Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find limit without using l'Hospital rule $\lim\limits_{x\rightarrow0}\frac{\tan x-\sin x}{x^3}$ How to find the following limit without using l'Hospital rule
$$\lim_{x\rightarrow0}\frac{\tan x-\sin x}{x^3}$$
Using l'Hospital I got $1\over2$. Thanks for your help.
| $$
\begin{aligned}
\lim _{x\to 0}\left(\frac{\tan \left(x\right)-\sin \left(x\right)}{x^3}\right)
& = \lim _{x\to 0}\left(\frac{\left(x+\frac{x^3}{3}+o\left(x^3\right)\right)-\left(x-\frac{x^3}{3!}+o\left(x^3\right)\right)}{x^3}\right)
\\& = \lim _{x\to 0}\left(\frac{x+\frac{x^3}{3}-x+\frac{x^3}{3!}+o\left(x^3\right)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/816953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Probability in cards that $4$ people each get queen and king? A $52$ card deck is shuffled and then dealt out to $4$ people $($each person gets $13$ cards$)$. What is the chance that each person gets an a Queen & a King?
My attempt: I know there's $\binom{52}{13}$ different hands that can be shuffled out. The chances o... | There are $\binom{52}{13}\binom{52-13}{13}\binom{52-13-13}{13}$ ways of dealing 4 hands of 13 cards each.
Given that each hand has to have one king and queen, there are $\binom{52-8}{13-2}\binom{52-13-2-8}{13-2}\binom{52-13-2-13-2-8}{13-2}$ for the remaining cards, and $(4\cdot 3\cdot 2)\cdot (4\cdot 3\cdot 2)$ ways fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/818829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is the sum of the power series below? For $$\sum_{n=1}^{\infty}\frac{(n+2)}{n(n+1)}x^n$$
What is the sum of it?
| Since
\begin{align}
\frac{n+2}{n(n+1)} = \frac{2}{n} - \frac{1}{n+1}
\end{align}
then the series is as follows.
\begin{align}
\sum_{n=1}^{\infty} \frac{(n+2) \ x^{n} }{ n(n+1)} &= 2 \sum_{n=1}^{\infty} \frac{x^{n}}{n} - \sum_{n=2}^{\infty} \frac{x^{n-1}}{n} \\
&= - 2 \ln(1-x) - \frac{1}{x} \left( - \ln(1-x) - x \right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/819514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Trying to evaluate integral$\int_0^\infty x \sqrt{1-e^{-x}}\,e^{-x}dx$ I am trying to integrating
$$
\int_0^\infty x \sqrt{1-e^{-x}}\,e^{-x}dx\equiv I
$$
but cannot get the answer, I would like a proof not a numerical answer. My attempt at proof:
$$
y=\sqrt{1-e^{-x}}\\
y(0)=0, \ y(\infty)=1\\
y^2=1-e^{-x}\\
2ydy=e^{... | By partial integration we have
$$
\int x \sqrt{1-e^{-x}}e^{-x}dx = \frac{2}{3} x (1-e^{-x})^{3/2} -\frac{2}{3}\int dx (1-e^{-x})^{3/2}
$$
now by letting $y=e^{-x}$, $-\ln y =x$, $dx = -dy / y$ we have
$$
\int dx (1-e^{-x})^{3/2} = \int \frac{-dy}{y}(1-y)^{3/2}
$$
now let $\sqrt{1-y} = \tau$, $y= 1-\tau^2$, $dy = -2\tau... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/820635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 1
} |
Find the limit of $ \lim_{x \to 7} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2} $ I need to evaluate the limit without using L'Hopital's rule.
$$\Large
\lim_{x \to 7} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}
$$
| Here is another solution, not using Taylor series, but based on multiplying by conjugates.
I wanted to see if it would work.
Note $x^3-y^3=(x-y)(x^2+xy+y^2)$
so $$\sqrt{x+2}-\sqrt[3]{x+20}=\frac{\sqrt{(x+2)^3}-(x+20)}{x+2+ \sqrt{x+2}\sqrt[3]{x+20}+\sqrt[3]{(x+20)^2}}
$$and
$$\sqrt{(x+2)^3}-(x+20)=\frac{(x+2)^3-(x+20)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/821855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Limit $\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}$ $$\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}=\lim_{x\rightarrow\infty}1-x+x\sqrt{\frac{2}{x^2}+\frac 2x+1}=\lim_{x\rightarrow\infty}1=1\neq2$$ as Wolfram Alpha state. Where I miss something?
| $1 - x + \sqrt{2 + 2x + x^2} = \dfrac{(1-x)^2 - (2 + 2x + x^2)}{1 - x - \sqrt{2 +2x + x^2}} = \dfrac{-1 - 4x}{1 - x - \sqrt{2 + 2x + x^2}} = \dfrac{-\dfrac{1}{x} - 4}{\dfrac{1}{x} - 1 - \sqrt{\dfrac{2}{x^2} + \dfrac{2}{x} + 1}} \to \dfrac{-4}{-1-1} = 2$ when $x \to \infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/822017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Sum Involving Bernoulli Numbers : $\sum_{r=1}^n \binom{2n}{2r-1}\frac{B_{2r}}{r}=\frac{2n-1}{2n+1}$ How can we prove that
$$\sum_{r=1}^n \binom{2n}{2r-1}\frac{B_{2r}}{r}=\frac{2n-1}{2n+1}$$
where $B_{2r}$ are the Bernoulli numbers?
$$\begin{array}{c|c|c|} n & \frac{2n-1}{2n+1} & \sum_{r=1}^n \binom{2n}{2r-1} \frac{B_{2... | This identity also has a proof using the technique of annihilated
coefficient extractors (ACE).
First observe that it is equivalent to
$$\sum_{r=1}^n \frac{2n+1}{2r} {2n\choose 2r-1} B_{2r}
= n - \frac{1}{2}.$$
The left simplifies to
$$\sum_{r=1}^n {2n+1\choose 2r} B_{2r}.$$
Introduce the following generating func... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/822466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Given $2^7 \equiv 2 \mod n$ and $3^7 \equiv 3 \mod n$, prove for all $a$ in $\mathbb{Z}$: $a^7 \equiv a \mod n$ Let $n$ be a positive whole number. Given $2^7 \equiv 2 \mod n$ and $3^7 \equiv 3 \mod n$, prove for all $a$ in $\mathbb{Z}$: $a^7 \equiv a \mod n$, without using a computer.
$n$ must be greater than 3 and l... | From the given congruences, $n$ is a common factor of
$$\eqalign{
2^7-2=2(2^3+1)(2^3-1)&=2\times3^2\times7\cr
\hbox{and}\quad 3^7-3=3(3^3+1)(3^3-1)&=2^3\times3\times7\times13\ ,\qquad\cr}$$
so $n\mid 2\times3\times7$. Since for any $a$ we have
$$\eqalign{
a^2\equiv a\pmod2\quad\Rightarrow\quad a^7\equiv(a^2)^3a\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/822796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integration by parts - $\int \ln (2x+1) \text{dx}$ Use integration by parts to find
$\int \ln (2x+1) \text{dx}$.
So far I have:
$$x\ln(2x+1)-\int\dfrac{2x}{2x+1}dx+c$$
Using integration by substitution to find the integral
$$u=2x+1\Rightarrow\text{du}=2\text{dx}$$
$$\int\dfrac{2x}{2x+1}\cdot\dfrac{1}{2}\text{du}=\int x... | You made a mistake when evaluating the last integral. One approach to evaluate the integral of $\tfrac{2x}{2x+1}$ is to write it as: $$\eqalign{\int\dfrac{2x}{2x+1}\mathrm dx&=\int\dfrac{-1+(2x+1)}{2x+1}\mathrm dx\\&=\int\left[\dfrac{-1}{2x+1}+\dfrac{2x+1}{2x+1}\right]\mathrm dx\\
&=\int\left[\dfrac{-1}{2x+1}+1\right]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/823162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Prove the continuity and differentiability of parametric integration $$F(\alpha )=\int_{0}^{+ \infty } \frac{\cos x}{1+(x+\alpha )^{2} } dx$$
Prove the function F is continuous and differentiable on the interval $[0, +\infty )$
| $|F(\alpha) - F(\beta)| \leq \displaystyle \int_{0}^\infty \left|\dfrac{cosx}{1+(x+\alpha)^2} - \dfrac{cosx}{1+(x+\beta)^2}\right|dx \leq |\beta - \alpha|\cdot \displaystyle \int_{0}^\infty \dfrac{(x+\alpha)+(x+\beta)}{(1+(x+\alpha)^2)(1+(x+\beta)^2)}dx \leq |\alpha - \beta|\cdot \left(\displaystyle \int_{0}^\infty \df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/823436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving the exponential inequality: $x^y+y^x\gt1$ How can the following inequality be proven?
$$x^y+y^x\gt1$$
for $(x\gt0,y\lt1)$
| Use this Bernoulli inequality
$$(1+x)^a\le 1+ax,0<a<1,x>-1$$
It is clear we only prove $0<x,y<1$
then
$$x^y=\dfrac{1}{\left(\dfrac{1}{x}\right)^y}=\dfrac{1}{\left(1+\dfrac{1-x}{x}\right)^y}>\dfrac{1}{1+\dfrac{1-x}{x}\cdot y}=\dfrac{x}{x+y-xy}>\dfrac{x}{x+y}$$
Similarly
$$y^x>\dfrac{y}{x+y}$$
so
$$x^y+y^x>\dfrac{x}{x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/827754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Solving $x^2 - x - 1 > 0$ I am having problems understanding how to solve:
$ x^2 - x - 1 > 0 $.
Any help would be much appreciated.
| Since $x^2-x-1=\left(x-\frac{1}{2}\right)^2-\frac{5}{4}$ we must have $|x-\frac{1}{2}|>\frac{\sqrt{5}}{2}$ so $x<\frac{1-\sqrt{5}}{2}$ or $x>\frac{1+\sqrt{5}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/828889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
How do you solve two equal absolute value expressions? I'm having trouble understanding how the following is solved.
$$|x+1| = |x-2|$$
| So we know what absolute value means. It means that we want the positive version of $x+1$ and $x-2$.
So we don't know what the signs of $x-2$ and $x+2$ are.
Thus, we can make four different equalities, of which $0$, $1$, $2$, $3$, or $4$ may be true. There may be multiple values of $x$ or there might not be any value o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/829876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding the inhomogeneous solution $x_{n+2} = x_{n+1} + 20x_n + n^2 + 5^n \text{ with } x_0 = 0 \text{ and } x_1 = 0$
How would I find the inhomogeneous solution for this since the homogenous solution is 0 given initial conditions?
| Define $y_n = x_{n+1} - 5x_n$
Then $y_{n+2} = y_{n+1} + 20 y_n + ((n+1)^2-5n^2) + 5^{n+1} - 5^{n+1} = y_{n+1} + 20 y_n + (-4n^2+2n+1)$.
There, that "erased" the $5^n$ term in the recurrence. The $n^2.1^n$ term spilled a bit, but no biggie, we can do the same to those, by putting $z_n = y_{n+1} - y_n$, then $u_n = z_{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/831905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proving Definite Integral Strong Inequation I got the following question:
Prove:
$$\int^{1}_0 \frac{x^9}{\sqrt{1+x}} < \frac{1}{10}$$
I proved the weak inequation like so:
$x \geq 0 \Rightarrow \sqrt{1+x} \geq 1$ (monotonicity of square root)
$\Rightarrow \frac{x^9}{\sqrt{1+x}} \leq x^9$
$\Rightarrow \int^{1}_0 \frac{x... | If you like something stronger, consider that $\frac{1}{\sqrt{1+x}}$ is a convex function over $[0,1]$, hence:
$$\frac{1}{\sqrt{1+x}}\leq 1-\left(1-\frac{1}{\sqrt{2}}\right)x\tag{1}$$
and:
$$\int_{0}^{1}\frac{x^9}{\sqrt{1+x}}\,dx\leq \int_{0}^{1}x^9\,dx-\left(1-\frac{1}{\sqrt{2}}\right)\int_{0}^{1}x^{10}\,dx = \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/832117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that limit $\lim_{n\to\infty}\sqrt{4+\frac{1}{n^2}}+\sqrt{4+\frac{2}{n^2}}+\cdots+ \sqrt{4+\frac{n}{n^2}}-2n=\frac{1}{8}$
Let $$a_{n}=\sqrt{4+\dfrac{1}{n^2}}+\sqrt{4+\dfrac{2}{n^2}}+\cdots+
\sqrt{4+\dfrac{n}{n^2}}-2n,$$
show that
$$\lim_{n\to\infty}a_{n}=\dfrac{1}{8}$$
My attempt:
Since
$$\sqrt{4+\dfrac{i}... | Let $0 < \alpha < \frac{1}{4}$ then for $n$ large enough and $k \in \{1,\ldots,n\}$ $$2 + \frac{\alpha \, k}{n^2} < \sqrt{4+\frac{k}{n^2}} < 2 + \frac{k}{4 n^2}.$$ The limit now follows from $$1+2+\ldots+n = \frac{n(n+1)}{2}$$ and taking $\alpha \uparrow \frac{1}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/833824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 5
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Find all positive integer solutions $(x,y,z)$ that satisfy $5^x \cdot 7^y +4= 3^z$? This is another contest math-problem.
The only problem that I cannot find the way to tackle this problem.
Can anybody try to provide the solution to solve this problem?
Thanks
| Given; $5^x \cdot 7^y +4= 3^z$
we get, $(-1)^x +1 \equiv 0 \pmod3$ hence $x$ must be odd, again, $(-1)^y \equiv (-1)^z \pmod 4$ hence $y$ and $z$ have the same parity. Again $-1 \equiv 3^z \pmod 5$ or $1 \equiv 3^{2z} \pmod 5$ hence by Euler-Fermat theorem $2|z$. Again since $y$ and $z$ are even we have, $5^x \equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/835225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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From any arbitrary point $P$ on $y =\cos x$ tangents $PA$ and $PB$ are drawn to a circle which passes through From any arbitrary point $P$ on $y =\cos x$ tangents $PA$ and $PB$ are drawn to a circle which passes through the points $(1,0)$ and $(3,0)$ and touches the circle $x^2+y^2-2x-8=0$ and have its centre in first... | Given Circle
The fixed circle described in your problem is actually
$$(x-2)^2+(y-\frac{\sqrt{5}}{2})^2=\frac{9}{4}$$
This is because its tangent circle is $(x-1)^2+y^2=9$ is centered in $(1,0)$ which lies on the given circle, which means that its diameter is equal to the radius of the tangent circle, so it is $\frac{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/836208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Linear inequalities with one unknown $$3x+\frac{2}{4} \leq x+\frac{7}{2}$$ What is the solution to the inequality?
I started multiplying both sides by 4
which gave me $3x+2\leq14$.
Then I subtracted two from both sides obtaining
$$3x\leq12$$
which then I divided by 3 to obtain the solution
$$x\leq4\ .$$
This solution d... | Just subtract $x+\frac{2}{4}$ on both sides...
$$3x+\frac{2}{4} \leq x+\frac{7}{2} \iff 2x \leq \frac{7}{2}-\frac{2}{4} = \frac{14}{4}-\frac{2}{4} = \frac{12}{4} = 3$$
and finally divide both sides by $2$ to find $x \leq \frac{3}{2}$.
Note:
I started multiplying both sides by $4$ which gave me $3x+2\leq 14$.
You shou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/836604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Easier Proof of $\sin{3\theta} + \sin\theta = 2\sin{2\theta}\cos\theta$ I am curious to see whether anybody can give me a proof that takes less steps.
Here is how I did it:
$$\sin{3\theta} + \sin\theta = 2\sin{2\theta}\cos\theta$$
LHS $$\eqalign{\sin(2\theta + \theta) + \sin\theta &= \sin2\theta\cos\theta + \cos2\theta... | Use the trigonometric identity about $\sin A + \sin B= 2\sin \frac{A+B}{2} \cos\frac{A-B}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/837430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Expressing sums of powers I am trying to solve this question but I'm not really understanding how to continue, I would greatly appreciate some kind of tip.
The Question
The formula for $1^2 + \cdots + n^2$ may be derived as follows. We begin with the formula $$(k+1)^3 - k^3 = 3k^2 + 3k + 1$$
Writing this formula for... | The formula is the Binomial Theorem, or just multiplication: $(k+1)^3=k^3+3k^2+3k+1$. This can be rewritten as
$$(k+1)^3-k^3=3k^2+3k+1.$$
The corresponding formula for fourth powers is $(k+1)^4=k^4+4k^3+6k^2+4k+1$. It yields the identity
$$(k+1)^4-k^4=4k^3+6k^2+4k+1.$$
That can be used to give a telescoping argument ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/838102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solvability of a variant Fermat equation If
$p$
is an odd prime,
are there any pairwise coprime integers
$a, b, c > 0$
such that
$$(a-b+c)^{p} + (a+b-c)^{p} = (a+b+c)^{p}$$
[a proof without Fermat's last theorem]
| Fermat's Last Theorem states that there are no non-trivial (i.e. $x,y,z\neq 0$) positive integer solutions to $x^n+y^n=z^n$ for $n\geq 3$. The first odd prime is 3, so there are no solutions to your equation.
The only possibilities are for at least one of $a-b+c$, $a+b-c$, or $a+b+c$ to be non-positive. If one is zer... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/839566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Extrema of $f(x)=\frac{\sin (5x)} 5 - \frac{2\sin(3x)} {3} + \sin (x)$. (a) I need help in finding maxima and minima of the following funcion:
$$f(x)=\frac{\sin (5x)} 5 - \frac{2\sin(3x)} {3} + \sin (x)$$
therefore I need to find the roots of $f'(x)=\cos(5x)-2\cos(3x)+\cos(x)$.
(b) I need to find the minima of $$f(x)=4... | Hints:
a) $\cos(5x) \equiv 16\cos^5(x)-20\cos^3(x)+5\cos(x)$
$\cos(3x) \equiv 4\cos^3(x)-3\cos(x)$
So, we've got:
Now, let $c \equiv \cos(x)$.
We've got:
$$[16c^5-10c^3+5c]-2[4c^3-3c]+c=0 \iff \boxed{2c\underbrace{[8c^4-9c^2+6=0]}_{\textrm{quadratic in} \ c^2}} \tag{A}$$
Solve $(A)$!
b) Solving this analytically is ver... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Real part of Complex Function I've this function
$$f(k,\theta) = \frac{1}{k}\frac{1}{\cot\delta_0(k) -i }$$
and i know that $k\cot\delta_0(k) = -\frac{1}{a} + \frac{1}{2}r_ek^2 + \cdots$
it is an expansion.
How can i get that the $\Re{f(k,\theta)} = -a + a^2(a-\frac{1}{2}r_e)k^2$ ?
| You probably already solved this by yourself, but to close off this question with an answer, I followed through on my comment.
$$
\Re f(k,\theta) = \frac{1}{k}\frac{\cot \delta_0(k)}{\cot^2 \delta_0(k) +1}
$$
multiplying top and bottom by $k^2$ we obtain
$$
\Re f(k,\theta) =\frac{k\cot \delta_0(k)}{k^2\cot^2 \delta_0(k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/840872",
"timestamp": "2023-03-29T00:00:00",
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Given a sequence $ (a_{n})_{n \in \mathbb{N}} $, find $ \sum_{n = 1}^{\infty} \frac{1}{a_{n} + 2} $. I would appreciate it if somebody could help me with the following problem:
If the sequence $ (a_{n})_{n \in \mathbb{N}} $ satisfies
$$
a_{1} = 1
\quad \text{and} \quad
\forall n \in \mathbb{N}, n \ge 2: \quad
a_{n} ... | Since $a_1=1$ and
$$
a_{n+1}=\prod_{k=1}^n\left(1+\frac{a_k}2\right)\tag{1}
$$
we have that
$$
a_n\ge1\tag{2}
$$
and
$$
a_{n+1}=a_n\left(1+\frac{a_n}2\right)\tag{3}
$$
Subtracting $a_n$ from both sides of $(3)$ gives
$$
a_{n+1}-a_n=\frac{a_n^2}2\tag{4}
$$
and since $a_1=1$, $(1)$ and $(2)$ show that
$$
a_n\ge\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/842273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Contour integral $\int^\pi_{-\pi}(a-\cos\theta)^b\exp(c\cos\theta)d\theta$ assuming $a>1$, $b>0$, $c>0$ Under the condition $a>1$, $b>0$, $c>0$, is there any good function to express the following integral?
$$
\int^\pi_{-\pi}\left(a-\cos\theta\right)^b\exp\left(c\cos\theta\right)d\theta
$$
I guess some hypergeometric f... | $\int_{-\pi}^\pi(a-\cos\theta)^be^{c\cos\theta}~d\theta$
$=\int_{-\pi}^0(a-\cos\theta)^be^{c\cos\theta}~d\theta+\int_0^\pi(a-\cos\theta)^be^{c\cos\theta}~d\theta$
$=\int_\pi^0(a-\cos(-\theta))^be^{c\cos(-\theta)}~d(-\theta)+\int_0^\pi(a-\cos\theta)^be^{c\cos\theta}~d\theta$
$=\int_0^\pi(a-\cos\theta)^be^{c\cos\theta}~d... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Trigonometric Triangle Equality $A, B, C$ are the angles of a triangle then $tan^2(A/2)+tan^2(B/2)+tan^2(C/2)$ is
always greater than what integral value.
| Since $A,B,C$ are angles of a triangle, we have $0<\dfrac{A}{2},\dfrac{B}{2},\dfrac{C}{2}<\dfrac{\pi}{2}$. For this range of values, $\tan^2 x$ is a convex function. Hence, from Jensen's inequlaity,
$$\tan^2\left(\frac{1}{3}\frac{A}{2}+\frac{1}{3}\frac{B}{2}+\frac{1}{3}\frac{C}{2}\right) \leq \frac{1}{3}\tan^2\frac{A}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/842881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Limit of a matrix multiplication How can I calculate this limit:
$\displaystyle\lim_{n\to\infty}\begin{bmatrix}0.9 & 0.2\\0.1 & 0.8\end{bmatrix}^n$
What is the tool that i need to aply? eigenvalues and eigenvectors? diagonalization? canonical form?
(This came in a contest and was the only problem i cannont have an ide... | Because that is a stochastic matrix you just need to fine the fixed probability vector or steady state vector.
$$(t1,t2)\cdot\left(
\begin{array}{cc}
0.9 & 0.1 \\
0.2 & 0.8 \\
\end{array}
\right)=(t1,t2)$$
Notice I have transposed A for convenience and this yields the simultaneous set of equations
$$.9 \cdot t1+.2 \c... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
} |
Diagonalization of a Toeplitz matrix Let $0<\lambda\leq1$ so that the $n \times n$ matrix
$$\Sigma = \begin{pmatrix}
1&1-\lambda& \cdots &1-\lambda\\
1-\lambda&\ddots&\ddots& \vdots\\
\vdots &\ddots&\ddots&1-\lambda\\
1-\lambda&\cdots&1-\lambda&1\\
\end{pmatrix}$$
is positive definite. I believe we can orthogonally dia... | We can rewrite $\Sigma$ as $(1 - \lambda)M + \lambda I$, where $M$ is given by
$$
M = \pmatrix{
1&\cdots & 1\\
\vdots &\ddots&\vdots\\
1&\cdots &1
}
$$
It follows that $V^T\Sigma V$ will be diagonal iff $V^T M V$ will be diagonal. And so, it suffices to show that
$M = VDV^T$ with
$$
D = \pmatrix{
0&&&\\\
&\ddots&&\\... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Apostol (6.22.10): Finding $\int \frac{\arcsin x}{x^2} dx$ I'm trying to solve another integral from Apostol (Chapter 6, Section 6.22, Question 10) which says to show the following:
$$
\int \frac {\arcsin x}{x^2}dx = \log|{\frac {1-\sqrt{1-x^2}}{x}}| - \frac {\arcsin x}{x} +C
$$
I have tried the following:
With $\int \... | Note that $(1+\sqrt{1-x^2})(1-\sqrt{1-x^2}) = 1 - (1-x^2) = x^2$, so
$$\log\left|\dfrac{1 + \sqrt{1-x^2}}{x}\right| = \log \left|\dfrac{x}{1-\sqrt{1-x^2}}\right| = - \log \left|\dfrac{1-\sqrt{1-x^2}}{x}\right|$$
You and Apostol have the same answer after all.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/849163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Distribution of a binomial variable squared If I know $X$ is a binomial random variable, how can I find the distribution of $X$ squared (I know that $P(Y=y=x^2) = p(X=x)$ but does this distribution have a standard name)? In particular, how can I find its expected value?
Thanks!
EDIT:
Thank you all! (-:
| $\begin{align}
\text{E}(X^2) & = \sum\limits_{k = 0}^n k^2P[X = k]\\
& = \sum\limits_{k = 0}^n k^2\,^nC_kp^kq^{n - k}\\
& = \sum\limits_{k = 1}^n k^2\, \dfrac{n!}{k! (n - k)!} p^kq^{n - k}\\
& = \sum\limits_{k = 1}^n k \dfrac{n!}{(k - 1)! (n - k)!} p^kq^{n - k}\\
& = \sum\limits_{k = 1}^n (k - 1 + 1) \dfrac{n!}{(k - 1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/849196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 2
} |
The convergence of $\sqrt {2+\sqrt {2+\sqrt {2+\ldots}}}$ I would like to know if this sequence converges $\sqrt {2+\sqrt {2+\sqrt {2+\ldots}}}$.
I know this sequence is increasing monotone, but I couldn't prove it's bounded.
Thanks
| As written it's not clear this is a sequence at all but I'm assuming your sequence is
$\sqrt{2}$, $\sqrt{2 + \sqrt{2}}$, $\sqrt{2 + \sqrt{2+ \sqrt{2}}}$, $\sqrt{2 + \sqrt{2+ \sqrt{2+\sqrt{2}}}}$, $\sqrt {2+\sqrt {2+\sqrt {2+ \sqrt{2 +\ldots}}}}$.
Then it should be obvious that each entry in the sequence is positive and... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
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Rewriting this complex square root for some reason I can to figure out how to rewrite this square root.
I have:
$\sqrt{2+i}$
And I need to rewrite it into:
$\frac{\sqrt{2(\sqrt{5} + 2)} + \sqrt{-2(\sqrt{5} - 2)}}{2}$
Can anybody show me I to do this? I've been trying for an hour now..
| One can use trigonometry for sure but I thing the fastest way to solve this is just to solve the quadratic.
$$(x+yi)^2=2+i$$ so
$$x^2-y^2=2$$
$$2xy=1$$
and $$(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2=4+1=5$$
So $$x^2+y^2=\sqrt{5}$$
(positive since $x$ and $y$ real)
and
$$x^2=\frac{\sqrt{5}+2}{2}$$
$$y^2=\frac{\sqrt{5}-2}{2}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/849747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$ \Big(\dfrac{x^7+y^7+z^7}{7}\Big)^2=\Big(\dfrac{x^5+y^5+z^5}{5}\Big)^2\cdot\Big(\dfrac{x^4+y^4+z^4}{2}\Big) $ I have a question. I tried so to solve it, but there is a problem.
that is i don't have any idea to findout how can i work with degrees 4,5,7 ...
this is the problem :
let $ x , y $ and $ z $ three real number... | If $x=y=z=0$, then it is trivial.
WLOG, $x\ne 0$ (at least one of variables $\ne 0$).
Denote
$y=ax$,
$z=-x(1+a)$.
It is enough to prove, that
$$
\left(\frac{1^7+a^7-(1+a)^7}{7}\right)^2 = \left( \frac{1^5+a^5-(1+a)^5}{5} \right)^2 \cdot \left(\frac{1^4+a^4+(1+a)^4}{2}\right);
$$
$$
\left(\frac{7a+21a^2+35a^3+35a^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/851985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 0
} |
How to prove $\frac{1}{4}(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a})\ge \sqrt[4]{\frac{a^4+b^4+c^4+d^4}{4}}$ Let $a,b,c,d>0$, show that
$$\dfrac{1}{4}\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)\ge \sqrt[4]{\dfrac{a^4+b^4+c^4+d^4}{4}}$$
I know this is interesting inequality,and... | The above answer by johannesvalks "is not yet completed", as johannesvalks himself points out. Here is a new answer without calculus.
Rewrite the question as follows: Show that
$$
f = (a b c d)^4\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)^4 - 64 (a b c d)^4\left({{a^4+b^4+c^4+d^4}}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/852070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 1,
"answer_id": 0
} |
Max. and Min. value of $|z|$ in $\left|z+\frac{2}{z}\right| = 2\,$ If $z$ is a complex no. such that $\displaystyle \left|z+\frac{2}{z}\right| = 2\,$ Then find max. and min. value of $\left|z\right|$.
$\bf{My\; Try:}$ Given $\displaystyle \left|z+\frac{2}{z}\right| = 2\Rightarrow \left|z+\frac{2}{z}\right|^2 = 2^2=4$.... | $$|z|-\frac{2}{|z|}\le \left|z+\frac{2}{z}\right|=2 \implies\\ |z|^2-2|z|-2\le0\implies\\ 1-\sqrt{3}<0\le|z|\le1+\sqrt{3}$$
Maximum value is $1+\sqrt{3}$ provided we show that equality holds.
Recall that equality holds when $z$, $-\frac{2}{z}$ have same argument. Hence $\arg{z}=-\frac{\pi}{2}$.
Similarly using $\frac{... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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$2^{1/4} \times 4^{1/8} \times 8^{1/16} \times 16^{1/32} \times \ldots\to2$ $2^{1/4} \times 4^{1/8} \times 8^{1/16} \times 16^{1/32} \times \ldots\to2$
How can I explain this to a school student who doesn't know what a limit is?
| $$2^{1/4} \times 4^{1/8} \times 8^{1/16} \times 16^{1/32} \times \ldots=\prod^{\infty}_{n=1}(2^n)^{1/2^{(n+1)}}=x$$
Log by base 2 both sides and we have:
$$1/4+1/4+3/16+1/8+5/64..=\sum^{\infty}_{n=1} n/2^{(n+1)}$$
This sum is $1$. To show it we need to find
$$S=\lim \sum^{\infty}_{n=1} n/2^{(n+1)}=1$$
Let's multiply i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/857924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Volume of sphere inside a cylinder I have homework question:
Calculate the volume of $\{(x,y,z) \mid x^2+y^2+z^2\le (2a)^2\} \cap \{(x,y,z) \mid (x-a)^2 + y^2 \le a^2\}$
in two ways:
1) Polar coordinates
2) Spherical coordinates
How can I solve it?
Edit: What I tried:
$$2\int_0^{2\pi}\int_0^a r\sqrt{4a^2-x^2-y^2}\,dr\... | For cylindrical coordinates:
$$z=\pm \sqrt{4a^2-(x^2+y^2)}$$
Using polar coordinates
$$x=r\cos\phi \ , \ y=r\sin \phi $$
we end up with
$$z=\pm \sqrt{4a^2-r^2}$$
now, just integrate
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_0^{2a\cos \phi}\int_{-\sqrt{4a^2-r^2}}^{\sqrt{4a^2-r^2}}rdzdrd\phi=2\int_{-\frac{\pi}{2}}^{\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/857989",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show that the following series is less than $4 \pi^2 / 3$. Show: For any $k = 0,1,2,...$,
$$ \sum_{i=0}^{i=k} \frac{(k+1)^2}{(i+1)^2 (k-i+1)^2} \leq \frac{4 \pi^2}{3}. $$
| Note that
$$
\frac{1}{i+1}+\frac{1}{k-i+1}=\frac{k+2}{(i+1)(k-i+1)}
$$
So, (because $(a+b)^2\leq 2a^2+2b^2$):
$$
\frac{(k+2)^2}{(i+1)^2(k-i+1)^2}\leq\frac{2}{(i+1)^2}+\frac{2}{(k-i+1)^2}
$$
Adding we get
$$
\sum_{i=0}^{k}\frac{(k+2)^2}{(i+1)^2(k-i+1)^2}
\leq 4\sum_{i=0}^{k}\frac{1}{(1+i)^2}\leq 4\zeta(2)=\frac{2\pi^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/858885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find the Value of Integral Find the Value of $$\begin{align}I=\int_{0}^{1}\frac{\ln(x)\,dx}{1-x^2}\end{align}$$
I have tried like this: We have $$\begin{align}2I=\int_{0}^{1}\frac{\ln(x^2)\,dx}{1-x^2}=\int_{0}^{1}\frac{\ln(1-(1-x^2))\,dx}{1-x^2}\end{align}$$ So
$$2I=\begin{align}\int_{0}^{1}\frac{-(1-x^2)-\frac{(1-x^2)... | Let
$$ I(\alpha)=\int_{0}^{1}\frac{x^{\alpha}dx}{1-x^2}. $$
Then $I'(0)=I$. Now
\begin{eqnarray*}
I(\alpha)=\lim_{a\to 1^-}\int_{0}^{a}\sum_{n=0}^\infty x^{\alpha+2n}dx=\lim_{a\to 1^-}\sum_{n=0}^\infty \frac{1}{\alpha+2n+1}a^{\alpha+2n+1}
\end{eqnarray*}
and hence
\begin{eqnarray*}
I'(\alpha)&=&\lim_{a\to 1^-}\sum_{n=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/860091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Find the limit of $\lim\limits_{x \to+\infty} \frac{4x-3}{2x+5}$ and $\lim \limits_{x \to -\infty} \frac{2x^2-x+5}{4x^3-1}$ Hello guys help me with these:
1.) $\lim \limits_{x \to +\infty} frac{4x-3}{2x+5}$
$=\lim \frac{4-\frac{3}{x}}{2+\frac{5}{x}}$
$= \lim \frac{4-3\frac{1}{x}}{2+5\frac{1}{x}}$
=$\frac{\lim 4- \lim ... | $$\lim_{x \to -\infty} \frac{2x^2-x+5}{4x^3-1} =\lim_{x \to -\infty} \frac{2x^2(1-\frac{1}{2x}+\frac{5}{2x^2})}{4x^3(1-\frac{1}{4x^3})} =\lim_{x \to -\infty} \frac{(1-\frac{1}{2x}+\frac{5}{2x^2})}{2x(1-\frac{1}{4x^3})}$$
It is clear that
$$\lim_{x \to -\infty} \frac{1}{4x^3} =\lim_{x \to -\infty} \frac{1}{2x} = \lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/861236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\frac{1}{2}n^2-3n=\Theta{(n^2)}$ Show that $$\frac{1}{2}n^2-3n=\Theta{(n^2)}$$
$$$$
$\displaystyle{\frac{1}{2}n^2-3n=\Theta{(n^2)}: \\ \exists c_1, c_2 >0 , \ \ \exists n_0 \geq 1 \text{ such that } \forall n \geq n_0 \\ 0<c_1 n^2 \leq \frac{1}{2}n^2-3n \leq c_2 n^2}$
$$$$
That means that I have to find the ... | $$n\geqslant12\implies\frac14n^2\leqslant\frac12n^2-3n\leqslant\frac12n^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/861564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find expression for : $ S_n =\sum_{i=1}^{n} \frac{i}{i^4+i^2+1} $ I want to find a formula for the sum of this series using its general term.
How to do it?
Series
$$
S_n = \underbrace{1/3 + 2/21 + 3/91 + 4/273 + \cdots}_{n \text{ terms}}
$$
General Term
$$
S_n = \sum_{i=1}^{n} \frac{i}{i^4+i^2+1}
$$
| Using partial fractions, we get $\dfrac{i}{i^4+i^2+1} = \dfrac{\tfrac{1}{2}}{i^2-i+1} - \dfrac{\tfrac{1}{2}}{i^2+i+1}$.
Thus, $S_n = \displaystyle\sum_{i = 1}^{n}\dfrac{i}{i^4+i^2+1} = \dfrac{1}{2}\sum_{i = 1}^{n}\dfrac{1}{i^2-i+1} - \dfrac{1}{i^2+i+1}$.
Since $(i+1)^2-(i+1)+1 = i^2+i+1$, this sum telescopes to $S_n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/865442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Solving a trigonometric limit $\lim_{x\to\pi/6}\frac{1 - 2\sin{x}}{2\sqrt{3}\cos{x} - 3}$ First off, please excuse my n00bishness I have only just begun learning about algebraic manipulation of limits so this is probably a really dumb or obvious question.
I'm trying to solve the following limit:
$$
\lim_{x\to\pi/6}\fra... | $$\lim_{x\to\pi/6}\frac{1 - 2\sin{x}}{2\sqrt{3}\cos{x} - 3}$$
$$=-\frac2{2\sqrt3}\lim_{x\to\pi/6}\frac{\sin x-\sin\frac\pi6}{\cos x-\cos\frac\pi6}$$
$$=-\frac2{2\sqrt3}\lim_{x\to\pi/6}\frac{\sin x-\sin\frac\pi6}{x-\frac\pi6}\cdot\frac1{\lim_{x\to\pi/6}\dfrac{\cos x-\cos\frac\pi6}{x-\frac\pi6}}$$
$$=-\frac2{2\sqrt3}\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/865509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
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Pythagorean triple means Just out of curiosity, do there exist two positive integers whose arithmetic mean ($A $), geometric mean ($G $) and harmonic mean ($H $) constitute a Pythagorean triple?
That is, $A $, $G $ and $H $ are positive integers, and $H^2 + G^2 = A^2$.
| Let $a$ and $b$ be positive integers with $a \ge b$; then you're asking whether we can ever have
$$\left(\frac{2}{\frac 1 a + \frac 1 b}\right)^2 + \Big(\sqrt{ab}\Big)^2 = \left(\frac{a + b}{2}\right)^2$$
or upon some simplification,
$$\frac{4a^2 b^2}{(a+b)^2} + ab = \frac{(a + b)^2}{4}$$
Rearranging,
$$\frac{4a^2 b^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/865832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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$12\frac{\sin 45^\circ}{\sin 60^\circ}$ Need help breaking this down. Otherwise known as $12\dfrac{\left(\frac{1}{\sqrt2}\right)}{\left(\frac{\sqrt3}{2}\right)}$
How do you simplify this multi level fractional radical expression into $4\sqrt{6}$.
| $$12\dfrac{\left(\frac{1}{\sqrt2}\right)}{\left(\frac{\sqrt3}{2}\right)}=12\cdot\frac{1}{\sqrt2}\div\frac{\sqrt3}{2}=12\cdot\frac{2}{\sqrt{6}}=\frac{24}{\sqrt{6}}=\frac{24}{\sqrt{6}}\cdot\frac{\sqrt{6}}{\sqrt{6}}=\frac{24\sqrt{6}}{6}=4\sqrt{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/866442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Integration practice of $\int \frac{\sqrt{25-y^2}}{y}dy$ I need to solve $\int \frac{\sqrt{25-y^2}}{y}dy$.
I originally thought IBP, but that led to a very large and confusing algebra problem.
Then I started to look at the $\sqrt{25-y^2}$ and started to consider doing a $u$-substituion, and letting $u=\sin\theta$ and ... | Trigonometric substitutions are not necessary. Let $u^2 = 25 - y^2$, so that $2u \, du = -2y \, dy$. Then $$\begin{align*} \int \frac{\sqrt{25-y^2}}{y} \, dy &= - \int \frac{-y \sqrt{25-y^2}}{y^2} \, dy \\ &= - \int \frac{u \sqrt{u^2}}{25-u^2} \, du \\ &= \int \frac{u^2 - 25 + 25}{u^2 - 25} \, du \\ &= u + 25 \int \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/867085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Let $x,y,z$ be positive integers such that $\frac{1}{x}-\frac{1}{y}=\frac{1}{z}$. Let $h=\gcd(x,y,z)$, Prove that $hxyz,h(y-x)$ are perfect squares Let $x,y,z$ be positive integers such that $\frac{1}{x}-\frac{1}{y}=\frac{1}{z}$. Let $h=\gcd(x,y,z)$, Prove that $hxyz,h(y-x)$ are both perfect squares.
My attempt:
Let $x... | Let us consider
$$
\frac{1}{x} - \frac{1}{y} = \frac{1}{z},
$$
and
$$
h = \textrm{gcd}(x,y,z).
$$
Say that
$$
\begin{eqnarray}
x &=& h x',\\
y &=& h y',\\
z &=& h z',\\
\end{eqnarray}
$$
then we obtain
$$
\frac{1}{x'} - \frac{1}{y'} = \frac{1}{z'},
$$
where
$$
\textrm{gcd}(x',y',z') = 1.
$$
We can write
$$
\Big(y'-x'\... | {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Another integral for $\pi$ Here is a new integral for $\pi$.
$$\int_{0}^{1}\sqrt{\frac{\left\{1/x\right\}}{1-\left\{1/x\right\}}}\, \frac{\mathrm{d}x}{1-x} = \pi $$
where $\left\{x\right\}$ denotes the fractional part of $x$.
Do you have any proof?
| With the substitution $1/x=t$, the integral is:
$$I=\int_1^{\infty} \sqrt{\frac{\{t\}}{1-\{t\}}}\frac{dt}{t(t-1)}=\sum_{n=1}^{\infty}\int_n^{n+1}\sqrt{\frac{t-n}{n+1-t}}\frac{dt}{t(t-1)}$$
Next, use the substitution $\dfrac{t-n}{n+1-t}=x^2 \Rightarrow t=n+1-\dfrac{1}{1+x^2} \Rightarrow dt=\dfrac{2x}{(1+x^2)^2}\,dx$ i.e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/870680",
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"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 1
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QR transformation with Householder transformation It's a task i do to understand minimizing the error including the QR transformation with the help of Householder transformation. I think i really do something wrong but i dont get it running i hope you can help me.
Given the task to minimizing the error of
$$ A = \begin... | Based on the comments, this is probably what you're supposed to do.
To transform $A$ to the upper triangular form requires an application of two Householder transformations. Take
$$
Q_1=I-2\frac{v_1v_1^T}{v_1^Tv_1}=
\frac{1}{6}
\begin{bmatrix}
-3 & -3 & -3 & -3 \\
-3 & 5 & -1 & -1 \\
-3 & -1 ... | {
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"answer_count": 1,
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Denesting radicals like $\sqrt[3]{\sqrt[3]{2} - 1}$ The following result discussed by Ramanujan is very famous: $$\sqrt[3]{\sqrt[3]{2} - 1} = \sqrt[3]{\frac{1}{9}} - \sqrt[3]{\frac{2}{9}} + \sqrt[3]{\frac{4}{9}}\tag {1}$$ and can be easily proved by cubing both sides and using $x = \sqrt[3]{2}$ for simplified typing.
R... | A general identity: $$\sum_{k=0}^{3n-1}\frac{(-2)^{k/3}}{9^{1/3}} = \frac{1-(-2)^n}{3}\sqrt[3]{\sqrt[3]2-1}$$ so Ramanujan's case was $n=1$, yielding the famous $$\sqrt[3]{\sqrt[3]2-1}=\sqrt[3]{\frac 19}-\sqrt[3]{\frac 29}+\sqrt[3]{\frac 49}$$ and possibly sheds light on how he found other like $(2)$, $(3)$, $(5)$ and ... | {
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"source": "stackexchange",
"question_score": "86",
"answer_count": 3,
"answer_id": 1
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Find the $n$th term of $1, 2, 5, 10, 13, 26, 29, ...$ How would you find the $n$th term of a sequence like this?
$1, 2, 5, 10, 13, 26, 29, ...$
I see the sequence has a group of three terms it repeats: Double first term to get second term, add three to get third term, repeat.
What about: $2, 3, 6, 7, 14, 15, 30,... $?... | Denote the $n$-th term by $x_{n}$ and have a look at the terms with
odd index $n$. Based on what you remarked yourself:
$x_{1}=1$
$x_{3}=2x_{1}+3=2+3$
$x_{5}=2x_{3}+3=2\left(2+3\right)+3=2^{2}+2.3+3$
$x_{7}=2x_{5}+3=2\left(2^{2}+2.3+3\right)+3=2^{3}+2^{2}.3+2.3+3$
et cetera.
This starts to look like
$x_{2n+1}=2^{n}+\l... | {
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"answer_id": 1
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Show that $\frac{x^4 +7x^3+5}{4x+1}$ is big-theta($x^3$) I'm having trouble grasping how to set these types of problems. There are a lot of related questions but it's difficult to abstract a general procedure on finding constants that give the given function bounding constraints to make it big-theta(general function).... | Yes, you could take $c_1=\frac{1}{4}$, then you have the following:
$$ \frac{1}{4}x^3 \leq \frac{x^4 +7x^3+5}{4x+1} \Rightarrow \frac{4x^4+x^3}{4} \leq x^4+7x^3+5 \Rightarrow x^4+\frac{1}{4}x^3 \leq x^4+7x^3+5 \\ \Rightarrow \frac{27}{4}x^3 \geq -5 \Rightarrow x^3 \geq -\frac{20}{27} \Rightarrow x \geq - \sqrt[3]{\frac... | {
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"url": "https://math.stackexchange.com/questions/874300",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Proof of an inequality
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{\sqrt{a+b+c}+\sqrt{a}}{b+c} + \frac{\sqrt{a+b+c}+\sqrt{b}}{c+a} + \frac{\sqrt{a+b+c}+\sqrt{c}}{a+b} \geq \frac{9+3\sqrt{3}}{2\sqrt{a+b+c}}$$
| We need to prove that
$$\sum_{cyc}\frac{a+b+c+\sqrt{a(a+b+c)}}{b+c}\geq\frac{9+3\sqrt3}{2}$$ or
$$\sum_{cyc}\frac{a}{b+c}+\sum_{cyc}\frac{\sqrt{a(a+b+c)}}{b+c}\geq\frac{3+3\sqrt3}{2}.$$
But by C-S $$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^2}{ab+ac}\geq\frac{(a+b+c)^2}{2(ab+ac+bc)}\geq\frac{3}{2}.$$
Thus, it remains t... | {
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"source": "stackexchange",
"question_score": "4",
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Calculate $\frac{1}{5^1}+\frac{3}{5^3}+\frac{5}{5^5}+\frac{7}{5^7}+\frac{9}{5^9}+\cdots$ I'm an eight-grader and I need help to answer this math problem.
Problem:
Calculate $$\frac{1}{5^1}+\frac{3}{5^3}+\frac{5}{5^5}+\frac{7}{5^7}+\frac{9}{5^9}+\cdots$$
This one is very hard for me. It seems unsolvable. How to calcul... | This sum can be represented in the form $$S=\sum_{k=0}^\infty (2k+1)x^{2k+1}$$$$=>S=x\sum_{k=0}^\infty (2k+1)x^{2k}$$ where $x=\frac 15$. Now we look at the following geometric progression $S(x)=\sum_{k=0}^\infty x^{2k+1}$ where $|x|\lt 1$.hence $S(x)=\frac a{1-r}$ where a=x and $r=x^2$. Therefore $S(x)=\frac x{1-x^2}$... | {
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"source": "stackexchange",
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"answer_count": 4,
"answer_id": 1
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prove $\sum\limits_{cyc} \frac {a^3} {b+c+d} \geq \frac {1} {3}$
Show that if $a$, $b$, $c$ and $d$ are non-negatives and $ab+bc+cd+da=1$ then:
$$\sum\limits_{cyc} \frac {a^3} {b+c+d} \geq \frac {1} {3}$$
yet again it should be solved with Cauchy inequality.
thing i have done so far:
$(\sum\limits_{cyc} \frac {a... | This solution does not use Cauchy inequality, but I think it is interesting by its own.
Multiply the right-hand side by $ab+bc+cd+da$. Now the inequality is homogeneous so we can
assume $a+b+c+d=1$ instead of $ab+bc+cd+da = 1$ by multiplying a common constant to all variables. Now we have to show this.
$$\sum_{cyc} \fr... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Antiderivative of $\frac{1}{1+\sin {x} +\cos {x}}$ How do we arrive at the following integral
$$\displaystyle\int\dfrac{dx}{1+\sin {x}+\cos {x}}=\log {\left(\sin {\frac{x}{2}}+\cos {\frac{x}{2}}\right)}-\log {\left(\cos {\frac{x}{2}}\right)}+C\ ?$$
| Hint :
Rewrite the integrand as
\begin{align}
\frac{1}{1+\color{blue}{\sin x}+\color{red}{\cos x}}&=\frac{1}{\sin^2\frac x2+\cos^2\frac x2+\color{blue}{2\sin\frac x2\cos\frac x2}+\color{red}{\cos^2\frac x2-\sin^2\frac x2}}\\
&=\frac{1}{2\cos\frac x2\left(\sin\frac x2+\cos\frac x2\right)}\\
&=\frac{\sin\frac x2}{2\cos\f... | {
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"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Prove $\frac{\sin A}{\sec A+\tan A-1}+ \frac{\cos A}{\csc A+\cot A-1}=1$ $$\frac{\sin A}{\sec A+\tan A-1}+ \frac{\cos A}{\csc A+\cot A-1}=1$$
Prove that L.H.S.$=$R.H.S.
This type of questions always creates problem when in right hand side some trigonometry function is given then it is bit easy to think how to proceed ... | HINT
Simplify as follows : substitute $\sec \theta = \frac{1}{\cos \theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$. By this, we will get first term of LHS is
$\frac{\sin \theta \ \cos \theta }{1 + \sin \theta - \cos \theta }$. Similar if we substitute \second term of LHS, we will get LHS =
$\frac{\sin \th... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
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Calculating a cubic spline goes wrong I am trying to solve a old exam and really stuck at the cubic splines.
We have the function $f(x) = \cos^2(\frac{x}{2})$ and the points $x_0 = \frac{\pi}{2}$, $x_1=0$ and $x_2 = \frac{\pi}{2}$.
With some simple math we get the points:
$(-\frac{\pi}{2},\frac{1}{2}),(0,1),(\frac{... | I hope that the mistake I pointed out in a comment will help to straighten out the computation. Meanwhile, let me present a different approach, which may be more illuminating that "put these numbers into these formulas".
The given data points are symmetric about the $y$-axis. So will be the spline. In particular, the ... | {
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"source": "stackexchange",
"question_score": "1",
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How to show $n(n+1)(2n+1) \equiv 0 \pmod 6$? I've been asked to show that:
$n(n+1)(2n+1) \equiv 0 \pmod 6$
I found in a previous question that:
$n(n+1)$ was divisible by $2$ and resulted in an even number e.g
$n(n+1) \equiv 0 \pmod 2$
so I figured I needed to find:
$(2n+1) \equiv 0 \pmod 3$ in order to complete
$n(n+1)... | The numbers $2n$, $2n+1$, and $2n+2$ are three consecutive numbers, so one of them is divisible by $3$. If $2n+1$ is not, then one of $2n$ or $2n+2$ is. But if $2n$ is divisible by $3$, so is $n$. And if $2n+2$ is divisible by $3$, so is $n+1$. Thus one of $n$, $2n+1$, and $n+1$ is divisible by $3$.
| {
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"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Show that there is an angle $\theta$ such that $a=\cos\theta$ and $b=\sin\theta$ My problem is from Israel Gelfand's Trigonometry textbook.
Page 50. Exercise 3: Suppose that $\alpha$ is some angle. If $a=4\cos^3\alpha-3\cos\alpha$ and $b=3\sin\alpha-4\sin^3\alpha$, show that there is an angle $\theta$ such that $a=\cos... | To solve it using the approach the OP wants to take, it helps to use the abbreviations $c$ and $s$ for $\cos\alpha$ and $\sin\alpha$. With these abbreviations, the desired identity becomes
$$16(c^6+s^6)-24(c^4+s^4)+9(c^2+s^2)=1$$
The key thing we know is that $c^2+s^2=1$. Squaring this gives
$$1=(c^2+s^2)^2=c^4+2c^2s... | {
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Prove $\int_{\mathbb{R^{+}}} \frac{\sin^3 {(\pi x^2)} \cos {(4x^2)}}{x^5} dx=\frac{\pi}{32} (3\pi-4)^2$ How do you arrive at the result $$I=\displaystyle\int_{\mathbb{R^{+}}} \dfrac{\sin^3 {(\pi x^2)} \cos {(4x^2)}}{x^5} dx=\dfrac{\pi}{32} (3\pi-4)^2\ ?$$
Wolfram Alpha agrees numerically.
I tried replacing $\sin p$ by ... | Here I present another way to solve/generalize the problem. It is much easier. Define
$$ I_q(p)=\frac{1}{2}\int_0^\infty\frac{\sin^3(pt)\cos(qt)}{t^3}dt,p\ge0,q\ge0. $$
Clearly $I_q(0)=I_q'(0)=I_q''(0)=0$ and
\begin{eqnarray}
I_q''(p)&=&\frac{3}{8}\int_0^\infty\frac{(3\sin(3pt)-\sin(pt))\cos(qt)}{t}dt\\
&=&\frac{3}{16}... | {
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"url": "https://math.stackexchange.com/questions/881289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Evaluating $\int x^2 \sqrt{x^2-1} dx$ How do I evaluate the following indefinite integral?
$$\int x^2 \sqrt{x^2-1} dx$$
Through integration of parts, I have obtained
$$ \frac{x}{3}(x^2-1)^{3/2} - \frac{1}{3} \int (x^2-1)^{3/2} dx $$
I've attempted evaluating the second term through substitution, where
$$ x = \sec(u)$... | $$\int x^{2}\sqrt{x^{2}-1}\,dx\\
x=\cosh{y}\Rightarrow dx=\sinh{y}dy\\
\int x^{2}\sqrt{x^{2}-1}\,dx=\int \cosh^{2}{y}\sinh^{2}{y}\,dy=\frac{1}{4}\int\sinh^{2}{2y}\,dy\\
=\frac{1}{4}\int\frac{\cosh{4y}-1}{2}\,dy=\frac{\sinh{4y}}{2}-\frac{y}{8}+C\\
=\sinh{2y}\cosh{2y}-\frac{\cosh^{-1}x}{8}+C\\
=2(\cosh{y})(\cosh{2y})(\si... | {
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"answer_count": 7,
"answer_id": 6
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Why does the imaginary number $i$ satisfy $i\times 0=0$? Why does the imaginary number $i$ satisfy $i\times 0=0$?
I mean, we don't really know what $i$ is. How could we be sure about that? I think there's a reason behind why mathematicians decided that.
| Ignore this answer if you've never heard of matrix multiplication. Better yet, learn matrix multiplication and then read this answer!
The answer to your question depends on what definition of complex numbers you're using.
For example, I like to think of the complex numbers as matrices of the form
$$
\begin{pmatrix}
a ... | {
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"source": "stackexchange",
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"answer_count": 4,
"answer_id": 0
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Polygamma function series: $\sum_{k=1}^{\infty }\left(\Psi^{(1)}(k)\right)^2$ Applying the Copson's inequality, I found:
$$S=\displaystyle\sum_{k=1}^{\infty }\left(\Psi^{(1)}(k)\right)^2\lt\dfrac{2}{3}\pi^2$$ where
$\Psi^{(1)}(k)$ is the polygamma function.
Is it known any sharper bound for the sum $S$?
Thanks.
| Continuing from Olivier Oloa's answer,
$$ \begin{align} \sum_{k=1}^{\infty} \big(\psi^{(1)} (k) \big)^{2} &= \int_{0}^{1} \int_{0}^{1} \frac{\ln u \ln v}{(1-uv)(1-u)(1-v)} \ du \ dv \\ &= \int_{0}^{1} \frac{\ln v}{(1-v)^{2}} \int_{0}^{1} \left(\frac{\ln u}{1-u} - \frac{v \ln u}{1-vu} \right) \ du \ dv \\ &= \int_{0}^{... | {
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"url": "https://math.stackexchange.com/questions/882621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 4,
"answer_id": 3
} |
Integrate a division of polynomials Hi I have the following integral:
$$\int \frac{2x}{x^2+6x+3}\, dx$$
I made some changes like:
$$\int \dfrac{2x+6-6}{x^2+6x+3}\, dx$$
then I have:
$$\int \dfrac{2x+6}{x^2+6x+3}\, dx -\int\dfrac{6}{x^2+6x+3}\, dx$$
and thus: $$\ln(x^2+6x+3)-\int\dfrac{6}{x^2+6x+3}\, dx$$
Ok, I have de... | $$\frac{2x}{x^2+6x+3}=\frac{A}{x-(-3-\sqrt{6})}+\frac{B}{x-(\sqrt{6}-3)}$$
Find $A$ and $B$ and then:
$$\int \frac{2x}{x^2+6x+3} dx=\int \frac{A}{x-(-3-\sqrt{6})} dx+\int \frac{B}{x-(\sqrt{6}-3)} dx$$
| {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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prove $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq 3(a^2+b^2+c^2)$
If $a,b,c$ are positive real numbers and $a+b+c=1$,Prove: $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq 3(a^2+b^2+c^2)$$
Additional info:We can use AM-GM and Cauchy inequalities mostly.We are not allowed to use induction.
Things I have done so... | In this case you want to have quadratic terms in the RHS, so use Cauchy in the following form:
$$\left(\sum_{cyc} \frac{a^2}b \right) \left(\sum_{cyc} a^2b \right) \ge \left(a^2+b^2+c^2\right)^2 \tag{1}$$
Then it is enough to show that
$$a^2+b^2+c^2 \ge 3\sum_{cyc} a^2b \tag{2}$$
Homogenising,
$$\iff \left( a^2+b^2+... | {
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Evaluate $\int_0^{\sqrt{5}} \sqrt{4x^4 + 16 x^2} dx$ $$\int_0^{\sqrt{5}} \sqrt{4x^4 + 16 x^2} dx$$
The square root really got me confused here. I've tried using the standard trick with $x = x + \sqrt{x^2+4}$ but failed. That might still be the best approach.
What's the best way to solve this and how to do it?
| Hint
\begin{align}
\int^\sqrt{5}_0\sqrt{4x^4+16x^2} \ dx
&=2\int^{\sqrt{5}}_0 x\sqrt{x^2+4} \ dx \\
\end{align}
Now let $x=2\tan{u}$, then $dx=2\sec^2{u} \ du$.
\begin{align}
2\int^{\sqrt{5}}_0 x\sqrt{x^2+4} \ dx
&=8\int^{\arctan\left(\frac{\sqrt{5}}{2}\right)}_0 \tan{u}\sec^3{u} \ du\\
\end{align}
Use the substitutio... | {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Evaluate $\int x\sqrt{(a^2 - x^2)}dx$ I need to find $$\int x\sqrt{(a^2 - x^2)}dx$$
I tried putting $x=a cos(t)$ but I ended up getting a very complicated expression, so any tips?
| $-\frac{1}{3}(a^2-x^2)^\frac{3}{2}$ is one of the origional function because $\frac{d}{dx}\bigg(-\frac{1}{3}(a^2-x^2)^\frac{3}{2}\bigg)=x\sqrt{a^2-x^2}.$
Why did I come up with it because $\int x\sqrt{a^2-x^2}dx=\int\sqrt{a^2-x^2}d(x^2)=-\frac{1}{3}(a^2-x^2)^\frac{3}{2}$+C.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/886337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How can I calculate $AB$ in a quadrilateral figure $(ABCD)$ $AD=3cm$, $DC=4cm$, $BC=5cm$. The diagonals of this quad figure must also be perpendicular. Find $AB$.
| Let $O$ be the intersecting point between the two diagonals $AC$ and $BD$.
Then by Pythagoras' Theorem we have the following identities:
$$\begin{cases}
AO^2 + DO^2 = AD^2 = 9 & (1)\\
DO^2 + CO^2 = CD^2 = 16 & (2)\\
CO^2 + BO^2 = BC^2 = 25 & (3)\\
BO^2 + AO^2 = AB^2 & (4)
\end{cases}$$
Note that $(1) + (3) - (2) = (4)... | {
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How to find $\int_{0}^{\pi/2} \log ({1+\cos x}) dx$ using real-variable methods? How do you find the value of this integral, using real methods?
$$I=\displaystyle\int_{0}^{\pi/2} \log ({1+\cos x}) dx$$
The answer is $2C-\dfrac{\pi}{2}\log {2}$ where $C$ is Catalan's constant.
| $$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \ln (1+\cos x) d x &=\int_{0}^{\frac{\pi}{2}} \ln \left(2 \cos ^{2} \frac{x}{2}\right) d x \\
&=\int_{0}^{\frac{\pi}{2}} \ln 2 d x+2 \int_{0}^{\frac{\pi}{2}} \ln \left(\cos \frac{x}{2}\right) d x \\
&=\frac{\pi}{2} \ln 2+4 \int_{0}^{\frac{\pi}{4}} \ln (\cos x) d x \\
&=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/887069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 4
} |
Find $ \int \sqrt{\frac{1}{\theta^2}+ \frac{1}{\theta^4}} d\theta$ Any Ideas ! is this integrable function
| Just for fun, an alternate approach: If $x = 1 + \frac{1}{\theta^2}$, then $\frac{d x}{d \theta} = - \frac{2}{\theta^3} = \frac{1}{\theta} \frac{-2}{\theta^2}$. So, the integral of interest is transformed to
$$
\int \frac{1}{\theta} \sqrt{ 1 + \frac{1}{\theta^2} } d \theta
= -2 \int (x-1) \sqrt{ x } d x
$$
Where $u = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/887758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
For a triangle $ABC$, $a^2+b^2+c^2=8R^2$ then it is a right triangle? $ABC$ is a triangle, $a^2+b^2+c^2=8R^2$ then how do we prove it is a right triangle?
| Let $S$ be the area of the triangle.
Since we have $4RS=abc$ and
$$\begin{align}16S^2&=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)\\&=-(a^4+b^4+c^4)+2(a^2b^2+b^2c^2+c^2a^2)\end{align}$$
if $a^2+b^2+c^2=8R^2$, then we have
$$\begin{align} 0 &=\frac{a^2+b^2+c^2}{R^2}-8\\&=\frac{a^2+b^2+c^2}{a^2b^2c^2}(-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2c^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/889946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Simplifying a trigonometric identity Simplify $1 + \tan^2x$
My attempt:
$$\begin{align}1 + \tan^2x&\\
&= \frac{1}{1} + \frac{\sin^2x}{\cos^2x}\\
&= \frac{1(\cos^2x)}{1(\cos^2x)} +\frac{\sin^2x}{\cos^2x}\\
&=\frac{\cos^2x}{\cos^2x}+\frac{\sin^2x}{\cos^2x}\\
&=\frac{\cos^2x + \sin^2x}{\cos^2x\cos^2x}\\
&= \frac{\sin^2x}{... | Firstly, I think that everyone (including yourself) disagrees with the assertion that in general
$$\frac{a}{a}+\frac{b}{a}=\frac{a+b}{a^2}$$
by the distributive law
$$\frac{a}{a}+\frac{b}{a}=\frac{1}{a}(a+b)=\frac{a+b}{a}$$
Secondly, I think that $\tan^2x+1=\sec^2x$ should be considered to be just as basic of an identi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/892065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
any other method for evaluating $\int\limits \frac{ x^2+x+1 }{ \sqrt{x^2-x+1} } dx$? I tried below and its getting tedious :
$\begin{align}\\ \int\limits \frac{ x^2+x+1 }{ \sqrt{x^2-x+1} } dx &= \int\limits \frac{(2x-1)+ x^2-x+2 }{ \sqrt{x^2-x+1} } dx \\~\\ &= \int\limits \frac{(2x-1)dx}{ \sqrt{x^2-x+1} } + \int\... | HINT:
I think using Trigonometric substitution of
$\displaystyle x^2-x+1=\frac{(2x-1)^2+(\sqrt3)^2}4$
with $2x-1=\sqrt3\tan\theta$ won't be a bad way to start with
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/893756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Sum of all possible remainders when $2^n$, where n is a nonnegative integer, is divided by 1000
Let $R$ be the set of all possible remainders when a number of the
form $2^n$, $n$ a nonnegative integer, is divided by $1000$. Let $S$
be the sum of all elements in $R$. Find the remainder when $S$ is
divided by $100... | Well for the second solution they used Euler's theorem
$$2^{\varphi(125)}\equiv2^{100}\equiv1\pmod{125}$$
So either $2^{50}\equiv-1\pmod{125}$ or $2^{20}\equiv 1\pmod{125}$
First part can be proved by euler's theorem since $2^{100k+n}\equiv2^{100k}\cdot2^{n}\equiv1\cdot2^{n}\pmod{125}$ it means it cycles $\pmod{125}$ n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/895438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Closed form for the integral $\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx$ Here is a challenging one maybe some would like a go at.
Show that:
$$\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx=\frac{-9\pi^{4}}{256}+\frac{\pi^{3}}{32}\ln2+\frac{\pi^{2}}{6}G-\frac{1}{1536}\left[\psi_{3}\left... | Following the same idea of @Cody we have
$$I=\int_{0}^{1}\frac{x\ln^{3}x}{(1-x)(1+x^2)}\ dx+\int_{0}^{1}\frac{\ln^{2}x\ln(1+x)}{1+x^2}\ dx=K+J$$
\begin{align}
K&=\int_{0}^{1}\frac{x\ln^{3}x}{(1-x)(x^{2}+1)}\ dx\\
&=\frac12\int_0^1\frac{x\ln^3x}{1+x^2}\ dx-\frac12\int_0^1\frac{\ln^3x}{1+x^2}\ dx+\frac12\int_0^1\frac{\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/895660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
"answer_count": 2,
"answer_id": 0
} |
Solving $x^3+y^3=x^2y^2+1$ in non-negative integers I wanted to solve $x^3+y^3=x^2y^2+1$ in non-negative integers.
First I set $a=x+y$ and $b=xy$ to get $b^2+3ab+1=a^3$. View as a quadratic in $b$, the discriminant = $4a^3+9a^2-4$, which needs to be a perfect square.
Secondly, rearranging the quadratic in $b$ we get $4... | I think the following fact has been violated:-
In the quadratic equation, f(X) = 0, its discriminant ($\triangle$) should be free from X.
It is alright to have $\triangle = 4a^3 + 9a^2 – 4$, but rewriting it as $(2b + 3a)^2$ and use it to solve f(b) = 0 is not allowable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/896209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
System of equations in Lagrange multiplier problem Continuing from Confounding Lagrange multiplier problem:
I'm having trouble solving the system of equations below arisen from a Lagrange multiplier problem where we are to optimize $f(x,y,z) = 4x^2 + 3y^2 + 5z^2$ over $g(x,y,z) = xy + 2yz + 3xz = 6$.
$$ \begin{cases} 8... | Adding the equations $8x=\lambda(y+3z)$, $6y=\lambda(x+2z)$, and $10z=\lambda(2y+3x)$ gives
$\lambda(4x+3y+5z)=8x+6y+10z$, so either $\lambda=2$ or $4x+3y+5z=0$.
$\textbf{1)}$ If $\lambda=2$, we have $8x=y+6z$ and $6y=2x+4z$, so $4x-y-3z=0$ and $x-3y+2z=0$.
Then $3(4x-y-3z)-(x-3y+2z)=0\implies 11x-11z=0\implies z=x$, a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/896584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Tangent to curve $x^3+y^3=a^3$ meets it again. Tangent to curve $x^3+y^3=a^3$ at $(x_1,y_1)$ meets it again at $(x_2,y_2)$.How to prove that $$\frac{x_2}{x_1}+\frac{y_2}{y_1}+1=0$$
Since $y'=-\frac{x_1^2}{y_1^2}$
$$\frac{y_2-y_1}{x_2-x_1}=-\frac{x_1^2}{y_1^2}=\frac{x_1^2+x_1x_2+x_2^2}{y_1^2+y_1y_2+y_2^2}$$
Solving we ... | $$\frac{y_2-y_1}{x_2-x_1} = \frac{y_2^3-y_1^3}{x_2^3-x_1} \times \frac{x_2^2+x_1x_2+x_1^2}{y_2^2+y_1y_2+y_1^2} = (-)\frac{x_2^2+x_1x_2+x_1^2}{y_2^2+y_1y_2+y_1^2}$$
And you have written $$\frac{y_2-y_1}{x_2-x_1} = \left(\frac{x_2^2+x_1x_2+x_1^2}{y_2^2+y_1y_2+y_1^2}\right) $$
Which is wrong.
So we have
$$\frac{x_1^2}{y_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/897196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Initial value $\left ( \frac{dy}{dt} \right )+3y=11$, $y(0)=1$ I have never done an initial value problem, and would like some help on how to start this please.
| $$\frac{dy}{dt} + 3y = 11\ \ ... (1)
$$
homogeneous form is
$$
\frac{dy}{dt} + 3y = 0\\
\int \frac{dy}{y} = -3\int dt\\
\log{|y|} = -3t + C\\
y = \exp(-3t + C)\\
y = C\exp(-3t)\\
$$
variation of Parameters
C => u(t)
$$
y = u(t)\exp(-3t)\ \ ...(2)\\
\frac{dy}{dt} = \frac{du}{dt}\exp(-3t) - 3u\exp(-3t)\ \ ...(3)
$$
(2), ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/898873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Where did the negative answer come from? The question is to evaluate $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots }}}}$
$$x=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots }}}}$$
$$x^2=2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots }}}}$$
$$x^2=2+x$$
$$x^2-x-2=0$$
$$(x-2)(x+1)=0$$
$$x=2,-1$$
because $x$ is positive $x=2$ is the answer. bu... | By squaring both sides, you turn an equation with one solution into an equation with two solutions; you can end up creating an extraneous solution. In this case, you come to find that $-1$ satisfies $x^2 = 2 + x$, but does not satisfy $x = \sqrt{2 + \sqrt{2 + \cdots}}$, and so is extraneous.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/898964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 5,
"answer_id": 0
} |
A $n\cdot n$ square grid problem? I thought of this problem when I was playing a game called BINGO with my friend. The game basically is like this:
Suppose $2$ people are playing the game(can be played with any no of people though). Both make a $5\cdot 5$ square grid and and fill the numbers from $1$ to $25$ in the gri... | It depends on $n$ being odd or even so we'll deal with these separately:
Odd:
The minimum is as how you've drawn it for $n=5$. We have:
Cells down the first $\frac{n-1}{2}$ vertical lines: $\frac{n-1}{2} \times n$.
Extra cells across the first $\frac{n-1}{2}$ horizontal lines: $\frac{n-1}{2} \times \frac{n+1}{2}$.
Extr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/899262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
When does $2^n+n \mid 8^n+n$? How to find all positive integers $n$ such that $2^n+n$ divides $8^n+n$ ?
| Since $$2^n\equiv -n\pmod{2^n+n}$$ we deduce $$8^n = (2^n)^3 \equiv (-n)^3 \pmod {2^n+n}$$
So $2^n+n\mid 8^n+n$ if and only if $2^n+n\mid n-n^3$.
For $n\geq 10$, $2^n>n^3$ so $2^n+n$ cannot divide $n^3-n=-(n-n^3)$.
Clearly, if $n=0,1$, $n^3-n=0$ so $2^n+n\mid n^3-n$.
So you really only need to check additionally $n=2,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/901938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Why does $(a+b)^2= a^2+b^2 + 2ab$? Why is the $2ab$ there? When I was doing research on finding the derivative I came across something strange.
If $f(x) = x^2$ you find the derivative by going
$$\frac{f(x+h)^2-f(x)^2}{h}
=\frac{x^2+2xh+h^2-x^2}{h}.$$
Why is there the $2xh$? Can someone explain the logic behind this? I... | \begin{align*}
(a+b)(a + b) &= a(a + b) + b(a + b)\quad &(\text{addition distributes over multiplication})\\
&= aa + ab + ba + bb\quad &(\text{multiplication distributes over addition})\\
&= a^2 + ab + ba + b^2\quad &(xx = x^2)\\
&= a^2 + ab + ab + b^2\quad &(\text{multiplication is commutative})\\
&= a^2 + 2ab + b^2
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/902603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 9,
"answer_id": 6
} |
square of complex numbers I have this equation from here:
but it is not equal to:
$$(a + bi)^2 = a^2 + 2abi + (bi)^2.$$
could someone explain me what is the difference between this two calcultion?
| $|a+bi|^2 = a^2+b^2$, as you say. And $(a + bi)^2 = a^2 + 2abi + (bi)^2$, as you say.
You would expect that $|a+bi|^2 = |(a+bi)^2|$, and that's what happens:
$(a+bi)^2 = (a^2-b^2) + (2ab)i$
So |$(a+bi)^2| = \sqrt{(a^2-b^2)^2 + (2ab)^2} = \sqrt{a^4+b^4 + 2a^2b^2} = \sqrt{(a^2+b^2)^2} = a^2 + b^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/903785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How solve this equation $\sin x\cdot \sin20=2\sin(110-x) (\sin10)^2$ let
$0<x<90$, and such
$$\sin x\cdot \sin20=2\sin{(110-x)}(\sin10)^2$$
find the $x$
my idea: since
$$\sin x\cdot 2\sin10\cos10=2\sin(70+x)(\sin10)^2$$
so
$$\cot10=\dfrac{\sin(70+x)}{\sin x}$$
then How find it?
| $2\sin(x)\cdot\sin10\cdot\cos10=2[\cos20\cdot\cos x+\sin20\cdot\sin x]\cdot(\sin10)^2$
$\dfrac{\sin x\cdot\cos10}{\sin10}=\cos(20-x)$
$\dfrac{\cos10}{\sin10}=\dfrac{\cos(20-x)}{\sin x}$
$\Longrightarrow x=10$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/904160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Calculate $\int_{S^2}\frac{1}{\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}dS$ where $a^2+b^2+c^2<1$.
Let $a^2+b^2+c^2 < 1$ and $S^2$ be unit sphere in $R^3$. Calculate $$\int_{S^2}\frac{1}{\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}dS$$
Let $(x,y,z)=(\cos\theta \cos\phi,\cos\theta \sin\phi, \sin\theta)$. By definition,
$$\int_{S^2}\frac{1}{\... | taking my suggested simplification a little further, since $y^2+z^2=1-x^2$ on the surface of the sphere, we obtain:
$$
I = 2\pi \int_{-1}^1 \frac{1}{\sqrt{1-2xr+r^2}}dx
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/907439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
The general solution for this differential equation? Find the general solution of this differential equation:
$$ \frac{dy}{dx} = \frac{3x^5 y^3}{4} $$
Here's what I've done so far:
$ dy=\frac{3x^5 y^3 dx}{4} $
$ 4dy = 3x^5 y^3 dx $
$ \frac{4dy}{y^3} = 3x^5 dx $
Integrating both sides:
$ \int \frac{4dy}{y^3} = \int 3x^5... | Your equation is separable and you can rewrite it as $$\frac{dy}{y^3}=\frac{3}{4}x^5 dx$$ So, integrating both sides gives $$-\frac{1}{2 y^2}=\frac{x^6}{8}+C_1$$ which, after simplifications, write $$y^2=-\frac{4}{x^6+C_2}$$ that is to say $$y=\pm\frac{2}{\sqrt{C_3 -x^6}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/908382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of $\sum_{k=0}^n{n\choose k}^2u^k$ I am trying to evaluate the finite sum
\begin{equation}
f(u)=\sum_{k=0}^n{n\choose k}^2u^k,\quad 0<u\le1
\end{equation}
In an first attempt, I think of the generating function
\begin{equation}
(1+x)^n(u+x)^n = \sum_{k\ge0}{n\choose k}x^k\sum_{k\ge0}{n\choose k}u^kx^{n-k}=\c... | The Legendre polynomial $P_n$ happens to satisfy
$$P_n(x) = \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k}^2 (x - 1)^{k}(x + 1)^{n-k}.$$
Hence
$$(1-x)^n P_n\left(\frac{x-1}{x+1}\right) = \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k}^2 (2x)^{k} 2^{n-k} =\sum_{k=0}^n \binom{n}{k}^2 x^k.$$
To prove the earlier equation, note that Rodr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/912804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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The sequence $(a_0,a_1,a_2,\cdots,)$ satisfies $ a_{n+1}=a_n+2a_{n−1}$. What is $a_5$? Assume that the sequence $(a_0,a_1,a_2,\cdots,)$ satisfies the recurrence $\displaystyle a_{n+1}=a_n+2a_{n−1}$. We know that $a_0=4$ and $a_2=13$. What is $a_5$?
I got $a_1=5, a_3=23, a_4=49, a_5=95$
| If you want to calculate the $n$-th term, here is how:
$a_{n+1} - a_n - 2a_{n-1} = 0 \to x^2 - x - 2 = 0 \to (x-2)(x+1) = 0 \to x = -1, 2$. Thus the
$a_n = A(-1)^n + B2^n$. We have that: $a_0 = 4 \to a_1 = a_2 - 2a_0 = 13 - 2(4) = 5$. So:
$5 = a_1 = A(-1)^1 + B2^1 = -A + 2B$, and
$13 = a_2 = A(-1)^2 + B2^2 = A + 4B$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/916379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
A closed-form of product the gamma functions containing $\pi$ and $\phi$ Playing with gamma functions by randomly inputting numbers to Wolfram Alpha, I got the following beautiful result
\begin{equation}
\frac{\Gamma\left(\frac{3}{10}\right)\Gamma\left(\frac{4}{10}\right)}{\Gamma\left(\frac{2}{10}\right)}=\frac{\sqrt[... | In this answer, we derive Gauss's Multiplication Formula:
$$
\prod_{k=0}^{n-1}\Gamma\left(x+\frac kn\right)
=\sqrt{n2^{n-1}\pi^{n-1}}\frac{\Gamma(nx)}{n^{nx}}\tag{1}
$$
In this answer, we derive Euler's Reflection Formula:
$$
\Gamma(x)\Gamma(1-x)=\pi\csc(\pi x)\tag{2}
$$
Multiply $(2)$ by $\frac{\Gamma\left(\frac8{10}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/916853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Inequality: $x^2+y^2+xy\ge 0$ I want to prove that $x^2+y^2+xy\ge 0$ for all $x,y\in \mathbb{R}$.
My "proof": Suppose wlog that $x\ge y$, so $x^2\cdot x\ge x^2\cdot y\ge y^2\cdot y=y^3$ (because $x^2\ge 0$ so we can multiply both sides by it without changing the inequality sign) giving $x^3\ge y^3$. Substracting we ha... | Note that we can prove it by $$x^2+y^2+xy=\left(x+\frac y2\right)^2+\frac{3}{4}y^2\ge 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/920605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 0
} |
Prove $3(\sin x-\cos x)^4 + 6(\sin x+ \cos x)^2 + 4(\sin^6 x + \cos^6 x) -13 = 0$
Q) Prove that $3(\sin \theta-\cos \theta)^4 + 6(\sin \theta+ \cos \theta)^2 + 4(\sin^6 \theta + \cos^6 \theta) -13 = 0$
Source: Trigonometric Functions, Page 5.9, Mathematics XI - R.D. Sharma
My Attempt::
For writing convenience, let $\... | Alternative Simplification of $$\displaystyle \sin^6(x)+\cos^6(x)$$
Using $$\sin^2(x)+\cos^2(x)=1\Rightarrow \underbrace{\sin^2 x}_{a}+\underbrace{\cos^2 x}_{b}+\underbrace{(-1)}_{c}=0$$
Now Using The Formula, If $${a+b+c=0\;,}$$ Then $$a^2+b^2+c^3=3abc.$$
So $$\left(\sin^2x\right)^3+\left(\cos^2x\right)^3+(-1)^3=3\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/921345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 3
} |
How to Integrate $\int\frac{(x^2)}{\sqrt{7-x^2}}dx$. I am trying to Integrate $\int\frac{(x^2)}{\sqrt{7-x^2}}dx$ and I have worked this problem a couple times and keep getting the same answer. So I will show my process and please point my errors out.
$$a=\sqrt{7}\hspace{10pt} x=\sqrt{7}\sin\theta\hspace{10pt}dx=\sqrt{... | I suppose that the error is when you worked $$\frac{2\sin\theta\cos\theta}{2}=\sin\theta\cos\theta$$ You had $\sin\theta=\frac{x}{\sqrt 7}$, so $\cos^2\theta=1-\frac{x^2}{ 7}$, $cos \theta=\sqrt{1-\frac{x^2}{7}}$ and then $$\sin\theta\cos\theta=\frac{x}{\sqrt 7}\sqrt{1-\frac{x^2}{7}}=\frac{1}{7}x\sqrt{7-{x^2}} $$
I am... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/922192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Does the sequence converge, and to what? We have a sequence $\{a_n\}$
$$a_0 = 0$$
$$a_{n+1} = \frac{a_{n}}{2} + 1$$
Does it converge? And to what?
| Claim: For $n \geq 0$, $$a_{n}=\frac{2^{n}-1}{2^{n-1}}$$ We can show this inductively.
Begin with $n=0$ we have $$a_{0}=0=\frac{2^0-1}{2^{-1}} \ \ \checkmark$$ Now let's suppose the result holds for all $k$ where $0<k\leq n$. By definition we have $a_{k+1}=\frac{a_{k}}{2}+1$, and by our induction hypothesis we have $a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/922738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Volume of Solid Revolution region bounded by $$y=x$$ and $$y=x^2$$
a) find the volume of the solid of revolution formed by revolving R region about the line $x=2$.
b) find the volume of the solid of revolution formed by revolving R region about the line $y=b$ where $b > 1$ is a constant
c) Find any values of a for whic... | we have that the region is bounded by $y=x$ and $y=x^2$, finding the intersections
$\left.\begin{matrix}y=x\\y=x^2\end{matrix}\right\} x=x^2\iff x=0\vee x=1$
so, them intercepts on $(0,0)$ and $(1,1)$, then:
a) then the volume of revolution about the line $x=2$, we have that $x=y,x=\sqrt{y}$, then.
for $0\le y\le 1\Rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/922907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limit with trigonometric function $\lim_{x \to \frac{\pi}{4}} \frac{\tan^2(x)-1}{\cos(x)-\sin(x)}$ I have this limit, I have resolved it until a part but I'm stucked now.
$$\lim_{x \to \frac{\pi}{4}} \frac{\tan^2(x)-1}{\cos(x)-\sin(x)}$$
$$ \lim_{x \to \frac{\pi}{4}} \frac{\frac{\sin^2(x)}{\cos^2(x)}-\frac{\cos^2(x)}{\... | Going from line 4 to line 5, $(\sin^2x-\cos^2x)/(\cos x-\sin x)=-\sin x-\cos x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/923193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve the following inequality I have the following inequality $\frac {2x^2}{x+2} < x-2$. I tried to solve it the with the following steps.
step 1 $\frac {2x^2}{x+2} < x-2$
step 2 $\frac {2x^2}{x+2} - (x-2) < 0$
step 3 $\frac {2x^2}{x+2} - \frac {(x-2)(x+2)}{1(x+2)} < 0$
step 4 $\frac {2x^2}{x+2} - \frac {x^2-2^2}{x+2}... | I would like to suggest another way: you can define two cases, one where $x>-2$, the other where $x<-2$, and multiply the initial inequality by $(x+2)$.
It is then straightforward to get the final result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/923478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Prove that $x+\frac{1}{2x}-\frac{1}{8x^3}<\sqrt{x^2+1}This is a math problem from the German Math Olympiad, but in this case I do not know where to start, probably because I do not have enough intuition regarding inequalities.
However, I tried to apply the standard AM-GM equation which didn't help, and in general I cou... | It might be helpful to note that
$$\begin{align}\sqrt{x^2+1}&=x\sqrt{1+\frac 1{x^2}}\\
&=x\left(1+\frac 1{x^2}\right)^{\frac 12}\\
&=x\left[1+\frac12\cdot \frac 1{x^2}+\frac {\frac 12\cdot -\frac 12}{1\cdot 2}\frac1{4x^4}+\cdots \right]\\
&=x+\frac 1{2x}-\frac 1{8x^3}+\cdots \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/926386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
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.