\tan\frac{\theta }{2}\mathrm{d}\theta =4\int_{0}^{\pi /2}x\ln \tan x\mathrm{d}x$$
then use the definition of Lobachevskiy Function(You can see this in table of integrals,series,and products,Eighth Edition by Ryzhik,page 900)
$$\mathrm{L}\left ( x \right )=-\int_{0}^{x}\ln\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
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Choosing a substitution to evaluate $\int \frac{x+3}{\sqrt{x+2}}dx$ Is there any other value you can assign to the substitution variable to solve this integral?
$$\int \frac{x+3}{\sqrt{x+2}}dx$$
Substituting $u = x + 2$:
$$du = dx; u +1 = x+3 ,$$
and we get this new integral that we can then split into two different ... | let's make it easier than that!
Use this:
$$x + 2 = t^2 ~~~~~~~~~~~ x+3 = t^2 + 1 ~~~~~~~ \text{d}x = 2t\ \text{d}t$$
Obtaining
$$I = \int\frac{t^2 + 1}{t}\ 2t\ \text{d}t = 2\int t^2 + 1\ \text{d}t = \frac{2}{3}t^3 + 2t$$
Coming back to $x$, having $t = \sqrt{x+2}$ and you'll have
$$I = \frac{2}{3}\sqrt{x+2}(x+5)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Bounded sequence $a_n=\sqrt{4+2 \sqrt{4+\cdots+2 \sqrt{4+2 \sqrt{4+2 \sqrt{4+4}}}}}$ Let
$$a_n=\sqrt{4+2 \sqrt{4+\cdots+2 \sqrt{4+2 \sqrt{4+2 \sqrt{4+4}}}}}$$
the sign $\sqrt{}$ occurs $n$ times.
a) Prove, that $a_n< \sqrt{5}+1$ for all $n$.
b) Find $\lim_{n\rightarrow \infty } a_n$
Author O.Kukush
| You can prove (a) by mathematical induction. $a_1$ is less than $1+\sqrt5$ by direct comparison as $a_1 = \sqrt8 < 3$ while $1+\sqrt5>3$. Let $a_n < 1+\sqrt5$. Then $4+2a_n < 6+2\sqrt5$ and taking square root on both sides (that both are positive is trivial) we obtain $a_{n+1} < 1+\sqrt5$.
It also follows that $(a_n)$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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If $p$ is a prime other than 2 or 5, prove that $p$ must be one of the forms $10k + 1$, $10k + 3$, $10k + 7$, or $10k + 9$ If $p$ is a prime other than 2 or 5, prove that $p$ must be one of the forms $10k + 1$, $10k + 3$, $10k + 7$, or $10k + 9$
-The section we are covering is on the division algorithm, although I am u... | So you wish to express $10k + c$ as $bq + r$ with $0 \leq r < b$. If $r = 0$ and $10k + c \neq b$, that means $10k + c$ and $b$ share a common prime factor, and if $b$ is the prime then $10k + c$ is composite.
Try it with $b = 2$, which you know is a prime number. Then:
*
*$10k + 0 = 2(5k) + 0$
*$10k + 2 = 2(5k + 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1635245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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This simple algebra question \begin{cases}
3x + y = \frac{12}{y^2}\\
3y + x = \frac{15}{x^2}
\end{cases}
$x+y = ?$
I tried adding side by side but I found only $4x+4y = \dfrac{15}{x^2} + \dfrac{12}{y^2}$ so I couldn't find anything. Can someone solve it?
| *
*Multiply the first equation by $y^2$: $3xy^2+y^3 = 12$
*Multiply the second equation by $x^2$: $3x^2y+x^3=15$
*Add them up to get $3xy^2+y^3+3x^2y+x^3=(x+y)^3=27.$
Hence, $x+y=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1635745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What is the number of ordered triplets $(x, y, z)$ such that the LCM of $x, y$ and $z$ is ...
What is the number of ordered triplets $(x, y, z)$ such that the LCM of $x, y$ and $z$ is $2^33^3$ where $x, y,z\in \Bbb N$?
What I tried :
At least one of $x, y$ and $z$ should have factor $2^3$ and at least one should hav... | We use Inclusion/Exclusion.
First we find the number of (positive) triples in which each entry divides $2^33^3$. At each of $x$, $y$, $z$ we have $(4)(4)$ choices, for a total of $16^3$.
We want to subtract the number of such triples in which each entry divides $2^23^3$. There are $12^3$ such triples. There are also $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.