Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
What should be number of integral values of n? If the period of the function $\cos(nx)\sin(5x/n)$ is $3\pi$ then what should be number of integral values of $n$ ?
My approach :
I tried like period of $\cos(nx)$ is $2\pi$/n and $\sin(5x/n)$ is $2\pi n/5$
So the period should be L.C.M of $2\pi$/n and $2\pi n/5$.Which is... | You're given
$$\cos nx\cdot\sin\frac{5x}n=\cos(n(x+3\pi))\cdot\sin\frac{5(x+3\pi)}n$$
Using some trigonometry:
$$\cos(n(x+3\pi))\cdot\sin\frac{5(x+3\pi)}n=\cos(nx+3n\pi)\cdot\sin\left(\frac{5x}n+\frac{15\pi}n\right)=$$
$$=\overbrace{(-1)^n}^{=\cos3n\pi}\cos nx\left(\sin\frac{5x}n\cos\frac{15\pi}n+\sin\frac{15\pi}n\cos\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1276819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Trig substitution or part integration How can I integrate the following function:
$$\int \frac{(1-x^2)^{3/2}}{x^6}\,dx $$
? I've tried integration by parts but I think I'm doing in the wrong way.
I need some help!
| $$\int \frac{(1-x^2)^{3/2}}{x^6}\,dx $$
Substitute $x=\sin(u)$ and $dx=\cos(u) du$. Then $(1-x^2)^{3/2}=(1-\sin^2(u))^{3/2}=\cos^3(u)$ and $u=\sin^{-1}(x)$
$$\int \frac{(1-x^2)^{3/2}}{x^6}\,dx =$$
$$\int \cot^4(u)\csc^2(u) du$$
For the intergrand $\cot^4(u)\csc^2(u)$ substitute $s=\cot(u)$ and $ds=-\csc^2(u) du$
$$\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1277459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Logarithmic Differentiation equation, Help! So, I have to differentiate this via $\log$. I am still learning, so please be patient, I will try to explain everything I did. Please tell me if it is correct.
$$y=\frac{(x+3)^4(2x^2+5x)^3}{\sqrt{4x-3}}$$
$$\ln(y) = \ln\left((x+3)^4(2x^2+5x)^3)\right) - \ln\left(\sqrt{4x-3}\... | You've left out a plus sign in the last line. It should read
$$\ln(y) = 4\ln(x+3) + 3\ln(2x^2+5x) - \frac{1}{2} \ln(4x-3).$$
Then you can still factor the second term to get
\begin{align*}
\ln(y) &= 4\ln(x+3) + 3\ln(x(2x+5)) - \frac{1}{2} \ln(4x-3)\\
&=4\ln(x+3) + 3\ln(x)+3\ln(2x+5) - \frac{1}{2} \ln(4x-3).
\end{align*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1279571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
What is indefinite $ \int\frac{\sin\left(x\right)\cos\left(x\right)}{\sqrt{3 - x^{4}}}{\rm d}x . $ $$
\int\frac{\sin\left(x\right)\cos\left(x\right)}{\sqrt{3 - x^{4}}}{\rm d}x .
$$
Hello.
Today, in exam, we had to evaluate this integral. Noone was able to do it. Appreciate any help.
| As said in comments, the integral you posted does not seem to be doable.
Let us admit a typo in the problem and let us consider $$I=\int\frac{\sin\left(x\right)\cos\left(x\right)}{\sqrt{3 - \sin^4(x)}}\,{\rm d}x $$ The denominator looking more or less as $\sqrt{1-y^2}$, I thought that $\sin^{-1}(...)$. So, let us try a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1279974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve the inequality $2\left|x+\frac{1}{4}\right| < 9$
I keep trying to figure this out and I can't.
I tried to split up the absolute value first.
$2 (x+\frac{1}{4}) < 9$
$2+x+\frac{1}{4} < 9$
subtract $2$ from both sides? $x+\frac{1}{4} < 7$
and
$2 -(x+\frac{1}{4}) < 9$
$2-x-\frac{1}{4} < 9$
subtract $2$ from both s... | $2|x+0,25| < 9 \rightarrow |x+0,25| < 4,5 \rightarrow |x| < 4,25 \rightarrow -4,25 < x <4,25$
You can negate $0,25$ from each side, since it is positive, and the absolute value won't effect it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1280776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What's wrong?: Find the infinite sum $S = 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} + \ldots$ The answer I got by hand is not the same to the one I found using a spreadsheet.
$\displaystyle S = 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} + \ldots$
$\displaystyle \frac{1}{4}S =... | When you write "Using Excel the answer is $\frac{5}{3}$", I think what you mean is that you have constructed a model of the problem in Excel that suggests the answer is $\frac{5}{3}$. Your Excel model of the problem is wrong (as calculation by hand of the first few terms of of the sum shows). So the issue is in your Ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1280948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
How to prove $\lim\limits_{(x,y)\to(0,0)}\frac{{x^3{y^2}}}{{{x^4} + {3y^4}}} = 0$? To prove that $$\lim\limits_{(x,y)\to(0,0)}\frac{{x^3{y^2}}}{{{x^4} + {3y^4}}} = 0$$
I start with
$$\left| {\frac{{{x^3}{y^2}}}{{{x^4} + 3{y^4}}}} \right| \leqslant \left| {\frac{{{x^3}{y^2}}}{{{x^4}}}} \right| = \left| {\frac{{{y^2}}}{... | Another approach is to use the fact that for $x\gt0$, we have $\left(\sqrt{x}-\frac1{\sqrt{x}}\right)^2\ge0\implies x+\frac1x\ge2$:
$$
\begin{align}
\left|\frac{x^3y^2}{x^4+3y^4}\right|
&=\frac{|x|}{\sqrt3}\frac{x^2(\sqrt[4]3\,y)^2}{x^4+3y^4}\\[9pt]
&=\frac{|x|}{\sqrt3}\frac1{\left(\raise{2pt}{\frac{x}{\sqrt[4]3\,y}}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Integration of $\frac{1}{\sin x+\cos x}$ I'm given this $\int\frac{1}{\sin x+\cos x}dx$.
My attempt,
$\sin x+\cos x=R\cos (x-\alpha)$
$R\cos \alpha=1$ and $R\sin \alpha=1$
$R=\sqrt{1^2+1^2}=\sqrt{2}$,
$\tan\alpha=1$
$\alpha=\frac{\pi}{4}$
So, $\sin x+\cos x=\sqrt{2}\cos (x-\frac{\pi}{4})$
$\int\frac{1}{\sin x+\cos x}d... | Another method would be to multiply by $\cos x-\sin x$ on the top and bottom to get
$\displaystyle\int\frac{\cos x-\sin x}{\cos 2x} dx=\int\frac{\cos x}{1-2\sin^2 x} dx-\int\frac{\sin x}{2\cos^2 x-1} dx$, and then use
$\displaystyle\int\frac{1}{1-2u^2}du=\frac{\sqrt{2}}{4}\ln\bigg|\frac{1+\sqrt{2}u}{1-\sqrt{2}u}\bigg|+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1287026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Integral solutions to $56u^2 + 12 u + 1 = w^3$ I would like to find all integer solutions to $$56u^2 + 12 u + 1 = w^3.$$ My computer thinks the only integral point is $(0,1).$
This problem arises from Integer solutions of $x^3 = 7y^3 + 6 y^2+2 y$? and is likely to be easier.
Note after getting answers: comes down to th... | Multiplying by $14^3$, we obtain
$$14^4\cdot 2^2 u^2 + 14^3 \cdot 12u + 14^3 = (14w)^3 \implies (14(28u+3))^2+980 = (14w)^3$$
This is a Mordell equation of the form $Y^2 = X^3-980$, which has $5$ solutions given by $(14,42)$, $(21,91)$, $(29,153)$, $(126,1414)$ and $(326,5886)$.
Of this only $(14,42)$ has $Y$ of the fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
how to prove by contradiction that any distance between a curve $x^4 - x^2 + y^4 - y^2 = 0$ and the origin is less than or equal to $\sqrt{2}$ Given a closed trajectory $x^4 - x^2 + y^4 - y^2= 0$
Prove that any distance between any point on the curve and the origin does not exceed $\sqrt2$ (ie, maximum distance from th... | We have
$$0 = x^4-x^2 + y^4-y^2 = \left(x^2-\dfrac12\right)^2 + \left(y^2-\dfrac12\right)^2 - \dfrac12 \implies \left(x^2-\dfrac12\right)^2 + \left(y^2-\dfrac12\right)^2 = \dfrac12$$
I trust you can finish off from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Taylor series $\ln(1+e^x)$ about $x=0$ What's the best way to determine this up $x^3$ terms?
I thought it would be to take the series for $\ln(1+x)$ and the series for $e^x$ up to $x^3$ and sub the second series into the first.
$$\ln(1+x) = x - \frac {x^2}{2} +\frac {x^3}{3} + \cdots$$
$$e^x = 1+ x + \frac {x^2}{2} +\... | Your approach won't quite work because the argument of the logarithm in $\log(1+e^x)$ is not near $1$ at $x=0$ (it's near $2$ instead). To fix this, note that
$$
\log(1+e^x)=\log 2 + \log\left(\frac{1}{2} + \frac{1}{2}e^x\right)=\log 2 + \log\left(1 + \frac{1}{2}x+\frac{1}{4}x^2 + \frac{1}{12}x^3+\ldots\right).
$$
The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the remainder when $(x+1)^n$ is divided by $(x-1)^3$
Find the remainder when $(x+1)^n$ is divided by $(x-1)^3$
I know that
\begin{equation*}
(1 + x)^n = 1 + nx +\frac {n(n-1)}2!\cdot x^2 +\frac {n(n-1)(n-2)}3! \cdot x^3 +...
\end{equation*}
but how can I use it to solve the above problem>Is there any other ... | By setting $y=x-1$, we just have to find the remainder of $(y+2)^n$ when divided by $y^3$.
By the binomial theorem:
$$(y+2)^n \equiv 2^n + n2^{n-1} y + n(n-1)2^{n-3} y^2\pmod{y^3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1295101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
With the aid of series, show that if $f(z)=\frac{\operatorname{cos}z}{z^2-(\pi/2)^2}$, then $f$ is an entire function. Prove that if
$$f(z)=
\begin{cases}
\frac{\operatorname{cos}z}{z^2-(\pi/2)^2}, & \text{when} \; z\neq \pm \pi/2, \\
-\frac{1}{\pi}, & \text{when} \; z=\pm \pi/2,
\end{cases}
$$
then $f$ is an entire f... | Taylor expanding $\cos z$ about $z=\frac{\pi}{2}$ one finds that:
\begin{equation}
\cos z=-\left(z-\frac{\pi}{2}\right)+\frac{1}{6}\left(z-\frac{\pi}{2}\right)^3+\ldots
\end{equation}
Similarly, Taylor expanding $\frac{1}{z+\frac{\pi}{2}}$ about $z=\frac{\pi}{2}$ one finds that:
\begin{equation}
\frac{1}{z+\frac{\pi}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1296896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
6 people are holding a show, one at a time, such that person $x$ has to go after person $y$ and person $z$. How many ways could the show be held? Let's say the people are called $a$, $b$, $c$, $x$, $y$, $z$
My initial thinking was to go by fixing "$x$" in a certain position, so:
$\underline {} \underline {} \underline ... | You can make your method work.
If $x$ is in the last position, then there are $5!$ ways to arrange the other letters.
If $x$ is in the fifth position, then there are $3$ ways to select the letter in the last position from among $a, b, c$ and $4!$ ways to arrange the letters before $x$. Therefore, there are $3 \cdot 4!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
If $x-2y+4=0$ and $2x+y-5=0$ are the sides of isosceles triangle having area $10$ sq unit .Equation of third side is? Okay, I know two sides of an isosceles triangle are equal . I have also taken out the intersection points of the lines given in the question . Other than this , I have no clue about how I will find out... | The lines: $x-2y+4=0$ & $2x+y-5=0$ are normal to each other thus the triangle is an isosceles right triangle. Thus you are looking for the third side representing the hypotenuse of isosceles right triangle. Let $a$ be length of two equal & perpendicular sides of isosceles right triangle then the area of triangle is giv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
"Rationalizing the denominator" of $1/(a + b\sqrt[3]{2} + c\sqrt[3]{4})$? If $(a, b, c) \in \mathbb{Q}^3 \setminus \{(0, 0, 0)\}$, so that $a + b\sqrt[3]{2} + c\sqrt[3]{4}$ is a nonzero element of $\mathbb{Q}(\sqrt[3]{2})$, is there a formula for $${1\over{a + b\sqrt[3]{2} + c\sqrt[3]{4}}}$$ as a rational linear combin... | We know that $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx).$$
Making the obvious substitutions, we get $$a^3+2b^3+4c^3-6abc=(a+b\sqrt[3]{2}+c\sqrt[3]{4})(a^2+b^2\sqrt[3]{4}+2c^2\sqrt[3]{2}-ab\sqrt[3]{2}-ac\sqrt[3]{4}-2bc),$$
so
$$\frac{1}{a+b\sqrt[3]{2}+c\sqrt[3]{4}}=\frac{(a^2-2bc)+(2c^2-ab)\sqrt[3]{2}+(b^2-ac)\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Rolling two dice, what is the probability that two consecutive $7$s happens earlier than a $12$? Alice and Bob are playing a game involving two dice. If a sum of 12 appears, Alice wins and they stop playing. If a 7 appears twice in a row, Bob wins and they stop playing. What is the probability that Bob wins this game?
... | Let's call the probability that Bob wins be $x$.
On the first roll of the two dice, there are three options:
*
*The dice show a sum of $12$ (probability $\frac{1}{36}$) and Alice would win immediately.
*The dice show a sum of $7$ (probability $\frac{1}{6}$) and this is a little more complicated (see next tree diagr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Given $\csc\theta=-\frac53$ and $\pi<\theta<\frac32\pi$, evaluate sine ,cosine, and tangent of $2\theta$ If $\csc\theta=\frac{-5}{3}$, what is the exact value of $\tan(2\theta)$, $\sin(2\theta)$, and $\cos(2\theta)$ on the interval of $\left(\pi, \frac{3\pi}{2}\right)$?
I think I'm getting the fraction negatives wrong.... | For the given interval of $\theta\in \left(\pi, \frac{3\pi}{2}\right)$, we have $$\csc\theta=-\frac{5}{3} $$$$\implies \sin\theta=-\frac{3}{5} $$$$\implies\theta=\sin^{-1}\left(\frac{-3}{5}\right) $$ $$\implies \color{green}{\theta=\pi+\sin^{-1}\left(\frac{3}{5}\right)} \quad (\pi<\theta<\frac{3\pi}{2})$$
Now, we have ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1301708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Help with limit of function How can I calculate the limit $$\lim_{x \to \infty} x^{3/2}( \sqrt{x+1}+ \sqrt{x-1}-2 \sqrt{x})$$
I had ideas like using Laurent series, but I dont think I am allowed since its an elementary course, I tried to play around with the terms but I didnt manage. Help anyone?
| Classic limits involving $\sqrt{1+h}$ and $\sqrt{1-h}$ are
\begin{eqnarray*}
\lim_{h\rightarrow 0^{+}}\frac{\left( \sqrt{1+h}-1-\frac{1}{2}h\right) }{%
h^{2}} &=&-\frac{1}{8}, \\
\lim_{h\rightarrow 0^{+}}\frac{\left( \sqrt{1-h}-1+\frac{1}{2}%
h\right) }{h^{2}} &=&-\frac{1}{8}.
\end{eqnarray*}
Therefore it suffices to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1302346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
why $\tan x = \frac{\sin x}{\cos x}$? and not $\tan x$ = opposite/adjacent? we know that $\tan x =\left(\frac{\text{opposite}}{\text{adjacent}}\right)$, but sometimes I see that $\tan x = (\frac{\sin x}{\cos x})$, is that the same thing or why it is different sometimes?
cause when $\tan x =\left(\frac{\text{opposite}}{... | Roughly, they are the same definition of tangent:
$$\begin{align} \require{cancel}
\tan = \frac{\sin}{\cos}
&= \frac{\left(\frac{\text{opposite}}{\text{hypotenuse}}\right)}
{\left(\frac{\text{adjacent}}{\text{hypotenuse}}\right)} \\
&= \left(\frac{\text{opposite}}{\text{hypotenuse}}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1302989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Laplace transform of function Assume that $f(u)=(\frac{b}{πu^3})^{1/2} e^{2b} e^{-bu} e^{-b/u}$, where $b>0.$ I am trying to calculate the Laplace transform $L\{f(u)\}(s)$ and then the $n_{th}$ derivative of this transform, $L^{n}(s),$ but i have difficulties in calculating the integrals, especially in the second one. ... | So you want to evaluate the integral
$$I = \int_0^{\infty} du \, u^{-3/2} e^{-(a u + b/u)} $$
where $a=s+b$. Let $u = v^2$; then
$$I = 2 \int_0^{\infty} \frac{dv}{v^2} e^{-(a v^2+b/v^2)} = 2 e^{2 \sqrt{a b}} \int_0^{\infty} \frac{dv}{v^2} e^{-(\sqrt{a} v+\sqrt{b}/v)^2}$$
Now let $y=\sqrt{a} v+\sqrt{b}/v$. Then
$$v ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
I think I can complete the square of any quadratic, is it true? (Any reason to ever use Quad. Formula?) I was taught that you could only complete the square of a quadratic if the coefficient on the $x^2$ term is 1.
However, playing a little bit with other quadratics, I've found that it's just not true. Based on the CT... | It seems like your questions have been answered, so this response is more of an addition. Completing the square is a useful rearrangement in and of itself - beyond the usage for solving quadratics. You may be aware of its application in placing the equation of a conic section in standard form.
For example, consider
$$x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 6,
"answer_id": 1
} |
Solutions to $(4x^2+\frac{16}3x)^{\sqrt {3-x}}=(4x^2+\frac{16}3x)^{\sqrt {2x+11}-\sqrt{x+2}}$ $$(4x^2+\frac{16}3x)^{\sqrt {3-x}}=(4x^2+\frac{16}3x)^{\sqrt {2x+11}-\sqrt{x+2}}$$
I found the solutions to be $0, -\frac32, -1, -\frac43$
I can't figure out why any of those wouldn't work, but my textbook apparently only sees... | $x = \frac 16$ is one of the solutions to $ 4x^2-\frac {16}{3}x = 1$
Three ways to establish equality.
1) $4x^2 + \frac{16}{3} x = 0$ whose roots sum to $-\frac 43 $
2) $4x^2 + \frac{16}{3} x = 1$ whose roots also sum to $-\frac 43 $
3) $\sqrt{3-x} = \sqrt{2x+11}- \sqrt{x+2}$ is equivalent to $x^2+\frac 52 x + \frac 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Prove factorial problem $\forall n\in\mathbb N$ $C_n=\frac{1}{n+1}\left(\begin{matrix}2n\\n\end{matrix}\right)$.
Prove that $C_{n+1}=\left(\begin{matrix}2n\\n\end{matrix}\right)-\left(\begin{matrix}2n\\n-1\end{matrix}\right)$
I tried many ways but got nothing. Appreciate any tips.
My attempt: By defining, $C_{n+1}= \... | You need to simplify the second term: $\binom{2n}{n-1}=\dfrac{(2n)!}{(n-1)!(n+1)!}= \dfrac{(2n)!}{\frac{n!}{n}\cdot (n!\cdot (n+1))}= \dfrac{n}{n+1}\cdot \binom{2n}{n} \Rightarrow C_n = \binom{2n}{n} - \dfrac{n}{n+1}\binom{2n}{n}= \left(1-\dfrac{n}{n+1}\right)\binom{2n}{n} = \dfrac{1}{n+1}\binom{2n}{n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1308530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $x^2 \equiv y^2 \pmod p$ if and only if $x \equiv y \pmod p$ or $x \equiv -y \pmod p$. Hint: $x^2-y^2 = (x+y)(x-y)$. This is the exercise verbatim:
An integer n is a square modulo p if there exists another integer x
such that $n \equiv x^2 \pmod p$. Prove that $x^2 \equiv y^2 \pmod p$ if and only if $x \equi... | if $x^2 \equiv y^2 \pmod p$ then $p \mid x^2 - y^2$ and so $p \mid (x-y)(x+y)$ And so by euclids lemma , since $p$ is prime then we have $p \mid (x-y)$ or $p \mid (x+y)$ and so $x \equiv y \pmod p$ or $x \equiv -y \pmod p$
On the other hand if $x \equiv y \pmod p$ or $x \equiv -y \pmod p$ then assume that $x \equiv y ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Why does the largest $x$ such that $a$, $b$ divided by $x$ leave the same remainder equal $a-b$? Suppose two numbers $a$ and $b$ as, $a=kq_1+r_1=3\times 17 + 1 = 52$ and $b = kq_2+r_2=3 \times 15 +1=46$.
It is clear that $52$ and $46$ leave the same reminder 1 when divided by $3$, because I designed them this way. But... | Here's my attempt at a step-by-step answer that does not use the notion of modular arithmetic.
Suppose two numbers $a$ and $b$ as, $a=kq_1+r_1=3\times 17 + 1 = 52$ and $b = kq_2+r_2=3 \times 15 +1=46$.
It is clear that $52$ and $46$ leave the same reminder 1 when divided by $3$, because I designed them this way. But ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1312201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 1
} |
Finding $\lim\limits_{x\to 0}(\frac{\sin(x)}{x})^{\frac{\sin(x)}{x-\sin(x)}}$ with and without L'Hopital's Rule. So, we have to find $\lim\limits_{x\to 0}(\frac{\sin(x)}{x})^{\frac{\sin(x)}{x-\sin(x)}}$ with and without L'Hopital's Rule.
My Work: Let $\lim\limits_{x\to 0}(\frac{\sin(x)}{x})^{\frac{\sin(x)}{x-\sin(x)}}... | You need to work a bit more. Let's continue it as follows
\begin{align}
\log L &= \lim_{x \to 0}\frac{\sin x}{x - \sin x}\log\left(\frac{\sin x}{x}\right)\notag\\
&= \lim_{x \to 0}\frac{\sin x}{x - \sin x}\log\left(1 + \frac{\sin x}{x} - 1\right)\notag\\
&= \lim_{x \to 0}\frac{\sin x}{x - \sin x}\cdot\left(\dfrac{\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1312894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Minimal Polynomial of $\sqrt{2}+\sqrt{3}+\sqrt{5}$ To find the above minimal polynomial, let
$$x=\sqrt{2}+\sqrt{3}+\sqrt{5}$$
$$x^2=10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}$$
Subtracting 10 and squaring gives
$$x^4-20x^2+100=4(31+2\sqrt{60}+2\sqrt{90}+2\sqrt{150})$$
$$x^4-20x^2+100=4(31+4\sqrt{15}+6\sqrt{10}+10\sqrt{6})$$
$... | An elementary way
$(x-\sqrt2)^2=(\sqrt3+\sqrt5)^2 \implies x^2+2-2x\sqrt2=8+2\sqrt{15}\implies(x^2-6-2x\sqrt2)^2=((x^2-6)^2-4x(x^2-6)\sqrt2+8x^2)^2=60\implies((x^2-6)^2+8x^2-60)^2=(4x(x^2-6)\sqrt2)^2$
This resulting polynomial of $\mathbb{Q}[x]$ is obviously of degree 8
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 3
} |
Prove that $f(n)=(n!)^{\frac{1}{n}}-\frac{n+1}{2}$ is a monotone decreasing sequence Prove that the sequence $f(n)=(n!)^{\frac{1}{n}}-\frac{n+1}{2}$ is a monotone decreasing sequence for $n>2.$
We have to show that $f(n+1)<f(n)$ for all $n>2.$ We have
$$
f(n+1)-f(n)=((n+1)!)^{\frac{1}{n+1}}-\frac{n+2}{2}-\left((n!)^{\f... | Possible Proof by Induction.
Base case $n =3$, $$f(3) = (3!)^\frac{1}{3} -\frac{3 + 1}{2}$$
and so $$f(3) = (6)^\frac{1}{3} - 2 \approx -0.18$$ and now
$$f(4) = (4!)^\frac{1}{4} -\frac{4 + 1}{2}$$
and so $$f(4) = (24)^\frac{1}{4} -2.5 \approx -2.9$$
and so we have $f(4) < f(3)$ and so the base case works. Now assume i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is there a way to solve for pythagorean triples with one side length equalling 1? When you have one side of a right triangle fixed, is there a trend in pythagorean triples? For instance if I fix one side of a triangle at 1 unit what will the other side have to equal to get pythagorean triplets for the first 100 values?... | You cannot do it with Pythagorean triples but you can do it with the sum of three cubes.
$$1^3+6^3+8^3=9^3=729$$
$$1^3+71^3+138=144^3=2985984$$
$$1^3+135^3+138^3=172^3=5088448$$
$$1^3+242^3+720^3=729^3=387420489$$
$$1^3+372^3+426^3=505^3=128787625$$
$$1^3+426^3+486^3=577^3=192100033$$
$$1^3+566^3+823^3=904^3=738763264$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Polar form to cartesian Let $\Gamma$ be a circle that passes through the origin. Show that we can find real numbers $s$ and $t$ such that $\Gamma$ is the graph of
$r = 2s \cos (\theta + t).$
I know this has to be converted to a cartesian equation, but how do I do this, and what do I do after?
Thanks
| A circle that passes through the origin can be expressed as
$$(x-a)^2+(y-b)^2=a^2+b^2,$$
i.e.
$$x^2-2ax+y^2-2by=0$$
for some $a,b\in\mathbb R$ such that $(a,b)\not=(0,0)$.
Here, setting $x=r\cos\theta,y=r\sin\theta$ gives you
$$r^2\cos^2\theta-2ar\cos\theta+r^2\sin^2\theta-2br\sin\theta=0,$$
i.e.
$$r=2a\cos\theta+2b\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A unusual inequality about function $\ln$ These day,I met a unusual inequality when I solve a difficult problem, and proving the inequality means I have done the work!
Could you show me how to prove it or deny it? By the way, I believe that it's true!
Prove that, for all $t > 0$,
\begin{align*}
&4\ln t\ln (t + 2) - \l... | We need to show that for $t> 0$,
$$4\ln t \ln (t + 2) + 4 \ln (t + 1) \ln (t + 3) > \ln t \ln ( t + 1) + 3\ln t \ln (t + 3) + 3\ln (t + 1) \ln (t + 2) $$
Case $t> 1$
Note that $\ln t, \ln (t+1)$ and $\ln (t+2), \ln(t+3)$ are similarly ordered, so by Rearrangement
$$\ln t \ln(t+2)+\ln (t+1) \ln(t+3) > \ln t \ln(t+3) +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
How to integrate$ I=\int\ln\left(\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}\right)\ \mathrm dy $ I am stuck with the integration
$$
I=\int\ln\left(\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}\right)\ \mathrm dy
$$
I got this from the question from the book
"Field and wave electromagnetics, Cheng, 2nd, Problem 3-18.
I tried t... | Following @ClaudeLeibovici, we have
$$I=y\log\left(\frac{L}2+\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}\right)-J$$
where
$$\begin{align}
J&=\int\frac{y^2}{\left(\frac{L}2+\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}\right)\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}}dy\\\\
&=-\int\frac{y^2\left(\frac{L}2-\sqrt{\left(\frac{L... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1320743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
How can I compare numbers raised to a square root For example: $3^\sqrt5$ versus $5^\sqrt3$
I tried to write numbers as this:
$3^{5^{\frac{1}{2}}}$ and then as
$3^{\frac{1}{2}^5}$
But this method gives the wrong answer because $a^{(b^c)} \ne a^{bc}$
| user134824's approach, but with the $\sqrt{3}$ instead, is a bit nicer:
$$
\begin{align}
3^\sqrt{5} &\text{ vs. } 5^\sqrt{3}\\
(3^\sqrt{5})^\sqrt{3} &\text{ vs. } (5^\sqrt{3})^\sqrt{3}\\
3^{\sqrt{5}\sqrt{3}} &\text{ vs. } 5^{\sqrt{3}\sqrt{3}}\\
3^\sqrt{15} &\text{ vs. } 5^3\\
\end{align}
$$
Now notice that since $\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1321704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Generating Functions Partitions Let $U(x)=\sum_{n=0}^{\infty} u_nx^n$, where $u_n$ is the number of partitions of $n$ into at most two parts. For example, $u_4=3$ because $4$ can be partitioned into at most two parts as $4$, $3+1$, or $2+2$. Use the convention that $u_0=1$.
Then $\frac 1{U(x)}$ is a polynomial. What po... | Hint: It is pretty easy to see that
$$
u_n=\left\lfloor\frac n2\right\rfloor+1
$$
This will be a useful series here
$$
\begin{align}
\frac1{(1-x)^2}
&=\frac{\mathrm{d}}{\mathrm{d}x}\frac1{1-x}\\
&=\frac{\mathrm{d}}{\mathrm{d}x}\left(1+x+x^2+x^3+\dots\,\right)\\[6pt]
&=1+2x+3x^2+\dots
\end{align}
$$
Thus,
$$
\begin{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Probability that the total of three rolls of a fair die is odd A fair die is thrown three times. Events A, B, C are defined as follows:
A:The total score is an odd number.
B:A six appears as the first throw
C:The total score is 13.
My question is how to find the probability of event A?
| Here is how to do it for two dice. Consider the following table:
$$
\begin{align*}
2 &= 1+1 \\
3 &= 1+2=2+1 \\
4 &= 1+3=2+2=3+1 \\
5 &= 1+4=2+3=3+2=4+1 \\
6 &= 1+5=2+4=3+3=4+2=5+1 \\
7 &= 1+6=2+5=3+4=4+3=5+2=6+1 \\
8 &= 2+6=3+5=4+4=5+3=6+2 \\
9 &= 3+6=4+5=5+4=6+3 \\
10 &= 4+6=5+5=6+4 \\
11 &= 5+6=6+5 \\
12 &= 6+6
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Counting the number of solutions of equation $x^2 + y^2 = 1$ over $\Bbb Z/p$ List proofs of the fact that the number of solutions to $x^2 + y^2 = 1$ over $\Bbb Z/p$, where $p$ is a prime $\neq 2$, is $p-(-1)^{\frac{p-1}2}$. I thought of two.
I write one below.
| Either $p = 2$ or $p \equiv 1 \pmod{4}$ or $p \equiv 3 \pmod{4}$.
If $p = 2$ then $f(x) = x^2$ is a bijection so there is always a unique $y = f^{-1}(1 - x^2)$. Therefore the number of solutions is $p$.
If $p \equiv 1 \pmod{4}$ then $(-1 \mid p) = (-1)^{\frac {p-1} 2} = 1$, so there is an $i \in \Bbb Z/p$ for which $i^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Please help to find the formula for a relation I'm trying to find the formula for the following relation:
$ x_1 + x_2 + x_3 + x_4 = n $
where:
$ 0 \leq x_1 \leq 3$
$ 0 \leq x_2 \leq 3$
$ x_3 \geq 0 $
$ x_3 \geq 0 $
Let $a_n$ be the number of different compositions of $n$ items.
Here is the generating-function I've ... | I am assuming that
the desired relation is
$(1+x+x^2+x^3)^2(\frac{1}{(1-x)^2})
=\sum_{n=0}^{\infty} a_n
$.
Then
$\begin{array}\\
\sum_{n=0}^{\infty} a_n
&=(1+x+x^2+x^3)^2(\frac{1}{(1-x)^2})\\
&=(\frac{1-x^4}{1-x})^2(\frac{1}{(1-x)^2})\\
&=\frac{(1-x^4)^2}{(1-x)^4}\\
&=(1-x^4)^2(1-x)^{-4}\\
&=(1-x^4)^2\sum_{n=0}^{\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find $\lim _{ n\to \infty } \frac { ({ n!) }^{ 1\over n } }{ n } $? How to find $\lim _{ n\to \infty } \frac { ({ n!) }^{ 1\over n } }{ n } $ ?
I tried taking using logarithm to bring the expression to sum form and then tried L Hospital's Rule.But its not working.Please help!
This is what wolfram alpha is showi... | Note
\begin{align}\frac{(n!)^{1/n}}{n} &= \left[\left(1 - \frac{0}{n}\right)\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)\cdots \left(1 - \frac{n-1}{n}\right)\right]^{1/n}\\
&= \exp\left\{\frac{1}{n}\sum_{k = 0}^{n-1} \log\left(1 - \frac{k}{n}\right)\right\}
\end{align}
and the last expression converges to
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1327575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How do you prove $\sum \frac {n}{2^n} = 2$? How do you prove $$\sum_{n=1}^{\infty} \frac {n}{2^n} = 2\ ?$$
My attempt: I have been trying to find geometric series that converge to 2 which can bind the given series on either side. But I am unable to find these. Is there a general technique to find the sum? This is a hig... | Here is a simple way to see this. It ignores technical aspects of rearranging infinite series.
Write down the series as follows:
$$
\quad \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \frac{4}{2^4} + \dots \\
= \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \dots \\
\qquad + \frac{1}{2^2} + \frac{1}{2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1330493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
Is this combinatorial identity a special case of Saalschutz's theroem? When I solved a question, the following combinatorial identity was used
$$
\sum_{k=0}^{n}(-1)^k{n\choose k}{n+k\choose k}{k\choose j}=(-1)^n{n\choose j}{n+j\choose j}.
$$
But to prove this identity is difficulty for me. I think it may be a special c... | Remark. Incorporates a correction by Markus Scheuer.
Suppose we seek to evaluate
$$S_j(n) = \sum_{k=0}^n (-1)^k {n\choose k}
{n+k\choose k} {k\choose j}$$
which is claimed to be
$$(-1)^n {n\choose j}{n+j\choose j}.$$
Introduce
$${n+k\choose k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+k}}{z^{k+1}} \; dz$$
a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
} |
The number of roots common between the two equations is
The number of roots common between the two equations
$x^3+3x^2+4x+7=0$ and $x^3+2x^2+7x+5=0$ is
$\color{green}{a.)\ 0 } \\~\\
b.)\ 1 \\~\\
c.)\ 2 \\~\\
d.)\ 3 \\~\\ $
i tried to solve both equations by subtracting then
$x^3+3x^2+4x+7-(x^3+2x^2+7x+5)=0 \\
x^2-3... | Considering $$f(x)=x^3+3x^2+4x+7$$ $$g(x)=x^3+2x^2+7x+5$$ you could notice that their derivatives never cancel in the real domain. So, $f(x)=0$ has only one real root and same for $g(x)=0$. So, the maximum number of common roots is $1$.
Now, inspection :
*
*$f(-3)=-5$, $f(-2)=3$; so the root for $f(x)=0$ is somewhe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1332079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
When will it diverge? When will it converge?
Test for what $x\in \mathbb{R}$ the series
$\sum_{n=0}^\infty nx^n$
converges and for what $x\in \mathbb{R}$ it diverges. Determine the limit of sequence for the case of the convergence.
We started dealing with series since the beginning of the week and I tried myself on t... | If $x>1$, the terms don't tend to zero as $n \to \infty$, which is required for convergence (there's no point where the value stops jumping around too much as you take more and more terms).
If $x<1$, notice the following trick: we know that if
$$ S_n = 1 + x + \dotsb + x^n, $$
then
$$ xS_n = x + x^2 + \dotsb + x^{n+1},... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Find all $x$ such that $2^x,2^{x^2}$ and $2^{x^3}$ form $3$ terms of an A.P. I know that if $a,b,c$ are in Arithmetic Progression, then $2b=a+c$, but in this case, I am unable to solve for $x$. Hints are appreciated. Thanks.
| HINT:
$$2^x(1+2^{x^3-x})=2^{x^2+1}\iff 1+2^{x^3-x}=2^{x^2-x+1}$$
If $x^3-x>0,1+2^{x^3-x}>1$ is odd unlike $2^{x^2-x+1}$ as $x^2-x+1>0$ for real $x$
So, $x^3-x$ must be $0$
and consequently, $2=2^{x^2-x+1}\implies 1=x^2-x+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding the maximum value of the function $( x^2-x+1)^{ 1/3}$ on the interval $[0,1]$ While finding the maximum value of the function $( x^2-x+1)^{ 1/3}$ on the closed interval $[0,1]$ the point where the derivative is $0$ is $1/2$
and $f(1/2)=(3/4)^{1/3}$.
Then, for $f(0)=1$ and for $f(1)=1$... Now here I have a pr... | If $f(x) = (x^2 - x - 1)^{1/3}$, the derivative is
$$
f'(x) = \frac{1}{3}(x^2 - x - 1)^{-2/3}(2x-1).
$$
So, indeed, the derivative is zero exactly at $x = 1/2$. Now to find the absolute minimum and maximum values, you evaluate the function at the endpoints of the interval and at the critical points, so
$$
\begin{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Taylor series expansion of function $f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}}$ Determine the Taylor series expansion of function $f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}}$.
$f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}} = \arcsin(x)\frac{1}{\sqrt{1-x^2}}$
It is known:
(1.) $\arcsin(x) = \sum_{n=0}^\infty\frac{(2n-1)!!x^{2n+1}}{2... | I would substitute $x = \sin(x)$ and simplify to $1/\cos(x)$ and proceed from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding value of an expression If $x^2-3x-1=0$ then find the value of $(x^6+1)/x^3$ I tried to solve the quadratic but it became too complicated any way of doing this without a calculator
| A variant on the answers already given:
$$\begin{align}
x^2-3x-1=0&\implies x-{1\over x}=3\\
&\implies x^2-2+{1\over x^2}=9\\
&\implies x^2+2+{1\over x^2}=13\\
&\implies x+{1\over x}=\pm\sqrt{13}
\end{align}$$
and
$$\begin{align}
{x^6+1\over x^3}&={(x^2+1)(x^4-x^2+1)\over x\cdot x^2}\\
&=\left(x+{1\over x} \right)\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
I need help with a Finite Series Problem:
Find the sum to $n$ terms of
\begin{eqnarray*}
\frac{1}{1\cdot 2\cdot 3} + \frac{3}{2\cdot 3\cdot 4} + \frac{5}{3\cdot 4\cdot 5} +
\frac{7}{4\cdot 5\cdot 6}+\cdots \\
\end{eqnarray*}
Answer:
The way I see it, the problem is asking me to find this series:
\begin{eqnarray... |
The good old way: compute a few terms and find a pattern.
$$\frac2{12},\frac7{24},\frac{15}{40},\frac{26}{60},\frac{40}{84}\cdots$$
By finite differences, we find that the numerators are $\dfrac{n(3n+1)}2$ and the denominators $2(n+1)(n+2)$.
Check:
$$S_1=\frac2{12}$$and
$$S_n-S_{n-1}=\frac{n(3n-1)}{4(n+1)(n+2)}-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
What is the remainder when $6\times7^{32} + 7\times9^{45}$ is divided by $4$?
What is the remainder when $6\times7^{32} + 7\times9^{45}$ is divided by $4$ ?
$7 \equiv 3 \pmod 4$
$7^2 \equiv 9 \pmod 4\equiv 1 \pmod 4$
$(7^2)^{16} \equiv 1^{16} \pmod 4$
i.e $7^{32} \equiv 1 \pmod 4$
Similarly $9 \equiv 1 \pmod 4$ imp... | $$
6 \cdot 7^{32} + 7 \cdot 9^{45} \equiv 6 \cdot 1 + 7 \cdot 1 \equiv 6+7 \equiv 13 \equiv 1 \mod 4
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
} |
Integral gaussian hypergeometric function How can we define integral with interval $[b,\infty)$
$$
\begin{align}
C(b,\alpha) & = \int_b^\infty \frac{1}{1+w^{\alpha/2}}\,\mathrm{d}w \\[8pt]
& = 2\pi/\alpha \csc(2\pi/\alpha)-b_2 F_1 (1,2/\alpha;(2+\alpha)/\alpha;-b^{\alpha/2})
\end{align}
$$
where $_2F_1(\cdot)$ is the G... | I rename the parameter for convenicence:
$$I(a,b)=\int_b^\infty \frac{dw}{1+w^a}=\frac{1}{a} \int_{b^a}^\infty \frac{u^{1/a-1}du}{1+u}=$$
$$=\frac{1}{a} \int_{0}^\infty \frac{u^{1/a-1}du}{1+u}-\frac{1}{a} \int^{b^a}_0 \frac{u^{1/a-1}du}{1+u}=$$
$$=\frac{1}{a} B\left(\frac{1}{a},1-\frac{1}{a} \right)-\frac{b}{a}\int^{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\int _0^\infty \frac{\ln x}{(x^2+1)^2}dx$ Calculate $$\int _0^\infty \dfrac{\ln x}{(x^2+1)^2}dx.$$
I am having trouble using Jordan's lemma for this kind of integral. Moreover, can I multiply it by half and evaluate $\frac{1}{2}\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$?
| Here is how to do it using complex analysis.
First of all in this case you can't compute $\frac{1}{2}\int_{-\infty}^\infty \frac{\ln x}{(x^2+1)^2}$ since it does not equal your integral (why?).
Now take $R>1$, $r<1$ and $\gamma$ a "keyhole" contour as shown below
Lets take the branch cut of the logarithm with domain ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 0
} |
Differentiate the Function: $g(y)=ln\frac{(2y+1)^5}{\sqrt{y^2+1}}$ $g(y)=ln\frac{(2y+1)^5}{\sqrt{y^2+1}}$
$g(y)= ln(2y+1)^5-ln\sqrt{y^2-1}$
g'(y)=$\frac{5(2y+1)^4\cdot (2)}{(2y+1)^5}-\frac{2(y^2+1)(2y)}{(y^2+1)^2}$
At this point I can cancel the (2y+1) from the numerator and denominator of the first rational equation a... | HINT:
Write
$$\frac{d}{dy}\left(\log \sqrt{y^2+1}\right)=\frac{d}{dy}\left(\frac12\log (y^2+1)\right)=\frac12\left(\frac{2y}{y^2+1}\right)$$
To address the OP's comment, here $g(y)=y^2+1$ and $g'(y)=2y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do I show that:if$p$ is prime $>5$ then $p^4-20p^2+19$ is always divisible by $180$.? Is there someone who can show me How do i show that :If $p$ is a prime number greater than $5$ then : $$p^4-20p^2+19$$ is always divisible by $180$.
Note : i think should factor $p^4-20p^2+19=$ as:$p^4-20p^2+19 = (p^2-1)(p^2-1... | $180=5\cdot9\cdot4$
For any integer $p, p^4-20p^2+19\equiv p^4-2p^2+1\pmod9$
Now $p^4-2p^2+1=(p^2-1)^2$
For $(p,3)=1,p\equiv\pm1\pmod3\implies p^2\equiv1\pmod3\iff p^2-1\equiv0$
and for any integer $p, p^4-20p^2+19\equiv p^4-1\pmod{20}$
$\implies p^4\equiv1\pmod5$ by Fermat's Little Theorem for $(5,p)=1$
Now if $(p,2)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Summation Theorem how to get formula for exponent greater than 3 I'm studying in the summer for calculus 2 in the fall and I'm reading about summation. I'm given these formulas:
\begin{align*}
\sum_{i=1}^n 1 &= n, \\
\sum_{i=1}^n i &= \frac{n(n+1)}{2},\\
\sum_{i=1}^n i^2 &= \frac{n(n+1)(2n+1)}{6},\\
\sum_{i=1}^n i^3 &... |
The sum of the $k$-th powers of numbers $1$ to $n$ can also be derived by means of formal power series.
\begin{align*}
S_k(n):=\sum_{j=1}^n j^k\qquad\qquad n,k\geq 1
\end{align*}
In order to do so we encode for fixed $k$ the sequence
$\left(S_k(n)\right)_{n\geq 0}$ by generating functions $$A(z)=\sum_{n=0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1353294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 9,
"answer_id": 6
} |
$y=-2x+k$ and curve $y=\frac{2}{x-3}$ A straight line has equation $y=-2x+k$, where $k$ is a constant, and a curve has an equation $y=\frac{2}{x-3}$.
i)Show that the $x$-coordinates of any points of intersection of the line and curve are given by the equation $2x^2-(6+k)x+(2+3k)=0$
ii)Find the two values of $k$ for whi... | In this answer, I suppose that $A,B$ are the points where the line $y=-2x+k$ is tangent to the curve.
For $k=2$, $2x^2-(6+k)x+3k+2=0\Rightarrow 2x^2-8x+8=0,$ i.e. $2(x-2)^2=0$. So, $x=2$, and $y=\frac{2}{x-3}=\frac{2}{2-3}=-2$. So, we know that $(2,-2)$ is the point where the line $y=-2x+2$ is tangent to the curve.
Fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1353795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\sum_{i\in\mathbb{N}}(n-2i)^k\binom{n}{2i+1}$
Find $$\sum_{i\in\mathbb{N}}(n-2i)^k\binom{n}{2i+1},$$ where both $n$ and $k$ are natural numbers.
I know the following identity:
$$
\sum_{i\in\{0\}\cup\mathbb{N}}i(i-1)\cdots(i-p)\binom{n}{2i+1}=(n-p-2)(n-p-3)\cdots(n-2p-2)2^{n-2p-3}.
$$
But I am not sure whether ... | Suppose we seek to evaluate
$$\sum_{q=0}^n (n-2q)^k {n\choose 2q+1}.$$
We observe that
$$(n-2q)^k = \frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{k+1}} \exp((n-2q)z) \; dz.$$
This yields for the sum
$$\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{k+1}}
\sum_{q=0}^n {n\choose 2q+1} \exp((n-2q)z) \; dz
\\ = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1353963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
How to get the foci / focus Hyperbola How to find the lower and upper focus? Hyperbola
I started with this
$$ 9x^2 + 54x - y^2 + 10y + 81 = 0 $$
and broke it down to
$$ \frac{9(x+3)^2}{25} - \frac{(y-5)^2}{25} = -1 $$
center = (-3,5)
Lower Vertex = (-3,0)
Upper Vertex = (-3,10)
How to get the foci?
foci / focus = (... | We have $a^2=\frac{25}{9}$ because we have
$$\frac{(x+3)^2}{\dfrac{25}{9}}-\frac{(y-5)^2}{25}=-1.$$
The foci of a hyperbola
$$\frac{(x-m)^2}{a^2}-\frac{(y-n)^2}{b^2}=-1$$
are
$$\left(m,n\pm\sqrt{a^2+b^2}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1355684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is there anyway to show $\left| {\frac{{\cos x - \cos y}}{{x - y}}} \right| \le 1$ other than taking derivatives? The purpose is to show $\left| {\frac{{\cos x - \cos y}}{{x - y}}} \right| \le 1$ for any $x,y\in\Bbb{R}$.
Taking partial derivatives with respect to $x,y$ respectively $\frac{{\partial \frac{{\cos x - \co... | The only "analytic" inequality needed,
aside from basic trigonometry,
is
$|\sin(x)| \le |x|
$.
Let
$x = y+h$,
so
$\begin{array}\\
\cos(x)-\cos(y)
&=\cos(y+h)-\cos(y)\\
&=\cos(y)\cos(h)-\sin(y)\sin(h)-\cos(y)\\
&=\cos(y)(\cos(h)-1)-\sin(y)\sin(h)\\
\end{array}
$
We now use
$|a\cos(y)+b\sin(y)|
\le \sqrt{a^2+b^2}
$.
To s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Limit of a product I While reviewing old problems in American Mathematical Monthly the following problem was encountered. What are some methods to solving the problem ?
Proposed by L. S. Johnston, 1929.
Consider the infinite sequence $\{ a_{n} \}$ of real positive numbers with the recurrent relation
\begin{align}
a... | L. S. Johnston's solution:
obtained from: American Mathematical Monthly vol 36, issue 4, 1929, p. 235, problem 3313.
If $a_{1} = 1$ it is evident that $a_{k} = 1$ for all $k$'s, and that
$\lim_{n\to\infty} \, \prod_{k=1}^{n} \{a_{k}\} =1$.
Consider next the case for which $a_{1} > 1$ and set $a_{1} = \sec(\omega)$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1357855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show that $\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{1\cdot 2\cdot 3\cdot \ldots \cdot n}\leq 2^{n}$ for all $n\in\mathbb{N}$.
Show that $$\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{1\cdot 2\cdot 3\cdot \ldots \cdot n}\leq 2^{n}\qquad (n\in \mathbb{N}).$$
I want to show the last step, that is, the inductiv... | Clearly, $2r-1<2r$ $$\implies\prod_{r=1}^n\dfrac{2r-1}{2r}<1$$
$$\implies\prod_{r=1}^n(2r-1)<\prod_{r=1}^n(2r)=2^n\prod_{r=1}^nr$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Finding the number of ways to pick ${n}$ marbles from a jar Problem:
А jar contains 8 blue marbles, 6 green marbles, and 4 red marbles. Five marbles are selected at random, all at once. In how many ways can:
A.) two red and three blue marbles be obtained?
B.) two green and two blue marbles be obtained?
C.) two red ma... | We assume (as we are expected to) that the marbles are distinguishable.
A.) There are $\binom{4}{2}$ ways to choose the reds. For each of these ways, there are $\binom{8}{3}$ ways to choose the blues, for a total of $\binom{4}{2}\binom{8}{3}$.
B.) We interpret the question as asking for exactly $2$ green and exactly $2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simplification of $\sqrt{2\cosh(x)+2}$ I do not understand the simplification
$$
\sqrt{2\cosh(x)+2}=2\cosh(x/2)
$$
More generally, I do not understand why
$ \sqrt{a\cosh(x)+a}=b\cosh(x/2)$
What is the relationship between $a$ and $b$?
How does adding $a$ remove the radical?
| We have that: $$\cosh(2x) = \cosh^2x + \sinh^2x = 2\cosh^2x - 1,$$since $\cosh^2x - \sinh^2x=1$. In that identity, we make $x \mapsto x/2$, so we obtain: $$\cosh x = 2 \cosh^2\left(\frac x 2\right)-1 \implies \cosh\frac{x}{2} = \sqrt{\frac{1+\cosh x}{2}}.$$Multiplying both sides by $2$ we have: $$2 \cosh \frac x 2 = 2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proving $2p +1 \mid 2^p + 1$ The following theorem is well known and already proven by Lagrange 1775:
Let $p \equiv 3 \pmod{4}$ be prime. $2p+1$ is also prime if and only if $2p+1 \mid 2^p - 1$.
But how can we prove this:
Let $p \equiv 1 \pmod{4}$ be prime. $2p+1$ is also prime if and only if $2p+1 \mid 2^p + 1$.
Th... | Let $p$ be a positive prime such that $p\equiv 1\pmod{4}$. If $2p+1$ is prime, then $$\left(\frac{2}{2p+1}\right)=(-1)^{\frac{(2p+1)^2-1}{8}}=(-1)^{\frac{p(p+1)}{2}}=-1\,.$$ Hence, $2^p\equiv -1\,\big(\text{mod }(2p+1)\big)$. Therefore, $2p+1$ divides $2^p+1$. Conversely, suppose now that $2p+1$ divides $2^p+1$. L... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
About $(x^3 - 4)^2 - x^6 + 2x^5 = 2x^5 -8x^3 + 16$ Studying polynomials I got the follows:
$$
(x^3 - 4)^2 - x^6 + 2x^5 = 2x^5 -8x^3 + 16
$$
I can't understand from where we got this $-8x^3$.
I got to simplify this polynomial just to:
$$
2x^5 + 16
$$
Can someone help me understand from where we got the expression $-... | $$\begin{align}
(x^3 - 4)^2 - x^6 + 2x^5 &= x^3\cdot x^3 + (-4)^2 + 2\cdot (-4) \cdot x^3 -x^6+ 2x^5\\
&=x^6-x^6 + 16 -8x^3 + 2x^5 = 2x^5 -8x^3 + 16
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Expected number of coin flips for heads The expected number of coin flips to get one heads is $2$. What is wrong with this argument?
There is a $1/2$ chance of getting $H$, $1/4$ chance of getting $TH$,
$1/8$ chance of getting $TTH$, etc. so the expected value of flips is
$$\frac{1}{2} + \frac{2}{4} + \frac{3}{8} ... | you need to consider the event $A_k$ ={Head shows up in the $k$ flip}
$$P(A_k) = P(T(k-1 \text{times} ) H) = 2^{-k}$$
The expected value of $X = \sum_k k \chi_{A_k}$ which thakes the value $k$ when $A_k$ occurs is
\begin{align}\Bbb{E}[X] = \sum_k k2^{-k} &= \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots & = 1\\
& \qq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1361468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Polynomials: Linking equations via their roots The equation $ax^3+bx^2+cx+d=0$ has solutions $1,p$ and $q$. The equation $x^3+sx^2+tx+r=0$ has solutions $1, 1/p$ and $1/q$. Show that $r=a/d$ , and find an expression for $s$ in terms of $c$ and $d$ , simplifying your answer.
| $ax^3+bx^2+cx+d=0\ \ \ \ (1)$
$x^3+sx^2+tx+r=0\ \ \ \ (2)$
Clearly the roots of $(1)$ are reciprocal of those of $(2)$
If $y=1/x\iff x=1/y$
Putting the values of $x$ in $(1),$
$a+by+cy^2+dy^3=0\iff dy^3+cy^2+by+a=0\ \ \ \ (3)$ whose roots will be $1,1/p,1/q$
So, $(3)$ will be same as $(2)$
Then the ratios of the coe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the matrix $\mathbf{A}$ if $A\binom{7}{-1} = \binom{6}{2}.$ Find the $2\times2$ matrix $A$ where $A^2=A$ and
$$A\begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}.$$
I tried plugging in: $A= \begin{pmatrix}a&b\\c&d\end{pmatrix}$ but that became messy very quickly. I got the equations:
$7a... | Write out the equations and use a simple linear solver to find:
${1 \over 10}\left(
\begin{array}{cc}
9 & 3 \\
3 & 1 \\
\end{array}
\right)
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Exponential diophantine equation: $2p^2-6p+7=3^n$. I'm trying to prove that the only integer positive solutions are $(n=1,\ p=1)$ and $(n=3,\ p=5)$. Is there a simple way to do that?
| Let $p=a$ be any integer. $(2a-3)^2+5=2\cdot 3^n$. Now $3$ cases:
*
*$n=3k$. Then $(2(2a-3))^2+20=\left(2\cdot 3^k\right)^3$. But $x^2+20=y^3$ has $2$ integral solutions $(x,y)=(\pm 14,6)$ (http://tnt.math.se.tmu.ac.jp/simath/MORDELL/), so $(a,k)=(5,1),(-2,1)$, so $(a,n)=(5,3),(-2,3)$.
*$n=3k+1$. Then $(6(2a-3)))^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Induction Proof using factorials Recall that for $n \in N$, $n! = 1 \cdot 2 \cdots n$.
Prove the following for each $n \in N$:
$$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \cdots + \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}$$
I understand how to do the proof, but in the inductive step I am facing some difficulty proving... | HINT: we have to show that $$\frac{(k+1)}{(k+2)!}-\frac{1}{(k+1)!}=-\frac{1}{(k+2)!}$$ this is true since $$\frac{(k+1)}{(k+1)!(k+2)}-\frac{1}{(k+1)!}=\frac{1}{(k+1)!}\left(\frac{k+1}{k+2}-1\right)=\frac{1}{(k+1)!}\left(\frac{k+1-k-2}{k+2)}\right)=-\frac{1}{(k+1)!(k+2)}=-\frac{1}{(k+2)!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $\frac{a+b}{b+c}=\frac{c+d}{d+a}$ then.. If $\frac{a+b}{b+c}=\frac{c+d}{d+a}$ then
(A) $a=c$
(B) either $a=c$ or $a+b+c+d=0$
(C) $a+b+c+d=0$
(D) $a=c$ and $b=d$
I solved $\frac{a+b}{b+c}=\frac{c+d}{d+a}$ and got $a(a+b+d)=c(c+b+d)$ and so I thought that (A) is the correct option. But the correct answer is (B).
I'm h... | If $a+b+c+d=0$, then $(a+b)=-(c+d)$ and $(b+c)=-(a+d)$, therefore $$\frac{a+b}{b+c}=\frac{-(c+d)}{-(a+d)} = \frac{c+d}{a+d}$$
provided that we don't divide by zero anywhere.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Differentiate the Function: $y=2x \log_{10}\sqrt{x}$ $y=2x\log_{10}\sqrt{x}$
Solve using: Product Rule $\left(f(x)\cdot g(x)\right)'= f(x)\cdot\frac{d}{dx}g(x)+g(x)\cdot \frac{d}{dx}f(x)$
and $\frac{d}{dx}(\log_ax)= \frac{1}{x\ \ln\ a}$
$(2x)\cdot [\log_{10}\sqrt{x}]'+(\log_{10}\sqrt{x})\cdot [2x]'$
$y'=2x\frac{1}{\sqr... | Take $\log_{10}x$ as $\log x$.
$$y=2x\log\sqrt{x}=x\log x=\log x^x$$
$$10^y=x^x \implies y\ln10=x\ln x \implies y=\frac{x\ln x}{ \ln10}$$
Then
$$y'=\frac{1}{\ln10}\times(\ln x + 1)=\log x + \frac{1}{\ln 10}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
} |
Convergence of sequence: $f(n+1) = f(n) + \frac{f(n)^2}{n(n+1)}$ I am trying to work out a problem on some old analysis qual exam, and managed to reduce it to this question, but I can't seem to figure out this final step:
Consider the sequence defined by the recurrence relation $$f(n+1) = f(n) + \frac{f(n)^2}{n(n+1)}$$... | We can answer this question in the same manner as this answer.
Since $f(1)=1$ and
$$
f(n+1)=f(n)+\frac{f(n)^2}{n(n+1)}\tag{1}
$$
we have that $f(n)$ is increasing and inductively we have that
$$
1\le f(n)\le n\tag{2}
$$
Furthermore, $(1)$ implies that
$$
\begin{align}
&\left(\frac1{f(n+1)}-\frac1{n+1}\right)-\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Wanted : for more formulas to find the area of a triangle? I know some formulas to find a triangle's area, like the ones below.
*
*Is there any reference containing most triangle area formulas?
*If you know more, please add them as an answer
$$s=\sqrt{p(p-a)(p-b)(p-c)} ,p=\frac{a+b+c}{2}\\s=\frac{h_a*a}{2}\\s=... | Incircle bisectors $d_a,d_b,d_c$,
mentioned in
incircle-bisectors-and-related-measures
along with the radii of corresponding incircles $r,r_a,r_b,r_c$
provide a whole lot of expressions for the area $S$ of $\triangle ABC$.
Recall that cevian $AD_a$
splits the triangle $ABC$ into a pair of triangles
$T_1=\triangle ABD_a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 13,
"answer_id": 9
} |
find $\frac{dy}{dx}$ in terms of $x$ and $y$ of $x^2y^2=\frac{(y+1)}{(x+1)}$ - basic question
Find $\frac{dy}{dx}$ in terms of $x$ and $y$ of $x^2y^2=\frac{(y+1)}{(x+1)}$
Ok so using the product rule on the LHS and the quotient rule on the RHS I differentiated both sides of the equation and got:
$2xy(x\frac{dy}{dx}+y... | $x^2y^2=(y+1)/(x+1)$
Take log of both sides,
$log(x^2y^2)=log \frac{y+1}{x+1}$
$log(x^2)+log(y^2)=log(y+1)-log(x+1)$
$2\times log(x)+2\times log(y)=log(y+1)-log(x+1)$
Now differentiate both sides wrt $x$,
$\frac{2}{x}+\frac{2}{y}\times\frac{dy}{dx}=\frac{1}{y+1}\times\frac{dy}{dx}-\frac{1}{x+1}$
$\frac{2}{y}\times\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$
Given $|x|<1 $ prove that $\\1+2x+3x^2+4x^3+5x^4+...=\frac{1}{(1-x)^2}$.
1st Proof: Let $s$ be defined as
$$
s=1+2x+3x^2+4x^3+5x^4+\cdots
$$
Then we have
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s... | Here is another variation.
Assuming the geometric series $\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$ is known, we consider functions $f,g:(-1,1) \rightarrow \mathbb{R}$
\begin{align*}
f(x)&=\sum_{n=0}^{\infty}(n+1)x^n\qquad\qquad g(x)=\frac{1}{(1-x)^2}
\end{align*}
We obtain
\begin{align*}
\int f(x) dx&=\int\left(\sum_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 15,
"answer_id": 4
} |
Minimum value of $a+b$ If the graph of $f(x)=2x^3+ax^2+bx$ intersects the $x$-axis at three distinct points, then what is minimum value of $a+b$? Here $a$ and $b$ are natural numbers.
My attempt:
As the graph intersects the $x$-axis at three distinct points, it has $2$ local maxima/minima. Let these $2$ local maxima/mi... | For intersection of the curve $f(x)=2x^3+ax^2+bx$ with the x-axis, we have $$f(x)=0 \implies 2x^3+ax^2+bx=0$$ $$\implies x(2x^2+ax+b)=0 \implies x=0\ \text{&}\ 2x^2+ax+b=0 $$ Now, for intersection of curve $f(x)$ with the x-axis at three distinct points, the quadratic equation: $2x^2+ax+b=0$ must have two distinct roo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
given $2f(x) + f(1-x) = x^2$ find $f(-5)$ I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.
A function $f$ has property that $2f(x)+ f(1-x) = x^2$ for any number of x. What is the number for $f(-5)$?
(A) $\frac {15}4$ (B) $\frac... | Hint
If (this was the original post) $$2f(x) - f(1-x) = x^2$$ then $$2f(1-x) - f(x) = (1-x)^2$$ So, multiplying the first by $2$ and adding to the second, we have $$3f(x)=2x^2+(1-x)^2$$
Edit
By the way, the answer for $x=-5$ is not in the list of choices and your answer is perfectly correct.
You would get $\frac {34}3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Proving that $1\cdot3+3\cdot5+5\cdot7+\cdots+(2n-1)(2n+1)={n(4n^2+6n-1) \over 3}$ by induction for $n\geq 1$
Prove using mathematical induction that
$$1\cdot3+3\cdot5+5\cdot7+\cdots+(2n-1)(2n+1)= {n(4n^2+6n-1) \over 3}.$$
Step 1: If we assume that the equation is true for a natural number, $n=k$, then we get
$$1\cdot... | You have this line: $$1\cdot3+3\cdot5+5\cdot7+\cdots+(2k-1)(2k+1)+(2k+1-1)(2k+1+1) $$$${}= {k(4k+1^2+6k+1-1) \over 3}$$
That should be: $$1\cdot3+3\cdot5+5\cdot7+\cdots+(2k-1)(2k+1)+(2(k+1)-1)(2(k+1)+1)$$$${} = {k(4k^2+6k-1) \over 3}+(2(k+1)-1)(2(k+1)+1)$$
First of all, I'm just doing literally the same thing to both s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Solve the equation $(m^2-m-2)x=m^2+4m+3$ Here's how I solve it
I think that m is the variable (am I right?).
Then
$$m^2x-mx-2x-m^2-4m-3=0$$
$$m^2(x-1)-m(x+4)-(2x+3)=0$$
$$D=x^2+8x+16+4(x-1)(2x+3)$$
$$=x^2+8x+16+4(2x^2-2x+3x-3)$$
$$=9x^2+12x+4$$
$$=(3x+2)^2$$
$$m=\frac{x+4\pm (3x-2))}{2(x-1)}$$
$$m_1=\frac{4x+2}{2x-2... | Assuming you want to solve for $m$, the equation can immediately be rewritten as:
$(m-2)(m+1)x = (m+3)(m+1)$
Now it should be obvious that $m=-1$ is a solution as both LHS and RHS become zero (and equal) for this value of $m$.
Now consider the case $m \neq -1$. It is now permissible to divide both sides by $m+1$, simpl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Simplify Square Root Expression $\sqrt{125} - \sqrt{5}$ $\sqrt{125}-\sqrt5$ simplify it.
I thought it would be $\sqrt {5\cdot5\cdot5}-\sqrt 5$ which would be the square root of 25 which is 5 but it is not.
Can you show how to simplify this?
| Starting from where you left off:
$$
\begin{eqnarray}
\sqrt{125}-\sqrt5
&=& \sqrt {5\cdot5\cdot5} - \sqrt 5 \\
&=& \sqrt{5} \cdot \sqrt{5} \cdot\sqrt{5}-\sqrt 5 \\
&=& \sqrt{5} (\sqrt{5} \cdot\sqrt{5} - 1) \\
&=& \sqrt{5} (5 - 1) \\
&=& 4 \sqrt{5}
\end{eqnarray}
$$
If the question were $\sqrt{125} \div \sqrt5$, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Why are sums of powers of 2 able to give all numbers? It is known that
If we sum up a combination of numbers that are positive powers of 2(starting from 0 to infinity), we can get any number possible.
(Correct me if this is wrong).
Can anyone give a proof and explain how this works? Please try to make it understanda... | Here's an example, which can be easily expanded to the general case. Let's say we want to write $21$ as the sum of distinct powers of two. ("Distinct" meaning that they're all different.)
Well, if we drop the distinctness condition, it's easy:
$21=1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1$
Let's try to clean this up a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
find x in $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$
Which one satisfies the equation $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$
(A)$27$ (B)$32$ (C)$45$ (D)$52$ (E)$63$
let $a = 6+\sqrt x , b=6-\sqrt x$
cube each side
\begin{align}
(\sqrt[3]a + \sqrt[3]b)^3 &= (\sqrt[3]3)^3 \\
(\sqrt[3... | Step 1 (conjecture): there is some $e>0$ such that
$$
6+\sqrt{x}=(e+\sqrt[3]{3}/2)^3,\quad 6-\sqrt{x}=(-e+\sqrt[3]{3}/2)^3.\tag{$*$}
$$
You can see that if such an $e$ exists then the equality $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$ is satisfied. Solving for $e$ is simple:
$$
12=(6+\sqrt{x})+(6-\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
Show that $\int_{\pi}^{\infty} \frac{1}{x^2 (\sin^2 x)^{1/3}} dx$ is finite. Show that $\int_{\pi}^{\infty} \frac{1}{x^2 (\sin^2 x)^{1/3}} dx$ is finite. I've been trying to use Holder inequality but it seems I can't get the right combination of $p$ and $q$. Maybe I'm on the wrong track?
| We have:
$$ \int_{0}^{\pi}\frac{dx}{\left(\sin^2 x\right)^{\frac{1}{3}}}=\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{6}\right)}{\Gamma\left(\frac{2}{3}\right)}$$
hence:
$$\begin{eqnarray*} \left|\int_{\pi}^{+\infty}\frac{dx}{x^2\left(\sin^2 x\right)^{\frac{1}{3}}}\,dx\right|&\leq& \pi\left(\frac{1}{\pi^2}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Find the maximum value of the fraction
Let $a$ and $b$ be positive integers satisfying $\frac{ab+1}{a+b}<\frac{3}{2}$. The maximum possible value of $\frac{a^3b^3+1}{a^3+b^3}$ is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.
Trial and error makes the job very easy, but it isn't ... | Try manipulating the first inequality to define $a$ in terms of $b$ (or vice versa)
$$ \frac{ab+1}{a+b} < \frac{3}{2} \Rightarrow 2ab+2 < 3a+3b$$
$$ \Rightarrow 2ab - 3a < 3b-2 $$
$$ \Rightarrow a < \frac{3b-2}{2b-3}$$
Notice that if $a=1$, then the second fraction involving $a$ and $b$ would become
$$ \frac{(1)^3b^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Evaluating the indefinite integral $\int\sqrt{16-9x^2}\,dx$ I need to solve the integral below, but I just can't figure how.
$$\int \sqrt{16-9x^2}\,dx$$
I have tried to replace $9x^2$ with $16\sin^2\theta$. I get to a point where I have the function below. Please let me know whether I'm on the right track, and please... | Hint: $$\cos^2\theta=\frac{1}{2}(1+\cos(2\theta))$$
Edit: Also, after some calculations, I am pretty sure you made an error (can anyone double check this?)
You have got $\displaystyle\frac{4}{3}\int\cos^2\theta d\theta$, but the substitution $9x^2=16\sin^2\theta$ gives us the following:
$$\sqrt{16-9x^2}=\sqrt{16-16\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 0
} |
Find the following probability A bowl contains 16 chips, of which 6 are red, 7 are white and 3 are blue. If four chips are taken at random and without replacement, find the probability that there is at least 1 chip of each colour.
Can someone please give me a hint?
Thank you!
| I think barak manos is doubly counting some selections. Following Marconius's tip, we have
\begin{align}
\binom{6}{2}(7)(3) + (6)\binom{7}{2}(3) + (6)(7)\binom{3}{2}
& = (15)(7)(3) + (6)(21)(3) + (6)(7)(3) \\
& = 315 + 378 + 126 = 819
\end{align}
different ways to select chips to satisfy the condition. Note t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1379580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Closed-form of $\int\limits_0^1\left(\frac{\left(x^2+1\right)\arcsin(x)}{\sqrt{1-x^2}}+2x\ln\left(x^2+1\right)\right)\frac{\ln x}{x^3+x}\,dx$ I've conjectured the following closed-form:
$$
I = \int\limits_0^1\left(\frac{\left(x^2+1\right)\arcsin(x)}{\sqrt{1-x^2}}+2x\ln\left(x^2+1\right)\right)\frac{\ln x}{x^3+x}\,dx = ... | Apply the obvious substitution $x\mapsto\sin{x}$ to the first integral
$$I=\int^\frac{\pi}{2}\limits_0\frac{x\ln(\sin{x})}{\sin{x}}\ {\rm d}x+\int\limits^1_0\frac{2\ln{x}\ln(1+x^2)}{1+x^2}\ {\rm d}x$$
The latter integral has been addressed here and is equivalent to
\begin{align}
\int\limits^1_0\frac{2\ln{x}\ln(1+x^2)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
Finding indefinite integral $\int \frac{x^2}{(x \sin x + \cos x)^2} dx $ I need hint in finding the integral of
$$\int \frac{x^2}{(x \sin x + \cos x)^2} dx $$
I tried dividing the term by $x^2\cos^2x$ and then substituting $\tan x$.
| Consider the following $$f(x)=\frac{\sin x-x\cos x}{x\sin x+\cos x}$$
$$\implies f'(x)=\frac{d}{dx}\left(\frac{\sin x-x\cos x}{x\sin x+\cos x}\right)=\frac{(x\sin x+\cos x)\frac{d}{dx}(\sin x-x\cos x)-(\sin x-x\cos x)\frac{d}{dx}(x\sin x+\cos x)}{(x\sin x+\cos x)^2}$$ $$ =\frac{(x\sin x+\cos x)(x\sin x)-(\sin x-x\cos x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How come $\ n\ $ always divides at least one of the item of the sequence? Given positive integer$\ \displaystyle n,\ $ the sequence is:
$\displaystyle 2^n$
$\displaystyle 2^n - 2^{n-1}$
$\displaystyle 2^n - 2^{n-1} + 2^{n-2}$
$\displaystyle 2^n - 2^{n-1} + 2^{n-2} - 2^{n-3}$
...
$\displaystyle 2^n - 2^{n-1} + 2^{... | We need only to check $n$ odd, since the conjecture is true for $n$ being powers of 2, for all terms are even but the last one $S(0)=even\pm 1=odd$.
a) $S(n)=2^n=1\cdot 2^{n}$
b) $S(n-1)=2^n-2^{n-1}=1\cdot 2^{n-1}$
c) $S(n-2)=2^{n-1}+2^{n-2}=3\cdot 2^{n-2}$
d) $S(n-3)=3\cdot 2^{n-2}-2^{n-3}=6\cdot 2^{n-3}-2^{n-3}=5\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Take an example of Integrate of root I want to solve an example like this : $\int_{0}^{4}\sqrt{4^2-x^2}\ dx$ according to this equation :$$\int \sqrt{a^2-x^2}\ dx= \frac{1}{2}\left(x\sqrt{a^2-x^2}+a^2\sin^{-1}\left(\frac{x}{a}\right)\right)$$
I have problem with that sine . I want to see the solving step by step.
| Notice, applying the standard formula we get
$$\int_{0}^{4}\sqrt{4^2-x^2}dx=\frac{1}{2}\left(x\sqrt{4^2-x^2}+4^2\sin^{-1}\left(\frac{x}{4}\right)\right)_{0}^{4}$$ $\color{red}{\text{applying the limits}}$ $$=\frac{1}{2}\left(4\sqrt{4^2-4^2}+4^2\sin^{-1}\left(\frac{4}{4}\right)-(0)\sqrt{4^2-(0)^2}-4^2\sin^{-1}\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Expressing the integral in terms of the original variable In evaluating the integral:
$$ \int{dx\over(a^2-x^2)^{3/2}} $$ or $$ \int{dx\over(a^2-x^2)^{1/2}\ (a^2-x^2)}$$
Let $ x=a\sin\theta $ and $ dx=a\cos\theta\ d\theta $. Then
$$ \int{{a\cos\theta\ d\theta}\over{a\cos\theta\ (a^2-a^2\sin^2\theta)}} = {1\over a^2}\in... | You can form a right angled triangle from your substitution $x = a\sin\theta$. This'll give the hypotenuse as $a$, opposite side as $x$ and by Pythagoras' theorem, your adjacent side will have length $\sqrt{a^2-x^2}$. $$ \implies \tan \theta \ = \ \dfrac{\text{opposite}}{\text{adjacent}} \ = \ \dfrac{x}{\sqrt{a^2-x^2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find min & max of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$ Problem: Find the maximum and minimal value of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$.
Since $x^2 > x$ (edit $x^2 \geq x$) for all $x \in \mathbb{R}$, $f$ is bowl-ish with a minimal value in the bottom.
This is a critical point which means t... | Think of a unit circle centred at the origin, and consider another circle centred at $(-\frac12,-\frac12)$. Expand the latter circle from a small size until it just touches the unit circle, at $(-\frac12\sqrt2,-\frac12\sqrt2)$. Then expand it further until it just touches the unit circle again, at $(\frac12\sqrt2,\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 6
} |
Finding a point using complex geometry
In the Cartesian plane let $A = (1,0)$ and $B = \left( 2, 2\sqrt{3} \right)$. Equilateral triangle $ABC$ is constructed so that $C$ lies in the first quadrant. Let $P=(x,y)$ be the center of $\triangle ABC$. Then $x \cdot y$ can be written as $\tfrac{p\sqrt{q}}{r}$, where $p$ and... | Let
$$\omega=\frac{1+i\sqrt{3}}{2},\qquad \overline{\omega}=\frac{1-i\sqrt{3}}{2}$$
be primitive sixth roots of unity. Given $A=1,B=2+2i\sqrt{3}$, in order that $ABC$ is an equilateral triangle, $C$ must be the image of $B$ with respect to a $\pm 60^\circ$ rotation with centre in $A$, hence:
$$ C = \frac{1\pm i\sqrt{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How can I find an ODE equation from $dy/dx$ What is the ODE satisfied by $y=y(x)$
given that $$\frac{dy}{dx} = \frac{-x-2y}{y-2x}$$
I understand that I need to get it in some form of $\int \cdots \;dy = \int \cdots \; dx$, but am not sure how to go about it.
| We have, $$\frac{dy}{dx} = \frac{-x-2y}{y-2x}$$
Let $y=ux\implies \frac{dy}{dx}=x\frac{du}{dx}+u$ $$u+x\frac{du}{dx}=\frac{-x-2ux}{ux-2x}$$
$$u+x\frac{du}{dx}=\frac{2u+1}{2-u}$$ $$x\frac{du}{dx}=\frac{2u+1}{2-u}-u$$ $$x\frac{du}{dx}=\frac{1+u^2}{2-u}$$ $$\frac{(2-u)du}{1+u^2}=\frac{dx}{x}$$ Integrating both the sides, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
A puzzling step in a solution for "Find $\sin(x)$ and $\cos(x)$, if $a\sin(x)+b\cos(x)=c$" A textbook problem:
Find $\sin(x)$ and $\cos(x)$, if $a\sin(x)+b\cos(x)=c$
The solution from the textbook:
Let's divide each term of this equation by $\sqrt{a^2+b^2}$:
$$\frac{a}{\sqrt{a^2+b^2}}\sin(x)+\frac{b}{\sqrt{a^2+b^2... | Using Anurag A's hint,
$$\sin(x-\phi)=\pm\sqrt{1-\cos^2(x-\phi)}$$
Since
$$\left(\frac{a}{\sqrt{a^2+b^2}}\right)^2+\left(\frac{b}{\sqrt{a^2+b^2}}\right)^2=1,$$
$$\sin(x-\phi)=\pm\sqrt{\left(\frac{a}{\sqrt{a^2+b^2}}\right)^2+\left(\frac{b}{\sqrt{a^2+b^2}}\right)^2-\left(\frac{c}{\sqrt{a^2+b^2}}\right)^2};$$
$$\sin(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proving $(1 + \frac{1}{n})^n < n$ for natural numbers with $n \geq 3$. Prove with induction on $n$ that \begin{align*} \Bigl(1+ \frac{1}{n}\Bigr)^n < n \end{align*} for natural numbers $n \geq 3$.
Attempt at proof: Basic step. This can be verified easily.
Induction step. Suppose the assertion holds for $n >3$, then we... | For $n=3$ we have
$(1 + 1/n)^{n} < n$.
If $n \geq 1$ such that
$(1 + 1/n)^{n} < n$,
then
$$
(1 + \frac{1}{n+1})^{n+1} < (1 + \frac{1}{n})^{n+1} < n(1 + \frac{1}{n}) = n+1,
$$
qed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
For positive integers $x, y, k$, prove that $4^x (4^y+1)=k(k+1)$ implies $x = y$ In the proof that I read, even $k$ implies $4^x=k$ and $4^y+1=k+1$. I am wondering why we don't need to factorize $4^y+1$ into $pq$, such that $p, q > 1$, and solve for $4^x p=k$, $q=k+1$.
| Since either $k$ or $k + 1$ is even, and the other is odd, then $k = 4^x$ or $k + 1 = 4^x$. Assuming the latter for sake of contradiction, this implies that $k \equiv -1 \mod 4$, but also that $k = 4^y + 1$. This is a contradiction, so $k = 4^x$.
Therefore $k + 1 = 4^y + 1 = 4^x + 1$, hence $4^x = 4^y \implies x = y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Prove that the trigonometric equation has no solution Prove that the trigonometric equation $$\frac{\sin^3 x}{1-\sin x}+\frac{\cos^3 x}{1-\cos x}=-1$$ has no solution. I tried applying $T2's$ lemma to contradict but could only do so for the first and third quadrant values if $x$. There must be some good proof without t... | $$\frac{\sin^3 x}{1-\sin x}+\frac{\cos^3 x}{1-\cos x}=-1$$
Obviously $\sin x\neq 1$ and $\cos x\neq 1$.
$$\sin^3 x(1-\cos x)+\cos^3 x(1-\sin x)=-1 -\sin x\cos x+\sin x+\cos x$$
$$\sin^3 x+\cos^3 x-\sin x\cos x(\sin^2 x+\cos^2 x)=-1 -\sin x\cos x+\sin x+\cos x$$
$$\sin^3 x+\cos^3 x-\sin x-\cos x=-1 $$
$$(\sin x+\cos x)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 2
} |
Sequence of $1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5$ ‘A sequence is formed by writing the integers the corresponding number of times as follows : 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, … What is the 800 th term in this sequence?’
| This is the same as asking for (the index of) the smallest triangular number that is no less than 800 (a triangular number is a positive integer of the form $\frac{n(n+1)}{2}$).
Solving $\frac{x(x+1)}{2} = 800$ gives $x=39.50312, x=-40.50312$.
Since $\frac{39 \times 40}{2} = 780$ and $\frac{40 \times 41}{2} = 820$ we s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to integrate $\int (x-1)\sqrt{x} \, \text{d}x$ How do I find this integral:
$$\int (x-1)\sqrt{x} \, \text{d} x$$
I thought to use use substitution, but am not sure what I should use as $u$.
| $$\int { \left( x-1 \right) \sqrt { x } dx } =\int { \left( { x }^{ \frac { 3 }{ 2 } }-{ x }^{ \frac { 1 }{ 2 } } \right) dx= } \int { { x }^{ \frac { 3 }{ 2 } } } dx-\int { { x }^{ \frac { 1 }{ 2 } } } dx=\\ =\frac { 2 }{ 5 } { x }^{ \frac { 5 }{ 2 } }-\frac { 2 }{ 3 } x^{ \frac { 3 }{ 2 } }+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.