Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Calculate the sum of $\left\lfloor \sqrt{k} \right\rfloor$ I'm trying to calculate for $n\in \mathbb{N}$ the following sum :
$\sum_{k=1}^{n^2}\left\lfloor \sqrt{k} \right\rfloor$.
I tried putting in the first terms, which gave me
$\sum_{k=1}^{n^2}\left\lfloor \sqrt{k} \right\rfloor=(1+2+3+\cdots+n)+\left\lfloor \sqrt{2... | Put
$$k= j^2+i$$
so that$j,i$ shall span
$$
k \in \left[ {1,n^2 } \right]\quad \Rightarrow \quad \left\{ \begin{array}{l}
j \in \left[ {1,n} \right] \\
i \in \left[ {0,\left( {j + 1} \right)^2 - j^2 } \right) = \left[ {0,2j} \right] \\
\end{array} \right.
$$
and clearly it is
$$
\left\lfloor {\sqrt k } \right\rf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4507417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving $x+y+z=4$, $x^2+y^2+z^2=14$, $x^3+y^3+z^3=34$
Solve the system $$\begin{equation} \label{equation1}
\begin{split}
x+y+z=4 \\
x^2+y^2+z^2=14 \\
x^3+y^3+z^3=34
\end{split}
\end{equation}$$
My work:
I found out that $$xy+yz+xz=1$$ and $$x^2y+x^2z+y^2x+y^2z+z^2x+z^2y=22$$
After this I'm stuck. Any help is gr... | There is always the way of a direct computation. Substituting $z=4-x-y$, the other two equations are $f(x,y)=0$ and $g(x,y)=0$, and the resultant of $f$ and $g$ with respect to $y$ yields
$$
(x + 1)(x - 2)(x - 3)=0.
$$
So we have $x=-1$ or $x=2$ or $x=3$. To be more precise,
\begin{align}
f(x,y)& =x^2 + xy - 4x + y^2 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4507702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
find the Laurent series expansion of $f(z) = \frac{1}{(z + 1)(z +3)}$ for the region $0 \lt |{z + 1}| \lt 2$ I have to find the Laurent series expansion of $$f(z) = \frac{1}{(z + 1)(z +3)}$$ for the region $$0 \lt |{z + 1}| \lt 2$$
Using partial fration f(z) can written as:
$$f(z) = \frac{1}{2} \frac{1}{z + 1} - \frac{... | It should rather be
$$f(z)=\frac{1}{2}\frac{1}{z+1}-\frac{1}{4} \frac{1}{1+\frac{(z+1)}{2}}$$
$$\implies f(z)=\frac{1}{2}\frac{1}{z+1}-\frac{1}{4} \sum_{k=0}^{\infty} (-1)^k \left(\frac{z+1}{2}\right)^k.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Show that $f(x)=\cos x-\sin x$ is bounded Show that $$f(x)=\cos x-\sin x$$ is bounded.
We know that $$-1\le\cos x\le1\\{-1}\le\sin x\le1,$$ so I tried to subtract these two inequalities, but it seems that it doesn't work as we get: $$0\le\cos x-\sin x\le0$$
I also noticed: $f(x)=\cos x-\sin x\ge\cos x-(-1)=\cos x+2$ an... | You can write
$$\begin{cases}-1\le\cos x\le1\\{-1}\le\sin x\le1\end{cases}\\
\implies \begin{cases}-1\le\cos x\le1\\{1}\ge-\sin x\ge -1\end{cases}
$$
We see that it is not possible to sum these inequalities directly from side by side.
We can try the following way to subtract the two inequalities.
Let $a≥x≥b \wedge c≥y≥... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove $ \int_{0}^{\infty} {x}^{a - 1} \sin \left(\frac{\pi}{2} a - b x\right) \frac{r}{{x}^{2} + {r}^{2}} d x = \frac{\pi}{2} {r}^{a - 1} {e}^{- br} $ I am attempting to calculate the following integral from Whittaker and Watson chapter 6 using the residue theorem. Show that if $a \in \left(0 , 2\right)$, $b > 0$, and ... | Let contour $C=C_1+C_2+C_3+C_4$, where $C_1$ is the positive real axis, from $\epsilon$ to $\infty$. $C_2$ is large semi-circle on the upper half plane, with radius $R$, $C_3$ is negative real axis, from $-\infty$ to $-\epsilon$. $C_4$ is small semi-circle on the upper half plane, with radius $\epsilon$.
On $C_1$,
$$I_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4517635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Obtaining $\zeta(4)$ from specific derived expression I started from the expansion of cot(x) by the Mittag-Leffler theorem, where:
$$\cot(x) = \frac{1}{x} + \sum_{k = 1}^{\infty} \frac{2x}{x^{2} - k^2 \pi^2}.$$
After splitting up $\cot(x)$, taking the derivative of both sides, and some algebra, I eventually was left w... | Expanding the internal passages:
\begin{align}
\frac{\text{d}}{\text{d}x} \left(4x^2\left[\sum_{k = 1}^{+\infty} \frac{1}{x^2 - k^2\pi^2}\right]^2\right) & = 8x \left[\sum_{k = 1}^{+\infty} \frac{1}{x^2 - k^2\pi^2}\right]^2 + 4x^2 \frac{\text{d}}{\text{d}x} \left[\sum_{k = 1}^{+\infty} \frac{1}{x^2 - k^2\pi^2}\right]^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Alternating sum of binomial coefficients $\sum_{k=0}^{49}(-1)^k \binom{99}{2k} = -2^{49}$ How can you prove that $$\sum_{k=0}^{49} \binom{99}{2k}(-1)^k = -2^{49}?$$
A more general formula seems to be $\sum_{k=0}^{n} \binom{2n + 1}{2k}(-1)^k$
For $n = 0:$ it equals $2^0$
For $n = 1: -2^1$
For $n = 2: -2^2$
For $n = 3: 2... | Use binomial theorem as
$$F(x)=\frac{(1+x)^{2n+1}+(1-x)^{2n+1}}{2}=\sum_{k=0}^{n} {2n+1 \choose 2k} x^{2k}$$
Then take$x=i$, to get
$$\sum_{k=0}^{n} {2n+1 \choose 2k} (-1)^{2k}=F(i)=\frac{(1+i)^{2n+1}+(1-i)^{2n+1}}{2}=2^{n+1/2}\cos[(2n+1)\pi/4]n$$
where we have used $(1\pm i)=\sqrt{2}[\cos(\pi/4)\pm i\sin(\pi/4)]$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving definite integral $\int_{0}^{1} \tan^{-1}(1-1/x)dx$ $$\int_{0}^{1} \tan^{-1}\left(1-\frac1x\right)dx$$
Here's what I have done so far. (the answer is given as $-\pi/4$)
Let
$$
I = \int_{0}^{1}\tan^{-1}\left(1-\frac1x\right)dx = \int_{0}^{1}\tan^{-1}\left(\frac{x-1}x\right)dx.
$$
Since, $\int_{0}^{1} f(x)dx = \i... | Noticing that
$$
\tan \left[\tan ^{-1}\left(\frac{x-1}{x}\right)+\frac{\pi}{4}\right]=\frac{\frac{x-1}{x}+1}{1-\frac{x-1}{x}}=2 x-1
$$
For $x\in (0,1)$, $$
\tan ^{-1}\left(\frac{x-1}{x}\right)+\frac{\pi}{4}=\tan ^{-1}(2 x-1)
$$
Integrating both sides from $0$ to $1$ yields
$$
\begin{aligned}
\int_{0}^{1} \tan ^{-1}\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4519205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
Solving $y' = \frac{x^2 + 2xy - y^2 - 2}{x^2 - 2xy - y^2 + 2}$ without tricks I would like to solve the equation:
$y' = \frac{y^2 - 2xy - x^2 + 2}{y^2 + 2xy - x^2 - 2} \tag{1}$
From that, we have:
$y'(y^2 + 2xy - x^2 - 2) = y^2 - 2xy - x^2 + 2 \implies $
$\frac{(2x + 2yy')(x+y) - (x^2 + y^2 + 2)(1 + y')}{(x+y)^2} = 0... | $$y' = \frac{y^2 - 2xy - x^2 + 2}{y^2 + 2xy - x^2 - 2} $$
$$\dfrac {dy}{y^2 - 2xy - x^2 + 2} = \dfrac{dx}{y^2 + 2xy - x^2 - 2}$$
$$\dfrac {d(x+y)}{-y^2 +x^2} = \dfrac{d(x-y)}{- 2xy +2}$$
Substitute $u=x+y; v=x-y$:
$$\dfrac {du}{2uv} = \dfrac{dv}{v^2-u^2 +4}$$
$$ 2uv\dfrac{dv}{du}=v^2-u^2 +4$$
$$ u\dfrac{dv^2}{du}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4521471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show inequality for positive real numbers If $x,y$ are positive real numbers then we have that $$ \frac{1}{\sqrt{x+y}}<\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}$$ right? But how can we show that?
I have tried the following but I don't think that this is the way we should go.
\begin{align*}\frac{1}{\sqrt{x + y}} < \frac{1}{... | $f(x) = 1/\sqrt{x}$ is strictly convex on $(0, \infty)$, so that
$$
f\left( \frac{x+y}{2}\right) < \frac 12 \left( f(x) + f(y) \right) \, ,
$$
for $0 < x < y$, this gives the stronger inequality
$$
\frac{1}{\sqrt{x+y}}<\frac{1}{2\sqrt 2}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right) \, .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Double integral $\int_{0}^{1}\int_{0}^{1}\frac{x^2y^3\log{(x)}\log{(y)}}{(1-x^2)(1-y^2)(1-x^2y^2)}dxdy$ I'm trying to evaluate this double integral:
$$I=\int_{0}^{1}\int_{0}^{1}\frac{x^2y^3\log{(x)}\log{(y)}}{(1-x^2)(1-y^2)(1-x^2y^2)}dxdy$$
Here is the closed-form:
$$\frac{91π^4}{11520}+\frac{21}{32} ζ(3)-\frac{7}{8} ζ... | It is nice to have arrives at $$I=\int_{1}^{\infty}\int_{1}^{\infty}\frac{\log{(x)}\log{(y)}}{x(1-x^2)(1-y^2)(x^2y^2-1)}dx\,dy$$
$$J=\int\frac{\log{(x)}\log{(y)}}{x(1-x^2)(1-y^2)(x^2y^2-1)}dx$$
$$\frac{2 \left(y^2-1\right)^2}{\log (y)}\,J=-y^2 \text{Li}_2(-x y)-y^2 \text{Li}_2(x y)-\text{Li}_2(1-x)+\text{Li}_2(-x)+y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to solve $(\sqrt{x}+7)(\sqrt{x}-1)=\frac{105}{4}$? I'm having difficulties solving $(\sqrt{x}+7)(\sqrt{x}-1)=\frac{105}{4}$. I guess I'm missing a crucial point. The solution is $x=\frac{49}{4}$. What are the steps?
| As the user Dark Magician said, we can use the substitution $t=\sqrt{x}$. We get the following from using the quadratic formula:
\begin{align}
(t+7)(t-1)&=\frac{105}{4}\\
t^2-t+7t-7&=\frac{105}{4}\\
t^2+6t-7-\frac{105}{4}&=0\\
t^2+6t-\frac{133}{4}&=0\\
4t^2+24t-133&=0\\
t&=\frac{-24\pm\sqrt{24^2-4\times4\times(-133)}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4523437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Excircles of a right triangle
Given a $\triangle ABC$ such that the $A$ excircle has a radius $r_A$ and the $B$ excircle has a radius $r_B$. Prove that if $r_A + r_B = AB$ then $\triangle ABC$ is a right triangle.
I called $D$ the point where the $A$ excircle touches the side $BC$, $E$ the point where the $B$ excircl... | The formula for the ex-radius is
$ r_A = \dfrac{ \Delta } {s - a} $
$ r_B = \dfrac{ \Delta }{ s- b} $
where $ \Delta $ is the area of the triangle, and $s = \frac{1}{2}(a + b + c) $
Given in the problem
$ r_A + r_B = AB = c $
Therefore,
$ \Delta \left( \dfrac{1}{s-a} + \dfrac{1}{s - b} \right) = c $
so that
$ \Delta ( ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4523574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Sum of two independent Gaussian r.v.s. given that one is negative Let $X \sim \mathcal N(0, \sigma_x^2)$ and $Y \sim \mathcal N(0, \sigma_y^2)$ be two independent r.v.s.
Is there a simple solution for $p(X+Y = a ~|~ X \leq 0)$?
I'm aware that without the conditioning on $X \leq 0$ we have that $Z = X+Y \sim \mathcal N(... | In your third equality line, the integrand is zero as these are absolutely continuous random variables. Dont you get an answer of zero just from that?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4523858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why $(2m)^2 + (m^2 - 1)^2 = (m^2 + 1)^2$ results in pythagorean triples? As you increase the value of n, you will generate all pythagorean triples whose first square is even. Is there any visual proof of the following explicit formula and where does it come from or how to derive it?
$(2n)^2 + (n^2 - 1)^2 = (n^2 + 1)^2$... | Here's a visual proof:
(This space intentionally left blank.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4531998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
A is an n by n matrix such that $A^3=2I$. A is an n by n matrix such that $A^3 = 2I$.
(1) Is $A$ invertible? If it is, then what is $A^{-1}$?
(2) Is $A+I$ invertible? If it is, then what is $(I+A)^{-1}$
The original version of the question is not in English, so maybe there are some grammar mistakes, sorry in advance.
| A matrix $B$ that satisfies $AB=BA = I$ is an inverse of $A$.
Since $A ({1 \over 2} A^2) = I$ we see that $A^{-1} = {1 \over 2} A^2$.
Suppose $x^3 = 2$, then we see that
${1 \over x+1} = {1 \over x} (1-{1 \over x} + {1 \over x^2} - {1 \over x^3} + {1 \over x^4} - {1 \over x^5} +\cdots) $, and using $x^3 = 2$ we get
$=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4533719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Relationship between e and sine I am reading a physics text about scattering but I am stuck with the maths.
I think the solution might be related to Euler's formula where:
$$\frac{e^{i\theta} - e^{-i\theta} }{2i} = \sin\theta$$
The text says that:
$$I(Q) = \bigg|\sum_{n=0}^{N-1}b e^{ikna\sin\theta}\bigg|^2$$
and
$$\sum... | So first, we have the identities
$$|x+iy|^2 = x^2 + y^2 \text{ for } x,y\in\mathbb{R}$$
$$\left|\frac{x}{y}\right| = \frac{|x|}{|y|}$$
$$1-\cos(x) = 2\sin^2(x/2)$$
$$\cos^2(x) + \sin^2(x) = 1$$
$$\sum_{j=0}^{N-1} r^j = \frac{1-r^N}{1-r}$$
Also for any $x$,
$$
\begin{align*}
\left|1-e^{ix}\right| &= |1-\cos(x)-i\sin(x)|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4536833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to find the value of $\sqrt{\Delta}$ for the given question? If it is given that $\alpha , \beta, \gamma$ are the roots of $x^3+x^2-2x+1=0$, then find the value $\sqrt{\Delta}$, where $$
\Delta=\left|\begin{array}{ccc}
\gamma^2 & \beta^2 & 2 \beta \gamma-\alpha^2 \\
2 \alpha \gamma-\beta^2 & \alpha^2 & \gamma^2 \\
... | By confort put $a,b,c$ for the three roots. You do have
$$a+b+c=-1\\ab+ac+bc=-2\\abc=-1\tag1$$ Besides $$(a+b+c)^3=a^3+b^3+c^3+3a^2(b+c)+3b^2(a+c)+3c^2(a+b)+6abc\\(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)\tag2$$ From $(2)$ and $(1)$
$$-1=a^3+b^3+c^3+3a^2(-1-a)+3b^2(-1-b)+3c^2(-1-c)-6$$ then $$2(a^3+b^3+c^3)=-3(a^2+b^2+c^2)-5=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4540737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int \sqrt{x^2 + a^2}\mathrm{d}x$ I tried using integration by parts and could get to this point:
\begin{align}
\int \sqrt{x^2 + a^2}\mathrm{d}x &= x\sqrt{x^2+a^2} - \int x\frac{2x}{\sqrt{x^2 + a^2}}\mathrm{d}x
\\\\ &= x\sqrt{x^2+a^2} -2\int \sqrt{x^2 + a^2}\mathrm{d}x + \int \frac{2a^2}{\sqrt{x^2 + a^2}}\ma... | W.l.o.g. we can assume that $a>0$ (because, otherwise one can replace $a$ with its absolute value)
Setting
$$
x=a\tan(u),
$$
we have
$$
dx=a\sec^2(u)du,
$$
so we get
\begin{eqnarray}
\int\sqrt{x^2+a^2}dx
&=&\int a\sec^2(u)\sqrt{a^2\tan^2(u)+a^2}du\\
&=&\int a\sec^2(u)\sqrt{a^2(1+\tan^2(u))}du\\
&=&\int a\sec^2(u)\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4545307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to evaluate this improper integral $\int_{-\infty}^\infty x^2 e^{-x^2}\cos x \, dx$? How to evaluate
$$
\int_{-\infty}^{\infty} x^2 e^{-x^2} \cos x \, dx?
$$
My Attempt:
Since the integrand is an even function, we have
$$
\int_{-\infty}^{\infty} x^2 e^{-x^2} \cos x \, dx = 2 \int_0^{\infty} x^2 e^{-x^2} \cos x \, ... | How to solve!
The actual result is different:
$$ \int_{-\infty}^{\infty} x^{2} \cdot e^{-x^{2}} \cdot \cos(x) ~ \mathrm{d}x = \frac{\sqrt{\pi}}{4 \cdot \sqrt[4]{e}} \approx 0.345097 $$
The easiest way to solve the problem would be to form one od the antiderivatives:
(You can easily do this by relating the trigonometric... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4546484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 2
} |
How to find the number of integral solutions of a given equation analytically? I was given $$\tan^{-1} x+\cot^{-1} y= \tan^{-1} 3$$
and asked to find the number of integral solutions.
I was able to find two, which are $(1,2) ; (2,7)$ which was obtained by solving the given equation by taking the tangent of both sides ... | According to your description of the problem, you got at least this:
$$
\frac{x+y^{-1}}{1-xy^{-1}} = 3
$$
Multiply numerator and denominator of the left side with $y$:
$$
\frac{xy+1}{y-x} = 3
$$
which means
$$
xy+1 = 3y-3x
$$
or
$$
(x-3)(y+3)=-10
$$
Now you only have to check in how may ways $-10$ can be expressed as p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4546803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
volume of the region in the first octant bounded by the coordinate planes and the planes The problem requires me to find the volume of the region in the first octant bounded by the coordinate planes and the planes $x+z=1$, $y+2z=2$, and here is my setup:
$$
\begin{aligned}
&\phantom{\implies}x+z=1\\
&\implies z=1-x\\
\... | The region is given by the inequalities $x,y,z\geq 0$, $x+z\leq 1,y+2z\leq 2$. It is pyramid with a rectangular base of sides $1$ and $2$, which lays in the $xy$-plane, and of height $1$ because the apex is at $(0,0,1)$. By elementary geometry the volume of this pyramid is $V=\frac{B\cdot h}{3}=\frac{(1\cdot 2)\cdot 1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $y = \frac{{a{x^2} - 3x + 5}}{{5{x^2} - 3x + a}}$ If $y = \frac{{a{x^2} - 3x + 5}}{{5{x^2} - 3x + a}}$, then number of possible integral values of 'a' for which y may be capable of all values $\forall x \in R$ is
(1) 4
(2) 5
(3) 6
(4) none of these
My approach is as follow $5{x^2} - 3x + a > 0 \Rightarrow 9 - 2... | Cross-multiply and simplify to get: $$(5y-a)x^2+(3-3y)x+(ay-5)=0\tag{1}$$
The statement “y may be capable of all values…” suggests that $(1)$ has at least one real solution of $x$ for all $y$. Thus, the discriminant of $(1)$ must always be non-negative $\forall y, a\in\mathbb R$: $$(3-3y)^2-4(ay-5)(5y-a)\geq 0$$ which ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
prove that $|x^y - y^x| > 2$
Prove that for any integers $x\neq y ,x,y>2, |x^y-y^x|>2$.
I know that the function $f(x) = \dfrac{\ln x}{x}$ is increasing for $x<e$ and decreasing for $x > e$. So $|x^y-y^x| = x^y - y^x \ge 1$ if $x < y$. I was thinking of proving something involving ratios (e.g. $f(x)/f(y)$ for $x < y$... | Letting $f(x)=x^{1/x}$ then for integers $2<x<y,$ $$\begin{align}y^x-x^y&=f(x)^{xy}-f(y)^{xy}\\&=(f(y)-f(x))\sum_{k=0}^{xy-1} f(x)^k f(y)^{xy-1-k}\\&>xyf(y)^{xy-1}(f(y)-f(x)).\end{align}$$
the last step because $f$ is decreasing on $[3,\infty)$ and thus $f(x)>f(y).$
Now, $f(x)-f(y)=f'(c)(x-y)$ for some $c\in(x,y).$ And... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4550484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Problem regarding solving logarithm without a calculator I have a question regarding the answer provided from this question, which is solving $\log$ without a calculator.
This was one of the solutions and I am having a hard time understanding it.
$$\log \left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}... | There is formula the solver knew very well and was comfortable with.
$\ln(1+x) = x -\frac {x^2}2 + \frac {x^3}3- \frac {x^4}4 +......$
From this they conclude two things
$\ln \frac{1+x}{1-x}=\ln(1+x) - \ln (1-x) = (x -\frac {x^2}2 + \frac {x^3}3- \frac {x^4}4 +......) - (-x -\frac {x^2}2 - \frac {x^3}3- \frac {x^4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4554682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Need Help Proving An Infinite Series Relating The Gamma Function And Riemann Zeta In a paper entitled 'On Question 330 Of Professor Sanjana' (by Ramanujan), Ramanujan goes on to define a function $f(p)$ as follows
$$f(p) = \frac{1}{1+p} + \frac{1}{2} \frac{1}{3+p} + \frac{1 \cdot 3}{2 \cdot 4} \frac{1}{5+p} + ...$$
$$=... | If another approach is allowed, I would propose the following. First, we note that
$$S(p)=\frac{1}{1+p} + \frac{1}{2} \frac{1}{3+p} + \frac{1 \cdot 3}{2 \cdot 4} \frac{1}{5+p} + ...=\sum_{k=0}^\infty\frac{1}{2k+1+p}\frac{(2k-1)!!}{(2k)!!}$$
$$=\sum_{k=0}^\infty\frac{1}{2k+1+p}\frac{(2k-1)!!(2k)!!}{\big((2k)!!\big)^2}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4554998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $0$ is a regular value of $F$
How can i show that $0$ is a regular value of $F:\mathbb{R}^{3} \rightarrow \mathbb{R}^2$ given by
$$F(x,y,z)=(x^3+y^3+z^3-1,z-xy)$$
I compute the diferential of $F$ obtaining:
$$dF=\pmatrix{ 3x^2 & -y\cr 3y^2 & -x\cr 3z^2 &1}$$
and i know that if $det[(dF)(dF)^t] \neq 0$ then $... | To solve this problem, you have to play with the equations $x^3+y^3+z^3 = 1$ and $z=xy$.
Let $Z = F^{-1}(0)$.
Let $(x,y,z)\in Z$, and consider the matrix of $dF_{(x,y,z)}$ in the canonical basis, which is
$$
\mathrm{Mat}_{\mathrm{can}}(dF_{(x,y,z)}) = \begin{pmatrix}3x^2 & 3y^2 & 3z^2 \\ -y & -x & 1 \end{pmatrix} = \be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the series $\sum\limits_{x=0}^{\infty} \binom{4+x}{x} \frac{1}{2^x}$ I am trying to find the value of
$$\sum\limits_{x=0}^{\infty} \binom{4+x}{x} \frac{1}{2^x}$$
and using the formula for $\binom{n}{r}$, I have simplified it to
$$\sum\limits_{x=0}^{\infty} \frac{(x+4)(x+3)(x+2)(x+1)}{24\cdot 2^x}$$
but I have ... | The problem is: Evaluate $\sum\limits_{x=0}^{\infty} \binom{4+x}{x} \frac{1}{2^x}$
Let $$S(n) = \sum\limits_{x=0}^{\infty} \binom{n+x}{x} \frac{1}{2^x}$$
So we have $$S(4) = \sum\limits_{x=0}^{\infty} \binom{4+x}{x} \frac{1}{2^x}$$
$$S(3) = \sum\limits_{x=0}^{\infty} \binom{3+x}{x} \frac{1}{2^x}$$
$$S(4) - S(3) = \sum\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4558641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $(\sin x + \cos x)(6 - \sin x)<9$ Is there any elementary way to prove that $(\sin x + \cos x)(6 - \sin x)<9$?
I've noticed that $(\sin x + \cos x)$ has to be positive so $x \in\left(-\dfrac{\pi}{4}, \dfrac{3\pi}{4}\right)$ and then $(6 - \sin x)\in\left(6-\dfrac{\sqrt2}{2},6+ \dfrac{\sqrt2}{2}\right)$ but s... | $\begin{equation} \begin{split}
(\sin x + \cos x)(6 - \sin x) & = 6 \cos x + 6 \sin x - \sin^2 x - \sin x \cos x \\ &= 6 \cos x + 6 \sin x - \dfrac{1}{2} (1 - \cos(2x) ) - \dfrac{1}{2} \sin(2x)\\
&= 6 \cos x + 6 \sin x + \dfrac{1}{2} (\cos(2x) - \sin(2x) ) - \dfrac{1}{2} \\
&\le 6 \sqrt{2} + \dfrac{1}{2} \sqrt{2} - \df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4560624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Do we have a closed form for $\int_0^{\infty} \frac{\ln t}{\left(1+t^2\right)^n} d t $? Latest Edit
We are glad to see there are 4 alternative solutions which give the same closed form to the integral:
$$\boxed{\int_0^{\infty} \frac{\ln t}{\left(a^2+t^2\right)^n} d t = \frac{a^{1-2n}\sqrt{\pi}\Gamma\left(n-\frac{1}{2}\... | Assuming that $n$ is a positive integer, using hypergeometric functions$$I_n=\int \frac{\log(t)}{\left(1+t^2\right)^n}\, dt=t \log (t) \, _2F_1\left(\frac{1}{2},n;\frac{3}{2};-t^2\right)-t \,
_3F_2\left(\frac{1}{2},\frac{1}{2},n;\frac{3}{2},\frac{3}{2};-t^2
\right)$$
$$J_n=\int_0^\infty \frac{\log(t)}{\left(1+t^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4562577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Confusion: When can I preform operation of infinity in limit (without using the explanation of Epsilon Delta Definition) Question 1:
$$\lim_{x\to\infty}\frac{1}{\sqrt{x+1}+\sqrt{x}}$$
Knowing this:
$$\lim_{x\to\infty}\frac{1}{x}$$
The denominator of question 1 is $\infty$, therefore, $\lim_{x\to\infty}\frac{1}{\sqrt{x+... | I usually substitute $-\frac{1}{M}$ for $x$, in case $x\rightarrow 0^{-}$ and $\frac{1}{M}$ in case $x\rightarrow 0^{+}$ and then let $M\rightarrow\infty$. For example,
$$
\lim_{x\rightarrow 0^{-}}\frac{1+2^{\frac{1}{x}}}{3+2^{\frac{1}{x}}}=\lim_{M\rightarrow\infty}\frac{1+2^{-M}}{3+2^{-M}}=\lim_{M\rightarrow\infty}\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4568102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
If in a $\Delta ABC$, $\sqrt 3 \sin C = 2\sec A - \tan A$ and $\angle C = {\lambda ^0}$, find the value of $\lambda$ If in a $\Delta ABC$, $\sqrt 3 \sin C = 2\sec A - \tan A$ and $\angle C = {\lambda ^0}$, find the value of $\lambda$
My approach is as follow
$\sqrt 3 \sin C = 2\sec A - \tan A \Rightarrow \sqrt 3 \sin C... | Another approach:
Assume $0\le x\lt1$, then:
$$4x^2+1\ge4x\implies 4-4x+x^2\ge 3-3x^2\implies \frac {(2-x)^2}{1-x^2}\ge3\implies \frac {2-x}{\sqrt {1-x^2}}\ge \sqrt3;$$
notice the equality holds when $x=\frac{1}{2}.$ Now, we have:
$$\sqrt3 \sin C=\frac{2}{cos A}-\frac{\sin A}{\cos A}=\frac {2-\sin A}{\sqrt {1-\sin^2 A}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4568759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find the length PD in the figure below. For reference:
Find the length PD in the figure below.
ABCDEF is a regular hexagon
AB = 10cm
AQE is a sector( center at F)
PD is tangent of sector
My progresss:
$\triangle DEH:10.sin60^o = \frac{DF}{2} \therefore
DF = 10\sqrt3$
$\triangle DFQ :DF^2 = DQ^2+10^2 \implies DQ = \sq... | We can, without loss of generality, set up a coordinate system so that the center of the hexagon is at (0, 0) and the top and bottom are parallel to the x-axis. Then A= (-10 sin(30), -10 cos(30)), B= (10 sin(30), -10 cos(30)), C= (10, 0), D= (10 sin(30), 10 cos(30)), E= (-10 sin(30), 10 cos(30)), and F= (-10, 0).
The ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4569787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$z^4=\bar z$ complex number equation Find all $z\in\mathbb{C}$ that satisfy $z^4=\bar z$ and display them on the complex plane.
I started with the exponential forms of the two and got to $r^3e^{i4\varphi}=e^{-i\varphi}$. Is it wrong to assume, that $r=1$ and $4\varphi =-\varphi \Rightarrow \varphi=0$?
| An alternative route
Let $z=x+yi$. Then, $$(x+yi)^4=x-yi.$$ Expansion on the left side yields $$x^4-6x^2y^2+y^4+(4x^3y-4xy^3)i=x-yi$$. This gives us two equations:
$$x^4-x-6x^2y^2+y^4=0$$ and $$y+4x^3y-4xy^3=0$$
The bottom equation simplifies when you divide by $y$ (assuming $y\ne 0$):
$$1+4x^3-4xy^2=0.$$
This is quadr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4571874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Prove that: $\sqrt{\frac{4x^2+y^2}{3x^2+yz}}+\sqrt{\frac{4y^2+z^2}{3y^2+xz}}+\sqrt{\frac{4z^2+x^2}{3z^2+xy}}\ge\frac{3\sqrt{5}}{2}$
Let $x$, $y$ and $z$ be positive numbers. Prove that:
$$\sqrt{\frac{4x^2+y^2}{3x^2+yz}}+\sqrt{\frac{4y^2+z^2}{3y^2+xz}}+\sqrt{\frac{4z^2+x^2}{3z^2+xy}}\ge\frac{3\sqrt{5}}{2}.$$
This prob... | (WARNING: this proof is very bad)
First, define two sequences $a_i$ and $b_i$ where $a_1=4x^2+y^2,a_2=4y^2+z^2,a_3=4z^2+x^2$ and $b_1=\frac{1}{3x^2+yz},b_2=\frac{1}{3y^2+xz},b_3=\frac{1}{3z^2+xy}$.
Since we have two three term sequences, we may now use Holder's inequality:
$$\sqrt{4x^2+y^2+4y^2+z^2+4z^2+x^2}=\sqrt{5x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4575195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Show that $\prod_{k=1}^\infty \frac{2k+1}{2\pi}\sin{\left(\frac{2\pi}{2k+1}\right)}\sec{\left(\frac{\pi}{k+2}\right)}=\frac{\pi}{2}$ Show that
$$\prod\limits_{k=1}^\infty \frac{2k+1}{2\pi}\sin{\left(\frac{2\pi}{2k+1}\right)}\sec{\left(\frac{\pi}{k+2}\right)}=\frac{\pi}{2}$$
Context:
Inspired by this question, I conside... | Let
$$K = \prod_{k=3}^\infty \frac1{\cos\left(\frac\pi k\right)}$$
As Bouwkamp points out under First method in this paper,
$$\cos(x) = \frac{\sin(2x)}{2x} \cdot \frac x{\sin(x)} \tag{1}$$
Then
$$\begin{align*}
\frac1K &= \prod_{k=3}^\infty \cos\left(\frac\pi k\right) \\[1ex]
&= \prod_{k=3}^\infty \frac{\sin\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4575371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 0
} |
Find all solutions for $\begin{cases} x^{\frac{1}{4}}+y^{\frac{1}{5}}=5\\x^{\frac{1}{2}}+y^{\frac{2}{5}}=13\end{cases}$ I've been struggling with a quite difficult problem the past few days. This is my class's 'Exponentials and Logarithms' unit and my peers and I are stumped. Here it is:
Find all integer solutions to ... | Given
$$\begin{cases} x^{\frac{1}{4}}+y^{\frac{1}{5}}=5\\x^{\frac{1}{2}}+y^{\frac{2}{5}}=13\end{cases}$$
then let $t = x^{1/4}$ and $u = y^{1/5}$ to obtain
$$ \begin{cases} t + u = 5 \\ t^2 + u^2 = 13 \end{cases}. $$
Square the first equation, and use the second, to obtain
$$25 = t^2 + u^2 + 2 u t = 13 + 2 u t $$
gives... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4575954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find a plane M such that $l_1\in M (l_1$ is contained in $M$), and $M$ is parallel $l_2$. $l_{1} :
\left\lbrace
\begin{array}{lcl}
2 x + 3 y + 4 z = 7 && \mbox{ } \\
4 x + y - 2 z = -1 && \mbox{ } \\
\end{array}
\right.$
$l_{2} : \frac{x-2}{20}=\frac{y-2}{10}=\frac{z +1 }{5}$
I have to find a plane M such that $... | You get the direction vector of $l_1: v_1 = (-1,2,-1)$
and the direction vector of $l_2: v_2=(4,2,1)$
You can just cross them to get the normal vector of M. Since both $l_1,l_2$ are $\perp$ to normal vector of M. So
$N_m=v_1\times v_2 = (4,-3,-10)$
Then M is $4x-3y-10z=d$, find d and you are done. (Use any point on $l_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4576336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Arrange $m$ H's and $n$ T's to have exactly $k$ H-runs In an arrangement of $m$ H's and $n$ T's, an uninterrupted sequence of one kind of symbol is called a run. (For example: HHHTHHTTTH is a sequence of $6$ H's and $4$ T's which opens with an H-run of length $3$, followed by a T-run of length $1$,an H-run of length $2... | As you observed, the number of $T$-runs is either $k - 1$, $k$, or $k + 1$. Let's consider cases.
Case 1: $k$ $H$-runs and $k - 1$ $T$-runs.
For this to occur, there must be a run of at least one $T$ in each of the $k - 1$ spaces between runs of at least one $H$. If $x_i$ is the number of $H$s in the $i$th run, then
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4577766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that the minimum value of all products of two values is less than $-\frac{1}{n}$.
Let $x_1, x_2, \dots, x_n$ be $n$ real numbers, where $n \geq 3$ is an odd natural number. We know that:
$$x_1^2 + x_2^2 + \dots + x_n^2 = 1$$
and:
$$x_1^3 + x_2^3 + \dots + x_n^3 = 0$$
Prove that the minimum value of a product of ... | Denote $a = \underset{1\le i\le n}{\min}{x_i}$ and $b = \underset{1\le i\le n}{\max}{x_i}$, then
$$\begin{align}
\sum_{i=1}^n (x_i^3-a^3)(x_i^3-b^3)\le0 &\Longleftrightarrow \sum_{i=1}^n x_i^6-\underbrace{\left( \sum_{i=1}^n x_i^3 \right)}_{=0}(a^3+b^3)+na^3b^3\le 0 \\
&\Longleftrightarrow a^3b^3\le -\frac{1}{n}\sum_{i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4580232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Any way to solve $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} = \sqrt{x+7}$? I was solving a radical equation $x+ \sqrt{x(x+1)} + \sqrt{(x+1)(x+2)} + \sqrt{x(x+2)} = 2$. I deduced it to $\sqrt{x } + \sqrt{x+1} + \sqrt{x+2} = \sqrt{x+7}.$
Answer is $\frac1{24}$.
The first equation has two solutions however the latter one has on... | Given the equation
$$ \sqrt{x} + \sqrt{x+1} + \sqrt{x+2} = \sqrt{x+7} $$
consider it in the form
$$ \sqrt{x} + \sqrt{x+1} = \sqrt{x+7} - \sqrt{x+2} $$
then by squaring both sides the result is
$$ \sqrt{x (x+1)} = 4 - \sqrt{(x+2)(x+7)}. $$
Now, with this equation written as $4 = \sqrt{x (x+1)} + \sqrt{(x+2)(x+7)}$ then,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4581999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Show that $2 \operatorname{Re}(z-1/z) > 1/|z|^2 - 1.$
Let $n$ be a positive integer and let $z$ be a root of $x^n-x-1.$ Show that $2 \operatorname{Re}(z-1/z) > 1/|z|^2 - 1$.
Solution. Write $z = re^{i\theta}, r > 0$ as $0$ is obviously not a root. Then $$1+2r\cos \theta = |z+1|^2-r^2 = r^{2n} - r^2.$$ We need to show... | We have to prove that $$r^{2n}+1>r^{2n-2}-r^2\tag{1}$$ which is easy to see since $r^{2n}>r^{2n-2}$ (when $r>1$, that is) and so the inequality is proven for $r>1$. When $r<1$ we have $r^{2n-2}<r^2$ so the right side of $(1)$ is negative and the left side is positive. Therefore your inequality is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4583050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
how to solve $\int\sqrt{1-16cos^2(x)}\ dx$ How can we solve this integral?
$\int\sqrt{1-16cos^2(x)}\ dx$
I tried to make some substitutions, but I failed to solve it.
For example, I took the substitution
$u=4cos(x)$
$dx=-\frac{du}{\sqrt{16-u^2}}$
hence, the integration will be
$\int-\frac{\sqrt{1-u^2}}{\sqrt{16-u^2}}du... | You can easily represent the antiderivative using the indefinite elliptic integrals of the second kind (Jacobi-Form) and complex numbers:
$$
\begin{align*}
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \int_{}^{} \sqrt{1 - 16 \cdot \left(1 - \sin\left( x \right)^{2} \right)} \, \operat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4585138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Right triangle with its perimeter and median and altitude A right triangle $ABC$ is given with $\measuredangle ACB=90^\circ$. If the perimeter of the triangle is $72$ and the difference between the lengths of the median and the altitude to the hypotenuse is $7$, find the area.
Let $CD=x$ $(x>0)$, then $CM=x+7$. By the... | Let $a = BC$, $b = AC$, $c = AB$ and $S$ the area of the triangle $ABC$ then $S = \dfrac{a b}{2}$.
*
* We have $a + b + c = 72$ then $a + b = 72 - c$ so :
$$(72 - c)^2 = (a + b)^2 = a^2 + b^2 + 2 a b = c^2 + 4 S$$
We deduce that :
$$4 S = (72 - c)^2 - c^2 = 5184 - 144 c \quad (1)$$
* We have :
$$S = \dfrac{1}{2} CD ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4586439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Prove that: $\sqrt [3]{36}<\ln 28<\sqrt [3]{37}$
Prove that:
$$\sqrt [3]{36}<\ln 28<\sqrt [3]{37}$$
This inequality is the result of an integral representation/inequality.
I lost access to the article that mentioned this inequality. Now I want to prove it myself. Using series is the standard way. However, this inv... | For the left inequality:
We have
$$\ln 28 = 5 \ln 2 - \ln(8/7)
\ge 5\ln 2 - 1/7$$
where we have used $\ln(1 + u) \le u$ for all $u \ge 0$.
From
$$0 \le \int_0^1 \frac{x^2(1 - x)^3}{1 + x}\,\mathrm{d} x = 8\ln 2 - \frac{83}{15},$$
we have
$\ln 2 \ge \frac{83}{120}$.
(Note: Alternatively,
we may use $\ln \frac{1+x}{1-x}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4589035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
$X$ has pdf $f(x)=c(x+1)$ Find pdf of $Y=X^2$.
Let $X$ be continuous random variable with pdf $f(x)=c(x+1)$ for
$-2<x<1$. Determine the pdf of $Y=X^2$.
I actually asked this question earlier but the second part of the question didn't make sense. I just wanted to verify if my solution for this question is correct.
$\... | As @Alex said, $c(x+1)$ becomes both negative, positive value on $-2 < x < 1$, which makes $f$ not a pdf.
To make $f$ a pdf, you should separate $c$ within the range with proper $c$'s.
i.e. $c_1(x+1)$ in ($-2 < x < -1$), $c_2(x+1)$ in ($-1 < x < 1$)
Let's ignore the negative value problem and consider the integration ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4592586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ I would like to solve the equation $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ analytically, without using Wolfram Alpha. I have tried several substitutions, including $x=\cos(t)$, $\sqrt{1-x^2}=t$, but they all fail and eventually result in a quartic polynomial which i... | Let us make the substitution $y := \sqrt{1-x^2}$; then we get the system of polynomial equations
\begin{align*}
x^2 + y^2 & = 1, \\
x(1+y) & = y.
\end{align*}
From here, if we solve the second equation for $y$, $y = \frac{x}{1-x}$, and substitute into the first equation, it is straightforward to get the polynomial equa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4593991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Does the sequence of positive roots of $x^n+x^{n-1}+\cdots+x-1=0$ converge? For a positive integer $n$, let $a_n$ denote the unique positive real root of the equation
$$x^n+x^{n-1}+\cdots+x-1=0.$$
Then what can we say about the sequence $\{a_n\}$?
We can observe that all $a_n<1$ so if $\lim a_n$ exists, let's say at $L... | Clearly this unique positive $x$ lies in the interval $\big(\frac12,1\big)$. Also,
$$
x^n+\cdots+x-1=0 \quad\Longrightarrow\quad (x-1)(x^n+\cdots+x)-x+1=0
\quad\Longrightarrow\quad x^{n+1}-2x+1=0.
$$
Since $x$ depends on $n$, we denote it as $x_n$.
If we set $x_n=\frac12+u_n$, then $u_n\in \big(0,\frac12\big)$, and $u_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Find the real roots of the equation: $\sqrt [4]{x}-\sqrt [4]{2-x}=1$
Find the real roots of the following equation: $$\sqrt [4]{x}-\sqrt [4]{2-x}=1$$
This is my textbook contest exercise.
$$\sqrt [4]{x}=\sqrt [4]{2-x}+1$$
$$x=\big(\sqrt [4]{2-x}+1\big)^4$$
I know that
$$(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4$$
So, we wi... | Let $x=u^4,\;2-x=v^4$ with the restriction $u,v≥0$, then we have:
$$\begin{align}&\begin{cases}u-v=1\\u^4+v^4=2\end{cases}
\\\implies &v^4+(v+1)^4=2\end{align}$$
Then, using the key substitution $v+\frac 12=y$ leads to:
$$\begin{align}&\left(y-\frac 12\right)^4+\left(y+\frac 12\right)^4=2\\
\implies &16y^4+24y^2-15=0\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Indefinite integral $\int\ln(1−x)dx$ by parts I have an exercise that explicitly asks for the indefinite integral of the following equation using integral by parts.
$$
\int\ln(1-x)\,dx
$$
I tried to define: $ u = \ln(1-x) $ so,
$$
\frac{\,du}{\,dx} = \frac{-1}{1-x}\\
\,du = \frac{-\,dx}{1-x}
$$
and, $ \,dv = \,dx $ res... | The mistake was to consider that $\int{\frac{\,dx}{1-x}}$ is simply $\ln(1-x)$ when the correct path of continuing is using the substitution method resulting in $-\ln(1 - x) + c$:
$ t = x - 1 $
$ \frac{\,dt}{\,dx} = -1$
$ \,dt = -\,dx$
$ \,dx = -\,dt$
$$
\int{\frac{-\,dt}{t}} = -1\int{\frac{\,dt}{t}} = -1\ln(1-x) + c
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4595973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to show that $\int_0^\pi \arcsin{\left(\frac{\sin{x}}{\sqrt{5/4-\cos{x}}}\right)}dx=\frac{\pi^2}{4}$? I am trying to show that
$$\int_0^\pi \arcsin{\left(\frac{\sin{x}}{\sqrt{5/4-\cos{x}}}\right)}dx=\frac{\pi^2}{4}$$
Context: I was working on another question ("Attempt $2$") and miscopied an integral, so I was tryi... | One can show that $\sin(x)/\sqrt{5/4-\cos(x)}$ increases from $0$ to $1$ for $x$ going from $0$ to $\pi/3$ and decreases from $1$ to $0$ for $x$ going from $\pi/3$ to $\pi$. Then the integral can be separated into
\begin{align}
I&=\int_0^\pi \arcsin{\left(\frac{\sin{x}}{\sqrt{5/4-\cos{x}}}\right)}\,dx\\
&=\int_0^{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4597839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
Converting a differential equation to a hypergeometric equation I would like to transform the following differential equation into a hypergeometric equation:
${z^2}\frac{{{d^2}W}}{{d{z^2}}} + z\frac{{{d^{}}W}}{{d{z^{}}}} + \left( {{A^2} - {B^2}\left( {1 - {C_{}}z + D\frac{{z\,}}{{{{\left( {1 - z} \right)}^{}}}}} \righ... | Using
\begin{align}
w &= x^p \, (1-x)^q \, y \\
w' &= x^p \, (1-x)^q \, \left(y' + \left(\frac{p}{x} - \frac{q}{1-x}\right) \, y \right) \\
w'' &= x^p \, (1-x)^q \, \left(y'' + \left(\frac{2 p}{x} - \frac{2 q}{1-x}\right) \, y' + \left(\frac{p(p-1)}{x^2} - \frac{2 p q}{x(1-x)} + \frac{q(q-1)}{(1-x)^2} \right) \right)
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4598360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating the limit $\lim_{x\to\infty}\left(\frac {x^5+\pi x^4+e}{x^5+ex^4+\pi}\right)^x$ How can I evaluate the following limit?
$$\lim_{x\to\infty}\left(\frac {x^5+\pi x^4+e}{x^5+ex^4+\pi}\right)^x$$
To solve this limit, I divided the numerator and denominator by $x^5$ and I got
$$\lim_{x\to\infty}\left(\frac {1+\fr... | $$\lim_{x\to\infty}\left(\frac {1+\frac {1}{x}\pi+\frac {1}{x^5}e}{1+\frac {1}{x}e+\frac {1}{x^5}\pi}\right)^x$$
Arrived at this point, you can use geometric series:
$$\left(\frac {1+\frac {1}{x}\pi+\frac {1}{x^5}e}{1+\frac {1}{x}e+\frac {1}{x^5}\pi}\right)^x\sim\left(\left(1+\frac {1}{x}\pi+\frac {1}{x^5}e\right)\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4599278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proving $3^n + 4^n + 5^n$ is divisible by $12$ for all positive odd integers $n$ by induction I am trying to prove by induction that $\sum_{i=3}^{5} i^n$ is divisible by $12$ for all positive odd integers $n$. I heard that the solution might involve normal induction, where you assume that it works for the lowest positi... | We set $n = 2k - 1$ and carry an induction over $k$. For the base case, $k = 1$, $n = 1$ and $3 + 4 + 5 = 12$, which is clearly divisible by 12. Assume true for $k$, and consider the case for $k + 1$. We have.
\begin{align*}
3^{2k + 1} + 4^{2k + 1} + 5^{2k + 1} & = 9\cdot 3^{2k-1} + 16\cdot 4^{2k-1} + 25\cdot 5^{2k-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4599709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Showing $\sum_{cyc}\frac{a^2+bc}{b+c}\geq a+b+c$ for positive $a$, $b$, $c$ The following is an inequality which I have trouble solving:
$$\frac{a^2+bc}{b+c}+\frac{b^2+ac}{c+a}+\frac{c^2+ab}{a+b}\geq a+b+c$$
($a, b, c>0$)
I tried multiplying LHS and RHS by 2 and then use
$${2(a+b+c)}={(a+b)+(b+c)+(c+a)}$$
and move to L... | Another way.
$$\sum_{cyc}\frac{a^2+bc}{b+c}-a-b-c=\sum_{cyc}\left(\frac{a^2+bc}{b+c}-\frac{b+c}{2}\right)=$$
$$=\frac{1}{2}\sum_{cyc}\frac{2a^2-b^2-c^2}{b+c}=\frac{1}{2}\sum_{cyc}\frac{a^2-b^2-(c^2-a^2)}{b+c}=$$
$$=\frac{1}{2}\sum_{cyc}(a^2-b^2)\left(\frac{1}{b+c}-\frac{1}{a+c}\right)=\frac{1}{2}\sum_{cyc}\frac{(a-b)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4600409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Let $f(x)=1+2\cos x +3 \sin x$. If real number $a, b, c$ are such that $af(x)+bf(x+c)=1$ holds for any $x \in \mathbb R$, then find $a, b, c$
Let $f(x)=1+2\cos x +3\sin x$. If real number $a, b, c$ are such that $af(x)+bf(x+c)=1$ holds for any $x \in \mathbb R$, then find $a, b, c$
Solution given as in book:
since ... | Note that $3\sin x+2\cos x=\sqrt{13}\sin (x+\alpha)$ where $\tan \alpha=\frac{2}{3}$.
Now we get $$a+b+\sqrt{13}(a\sin(x+\alpha)+b\sin(x+c+\alpha))=1$$
for all $x\in \mathbb R$.
Now if we set $x=-\alpha$ and $x=\pi-\alpha$, then $a+b=1$ and $\sin c=0$ so $c=k\pi$ for some integer $k$.
Plugging $x=\frac{\pi}{2}-\alpha$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Frullani like Trig integral $$\int_{0}^{\infty}\frac{\left|\cos\left ( x-\frac{\pi}{4} \right ) \right|- \left|\cos\left ( x+\frac{\pi}{4} \right ) \right| }{x}dx=\sqrt{2}\ln(1+\sqrt{2})$$ The above integral seems to look like a frullani type integral and has a closed form in terms of natural log. I tried to indefinite... | $$I=\int_{0}^{\infty}\frac{\left|\cos\left ( x-\frac{\pi}{4} \right ) \right|- \left|\cos\left ( x+\frac{\pi}{4} \right ) \right| }{x}dx=\int_{0}^{\infty}\frac{\left|\cos\left ( \frac{x}{2}-\frac{\pi}{4} \right ) \right|- \left|\cos\left ( \frac{x}{2}+\frac{\pi}{4} \right ) \right| }{x}dx$$
$$=\frac{1}{\sqrt2}\int_0^\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
Show that the series $\sum^\infty_{k=1}\frac{1}{k^4}$ is bounded. Show that the series $\sum^\infty_{k=1}\frac{1}{k^4}$ is bounded.
What I've tried: Since
$\frac{1}{2^4}+\frac{1}{3^4}\leq \frac{1}{2^4}+\frac{1}{2^4}=\frac{1}{2^3}$
$\frac{1}{4^4}+\frac{1}{5^4}+\frac{1}{6^4}+\frac{1}{7^4}\leq \frac{1}{4^4}+\frac{1}{4^4}+... | \begin{equation}
\sum_{k=1}^{\infty}\frac{1}{k^4}
\leq \sum_{k=1}^{\infty}\frac{1}{k^2}
= 1 + \sum_{k=1}^{\infty}\frac{1}{\left(k + 1\right)^2}
\leq 1+\int_{1}^{\infty}\frac{1}{x^2} dx
= 1 + 1
= 2
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4606187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Question to first order PDE with Characteristics Method We have the given problem
\begin{align}
-2yu_{x} + u_{y} + 2yu &= 2y \tag 1 \\
u(1,y) &= 1 + e^{-1-2y^2} \tag 2
\end{align}
where $x > 0, y \in \mathbb{R^*}$. We can modify the formula $(1)$ to get
\begin{align}
-2yu_{x} + u_{y} &= 2y - 2yu \\
\implies -2yu_{x} +... | Solution
\begin{align}
-2yu_{x} + u_{y} &= 2y - 2yu \\
\implies -2yu_{x} + u_{y} &= 2y(1-u) \\
\implies -u_{x} + \frac{u_{y}}{2y} &= 1-u \\
\implies u_{x} - \frac{u_{y}}{2y} &= u-1
\end{align}
so our problem would be
\begin{align}
u_{x} - \frac{u_{y}}{2y} &= u-1 \tag 3 \\
u(1,y) &= 1 + e^{-1-2y^2} \tag 4
\end{align}
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4607052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
In a triangle $\triangle ABC$, the sum of two sides is $x$, and their product is $y$. If $x^2-c^2=y$, find $\frac{r}{R}$ in terms of $x, c, y$ This problem comes from a previous JEE Advanced examination, the problem is as follows:
In a triangle $\triangle ABC$, the sum of two sides is $x$, and their product is $y$. If... | From Law of Cosine, replace cosine with accurate haversine.
$c^2 = a^2 + b^2 - 2ab×\cos(C) = (a-b)^2 + 4ab×hav(C)$
Let $\displaystyle z = ab×hav(C) = \frac{(c-(a-b)(c+(a-b))}{4}$
$→ z = (s-a)(s-b)$
$→ (a\,b-z) = s(s-c)$
$Δ = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(ab-z)\,z}$
If c is the smallest side, (a-b) is exact, RHS area... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4607869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
What is the concrete mathematical argument why $a^3+a=b^3+b\Rightarrow a=b$ is true in $\mathbb{R}$? The question arose from the following proof:
$f(x)=x^3+x+1$ is injective:
Let $a,b\in \mathbb{R}$ and $f(a)=f(b)$:
$f(a)=f(b)\Leftrightarrow a^3+a+1=b^3+b+1\Leftrightarrow a^3+a=b^3+b \Rightarrow a=b$
Is $a^3+a=b^3+b\Ri... | Since
$\quad a^3+a=b^3+b \text{ implies } (a = 0 \land b = 0) \; \displaystyle{\Large \lor} \; \frac{a^2+1}{b^2+1} = \frac{b}{a}$
we can write, with $a \ne 0$
$$a^3+a=b^3+b \; \text{ iff } \; \frac{a^2+1}{b^2+1} = \frac{b}{a}$$
Examining the fraction you must have that both $a$ and $b$ are of the same sign.
Again, exa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proving that $(1+1/x)\tan^{-1} x > 1$, when $x> 0$ Monotonicity of a function is an itneresting method to prove inequalities. However, one needs to construct a suitable function $f(x)$ in the given domain. In this, out of several choices one particular form of $f(x)$ may suceed. Here, we prove that $$(1+1/x)\tan^{-1}... | Dividing both sides by $(1+1/x)$,
$$
\arctan x > \frac x{x+1}
$$
Then we have
$$
\begin{align}
x > \tan \frac x{x+1}
\end{align}
$$
Applying inequalities about $\sin$ and $\cos$ i.e. $\sin \theta < \theta$ and $\cos \theta > 1 - 1/2\theta^2, \forall \theta \in (0,1)$,
$$
\begin{align}
\tan \frac x{x+1} & = \frac{\sin \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4610536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Proving that $\lim_\limits{x \to \infty} x\cdot \ln\left(1 + \frac{1}{x}\right) = 1$ using Taylor series We need to prove that $\lim_\limits{x \to \infty} \ln((1 + \frac{1}{x})^x) = \lim_\limits{x \to \infty} x \cdot \ln(1 + \frac{1}{x}) = 1$ and would like to use Taylor series.
To do this, when we expand $\ln(1 + \fra... | Using Taylor series here is like the proverbial killing a fly with a sledgehammer. Set $h = 1/x$, so the condition $x \to \infty$ is the same as $h \to 0^+$ and
$$
x\ln\left(1+\frac{1}{x}\right) = \frac{\ln(1+h)}{h} = \frac{\ln(1+h)-\ln 1}{h}.
$$
Therefore
$$
\lim_{x \to \infty} x\ln\left(1+\frac{1}{x}\right) =
\lim_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4610943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove $\sum_{i=1}^{n} \frac{1}{(i²+i)^\frac{3}{4}}>2- \frac{2}{\sqrt{n+1}}$
Let $i\in\mathbb{N}$ $$\sum_{i=1}^{n} \frac{1}{(i²+i)^\frac{3}{4}}>2-\frac{2}{\sqrt{n+1}}$$
A.M.-G.M. inequality makes this problem complex to solve.
It was true for $n=1,2,3,4$ then solving this became difficult. How can I solve this?
| We first observe that
$$\frac{2}{\sqrt{n}} - \frac{2}{\sqrt{n+1}} = \frac{2}{\sqrt{n(n+1)}(\sqrt{n} + \sqrt{n+1})}
< \frac{2}{\sqrt{n(n+1)}\cdot 2[n(n+1)]^{\frac{1}{4}}} = \frac{1}{(n^2+n)^{\frac{3}{4}}}$$
and then use this to argue
\begin{align*}
\displaystyle\sum_{i=1}^{n} \frac{1}{(i^2+i)^\frac{3}{4}} > \displaysty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4613604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Given positive real numbers $x_1,x_2,\ldots,x_n$ such that $\sum_{k=1}^{n}{\frac{1}{x_k+1}}=n-2$ , how to prove the following inequality? Given positive real numbers $x_1,x_2,\ldots,x_n$ such that $\sum_{k=1}^{n}{\frac{1}{x_k+1}}=n-2$ , prove that
$$
\sum_{i=1}^{n}{\frac{1}{x_i}}+\frac{n(n-2)}{\sum_{i=1}^{n}{x_i}}\geqs... | Let $$\sum_{i=1}^{n}{\frac{1}{x_i}}+\frac{n(n-2)}{\sum\limits_{i=1}^{n}{x_i}}<(n-1)(n-2)$$ for some values of $x_i$, $x_i=ka_i,$ where $k>0$ and
$$\sum_{i=1}^{n}{\frac{1}{a_i}}+\frac{n(n-2)}{\sum\limits_{i=1}^{n}{a_i}}=(n-1)(n-2).$$
Thus, $$\frac{1}{k}(n-1)(n-2)<(n-1)(n-2),$$ which gives $k>1$ and we obtain:
$$n-2=\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4619365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_{0}^{1}\frac{K(x)\ln\left(1-x^2\right)}{\sqrt{1-x^2}}\text{d}x,\int_{0}^{1}\frac{xK(x)^2\ln\left(1-x^2\right)}{\sqrt{1-x^2}}\text{d}x$ I am recently interested in integrals containing an elliptic integral $K(x)$, which is defined by $\int_{0}^{1} \frac{1}{\sqrt{1-t^2}\sqrt{1-x^2t^2} }\text{d}t$ for $|x|... | For the first integral we can start by substituting $1-x^2\to x^2$ followed by expanding the denominator into power series.
$$I=\int_0^1 \frac{K(x)\ln (1-x^2)}{\sqrt{1-x^2}}dx=2\int_0^1 \frac{K'(x)\ln x}{\sqrt{1-x^2}}dx$$
$$=2\sum_{k=0}^\infty \frac{\binom{2k}{k}}{2^{2k}}\int_0^1 x^{2k}K'(x)\ln x\,dx$$
We have:
$$\int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Prove for a,b,c,d > 0 $\frac{a^3}{b+c+d} + \frac{b^3}{a+c+d} + \frac{c^3}{a+b+d} + \frac{d^3}{a+b+c} \ge \frac{1}{3}$ where $ab + bc + cd + da = 1$ The task requires for Cauchy-Schwarz inequality to be used.
My attempt: Using Titu's Lemma (direct consequence of the inequality), I got:
$\frac{a^4}{a(b+c+d)} + \frac{b^4}... | Continuing from where you left off, note the condition is equivalent to $(a+c)(b+d)=1$. Then reduce the numerator using
$$
a^2+b^2+c^2+d^2 = (a+c)^2+(b+d)^2-2(ac+bd)\ge 2(a+c)(b+d)-2(ac+bd) =2-2(ac+bd)
$$
You now need to show $2[1-(ac+bd)]^2/(1+ac +bd)\ge 1/3$, which is equivalent to $ ac+bd\le 1/2$. Can you take it fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove $a+b+c \geq \frac{3+\sqrt{7}}{2}$ for $a,b,c \in [0, \frac32]$ with $a^2+b^2+c^2+abc=4$
Let $a,b,c \in \left[0;\frac{3}{2}\right]$ and $a^2+b^2+c^2+abc=4$. Prove that
$$a+b+c \geq \dfrac{3+\sqrt{7}}{2}.$$
Source: This is a math problem that my teacher gave me $3$ months ago (the submission deadline has expired)... | Rewriting the inequality as :
$$\frac{1}{2}\left(-bc+\sqrt{b^2c^2-4\left(b+c\right)^{2}+8bc+16}\right)+b+c \geq \dfrac{3+\sqrt{7}}{2}$$
We make :
$$b+c=\operatorname{constant}=C$$
The function :
$$g(bc)=-bc+\sqrt{b^2c^2-4\left(C\right)^{2}+8bc+16}$$
Is increasing as $bc$ increases with $-bc+\sqrt{b^2c^2-4\left(C\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4621391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solving y(t) from a differential equation Given the initial problem
$$\frac{3y^2-t^2}{y^5} \frac{dy}{dt} + \frac{t}{2y^4}=0, y(1) =3 $$
I got the solution to be
$$\frac{t^2}{4y^4}-\frac{3}{2y^2} = -\frac{53}{324} $$
But how do I write it in terms of $y(t)$
| Letting $P:=\frac{t}{2y^4}$ and $Q:=\frac{-t^2+3y^2}{t^5}$ then $P_y=Q_t$ so the non linear DE is exact and we have the implicit solution $\frac{-3}{2y^2}+\frac{t^2}{4y^4}=k$ for an arbitrary constant $k$. We have for $y,k\not=0$ that $-\frac{3}{2y^2}+\frac{t^2}{4y^4}=k$ give $t^2-6u-4ku^2=0$ with $u:=y^2$. Completing... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4624658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Natural log : Has this proof jumped a step? This proof in Nevanlinna and Paatero's Introduction to Complex Analysis (1969) is meant to show that if $$c_n=\left(1+\frac{z}{n}\right)^n$$ $(n=1,2,\dots,$ and $z = x + iy)$ then $$e^x = \liminf_ {n}|c_n|$$ ($|.|$ denotes the absolute value (corrected from "the real part").... | This may be the least troublesome derivation.
For $t > 0,$
$$
\log(1 + t) = \int_1^{1 + t}\frac{du}u < \int_1^{1 + t}du = t,
$$
and
$$
\log(1 + t) = \int_1^{1 + t}\frac{du}u >
\int_1^{1 + t}\frac{du}{1 + t} = \frac{t}{1 + t} =
t - \frac{t^2}{1 + t} > t - t^2,
$$
so
$$
0 < t - \log(1 + t) < t^2.
$$
We don't need the det... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4626977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Factor $1-64m^6$ under $\mathbb{Q}$ Factor $1-64m^6$ under $\mathbb{Q}$. We can employ the well known formula: $a^3-b^3 = (a-b)\cdot(a^2+ab+b^2)$.
$1-64m^6 = (1)^3-(4m^2)^3 = (1-4m^2)(1+4m^2+16m^4)=(1+2m)(1-2m)(1+4m^2+16m^4)$.
However, the answer is provided as $(1+2m)(1-2m)(1-2m+4m^2)\cdot(1+2m+4m^2)$.
I cannot unders... | $$1+4m^2+16m^4$$
$$=1+2\cdot 4m^2+16m^4-4m^2$$
$$=(1+4m^2)^2-(2m)^2$$
$$=(1-2m+4m^2)(1+2m+4m^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4627483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove $\frac{ab^2+2}{a+c} +\frac{bc^2+2}{b+a} +\frac{ca^2+2}{c+b} \geq \frac{9}{2}$ for $a,b,c\geq1$ Prove $\dfrac{ab^2+2}{a+c} +\dfrac{bc^2+2}{b+a} +\dfrac{ca^2+2}{c+b} \geq \dfrac{9}{2}$ for $a,b,c\geq1$.
I tried using the Titu Andreescu form of the Cauchy Schwarz inequality and got to this point:
$\dfrac{(b\sqrt{a}+... | Hint for $a,b,c,x,y,z\in[1,\infty)$:
Let $x,y,z\in[1,\infty) $ then we have the inequality (tangent line method):
$$f\left(x\right)=\frac{x^{2}y+2}{y+z}\geq f'(1)(x-1)+f(1)$$
remark that again the inequality of the tangent line method is greater than the following inequality :
$$f\left(x,y,z\right)=\frac{\frac{3}{2}\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4628757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
Confusion while factoring quadratics. I recently factored the quadratic $3x^2+4x-4$, the results where $x = -2$; $x = 2/3$. My question is why is it incorrect to write the factors of the quadratic as $(x+2)(x-2/3)$? The correct way to write it is $(x+2)(3x-2)$, but why? What is the difference in $(x-2/3)$ and $(3x-2)$,... | Observe that
\begin{align*}
3x^2 + 4x - 4 & = 3x^2 - 2x + 6x - 4\\
& = x(3x - 2) + 2(3x - 2)\\
& = (x + 2)(3x - 2)\\
& = 3(x + 2)\left(x - \frac{2}{3}\right)\\
& \neq (x + 2)\left(x - \frac{2}{3}\right)\\
& = x^2 + \frac{4}{3}x - \frac{4}{3}
\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4629268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Proving $\mathbb{Q}(\sqrt{2},\sqrt{5})=\mathbb{Q}(\sqrt{2}+\sqrt{5})$ using matrices I am trying to show that $\mathbb{Q}(\sqrt{2},\sqrt{5})=\mathbb{Q}(\sqrt{2}+\sqrt{5})$. I have calculated the minimal polynomial $x^4-14x^2+9$ and I know that its basis as a $\mathbb{Q}$-vector space is $\{1,
\sqrt{2}, \sqrt{5}, \sqrt... | As Mariano notes in the comments, there's a much quicker argument using degrees, but here's the matrix argument. (A nice bonus of this argument is that you can explicitly express the basis elements for $\mathbb{Q}(\sqrt{2}, \sqrt{5})$ in terms of powers of $\alpha$.)
Let $\alpha = \sqrt{2} + \sqrt{5}$. Note that $\alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4631378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
sum of 100 terms of logarithmic expression
Calculate value of
$\displaystyle \sum^{100}_{k=1}\ln\bigg(\frac{(2k+1)^4+\frac{1}{4}}{16k^4+\frac{1}{4}}\bigg)$
My try :: $\displaystyle x^4+4y^4$
$=(x^2+2xy+2y^2)(x^2-2xy+2y^2)$
So sum $\displaystyle \sum^{100}_{k=1}\ln\bigg(\frac{1+4(2k+1)^4}{1+4(2k)^4}\bigg)$
$\display... | Let's do bit by bit.
$$\frac{(2k+1)^4+\frac{1}{4}}{16k^4+\frac{1}{4}}$$
$$=\frac{(k+\frac{1}{2})^4+\frac{1}{64}}{k^4+\frac{1}{64}}$$
Now,
$$\ln\Big( \frac{(k+\frac{1}{2})^4+\frac{1}{64}}{k^4+\frac{1}{64}} \Big)$$
$$=\ln\Big( (k+\frac{1}{2})^4+\frac{1}{64} \Big)-\ln\Big( k^4+\frac{1}{64} \Big)$$
Now bring the summation.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4632826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Random Variables as Sums of Different Random Variables for $R = \sqrt{X^2 + Y^2}$ I am trying to understand the derivation of the PDF of $$R = \sqrt{X^2 + Y^2}$$
I am looking at some derivations and some derivations say that if $X$ and $Y$ are independent, then we can say that $$f_{X,Y} = f_X f_Y$$ which is just the la... | Now let me suppose that $X$ and $Y$ are valued in $\mathbb{R^+} = (0, \infty)$ and $S \colon = X/Y$.
Then we have
$$
\begin{cases}
X = \frac{RS}{\sqrt{1+S^2}} \\
Y = \frac{R}{\sqrt{1+S^2}}
\end{cases}
$$
Denote the density of $(X, Y)$ and $(R, S)$ by $f(x, y)$ and $g(r, s)$ respectively.
We can prove (maybe has been p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4635544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_0^\pi \left(\frac{\sin{2x}\sin{3x}\sin{5x}\sin{30x}}{\sin{x}\sin{6x}\sin{10x}\sin{15x}}\right)^2 dx$ (from MIT Integration Bee 2023) This is from the final round of MIT Integration Bee 2023.
$$\int_0^\pi \left(\frac{\sin{2x}\sin{3x}\sin{5x}\sin{30x}}{\sin{x}\sin{6x}\sin{10x}\sin{15x}}\right)^2 dx$$
The g... | $$A=\frac{\sin (2 x)\, \sin (3 x)\, \sin (5 x) \,\sin (30 x) }{\sin (x) \,\sin (6 x) \,\sin (10 x) \,\sin (15 x) }$$ Simplifying $$A=-1-2 \cos (2 x)+2 \cos (6 x)+2 \cos (8 x)$$ Squaring and simplifying again
$$A^2=7+8 \cos (2 x)-2 \cos (4 x)-8 \cos (6 x)-8 \cos (8 x)-4 \cos (10
x)+$$ $$2 \cos (12 x)+4 \cos (14 x)+2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4636662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
prove $\forall x \in (0, \frac{\pi}{2})$: $x^2 \cos^2 (x) < 2$ Prove $\forall x \in (0, \frac{\pi}{2})$: $x^2 \cos^2 (x) < 2$
It is clear, that:
*
*$x^2 \cos^2 (x) \to 0, x \to \frac{\pi}{2}$, since cosine goes to 0.
*$x^2 \cos^2 (x) \to 0, x \to 0$, since $x$ goes to 0.
Of course, we can split $(0, \frac{\pi}{2})... | I shall give you the Hint:
*
*Assume that $f(x)=x^2\cdot \cos^2x$, and then prove, $$f'(x)=2x\cos x(\cos x-x\sin x)$$
*Find the maxima of $f(x)$ between $\left[0,{\pi\over2}\right]$. As $f(0)=f({\pi\over2})=0$ so there must be peak between $\left[0,{\pi\over2}\right]$. Hence, $f'(x)=0$ has the maxima. And for your ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4645115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find all $x$ for that $x^2 + (x+1)^2$ is a square How to find all natural $x$ for that $x^2 + (x+1)^2$ is a perfect square?
| Suppose $x^2 + (x+1)^2 = y^2$. We can rewrite it as $(2x+1)^2 + 1 = 2y^2$ or $(2x+1)^2 - 2y^2 = -1$.
If $z=2x+1$ then we have $z^2 - 2y^2 = -1$. This is Pell's equation.
Wikipedia article shows how to solve it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
How to solve a cyclic quintic in radicals? Galois theory tells us that $\frac{z^{11}-1}{z-1} = z^{10} + z^9 + z^8 + z^7 + z^6 + z^5 + z^4 + z^3 + z^2 + z + 1$ can be solved in radicals because its group is solvable. Actually performing the calculation is beyond me, though - here what I have got so far:
Let the roots be... | The solution of a solvable quintic,
$$x^5 + ax^4 + bx^3 + cx^2 + dx + e = 0$$
can be given by,
$$x = \frac{1}{5}\left(-a+z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}\right)$$
and the $z_i$ are the roots of the quartic. However, there is also a form where one takes a fifth root only once.
$p=11$
Let,
$$x^5 +x^4 −4x^3 −3x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 3
} |
Proof that $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ Why is $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$ $\space$ ?
| $\dfrac{k+1}2-\dfrac{k-1}2=1 \implies\dfrac{k(k+1)}2-\dfrac{(k-1)k}2=k$
$\implies \sum_{k=1}^n\Bigg(\dfrac{k(k+1)}2-\dfrac{(k-1)k}2\Bigg)=\sum_{k=1}^nk\implies\dfrac{n(n+1)}2-\dfrac{1(1-1)}2=\sum_{k=1}^nk$
$\implies\sum_{k=1}^nk=\dfrac{n(n+1)}2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "136",
"answer_count": 36,
"answer_id": 9
} |
Is it possible to 'split' coin flipping 3 ways? When flipping a coin to make important decisions in life you can flip once to choose between 2 possible outcomes. (Heads I eat cake, Tails I eat chocolate!)
You can also flip twice to choose between 4 outcomes. (Heads-Heads, Tails-Tails, Heads-Tails, Tails-Heads)
Can you ... | Here's my method. Like voltrevo's method the expected number of tosses is also only $2\frac{2}{3}$.
In order to explain this easily, lets say there are three people named p1, p2 and p3. If the coin comes up heads on the first flip then p3 looses and p1 and p2 flip to decide who wins. However, if it comes up tails on t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 9,
"answer_id": 8
} |
Cartesian Equation for the perpendicular bisector of a line
Find the Cartesian equation for the perpendicular bisector of the line joining A(2,3) and B(0,6)
How do I do this?
Thank you!
| Notice, the slope of the line joining the given points $(2, 3)$ & $(0, 6)$
$$=\frac{6-3}{0-2}=\frac{-3}{2}$$ Now, the slope of the perpendicular bisector $$=\frac{-1}{\text{Slope of line}\space AB}=\frac{-1}{\frac{-3}{2}}=\frac{2}{3}$$ Now, the perpendicular bisector will pass through the mid-point of the line joining ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Solving separable differential equation Seems straight-forward but I've been unable to get it right. Here are my steps:
$$y'(x) = \sqrt{-2y(x) + 28},\hspace{20 pt} y(-4)=-4$$
$$\int {1 \over \sqrt{28-2y} }\hspace{2 pt}\text{d}y = \int \text{d}x$$
$$-\sqrt{28-2y} = x + c$$
$$(28-2y) = (x+c)^2$$
$$y = -1/2x^2 - cx - c^2/... | So what we want here is a particular solution to our ODE given our condition: $y(-4) = -4$
$$y'(x) = \sqrt{-2y(x) + 28},\hspace{20 pt} y(-4)=-4$$
$$\Rightarrow \dfrac{dy}{dx} = \sqrt{-2y(x) + 28}$$
$$\Rightarrow \int {1 \over \sqrt{28-2y} }\hspace{2 pt}\text{d}y = \int {1}~\text{d}x$$
$u = 28-2y$
$du = -2dx$
$dx = -\df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/5007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Find the sum to n terms of the series $\frac{1} {1\cdot2\cdot3\cdot4} + \frac{1} {2\cdot3\cdot4\cdot5} + \frac{1} {3\cdot4\cdot5\cdot6}\ldots $ Find the sum to n terms of the series $\frac{1} {1\cdot2\cdot3\cdot4} + \frac{1} {2\cdot3\cdot4\cdot5} + \frac{1} {3\cdot4\cdot5\cdot6}\ldots $
Please suggest an approach for ... | This is similar to what has been said by Branimir, but shows how we can extend the result to
$$\sum_{k=1}^n {1 \over k(k+1) \cdots (k+m)}, \qquad m \in \mathbb{N}.$$
We can build up the result from the identities
$${1 \over k(k+1)} = {1 \over k} - { 1 \over k+1}, \qquad (1)$$
$${1 \over k(k+1)(k+2)} = {1 \over 2} \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/5558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
For $n \in \mathbb{N}$ $\lfloor{\sqrt{n} + \sqrt{n+1}\rfloor} = \lfloor{\sqrt{4n+2}\rfloor}$ This is Exercise 3.20 from Apostol's book. Many of them seem tough and here is one of them which I am struggling with.
For $n \in \mathbb{N}$, prove that this identity is true: $$\Bigl\lfloor{\sqrt{n} + \sqrt{n+1}\Bigl\rfloor} ... | First a kooky Lemma:
For $x > 1$ then $\sqrt{1 - \frac 1 x} + \sqrt{1 + \frac 3 x} > 2$
[$\sqrt{1 - \frac 1 x} + \sqrt{1 + \frac 3 x} > 2$ iff
$1 - \frac 1 x + 1 + \frac 3 x + 2\sqrt{(1 - \frac 1 x)(1 + \frac 3 x)} > 4$ iff
$2 + \frac 2 x + 2\sqrt{(1 + \frac 2 x -\frac 2 {x^2}} > 4$ iff
$1 + \frac 2 x -\frac 2 {x^2} >... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/6087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 2
} |
Simple Proof by induction: $9$ divides $n^3 + (n+1)^3 + (n+2)^3$ I'm trying to prove using induction that $9$ divides $n^3 + (n+1)^3 + (n+2)^3$ whenever $n$ is a non-negative integer. So far, I have:
*
*Base case: $P(1) = (1) + (8) + (27) = 36, 36$ can be divided by $9$ so the base case is valid
*Inductive step: le... | Here's an all-purpose technique for proving your kind of induction problems.
You are trying to prove using induction that 9 divides $n^3 + (n+1)^3 + (n+2)^3$ whenever n is a non-negative integer.
You have already demonstrated the base case, so here's the inductive step.
Assuming it is true for n, is it also true for n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/6805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 0
} |
How to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$? I am trying to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$ but am not able to think something except induction,which is of-course not necessary (I think) here, so I am inquisitive to prove this in a more genera... | Use Gauss' trick for summing a sequence: Combine the sum term-by-term with its reversed-order self, noting that the binomial coefficients are symmetric.
$$S = \sum_{r=0}^n r { n \choose r } = \sum_{r=0}^n (n-r) {n \choose n-r }$$
So,
$$\begin{eqnarray}
2 S &=& \sum_{r=0}^n \left( r {n\choose r} + (n-r) {n \choose n-r}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/7757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 10,
"answer_id": 6
} |
Number of digits from $1$ to $n$ Let $n$ be a natural number $k$ digits. Show that the quantity $Q$ of digits required to write the natural numbers from $1$ to $n$ is:
$Q = k(n+1) - \underbrace{111\ldots11}_{k\textrm{ digits}}$
| As this question is now 14 hours old, I will give an answer.
First sum the total number of digits of all the natural numbers $ \le k-1.$ Call this sum $S.$
There are $9 \times 10^{i-1}$ natural numbers of length $i,$ so the total number of digits of natural numbers of length $i$ is $9i \times 10^{i-1}.$ Hence
$$ S=9(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/8576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What type of triangle satisfies: $\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $ If in a $\displaystyle\bigtriangleup$ ABC, $\displaystyle\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $, then $\displaystyle\bigtriangleup$ ABC is of which type ?
| I tried a geometry proof for this and thought it was worth sharing it here.
I am unable to draw the figure. I would appreciate if some one could take the effort to draw the picture or let me know what is the easiest way to draw such pictures.
Consider the triangle $ABC$. Let the angle bisector of $A$ meet $BC$ at $D$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/10545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Baffling identity: $\prod \log_{10} \tan = \sum \log_{10} \tan$ I am quite a baffled now, I am not getting by how it can be written that :
\begin{align*}
&\log_{10} \tan 40^\circ \cdot \log_{10} \tan 41^\circ \cdot \log_{10} \tan 42^\circ \cdot\log_{10} \tan 43^\circ \cdots \log_{10} \tan 50^\circ \\
&= \log_{10} \tan... | HINT: Observe that $$\log_{10} \tan{40} + \log_{10} \tan{50} = \log_{10} \sin{40} - \log_{10} \cos{40} + \log_{10} \sin{50} - \log_{10} \cos{50}$$ Now use the fact that $\sin(90 - x) =\cos{x}$ and see that they cancel out.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/10868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
AM-GM application How to show that
$$ \displaystyle (a_1 + a_2 + a_3)(\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3}) \ge 9 $$
$a_1,a_2,a_3$ are all of same algebraic sign.
| WLOG we may assume $a_{1}\leq a_{2}\leq a_{3}$.
$$\dfrac{a_{1}}{a_{1}}+\dfrac{a_{2}}{a_{2}}+\dfrac{a_{3}}{a_{3}}=3\tag{1}$$
It is not hard to see that $a_{1},a_{2},a_{3}$ and $\dfrac{1}{a_{1}},\dfrac{1}{a_{2}},\dfrac{1}{a_{3}}$ are oppositely sorted
Using Rearrangement Inequality we find
$$\dfrac{a_{1}}{a_{2}}+\dfrac{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/13029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Prove a number is composite How can I prove that $$n^4 + 4$$ is composite for all $n > 5$?
This problem looked very simple, but I took 6 hours and ended up with nothing :(. I broke it into cases base on quotient remainder theorem, but it did not give any useful information.
Plus, I try to factor it out:
$$n^4 - 16 + 2... | You can factor out $n^{4}+4$ algebraically by finding the four roots of $n^{4}+4=0$.
Since $n^{4}+4=0\Leftrightarrow n^{4}=4e^{i\pi }$, we have
$$\begin{eqnarray*}
n &=&4^{1/4}e^{i (\pi +2k\pi)/4}\quad k=0,1,2,3 \\
&& \\
n &=&\sqrt{2}e^{i\pi /4 }=1+i\quad \left( k=0\right)
\\[2ex]
n &=&\sqrt{2}e^{i 3\pi /4 }=-1+i\quad... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/21146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 3
} |
having trouble with limit of integral How do I solve the following?
$$\lim_{x\to 0} \int_0^1 \cos\left(\frac{1}{xt}\right)\, dt$$
| Here's another take. Since $\cos \left(\frac{1}{xt}\right)$ is an even function of $x$, we can take the limit from the right or from the left, and the result is the same. So assume $x > 0$. Then apply the $u = 1/(xt)$ substitution as in joriki's answer, followed by integration by parts in the opposite direction from... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/23433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Help solving $\int {\frac{8x^4+15x^3+16x^2+22x+4}{x(x+1)^2(x^2+2)}dx}$ $\displaystyle\int {\frac{8x^4+15x^3+16x^2+22x+4}{x(x+1)^2(x^2+2)}\,\mathrm{d}x}$
I used partial fractions, solved $A = 2, C = 3$.
$$\frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2} +\frac{(Dx+E)}{(x^2+2)}$$
\begin{align*}
&8x^4+15x^3+16x^2+22x+4\\
... | Below I show how the Heaviside cover-up method generalizes to handle nonlinear denominators. With numerator $\rm\:f(x)\ =\ 8x^4+15x^3+16x^2+22x+4\:,\: $ the undetermined partial fraction is
$$\rm\frac{f(x)}{x(x+1)^2(x^2+2)}\ =\ \frac{a}{x}\, + \,\frac{b\, (x+1) + c}{(x+1)^2}\, + \,\frac{d\, x+e}{x^2+2}$$
To find $\,\rm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/23484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 1
} |
What is the remainder of $16!$ is divided by 19? Can anyone share me the trick to solve this problem using congruence?
Thanks,
Chan
| A more explicit answer would go like this:
By Wilson's Theorem, $18!$ is congruent to $-1 \pmod {19}$
$$18! = (18\cdot 17)(16!)$$
Then $(18\cdot 17)(16!)$ is congruent to $-1 \pmod{19}$
But note that $18$ is congruent to $-1 \pmod{19}$ and $17$ is congruent to $-2 \pmod{19}$
Then it follows that $(18\cdot 17)(16!)$ is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/23809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$ $$y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$$
$$\frac{dy}{dx}=\frac{5}{6}x^{4}-\frac{3}{10x^{4}}$$
squaring this $$=\frac{25}{36}x^{8}+\frac{9}{100x^{8}}$$
Plugging into the formula $$ds=\sqrt{1+\left( \frac{dy}{dx}\ri... | As Américo Tavares pointed out, you are missing the term $-1/2$ due to the crossterm when squaring $d y/ dx$. The integral you want to solve is
$$\int_1^2 dx \,\sqrt{\frac{1}{2}+ \frac{25 x^8}{36} + \frac{9}{100 x^8}}
= \int_1^2 dx \, \sqrt{\frac{(9+25 x^8)^2}{900 x^8}}$$
$$ = \int_1^2 dx \,\frac{9+25 x^8}{30 x^4} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/24045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Convergence of Sum over Integer Lattice Does the sum
$$\sum_{z \in \mathbb{Z}^3\setminus \{(0,0,0)\}} \left( \frac{1}{|{\bf x} - {\bf z}|^2} - \frac{1}{|{\bf z}|^2} \right)$$
converge pointwise or even uniformly for $\varepsilon < |{\bf x}| < 1-\varepsilon$?
| It seems to converge conditionally, with convergence if you sum over a sphere centered at the origin and then let the radius of the sphere increase without bound. Let ${\bf x}=(0,0,d)$ and ${\bf z}=(x,y,z)$ (sorry for reusing $x$). Then $$\begin{align}\sum_{z \in \mathbb{Z}^3\setminus \{(0,0,0)\}} \left( \frac{1}{|{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/24455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate an expresssion $$\frac{(z x^{-1} y)^5 y^5}{x^{-4} z^{-4}}$$
How would I evaluate this if $x = 10$, $y = -3$ and $z = 3$ ? I would like a step-by-step solution to help me fully understand it.
| First, use the fact that $x^{-n}=\frac{1}{x^n}$ to find $$\frac{(z x^{-1} y)^5 y^5}{x^{-4} z^{-4}}=(zx^{-1}y)^5 \cdot \frac{y^5}{x^{-4}z^{-4}}=\left(\frac{zy}{x}\right)^5\cdot x^4z^4y^5 $$ Next, $$\left(\frac{zy}{x}\right)^5\cdot x^4z^4y^5=\frac{z^5y^5}{x^5}\cdot x^4z^4y^5=z^9y^{10}x$$ Can you finish from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/24884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Question regarding solving polynomial of congruence? In my textbook, they said:
$$2x^{3} + 7x - 4 \equiv 0 \pmod{5}$$
The solution of this equation are the integers with $x \equiv 1 \pmod{5}$, as can be seen by testing $x = 0, 1, 2, 3,$ and $4.$
And I have no clue how do they had $x \equiv 1 \pmod{5}$. I tested as they... | $$\begin{aligned}
2x^3+7x-4 \equiv 0 ( mod 5)\\
2x^3+2x-4 \equiv 0 ( mod 5)\\
2(x-1)(x^2+x+2) \equiv 0 ( mod 5)
\end{aligned}$$
Now you have two solutions $x-1 \equiv 0 (mod5)$ or $x^2+x+2 \equiv 0 (mod5)$.
You can continue and verify if $x \equiv 1 ( mod 5)$ is the only solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/26152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"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.