Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Very indeterminate form: $\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x \longrightarrow (\infty-\infty)^{\infty}$ Here is problem:
$$\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x$$
The solution I presented in the picture below was made by a Mathematics Teacher
I tried to solve t... | By squaring, we can verify that
$$
x\le\sqrt{x^2+3}\le x\left(1+\frac3{2x^2}\right)\tag1
$$
and
$$
x+1\le\sqrt{x^2+2x+3}\le(x+1)\left(1+\frac1{x(x+1)}\right)\tag2
$$
Adding $(1)$ and $(2)$ gives
$$
2x+1\le\sqrt{x^2+2x+3}+\sqrt{x^2+3}\le(2x+1)\left(1+\frac3{2x^2}\right)\tag3
$$
Multiplying numerator and denominator by $... | {
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"url": "https://math.stackexchange.com/questions/3185610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 3
} |
Find the fifth expression of Taylor's for $\sin (\tan x)$ around $x=0$
Find the fifth expression of Taylor's for $\sin (\tan x)$ around $x=0$
My try:
$$\sin x=x+r_{1}(x), r_{1}(x)=o(x)$$ $$\tan x=x+\frac{x^3}{3}+\frac{2}{15}x^5+r_{2}(x), r_{2}(x)=o(x^5)$$ So: $$\sin \tan x=(x+\frac{x^3}{3}+\frac{2}{15}x^5)+r_{3}(x+\f... | Taylor's expansions up to the same order can be composed. So you have to consider the Taylor polynomial of degree $5$ for $\sin x$, and substitute $x$ with the degree $5$ Taylor polynomial for $\tan x$, truncating the successive powers of the latter polynomial at degree $5$.
As a start-up, here is the computation for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit for $e$ and $\frac{1}{e}$ My question concerns the derivation of this: $$e^r = \lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n \ \ ...(1).$$
One of the definitions of $e$ is as follows:
$$e = \lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n.$$
Then, textbooks usually derive equation (1) in the ... | \begin{align*}\lim_{n\to\infty}\left( 1 - \frac{1}{n} \right)^n &= \lim_{n\to\infty}\left(\frac{n-1}{n}\right)^n = \lim_{n\to\infty}\left(\frac{1}{\frac{n}{n-1}}\right)^n = \lim_{n\to\infty}\frac{1}{\left(\frac{n}{n-1}\right)^n} \\ &= \lim_{n\to\infty}\frac{1}{\left(1 + \frac{1}{n-1}\right)^n} = \lim_{n\to\infty} \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 1
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a tricky trig questions from Step If $\theta + \phi + \psi = \pi/2$, show that $\sin^2 \theta + \sin^2 \phi + \sin^2 \psi + 2 \sin \theta \sin \phi \sin \psi = 1$.
By taking $\theta = \phi =\pi/5$ in this equation, or otherwise, show that $\sin(\pi/10)$ satisfies the equation
$$8x^3 + 8x^2 − 1 = 0$$
I got stuck in the ... | $$\begin{align*}& \sin^2 \theta + \sin^2 \phi + \sin^2 \psi + 2 \sin \theta \sin \phi \sin \psi
\\& = \sin^2 \theta + \sin^2 \phi + \sin^2 \left ( \frac {\pi} {2} - (\theta + \phi) \right ) + 2 \sin \theta \sin \phi \sin \left (\frac {\pi} {2} - (\theta+ \phi) \right ) \\ & = \sin^2 \theta + \sin^2 \phi + \cos^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3195445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Solve for x: $x^3-\lfloor x\rfloor=5$
Solve for x:$$x^3-\lfloor x\rfloor=5$$
My Attempt:
$x^3-5=\lfloor x\rfloor$
Now, $x-1<\lfloor x\rfloor\leq x$
$x-1<x^3-5\leq x$
Not able to proceed from here
| Write $\lfloor{x}\rfloor$ as $x-\epsilon$ where $\epsilon$ is the fractional part, so $0\le\epsilon<1$. Now consider the function $f(x)=x^3-x-5+\epsilon$, note that $f$ has at least one root since $f$ is cubic polynomial. Now differentiate and get $f’(x)=3x^2-1$ which has two roots $\pm\frac{1}{\sqrt{3}}$.
Now:
$f(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3195877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Decomposing bivariate quadratic form into sum of two squares I would like to decompose
$$ax_1^2 + bx_2^2 + 2cx_1x_2$$
into two expressions, each involving only one variable. I'm trying to use a transform like $x_1 = x_+ + x_-$ and $x_2 = x_+ - x_-$ to hopefully get something like
$$\left( bx_+ + \frac ac \right)^2 + \l... | You can write quadratic expressions as $x^T Q x$; in this case $x^T = (x_1, x_2), Q = \begin{pmatrix}a & c \\ c & b\end{pmatrix}$.
Then, if you write $Q$ as $Q = A^TDA$, with $D$ diagonal, you have $x^TQx = x^TA^TDAx = (Ax)^TD(Ax)$: $Ax = (y_1, y_2)^T$ are the new variables you are searching, $D = \begin{pmatrix}d_1 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3196796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Trigonometric and Exponential Integration
If $\displaystyle I_{n} =\int^{\infty}_{\frac{\pi}{2}}e^{-x}\cos^{n}(x)dx.$ Then $\displaystyle \frac{I_{2018}}{I_{2016}}$ is
Try: using by parts
$$I_{n}=\int^{\infty}_{\frac{\pi}{2}}e^{-x}\cos^{n}(x)dx$$
$$I_{n}=-\cos^{n}(x)\cdot e^{-x}\bigg|^{\infty}_{\frac{\pi}{2}}-n\int... | As you say,
$$
\frac{I_n}{n} = - \int_{\pi/2}^{\infty} \mathrm{e}^{-x} \cos^{n-1}(x)\sin(x)\ \mathrm{d}x
$$
I believe the trick is to use integration by parts again, by differentiating the term multiplying the exponential, that is
\begin{align*}
\frac{\mathrm{d} }{\mathrm{d} x} \, \cos^{n-1}(x)\sin(x) &= \cos^n(x) - (n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3200195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solve this equation using Bernoulli's equation How do you solve this equation:
$$\frac{dy}{dx} + \frac{1}{xy} = y^3$$
using Bernoulli's equation.
I've tried everything but I think there may be an error with the actual equation.
| Let $u=y^2$ ,
Then $\dfrac{du}{dx}=2y\dfrac{dy}{dx}$
$\therefore\dfrac{1}{2y}\dfrac{du}{dx}+\dfrac{1}{xy}=y^3$
$\dfrac{du}{dx}=2y^4-\dfrac{2}{x}$
$\dfrac{du}{dx}=2u^2-\dfrac{2}{x}$
Which reduces to a Riccati ODE.
Let $u=-\dfrac{1}{2v}\dfrac{dv}{dx}$ ,
Then $\dfrac{du}{dx}=\dfrac{1}{2v^2}\left(\dfrac{dv}{dx}\right)^2-\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3201072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $a$ and $b$ are rational numbers If $a+b$, $a^2+b$ and $b^2+a$ are rational numbers and $a+b\neq 1$ then $a$ and $b$ are rational.
I try, sum the expresions but I only got that $a^2+b^2$ and $ab$ are rational.
Any suggestion?
| Write : $a^2 + b - (b^2 + a) =(a-b)(a+b-1)$ , this is rational. Since $a+b$ is rational, so is $a+b-1$. It is non-zero, therefore $a-b$ is rational being the quotient of two rationals, the latter non-zero.
Of course, if $a+b$ and $a-b$ are rationals, so are $a$ and $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3202830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving a linear system of reciprocals. Solve for $\begin{cases}\frac{1}{x} +\frac{1}{y}+\frac{1}{z}=0\\\frac{4}{x} +\frac{3}{y}+\frac{2}{z}=5\\\frac{3}{x} +\frac{2}{y}+\frac{4}{z}=-4\end{cases}$
I turn the equations into $\begin{cases}yz+xz+xy=0\\4yz+3xz+2xy=5xyz\\3yz+2xz+4xy=-4xyz\end{cases}$
Not sure if I am doing f... | how about this, let:
$$X=\frac1x,\,Y=\frac1y,\,Z=\frac1z$$
and it is much easier to solve the following:
$$\begin{pmatrix}
1&1&1\\
4&3&2\\
3&2&4
\end{pmatrix}
\begin{pmatrix}
X\\
Y\\
Z
\end{pmatrix}=
\begin{pmatrix}
0\\5\\-4
\end{pmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3205742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove that $\sum_\text{cyc}\frac{a}{b^2}\ge 3\sum_{cyc}\frac{1}{a^2}$ for $a,b,c>0$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$
$a$, $b$ and $c$ are three positives such that $\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = 1$. Prove that $$\dfrac{a}{b^2} + \frac{b}{c^2} + \frac{c}{a^2} \ge 3 \cdot \left(\frac{1}{a... | Let $x=\frac{1}{a}$ , $y=\frac{1}{b}$ and $z=\frac{1}{c}$, so $\sum_{cyc} x=1$. Then we have
$(LHS-RHS)\cdot xyz=(\sum_{cyc}x)\cdot(\sum_{cyc}y^3z)-3\sum_{cyc}x^3yz=\sum_{cyc}x^4y+\sum_{cyc}x^3y^2-2\sum_{cyc}x^3yz$
$=\frac{1}{7}\sum_{cyc}((7x^4y+4z^3x^2+2x^3y^2+y^3z^2)-14x^3yz)\geq0.$
The last inequality come from t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3207885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do i integrate $\int \frac{dx}{(2x+3)\sqrt{(x^2+3x+2})}$? Integrate $\int \frac{dx}{(2x+3)\sqrt{(x^2+3x+2})}$
I put $x^2+3x+2=t,$ and notice that $2x+3 dx=dt$, but the $dx$ is above! Please help me!
| Found a three line solution. Do not always be fooled by the quadratics lol.
If we let the original integral be $I$,note that $I=\int \frac{dx}{(2x+3)\sqrt{\frac{1}{4}(4x^2+12x+8})}=\int \frac{dx}{(2x+3)\frac{1}{2}\sqrt{(4x^2+12x+8})}=\int \frac{dx}{(2x+3)\frac{1}{2}\sqrt{(2x+3)^2-1}} $.
Now you know what...Put $2x+3=z ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3209142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that ${{2m} \choose {m}} \leq \frac{2^{2m}}{\sqrt{2m}}$. The question is as it is in the title:
Show that
$${{2m} \choose {m}} \leq \frac{2^{2m}}{\sqrt{2m}}$$
for all $m \in \mathbb{N}$.
I've had various attempts at this question but it never seems to lead anywhere fruitful. The hint we have been given is in th... | This is an incomplete solution, maybe somebody else can see where to go from here.
Use induction with induction statement
$$P(m): {{2m} \choose {m}}\leq \dfrac{2^{2m}}{\sqrt{2m}}$$
Such that the base $(m=1)$ case is
$$P(1): {{2(1)} \choose {(1)}}\leq \dfrac{2^{2(1)}}{\sqrt{2(1)}}$$
This is easily shown because ${{2}\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3209660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
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$1+\sqrt{2}$ seems special: its powers quickly approach integers $ \alpha = 1+\sqrt{2}$ has some interesting properties. One of which is that if you look at powers
\begin{align*}
\alpha^2 &= 3+2\sqrt{2} \approx 5.8284 \\
\alpha^3 & \approx 14.0710 \\
\alpha^4 & \approx 33.9705 \\
& \vdots \\
\alpha^{10} & \approx 6725.... | The key fact here turns out to be that the conjugate (in this case in $\Bbb Q[\sqrt{2}]$) of $1 + \sqrt{2}$, namely, $1 - \sqrt{2}$, has absolute value less than $1$.
If we expand
$$(1 + \sqrt{2})^n + (1 - \sqrt{2})^n$$
and apply the Binomial Theorem to both of the two powers, we find all of the nonintegral terms canc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Prove $x^4+y^4+x^2+y^2+x^3y+y^3x\geq 0$ I tried 2 ways, first, take $x$ and $y$ $\geq0$ then obviously true. Take $x$ and $y$ both $\leq0$ same thing.
Now $y<0<x$ and wlog $|y|\leq x,$ then for big $x$ small $|y|$ we have $-x^4\leq x^3y$ and $-x^2\leq y^3x$ but if both $|y|$ and $x$ are big I got stuck.
So I decided to... | EDIT
As my first answer was mistaken, a correction is coming now.
$$\begin{aligned}x^4+y^4+x^2+y^2+x^3y+xy^3&=\underbrace{x^4+y^4{\color{red}{-2x^2y^2}}}+x^2+y^2\underbrace{{\color{red}{+2x^2y^2}}+xy(x^2+y^2)}\\
&= (x+y)^2(x-y)^2 +x^2+y^2 \quad +xy(x+y)^2 \\
&=(x+y)^2\left((x-y)^2+xy\right)+x^2+y^2\\&=(x+y)^2(x^2-xy+y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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a inequality with condition For $a,b,c \ge 0$ and $a+b+c \ge m$, I have to prove that $ 6(a^3+b^3+c^3)+9abc \ge m^3$. By power mean inequality
$ 6(a^3+b^3+c^3) \ge \frac {2(a+b+c)^3}{3} \ge \frac{2m^3}{3}$
Then I stuck
| By Schur's inequality (look at https://en.wikipedia.org/wiki/Schur%27s_inequality for its statement and a standard proof), the basic inequality $3(a^2+b^2+c^2)\geq (a+b+c)^2$ (which can be proved by Cauchy-Schwarz if we write $3=1^2+1^2+1^2$) and the non-negativity of $a,b$ and $c$ we have that
$$a^3+b^3+c^3+3abc\geq a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maclaurin expansion of $\exp(x)/(1-\exp(x))^2$ I am trying to expand following function $f(x)=\frac{e^x}{(1-e^x)^2}$ using the Maclaurin series.
According to wolframalpha it should be:
$\frac{1}{x^2}-\frac{1}{12}+\frac{x^2}{240}-\frac{x^4}{6048}+O(x^6)$
However, I am struggling to arrive at this result following a stan... | Bernoulli numbers $B_n$ are defined in a way that:
$$\frac{x}{e^x-1}= \sum_{n=0}^{\infty} \frac{B_n x^n}{n!}\tag{1}\label{eq1}$$
Divide \eqref{eq1} by $x$:
$$\frac{1}{e^x-1}= \sum_{n=0}^{\infty} \frac{B_n x^{n-1}}{n!}\tag{2}\label{eq2}$$
Calculate the first derivative of both sides of the \eqref{eq2}:
$$-\frac{e^x}{(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3214465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Double sum factorial manipulation $$\sum_{B = 0}^{n-1} \sum_{A = 0}^{n-B-1} \frac{(n-1)!}{B!(n-B-1)!} \frac{(n-B-1)!}{A!(n-B-A-1)!} \frac{A}{A+B+1}$$
This is driving me nuts! Is there anyway to reduce
$$\sum_{B = 0}^{n-1} \sum_{A = 0}^{n-B-1} \frac{(n-1)!}{A!B!(n-A-B-1)!} \frac{A}{A+B+1}$$
beyond
$$\sum_{B = 0}^{n-1} \... | Wolfram Alpha found this sequence which seems to match the sum for all positive integers...
$$ a_n = \frac{3^{n-1}}{2} - \frac{3^n - 1}{4n} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3217254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Given a scalar field, how to show that the eigenvalues of its Hessian are bounded?
Show that the eigenvalues of the Hessian of $$f(x_1,x_2) := x_1^2+x_1x_2+x_2^2+\ln (1+2e^{x_2})$$ are bounded, i.e., $$1 \leq \lambda_{\min} \left(\nabla^2 f(x)\right) \leq \lambda_{\max}\left(\nabla^2 f(x)\right) \leq 4$$
What I tried... | Hint: Show that $y$ is bounded for all $x_2\in\mathbb{R}$:
Part 1:
\begin{align}y=\frac{2e^{x_2}}{(1+2e^{x_2})^2}< \frac{1+2e^{x_2}}{(1+2e^{x_2})^2}=\frac{1}{1+2e^{x_2}}<1\end{align}
Part 2:
Observe that
\begin{align} \frac{2e^{x_2}}{(1+2e^{x_2})^2}=\frac{-1+1+2e^{x_2}}{(1+2e^{x_2})^2}=-\frac{1}{(1+2e^{x_2})^2}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3223466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find maximize of the function $\frac{a}{1+a^2}+\frac{b}{1+b^2}-\frac{1}{c^2+1}$ Let $a,b\in R^+$ such that $ab+bc+ca=1$. Find the maximize of $$P=\frac{a}{1+a^2}+\frac{b}{1+b^2}-\frac{1}{c^2+1}$$
By Wolframalpha i can see that if $a=b=2-\sqrt 3;c=\sqrt 3$ we will have $P=\dfrac 1 4$
WLOG $a\le b$. I proved that $$P=f(... | Let $a=\tan\frac{\alpha}{2}$, $b=\tan\frac{\beta}{2}$ and $c=\tan\frac{\gamma}{2},$ where $\left\{\frac{\alpha}{2},\frac{\beta}{2},\frac{\gamma}{2}\right\}\subset(0^{\circ},90^{\circ}).$
Thus, $\alpha+\beta+\gamma=180^{\circ}$ and
$$P=\frac{1}{2}\sin\alpha+\frac{1}{2}\sin\beta-\cos^2\frac{\gamma}{2}=\sin\frac{\alpha+\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Sum of Fourth Powers of cosine series has closed form solution. A problem posed in the 1988 Irish Mathematical Olympiad asks for a proof of the following
$$\sum\limits_{k=1}^{n} \cos^{4}\Big(\frac{k\pi}{2n+1}\big) = \frac{6n-5}{16}$$
Can anyone give me a heads-up on how to proceed to prove this very interesting result... | Using the hints given in answers, we can make the result more general
$$\color{blue}{\sum\limits_{k=1}^{n} \cos^{2p}\Big(\frac{k\pi}{2n+1}\big) = 4^{-p} \binom{2 p}{p}\,n+4^{-p} \binom{2 p-1}{p}-\frac{1}{2}}$$ and get the beautiful
$$\left(
\begin{array}{cc}
p & \sum\limits_{k=1}^{n} \cos^{2p}\Big(\frac{k\pi}{2n+1}\bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the Minimum Value of $x^2+y^2$ Given:
$x+y=2\sin{a}-\cos{b}\\
xy=2\cos{a}+\sin{b}$
Find the minimum value of $x^2+y^2$.
Attempt:
$\begin{aligned}
x^2+y^2&=(x+y)^2-2xy\\
&=(2\sin{a}-\cos{b})^2-2(2\cos{a}+\sin{b})\\
&=4\sin^2{a}-4\sin{a}\cos{b}+\cos^2{b}-4\cos{a}-2\sin{b}
\end{aligned}$
$\begin{aligned}
f_{a}(a,b)&=... | Write $b:={\pi\over2}+c$. Then the equations are
$$x+y=2\sin a+\sin c,\qquad xy=2\cos a+\cos c\ ,$$
and instead of $f$ you now have the function
$$g(a,c)=(2\sin a+\sin c)^2-4\cos a-2\cos c$$
representing $x^2+y^2$. One sees that for $a=c=0$ the function $g$ assumes the value $-6$, but it is impossible for $g$ to assume... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3231536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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$\sin(\theta) + \sin(5\theta) = \sin(3\theta)$ . Find number of solutions and the solutions for this equation in $[0,\pi]$ I tried to solve the equation by doing this-
$$2\sin(\frac{\theta+5\theta}{2})\cos(\frac{\theta-5\theta}{2})=\sin(3\theta)\\
2\sin(3\theta)\cos(2\theta) = \sin(3θ)\\
\cos(2θ) = \frac{1}{2}\\
... | $$\sin3\theta=2\sin\dfrac{5\theta+\theta}2\cos\dfrac{5\theta-\theta}2$$
$$\implies\sin3\theta(2\cos2\theta-1)=0$$
If $\sin3\theta=0,3\theta=m\pi$ where $m$ is any integer
We need $0\le\dfrac{m\pi}3\le\pi\iff 0\le m\le3$
If $2\cos2\theta-1=0,\sin^2\theta=\sin^2\dfrac{\pi}6$
$\theta=n\pi\pm\dfrac{\pi}6$
We need $0\le n\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3231634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Show power series converges for every $x$.
Let $$f(x) = 1 + a_{1}x^{1}+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+...$$ be a solution of the differential equation $f'(x)=xf(x).$
Now I need to explain that the power series that define $f(x)$ converges for every $x$.
This problem is introduced by two smaller proofs which I will ... | We can write the differential equation as
$$
\sum_{n\ge1}na_{n}x^{n-1}=\sum_{n\ge0}a_nx^{n+1}
$$
Changing indices, we obtain
$$
a_1+\sum_{n\ge1}(n+1)a_{n+1}x^{n}=\sum_{n\ge1}a_{n-1}x^n
$$
so that $a_1=0$ and
$$
(n+1)a_{n+1}=a_{n-1}
$$
Therefore all odd terms are zero, and $(n+2)a_{n+2}=a_n$ for even $n$, starting with ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Long division of $\frac{3x^3-x^2-13x-13}{x^2-x-6}$ I'm self-studying from Stroud & Booth's amazing textbook "Engineering Mathematics", and am on the "Partial Fractions" chapter. As part of an exercise I need to do long division of two polynomial equations.
The problem is, long division of polynomials was never explaine... | For long division, you essentially keep reducing the degree of the numerator until it's lower than the denominator.
In the example you gave, $\displaystyle \frac{3x^3 - x^2 - 13x - 13}{x^2 - x- 6}$, we first look at the leading term, $3x^3$. We now want to remove this from the numerator, so we multiply $x^2-x-6$ by $3x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3235798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Value of the binomial series $\sum_{i=0}^k \frac{{2i \choose i}}{4^i}$ Some time ago a question was asked here regarding the value of the sum $$\sum_{i=0}^k \frac{{2i \choose i}}{4^i}$$.
But it was deleted later by the OP. I went around it but didn't find a solution.
Some common combinatoric identities I know of are $... | We prove that $$\sum_{k=0}^{n} \frac{{2k \choose k}}{4^k} = \frac{(2n+1)!}{n!^24^n}.$$
We proceed by induction. The base case is easily verifiable. Denote the RHS of the above equation as $a_n$.
Now consider $$a_{n+1} = \frac{(2n+3)!}{(n+1)!^24^{n+1}} = \frac{(2n+2)(2n+3)}{4(n+1)^2} \cdot \frac{(2n+1)!}{n!^24^n} = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3237320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Factoring $(x+a)(x+b)(x+c)+(b+c)(c+a)(a+b)$ and use the result to solve an equation I managed to prove that $(x+a+b+c)$ is a factor of
$$(x+a)(x+b)(x+c)+(b+c)(c+a)(a+b)$$
Then I was asked to use the result to solve
$$(x+2)(x-3)(x-1)+4=0$$
I know by comparison, $a=2, b=-3, c=-1$, and thus $(x-2)$ is a factor, but I can'... | It is good that you managed to factor!
Here is one way to do it:
$$(x+a)(x+b)(x+c)+(b+c)(c+a)(a+b)=\\
x^3+(a+b+c)x^2+(ab+bc+ca)x+abc+\\
2abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2=\\
x^2(x+a+b+c)+(ab+bc+ca)x+\\
ab(a+b+c)+ac(a+b+c)+bc(a+b+c)=\\
x^2(x+a+b+c)+(ab+bc+ca)x+\\
(a+b+c)(ab+bc+ca)=\\
x^2(x+a+b+c)+(ab+bc+ca)(x+a+b+c)=\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Rolling two dice: different probabilities Today my professor said that if I roll two identical dice at the same time then there will be $21$ outcomes and the probability of getting sum of $7$ is $1/7$:
\begin{array}{c c c c c c}
\{1,1\}, & \{1,2\}, & \{1,3\}, & \{1,4\}, & \{1,5\}, & \{1,6\},\\
& \{2,2\}, & \{2,3\}, & \... | If you roll two dice then there are $6*6=36$ possible outcomes.
6 of those outcomes add up to 7.
So yes , the probability of rolling the sum of 7 is $\frac{6}{36}=\frac{1}{6}$
But in your first statement you considered $(2,1)$ and $(1,2)$ as one outcome. These are Different possible outcomes. If you consider those as o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to "see" this algebraic transformation used to derive the exponential form of the hyperbolic cosine? In this page, the exponential form of cosh is derived from the geometric definition (which is based on the unit hyperbola). Here is one step of the derivation, fully expanded out for novices like me:
$$a = ln(b + \s... | Lets call $c=\sqrt{(b^2-1)}$. Note that $b^2-c^2=1$ so $(b+c)(b-c)=1$ and $b-c={1\over b+c}$ , so that $b+c +{1\over b+c} = b+c + b-c =2b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Definite integral of $1/(2\sin^4x + 3\cos^2x)$ I have $f = \frac 1 {(2\sin^4x + 3\cos^2x)}$ which area should be calculated from $0$ to $\frac{3\pi}2 $.
I noticed that $$\int_0^{\frac{3\pi}2} f \,dx= 3\int_0^{\frac{\pi}2} f \,dx$$
I tried to calculate this integral with the Weierstrass substitution:
$t = \tan{\frac x2}... | Thank you, Dr Zafar Ahmed DSc, for a good idea, but your answer that I calculated is not correct. I tried to solve it myself with getting these coefficients (A, B, etc.). This is my solution that based on yours:
$$ \int_0^\infty \frac {1+\frac1{t^2}}{2t^2+\frac3{t^2}+3} \,dt$$
I. We can write $2t^2+\frac3{t^2}+3$ as
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3241362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Prove $0.9999^{\!101}<0.99<0.9999^{\!100}$ Prove
$$0.9999^{\!101}<0.99<0.9999^{\!100}$$
I think its original idea is
$$(1-x)^{(1-\frac{1}{x})}<(1+x)^{(\frac{1}{x})} \tag{I can't prove!}$$
For $x=100^{\!-1}\,\therefore\,(1-x)^{(1-\frac{1}{x})}<(1+x)^{(\frac{1}{x})}\,\therefore\,0.99^{\!-99}<1.01^{\!100}$.
Furthermore $... | $$0.99<0.9999^{100}$$ it's
$$1-\frac{1}{100}<\left(1-\frac{1}{100^2}\right)^{100},$$ which is true by Bernoulli:
$$\left(1-\frac{1}{100^2}\right)^{100}>1-100\cdot\frac{1}{100^2}=1-\frac{1}{100}.$$
$$0.9999^{101}<0.99$$ it's
$$\left(1-\frac{1}{100^2}\right)^{101}<1-\frac{1}{100}$$ or
$$\left(1-\frac{1}{100^2}\right)^{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3242298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Prove a summation equals to one How to prove that
$$
S = \sum_{k=0}^{K-1} \binom{k+K-1}{K-1} \frac{a^k b^K + a^K b^k}{(a+b)^{k+K}} \mathop{=}\limits^{?} 1,
$$
For $K=2$, I have S = $\frac{a^2+b^2}{(a+b)^2} + \frac{2ab(a+b)}{(a+b)^3}$, which can be boiled down as $S = \frac{(a+b)(a^2+b^2+2ab)}{(a+b)^3} = 1$.
Also, Wolf... | Write $x={a\over a+b},\ J=K-1$, and
$$f(x,J) = (1-x)^{J+1}\sum_{k=0}^J{J+k\choose k}x^k$$ Then we want to show $$f(x,J)+f(1-x,J)=1,\ J=0,1,2,\dots\tag{1}$$
We proceed by induction on $J$. When $J=0$, $(1)$ says $x+(1-x)=1.$
Suppose that $J>0$ and that $(1)$ is true for $J-1$. Then
$$\begin{align}
f(x,J)&=(1-x)^{J+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that the only solution to $X^4+4=pY^4$. is $X=\pm 1,Y=\pm 1,p=5$ for $p \ge 3$ prime.
Let $p$, be a prime, $p\ge 3$. Consider the equation $X^4+4=pY^4$. Prove that the only solution is $X=\pm 1,Y=\pm 1,p=5$
Hint: $x^4+4=(x^2-2x+2)(x^2+2x+2)$. I've tried Fermat's Little Theorem and the linear Diophantine Equatio... | Suppose $x,y\geq 0$ and let $g:= \gcd(x^2-2x+2,x^2+2x+2)$
Let prime $q\mid g$ then $q\mid 4x$. If $q\ne 2$ then $q\mid x$ but then $$q\mid x^2+2x \implies q\mid 2 \implies q=2$$
So the only prime which can divide $ \gcd(x^2-2x+2,x^2+2x+2)$ is 2.
*
*$\color{red}{Case\; 1}$: $2\mid g$ then $x$ is even, so $x=2z$ th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Determine the convergence/divergence of $\sum\limits_{n=1}^{\infty}\left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)$ by comparison test It's simple to evaluate the sum as follows
\begin{align*}
\sum_{n=1}^{\infty}\left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)&=\lim_{n \to \infty}\sum_{k=1}^{n}\left[\left(\sqrt{k+2}-\sqrt{k+1... | Just an alternative answer and using a different test from
$$\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}=\\
(\sqrt{n+2}-\sqrt{n+1})-(\sqrt{n+1}-\sqrt{n})=\\
\frac{1}{\sqrt{n+2}+\sqrt{n+1}}-\frac{1}{\sqrt{n+1}+\sqrt{n}}$$
and using alternating series test the series converges, where $|a_n|=\frac{1}{\sqrt{n}+\sqrt{n-1}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Checking a proof that the equality $ax^2+bx+c=0$ is never true for integer $a , b$ and $c$ if $ x = 2^{\frac{1}{3}} $ The solution of $ax^2+bx+c=0$ is $x=-\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}$ . Since $ x^3=2$ then after raising both sides to the third power we get $$ 2=-\frac{b^3}{8a^3}+\frac{\left(b^2-4ac\right)^{\... | hint: Write $$a\cdot2^{2/3}+b\cdot2^{1/3}=-c$$ and raise to the power three both sides.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3251941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
The area enclosed by the locus of point C Triangle ABC is such that AB = 4, BC = 2, and AC = 3. If vertex A is confined to the x-axis and vertex B is confined to the y-axis, what is the area of the region enclosed by the locus of all points point C could possibly be?
| You have some sort of a "do-nothing machine"
The midpoint of $AB$ traces a circle.
Lets start with $B = (0,0), A = (4,0)$ Where is $C$
Area $= \sqrt {(\frac 92)(\frac 52)(\frac 32)(\frac 12)} = \frac {\sqrt {135}}{4}$
$h = \frac {\sqrt {135}}{8}$
And from Pythagoras theorem we get that the altitude splits the base i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $6|a+b+c $ if and only if $6|a^3 +b^3+c^3$
Question-Prove that $6|a+b+c $ if and only if $6|a^3 +b^3+c^3$
I was playing around with the formulae
$$(a+b+c)^3=a^3 +b^3+c^3+3(a+b)(b+c)(c+a)$$ and,
$$a^3 +b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
and noted that if $(a+b+c)\equiv0$(mod 6)$\implies a^3 +b^3+c... | Among $a$, $b$, $c$ there are at least two even or two odd numbers. In any case $(a+b)(b+c)(c+a)$ is even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Factorize $x^4 + y^4 -x^2 y^2$ over $\mathbb C$ As in the title, I'd like to factorize $x^4 + y^4 - x^2y^2$ into irreducible factors over $\mathbb C$ (i.e. linear factors).
Attempts:
First I tried doing
$$x^4 + y^4 - x^2y^2 = (x^2 - y^2)^2 + x^2y^2 \\ = (x^2 - y^2) - (ixy)^2 \\ = (x^2 - y^2 - ixy)(x^2 - y^2 + ixy) $$
A... | Note that $x^2 - y^2 - ixy = y^2 \left( (\frac{x}{y})^2 - i(\frac{x}{y}) -1 \right)$, which is simply a quadratic in $(\frac{x}{y})^2$, multiplied by $y^2$. Factorise the quadratic and then multiply $y^2$ back in. Do the same with the other factor to finish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3258328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Solve the inequality $ 2x^{2} + x - 4 \sqrt{2x^{2} + x + 4} < 1 $
Solve the inequality
$$ 2x^{2} + x - 4 \sqrt{2x^{2} + x + 4} < 1 $$
Attempt:
I get that $2x^{2} + x + 4 > 0$ for all $x$. Let $y = 2x^{2} + x$, then the inequality becomes
$$ y - 4 \sqrt{y + 4} < 1 $$
$$ 0 < \frac{1 + 4 \sqrt{y+4} - y}{y}$$
so from ... | Let $y= \sqrt{2x^{2} + x + 4} \geq 0$, (notice that $2x^2+x+4>0$ for all real $x$) then we get $$
y^2-4-4y<1\implies y^2-4y-5 = (y-5)(y+1)<0$$
So $$y\in (-1,5)$$
and so
$$ \sqrt{2x^{2} + x + 4} < 5 \implies 2x^2+x+4<25$$
so $$2x^2+x-21= (2x+7)(x-3)<0 \implies x\in (-{7\over 2},3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3259493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Least value of $ab-cd$ is
If $a,b,c,d\in\mathbb{R}$ and $a^2+b^2=c^2+d^2=4$
$ac+bd=0$. Then least value of $ab-cd$ is
Plan
$(ac+bd)^2+(ab-cd)^2=a^2c^2+b^2d^2+a^2b^2+c^2d^2\geq 0$
$a^2(b^2+c^2)+d^2(b^2+c^2)\geq 0$
How do i solve it Help me please
| Geometric interpretation of conditions above with $U=\binom{a}{b}$, $V=\binom{c}{d}$ give :
$\|U\|=\|V\|=2$ and dot product $U . V=0$, i.e., $U$ and $V$ are orthogonal.
Thus $a = 2\cos \alpha, b=2\sin \alpha, c = 2\cos \beta, d=2\sin \beta \tag{1}$
for a certain $\alpha$ ($-\pi < \alpha \leq \pi$) with $\beta=\alpha\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3260685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Unable to solve $ \int \frac{x + \sqrt{2}}{x^2 + \sqrt{2} x + 1} dx $? This comes from a bigger problem :-
$$ \text{Evaluate } \int\frac{dx}{1+x^4} $$
After making $ \int \frac {dx}{1+x^4} = \frac{dx}{(1+x^2)^2 - (\sqrt{2}x)^2} $ and then applying partial fraction method, I got :-
$$ \int \frac{dx}{1 + x^4} =\frac{1}{... | $x+\sqrt2 = \frac{1}{2}(2x+2\sqrt2) = \frac{1}{2}(2x+\sqrt2)+\frac{1}{2}\sqrt2$
Let $x^2+\sqrt2 \ x+1 = t \implies (2x+\sqrt2)dx = dt $
So, the integral is
$$I = \frac{1}{2}\int\frac{dt}{t}+\frac{1}{\sqrt2}\int\frac{dx}{(x^2+\sqrt 2 \ x+1)}$$
The first part gets evaluated into $\frac{1}{2}\ln t$. Convert the second pa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3263899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Inequality for optimization. I found this problem:
Find the minimum value of:
$f(a,b,c)=a+b+c+\frac{9a}{(a+2b+c)^2}+\frac{9b}{(b+2c+a)^2}+\frac{9c}{(c+2a+b)^2}$
Where $a,b,c >0$.
After looking for some minimum values, I think the lowest one is for the equality case $a=b=c=3/4$.Obviously,
$f(a,a,a)=3a+\frac{27}{16a} \ge... | By AM-GM and C-S we obtain:
$$\sum_{cyc}\left(a+\frac{9}{(a+2b+c)^2}\right)\geq6\sqrt{\sum_{cyc}a\sum_{cyc}\frac{a}{(a+2b+c)^2}}\geq6\sum_{cyc}\frac{a}{a+2b+c}=$$
$$=6\sum_{cyc}\frac{a^2}{a(a+2b+c)}\geq\frac{6(a+b+c)^2}{\sum\limits_{cyc}a(a+2b+c)}=\frac{6(a+b+c)^2}{\sum\limits_{cyc}(a^2+3ab)}\geq\frac{9}{2},$$ where th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3267510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Show that $\left| z + \sqrt{z^2 + c^2} \right| > c$ for $\operatorname{Re}(z) > 0$ I have encountered this seemingly obvious yet bizarre-to-show inequality: $$\left| z + \sqrt{z^2 + c^2} \right| > c$$ for $z \in \Bbb C$ such that $\operatorname{Re}(z) > 0$ and $c > 0$. I've tried the reverse triangular inequality but i... | I'll assume that the branch of $\sqrt{z^2 + c^2}$ is chosen such that $\operatorname{Re} \sqrt{z^2 + c^2} > 0$ (otherwise the claim is wrong).
Then
$$
\left| z + \sqrt{z^2 + c^2} \right|^2 = |z|^2 + |z^2+c^2| + 2 \operatorname{Re} \left( \bar z \cdot\sqrt{z^2 + c^2}\right) \\
\ge |z|^2 + (c^2 - |z|^2) + 2 \operatorna... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3267936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$ given $\sin \alpha+\sin \beta+\sin\gamma=0=\cos\alpha+\cos\beta+\cos\gamma$ I am supposed to find the value of $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$ and I have been provided with the information that $\sin \alpha+\sin \beta+\sin\gamma=0=\cos\alpha+\cos\beta+\cos\gamma$.... | From Clarification regarding a question,
$\alpha-\beta=\dfrac{2\pi}3+2\pi a,\beta-\gamma=\dfrac{2\pi}3+2\pi b$
$\implies\alpha-\gamma=\dfrac{4\pi}3+2\pi c=-\dfrac{2\pi}3+2\pi d$
If $S=\sin^2\alpha+\sin^2\beta+\sin^2\gamma$
$$2S=3-(\cos2\alpha+\cos2\beta+\cos2\gamma)$$
Method$\#1:$
$$\cos2\alpha+\cos2\beta=\cos2\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3276737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Solve$(10+6\sqrt3)^{\frac{1}{3}}-(-10+6\sqrt3)^{\frac{1}{3}}$ We need to solve the following equation $y=(10+6\sqrt3)^{\frac{1}{3}}-(-10+6\sqrt3)^{\frac{1}{3}}$ and it is equal to 2 while I am getting the value in excel I am not able to solve it manually eventhough the values are conjugate
I tried $y=a-b$
$a=(10+6\sqrt... | Since $(1\pm\sqrt3)^3=10\pm6\sqrt3$,$$\sqrt[3]{10+6\sqrt3}+\sqrt[3]{10-6\sqrt3}=1+\sqrt3+1-\sqrt3=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3277692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Consider the polynomial expression $P(x) = (x+1)^5 + x$ Consider the polynomial expression $P(x) = (x+1)^5 + x$
*
*Find the quotient of $P(x)$ divided by $x^2 + x + 1$
*Prove that for each positive integer $n$, the integer number $P(n)$ is divisibile by at least two different prime numbers.
I've done the first tas... | *
*If $a,b$ are a roots of $x^2+x+1=0$ then $a^3=b^3 =1$ and by Vieta we have $a+b=-1$, so $a+1 =-b$. Now since $$p(a) = (a+1)^5+a = -b^5+a = -b^2+a = (b+1)+a = 0$$
and the same for $p(b)=0$ we see that $x^2+x+1$ divide $p$, so the remainder is $0$.
*
*Since for $n\geq 2$ we have $p(n)=\underbrace{(n^2+n+1)}_{\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find every $a>0$ such that $\int_2^{\infty} \frac{x^6(1+\cos^2 x)}{(x-1)^2(x+1)^2x^a} dx$ converges We know that $\forall x \in [2,+\infty)$,
$$\frac{x^6(1+\cos^2 x)}{(x-1)^2(x+1)^2x^a} \leq \frac{2x^6}{(x-1)^2(x+1)^2x^a}$$
And it's easy to prove that $\int_2^{\infty} \frac{2x^6}{(x-1)^2(x+1)^2x^a} dx$ converges if an... | It is true that $\frac{x^6(1+\cos^2 x)}{(x-1)^2(x+1)^2x^a} \leq \frac{2x^6}{(x-1)^2(x+1)^2x^a}$. What is also true is that $\frac{x^6}{(x-1)^2(x+1)^2x^a} \leq \frac{x^6(1+\cos^2 x)}{(x-1)^2(x+1)^2x^a}$. Altogether, this inequality is $$\frac{x^6}{(x-1)^2(x+1)^2x^a} \leq \frac{x^6(1+\cos^2 x)}{(x-1)^2(x+1)^2x^a} \leq ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find GCD of those two complex polynomials? I have a polynomial
$$f(x) = i(x^2-1)^3+(x^2+1)^3-8x^3$$
I want to check if this has repeated roots. To do so, I'll find greatest common divisor (euclidean algorithm) of $f(x)$ and its derivative $f'(x)$.
$$f(x) = i(x^2-1)^3+(x^2+1)^3-8x^3 = i(x^6-3x^4+3x^2-1)+(x^6+3x^... | By inspection, $f(1)=0$ so $x-1$ is a factor of $f(x)$.
Since $f'(x)= 3i(x^2-1)^2(2x)+3(x^2+1)^2(2x)-24x^2$, we also have that $f'(1)=0$, so $x-1$ is a factor of $f'(x)$.
This means $(x-1)^2$ is a factor of $f(x)$.
One also has that by inspection $f(i)=0$ so $x-i$ is a factor of $f(x)$.
We also have that $f'(i)=0$, so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Exponential equations, solve for x I'm preparing for uni entrance exam. I've been struggling with this problem for about 90 minutes, tried everything I could think of. Can anybody explain how to solve for x step by step?
$$3^{\frac{x-1}{2}}-2^{\frac{x+1}{3}}=2^{\frac{x-2}{3}}+3^{\frac{x-3}{2}}$$
| Bring same exponents to one side:
$$3^{\frac{x-1}{2}}-3^{\frac{x-3}{2}}=2^{\frac{x+1}{3}}+2^{\frac{x-2}{3}}$$
Factor out $3^{\frac{x-3}2}$ on the LHS and $2^{\frac{x-2}3}$ on the RHS:
$$3^{\frac{x-3}2}(3-1)=2^{\frac{x-2}3}(2+1)$$
Bring the $2$ and $3$ over again:
$$3^{\frac{x-3}2}\cdot2=2^{\frac{x-2}3}\cdot3$$
$$3^{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Completing the square (diagonal matrix/bilinear form) Let $A=\begin{pmatrix} 2 & 2 & 5 \\ 2 & 1 & 3 \\ 5 & 3 & \frac{17}{2} \end{pmatrix} \in M_3(\mathbb{R})$.
I want to find an invertible matrix $C$ such that $C^TAC=\begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$, since there are three eigenvalues ... | $$2x^2+4xy+10xz+y^2+6yz+\frac{17}2z^2=2\left(x+y+\frac52z\right)^2-y^2-4yz-4z^2=$$
$$=2\left(x+y+\frac52z\right)^2-\left(y+2z\right)^2$$
and there you go: one positive, one negative, one zero...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3284942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For $ f_{n+1} = \sin f_n(x) $, $f_n(x) \gt f_2(x)/(n-1) $?
Let $f_0(x) = x, f_{n+1} = \sin f_n(x) (0 < x < \pi)$.
Then for $n \ge 3$,
$$f_n(x) \gt \frac{f_2(x)}{n-1} ? $$
Here is what I have shown:
For $x > 0$, $\sin x > x - x^3 / 6$.
So to show the proposition, it suffices to show that, by induction, for $n$, $n... | Analogously to $\sin(x)<x-x^3/6$, we have $\sin(x)>x-x^3/6+x^5/120$ (one additional term in the Taylor Series). Additionally, $\sin([0, 1])\subseteq [0,1]$ so that $0\leq f_n(1)\leq 1$ for all $n$.
We prove by induction that $nf_n(1)^2<6$. Base case: $f_1(1)^2=\sin(1)<6$. Now let $n\geq 2$ and assume the statement hold... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3285832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Roots of $z^4 + 2z^3 + 20z + 12 = 0$
Find the roots of $$z^4 + 2z^3 + 20z + 12 = 0.$$
I still haven't been able to solve it after hours of trying. There are no rational roots and factoring seems impossible. Who of you, mathematical geniuses, can help me? :-)
| The rational roots test fail so there is no linear real factor but there may be some $(z^2 + az + b)(z^2 + cz + d)$ where
$a+c = 2$ and $b+ ac + d =0$ and $ad+bc = 20$ and $bd=12$
which .... isn't obvious but.... oh heck. If we assume $\{b,d\} = \{\pm 1,12\}, \{\pm 2,6\}, \{\pm 3,4\}$ then $ac =-b-d= \mp 13, \mp 8, \m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3286046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Finding $k$ such that $(x^2 + kx + 1)$ is a factor of $(x^4 - 12 x^2 + 8 x + 3)$
$(x^2 + kx + 1)$ is a factor of $(x^4 - 12 x^2 + 8 x + 3)$ . Find $k$
....couldnt figure out how to find $k$
I tried assuming $(x^2 + kx + 1)= (x - 1)^2 $ where $k = (-2) $ comsidering that $(x-1) $ is a factor of the above polynomial...... | We factor $x^4-12x^2+8x+3$.
We see, that $x=1$ is a root of $f(x)=x^4-12x^2+8x+3$, since $f(1)=1-12+8+3=0$
After long division $(x^4-12x^2+8x+3)\div (x-1)=x^3+x^2-11x-3$
We see, that x=3 is a root of the polynomial $g(x)=x^3+x^2-11x-3$
Again after long division we achieve:
$(x^3+x^2-11x-3)\div (x-3)=x^2+4x+1$
Which giv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3288779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If I know that $b$ is a multiple of $32$ then how can i show that $b+5$ and $b^2 + 7$ are coprime? I'm using the polynomial division method and i got a remainder of 32 but how do I explain in words or mathematically how this justifies my solution.
The previous question has answers based on Properties of Divisibility bu... | Let $d = gcd(b+5, b^{2}+7) = gcd(32k + 5, 32^{2}k^{2} + 320k + 32)$, for some integer $k$. Hence, $d | (b^{2} + 7) - b$, which implies $d | b^{2} + 7 - (b + 5)r$, when $r \in Z.$
Now, we want to choose $r$ such that $(b + 5)r = b^{2} + c$, for some integer $c$; specifically, let $r = (b - 5)$ as then $(b + 5)r$ will re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3289844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
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The number of triples that sum to a constant Problem:
How many triples are there of the form $(x_0,x_1, x_2)$ where $x_0 \in I$, $x_1 \in I$, $x_2\in I$ $x_0 \geq 0$,
$x_1 >= 0$, $x_2 >= 0$ and $n = x_0 + x_1 + x_2$ where $n \in I$?
Answer:
Let $c(n)$ be the number of tuples we can have for a given $n$. For $n = 0$, th... | This is a prototypical “stars and bars” problem. Here we have $n$ stars and $2$ bars, with $\binom{n+2}2 = {(n+2)(n+1)\over2}$ ways to choose among the $n+2$ possible positions for the two bars.
Another way: The number of partitions of $n$ into three non-negative integers is equal to the coefficient of $z^n$ in the for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof: If $x^n = y^n$ and n is even, then $x=y$ or $x=-y$. This is a problem from Spivak's Calculus 4th ed., Chapter 1, Problem 6(d)
Proof: If $x^n = y^n$ and n is even, then $x=y$ or $x=-y$.
I tried to prove it in the following way but I'm not sure if the proof makes sense, and the author uses another method.
Proof:... | Bear with me.
If $0 \le x < y$ then $x^n < y^n$.
Pf: By induction. Base case is $x < y$. Induction step is If we assume $x^{n} < y^{n}$ then $x^{n+1} = x*x^{n} < x*y^{n} < y*y^{n-1} = y^{n+1}$.
So if we know $x$ and $y$ are non negative and we are told $x^n = y^n$ then we know $x = y$. (By contradiction. If $x \ne y$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3295141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to manipulate $\frac{1}{\sin x}-\frac{1}{x}$ into $\sum_{k\geq 1}\left(\frac{1}{x-k\pi}+\frac{1}{x+k\pi}\right)(-1)^k$? 1) How do you get to $\sum_{k\geq 1}\left(\frac{1}{x-k\pi}+\frac{1}{x+k\pi} \right)(-1)^k$ from $\frac{1}{\sin x}-\frac{1}{x}$? (as demostrated here)
If $\pi\cot\pi x=\frac{1}{x}+\sum_{k\geq 1}\... | Not a full solution, but an attempt to use special functions here instead of complex analysis.
We are interested in the series:
$$f(x)=\frac{2x}{\pi^2} \sum_{k=1}^\infty \frac{(-1)^{k+1}}{(k-x/\pi)(k+x/\pi)} $$
Let's consider an auxiliary function:
$$g(x,y)= -\frac{x^2}{\pi^2} \sum_{k=0}^\infty \frac{y^k}{(k-x/\pi)(k+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3299385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I solve the equation $2\sin(3x)\cos(4x)=1$? How do I solve the equation $2\sin(3x)\cos(4x)=1$? Normally, I'd use the $\sin(2x)=2\sin(x)\cos(x)$ rule, but here you have two different values for $x$, so I'm not sure how to add them together.
| Let $f(x)=2\sin(3x)\cos(4x)-1$ and solve $f(x)=0$.
The strategy will be to solve this using a Chebychev polynomial of the first type.
First, get a similar function entirely in terms of cosine by shifting the graph of $f$ horizontally to the right by $\frac{\pi}{2}$ units. This will be an even function with symmetric s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3301492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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How can I prove that $\frac{n^2}{x_1+x_2+\dots+x_n} \le \frac{1}{x_1}+ \frac{1}{x_2} +\dots+ \frac{1}{x_n}$?
How can I prove that $\frac{n^2}{x_1+x_2+\dots+x_n} \le \frac{1}{x_1}+ \frac{1}{x_2} +\dots+ \frac{1}{x_n}$?
im trying to use AM-GM
$\sqrt[n]{ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}}
\le ... | The inequality you are interested in is equivalent $$\frac{1}{\frac{x_1+x_2+\dots+x_n}{n}} \le \frac{\frac{1}{x_1}+ \frac{1}{x_2} + \dots+ \frac{1}{x_n}}{n}.$$which is just AM-HM inequality.
You can also prove it using CS inequality.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Possible Jordan Canonical Forms: Intuition As I was reviewing linear algebra before I head off to grad school in the fall, I came across a question about Jordan Canonical Forms. It reads:
"Suppose that A is a square complex matrix with characteristic polynomial $c_A(x) =
(x−1)^4(x+ 3)^5$.
Assume also that $A−I$ has nul... | We have a matrix $A$ whose characteristic polynomial is
$$
\chi_A(t)=(t-1)^4(t+3)^5
$$
This immediately tells us that $A$ is a $9\times 9$ matrix whose eigenvalues are $1$ and $-3$. Moreover, the algebraic multiplicities of these eigenvalues are
\begin{align*}
\operatorname{am}_A(1) &= 4 & \operatorname{am}_A(-3)=5
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that if $x \in \mathbb R$ and $x>2$ then $y+ \frac{1}{y} = x$ will have real solution.
Prove that for real number $x$, if $x > 2$ then there is real number $y$ such that $y + \frac{1}{y} = x$
My attempt:
Rewriting equation, we have:
$$\tag1y + \frac{1}{y} = x$$
$$\tag2 \frac{y^2+ 1}{y} = x$$
$$\tag3 y^2+ 1 = y... | Multiplying both sides by $y$ and putting the result in standard form, we have
$$y^{2} - xy + 1 = 0.$$
Hence, the discriminate of this quadratic equation is $x^{2} - 4$; and so, if $x > 2$, we certainly get two distinct real solutions. But also, if $x < -2$ we may draw the same conclusion as well. Finally, if $x = \pm ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Can I split $\frac{1}{a-b}$ into the form $f(a)+f(b)$? I was wondering if I can split apart the following fraction
\begin{align}
\frac{1}{a-b}
\end{align}
into the form:
\begin{align}
f(a)+f(b)
\end{align}
where $f(a)$ and $f(b)$ is some function in terms of $a$ and $b$
| In fact, we can't even split it as
$$\frac{1}{a-b}=f(a)+g(b)$$
for any functions $f,g$.
Suppose instead that functions $f,g$ from $\mathbb{R}$ to $\mathbb{R}$ are such that
$$f(a)+g(b)=\frac{1}{a-b}$$
for all $a,b\in\mathbb{R}$ with $a\ne b$.
Then we would have
\begin{align*}
&
\begin{cases}
f(x+1)+g(x)={\Large{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3308043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
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How is the 'Answer' achieved in the 'Problem' after x,y-axis reflection? Problem: $f(x) = x^3 - x^2 + x – 1$ is reflected in both the $x$-axis and $y$-axis. What is the final equation to represent this transformation?
Answer: $x^3 + x^2 + x +1$
My solution: let $g(x)=-f(x)$ << reflection across $x$-axis and $h(x)$ be ... | $$-f(-x) = - [(-x)^{3} - (-x)^{2} + (-x) - 1 ] = - [-x^{3} - x^{2} - x - 1] = x^{3} + x^{2} + x + 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3309012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find $xyz$ if $x-\frac{1}{x}=y$, and $y-\frac{1}{y}=z$, and $z-\frac{1}{z}=x$ If
$$x-\frac{1}{x}=y, \qquad y-\frac{1}{y}=z, \qquad z-\frac{1}{z}=x$$
find the value of $xyz$.
This is how far I proceeded:
$x+y+z=z-1/z+x-1/x+y-1/y=>1/x+1/y+1/z=0
=>xy+yz+zx=0$
Also from question,
$x^2-1=xy,y^2-1=yz,z^2-1=zx$.
Adding $x^2... | Note that adding up the three equations gives you:
$$x+y+z =x+y+z - \frac{1}{x} - \frac{1}{y} - \frac{1}{z}$$
, so $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$$
Try it from there
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Value of $\lim\limits_{x\to-\infty}{(4x^2-x)^{1/2} +2x}$ The value of the $$\lim\limits_{x\to-\infty}{(4x^2-x)^{1/2} +2x}$$ is?
The answer given is $1/4$.
I rationalized and got $$\lim\limits_{x\to-\infty} \frac {-x}{|x|[(4- \frac {1}{x})^{1/2}-2]}$$ how to proceed further?
| \begin{align*}
(4x^2-x)^{1/2}+2x
&=\sqrt{4x^2-x}+2x
\\&=\frac{(\sqrt{4x^2-x}+2x)(\sqrt{4x^2-x}-2x)}{\sqrt{4x^2-x}-2x}
\\&=\frac{(4x^2-x)-4x^2}{\sqrt{4x^2-x}-2x}
\\&=\frac{4x^2-x-4x^2}{\sqrt{4x^2-x}-2x}
\\&=\frac{-x}{\sqrt{4x^2-x}-2x}
\\&=\frac{-x}{\sqrt{(2x)^2(1-\frac{1}{4x})}-2x}
\\&=\frac{-x}{2|x|\sqrt{1-\frac{1}{4x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 2
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Solve the equation $4 \sin x \cos (2x) \sin (3x) = 1$ for $0^{\circ} < x < 180^{\circ}$ I have tried using identities but none of them worked for me.
Eg: $4 \sin x(1 - 2 \sin^2 x)(3 \sin x - 4 \sin^3 x) = 1$
| Hint: It is
$$4\sin(x)\cos(2x)\sin(3x)-1=$$$$-2 \sin \left(\frac{\pi }{4}-2 x\right) \sin \left(2
x+\frac{\pi }{4}\right) (2 \cos (2 x)-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solve the following equation: $\sin x \cos x = \frac{1}{2}$ I am required to solve the following equation:
$$\sin x \cos x = \frac{1}{2}$$
My attempt:
Rewriting $\cos x$
$$\sin x \sqrt{1 - \sin^2 x} = \frac{1}{2}$$
Squaring both sides
$$\bigl(\sin x \sqrt{1 - \sin^2 x}\bigr)^2 = \bigl(\frac{1}{2}\bigr)^2$$
$$\sin^2 x... | I am now going to add one more for fun and wish you can enjoy it. $$
\begin{aligned}
&(\cos x+\sin x)^{2}=1+2 \sin x \cos x=2 \\
\Rightarrow & \cos x+\sin x=\pm \sqrt{2} \ldots(1) \\
&(\cos x-\sin x)^{2}=1-2 \sin x \cos x=0 \\
\Rightarrow & \cos x-\sin x=0 \quad \cdots(2)
\end{aligned}
$$
(1) $\pm$ (2) yields $$\quad \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3314378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 7
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Showing $\sqrt[3]{\cos\frac{2\pi}{9}}+\sqrt[3]{\cos\frac{4\pi}{9}}+\sqrt[3]{\cos\frac{8\pi}{9}} = \sqrt[3]{\frac{3\sqrt[3]9-6}{2}} $
Show that
$$\sqrt[3]{\cos\frac{2\pi}{9}}+\sqrt[3]{\cos\frac{4\pi}{9}}+\sqrt[3]{\cos\frac{8\pi}{9}} = \sqrt[3]{\frac{3\sqrt[3]9-6}{2}} $$
I tried to find a polynomial that had its roots,... | Sketch of the proof using @saulspatz ninth degree equation above, though all computations that follow from that should be easily done by trigonometry too:
Let $a,b,c, \omega a, \omega b, \omega c, \omega^2 a, \omega^2 b, \omega^2 c$ the roots of the 9th degree equation $8x^9-6x^3+1=0$, where $a,b,c$ are the real ones, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3315333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Determining the range of $6^n+ 6^{-n} +3^n +3^{-n}+2$
I have to solve for range of the function $$6^n+ 6^{-n} +3^n +3^{-n}+2$$
The textbook solves it as
$$\left(\sqrt{6^n} -\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} -\sqrt{ 3^{-n}} \right)^2 +6 \tag{1}$$
i.e., $$(a-b)^2+(a-b)^2$$
which will always be greater tha... | Note that:
$$6^n+ 6^{-n} +3^n +3^{-n}+2=\left(\sqrt{6^n} -\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} -\sqrt{ 3^{-n}} \right)^2 +6 \ \ \text{(textbook)}\\
6^n+ 6^{-n} +3^n +3^{-n}+2=\left(\sqrt{6^n} +\sqrt{ 6^{-n}} \right)^2 + \left(\sqrt{3^n} +\sqrt{ 3^{-n}} \right)^2 \overbrace{\require{cancel}\cancel{\color{red}+}}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What are the solutions for $ n(n+1)=p^2$ for n belongs to $N$ The following is my approach:
$ n^2+n =p^2$
$ n^2+n+\frac{1}{4} = p^2 + \frac{1}{4}$
$ (n+\frac{1}{2})^2 = p^2+\frac{1}{4}$
$ (n+\frac{1}{2}-p) (n+\frac{1}{2}+p) = \frac{1}{4}$
I am not able to proceed further from here. Any suggestion on what to do... | There are only $2$ integer solutions. Neither are greater than $0$.
Assume $n\ge0$.
Without loss of generality, $p\ge0$. Then since $n\ge0$, $p^2=n^2+n\implies p\ge n$. Note that
$$
n(n+1)=p^2\implies n=p^2-n^2\tag1
$$
If $p=n$, then $n=p^2-n^2=0$.
If $p\gt n$, then
$$
n(n+1)=p^2\implies n=\overbrace{\ (p-n)\ }^{\ge1}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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On completing the square and symmetry The quadratic form $f(x,y)=x^2+xy+y^2$ appears quite often on the site here. It is provably positive definite by completing the square: $$f(x,y)=\left(x+\frac y2\right)^2+\frac {3y^2}{4}$$
and this is the only analysis ever really given. I have never seen the symmetric version $$f(... | It is revealing to recast both expressions in terms of linear algebra. We may write $$f(x,y)=x^2+xy+y^2 \equiv v^\top A v = \begin{pmatrix} x & y \end{pmatrix} \begin{pmatrix} 1 & 1/2 \\ 1/2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}.$$
If we look to the expression $f(x,y)=(x+y/2)^2+\dfrac34 y^2$, this may ... | {
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"url": "https://math.stackexchange.com/questions/3318597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Prove inequality with logarithms:$\log_{a}{\frac{a+b}{2}} + \log_{b}{\frac{a+b}{2}} \ge 2$ Let $ a, b \in (1, \infty)$. Prove that:
$$\log_{a}{(\frac{a+b}{2})} + \log_{b}{(\frac{a+b}{2})} \ge 2$$
I tried switching the bases on each of the logaritm but I got stuck:
$$\frac{\log_{\frac{a+b}{2}}{(ab)}}{\log_{\frac{a+b}{2}... | By C-S and AM-GM
$$\log_a\frac{a+b}{2}+\log_b\frac{a+b}{2}=\frac{1}{\ln_{\frac{a+b}{2}}a}+\frac{1}{\ln_{\frac{a+b}{2}}b}\geq$$
$$\geq\frac{(1+1)^2}{\log_{\frac{a+b}{2}}a+\log_{\frac{a+b}{2}}b}=\frac{4}{\log_{\frac{a+b}{2}}ab}\geq\frac{4}{\log_{\frac{a+b}{2}}\left(\frac{a+b}{2}\right)^2}=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3318645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Simplify $\sqrt{10 + \sqrt{24} + \sqrt{40} + \sqrt{60}} $ Simplify
$$ \sqrt{10 + \sqrt{24} + \sqrt{40} + \sqrt{60}} $$
Attempt:
$$ \sqrt{10 + \sqrt{24} + \sqrt{40} + \sqrt{60}} = \sqrt{10 + 2\sqrt{6} + 2\sqrt{10} + 2\sqrt{15}} = \sqrt{10 + 2(\sqrt{6} + \sqrt{10} + \sqrt{15})} $$
let $X =\sqrt{10 + \sqrt{24} + \sqrt{... | You can continue so:
$$x^2=10 + 2(\sqrt{6} + \sqrt{10} + \sqrt{15})=2+5+3+ 2(\sqrt{6} + \sqrt{10} + \sqrt{15})=(\sqrt2+\sqrt5+\sqrt3)^2.$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Is this true for all polynomials I took $3$ random polynomials with non zero roots one having even degree and two having odd degrees
*
*$f(x)=\color{red}{4}x^2-(4\sqrt3+12)x+12\sqrt3$ having roots $\color{blue}{3,\sqrt3}$ and leading coefficient $\color{red}{4}$ and calculated values of $xf'(x)$$(f'(x)$ is the deriv... | This results from partial fraction decomposition.
Suppose $g(x)$ has no repeated roots, and without loss of generality is monic (leading coefficient $1$). Then it factors as $\displaystyle g(x)=\prod_{r}(x-r)$ over all roots $r$ and
$$ \frac{1}{g(x)}=\sum_{r}\frac{c(r)}{x-r} $$
for some constants $c(r)$, one for each r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3325284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Let $W = \textit{span}\{\mathbf{w}_1,\mathbf{w}_2,\mathbf{w}_3\}$... Let $W = \text{span}\{\mathbf{w}_1,\mathbf{w}_2,\mathbf{w}_3\}$, where \begin{equation*}
\mathbf{w}_1 = \begin{pmatrix}
1 \\
2 \\
1
\end{pmatrix}, \quad \mathbf{w}_2 = \begin{pmatrix}
1 \\
0 \\
1
\end{pmatrix}, \quad \mathbf{w}_3 = \begin{pmatrix}
2 \... | For $2)$, you could take the cross product of $(1,2,1)$ and $(1,0,1)$.
Get $\begin {vmatrix} i&j&k\\1&2&1\\1&0&1\end{vmatrix}=(2,0,-2)$.
For $3)$, given $\vec x=(x_1,x_2,x_3)$, we need to solve $\begin {pmatrix}1&1&2\\2&0&0\\1&1&-2\end{pmatrix}\vec y=\vec x$.
Thus, $\vec y=\begin {pmatrix}1&1&2\\2&0&0\\1&1&-2\end{pma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to solve this? (Argument of the complex number in complex plane) Let the $z \in C$ s.t. $|z-10i|= 6 $
$\newcommand{\Arg}{\operatorname{Arg}}$
Say the $\theta = \Arg(z)$
Find the maximum and minimum value of the $8\sin \theta + 6\cos \theta$
My trial)
Trying to solve by picture and graph, I considered $8i\sin \thet... | Let $z=r(\cos\theta+i\sin\theta).$
Thus, $$r^2\cos^2\theta+(r\sin\theta-10)^2=36$$ or
$$\sin\theta=\frac{r^2+64}{20r}\geq\frac{2\sqrt{64r^2}}{20r}=\frac{4}{5},$$ which gives
$$\arcsin\frac{4}{5}\leq\theta\leq\pi-\arcsin\frac{4}{5}.$$
Also, write $$8\sin\theta+6\cos\theta=10\sin\left(\theta+\arccos\frac{4}{5}\right).$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
If $a,b \in \mathbb{R}$ and $a+b\ge 0$, prove that $(a^2+b^2)^3\ge 32(a^3+b^3)(ab-a-b)$
If $a,b \in \mathbb{R}$ and $a+b\ge 0$, prove that $(a^2+b^2)^3\ge 32(a^3+b^3)(ab-a-b)$
This question in my opinion is difficult and have tried many things. I tried to expand both sides but that will not help since I will not be a... | Equivalently, you want that
$$
f = \frac18((a^2+b^2)^3 - 32(a^3+b^3)(ab-a-b)) \ge 0
$$
Let $a = x+y$ and $b = x-y$. Then we have $x \ge 0$ and arbitrary $y$.
Putting in these variables you have that
$$
f = (x - 4)^2 x^4 + y^2 (3 ((x- 8/3)^2 + 80/9) x^2 + y^2 (x (3 x + 24) + y^2))
$$
which is clearly $\ge 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3332108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Solving $\frac53\left(3-\frac{x}{5}\right) = x$ Although I had already solved the following equation, I can't figure out why I fail when passing thru a $15$ denominator I get a bad result.
$$\frac53\cdot\left(3-\frac{x}{5}\right) = x$$
Transformation to get $15$ denominator:
$$\begin{align}
\frac{5\cdot 5(15\cdot 3-3x)... | Your first line of working is wrong because
$$\frac53\left(3-\frac{x}5\right)\ne\frac{5\cdot5(15\cdot3-3x)}{15}$$
you should actually get
$$\frac53\left(3-\frac{x}5\right)=\frac{5\cdot5}{15}\cdot\frac{15\cdot3-3x}{15}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3332828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
The equation of the line cutting the circle at $A$ and $B$ given that $AB=\sqrt{2}$ Given the circle $C: 2x^2 + 2y^2 + 2x + 2y - 13 = 0$. Line $L$, with slope $m$ and passing through the point $P(0,2)$, cuts the circle at points $A$ and $B$ such that $AB=\sqrt{2}$.
*
*Find the equation of $L$.
*Find the equation o... | Circle: $(x+0.5)^2 + (y+0.5)^2 = 7 = r^2$
Distance P (0,2) to center C (-0.5,-0.5) = $\sqrt{0.5^2 + 2.5^2} = \sqrt {6.5}$
Distance AB to center = $\sqrt{r^2 - (\frac{AB}{2})^2} = \sqrt{7-{2 \over 4}}= \sqrt {6.5}$
Thus, AB perpendicular to line PC, with slope, $m = -\frac{0.5}{2.5} = -0.2$
Line L: $y = -x/5 + 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3334232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How can you show by hand that $ e^{-1/e} < \ln(2) $? By chance I noticed that $e^{-1/e}$ and $\ln(2)$ both have decimal representation $0.69\!\ldots$, and it got me wondering how one could possibly determine which was larger without using a calculator.
For example, how might Euler have determined that these numbers ar... | Here's a method that only uses rational numbers as bounds, keeping to small denominators where possible. It is elementary, but finding sufficiently good bounds requires some experimentation or at least luck.
We instead establish the inequality $$-\log \log 2 < \frac{1}{e}$$ of positive numbers, which we can see to be e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3338415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 0
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Primitive Root Modular Arithmetic Question 3)Given a prime $p$ and an integer $a$, we say that $a$ is a $\textit{primitive root} \pmod p$ if the set $\{a,a^2,a^3,\ldots,a^{p-1}\}$ contains exactly one element congruent to each of $1,2,3,\ldots,p-1\pmod p$.
For example, $2$ is a primitive root $\pmod 5$ because $\{2,2^2... | Modulo $7$, the powers of $1$ are $1,1,1,1,1,1,$ so $1$ is not a primitive root; the powers of $2$ are $2,4,1,2,4,1,$ so $2$ is not a primitive root; and the powers of $3$ are $3,2,6,4,5,1$, so $3$ is a primitive root. Can you take it from here? Test whether $4,5,$ and $6$ are primitive roots, and then compute the su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3338752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral of $\frac{1}{\sin^2(x)\cos^2(x)}$ $$ I = \int\frac{1}{\sin^2(x)\cos^2(x)}$$
I have tried the following:
$$\sin^2x = \sin x \cdot \sin x = \frac{1}{2}(1 - \cos2x)$$
$$\cos^2x = \cos x \cdot \cos x = \frac{1}{2}(1 + \cos2x)$$
The integral becomes:
$$I = 4\int\frac{1}{1 - \cos^2(2x)} = 4\int\frac{1}{\sin^2(2x)}$... | \begin{align*}
I & = \int\frac{1}{\sin^2(x)\cos^2(x)} \, dx\\
&= \int\frac{\sin^2x+\cos^2x}{\sin^2(x)\cos^2(x)} \, dx\\
&= \int\frac{1}{\cos^2(x)} \, dx + \int\frac{1}{\sin^2(x)} \, dx\\
&= \int\sec^2(x) \, dx + \int\csc^2(x) \, dx\\
&=\tan x -\cot x+c
\end{align*}
But this is same as your answer, so nothing wrong with... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3339030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Geometry - Rectangle ABCD with inside point E. Find the least possible value for sum of interger distances from E to 4 vertices.
The point E lies within the rectangle ABCD.
If the distances from the vertices to E are all distinct integers, what is the least possible value of AE + BE + CE + DE?
| We want
$$\left\{
\begin{align*}
a^2&+c^2&=p^2& \hspace{4em} (1)\\
a^2&+d^2&=q^2 &\hspace{4em} (2)\\
b^2&+c^2&=r^2 &\hspace{4em} (3)\\
b^2&+d^2&=s^2 &\hspace{4em} (4)
\end{align*}
\right.$$
where $a+b$ and $c+d$ are the side lengths of the rectangle, and $p,q,r,s$ are the distances to the vertices. We want to find a w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Bernoulli Numbers are rationals? There is a explicit form that admit Bernoulli numbers rationals but there is another definition where the Bernoulli numbers are $B_n$, such that $\displaystyle \frac{x}{e^x-1}= \sum_{n=0}^\infty B_n \frac{x^n}{n!}$.
How can I prove that?? or it's equivalence?
| We have
$$\frac{x}{e^x-1} = \frac{x}{x + x^2/2 + x^3/6 + x^4/24 + \cdots} = \frac{1}{1 + x/2 + x^2/6 + x^3/24 + \cdots}.$$
Now, if we let $f(x) := \frac{x}{2} + \frac{x^2}{6} + \frac{x^3}{24} + \cdots$, then
$$\frac{x}{e^x-1} = \frac{1}{1 + f(x)} = 1 - f(x) + (f(x))^2 - (f(x))^3 - \cdots.$$
In this expansion, each powe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $k^{th}$ power of a square matrix I am trying to find the $A^{k}$, for all $k \geq 2$ of a matrix, \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix}
My approach:
$A^{2}=\begin{pmatrix} a^2 & ab+b \\ 0 & 1 \end{pmatrix}$
$A^{3}=\begin{pmatrix} a^3 & a^{2}b+ab+b \\ 0 & 1 \end{pmatrix}$
$A^{4}=\begin{pmatrix} a^4 ... | To add on to lhf's hints, do you know induction? If so, try to find patterns in each of the 4 entries individually once you have corrected your examples.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3345771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Limit of function $f(x) = \sqrt{(xa + d)^2 + x^2 b^2} - \sqrt{(xa - d)^2 + x^2 b^2}$
I tried to calculate limit when $x$ goes to infinity for the following function
$$f(x) = \sqrt{(xa + d)^2 + x^2 b^2} - \sqrt{(xa - d)^2 + x^2 b^2}$$
where $a$, $b$, $d$ are some positive constants.
It's easy to see that terms b... | Let $t=1/(2 d x)$. Then the limit of interest can be written as
$$\lim_{t\to 0^{\pm}}\left(\sqrt{(2a d/t + d)^2 + b^2 d^2/t^2} - \sqrt{(2a d/t - d)^2 + b^2 d^2/t^2}\right)\\
\hspace{4cm}=2d\,\text{sgn}(t)\times \lim_{t\to 0^{\pm}}\frac{1}{t}\left[\sqrt{(a+t/2)^2+b^2}-\sqrt{(a-t/2)^2+b^2}\right].$$
The definition of th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3346300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
An identity containing $\exp\left(i\cdot \frac{2k\pi}{n} \right)$ I found the following equation without proof.
$$\frac{1}{(1+x)^n - 1} = \frac{1}{n}\sum_{k=0}^{n-1}\frac{a(k,n)}{x + 1 – a(k,n)}$$
where $$a(k,n) = \exp\left(i \cdot \frac{2k\pi}{n} \right); \quad n = 1,2,\ldots; \quad k=0,1,2,\ldots,n-1.$$
Note that $(a... | We have the following polynomial with its roots:
$$P(x)= (x+1)^n-1=\prod\limits_{j=0}^{n-1}\left(x-x_j\right)$$
From which
$$P'(x)=n(x+1)^{n-1}=\sum\limits_{k=0}^{n-1}\prod\limits_{j=0,j\ne k}^{n-1}\left(x-x_j\right)$$
then
$$\color{blue}{\frac{P'(x)}{P(x)}=\sum\limits_{k=0}^{n-1}\frac{1}{x-x_k}} \Rightarrow \\
\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3346978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Is it possible to solve such an unlinear matrix equation? I have a matrix equation of the form:
$Sx=x^TKxp$, while $
S = \begin{pmatrix}
0 & a & a \\
b & 0 & b \\
c & c & 0 \\
\end{pmatrix}
$, $a$ and $b$ are real, known numbers, $K = \begin{pmatrix}
0 & \frac{1}{2} & \frac{1}{2} \\
\frac{1}... | Let $u=x^TKx=rs+st+tr$ (1).
$Sx=up$ is then a straightforward equation which gives $r,s,t$ in terms of $u$.
Substitute into (1) and you have a quadratic equation for $u$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3348137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is there a substitute that will eliminate terms of the degree $3$ and $2$ from the quartic polynomial? The standart form of the quartic equation:
$$ax^4+bx^3+cx^2+dx+e=0$$ where $$a\neq 0.$$
The purpose of the question is not to solve quartic. Resources related to the solution are sufficiently available. Just about s... | We can assume that $a=1$ by dividing by it, so start with
$$ x^4 + bx^3 + cx^2 + dx + e = 0 $$
(there's no need for $f$, either).
We will use a more general substitution known as a Tschirnhaus substitution: put
$$ y = x^2 + \lambda x + \mu . $$
Then
$$ y^2 = x^4 + 2\lambda x^3 + 2(\mu+\lambda^2) x^2 + 2\lambda\mu x + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How can I solve $\int e^{2\theta} \sin(3\theta)\, d\theta$ with integration by parts? $\int e^{2\theta}\sin(3\theta)d\theta$ seems to be leading me in circles. The integral I get when I use integration by parts, $\int e^{2\theta}\cos(3\theta)d\theta$ just leads me back to $\int e^{2\theta}\sin(3\theta)d\theta$. I am no... | Hint
In case integration by parts is not mandatory,
find
$$\dfrac{e^{2x}(a\cos3x+b\sin3x)}{dx}$$ and compare with $e^{2x}\sin3x$ to find the values of $a,b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3351012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
If $A+B+C=180$, then $\frac{\sin2A+\sin2B+\sin2C}{\cos A+\cos B+\cos C-1}=8\cos\frac A2 \cos\frac B2 \cos\frac C2$ Then $2A+2B+2C =360$
So $$\sin 2C=-\sin(2A+2B)$$
Putting that in the equation
$$\frac{2\sin(A+B)\sin(A-B)-2\sin(A+B)\cos(A+B)}{\cos A+\cos B-\cos(A+B)+1}$$
$$\frac{2\sin(A+B)[\sin(A-B)-\cos(A+B)]}{\cos A+... | $\begin{align}
\sin(2a) + \sin(2b) + \sin(2c) &= 2 \sin(a+b) \cos(a-b) - \sin(2a+2b) \cr
&= 2\sin(a+b)\;\{ \cos(a-b) - \cos(a+b) \} \cr
&= 2\sin(c)\;\{2 sin(a) sin(b)\} \cr
&= 4\sin(a)\sin(b)\sin(c) \cr
{\sin(2a) + \sin(2b) + \sin(2c) \over 8 \cos{a\over2}\cos{b\over2}\cos{c\over2}}
&= 4 \sin{a\over2} \sin{b\over2} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3353872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Solve $\tan x =\sec 42^\circ +\sqrt{3}$ For the trigonometric equation,
$$\tan x =\sec 42^\circ+\sqrt{3}$$
Find the angle $x$, where $0<x<180^\circ$.
I tried to solve for an unknown angle $x$ in a geometry problem with a trigonometric approach. I ended up with the trig equation above. Without hesitation, I reached my c... | Golden triangles, I use these mnemonic:
$\;\displaystyle \cos 36° = \frac{\sqrt{2+\frac{1}{ϕ}}}{2} = \frac{ϕ}{2}\;,\;
\sin 36° = \frac{\sqrt{2-\frac{1}{ϕ}}}{2}$
$\;\displaystyle \cos 18° = \frac{\sqrt{2+ϕ}}{2} \qquad\;\;,\;
\sin 18° = \frac{\sqrt{2-ϕ}}{2} = \frac{1}{2ϕ}$
$\displaystyle \cos (60°-18°) =
\frac{1}{2} ×... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3354759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
If $\sin\alpha+\sin\beta=a$ and $\cos\alpha-\cos\beta=b$, then what is $\tan(\frac{\alpha-\beta}{2})$? If I square both the equations
$$2+2\sin(\alpha-\beta)=a^2+b^2$$
$$\sin(\alpha-\beta)=\frac{a^2+b^2-2}{2}$$
Since $\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$, then $$\sin(\alpha-\beta)=\frac{2\tan\frac{\alpha-\b... | Denote $x = \dfrac{\alpha + \beta}{2}$ and $y = \dfrac{\alpha - \beta}{2}$.
Question: Are the any relations between $x, y$ and $\alpha, \beta$?
Answer:
Yes. $$x - y = \dfrac{\alpha + \beta}{2} - \dfrac{\alpha - \beta}{2} = \dfrac{2\beta}{2} = \beta\\x + y = \dfrac{\alpha + \beta}{2} + \dfrac{\alpha - \beta}{2} = \dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3356271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
About $e^{i z} = \cos z + i \sin z$ in Michael Spivak "Calculus 3rd Edition". I am reading "Calculus 3rd Edition" by Michael Spivak.
The author wrote as follows (p. 555):
Moreover, if we replace $z$ by $i z$ in the series for $e^z$, and make a rearrangement of the terms (justified by absolute convergence), something p... | Your statement:
But I think the author didn't use a rearrangement of the terms at all.
He did re-arrange the terms as said...See:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3357381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Integration using Trig Substitution I was given the following problem: $$\int\sqrt{1-7w^2}\ dw$$ I used the sin substitution - getting $w=\frac1{\sqrt{7}}\sin\theta$. I then needed to change the $dw$ to a $d\theta$, so I got this: $dw=\frac1{\sqrt{7}}\cos\theta\ d\theta$. My new problem looks like this: $$\int\sqrt{1-7... | It may be more desirable to integrate directly without substitution.
$$I= \int \sqrt{1-x^2}dx = x\sqrt{1-x^2} + \int \frac{x^2}{\sqrt{1-x^2}}dx $$
$$=x\sqrt{1-x^2} -I + \int \frac{dx}{\sqrt{1-x^2}}=x\sqrt{1-x^2} -I + \sin^{-1}x+C$$
Thus,
$$I = \frac 12 \left( x\sqrt{1-x^2}+ \sin^{-1}x \right)+C $$
With $x=\sqrt{7}w$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3359185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How do I find all complex numbers that are solutions of $z^3 +8i=0$? My workbook has answers in the standard form $a+bi$. I would assume that to solve this, I would expand the complex number $z$ into trigonometric form to deal with that exponent. This is what I have so far using De Moivre's Theorem:
$(r^3(\cos(3\theta)... | $(r^3(\cos(3\theta)+i\sin(3\theta)) +8i=0$
So continue:
$(r^3(\cos(3\theta)+i\sin(3\theta)) = -8i= 8(0 + i(-1))$
$(r^3(\cos(3\theta)+i\sin(3\theta)) = 8((\cos \frac {3\pi}2+2k\pi) + i\sin(\frac{3\pi}2+2k\pi))$
So $r^3 = 8$ and $3\theta = \frac {3\pi}2 + 2k\pi$
So $r =2$ and $\theta = \frac \pi 2 + \frac 23k\pi$
So the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3361524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 6
} |
Solving $\sin 3\theta=1/2$ on the interval $[0,2\pi]$. I don't understand where some solutions came from. I am learning precalculus, and I understand how to obtain first two solutions, but I don't understand where did last four solutions came from:
All values of $\theta$ in the interval $[0,2\pi]$ that satisfy $\sin 3... | $$\theta \in \left[0, 2\pi\right]\implies 3\theta \in \left[0, 6\pi\right]$$
$$\sin 3\theta = \dfrac{1}{2}$$
$$3\theta = \frac{\pi}{6}+2n\pi, \dfrac{5\pi}{6}+2n\pi, n \in \Bbb Z$$
But $3\theta \in \left[0, 6\pi\right]$, so $$3\theta = \dfrac{\pi}{6},\dfrac{5\pi}{6},\dfrac{13\pi}{6},\dfrac{17\pi}{6},\dfrac{25\pi}{6},\df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3361862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Trying to find $c_{1}$ , $c_{2}$ for Big theta notation. Need to find $c_{1},$ $c_{2}$ and $n_{0}$.
\begin{equation}
c_{1}n^3 \leq \frac{n^3}{100} - 100n^2 - 100n + 3\leq c_{2}n^3
\end{equation}
\begin{equation}
c_{1} \leq \frac{1}{100} - \frac{100}{n} - \frac{100}{n^2} + \frac{3}{n^3} \leq c_{2}
\end{equation}
\begin{... | The first step is to find a suitable $n_0.$
You know that you need
$$
0 < c_1 \leq \frac{1}{100} - \frac{100}{n} - \frac{100}{n^2} + \frac{3}{n^3} \leq c_2.
$$
Ignore the $c_1$ and $c_2$ for a moment, and you may see that you need
$$
0 < \frac{1}{100} - \frac{100}{n} - \frac{100}{n^2} + \frac{3}{n^3}.
$$
This must be ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to calculate the asymptotes to $y^2+2y-4x^2=0$ Calculate the center and asymptotes to the hyperbola $y^2+2y-4x^2=0$, aswell as the intersections of the actual coordinate axis.
The hyperbola takes the form $\left ( \frac{x}{\frac{1}{\sqrt{4}}}\right )^{2}-\left ( {y+1}\right )^{2}=-1$ and I understand how to comple... | I think your hyperbola has the form
\begin{align*}y^2 + 2y {\color{red}+\color{red}1\color{red}-\color{red}1}-4x^2 =0 &\Leftrightarrow (y+1)^2 -1- 4x^2= 0 \\&\Leftrightarrow (y+1)^2 - 4x^2=1
\end{align*}
So $a = \frac{1}{2}$ and $b= 1$.
Then you should get the asymptotes
$$y = \pm 2x-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
find the value $\sum_{i=1}^{n}|a_{i}|-\sum_{1\le ilet $n>3$,if $a_{1}=a_{2}=\cdots=a_{n-1}=1$,and $a_{n}=-2$,find the following value
$$f_{n}=\sum_{i=1}^{n}|a_{i}|-\sum_{1\le i<j\le n}|a_{i}+a_{j}|+\sum_{1\le i<j<k\le n}|a_{i}+a_{j}+a_{k}|-\cdots+(-1)^{n-1}|a_{1}+a_{2}+\cdots+a_{n}|$$
I have find the $n=4$,then
$$f_{4}... |
First note the following identities
${n}\choose{1}$ - $2$${n}\choose{2}$ + $3$${n}\choose{3}$ - ... $=0$
${n}\choose{1}$ - ${n}\choose{2}$ + ${n}\choose{3}$ - ... $=1$.
The sums which do not include $a_n$ add up to
${n-1}\choose{1}$ - $2$${n-1}\choose{2}$ + $3$${n-1}\choose{3}$ - ... $=0$.
The sums which do include $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3363430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.