Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Integrating $\int_{-\infty}^0e^x\sin(x)dx$ I ask if anyone could solve the following:
$$\int_{-\infty}^0e^x\sin(x)dx=?$$
I can visually see that it will converge and that it should be less than $1$:
$$\int_{-\infty}^0e^x\sin(x)dx<\int_{-\infty}^0e^xdx=1$$
But I am unsure what its exact value is.
Trying to find the defi... | Integrating by parts two times you get
$$\begin{align}
\int e^x \cdot \sin x \quad dx & = e^x \cdot \sin x - \int e^x \cdot \cos x \quad dx\\
& = e^x \cdot \sin x - \left( \int e^x \cdot \cos x \quad dx \right)\\
& = e^x \cdot \sin x - \left( e^x\cdot \cos x + \int e^x \cdot \sin x\quad dx\right)\\
& = e^x \cdot \sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1611368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Why is $\lim_\limits{x\to 0}\frac{\sin(6x)}{\sin(2x)} = \frac{6}{2}=3$? Why is $\lim_\limits{x\to 0}\frac{\sin(6x)}{\sin(2x)} = \frac{6}{2}=3$?
The justification is that $\lim_\limits{x\to 0}\frac{\sin(x)}{x} = 1$
But, I am not seeing the connection.
L'Hospital's rule? Is there a double angle substitution happening?
| The key is to use Hopital rule:
$$\begin{align}
\lim_\limits{x\to 0}\frac{\sin(x)}{x} &= \lim_\limits{x\to 0}\frac{\sin'(x)}{1}\\
&= \lim_\limits{x\to 0}\frac{\cos(x)}{1}\\
&= \lim_\limits{x\to 0}\frac{1}{1}\\
&=1
\end{align}
$$
hence
$$
\begin{align}
\lim_{x\to 0} \frac{\sin(6x)}{\sin(2x)} &= \lim_{x\to 0} \frac{\sin(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 6
} |
Find a basis and dimension of $W_1\cap W_2$ if the span is given Find a basis and dimension of $W_1\cap W_2$ where $W_1$ is a subspace generated by vectors $\left\{ \begin{bmatrix}
1 & 1 \\
0 & 0 \\
\end{bmatrix},\begin{bmatrix}
3 & 1 \\
-1 & 0 \\
\end{bmatrix}... | To find the matrices in $W_1\cap W_2$, you need to solve the equation
$$\begin{align}
a\begin{bmatrix} 1 & 1\\0 & 0\end{bmatrix}+b\begin{bmatrix}3 & 1\\-1 & 0\end{bmatrix} &=c\begin{bmatrix}1 & 1\\1 & 0\end{bmatrix}+d\begin{bmatrix}2 & -1 \\1 & -1\end{bmatrix}\\
\begin{bmatrix}a+3b & a+b\\-b & 0\end{bmatrix} & =\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplifying nested square roots ($\sqrt{6-4\sqrt{2}} + \sqrt{2}$) I guess I learned it many years ago at school, but I must have forgotten it. From a geometry puzzle I got to the solution
$\sqrt{6-4\sqrt{2}} + \sqrt{2}$
My calculator tells me that (within its precision) the result equals exactly 2, but I have no idea h... | Since
\begin{align}
6-4\sqrt{2}&=4-2\cdot 2\cdot\sqrt{2}+2\\
&=(2-\sqrt{2})^2
\end{align}
Where $2-\sqrt{2}>0$, it follows
$\sqrt{6-4\sqrt{2}}=2-\sqrt{2}$.
Then $$\sqrt{6-4\sqrt{2}}+\sqrt{2}=\color{blue}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
} |
Prove $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq \frac{a+b}{a+c}+\frac{b+c}{b+a}+\frac{c+a}{c+b}.$
Prove that for all positive real numbers $a,b,$ and $c$, we have $$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \geq \dfrac{a+b}{a+c}+\dfrac{b+c}{b+a}+\dfrac{c+a}{c+b}.$$
What I tried is saying $\dfrac{a}{b}+\dfrac{b}{c}+\dfra... | our inequality is equivalent to $$a^4c^2+a^2b^4+b^2c^4+a^3b^3+a^3c^3+b^3c^3\geq a^3bc^2+a^2b^3c+ab^2c^3+3a^2b^2c^2 $$we have $$a^3b^3+a^3c^3+b^3c^3\geq 3a^2b^2c^2$$ by AM-GM, and the rest is equivalent to
$$(a-b)c^2(a^3-b^2c)+(b-c)b^2(a^2b-c^3)\geq 0$$ if we assume that $$a\geq b\geq c$$
and if $$a\geq c\geq b$$ on ob... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
How to find a basis for $W = \{A \in \mathbb{M}^{\mathbb{R}}_{3x3} \mid AB = 0\}$ $B = \begin{bmatrix}
1 & 2 & 1 \\
1 & 3 & 1 \\
1 & 4 & 1
\end{bmatrix}$ and I need to find a basis for $W = \{A \in \mathbb{M}^{\mathbb{R}}_{3x3} \mid AB = 0\}$ .
I know that $AB = A\cdot\begin{bmatrix} 1\\1\\1 \end{bmatrix... | Just to simplify notation, instead of $AB=0$, let us have a look at the transpose $B^TA^T=0$.
So we are looking at the subspace
$$W'=\{A\in M_{3\times3}; CA=0\}$$
where $C=B^T=
\begin{pmatrix}
1 & 1 & 1 \\
2 & 3 & 4 \\
1 & 1 & 1 \\
\end{pmatrix}
$.
Let us denote by $\vec a_1$, $\vec a_2$, $\vec a_3$ the columns ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Closed form or approximation of $\sum\limits_{i=0}^{n-1}\sum\limits_{j=i + 1}^{n-1} \frac{i + j + 2}{(i + 1)(j+1)} (i + 2x)(j +2x)$ During the solution of my programming problem I ended up with the following double sum:
$$\sum_{i=0}^{n-1}\sum_{j=i + 1}^{n-1} \frac{i + j + 2}{(i + 1)(j+1)}\cdot (i + 2x)(j +2x)$$
where $... | $$
\begin{align}
&\sum_{i=0}^{n-1}\sum_{j=i+1}^{n-1}\frac{i+j+2}{(i+1)(j+1)}\cdot(i+2x)(j+2x)\tag{1}\\
&=\sum_{j=0}^{n-1}\sum_{i=j+1}^{n-1}\frac{i+j+2}{(i+1)(j+1)}\cdot(i+2x)(j+2x)\tag{2}\\
&=\sum_{i=0}^{n-1}\sum_{j=0}^{i-1}\frac{i+j+2}{(i+1)(j+1)}\cdot(i+2x)(j+2x)\tag{3}\\
&=\small\frac12\left(\sum_{i=0}^{n-1}\sum_{j=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
To find the sum of the series $\,1+ \frac{1}{3\cdot4}+\frac{1}{5\cdot4^2}+\frac{1}{7\cdot4^3}+\ldots$ The answer given is $\log 3$.
Now looking at the series
\begin{align}
1+ \dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot4^2}+\dfrac{1}{7\cdot4^3}+\ldots &=
\sum\limits_{i=0}^\infty \dfrac{1}{\left(2n-1\right)\cdot4^n}
\\
\log 3 ... | hint: You want to know $S(\frac{1}{2})$ whereas $S(x) = 2\displaystyle \sum_{n=1}^\infty \dfrac{x^{2n-1}}{2n-1}, |x| < 1$, $S'(x) = ...$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1616688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Real values of $x$ satisfying the equation $x^9+\frac{9}{8}x^6+\frac{27}{64}x^3-x+\frac{219}{512} =0$
Real values of $x$ satisfying the equation $$x^9+\frac{9}{8}x^6+\frac{27}{64}x^3-x+\frac{219}{512} =0$$
We can write it as $$512x^9+576x^6+216x^3-512x+219=0$$
I did not understand how can i factorise it.
Help me
|
Using factor theorem:
If $x-a$ divides $f(x)$, then $f(a)=0$.
As $(\frac{1}{2})^{9}=\frac{1}{512}$, it's worth to try $x=\pm \frac{1}{2}$.
Substituting $x=\frac{1}{2}$, LHS vanishes, so $\frac{1}{2}$ is one possible value.
Further factorize gets,
$$\left( x-\frac{1}{2} \right)
\left( x^{2}-\frac{x}{2}-\frac{3}{4} \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
To evaluate the given determinant Question: Evaluate the determinant
$\left|
\begin{array}{cc} b^2c^2 & bc & b+c \\
c^2a^2 & ca & c+a \\
a^2b^2 & ab & a+b \\
\end{array}
\right|$
My answer:
$\left|
\begin{array}{cc} b^2c^2 & bc & b+c \\
c^2a^2 & ca & c+a \\
a^2b^2 & ab & a+b \\
\end{array}
\right|= \left|
\begin... | This is probably the shortest way:-
$$
\begin{align}
\begin{vmatrix} b^2c^2 & bc & b+c \\\
c^2a^2 & ca & c+a \\\
a^2b^2 & ab & a+b
\end{vmatrix}
&=a^3b^3c^3\begin{vmatrix}a^{-1} &1 & b^{-1}+c^{-1}\\\ b^{-1} &1&a^{-1}+c^{-1}\\\ c^{-1} &1& a^{-1}+b^{-1}\end{vmatrix}\\
&=(abc)^3\left(\frac 1a+\frac 1b+\frac 1c\right)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that $2\sin^{-1}\sqrt x - \sin^{-1}(2x-1) = \frac{\pi}{2}$. Prove that $2\sin^{-1}\sqrt x - \sin^{-1}(2x-1) = \dfrac\pi2$.
Do you integrate or differentiate to prove this equality? If so, why?
| I thought it might be instructive to present a way forward without the use of calculus, but rather on the use of standard trigonometric identities.
Proceeding, we let $f(x,y) =2\arcsin(x)-\arcsin(y)$ and note that
$$\begin{align}
\sin(f(x,y))&=\sin(2\arcsin(x))\cos(\arcsin(y))-\sin(\arcsin(y))\cos(2\arcsin(x))\\\\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How to calculate $\int_a^bx^2 dx$ using summation? So for this case, we divide it to $n$ partitions and so the width of each partition is $\frac{b-a}{n}$ and the height is $f(x)$.
\begin{align}
x_0&=a\\
x_1&=a+\frac{b-a}{n}\\
&\ldots\\
x_{i-1}&=a+(i-1)\frac{b-a}{n}\\
x_i&=a+i\frac{b-a}{n}
\end{align}
So I pick left ... | The sum that is of concern is multiplied by $\frac1{n^3}$ and the limit therefore converges. Then, proceeding, we have
$$\sum_{i=1}^n(i-1)^2=\sum_{i=0}^{n-1} i^2$$
We can evaluate the sum on the right-hand side of $(1)$ using a number of approaches. The approach we present here exploits telescoping series. We note t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Prove that $3(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}) \ge 10 + 8\cdot \frac{a^2+b^2+c^2}{ab+bc+ca}$
For the positive real numbers $a, b, c$ prove that $$3\bigg(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\bigg) \ge 10 + 8\cdot \dfrac{a^2+b^2+c^2}{ab+bc+ca}$$
I did the following:
$$\begin{split}\dfrac{a^2+b^2+c^2}{... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, your inequality is a linear inequality of $w^3$.
Hence, it remains to prove it for an extremal value of $w^3$, which happens in the following cases.
*
*$w^3\rightarrow0^+$. In this case our inequality is obvious;
*$b=c=1$, which gives $(a-1)^2(2a+3)\geq0$. Done!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
proving 1/x is convex (without differentiating) I know that $\frac{1}{x}$ is convex when $x \in (0,\infty)$, this can be proven easily by showing that the second derivative is positive. However, I am finding difficulty showing it using the definition of convexity, in other words, for $\alpha \in [0,1]$ and $x_1, x_2 \... | We want to show that
$\frac{1}{a x + (1-a) y }
\leq \frac{a}{x}+\frac{1-a}{y}
$
The right side is
$ \frac{a}{x}+\frac{1-a}{y}
=\frac{ay+(1-a)x}{xy}
=\frac{a(y-x)+x}{xy}
$
and the left side is
$\frac{1}{a x + (1-a) y }
=\frac{1}{a (x-y) + y }
$
so we want
$\frac{1}{a (x-y) + y}
\le \frac{a(y-x)+x}{xy}
$.
Cross multipl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1622113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
} |
$\int \frac{2x}{9x^2+3}dx=?$ So this one seems very easy. And it really is (I guess). But I have an issue with this one. I solved it this way:
\begin{align}
& \int \frac{2x}{9x^2+3}dx=\frac{1}{6} \int \frac{x}{3x^2+1}dx=\frac{1}{6} \left( x\frac{\arctan(\sqrt3x)}{\sqrt{3}}-\int 1\cdot\frac{\arctan(\sqrt3x)}{\sqrt3} dx... | $$\int \frac{2x}{9x^2+3}dx$$
$$2\int \frac{x}{9x^2+3}dx$$
Apply Integral Substitution $u=9x^2+3\quad \:du=18xdx$
$$2\int \frac{1}{18u}du = 2\frac{1}{18}\int \frac{1}{u}du=2\frac{1}{18}\ln \left|u\right|=2\frac{1}{18}\ln \left|9x^2+3\right|$$
So
$$\int \frac{2x}{9x^2+3}dx=\color{red}{\frac{1}{9}\ln \left|9x^2+3\right|+C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Show that the following limit is $AB$. I am interested in showing the following:
$$
\lim\limits_{p\rightarrow 1/2} \frac{B}{1-2p}-\frac{A+B}{1-2p}\frac{r^B-1}{r^{A+B}-1}=AB.
$$
Here $A,B\in(0,\infty)$ and $r=\frac{1}{p}-1$.
This is for a probability theory problem involving a biased random walk, leaving an interval $(-... | Considering the expression $$Z=\frac{B}{1-2p}-\frac{A+B}{1-2p}\frac{r^B-1}{r^{A+B}-1}$$ start changing variable $p=\frac{1-x}{2}$; this makes $$Z=\frac{B}{x}-\frac{(A+B) \left(\left(\frac{2}{1-x}-1\right)^B-1\right)}{x
\left(\left(\frac{2}{1-x}-1\right)^{A+B}-1\right)}$$ and use the generalized binomial theorem or, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How do you solve B and C for $\frac{s-1}{s+1} \frac{s}{s^2+1} = \frac{A}{s+1} + \frac{Bs+C}{s^2+1}$? How do you solve B and C for $\frac{s-1}{s+1} \frac{s}{s^2+1} = \frac{A}{s+1} + \frac{Bs+C}{s^2+1}$ ?
$A = \left.\frac{s^2-s}{s^2+1} \right\vert_{s=-1} = \frac{1-(-1)}{1+1}=1$
| Cross-multiplying on the right hand side gives
$As^{2} + A + Bs^{2} + Bs + Cs + C$ on the numerator.
Plug A = 1
$(1 + B)s^{2} + (B + C)s + (1 + C) = s^{2} - s$. Equate the coefficients
$1 + B = 1 \rightarrow B = 0$
$B + C = -1 \rightarrow C = -1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What does $\bigotimes$ and $X^*$ mean? Can someone explain / link me to a linear algebra worked problem where I can see how these work. I've searched and given their statistics and matrix specialty uses, can't find any ready examples.
| $\bigotimes$ is the Kronecker product of two matrices. In this post I used the Kronecker product to understand the model matrices in mixed effects models.
$A^* = A^H = \bar A^T$ is the Hermitian or conjugate transpose of a matrix $A$.
A good example of the use of Hermitian matrices is in obtaining the inverse of a Four... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proving $\frac{1}{\sqrt{1-x}} \le e^x$ on $[0,1/2]$. Is there a simple way to prove $$\frac{1}{\sqrt{1-x}} \le e^x$$ on $x \in [0,1/2]$?
Some of my observations from plots, etc.:
*
*Equality is attained at $x=0$ and near $x=0.8$.
*The derivative is positive at $x=0$, and zero just after $x=0.5$. [I don't know how t... | We have
$$-2 \ln \sqrt{1-x}=-\ln(1-x)= \int_{1-x}^1\frac{dt}{t} \leqslant \frac{x}{1-x}.$$
For $0 \leqslant x \leqslant 1/2$, we have $2(1-x) \geqslant 1$ and
$$-\ln \sqrt{1-x} < \frac{x}{2(1-x)} \leqslant x.$$
Hence,
$$\frac{1}{\sqrt{1-x}} = \exp[-\ln(\sqrt{1-x})]\leqslant e^x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the value of : $\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}$ I saw some resolutions here like $\sqrt{x+\sqrt{x+\sqrt{x}}}- \sqrt{x}$, but I couldn't get the point to find
$\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}$.
I tried $\frac{1}{x}.(\sqrt{x+\sqrt{x+\sqrt{x}}})=\frac{\sqrt{x}}{x}\left(\s... | For an alternative answer, consider the infinitely nested radical expression $y = \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+...}}}}}$.
Evidently $y > \sqrt{x+\sqrt{x+\sqrt{x}}}$ for $x > 1$ and therefore $\frac{y}{x}$ is an upper bound for $\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}$.
Also $y > \sqrt{x}$ and therefore $\lim_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1629846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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If $x^2+3x+5=0$ and $ax^2+bx+c=0$ have a common root and $a,b,c\in \mathbb{N}$, find the minimum value of $a+b+c$ If $x^2+3x+5=0$ and $ax^2+bx+c=0$ have a common root and $a,b,c\in \mathbb{N}$, find the minimum value of $a+b+c$
Using the condition for common root, $$(3c-5b)(b-3a)=(c-5a)^2$$
$$3bc-9ac-5b^2+15ab=c^2+25a^... | Hint:
The roots of $x^2+3x+5$ are not real, and if $z\in \Bbb C$ is a root of a polynomial with real coefficients but $z$ is not real, then $\bar z$ is also a root of this polynimial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
induction to prove the equation $3 + 9 + 15 + ... + (6n - 3) = 3n^2$ I have a series that I need to prove with induction. So far I have 2 approaches, though I'm not sure either are correct.
$$3 + 9 + 15 + ... + (6n - 3) = 3n^2$$
1st attempt:
\begin{align*}
& = (6n - 3) + 3n^2\\
& = 3n^2 + 6n - 3\\
& = (3n^2 + 5n - 4) +... | If you insist on induction :
Base case $n=1\ :\ 6\times 1-3=3=3\times 1^2$
If you assume $$3+9+15+...+(6n-3)=3n^2$$
You can conclude
$$3+9+15+...+(6n-3)+(6n+3)=3n^2+6n+3=3(n+1)^2$$
completing the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Upper bound on integral: $\int_1^\infty \frac{dx}{\sqrt{x^3-1}} < 4$ I'm going through Nahin's book Inside Interesting Integrals, and I'm stuck at an early problem, Challenge Problem 1.2: to show that
$$\int_1^\infty \frac{dx}{\sqrt{x^3-1}}$$
exists because there is a finite upper-bound on its value. In particular, sho... | $$\begin{eqnarray*}\color{red}{I}=\int_{1}^{+\infty}\frac{dx}{\sqrt{x^3-1}}=\int_{0}^{+\infty}\frac{dx}{\sqrt{x^3+3x^2+3x}}&=&\int_{0}^{+\infty}\frac{2\, dz}{\sqrt{z^4+3z^2+3}}\\&\color{red}{\leq}&\int_{0}^{+\infty}\frac{2\,dz}{\sqrt{z^4+3z^2+\frac{9}{4}}}=\color{red}{\pi\sqrt{\frac{2}{3}}.}\end{eqnarray*}$$
A tighter ... | {
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"url": "https://math.stackexchange.com/questions/1631639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Find $\alpha^3 + \beta^3$ which are roots of a quadratic equation. I have a question.
Given a quadratic polynomial, $ax^2 +bx+c$, and having roots $\alpha$ and $\beta$. Find $\alpha^3+\beta^3$. Also find $\frac1\alpha^3+\frac1\beta^3$
I don't know how to proceed. Any help would be appreciated.
| First note that $\alpha^3+\beta^3=(\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2)$ and also note that $-\frac{b}{a}=\alpha+\beta$ and $\frac{c}{a}=\alpha\beta$ (do you see why?) We can make $$\alpha^2+2\alpha\beta+\beta^2=(\alpha+\beta)^2=\frac{b^2}{a^2}$$ so our final outcome will be \begin{align}
\alpha^3+\beta^3&=(\alp... | {
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"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Solve $z^4+2z^3+3z^2+2z+1 =0$ Solve $z^4+2z^3+3z^2+2z+1 =0$ with $z$: a complex variable.
Attempt at solving the problem:
We divide the polynom by $z^2$ and we get:
$z^2+2z+3+\dfrac{2}{z}+ \dfrac{1}{z^2}=0 $ $ $ We set $w=z+ \dfrac{1}{z}$
We now have $w^2+2w+5=0$
$\bigtriangleup = -16$
Let's find $\omega$ s... | $$z^4+2z^2+3z^2+2z+1=0\Longleftrightarrow$$
$$\left(z^2+z+1\right)^2=0\Longleftrightarrow$$
$$z^2+z+1=0\Longleftrightarrow$$
$$z^2+z=-1\Longleftrightarrow$$
$$z^2+z+\frac{1}{4}=-\frac{3}{4}\Longleftrightarrow$$
$$\left(z+\frac{1}{2}\right)^2=-\frac{3}{4}\Longleftrightarrow$$
$$z+\frac{1}{2}=\pm\frac{i\sqrt{3}}{2}\Longl... | {
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"source": "stackexchange",
"question_score": "3",
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How to minimize $ab + bc + ca$ given $a^2 + b^2 + c^2 = 1$? The question is to prove that $ab + bc + ca$ lies in between $-1$ and $1$, given that $a^2 + b^2 + c^2 = 1$.
I could prove the maxima by the following approach. I changed the coordinates to spherical coordinates:
$a = \cos A \\
b = \sin A \cos B \\
c = \sin A ... | Hint: $a^2+b^2+c^2+ab+bc+ca=\frac{1}{2}((a+b)^2+(b+c)^2+(c+a)^2)\geq 0$.
| {
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"source": "stackexchange",
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"answer_id": 0
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Can you find the maximum or minimum of an equation without calculus? Without using calculus is it possible to find provably and exactly the maximum value
or the minimum value of a quadratic equation
$$ y:=ax^2+bx+c $$
(and also without completing the square)?
I'd love to know the answer.
| We assume (for the sake of discovery; for this purpose it is good enough
if this is just an inspired guess)
that the curve $y = ax^2 + bx + c$ is symmetric around a vertical axis.
It's obvious this is true when $b = 0$, and if we have plotted
$y = ax^2 + bx + c$ for various other values of $a$, $b$, and $c$,
we may obs... | {
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"question_score": "2",
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Generalisation of an already generalised integral Inspired by these two questions:
Closed form for $\int_0^\infty {\frac{{{x^n}}}{{1 + {x^m}}}dx }$
Interesting integral formula
I ask whether the following integral has a closed form:
$$\int_0^\infty \frac{x^n}{(1+x^m)^\alpha} \text{d}x$$
For $\alpha \in \mathbb{N}$ and ... | $\text {Let } y=\dfrac{1}{1+x^{m}}, \textrm{then } x=\left(\dfrac{1}{y}-1\right)^{\frac{1}{m}} \textrm{ and } d x=\dfrac{1}{m} \left(\dfrac{1}{y}-1\right)^{\frac{1}{m-1}}\left(\dfrac{d y}{-y^{2}}\right).$
Simplifying gives
$$
\begin{aligned}
\int_{0}^{\infty} \frac{x^{n}}{\left(1+x^{m}\right)^{\alpha}} &=\frac{1}{m} \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634091",
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"source": "stackexchange",
"question_score": "1",
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Find a thousand natural numbers such that their sum equals their product The question is to find a thousand natural numbers such that their sum equals their product. Here's my approach :
I worked on this question for lesser cases :
\begin{align}
&2 \times 2 = 2 + 2\\
&2 \times 3 \times 1 = 2 + 3 + 1\\
&3 \times 3 \time... | A solution with four numbers different from 1 is:
$$16 \times 4 \times 4 \times 4 \times 1^{996} = 16 + 4 + 4 + 4 + (996 \times 1) = 1024$$
How was this found? $1024 = 2^{10}$ appeared to be a promising candidate for the sum and product because it's slightly larger than 1000 and has many factors. The problem then was ... | {
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"source": "stackexchange",
"question_score": "23",
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Find $\lim_{(x,y)\in S} \frac{x^2+ay^2}{x^2+by^2}$ as $(x,y) \to (0,0)$ for specified $S$. Find $$\lim_{\stackrel{(x,y)\to(0,0)}{(x,y)\in S}} \frac{x^2+ay^2}{x^2+by^2}$$ where $a,b\in \mathbb{R}$ and
*
*$S=S_1=\{(x,y)\in\mathbb{R}^2 \mid |y| < c|x|^p\}$ with $c > 0, p > 1$
*$S=S_2=\{(x,y)\in\mathbb{R}^2 \mid |y| < ... | Set $$f(x,y)=\frac{x^2+ay^2}{x^2+by^2}$$ I don't think polar coordinates help.
For starters, the limit of $f(x,0)$ as $x\to 0$ is then $1$, so if the limit of $f$ exists for $(x,y)\to (0,0)$, it must be $1$.
For case 1, $|y|<c|x|^p=o(x)$ (i.e., $|y|$ goes to zero faster than $x$) in $S_1$. We have also $ay^2=o(x^2)$ a... | {
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"source": "stackexchange",
"question_score": "1",
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Decompose $\frac{x^4 + 5}{x^5 + 6x^3}$ (partial fraction decomposition)
Write out the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficients.
$$\frac{x^4 + 5}{x^5 + 6x^3}$$
So I factored the denominator to be $x^3(x^2+6)$ and here is the answer i got:
$\f... | Question 1:
$$
x^{4}+5=f(x)\left(x^{2}+6\right)+x^{3} g(x)
$$
Let $ \quad \displaystyle \frac{x^{4}+5}{x^{3}\left(x^{2}+6\right)} \equiv \frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x^{3}}+\frac{f(x)}{x^{2}+6}$, then
$$x^{4}+5 \equiv A x^{2}\left(x^{2}+6\right)+B x\left(x^{2}+6\right)+C\left(x^{2}+6\right)+x^{3} f(x) \cdots ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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About Factorization I have some issues understanding factorization.
If I have the expression $x^{2}-x-7$ then (I was told like this) I can put this expression equal to zero and then find the solutions with the quadratic formula, so it gives me $x_{0,1}= 1 \pm 2\sqrt{2}$ then $$x^{2}-x-7 = (x-1-2\sqrt{2})(x-1+2\sqrt{2})... | For equations of the form $ax^2 + bx + c$ you can use the quadratic equation $x = \frac{-b \pm \sqrt{b^2-4ac }}{2a}$ to help you factor, like you were doing. The adjustment you need to make is to multiply your factorization by a.
In the example you had, $3x^2 -x - 2$, you found the roots were x = 1 and x $= \frac{-2}{... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Find bases for kernel and image of T where $T: P_2 \to M_2$ T is defined as $$T:P_2(\mathbb R) \to M_2 (\mathbb R) \ \text{where} \ T(ax^2 +bx+c)=\begin{pmatrix}-2a +c & b+c\\-3b-3c&6a-3c\\ \end{pmatrix}$$ and I need t find bases for $ Im(T) $ and $ker(T)$
I started with $im(T)$ and got
$$Im(T) = \begin{pmatrix}-2a +c... | Let the basis of $P_2$ is $(1,x,x^2)$ and the basis of $M_2$ is $(a_{11}, a_{12}, a_{21}, a_{22})$
Where
$a_{11}= \begin{pmatrix}1&0 \\ 0&0 \end{pmatrix}$
$a_{12}= \begin{pmatrix}0&1 \\ 0&0 \end{pmatrix}$
$a_{21}= \begin{pmatrix}0&0 \\ 1&0 \end{pmatrix}$
$a_{22}= \begin{pmatrix}0&0 \\ 0&1 \end{pmatrix}$
Then
$T(... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\frac{1}{n+1} + \frac{1}{n+3}+\cdots+\frac{1}{3n-1}>\frac{1}{2}$ Without using Mathematical Induction, prove that $$\frac{1}{n+1} + \frac{1}{n+3}+\cdots+\frac{1}{3n-1}>\frac{1}{2}$$
I am unable to solve this problem and don't know where to start. Please help me to solve this problem using the laws of inequa... | Any statement that needs to be proved for all $n\in\mathbb{N}$, will need to make use of induction at some point.
We have
\begin{align}
S_n & = \sum_{k=1}^n \dfrac1{n+2k-1} = \dfrac12 \left(\sum_{k=1}^n \dfrac1{n+2k-1} + \underbrace{\sum_{k=1}^n \dfrac1{3n-2k+1}}_{\text{Reverse the sum}}\right)\\
& = \dfrac12 \sum_{k=1... | {
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"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Proving $\tan A=\frac{1-\cos B}{\sin B} \;\implies\; \tan 2A=\tan B$
If $\tan A=\dfrac{1-\cos B}{\sin B}$, prove that $\tan 2A=\tan B$.
My effort:
Here
$$\tan A=\frac{1-\cos B}{\sin B}$$
Now
$$\begin{align}\text{L.H.S.} &=\tan 2A \\[4pt]
&=\frac{2\tan A}{1-\tan ^2A} \\[6pt]
&=\frac{(2-2\cos B)\over\sin B}{1-\frac{(1... | Continuing from your last step,
$$
LHS=\frac{\frac{2-2\cos B}{\sin B}}{1-\frac{(1-\cos B)^2}{\sin^2B}}\\
=\frac{\frac{2-2\cos B}{\sin B}}{\frac{\sin^2B-(1-\cos B)^2}{\sin^2B}} \\
=\frac{2-2\cos B}{\frac{\sin^2B-1-\cos^2B+2\cos B}{\sin B}}
$$
and then use that $1=\sin^2+\cos^2B$ to get
$$
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Find $\lim_{n \rightarrow \infty}\frac{1}{n} \int_{1}^{\infty} \frac{\mathrm dx}{x^2 \log{(1+ \frac{x}{n})}}$ Find:
$$\lim_{n \rightarrow \infty} \frac{1}{n} \int_{1}^{\infty} \frac{\mathrm dx}{x^2 \log{(1+ \frac{x}{n})}}$$
The sequence $\frac{1}{nx^2 \log{(1+ \frac{x}{n})}}=\frac{1}{x^3 \frac{\log{(1+ \frac{x}{n})}}{\... | I showed in THIS ANSWER, using only Bernoulli's Inequality the sequence $\left(1+\frac xn\right)^n$ is monotonically increasing for $x>-n$.
Then, we can see that for $x\ge 1$ and $n\ge1$, the sequence $f_n(x)$ given by
$$f_n(x)=n\log\left(1+\frac xn\right)$$
is also monotonically increasing. Therefore, a suitable do... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Intergate $\int \frac{x}{(x^2-3x+17)^2}\ dx$
$$\int \frac{x}{(x^2-3x+17)^2}\ dx$$
My attempt:
$$\int \frac{x}{(x^2-3x+17)^2}\ dx=\int \frac{x}{\left((x-\frac{3}{2})^2+\frac{59}{4}\right)^2}\ dx$$
let $u=x-\frac{3}{2}$
$du=dx$
$$\int \frac{u+\frac{3}{2}}{\left((u)^2+\frac{59}{4}\right)^2}\ du$$
How can I continue from... | Notice, $$\int \frac{x}{(x^2-3x+17)^2}\ dx$$
$$=\frac{1}{2}\int \frac{(2x-3)+3}{(x^2-3x+17)^2}\ dx$$
$$=\frac{1}{2}\left(\int \frac{(2x-3)}{(x^2-3x+17)^2}\ dx+3\int \frac{1}{(x^2-3x+17)^2}\ dx\right)$$
$$=\frac{1}{2}\left(\int \frac{d(x^2-3x+17)}{(x^2-3x+17)^2}+3\int \frac{1}{\left(\left(x-\frac{3}{2}\right)^2+\frac{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How to prove $1^3+5^3+3^3=153,16^3+50^3+33^3=165033,166^3+500^3+333^3=166500333,\cdots$? I saw on facebook some image on which these identities that I am going to write below are labeled as "amazing math fact" and on the image there are these identities:
$1^3+5^3+3^3=153$
$16^3+50^3+33^3=165033$
$166^3+500^3+333^3=166... | This deserves a shorter proof. Call $x=10^{k+1}$. Then we have:
$$\begin{array}{rcl}166\ldots 666&=&\frac{x}{6}-\frac{2}{3}\\500\ldots 000&=&\frac{x}{2}\\333\ldots 333&=&\frac{x}{3}-\frac{1}{3}\end{array}$$
Now: the sum of the cubes on the left-hand side is:
$$\color{red}{166\ldots 666^3+500\ldots 000^3+333\ldots 333^3... | {
"language": "en",
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"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
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Find the limit of fraction involving logarithms I am looking for a way to prove the following limit for integer $x$s:
$$\lim_{x\to\infty}{\frac{\log(x+2)-\log(x+1)}{\log(x+2)-\log(x)}}=\frac{1}{2}$$
I could find the result by using a computer program but I cannot formally establish the above equality.
| We will use the fundamental limit $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1\tag{1}$$ We have
\begin{align}
L &= \lim_{x \to \infty}\frac{\log(x + 2) - \log(x + 1)}{\log(x + 2) - \log x}\notag\\
&= \lim_{x \to \infty}\dfrac{\log\left(\dfrac{x + 2}{x + 1}\right)}{\log\left(\dfrac{x + 2}{x}\right)}\notag\\
&= \lim_{x \to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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How to derive the equation of tangent to an arbitrarily point on a ellipse?
Show that the equation of a tangent in a point $P\left(x_0, y_0\right)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, could be written as: $$\frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1$$
I've tried implicit differentiation $\to \frac{2x... | We first differentiate the ellipse equation w.r.t. $x$ to find the slope of the tangent $m$,
$$\dfrac{2x_0}{a^2}+\dfrac{2y_0m}{b^2}=0$$
$$m=-\dfrac{x_0}{y_0}\cdot\dfrac{b^2}{a^2}$$
Thus, the tangent is given by
$$\color{darkgreen}{y-y_0}=\color{blue}{-}\dfrac{\color{blue}{x_0}}{\color{darkgreen}{y_0}}\cdot\dfrac{\colo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Numerical evaluation of first and second derivative We start with the following function $g: (0,\infty)\rightarrow [0,\infty)$,
$$ g(x)=x+2x^{-\frac{1}{2}}-3.$$ From this function we need a 'smooth' square-root. Thus, we check $g(1)=0$,
$$g'(x)=1-x^{-\frac{3}{2}},$$ $g'(1)=0$ and
$$g''(x)=\frac{3}{2}x^{-\frac{5}{2}}\... | Setting $x^{1/2} = 1 + u$ (the content of Octania's comment) gives, for $x > 0$,
$$
g(x) = x + 2x^{-1/2} - 3
= \frac{x^{3/2} - 3x^{1/2} + 2}{x^{1/2}}
= \frac{(1 + u)^{3} - 3(1 + u) + 2}{1 + u}
= \frac{u^{2} (3 + u)}{1 + u}.
$$
Your definition of $f$ amounts to
$$
f(x) = u \sqrt{\frac{3 + u}{1 + u}}
= u \sqrt{1 ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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An inequality with the sinus in a triangle. I have solved this problem in a way, rather "inspired". I would like to have a solution found an easier way but I was unable so far.
Let $A,B,C$ the angles of a triangle $\triangle {ABC}$; prove that
$$(\sin^2 A+\sin^2 B+\sin^2 C)\le \frac14 \left(\frac{1}{\sin A}+\frac{1}{\... | Wee need to prove that
$$\sum\limits_{cyc}\frac{4S^2}{b^2c^2}\leq\frac{1}{4}\sum\limits_{cyc}\frac{bc}{2S}\sum\limits_{cyc}\frac{2S}{bc}$$ or
$$\frac{(a^2+b^2+c^2)\sum\limits_{cyc}(2a^2b^2-a^4)}{a^2b^2c^2}\leq\frac{(ab+ac+bc)(a+b+c)}{abc}$$ or
$$abc(ab+ac+bc)\geq(a^2+b^2+c^2)(a+b-c)(a+c-b)(b+c-a)$$ or
$$\sum\limits_{cy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $a^5 + b^5 + c^5$
Suppose we have numbers $a,b,c$ which satisfy the equations
$$a+b+c=3,$$
$$a^2+b^2+c^2=5,$$
$$a^3+b^3+c^3=7.$$
How can I find $a^5 + b^5 + c^5$?
I assumed we are working in $\Bbb{C}[a,b,c]$. I found a reduced Gröbner basis $G$:
$$G = \langle a+b+c-3,b^2+bc+c^2-3b-3c+2,c^3-3c^2+2c+\frac... | You have to use Newton Identities. See https://en.wikipedia.org/wiki/Newton%27s_identities
In general if you have $n$ variables $x_1\ldots.x_n$, define the polynomials
$$p_k(x_1,\ldots,x_n)=\sum_{i=1}^nx_i^k = x_1^k+\cdots+x_n^k,$$
and
\begin{align}
e_0(x_1, \ldots, x_n) &= 1,\\
e_1(x_1, \ldots, x_n) &= x_1 + x_2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Show that $a_n = 1 + \frac{1}{2} + \frac{1}{3} +\dotsb+ \frac{1}{n}$ is not a Cauchy sequence
Let $$
a_n = 1 + \frac{1}{2} + \frac{1}{3} + \dotsb + \frac{1}{n}
\quad (n \in \mathbb{N}).
$$
Show that $a_n$ is not a Cauchy sequence even though
$$
\lim_{n \to \infty} a_{n+1} - a_n = 0
$$
(Therefore $a_n$ does no... | We use the definition of Cauchy sequence to show the sequence $(a_n)$ is not Cauchy.
Let $\epsilon=1/2$. We will show that there does not exist an $m$ such that for any $n\gt m$, we have $|a_n-a_m|\lt \epsilon$.
For let $m$ be given, and let $2^k$ be the smallest power of $2$ that is $\gt m$. Let $n=2^{k+1}-1$. Then
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Triangle angles.
For $\vartriangle$ABC it is given that $$\frac1{b+c}+\frac1{a+c}=\frac3{a+b+c}$$ Find the measure of angle $C$.
This is a "challenge problem" in my precalculus book that I was assigned. How do I find an angle from side lengths like this? I have tried everything I can. I think I may need to employ the... | Multiply both sides by $a+b+c$ to get $$\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}=3$$
We get
$$\frac{a}{b+c}+\frac{b}{a+c}=1$$
$$a(a+c)+b(b+c)=(a+c)(b+c)$$
$$a^2+ac+b^2+bc=ab+ac+bc+c^2$$
$$a^2+b^2=ab+c^2$$
$$a^2-ab+b^2=c^2$$
By the cosine law, we have $a^2-2\cos(C)ab+b^2=c^2$. Hence $\cos(C)=\frac12$, so $C=60^\circ$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1656778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
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Find the sum of all $abc$
Let $T$ be the set of all triplets $(a,b,c)$ of integers such that $1 \leq a < b < c \leq 6$. For each triplet $(a,b,c)$ in $T$ , take number $a \cdot b \cdot c$. Add all these numbers corresponding to all the triplets in $T$. Prove that the answer is divisible by $7$.
Attempt
Let $S = \disp... | A triple $(a, b, c)$ of integers such that $1 \leq a < b < c \leq 6$ is one of the $\binom{6}{3} = 20$ three-element subsets of the set $\{1, 2, 3, 4, 5, 6\}$ because once we choose a subset there is only one way of placing the elements in increasing order. They are
\begin{array}{c c}
(a, b, c) & abc\\
(1, 2, 3) & 6\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1657517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find all square numbers $n$ such that $f(n)$ is a square number Find all the square numbers $n$ such that $ f(n)=n^3+2n^2+2n+4$ is also a perfect square.
I have tried but I don't know how to proceed after factoring $f(n)$ into $(n+2)(n^2+2)$. Please help me.
Thanks.
| Your factoring seems to appear difficult to solve.
Instead say $n=t^2$.
If $t>1$, note that $f(t^2)=t^6+2t^4+2t^2+4$, and that $(t^3+t+1)^2=t^6+2t^4+t^2+2t^3+2t+1>f(t^2) >t^6+2t^4+t^2=(t^3+t)^2$ (since $2t^3-t^2+2t-3>0$ from here)
If $t<-1$, note that $(t^3+t-1)^2=t^6+2t^4+t^2-2t^3-2t+1>f(t^2) >t^6+2t^4+t^2=(t^3+t)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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If $x^2+y^2-xy-x-y+1=0$ ($x,y$ real) then calculate $x+y$ If $x^2+y^2-xy-x-y+1=0$ ($x,y$ real) then calculate $x+y$
Ideas for solution include factorizing the expression into a multiple of $x+y$ and expressing the left hand side as a sum of some perfect square expressions.
| We have
$$2(x^2+y^2-xy-x-y+1)=(x-y)^2+(x-1)^2+(y-1)^2.$$ The right-hand side (for real $x$ and $y$) is equal to $0$ if and only if $x=y=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Evaluate the limit $\lim_{x\to \infty}( \sqrt{4x^2+x}-2x)$
Evaluate :$$\lim_{x\to \infty} (\sqrt{4x^2+x}-2x)$$
$$\lim_{x\to \infty} (\sqrt{4x^2+x}-2x)=\lim_{x\to \infty} \left[(\sqrt{4x^2+x}-2x)\frac{\sqrt{4x^2+x}+2x}{\sqrt{4x^2+x}+2x}\right]=\lim_{x\to \infty}\frac{{4x^2+x}-4x^2}{\sqrt{4x^2+x}+2x}=\lim_{x\to \infty}... | With the substitution $x=1/t$ (under the unrestrictive condition that $x>0$) you get
$$
\lim_{x\to \infty}(\sqrt{4x^2+x}-2x)=
\lim_{t\to0^+}\left(\sqrt{\frac{4}{t^2}+\frac{1}{t}}-\frac{2}{t}\right)=
\lim_{t\to0^+}\frac{\sqrt{4+t}-2}{t}
$$
which is the derivative at $0$ of $f(t)=\sqrt{4+t}$; since
$$
f'(t)=\frac{1}{2\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1660120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2$ if $a+b+c=0$ Found this lovely identity the other day, and thought it was fun enough to share as a problem:
If $a+b+c=0$ then show $$\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2.$$
There are,... | If we remember the well-known formulas
$(x+y)^3 - (x^3 + y^3) = 3xy(x+y)$
$(x+y)^5 - (x^5 + y^5) = 5xy(x+y)(x^2 + xy + y^2)$
$(x+y)^7 - (x^7 + y^7) = 7xy(x+y)(x^2 + xy + y^2)^2$
then your identity becomes the observation that there are equal powers of the $xy(x+y)$ and $(x^2+xy+y^2)$ factors on both sides. (Under the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 7,
"answer_id": 3
} |
$\lim_{x\to 0} (2^{\tan x} - 2^{\sin x})/(x^2 \sin x)$ without l'Hopital's rule; how is my procedure wrong?
please explain why my procedure is wrong i am not able to find out??
I know the property limit of product is product of limits (provided limit exists and i think in this case limit exists for both the functions)... | @vim suggested Taylor expansions:
\begin{align}\lim _{x \to 0}\frac{2^{\tan x} - 2^{\sin x}}{x^2 \sin x}
&=\lim _{x \to 0}\frac{2^{x+\frac{x^3}{3}+O(x^5)}-2^{x-\frac{x^3}{6}+O(x^5)}}{x^2(x+\frac{x^3}{3}+O(x^5))}\\
&=\lim _{x \to 0}\frac{\left(2^{x+O(x^5)}\right)\left(2^{\frac{x^3}{3}}-2^{\frac{x^3}{6}}\right)}{x^3\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to evaluate this limit? I have a problem with this limit, I have no idea how to compute it. Can you explain the method and the steps used(without L'Hopital if is possible)? Thanks
$$\lim _{x\to 0+}\left(\frac{\left[\ln\left(\frac{5+x^2}{5+4x}\right)\right]^6\ln\left(\frac{5+x^2}{1+4x}\right)}{\sqrt{5x^{10}+x^{11}}-... | Expand the denominator
$$
\sqrt{5}x^5 (\sqrt{1+x/5}-1)\sim \sqrt{5}x^5\left(\frac{x}{10}-\frac{x^2}{200}+\ldots\right)
$$
and the first term of the numerator
$$
\ln (5+x^2)=\ln5+\ln(1+x^2/5)=\ln5+x^2/5-x^4/50+\ldots
$$
$$
\ln(5+4x)=\ln 5+\log(1+4x/5)\sim \ln 5+4x/5-8x^2/25+\ldots
$$
Therefore
$$
\ln \left(\frac{5+x^2}{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Integrate $I= \int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x$
$$I= \int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x$$
My Endeavour :
\begin{align}I&= \int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x\\ &= \int \frac{x}{\sqrt{1+ x^4}}\,\mathrm d x - \int \frac{1}{x\sqrt{1+ x^4}}\,\mathrm d x\end{align}
\begin{align}\te... | $$\int\frac {1}{\sqrt{z^2-1}}dz=\operatorname{arcosh}z+c$$ not $\operatorname{arcsin z}+c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Where does the normal to the graph of $y = \sqrt x$ at the point $(a, \sqrt a)$ intersect the $x$-axis?
Where does the normal to the graph of $y = \sqrt x$ at the point $(a, \sqrt a)$ intersect the $x$-axis?
I know how to find equation with numbers, but got really confused with this one. If anybody could break it dow... | \begin{align*}
y' &= \frac{1}{2\sqrt{x}} \\
\text{slope of the normal} &=-2\sqrt{a} \\
y-\sqrt{a} &= -2\sqrt{a}(x-a) \\
2\sqrt{a}x+y &= \sqrt{a}(2a+1) \\
\frac{x}{a+\frac{1}{2}}+\frac{y}{\sqrt{a}(2a+1)} &= 1 \\
x\text{-intercept} &= a+\frac{1}{2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of $\lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
$\bf{My\; Try::}$ Here $(x+1)\;,(x+2)\;,(x+3)\;,(x+4)\;,(x+5)>0\;,$ when $x\righ... | Hint:
Your method can be completed like this
$$HM \le GM \le AM \implies$$
$$\frac5{\frac1{x+1}+\frac1{x+2}+\frac1{x+3}+\frac1{x+4}+\frac1{x+5}} \le \sqrt[5]{(x+1)(x+2)(x+3)(x+4)(x+5)} \le x+3$$
Subtracting $x$ throughout :
$$\frac{15x^4+170x^3+675x^2+1096x+600}{5x^4+60x^3+255x^2+450x+274} \le \sqrt[5]{(x+1)(x+2)(x+3)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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The Sum of the series $\sum\limits_{n=0}^{\infty}\frac{1}{n^2+3}$ I know how to get the sum of geometric series, but otherwise.
How do I get the sum of this series? Thank you.
$$\sum\limits_{n=0}^{\infty}\frac{1}{n^2+3}$$
| The Weierstrass product for the sine function gives:
$$ \forall z\in\mathbb{C},\quad\sin(z) = z\prod_{n\geq 1}\left(1-\frac{z^2}{n^2 \pi^2}\right)\tag{1} $$
hence by replacing $z$ with $iz$ we get:
$$ \forall z\in\mathbb{C},\quad\sinh(z) = \frac{e^z-e^{-z}}{2} = z\prod_{n\geq 1}\left(1+\frac{z^2}{n^2 \pi^2}\right)\tag{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$(5 + \sqrt{2})(2-\sqrt{2})=(11-7\sqrt{2})(2+\sqrt{2})$ - Unique factorization? Personal question : We know that $5 + \sqrt{2}$, $2-\sqrt{2}$, $11-7\sqrt{2}$ and $2+\sqrt{2}$ are irreductible in $\mathbb{Z}[\sqrt{2}]$ and that $$(5 + \sqrt{2})(2-\sqrt{2})=(11-7\sqrt{2})(2+\sqrt{2}).$$ Why this fact doesn't contradict t... | Unique factorization only means unique factorization upto units.
The units in ${\mathbb Z}[\sqrt{2}]$ are the elements of the form $(1 + \sqrt{2})^n$ with $n \in {\mathbb Z}$.
For your two factorizations:
$$5 + \sqrt{2} = (3 + 2\sqrt{2}) (11 - 7 \sqrt{2}) = (1 + \sqrt{2})^2 (11 - 7 \sqrt{2})$$
so they differ by a unit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1673519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Integral $\int \frac{\sqrt{x}}{(x+1)^2}dx$ First, i used substitution $x=t^2$ then $dx=2tdt$ so this integral becomes $I=\int \frac{2t^2}{(1+t^2)^2}dt$ then i used partial fraction decomposition the following way:
$$\frac{2t^2}{(1+t^2)^2}= \frac{At+B}{(1+t^2)} + \frac{Ct + D}{(1+t^2)^2} \Rightarrow 2t^2=(At+B)(1+t^2)+C... | Notice, you can continue from here without using partial fractions $$\int \frac{2t^2}{(1+t^2)^2}\ dt$$
$$=\int \frac{2t^2}{t^4+2t^2+1}\ dt$$
$$=\int \frac{2}{t^2+\frac{1}{t^2}+2}\ dt$$
$$=\int \frac{\left(1+\frac{1}{t^2}\right)+\left(1-\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}+2}\ dt$$
$$=\int \frac{\left(1+\frac{1}{t^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1674188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $a^{-1}b^2a=b^3$ and $b^{-1}a^2b=a^3$ then $a=b=e$
Let $G$ be a group and $a,b\in G$ such that
$a^{-1}b^2a=b^3$ and $b^{-1}a^2b=a^3$. Then prove that $a=b=e$, where $e$ is the identity of group.
If we are able to prove $ab = ba$ then we are done. But i am unable to show that.
Can we use somewhere that $\gcd(2,... | The first equation $a^{-1}b^2a=b^3 \Rightarrow a^{-1}b^{2k}a=b^{3k}$ for all $k \in {\mathbb Z}$, so
$$a^{-2}b^{-4}a^2 = a^{-1}b^{-6}a = b^{-9}.$$
The second equation gives $a^{-2}b^{-1}a^2=ab^{-1}$, so
$$a^{-2}b^{-4}a^2 = (ab^{-1})^4$$ and hence
$$b^{-9} = (ab^{-1})^4.$$
Since $b^9$ commutes with $b$ and $(ab^{-1})^4$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1676948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The indefinite integral $\int x^{13}\sqrt{x^7+1}dx$ I used integration by parts technique to solve this problem
so $u = \sqrt{x^7+1}$ ,$du = \frac{1}{2}\frac{7x^6}{\sqrt{x^7+1}}dx$
$v =\frac{x^{14}}{14}$ $dv=x^{13}dx$
then it becomes
$\frac{x^{14}}{14}\sqrt{x^7+1}-\int\frac{x^{14}}{28}\frac{7x^6}{\sqrt{x^7+1}}dx$
and... | The integrand is a special case of the more general form $$f(x;m,n,a) = x^{2n-1} (x^n + a)^{1/m}.$$ We observe that the choice $$u^m = x^n + a, \quad m u^{m-1} \, du = n x ^{n-1} \, dx$$ along with $x^{2n-1} = \frac{1}{n} x^n (nx^{n-1})$ gives $$\int f(x;m,n,a) \, dx = \frac{1}{n} \int (u^m - a) u \cdot mu^{m-1} \, du... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1679142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Probability of sum to be divisible by 7 6 fair dice are thrown simultaneously. What is the probability that the sum of the numbers appeared on dice is divisible by 7 ?
| After $n$ rolls, let $p_n$ be the probability that the sum is a multiple of $7$ after $n$ rolls.
Then we get the recursion: $$p_{n+1} = \frac{1}{6}(1-p_n)$$
That's because to get a multiple at roll $n+1$, you have to get a non-multiple of $7$ at roll $n$, and then exactly the right value.
Then start with $p_0=1$, and w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1680976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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2,3,7 is the only triple $\geq 2$ that satisfies this property The property being that "taking the product of two of the numbers and adding one yields a number that is divisible by the third".
Clearly, this holds for 2,3,7 since
*
*$2\cdot 3 + 1 = 1\cdot 7$
*$2\cdot7 + 1 = 5\cdot 3$
*$3\cdot 7 + 1 = 11\cdot 2$
b... | Equivalently $a$ divides $bc+1$ and cyclic ones. In particular, they are pairwise coprime. Then $abc$ divides $\prod_{\mathrm{cyc}}(ab+1)$, implying that $abc$ divides $1+\sum_{\mathrm{cyc}}ab$. Suppose wlog $a>b>c\ge 2$. If $c\ge 3$ then
$$
3ab \le abc \le 1+\sum_{\mathrm{cyc}}ab\le 1+(ab-1)+(ab-1)+ab<3ab,
$$
which i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Establish the identity $\frac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} = \sec\theta \cot\theta\$ Establish the identity:
$$\dfrac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} = \sec\theta \cot\theta$$
The first step I got was:
$$\sec\theta \cot\theta = \dfrac{\sec\theta \cot\theta\,\big(\cos\theta + \tan\... | On the one hand
\begin{align}
\dfrac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} &=
\dfrac{\dfrac{\cos\theta}{\sin\theta} + \dfrac{1}{\cos\theta}}{\cos\theta + \dfrac{\sin\theta}{\cos\theta}} =
\dfrac{\cos^2\theta+\sin\theta}{\sin\theta\cos\theta} \div\dfrac{\cos^2\theta+\sin\theta}{\cos\theta}
\\ &=
\dfrac{\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How to think about negative infinity in this limit $\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + 1}$
Question:
calculate:
$$\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + 1}$$
Attempt at a solution:
This can be written as:
$$\lim_{x \to -\infty} \frac{3 + \frac{1}{x}}{\sqrt{1 + \frac{3}{x}} + \sqrt{1 + \fra... | In fact:
$\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + 1}$
$ = \lim_{x \to -\infty} \frac{x*(3 + \frac{1}{x})}{(-x)(\sqrt{1 + \frac{3}{x}} + \sqrt{1 + \frac{1}{x^2}})}$
$ = \lim_{x \to -\infty} \frac{-3 - \frac{1}{x}}{\sqrt{1 + \frac{3}{x}} + \sqrt{1 + \frac{1}{x^2}}} = -\frac{3}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Integral $\int \sqrt{\frac{x}{2-x}}dx$ $$\int \sqrt{\frac{x}{2-x}}dx$$
can be written as:
$$\int x^{\frac{1}{2}}(2-x)^{\frac{-1}{2}}dx.$$
there is a formula that says that if we have the integral of the following type:
$$\int x^m(a+bx^n)^p dx,$$
then:
*
*If $p \in \mathbb{Z}$ we simply use binomial expansion, other... |
$$\int \sqrt{\frac{x}{2-x}}dx$$
Set $t=\frac {x} {2-x}$ and $dt=\left(\frac{x}{(2-x)^2}+\frac{1}{2-x}\right)dx$
$$=2\int\frac{\sqrt t}{(t+1)^2}dt$$
Set $\nu=\sqrt t$ and $d\nu=\frac{dt}{2\sqrt t}$
$$=4\int\frac{\nu^2}{(\nu^2+1)^2}d\nu\overset{\text{ partial fractions}}{=}4\int\frac{d\nu}{\nu^2+1}-4\int\frac{d\nu}{(\n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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What's the formula for this series for $\pi$? These continued fractions for $\pi$ were given here,
$$\small
\pi = \cfrac{4} {1+\cfrac{1^2} {2+\cfrac{3^2} {2+\cfrac{5^2} {2+\ddots}}}}
= \sum_{n=0}^\infty \frac{4(-1)^n}{2n+1}
= \frac{4}{1} - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} + \cdots\tag1
$$
$$\small
\pi = 3 + \cf... | Given the symmetric continued fraction found in this post
$$\frac{\displaystyle\Gamma\left(\frac{a+3b}{4(a+b)}\right)\Gamma\left(\frac{3a+b}{4(a+b)}\right)}{\displaystyle\Gamma\left(\frac{3a+5b}{4(a+b)}\right)\Gamma\left(\frac{5a+3b}{4(a+b)}\right)}=\cfrac{4(a+b)}{a+b+\cfrac{(2a)(2b)} {3(a+b)+\cfrac{(3a+b)(a+3b)}{5(a+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1689040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 0
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Proof of $Δv = c \cdot \tanh \left ( \frac{u}{c}\ln \left( \frac{m_0}{m_1}\right) \right) $
Question: The relativistic rocket equation is given below
$$ \frac{m_0}{m_1} = \left (\frac{1+\frac{Δv}{c}}{1-\frac{Δv}{c}} \right)^{\frac{c}{2u}} $$
Show that the equation rearranges to
$$ Δv = c \cdot \tanh \left ( \frac... | We are given the expression
$$\frac{m_0}{m_1}=\left(\frac{1+\frac{\Delta v}{c}}{1-\frac{\Delta v}{c}}\right)^{c/2u} \tag 1$$
Raising each side of $(1)$ to the $2u/c$ power yields
$$\left(\frac{m_0}{m_1}\right)^{2u/c}=\frac{1+\frac{\Delta v}{c}}{1-\frac{\Delta v}{c}}$$
whereupon solving for $\Delta v/c$ we find that
$$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove that $\sin(x) + \cos(x) \geq 1$ $\forall x\in[0,\pi/2]: \sin{x}+\cos{x} \ge 1.$
I am really bad at trigonometric functions, how could I prove it?
| Re-arrange to get $\cos(x) < 1 - \sin(x)$
We know that $\sin^2 (x) + \cos^2(x) = 1$ and therefor that $|\cos(x)| = \sqrt{1-\sin^2(x)}$
Now we want to know if $\sqrt{1-\sin^2(x)} < 1 - \sin(x)$
Now we square both sides: $1-\sin^2(x) < (1 - \sin(x))^2 = 1 - 2\sin(x) + \sin^2(x)$
Re-arranging again we get $0 < 2\sin^2(x) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1690381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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indefinite integration $ \int \frac { x^2 dx} {x^4 + x^2 -2}$ problem : $ \int \frac { x^2 dx} {x^4 + x^2 -2}$
solution : divide numerator and denominator by $x^2$
$ \int \frac { dx} {x^2 + 1 -\frac{1}{x^2}}$
Now whats the next step $?$
Am I doing right $?$
| HINT:
Setting $x^2=y,$
$$A=\frac{x^2}{x^4+x^2-2}=\dfrac y{y^2+y-2}=\dfrac y{(y+2)(y-1)}$$
As $y+2+2(y-1)=3y$
$$3A=\dfrac{y+2+2(y-1)}{(y+2)(y-1)}=\dfrac1{y-1}+\dfrac2{y+2}$$
Replace $y$ with $x^2$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving for $k$ when $\arg\left(\frac{z_1^kz_2}{2i}\right)=\pi$ Consider $$|z|=|z-3i|$$
We know that if $z=a+bi\Rightarrow b=\frac{3}{2}$
$z_1$ and $z_2$ will represent two possible values of $z$ such that $|z|=3$. We are given $\arg(z_1)=\frac{\pi}{6}$
The value of $k$ must be found assuming $\arg\left(\frac{z_1^kz_2... | $$z_1=\frac{3\sqrt{3}}{2}+\frac{3}{2}i$$
and
$$z_2=-\frac{3\sqrt{3}}{2}+\frac{3}{2}i$$
Thus,
$$z_1=3e^{\frac{i\pi}{6}}$$ and $$z_2=3e^{\frac{5i\pi}{6}}$$
$$i=e^{\frac{i\pi}{2}}$$
$$\arg\left(\frac{z_1^kz_2}{2i}\right)=\arg\left(\frac{3^{k+1}}{2}\frac{e^{\frac{ki\pi}{6}}e^{\frac{5i\pi}{6}}}{e^{\frac{i\pi}{2}}}\right)=... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the expected value of the largest of the three dice rolls?
You toss a fair die three times. What is the expected value of the largest of the three outcomes?
My approach is the following:
calculate the probability of outcome when $\max=6$, which is
$$P(\text{at least one $6$ of the three rolls}) = 1 - P(\text{... | Picture the cube of possible outcomes. The cells that represent a maximum of $6$ lie in a greedy half of the outer layer, which has $6^3-5^3=216-125=91$ cells in it.
The next layer represents max $5$, and has $5^3-4^3=125-64=61$ cells in it.
We can proceed in a similar manner and arrive at the sum of the whole cube:
$$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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Trigonometry conversion rules, why this way? We have domain $\,[0, 2\pi]\,$ and the following functions are given:
$$f(x)=\cos(2x) \text{ and } g(x)=\sin(x-\pi/3)$$Solve exactly: $\,f(x)=g(x)$
Why does one solve:
(right way)
$$\cos(2x)=\sin(x-\pi/3)\\\cos(2x)=\cos(\pi/2-(x-\pi/3))\\ \text{etc.}\ldots
$$
and not (which ... | You can choose to transform the sine into cosine or conversely.
First method
\begin{gather}
\cos2x=\cos\left(\frac{\pi}{2}-\left(x-\frac{\pi}{3}\right)\right)
\\[6px]
\cos2x=\cos\left(\frac{5\pi}{6}-x\right)
\\[6px]
2x=\frac{5\pi}{6}-x+2k\pi
\qquad\text{or}\qquad
2x=-\frac{5\pi}{6}+x+2k\pi
\\[6px]
x=\frac{5\pi}{18}+k\f... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to rearrange this? Can you please help me to derive the second equation from the first one? I'd really appreciate your help!
$$
-\frac{1}{w-s-d} + \frac{1+r}{1+z} \cdot \frac{1}{(1+r)s+(1+n)d} = 0 \tag{1}
$$
$$
(2 + z)(1 + r)s = (1 + r)(w - d)-(1 + z)(1 + n)d \tag{2}
$$
Many thanks!
| \begin{align}
-\frac{1}{w-s-d}+\frac{1+r}{1+z}\cdot \frac{1}{(1+r)s + (1+n)d} &= 0\\
\frac{1+r}{1+z}\cdot \frac{1}{(1+r)s + (1+n)d}&= \frac{1}{w-s-d}\\
\frac{1}{1+z}\cdot \frac{1}{(1+r)s + (1+n)d}&=\frac{1}{(w-s-d)(1+r)}\\
(1+z)[(1+r)s+(1+n)d] &= (w-s-d)(1+r)\\
(1+z)[(1+r)s+(1+n)d]+s(1+r) &= (w-d)(1+r)\\
(1+z)(1+r)s+s(... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve $ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $
Question: Solve $$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 0 \le x \le 360^{\circ} $$
My attempt:
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 6(\frac{1}{2} - \frac{\cos(2x)}{2}) + \sin(x)\cos(x) -(\frac{1}{2} + \frac{\cos(2x)}{2}) = 5 $$
$$ 3 - 3\cos(2x)+ \s... | It was a good work arriving to $$7\cos(2x) - \sin(2x) + 5 = 0$$ But the substitution to be used was just $t=\tan(\frac{2x}2)=\tan(x)$ which leads to the quadratic $$t^2+t-6=0$$ leading to $$(t-2)(t+3)=0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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If a, b, c are three natural numbers with $\gcd(a,b,c) = 1$ such that $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$ then show that $a+b$ is a square.
If a, b, c are three natural numbers with $\gcd(a,b,c) = 1$ such that $$\frac{1}{a} + \frac{1}{b}= \frac{1}{c}$$ then show that $a+b$ is a perfect square.
This can be simplifie... | Rewrite as $(a-c)(b-c)=c^2$. First we show that $a-c$ and $b-c$ are relatively prime. Suppose to the contrary that the prime $p$ divides $a-c$ and $b-c$. Then $p$ divides $c$ and therefore $a$ and $b$, contradicting the fact that $\gcd(a,b,c)=1$.
Since $a-c$ and $b-c$ are relatively prime, it follows that $a-c=s^2$ and... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Expansion and factorisation I have a little problems with a few questions here and I need help.. Thanks ...
*
*Factorise completely
$$9x^4 - 4x^2 - 9x^2y^2 + 4y^2 $$
My workings ..
$$ (3x^2+2x)(3x^2-2x) - y^2 (9x^2-4) = (3x^2 + 2x)(3x^2 -2x) - y^2 (3x+2)(3x-2) $$
*Factorise $3x^2 + 11x - 20$ and , hence Factori... | HINTS :
For 1.
The coefficients are $\pm 9,\pm 4$, so
$$9x^4-4x^2-9x^2y^2+4y^2=(9x^4-9x^2y^2)-(4x^2-4y^2)$$
is worth trying.
For 2.
Compare $3x^2+11x-20$ with $$11a-11b-20+3a^2+3b^2-6ab=11(a-b)-20+3(?)=3(?)+11(a-b)-20$$
For 3.
Note that $312=78\times 4$ and that $1002=2\times 501$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$ $=$ $\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}$
Question: Prove $\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$ $=$ $\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}$
My attempt:
$$\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$$
$$ \frac{\frac {1}{\cos(x)} - \frac{1}{\s... | Take the right hand side:
$$\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}=\frac{\frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}}{\frac{1}{\cos(x)} + \frac{1}{\sin(x)}} = \frac{\sin^2x+\cos^2x}{\sin x+\cos x}=\frac{1}{\sin x+\cos x} $$
and this is equal to the expression you found for the left hand side.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Unknown Inequality $$ \left( \sqrt{3}\sqrt{y(x+y+z)}+\sqrt{xz}\right)\left( \sqrt{3}\sqrt{x(x+y+z)}+\sqrt{yz}\right)\left( \sqrt{3}\sqrt{z(x+y+z)}+\sqrt{xy}\right) \leq 8(y+x)(x+z)(y+z)$$
I can prove this inequality, but i need know if this inequaliy is known...
| Here is a nice and simple proof (for $x,y,z\ge 0$ of course):
By Cauchy-Schwarz inequality:
$$\left( \sqrt{3}\sqrt{y(x+y+z)}+\sqrt{xz}\right)^2 \le (3+1)\big[(y(x+y+z) + xz\big] = 4(x+y)(y+z).$$
Similarly:
$$\left( \sqrt{3}\sqrt{x(x+y+z)}+\sqrt{yz}\right)^2 \le 4(x+z)(x+y)$$
$$\left( \sqrt{3}\sqrt{z(x+y+z)}+\sqrt{xy}\r... | {
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"timestamp": "2023-03-29T00:00:00",
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Let $x,y,z>0$ and $x+y+z=1$, then find the least value of ${{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}}$ Let $x,y,z>0$ and $x+y+z=1$, then find the least value of
$${{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}}$$
I tried various ways of rearranging and using AM > GM inequality. But I couldn't get it. I am n... | Using AM-GM inequality (for $n=6$):
Let $f(x,y,z)=\frac{x}{2-x}+\frac{y}{2-y}+\frac{z}{2-z}$. Take $x_1=\alpha(2-x), x_2=\alpha(2-y), x_3=\alpha(2-z), x_4=\frac{2}{(2-x)}, x_6=\frac{2}{(2-y)}+\frac{2}{(2-z)}$, where $\alpha=\frac{18}{25}$. Then by AM-GM inequality
$\frac{1}{6}\left(5\alpha+\frac{2}{(2-x)}+\frac{2}{(2-y... | {
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"url": "https://math.stackexchange.com/questions/1708395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Determine whether $p_n$ is decreasing or increasing, if $p_{n+1} = \frac{p_n}{2} + \frac{1}{p_n}$ If $p_1 = 2$ and $p_{n+1} = \frac{p_n}{2}+ \frac{1}{p_n}$, determine $p_n$ is decreasing or increasing.
Here are the first few terms:
$$p_2 = \frac{3}{2}, p_3 = \frac{3}{4} + \frac{2}{3} = \frac{17}{12}, p_4 = \frac{17}{24... | The sequence appears to be decreasing to $\sqrt2$. (It's the Newton-Raphson iteration for computing $\sqrt 2$ as the root of the function $f(x):=x^2-2$.) So show this in two steps:
(1) First prove by induction that $p_n\ge \sqrt2$ for every $n$. This follows from
$$p_{n+1}-\sqrt 2=\left({p_n\over2} +\frac1{p_n}\right)-... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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sum of the series $\frac{2^n+3^n}{6^n}$ from $n=1$ to $\infty$ Find the sum of the series $\sum_{n=1}^{\infty} \frac{2^n+3^n}{6^n}=?$
My thoughts: find $\sum_{n=1}^{\infty} 2^n$, $\sum_{n=1}^{\infty} 3^n$ and $\sum_{n=1}^{\infty} 6^n$ (although I don't know how yet...)
Then, $\sum_{n=1}^{\infty} \frac{2^n+3^n}{6^n}= \f... | Hint:
$$
\sum_{n=1}^{\infty} \frac{2^n+3^n}{6^n}
=\sum_{n=1}^{\infty} \frac{2^n}{6^n}
+\sum_{n=1}^{\infty} \frac{3^n}{6^n}
=\sum_{n=1}^{\infty} \left(\frac{2}{6}\right)^n
+\sum_{n=1}^{\infty} \left(\frac{3}{6}\right)^n
$$
| {
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Prove: $\csc a +\cot a = \cot\frac{a}{2}$
Prove: $$\csc a + \cot a = \cot\frac{a}{2}$$
All I have right now, from trig identities, is
$$\frac{1}{\sin a} + \frac{1}{\tan a} = \frac{1}{\tan(a/2)}$$
Where do I go from there?
| $\csc a+\cot a=\cot\frac{a}2$
L.H.S.
$\implies\large \frac{1}{\sin a}+\frac{\cos a}{\sin a} \\
\implies \large\frac{1+\cos a}{\sin a}\\
\implies \large\frac{1+2\cos^2\frac{a}{2}-1}{\sin a}\\
\implies \large\frac{2\cos^2\frac{a}{2}}{2\sin\frac{a}2\cos\frac{a}{2}}\\
\implies \cot \frac{a}{2}=\text{R.H.S}$
| {
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Probability Problem - Finding a pdf Below is a problem I did. However, I did not come up with the answer in book.
I am thinking that I might have the wrong limits for the integral. I am hoping
somebody can point out what I did wrong.
Bob
Problem
Let $Y = \sin X$, where $X$ is uniformly distributed over $(0, 2 \pi)$. Fi... | If $X$ is uniform in $(0,2\pi)$, then has cdf
$$
F_X(x)=\begin{cases}
0 & x\le 0\\
\frac{x}{2\pi} & x\in (0,2\pi)\\
1 & x\ge 2\pi
\end{cases}
$$
The random variable $Y=\sin (X)$ takes values on $(-1,1)$. Hence, $\Bbb P(Y\le y) = 0$ for $y \le -1$ and $\Bbb P(Y\le y) = 1$ for $y \ge 1$. Let now $y \in (-1, 1)$. We ... | {
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"timestamp": "2023-03-29T00:00:00",
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For each bounded subset in $\mathbb{C}$, $\lim_{n\rightarrow \infty}(1+x/n)^n=e^x$ uniformly For this question, it is defined that $e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$, $\forall x\in\mathbb{C}$.
Though I know why $\lim_{n\rightarrow \infty}(1+x/n)^n=e^x$ pointwise for any real $x$, I feel difficult to show that the ... | Consider the power series for $\log$ which converges in the complex plane if $|z| < 1:$
$$\log(1+z) = z - \frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} + \ldots.$$
Hence, if $|z| < n$ we have
$$\log(1+z/n) = \frac{z}{n} - \frac{z^2}{2n^2} + \frac{z^3}{3n^3} - \frac{z^4}{4n^4} + \ldots,$$
and, using the triangle inequal... | {
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"url": "https://math.stackexchange.com/questions/1714973",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Confusion about nth root of a number Let's say we want to find the 6th roots of 64. Then according to the method from my textbook:
$z^6=64$
$z^6=64+0i=64[\cos(0)+i\sin(0)]=64cis(0)=64cis(0+2k\pi)$
Then by De Moivre's theorem:
$z=\color{red}{64^{\frac{1}{6}}}cis(\frac{2k\pi}{6})$ $,k=0, 1, 2, 3, 4, 5$
$z=\color{red}{2}... | Notice:
$$z^6=64\Longleftrightarrow z^6=|64|e^{\arg(64)i}\Longleftrightarrow z^6=64e^{0i}\Longleftrightarrow$$
$$z=\left(64e^{2\pi ki}\right)^{\frac{1}{6}}\Longleftrightarrow z=2e^{\frac{\pi ki}{3}}$$
With $k\in\mathbb{Z}$ and $k:0-5$
So, the solutions are:
$$z_0=2e^{\frac{\pi\cdot0i}{3}}=2$$
$$z_1=2e^{\frac{\pi\cdot1i... | {
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On Ramanujan's curious equality for $\sqrt{2\,(1-3^{-2})(1-7^{-2})(1-11^{-2})\cdots} $ In Ramanujan's Notebooks, Vol IV, p.20, there is the rather curious relation for primes of form $4n-1$,
$$\sqrt{2\,\Big(1-\frac{1}{3^2}\Big) \Big(1-\frac{1}{7^2}\Big)\Big(1-\frac{1}{11^2}\Big)\Big(1-\frac{1}{19^2}\Big)} = \Big(1+\fra... | It can be transformed to next equation.
$$\displaystyle m=\frac{n^2}{n^2-1}\frac{a+1}{a-1}\frac{b+1}{b-1}\frac{c+1}{c-1}$$
so, function m decrease monotonously ,when all values are bigger than 2. If whatever m is, this equation has only finite solutions(when all are integer), $2\leq m\leq12(n=2,a=2,b=3,c=5)$.
| {
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Find the remainder of $9^2\cdot 13\cdot 21^2$ when divided by $4$
Find the remainder of $9^2\cdot 13\cdot 21^2$ when divided by $4$
How should I approach this type of questions?
Without calculator of course
I did this:
$9^2\cdot 13\cdot 21^2=81\cdot 13\cdot 441=81\cdot 5733=464,373=33\bmod 4=1 \bmod 4$
| $9^2*13*21^2$ is not a multiple of 4 or two.
Hence it cannot be 0 or 2 (mod 4)
$9^2*13*21^2$ can be written as
$9^2*21^2*(3^2+2^2) =$
$(9*21*3)^2 + (9*21*2)^2$
A sum of squares cannot give 3 (mod 4)
Hence it is 1 (mod 4)
| {
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How do I factorize this numerator? $$\lim_{x\to -3} \frac 1{x+3} + \frac 4{x^2+2x-3}$$
I have the solution I just need to know how i turn that into:
$$\frac {(x-1)+4}{(x+3)(x-1)}$$
I know this might be really simple but I'm not sure how to factorise the numerator.
Thanks in advance!
| You have
$$ x^2+2x-3 = (x-1)(x+3),$$
so
$$\frac{1}{x+3}+\frac{4}{x^2+2x-3}=\frac{1}{x+3}+\frac{4}{(x-1)(x+3)}=\frac{1(x-1) +4}{(x-1)(x+3)} $$
| {
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Solving the following trigonometric equation: $\sin x + \cos x = \frac{1}{3} $ I have to solve the following equation:
$$\sin x + \cos x = \dfrac{1}{3} $$
I use the following substitution:
$$\sin^2 x + \cos^2 x = 1 \longrightarrow \sin x = \sqrt{1-\cos^2 x}$$
And by operating, I obtain:
$$ \sqrt{(1-\cos^2 x)} = \dfrac... | Another way to solving this by the tangent half angle substitution $t = \tan \frac{x}{2}$ (or $x = 2 \arctan t$)
$$ \begin{align} \cos x & \rightarrow \frac{1-t^2}{1+t^2} \\
\sin x & \rightarrow \frac{2 t}{1+t^2} \end{align} $$
Your equation becomes
$$ \frac{2 t}{1+t^2} + \frac{1-t^2}{1+t^2} = \frac{1}{3} $$ which ha... | {
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Determine the point of intersection between $f(x) = x^2$ and its normal in the point $(a, a^2)$ Determine the point of intersection between $f(x) = x^2$ and its normal in the point $(a, a^2)$
Answer:
This should be easy enough...
$f'(x) = 2x$
The tangent line in the point $(a, a^2)$ is $y - a^2 = 2 (x - a) \rightarrow ... | The tangent slope of $f(x)$ at $x = a$ is $f'(a) = 2a$, so you will need to amend the tangent and normal line accordingly.
Normal line is $y - a^2 = -\dfrac{1}{2a}(x - a)$ or $y = -\dfrac{x}{2a}+\dfrac{1}{2}+a^2$.
Set $f(x) = y$, so $x^2 = -\dfrac{x}{2a}+\dfrac{1}{2}+a^2$ or $2ax^2 + x - (a + 2a^3) = 0$. Finish with th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1723628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can the system of equations be extracted from its solution? While I was solving the secondary school exam of 2014 I came across a question that states:
After solving those equations: $a_{1}x + b_{1}y = c_{1}$ and $a_{2}x + b_{2}y = c_{2}$, we found that x = $\frac{-7}{\begin{vmatrix}
3 & 1 \\
1 & -2
\end{vmatrix}}... | Presumably the denominator $\left|\matrix{3 & 1\cr 1 & -2}\right|$ indicates that the coefficient matrix $\pmatrix{a_1 & b_1\cr a_2 & b_2} = \pmatrix{3 & 1\cr 1 & -2}$. Then we have
$$ \pmatrix{c_1 \cr c_2} = \pmatrix{3 & 1\cr 1 & -2} \pmatrix{x\cr y\cr} = \pmatrix{3 & 1\cr 1 & -2} \pmatrix{1\cr 3\cr} = \pmatrix{6\cr ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1723888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is $\arctan(x) + \arctan(y)$ I know $$g(x) = \arctan(x)+\arctan(y) = \arctan\left(\frac{x+y}{1-xy}\right)$$
which follows from the formula for $\tan(x+y)$. But my question is that my book defines it to be domain specific, by which I mean, has different definitions for different domains:
$$g(x) = \begin{cases}\arct... | The fact, that $\tan x = \theta$ only implies $x = \arctan \theta + k \pi$, makes the fomular of $g(x)$ a little complicated.
First proof.
There are many good answers pointed by KonKan. The following one is based on the Lagrange mean value theorem. It is not the shortest proof, but the idea is easy.
Denote
$$F(y) = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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Find the sum $\sum _{n=1}^{\infty }\left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)$ $$\sum _{n=1}^{\infty }\left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)$$
On their own, all three are divergent, so I thought the best way would be to rewrite it as:
$$\frac{\sqrt{n+2}-2\sqrt{n+1}-\sqrt{n}}{\sqrt{n+2}-2\sqrt{n+1}-\sqrt{n}}\cdo... | $$\sqrt{n+2}-2\sqrt{n+1}+\sqrt n=\sqrt{n+2}-\sqrt{n+1}-(\sqrt{n+1}-\sqrt n)$$ $$=\dfrac1{\sqrt{n+2}+\sqrt{n+1}}-\dfrac1{\sqrt{n+1}+\sqrt n}=F(n+1)-F(n)$$
where $F(m)=\dfrac1{\sqrt{m+1}+\sqrt m}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Real solutions of $x^5+ax^4+bx^3+cx^2+dx+e=0$ If $2a^2＜5b$,prove that the equation $x^5+ax^4+bx^3+cx^2+dx+e = 0$
has at least one complex root。Thanks.
| I am assuming all coefficients are real.
If
$f(x)
= x^5+ax^4+bx^3+cx^2+dx+e
$,
then
$f'(x)
= 5x^4+4ax^3+3bx^2+2cx+d
$,
$f''(x)
= 20x^3+12ax^2+6bx+2c
$,
$f'''(x)
= 60x^2+24ax+6b
$.
If $f$ has 5 real roots,
then $f'$ has 4 real roots,
$f''$ has 3 real roots,
and
$f'''$ has 2 real roots.
The roots of $f'''$
are the roots ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is $2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots\infty$ equal to? I came across this question while doing my homework:
$$\Large 2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots\infty=?$$
$$\small\text{OR}$$
$$\large\prod\limits_{x=1}^{\... | The only thing left that we have to do is to evaluate the following infinite sum:
$$\sum\limits_{x=1}^{\infty} \frac{x}{2^{x+1}}$$
Dividing by $2$ would gives us $$\sum\limits_{x=1}^{\infty} \frac{x}{2^{x+2}}=\sum\limits_{x=2}^{\infty} \frac{x-1}{2^{x+1}}$$
Now, subtract this from the original equation. The limit shoul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1727174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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If $n = a^2 + b^2 + c^2$ for positive integers $a$, $b$,$c$, show that there exist positive integers $x$, $y$, $z$ such that $n^2 = x^2 + y^2 + z^2$.
If $n = a^2 + b^2 + c^2$ for positive integers $a$, $b$,$c$, show that there exist positive integers $x$, $y$, $z$ such that $n^2 = x^2 + y^2 + z^2$.
I feel that the pr... | If:
$$x=2ac$$
$$y=2bc$$
$$z=a^2+b^2-c^2$$
$$n=a^2+b^2+c^2$$
Then:
$$x^2+y^2+z^2=n^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How to compute this double integral I'm trying to show that $\int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}dxdy \neq \int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}dydx$ by computing these integrals directly.
I tried using polar coordinates with no success as the bounds of integration caused problems.
I also tried the subst... | Put $$f(x,y) = \arctan\left( \tfrac x y \right), \,\,\,\,\,\, (x,y) \in (0,1).$$ We see $$\frac{\partial f}{\partial y} = \frac{1}{1+\left(\tfrac x y \right)^2}\left( - \frac x {y^2} \right) = -\frac{x}{x^2 + y^2}.$$ Then $$\frac{\partial }{\partial x}\frac{\partial f}{\partial y} = - \frac{1}{x^2 + y^2} + \frac{x}{(x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1731441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find limit $\lim_{n \to \infty} n \left(n^2\log\left(1+\frac{1}{n^2}\right) + n \log\left(1-\frac{1}{n}\right)\right) $ Define $a_n$ as:
$$
a_n = \left(\left(1+\frac{1}{n^2}\right)^{n^2}\left(1-\frac{1}{n}\right)^n\left(1+\frac{1}{n}\right)\right)^n
$$
Now I want to calculate $\lim_{n \to \infty} a_n$. So, real questio... | Let $f_n$ be the sequence given by $f_n=n \left(n^2\log\left(1+\frac{1}{n^2}\right) + n \log\left(1-\frac{1}{n}\right)\right)$. We can evaluate the limit of $f_n$ by expanding the logarithm functions embedded in the limit of interest as
$$\begin{align}
\lim_{n\to \infty}f_n&=\lim_{n \to \infty} n \left(n^2\log\left(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1732579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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An inequality involving $\frac{x^3+y^3+z^3}{(x+y+z)(x^2+y^2+z^2)}$
$$\frac{x^3+y^3+z^3}{(x+y+z)(x^2+y^2+z^2)}$$
Let $(x, y, z)$ be non-negative real numbers such that $x^2+y^2+z^2=2(xy+yz+zx)$.
Question: Find the maximum value of the expression above.
My attempt:
Since $(x,y,z)$ can be non-negative, we can take $x=0$... | Let $P$ be the given expression we want to maximise.
From the hypothesis, we get $xy+yz+zx=\dfrac{1}{4}(x+y+z)^2$
So, $P=\dfrac{x^3+y^3+z^3}{2(x+y+z)(xy+yz+zx)}=\dfrac{2(x^3+y^3+z^3)}{(x+y+z)^3}$
$\qquad \;\;=2\left[\left(\dfrac{x}{x+y+z}\right)^3+\left(\dfrac{y}{x+y+z}\right)^3+\left(\dfrac{z}{x+y+z}\right)^3\right]$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Solve $x^2 = 2^n + 3^n + 6^n$ over positive integers.
Solve $x^2 = 2^n + 3^n + 6^n$ over positive integers.
I have found the solution $(x, n) = (7, 2)$. I have tried all $n$'s till $6$ and no other seem to be there.
Taking $\pmod{10}$, I have been able to prove that if $4|n$ that this proposition does not hold. Can y... | If $n=2k+1\ge 3$, then $x^2\equiv 3\mod 4$.
If $n=2k\ge 4$, then $(x+2^k)(x-2^k)=x^2-2^{2k}=6^{2k}+3^{2k}=3^{2k}(1+2^{2k})$.
We have $\gcd (x-2^k, x+2^k)=1$, then
$x+2^k\ge 3^{2k} \Rightarrow x-2^k\ge 3^{2k}-2^{2k+1}>2^{2k}+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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