Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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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 equal to $2\pi n/\gcd(n,5)$. However after this I'm not being able to proceed. Help please!
| 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:
$$\cos(n(x+3\pi))\cdot\sin\frac{5(x+3\pi)}n=\cos(nx+3n\pi)\cdot\sin\left(\frac{5x}n+\frac{15\pi}n\right)=$$
$$=\overbrace{(-1)^n}^{=\cos3n\pi}\cos nx\left(\sin\frac{5x}n\cos\frac{15\pi}n+\sin\frac{15\pi}n\cos\frac{5x}n\right)$$
Equalling the last expression to the left side of the first one above, we get:
$$\sin\frac{5x}n=(-1)^n\left(\sin\frac{5x}n\cos\frac{15\pi}n+\sin\frac{15\pi}n\cos\frac{5x}n\right)$$
As this must be true for any value of $\;x\;$ , we can choose say $\;x=0\;$ :
$$0=(-1)^n\left(0+\sin\frac{15\pi}n\right)\implies n\in\pm\{1,3,5,15\}$$
and we've reduced the possible choices for $\;n\;$ . Take it from here.
| {
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"url": "https://math.stackexchange.com/questions/1276819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 $ds=-\csc^2(u) du$
$$\int \cot^4(u)\csc^2(u) du=\int s^4 ds$$
The integral of $s^4$ is $\frac{s^5}{5}$
$$\int s^4 ds=-\frac{s^5}{5}+C$$
Substitute back for $s=\cot(u)$ and $u=\sin^{-1}(x)$
$$-\frac{s^5}{5}+C=-\frac{1}{5}\cot\left(\sin^{-1}x\right)^5+C$$
Simplify using $\cot(\sin^{-1}(z))=\frac{\sqrt{1-z^2}}{z}$
$$-\frac{1}{5}\cot\left(\sin^{-1}x\right)^5+C=\frac{\left(1-x^2\right)^{5/2}}
{5x^5}+C$$
So:
$$\int \frac{(1-x^2)^{3/2}}{x^6}\,dx=\frac{\left(1-x^2\right)^{5/2}}
{5x^5}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1277459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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.
$$y=\frac{(x+3)^4(2x^2+5x)^3}{\sqrt{4x-3}}$$
$$\ln(y) = \ln\left((x+3)^4(2x^2+5x)^3)\right) - \ln\left(\sqrt{4x-3}\right)$$
$$\ln(y) = 4\ln(x+3) 3\ln(2x^2+5x) - \frac{1}{2} \ln(4x-3)$$
aaaaaaaaand don't know what to do next, any help in the process or next step?
| 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*}
From that point, you can go ahead and differentiate to obtain
$$\frac{y'}{y}=\frac{4}{x+3}+\frac{3}{x}+\frac{6}{2x+5}-\frac{2}{4x-3}.$$
Then multiply the $y$ across to get
$$y'=\left(\frac{4}{x+3}+\frac{3}{x}+\frac{6}{2x+5}-\frac{2}{4x-3}\right)y.$$
Finally, you would like to express $y'$ as a function of $x$, rather than both $x$ and $y$, so substitute in your original equation
$$y=\frac{(x+3)^4(2x^2+5x)^3}{\sqrt{4x-3}}$$
on the right to obtain
$$y'=\left(\frac{4}{x+3}+\frac{3}{x}+\frac{6}{2x+5}-\frac{2}{4x-3}\right)\left(\frac{(x+3)^4(2x^2+5x)^3}{\sqrt{4x-3}}\right),$$
and then simplify to taste.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 change of variable such that $$\frac{\sin ^2(x)}{\sqrt{3}}=y$$ that is to say $$x=\sin ^{-1}\left(\sqrt[4]{3} \sqrt{y}\right)$$ $$dx=\frac{\sqrt[4]{3}}{2 \sqrt{y} \sqrt{1-\sqrt{3} y}}$$ $$\sin(x)=\sqrt[4]{3} \sqrt{y}$$ $$\cos(x)=\sqrt{1-\sqrt{3} y}$$ Now, replacing $$I=\frac{1}{2}\int\frac{1}{ \sqrt{1-y^2}}dy=\frac{1}{2} \sin^{-1}(y)=\frac{1}{2} \sin ^{-1}\left(\frac{\sin ^2(x)}{\sqrt{3}}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1279974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 sides? $-x-\frac{1}{4} < 7$
well that's not right. I suck at this. :c
Thanks for any helpers.
| $2|x+0,25| < 9 \rightarrow |x+0,25| < 4,5 \rightarrow |x| < 4,25 \rightarrow -4,25 < x <4,25$
You can negate $0,25$ from each side, since it is positive, and the absolute value won't effect it.
| {
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"url": "https://math.stackexchange.com/questions/1280776",
"timestamp": "2023-03-29T00:00:00",
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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.
$\displaystyle S = 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} + \ldots$
$\displaystyle \frac{1}{4}S = \hspace{8.5pt} \frac{1}{4} + \frac{3}{16} + \frac{7}{64} + \frac{15}{256} + \ldots$
$\displaystyle \frac{3}{4}S = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots \qquad \leftarrow S- \frac{1}{4}S$
For the Infinite Sum on the RHS $\displaystyle \left(S = \frac{a}{1-r}\right)$:
$\displaystyle a = 1$
$\displaystyle r = \frac{1}{2}$
Then
$\displaystyle \frac{3}{4}S = \frac{1}{1-\frac{1}{2}}$
$\displaystyle \frac{3}{4}S = 2$
$\displaystyle S = \frac{8}{3}$
Using Excel the answer is $\displaystyle \frac{5}{3}$, but I don't know where is the issue.
Thanks!!
| 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 Excel model. Debugging your Excel model is off topic for MSE.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1280948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 4
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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
$$\left| {\frac{{{x^3}{y^2}}}{{{x^4} + 3{y^4}}}} \right| \leqslant \left| {\frac{{{x^3}{y^2}}}{{{x^4}}}} \right| = \left| {\frac{{{y^2}}}{x}} \right|.$$
But I do not know how to show $| {\frac{{{y^2}}}{x}}|$ is bounded using the hypothesis that $0<|x|<\delta$, $0<|y|<\delta$ and $0<\sqrt{x^2+y^2}<\delta$ since the quarters powers of $x$ and $y$ are very difficult to manage. Even setting $\delta=1$ gets me nowhere.
| Another approach is to use the fact that for $x\gt0$, we have $\left(\sqrt{x}-\frac1{\sqrt{x}}\right)^2\ge0\implies x+\frac1x\ge2$:
$$
\begin{align}
\left|\frac{x^3y^2}{x^4+3y^4}\right|
&=\frac{|x|}{\sqrt3}\frac{x^2(\sqrt[4]3\,y)^2}{x^4+3y^4}\\[9pt]
&=\frac{|x|}{\sqrt3}\frac1{\left(\raise{2pt}{\frac{x}{\sqrt[4]3\,y}}\right)^2+\left(\frac{\sqrt[4]3\,y}{x}\right)^2}\\
&\le\frac{|x|}{2\sqrt3}
\end{align}
$$
| {
"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
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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, $\sin x+\cos x=\sqrt{2}\cos (x-\frac{\pi}{4})$
$\int\frac{1}{\sin x+\cos x}dx=\int \frac{1}{\sqrt{2}\cos (x-\frac{\pi}{4})}dx$
$=\frac{1}{\sqrt{2}}\int \sec (x-\frac{\pi}{4})dx$
$=\frac{1}{\sqrt{2}} \ln \left | \sec (x-\frac{\pi}{4})+\tan (x-\frac{\pi}{4}) \right |+c$
Am I correct? Is there another way to solve this integral? Thanks in advance.
| 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
$\displaystyle\int\frac{1}{1-2u^2}du=\frac{\sqrt{2}}{4}\ln\bigg|\frac{1+\sqrt{2}u}{1-\sqrt{2}u}\bigg|+C=\frac{\sqrt{2}}{2}\tanh^{-1}(\sqrt{2}u)+C$ to find both terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1287026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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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 the Mordell curve $Y^2 = X^3-980$, solutions below.
E_-00980: r = 2 t = 1 #III = 1
E(Q) = <(14, 42)> x <(126, 1414)>
R = 1.4319518662
10 integral points
1. (14, 42) = 1 * (14, 42)
2. (14, -42) = -(14, 42)
3. (21, 91) = -2 * (14, 42)
4. (21, -91) = -(21, 91)
5. (326, 5886) = 3 * (14, 42)
6. (326, -5886) = -(326, 5886)
7. (29, 153) = 1 * (14, 42) - 1 * (126, 1414)
8. (29, -153) = -(29, 153)
9. (126, 1414) = 1 * (126, 1414)
10. (126, -1414) = -(126, 1414)
| 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 form $14(28u+3)$. Hence, the only solution is $u=0$ and $w=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 the origin to the curve is $\sqrt2$)
I proved it using polar transformations but i'd rather use proof by contradiction without using polar transformation, so i tried
$$\begin{array}{l}
\left( {\exists \alpha ,\beta } \right){\rm{ }}{\alpha ^2} + {\beta ^2}>2\\
{\alpha ^4} - {\alpha ^2} + {\beta ^4} - {\beta ^2} = 0\\
\Rightarrow {\alpha ^4} + {\beta ^4} = {\alpha ^2} + {\beta ^2}\\
\Rightarrow {\left( {{\alpha ^2} + {\beta ^2}} \right)^2} - 2{\left( {\alpha \beta } \right)^2} = {\alpha ^2} + {\beta ^2}{\rm{ }}\left( {{\rm{Completing - Square}}} \right)\\
\Rightarrow {\left( {{\alpha ^2} + {\beta ^2}} \right)^2} = {\alpha ^2} + {\beta ^2} + 2{\left( {\alpha \beta } \right)^2}\\
\Rightarrow {\left( {a + b} \right)^2} = a + b + 2ab \le 2\left( {{a^2} + {b^2}} \right){\rm{ }}\left( {{\rm{Cauchy - Schwarz - Inequality}}} \right)\\
\Rightarrow a + b \le \left( {{a^2} + {b^2}} \right) + \left( {{a^2} + {b^2}} \right) - 2ab\\
\Rightarrow a + b \le \left( {{a^2} + {b^2}} \right) + {\left( {a - b} \right)^2}\\
\Rightarrow {\left( {a - b} \right)^2} \ge \left( {a + b} \right) - \left( {{a^2} + {b^2}} \right) \ge 0\\
\Rightarrow \left( {a + b} \right) \ge \left( {{a^2} + {b^2}} \right)\\
\Rightarrow \left( {{\alpha ^2} + {\beta ^2}} \right) \ge \left( {{\alpha ^4} + {\beta ^4}} \right)\\
\Rightarrow \left( {{\alpha ^2} + {\beta ^2}} \right) - \left( {{\alpha ^4} + {\beta ^4}} \right) \ge 0
\end{array}
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$$
Since the last equation includes 0, I cannot show any contradiction.
I think the problem is that I did not use my hypothesis, ${\alpha ^2} + {\beta ^2}>2.
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% Gaey4kaSIaeqOSdi2aaWbaaSqabeaacaaIYaaaaOGaaGOmaaaa!34DD!
$
I think this claim can be proven using contradiction by I'm just not sure how to use my hypothesis.
| 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.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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.
$$\ln(1+x) = x - \frac {x^2}{2} +\frac {x^3}{3} + \cdots$$
$$e^x = 1+ x + \frac {x^2}{2} +\frac {x^3}{6} + \cdots$$
This gave me $$\frac 5{6} + x + \frac {3x^2}{2} + \frac {2x^3}{3}$$
but this is not the answer in my book. help?
| 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
$$
\log(1+e^x)=\log 2 + \log\left(\frac{1}{2} + \frac{1}{2}e^x\right)=\log 2 + \log\left(1 + \frac{1}{2}x+\frac{1}{4}x^2 + \frac{1}{12}x^3+\ldots\right).
$$
Then use the Taylor series for the logarithm:
$$
\log 2 + \log\left(1 + \frac{1}{2}x+\frac{1}{4}x^2 + \frac{1}{12}x^3+\ldots\right)\\ = \log 2 + \left(\frac{1}{2}x+\frac{1}{4}x^2 + \frac{1}{12}x^3+\ldots\right) - \frac{1}{2}\left(\frac{1}{2}x+\frac{1}{4}x^2 +\ldots\right)^2 + \frac{1}{3}\left(\frac{1}{2}x + \ldots\right)^3 + \ldots \\
= \log 2 + \frac{1}{2}x + \left(\frac{1}{4}-\frac{1}{8}\right)x^2 + \left(\frac{1}{12}-\frac{1}{8}+\frac{1}{24}\right)x^3 + \ldots \\
= \log 2 + \frac{1}{2}x + \frac{1}{8}x^2 + O(x^4).
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
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 easier way to do it ?
| 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",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 function.
The general method of proving such results from the book is by using the fact that if a function has a power series representation, then it is analytic in the circle of convergence.
Hence, using this method, I first need to find the power series of $f$ when $z\neq \pm \pi/2$, and the proof is complete if I show that this series at $z=\pm \pi/2$, equals $-\frac{1}{\pi}$.
So I start with the Taylor series of the entire function $\operatorname{cos}z$, however, I'm having trouble dealing with the denominator $z^2-(\pi/2)^2$. How can I find the power series representation of this function?
I would greatly appreciate some help.
| 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}}=\frac{1}{\pi}-\left(\frac{z-\frac{\pi}{2}}{\pi^2}\right)+\ldots
\end{equation}
We therefore find that the Laurent series expansion of $f\left(z\right)$ about $z=\frac{\pi}{2}$ is given by:
\begin{equation}
\begin{aligned}
f\left(z\right)&=\frac{\cos z}{z^2-\left(\frac{\pi}{2}\right)^2}\\
&=\left(\cos z\right)\left(\frac{1}{z+\frac{\pi}{2}}\right)\left(\frac{1}{z-\frac{\pi}{2}}\right)\\
&=\left[-\left(z-\frac{\pi}{2}\right)+\frac{1}{6}\left(z-\frac{\pi}{2}\right)^3+\ldots\right]\left[\frac{1}{\pi}-\left(\frac{z-\frac{\pi}{2}}{\pi^2}\right)+\ldots\right]\left(\frac{1}{z-\frac{\pi}{2}}\right)\\
&=-\frac{1}{\pi}+\ldots
\end{aligned}
\end{equation}
where all terms but $-\frac{1}{\pi}$ vanish when $z=\frac{\pi}{2}$; therefore, this series at $z=\frac{\pi}{2}$ equals $-\frac{1}{\pi}$. This can similarly be done with a series around $z=-\frac{\pi}{2}$ where we expand $\frac{1}{z-\frac{\pi}{2}}$ instead of $\frac{1}{z+\frac{\pi}{2}}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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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:
$\underline {} \underline {} \underline {}\underline {}\underline {}\underline {x}$
Now for this configuration we have 5! combinations
$\underline {}\underline {}\underline {}\underline {}\underline {x}\underline {}$
For this one, after the $x$, only $a$, $b$ and $c$ can go, so that's $3 \cdot 4!$.
Similarly, I continued and got an answer in the $300$'s, which is not a possible given answer I have.
What's wrong with my method?
| 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 $3 \cdot 4!$ arrangements with $x$ in the fourth position as you found.
If $x$ is in the fourth position, then there are $3$ ways to select the letter in fifth position and two ways to select the letter in the sixth position from among $a, b, c$. There are $3!$ ways to arrange the letters before $x$. Thus, there are $3 \cdot 2 \cdot 3! = 6 \cdot 3!$ arrangements in which $x$ is in the fourth position.
If $x$ is in the third position, then there are $2!$ ways to arrange $x$ and $y$ in the first two positions and $3!$ ways to arrange $a$, $b$, and $c$ in the last three positions, giving $2! \cdot 3!$ arrangements in which $x$ is in the third position.
Since $y$ and $z$ must precede $x$, $x$ cannot be in the first two positions. Hence, the number of arrangements in which $x$ follows both $y$ and $z$ is
$$5! + 3 \cdot 4! + 6 \cdot 3! + 2! \cdot 3! = 240$$
That said, this method is less efficient than those of Andre Nicolas and JMoravitz.
Here is an alternate method:
There are six positions to fill. We have six ways of placing the $a$, five ways of placing the $b$, and four ways of placing the $c$. This leaves us with three positions to fill. We must place the $x$ in the final open position. There are $2!$ ways to place $y$ and $z$ in the remaining open positions. Thus, the number of ways the show can be held so that $x$ appears after both $y$ and $z$ is
$$P(6, 3) \cdot 1 \cdot 2! = 6 \cdot 5 \cdot 4 \cdot 1 \cdot 2 \cdot 1 = 240$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 equation of the third side of the isosceles triangle . Please help.
| 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 given as $$\frac{1}{2}(a)(a)=10 \implies a=\sqrt{20}=2\sqrt{5}$$ The equations of lines bisecting the angle (right angle here) between the given lines: $x-2y+4=0$ & $2x+y-5=0$ are given as $$\frac{x-2y+4}{\sqrt{1^2+(-2)^2}}=\pm \frac{2x+y-5}{\sqrt{2^2+1^2}}\implies x-2y+4=\pm(2x+y-5)$$$$ \implies x+3y-9=0 \quad \text{&} \quad 3x-y-1=0 $$
Case 1: Let the equation of third (unknown) line (representing the hypotenuse of isosceles right triangle) be $3x-y+c=0$ normal to the first right-angle bisector: $x+3y-9=0$. Now, solving: $3x-y+c=0$ & any given line say $2x+y-5=0$, we get the intersection point $\left(\frac{5-c}{5}, \frac{2c+15}{5} \right)$. Similarly, solving both the equations of the given lines, we get intersection point $\left(\frac{6}{5}, \frac{13}{5} \right)$. Now, calculating the length $a=2\sqrt{5}$ of equal sides of isosceles triangle using distance formula as
$$\sqrt{\left(\frac{6}{5}-\frac{5-c}{5}\right)^2+\left(\frac{13}{5}-\frac{2c+15}{5}\right)^2}=a=2\sqrt{5}$$$$ c^2+2c-99=0 \implies c=9 \quad \text{&} \quad c=-11$$ Thus by setting the values of $c$, we get two equations: $3x-y+9=0$ & $3x-y-11=0$ lying on either side of right-angled vertex of given isosceles (right) triangle.
Case 2: Let the equation of third (unknown) line (representing the hypotenuse of isosceles right triangle) be $x+3y+c=0$ normal to the second right-angle bisector: $3x-y-1=0$. Now, solving: $x+3y+c=0$ & $2x+y-5=0$, we get the intersection point $\left(\frac{c+15}{5}, \frac{-(2c+5)}{5} \right)$. Similarly, solving both the equations of the given lines, we get intersection point $\left(\frac{6}{5}, \frac{13}{5} \right)$. Now, calculating the length $a=2\sqrt{5}$ of equal sides of isosceles triangle using distance formula as
$$\sqrt{\left(\frac{6}{5}-\frac{c+15}{5}\right)^2+\left(\frac{13}{5}-\frac{-(2c+5)}{5}\right)^2}=a=2\sqrt{5}$$$$ c^2+18c-19=0 \implies c=1 \quad \text{&} \quad c=-19$$ Thus by setting the values of $c$, we get two equations: $x+3y+1=0$ & $x+3y-19=0$ lying on either side of right-angled vertex of given isosceles (right) triangle.
Thus, there are total four lines: $3x-y-11=0$, $3x-y+9=0$, $x+3y+1=0$ & $x+3y-19=0$ representing the third unknown side of the isosceles (right) triangle satisfying all the conditions provided in the question. Hence, you may select any of them as your choice.
| {
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"timestamp": "2023-03-29T00:00:00",
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"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 combination of $1$, $\sqrt[3]{2}$, and $\sqrt[3]{4}$?
| 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
$$\frac{1}{a+b\sqrt[3]{2}+c\sqrt[3]{4}}=\frac{(a^2-2bc)+(2c^2-ab)\sqrt[3]{2}+(b^2-ac)\sqrt[3]{4}}{a^3+2b^3+4c^3-6abc}.$$
For this specific problem, it is (more than) enough to find a polynomial $f(a,b,c)$ such that
$$f(a,b,c)(a+b+c)=g(a^3,b^3,c^3),$$
for some other polynomial $g$ (this way the cube roots disappear). Although clever, the above solution I have above does not completely explain motivation. If we can rationalize the denominator of a fraction $1/q$, then it follows that $q$ must be algebraic. However, it is well known that the sum of algebraic numbers is itself algebraic (the proof itself is a bit more involved). Since all cube roots are (clearly) algebraic, it must be possible to rationalize the denominator. To rationalize the denominator of a fraction $1/q$, consider its minimal polynomial
$$Q(q)=\sum q_n q^n=0.$$
Note that
$$q\left(\dfrac{Q(q)-q_0}{q} \right)=-q_0,$$
which is clearly rational. Thus, in the example given in this problem, it is enough to find the minimal polynomial of $a+b\sqrt[3]{2}+c\sqrt[3]{4}$, which is simple yet painstaking.
In the case the above was not enough, we provide yet another way to get motivation. We want to multiply some factors to the denominator, currently an element of $\mathcal{S} = \mathbb{Q}[\sqrt[3]{2}] \setminus \mathbb{Q}$, to turn it into an element of $\mathbb{Q}$. Consider the map $F$ on $\mathbb{Q}[\sqrt[3]{2}]$ that sends
$$a + b\sqrt[3]{2} + c\sqrt[3]{4} \mapsto a + b\sqrt[]{2}\omega + c\sqrt[3]{4}\omega^2,$$
where $\omega = e^{2\pi i/3}$. The value of $\omega$ as a third root of unity is chosen so that $f$ is additive, multiplicative, and bijective.
The elements of $\mathbb{Q}$ are distinguished by being fixed by $F$. No element of $\mathcal{S}$ is fixed by $F$, so multiplying the given denominator by
$$(a + b\sqrt[3]{2}\omega + c\sqrt[3]{4}\omega^2)(a + b\sqrt[3]{2}\omega^2 + c\sqrt[3]{4}\omega),$$
we immediately produce an element fixed by $F$. It is easy to check that the resulting denominator and numerator are the same as we had earlier.
| {
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"timestamp": "2023-03-29T00:00:00",
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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?
My thought was to draw a tree diagram, which I did, but I can't seem to wrap my head around the recursion that is in the problem. How do I put this into the tree? Thanks!
| 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 diagram, where let the probability that Bob would eventually win in this case be $a$ for the time being).
*The dice show a sum of some other number (probability $\frac{29}{36}$) and this would be a recursion back on itself. The key here is that if the number rolled wasn't a 7 or a 12, then the situation is exactly the same as it would be had the number not been rolled at all.
The crux is that the probability of Bob winning ($x$) is equal to:
$$x = \frac{1}{36}(0) + \frac{1}{6}(a) + \frac{29}{36}(x)=\frac{a}{6} + \frac{29x}{36} = \frac{6a+29x}{36}$$
Do you see where I got this from?
Now for the other tree diagram. This is assuming that the dice showed a sum of 7 on the first roll. Again, there are three possibilities:
*
*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 Bob would win immediately.
*The dice show a sum of some other number (probability $\frac{29}{36}$) and this would be a recursion back to the first tree (and the probability of Bob eventually winning would be $x$ again).
So, $a$, the probability that Bob wins given that they reached this tree, is:
$$a = \frac{1}{36}(0) + \frac{1}{6}(1) + \frac{29}{36}(x) = \frac{1}{6} + \frac{29x}{36} = \frac{6+29x}{36}$$
Substitute this into the other equation and solve for $x$:
$$x = \frac{6a+29x}{36} = \frac{6\left(\frac{6+29x}{36}\right)+29x}{36} = \frac{\frac{6+29x}{6}+29x}{36}=\frac{6+203x}{216}$$
$$216x = 6+203x$$
$$13x = 6$$
$$x = \boxed{\frac{6}{13}}$$
The probability that Bob wins is $\frac{6}{13}$ (and the probability that Alice wins is $1-\frac{6}{13}=\frac{7}{13}$).
| {
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"timestamp": "2023-03-29T00:00:00",
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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. I've used the sine formula($2\sin{\theta}\cos{\theta}$), the cosine formula ($2\cos^2{\theta}-1$) and the tangent formula, $\left(\frac{\sin(2\theta)}{\cos(2\theta)}\right)$
originally answering the problem I got $\sin(2\theta)=\frac{24}{25}$,
$\cos(2\theta)=\frac{7}{25}$, and $\tan(2\theta)=\frac{24}{7}$
| 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 $$\sin 2\theta=\sin2\left(\pi+\sin^{-1}\left(\frac{3}{5}\right)\right)$$$$=\sin\left(2\sin^{-1}\left(\frac{3}{5}\right)\right)$$ $$=\sin\left(\sin^{-1}\left(2\times \frac{3}{5}\times \frac{4}{5}\right)\right)$$$$=\sin\left(\sin^{-1}\left(\frac{24}{25}\right)\right)=\frac{24}{25}$$
$$\implies \cos 2\theta=\cos2\left(\pi+\sin^{-1}\left(\frac{3}{5}\right)\right)$$$$=\cos\left(2\sin^{-1}\left(\frac{3}{5}\right)\right)$$ $$=\cos\left(\sin^{-1}\left(2\times \frac{3}{5}\times \frac{4}{5}\right)\right)$$$$=\cos\left(\sin^{-1}\left(\frac{24}{25}\right)\right)$$$$=\cos\left(\cos^{-1}\left(\frac{7}{25}\right)\right)$$$$=\frac{7}{25}$$
$$\implies \tan 2\theta=\frac{\sin 2\theta}{\cos 2\theta}=\frac{\frac{24}{25}}{\frac{7}{25}}=\frac{24}{7}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1301708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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 write the expression as the sum of those classic
limits as follows
\begin{eqnarray*}
\frac{\sqrt{1+h}+\sqrt{1-h}-2}{h^{2}} &=&\frac{\left( \sqrt{1+h}-1-\frac{1}{2%
}h\right) +\left( \sqrt{1-h}-1+\frac{1}{2}h\right) }{h^{2}} \\
&=&\frac{\left( \sqrt{1+h}-1-\frac{1}{2}h\right) }{h^{2}}+\frac{\left( \sqrt{%
1-h}- 1+\frac{1}{2}h\right) }{h^{2}}
\end{eqnarray*}
Therefore
\begin{eqnarray*}
\lim_{h\rightarrow 0^{+}}\frac{\sqrt{1+h}+\sqrt{1-h}-2}{h^{2}}
&=&\lim_{h\rightarrow 0^{+}}\frac{\left( \sqrt{1+h}-1-\frac{1}{2}h\right) }{%
h^{2}}+\lim_{h\rightarrow 0^{+}}\frac{\left( \sqrt{1-h}- 1+\frac{1%
}{2}h\right) }{h^{2}} \\
&=&\left( -\frac{1}{8}\right) +\left( -\frac{1}{8}\right) =-\frac{1}{4}.
\end{eqnarray*}
(Each classic limit above can be computed by l'Hospital's rule (twice) for
example.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $\tan x =\left(\frac{\text{opposite}}{\text{adjacent}}\right)$:
$\tan x = \frac{1}{2}$ - for example
but sometimes is:
$\tan x = \frac{1}{5}$ or $ \frac{2}{5}$ - if hypotenuse is $= 5$
why is that?
| Roughly, they are the same definition of tangent:
$$\begin{align} \require{cancel}
\tan = \frac{\sin}{\cos}
&= \frac{\left(\frac{\text{opposite}}{\text{hypotenuse}}\right)}
{\left(\frac{\text{adjacent}}{\text{hypotenuse}}\right)} \\
&= \left(\frac{\text{opposite}}{\text{hypotenuse}}\right) \div
\left(\frac{\text{adjacent}}{\text{hypotenuse}}\right) \\
&= \left(\frac{\text{opposite}}{\cancel{\text{hypotenuse}}}\right) \times
\left(\frac{\cancel{\text{hypotenuse}}}{\text{adjacent}}\right) \\
&= \frac{\text{opposite}}{\text{adjacent}}
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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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. Can someone help in both of these integrals?
| 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
$$v = \frac{y\pm \sqrt{y^2-4 \sqrt{a b}}}{2 \sqrt{a}} $$
$$dv = \frac1{2 \sqrt{a}} \left (1 \pm \frac{y}{\sqrt{y^2-4 \sqrt{a b}}} \right ) dy $$
$$\frac1{v} = \frac{y\mp \sqrt{y^2-4\sqrt{a b}}}{2 \sqrt{b}} $$
Then the integral is
$$I = \frac{e^{2 \sqrt{a b}}}{4 \sqrt{a b^2}} \int_{\infty}^{2 (a b)^{1/4}} dy \left (1 - \frac{y}{\sqrt{y^2-4 \sqrt{a b}}} \right ) \left ( y+ \sqrt{y^2-4 \sqrt{a b}} \right )^2 e^{-y^2} \\ + \frac{e^{2 \sqrt{a b}}}{4 \sqrt{a b^2}} \int_{2 (a b)^{1/4}}^{\infty} dy \left (1 + \frac{y}{\sqrt{y^2-4 \sqrt{a b}}} \right ) \left ( y- \sqrt{y^2-4 \sqrt{a b}} \right )^2 e^{-y^2} $$
The splitting up of the integral into the above two pieces corresponds to the two branches of $v(y)$ defined above. We can simplify somewhat to obtain
$$\begin{align}I &= \frac{e^{2 \sqrt{a b}}}{2 \sqrt{a b^2}} \left [\int_{2 (a b)^{1/4}}^{\infty} dy \frac{y^3}{\sqrt{y^2-4 \sqrt{a b}}} e^{-y^2} - \int_{2 (a b)^{1/4}}^{\infty} dy\, y \sqrt{y^2-4 \sqrt{a b}} e^{-y^2} \right ] \\ &= \frac{e^{2 \sqrt{a b}}}{4 \sqrt{a b^2}} \left [\int_{4 \sqrt{a b}}^{\infty} dy \frac{y}{\sqrt{y-4 \sqrt{a b}}} e^{-y} - \int_{4 \sqrt{a b}}^{\infty} dy\, \sqrt{y-4 \sqrt{a b}} e^{-y} \right ]\\ &= \frac{e^{-2 \sqrt{a b}}}{\sqrt{b}} \int_0^{\infty} dy \, y^{-1/2} e^{-y}\\ &= \sqrt{\frac{\pi}{b}} e^{-2 \sqrt{a b}}\end{align}$$
To directly address your transform, just plug in $a=b+s$ and using your $f$, we get
$$F(s) = \int_0^{\infty} du \, f(u) e^{-s u} = e^{-2 \sqrt{b} (\sqrt{b+s}-\sqrt{b})} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 CTS algorithm, you just need to divide the coefficient of the $x$ term by twice the square root of the coefficient of the $x^2$ term.
So, if you have $ax^2 + bx + c$, your perfect square would be $(\sqrt{a}x + \frac{b}{2 \sqrt{a}})^2$
If $a$ is not a perfect square it could get nasty, but then you can just square the whole quadratic and go from there.
For example:
In the equation $5x^2 + 6x + 5 = 0$, we could do:
$25x^2 + 30x + 25 = 0$
$(5x+3)^2 = -16$
$5x+3 = \pm4i$
$x = \pm \frac{4i}{5} - \frac{3}{5}$
My questions are:
-Is this correct?
-Is there ever an advantage to using the quadratic formula?
-Are there quadratics that are unsolvable this way?
| 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
$$x^2 + 2x + 2y^2 + y + 3 = 6$$
This is an ellipse, but our equation does not tell us much information. Rearranging into standard form will reveal quite a bit of info. Standard form for a horizontal ellipse is given as
$$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$$
where $(h,k)$ is the center and $a$ and $b$ represent the semi-major and semi-minor axes respectively (the longest and shortest "radius" of this "eccentric circle").
In order to place our ellipse into standard form, we complete the square twice:
$$x^2 + 2x + 2y^2 + 4y + 3 = 6$$
$$\left(x^2 + 2x \color{red}{+1}\right) + 2\left(y^2 + 4y \color{red}{+4}\right)+ 3 = 6 \color{red}{+1} \color{red}{+8} $$
$$(x+1)^2+ 2(y+2)^2+ 3 = 15 $$
$$(x+1)^2+ 2(y+2)^2 = 12 $$
$$\frac{(x+1)^2}{12}+ \frac{(y+2)^2}{6} = 1 $$
We now see that we have an ellipse centered at $(h,k) = (-1,-2)$ with semi-major axis $a = \sqrt{12}$ and semi-minor axis $b=\sqrt6$. This matches the graph.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $-\frac43$ as valid...
Namely, the question asks for the sum of all the answers, and the correct answer is "$-\frac43$"
So am I wrong? If so, where? Am I missing a solution? Thanks.
Apparently $1/6$ is a solution as well... How do you get this? I still don't get their solution...
| $x = \frac 16$ 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 32 = 0 $ whose roots sum to $-\frac 52 $
| {
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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, $C_{n+1}= \dfrac{1}{n+2}\left(\begin{matrix}2n+2\\n+1\end{matrix}\right)=\dfrac{1}{n+2}\dfrac{(2n+2)!}{(n+1)!(n+1)!}=\dfrac{1}{n+2}*\dfrac{(2n+2)(2n+1)n!}{(n+1)^2n!n!}$
$=\dfrac{1}{n+2}\dfrac{2(2n+1)n!}{(n+1)n!}$
| You need to simplify the second term: $\binom{2n}{n-1}=\dfrac{(2n)!}{(n-1)!(n+1)!}= \dfrac{(2n)!}{\frac{n!}{n}\cdot (n!\cdot (n+1))}= \dfrac{n}{n+1}\cdot \binom{2n}{n} \Rightarrow C_n = \binom{2n}{n} - \dfrac{n}{n+1}\binom{2n}{n}= \left(1-\dfrac{n}{n+1}\right)\binom{2n}{n} = \dfrac{1}{n+1}\binom{2n}{n}$.
| {
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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 $x \equiv y \pmod p$ or $x \equiv -y \pmod p$.
Hint: $x^2-y^2 = (x+y)(x-y)$.
This is my attempt to solve it:
If $x \equiv y \pmod p$ then $\displaystyle \frac{x-y}{p} = q_1 \Rightarrow (x-y)=pq_1$.
If $x \equiv -y$ then $\displaystyle \frac {x+y}{p}=q_2 \Rightarrow (x +y)=pq_2$.
Then $x^2-y^2 = (x+y)(x-y)=(q_1p)\cdot(q_2p)=q_1q_2p^2$
If $p$ is to divide $ x^2-y^2 $ evenly, then it must also divide $q_1q_2p^2$, therefore $\displaystyle \frac {x^2-y^2}{p}=q_1q_2p$.
To satisfy the condition of divisibility by $p$, $q_1q_2p$ must be an integer. Since the assertion to be proven is "Prove that $x^2 \equiv y^2 \pmod p$ if and only if $x \equiv y \pmod p$ or $x \equiv -y \pmod p$.", we must show that the premise is true if either $q_1$ or $q_2$ is an integer for sure.
Is this line of reasoning correct up until this point?
How to guarantee that $q_1q_2p$ is an integer?
| 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 $x \equiv y \pmod p$ which implies that $p \mid x-y$ and so there exists an integer $k$ such that $pk = x-y$ Now you multiply both sides by $(x+y)$ to get $pk(x+y) = x^2 - y^2$ and let $t = k(x+y)$ and so you have that $pt = x^2-y^2$ and so $p \mid x^2 -y^2$ and so you get the result that $x^2 \equiv y^2 \pmod p$
Same thing you can do if you assume that $x \equiv -y \pmod p$ but you will just multiply by $(x-y)$ instead.
| {
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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 surprisingly however I design the numbers the largest $x$ which leaves the same reminder is $kq_1-kq_2=k(q_1-q_2)$. Why is that? In this case we have $52 = 6\times 8+ \color \red 4$ and $46 = 6\times 7 + \color \red 4$.
Now suppose there are three numbers $a$, $b$, $c$ and $x$(assuming $a>b>c \geq x$) such that $x$ leaves the same reminder when we divide each of $a,b$ and $c$ with it. $x$ is supposed to be the largest possible value that holds the assertion. Now $x$ is given by the H.C.F of $a-b, a-c$ and $b-c$. Why is that? How can we prove this mathematically?
| 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 surprisingly however I design the numbers the largest $x$ which leaves the same reminder is $kq_1-kq_2=k(q_1-q_2)$. Why is that? In this case we have $52 = 6\times 8+ \color \red 4$ and $46 = 6\times 7 + \color \red 4$.
Let $a$ and $x$ be positive integers. The integer quotient of $a$ and $x$ is the largest integer $q$ such that $qx\le a$. The remainder of $a$ on division by $x$ is $r=a-qx$; it necessarily satisfies $0\le r<x$.
Observe first that $kq_1-kq_2=(a-r)-(b-r)=a-b$, where $r$ is the common remainder of $a$ and $b$ on division by $k$. So your statement could be rephrased as "the largest $x$ that leaves the same remainder is $a-b$." We prove this.
Let $a>b$, and $x$ be positive integers. Let $q_a$ and $q_b$ be the corresponding integer quotients and let $r_a$ and $r_b$ be the corresponding remainders on division by $x$.
For $a$ and $b$ to have the same remainder on division by $x$ means $r_a=r_b$ which means that $a-q_ax=b-q_bx$. This implies that $a-b=x(q_a-q_b)$. So if $a$ and $b$ have the same remainder on division by $x$, then $a-b$ is an integer multiple of $x$.
Conversely, if $a-b$ is an integer multiple of $x$, then $a-b=qx$ and therefore $a=b+qx$ for some integer $q$. Since $r_a=a-q_ax$ and $r_b=b-q_bx$, we have
$$
r_a=b+qx-q_ax=r_b+q_bx+qx-q_ax=r_b+(q_b+q-q_a)x.
$$
We claim that $q_b+q-q_a=0$, and therefore $r_a=r_b$. This follows since $0\le r_a,r_b<x$, so $-x<r_a-r_b<x$. The only integer multiple of $x$ in this range is $0\cdot x$. We have shown that if $a-b$ is a multiple of $x$, then $r_a=r_b$.
So $a$ and $b$ have the same remainder on division by $x$ if and only if $a-b$ is divisible by $x$. The largest $x$ such that $a$ and $b$ have the same remainder on division by $x$ must therefore be the largest integer that divides $a-b$. This is $a-b$ itself.
Now suppose there are three numbers $a$, $b$, $c$ and $x$(assuming $a>b>c>x$) such that $x$ leaves the same reminder when we divide each of $a,b$ and $c$ with it. Now $x$ is given by the H.C.F of $a-b, a-c$ and $b-c$. Why is that? How can we prove this mathematically?
We build on the previous proof to prove this one. Any two of $a$, $b$, $c$ have the same remainder on division by $x$ if and only if their difference is divisible by $x$. So $a$, $b$, and $c$ all have the same remainder on division by $x$ if and only if all of the differences, $a-b$, $a-c$, and $b-c$ are divisible by $x$. The largest $x$ such that $a$, $b$, and $c$ all have the same remainder on division by $x$ is therefore the largest $x$ such that $x$ divides all three of $a-b$, $a-c$, and $b-c$. By definition of greatest common factor, this is the greatest common factor of $a-b$, $a-c$, and $b-c$.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $\lim\limits_{x\to 0}(\frac{\sin(x)}{x})^{\frac{\sin(x)}{x-\sin(x)}}=L$
Taking $\ln$ of both sides and bringing the exponent down.
$\lim\limits_{x\to 0}{\frac{\sin(x)}{x-\sin(x)}\ln(\frac{\sin(x)}{x})}=\ln(L)$
But it changes to undefined form? The answer (in my textbook) is $$\boxed{L=\frac1{e}}$$
| 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 x}{x} - 1\right)\cdot\dfrac{\log\left(1 + \dfrac{\sin x}{x} - 1\right)}{\dfrac{\sin x}{x} - 1}\notag\\
&= \lim_{x \to 0}\frac{-\sin x}{x}\cdot\dfrac{\log\left(1 + \dfrac{\sin x}{x} - 1\right)}{\dfrac{\sin x}{x} - 1}\notag\\
&= -1\cdot\lim_{t \to 0}\frac{\log(1 + t)}{t}\text{ (putting }t = \frac{\sin x}{x} - 1)\notag\\
&= -1\cdot 1 = -1
\end{align}
Hence we have $L = e^{-1} = 1/e$.
| {
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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
$$x^4-20x^2+100=4(31+2\sqrt{60}+2\sqrt{90}+2\sqrt{150})$$
$$x^4-20x^2+100=4(31+4\sqrt{15}+6\sqrt{10}+10\sqrt{6})$$
$$x^4-20x^2-24=40\sqrt{6}+24\sqrt{10}+16\sqrt{15}$$
$$x^4-20x^2-24=8(2\sqrt{6}+2\sqrt{10}+2\sqrt{15})+24\sqrt{6}+8\sqrt{10}$$
$$x^4-20x^2-24=8(x^2-10)+24\sqrt{6}+8\sqrt{10}$$
$$x^4-28x^2-104=24\sqrt{6}+8\sqrt{10}$$
Again, squaring both sides
$$x^8-56x^6+576x^4+5428x^2+10816=4096+765\sqrt{6}$$
But if I square again, I will get a degree 16 polynomial. Mathematica says the minimal polynomial is degree 8, which would make sense since elements of $\mathbb{Q}[\sqrt{2},\sqrt{3},\sqrt{5}]$ look like
$$a+b\sqrt{2}+c\sqrt{3}+d\sqrt{5}+e\sqrt{6}+f\sqrt{10}+g\sqrt{15}+h\sqrt{30}$$
Where am I making mistakes?
| 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
| {
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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
$$
f(n+1)-f(n)=((n+1)!)^{\frac{1}{n+1}}-\frac{n+2}{2}-\left((n!)^{\frac{1}{n}}-\frac{n+1}{2}\right)=((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}}-\frac{1}{2} <0,
$$
and reduce the problem to the following inequality
$$
((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}}<\frac{1}{2}.
$$
With Maple I calculate
$$
\lim\limits_{n \to \infty}(((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}})=\frac{1}{e}<\frac{1}{2},
$$
but in is not enought to prove the inequality. Any ideas?
| 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 it works for a value $k > 2$ that is $f(k+1) < f(k)$ now we try to prove that $f(k+2) < f(k+1)$
Given that $f(k+1) < f(k)$ we have that $$\large{(k!)^\frac{1}{k} -\frac{k+1}{2} > (k+1)!^\frac{1}{k+1} - \frac{k+2}{2}}$$
Now if we are able to deduce from here that $f(k+2) < f(k+1)$ then the answer is completed
Notice that
$$f(k+2) = (k+2)!^\frac{1}{k+2} - \frac{k+3}{2}$$
| {
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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? Is there a pattern to this?
| You cannot do it with Pythagorean triples but you can do it with the sum of three cubes.
$$1^3+6^3+8^3=9^3=729$$
$$1^3+71^3+138=144^3=2985984$$
$$1^3+135^3+138^3=172^3=5088448$$
$$1^3+242^3+720^3=729^3=387420489$$
$$1^3+372^3+426^3=505^3=128787625$$
$$1^3+426^3+486^3=577^3=192100033$$
$$1^3+566^3+823^3=904^3=738763264$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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.
$$r=2a\cos\theta+2b\sin\theta.$$
Since this can be written as
$$r=2\sqrt{a^2+b^2}\cdot \cos\theta\cdot \frac{a}{\sqrt{a^2+b^2}}-2\sqrt{a^2+b^2}\cdot\sin\theta\cdot\left(-\frac{b}{\sqrt{a^2+b^2}}\right)$$
setting $s=\sqrt{a^2+b^2},\cos t=\frac{a}{\sqrt{a^2+b^2}},\sin t=-\frac{b}{\sqrt{a^2+b^2}}$ gives you
$$r=2s\cos(\theta+t).$$
| {
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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) - \ln t\ln (t + 1) - 3\ln t\ln (t + 3)\\
+ &4\ln (t + 1)\ln (t + 3) - 3\ln (t + 1)\ln (t + 2) - \ln (t + 2)\ln \left( {t + 3} \right)>0.
\end{align*}
Let
$$f\left( t \right) = 4\ln t\ln \left( {t + 2} \right) - \ln t\ln \left( {t + 1} \right) - 3\ln t\ln \left( {t + 3} \right) + 4\ln \left( {t + 1} \right)\ln \left( {t + 3} \right) - 3\ln \left( {t + 1} \right)\ln \left( {t + 2} \right) - \ln \left( {t + 2} \right)\ln \left( {t + 3} \right),$$
We have
$$f'\left( t \right) = \frac{{2\left[ {{t^2}\ln t - 3{{\left( {t + 1} \right)}^2}\ln \left( {t + 1} \right) + 3{{\left( {t + 2} \right)}^2}\ln \left( {t + 2} \right) - {{\left( {t + 3} \right)}^2}\ln \left( {t + 3} \right)} \right]}}{{t\left( {t + 1} \right)\left( {t + 2} \right)\left( {t + 3} \right)}}.$$
Let
$$g\left( t \right) = {t^2}\ln t - 3{\left( {t + 1} \right)^2}\ln \left( {t + 1} \right) + 3{\left( {t + 2} \right)^2}\ln \left( {t + 2} \right) - {\left( {t + 3} \right)^2}\ln \left( {t + 3} \right),$$
we got
$$g'\left( t \right) = 2\left[ {t\ln t - 3\left( {t + 1} \right)\ln \left( {t + 1} \right) + 3\left( {t + 2} \right)\ln \left( {t + 2} \right) - \left( {t + 3} \right)\ln \left( {t + 3} \right)} \right].$$
And let
$$h\left( x \right) = t\ln t - 3\left( {t + 1} \right)\ln \left( {t + 1} \right) + 3\left( {t + 2} \right)\ln \left( {t + 2} \right) - \left( {t + 3} \right)\ln \left( {t + 3} \right),$$
we have
\begin{align*}
h'\left( x \right) &= \ln t - 3\ln \left( {t + 1} \right) + 3\ln \left( {t + 2} \right) - \ln \left( {t + 3} \right)\\
&= \ln \frac{{t{{\left( {t + 2} \right)}^3}}}{{{{\left( {t + 1} \right)}^3}\left( {t + 3} \right)}} = \ln \left[ {1 - \frac{{2t + 3}}{{{{\left( {t + 1} \right)}^3}\left( {t + 3} \right)}}} \right] < 0.
\end{align*}
However, it seems that there are no use!
| 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
$$\ln t \ln(t+2)+\ln (t+1) \ln(t+3) > \ln t \ln(t+3) + \ln (t+1) \ln (t+2)$$
So it is enough to show that
$$\ln t \ln (t + 2) + \ln (t + 1) \ln (t + 3) > \ln t \ln ( t + 1) $$
$$\iff \ln t(\ln (t + 2)-\ln(t+1)) + \ln(t+1)\ln (t + 3) > 0$$
which is obvious as $\ln$ is increasing.
Case $0< t \le 1$
We can write the inequality as
$$\ln (t + 1) \ln \frac{(t+3)^4}{(t+2)^3} > \ln t \ln \frac{(t+1)(t+3)^3}{(t+2)^4}$$
As the fractions in the arguments are $> 1$, the LHS is clearly positive while the RHS is negative from the $\ln t$ term.
| {
"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 to solve this equation using method of integration by parts, but my equation got worse.
I know the answer by Wolfram Alpha, but I can't get how.
| 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
$$\begin{align}
J&=\int\frac{y^2}{\left(\frac{L}2+\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}\right)\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}}dy\\\\
&=-\int\frac{y^2\left(\frac{L}2-\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}\right)}{(y^2+z^2)\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}}dy\\\\
&=-\frac{L}{2}\int\frac{y^2}{(y^2+z^2)\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}}dy+\int\frac{y^2}{(y^2+z^2)}dy\\\\
&=-\frac{L}{2}\int\frac{1}{\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}}dy+\frac{L}{2}z^2\int\frac{1}{(y^2+z^2)\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}}dy+y-z\arctan(y/z) \\\\
&=K_1+K_2+y-z\arctan(y/z)
\end{align}$$
where in $(1)$
$$K_1=-\frac{L}{2}\int\frac{1}{\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}}dy$$
and
$$K_2=\frac{L}{2}z^2\int\frac{1}{(y^2+z^2)\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}}dy$$
We can easily evaluate $K_1$ by making the substitution $y=\sqrt{\left(\frac{L}{2}\right)^2+z^2}\tan t$. Then,
$$K_1=-\frac{L}{2}\log\left(y+\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}\right)$$
where the term $\sqrt{\left(\frac{L}{2}\right)^2+z^2}$ is an integration constant that we omitted.
We effect the same substitution for $K_2$. Then, we have
$$\begin{align}
K_2&=\frac{L}{2}z^2\int\frac{1}{(y^2+z^2)\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}}dy\\\\
&=\frac{L}{2}z^2\int \frac{\cos t}{\left(\frac{L}{2}\right)^2+z^2\sin^2t}dt\\\\
&=z\arctan\left(\frac{(L/2)\sin t}{z}\right)\\\\
&=z\arctan\left(\frac{(L/2)y}{z\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}}\right)\\\\
\end{align}$$
Putting it all together reveals
$$\begin{align}
I&=y\log\left(\frac{L}2+\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}\right)\\\\
&-y+z\arctan(y/z)\\\\
&+\frac{L}{2}\log\left(y+\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}\right)\\\\
&-z\arctan\left(\frac{(L/2)y}{z\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}}\right)
\end{align}$$
| {
"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 $\sqrt{15} < 4$, $3^\sqrt{15} < 3^4$ (via monotonicity of $3^x$), and since $3^4 < 5^3$ ($81 < 125$), we can say $3^\sqrt{15}<5^3$ (via transitivity), and subsequently $3^\sqrt{5} < 5^\sqrt{3}$ (via monotonicity of $x^\dfrac{1}{\sqrt{3}}, x \ge 0$).
| {
"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 polynomial is it?
Can I use generating functions to solve this? If so, how?
| 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,
$$
\begin{align}
U(x)
&=1+x+2x^2+2x^3+3x^4+3x^5+\cdots\\[12pt]
&=(1+x)(1+2x^2+3x^4+\cdots\,)\\[6pt]
&=\frac{1+x}{(1-x^2)^2}
\end{align}
$$
| {
"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:
$$
\begin{align*}
2 &= 1+1 \\
3 &= 1+2=2+1 \\
4 &= 1+3=2+2=3+1 \\
5 &= 1+4=2+3=3+2=4+1 \\
6 &= 1+5=2+4=3+3=4+2=5+1 \\
7 &= 1+6=2+5=3+4=4+3=5+2=6+1 \\
8 &= 2+6=3+5=4+4=5+3=6+2 \\
9 &= 3+6=4+5=5+4=6+3 \\
10 &= 4+6=5+5=6+4 \\
11 &= 5+6=6+5 \\
12 &= 6+6
\end{align*}
$$
You can diligently count that there are $18$ ways of obtaining odd sums, so the probability is $18/36 = 1/2$.
This was the brute-force computation, but the fact that the answer is nice prompts us to look for a simple reason that the answer is half. Indeed, let $X,Y$ be the values of the two dice; I claim that no matter what the value of $X$ is, the probability that $X+Y$ is odd is exactly $1/2$. Why? If $X$ is odd, then $X+Y$ is odd if $Y=2,4,6$; while if $X$ is even, then $X+Y$ is odd if $Y=1,3,5$. Can you generalize this to three dice?
A perhaps more striking way is to consider the dice modulo $2$, which turns them into fair coins. Now the question, if $X,Y$ are chosen randomly from $\{0,1\}$, what is the probability that $X \oplus Y = 1$? It's not hard to check that the probability is $1/2$.
| {
"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 $i^2 = -1$. $x^2 + y^2 = 1 \iff (x+iy)(x-iy) = 1$. First, count the number of $(u,v) \in (\Bbb Z/p)^2$ for which $uv = 1$; clearly that's $p - 1$. For each $(u,v)$ we can set up a system of linear equations $$\left(\begin{array}\\1 & i \\ 1 & -i\end{array}\right)\left(\begin{array}\\x \\ y\end{array}\right) = \left(\begin{array}\\u \\ v\end{array}\right)$$ which has a unique solution for $(x,y)$ because $\det = -2i$. Therefore the number of solutions is $p - 1$.
If $p \equiv 3 \pmod{4}$ then $(-1 \mid p) = (-1)^{\frac{p-1}2} = -1$, so $-1$ is not a residue. Note that for any $x \not\in \{-1,+1\}$, $(\exists y: x^2 + y^2 = 1) \iff \not\exists y: x^2 - y^2 = 1$. This is because $(1 - x^2 \mid p) = (-1 \mid p)(x^2 - 1 \mid p) = -(x^2 - 1 \mid p)$. For this reason, the three sets $A = \{x \mid \exists y: x^2 + y^2 = 1, x^2 \neq 1\}$, $B = \{x \mid \exists y: x^2 - y^2 = 1, x^2 \neq 1\}$ and $C = \{-1,+1\}$ form a partition of $\Bbb Z/p$. This implies that $|A| + |B| + |C|= |\Bbb Z /p|$. $|\Bbb Z/p| = p$. We shall now count the elements of $B$. To do this, determine $|\{(x,y) \mid x^2 - y^2 = 1\}|$, which by a method used above is $p - 1$. Any given $x \in B$ corresponds to two $y$-values on $x^2 - y^2 = 1$ so $|B| = \dfrac{p-1 - 2}{2} = \dfrac{p-3}2$. Now we know $|A| = p - 2 - \dfrac{p-3}{2} = \dfrac{p-1}2$. Every $x \in A$ corresponds to two $y$-values for which $x^2 + y^2 = 1$. So $|\{(x,y) \mid x^2 + y^2 = 1\}| = 2|A| + 2 = p+1$.
| {
"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 made for $\{a_n\}$ sequence according to the limitations:
$ (1+x+x^2+x^3)^2\left(\frac{1}{(1-x)^2}\right) $
How to find the formula of $a_n$?
$ a_0 = 1 $
$ a_1 = 4 $
Regards.
| 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} \binom{-4}{n}(-1)^n x^n\\
&=(1-x^4)^2\sum_{n=0}^{\infty} (-1)^n\binom{n+3}{3}(-1)^n x^n\\
&=(1-2x^4+x^8)\sum_{n=0}^{\infty} \binom{n+3}{3} x^n\\
&=\sum_{n=0}^{\infty} \binom{n+3}{3} x^n
-2\sum_{n=0}^{\infty} \binom{n+3}{3} x^{n+4}
+\sum_{n=0}^{\infty} \binom{n+3}{3} x^{n+8}\\
&=\sum_{n=0}^{\infty} \binom{n+3}{3} x^n
-2\sum_{n=4}^{\infty} \binom{n-1}{3} x^{n}
+\sum_{n=8}^{\infty} \binom{n-5}{3} x^{n}\\
&=\sum_{n=0}^{7} \binom{n+3}{3} x^n
-2\sum_{n=4}^{7} \binom{n-1}{3} x^{n}
+\sum_{n=8}^{\infty} (\binom{n+3}{3}-2\binom{n-1}{3}+\binom{n-5}{3}) x^{n}\\
&=\sum_{n=0}^{3} \binom{n+3}{3} x^n
+\sum_{n=4}^{7} (\binom{n+3}{3}-2\binom{n-1}{3}) x^{n}
+\sum_{n=8}^{\infty} (\binom{n+3}{3}-2\binom{n-1}{3}+\binom{n-5}{3}) x^{n}\\
\end{array}
$
This gives all values of $a_n$.
The expressions for
$a_n$ for
$n \ge 8$
can be simplified,
because the
$n^3$ and, possibly,
the $n^2$ will drop out.
| {
"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 showing,but its not providing the steps!
BTW if someone can tell me a method without using integration, I'd love to know!
| 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
$$\exp\left\{\int_0^1\log(1 - x)\, dx\right\} = \exp(-1) = \frac{1}{e}.$$
Alternative: If you want to avoid integration, consider the fact that if $\{a_n\}$ is a sequence of positive real numbers such that $\lim\limits_{n\to \infty} \frac{a_{n+1}}{a_n} = L$, then $\lim\limits_{n\to \infty} a_n^{1/n} = L$.
Now $\frac{(n!)^{1/n}}{n} = a_n^{1/n}$, where $a_n = \frac{n!}{n^n}$. So
$$\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}}\cdot \frac{n^n}{n!} = \frac{n+1}{n+1}\cdot\frac{n^n}{(n+1)^n} = \left(\frac{n}{n+1}\right)^n = \left(\frac{1}{1 + \frac{1}{n}}\right)^n = \frac{1}{\left(1 + \frac{1}{n}\right)^n}.$$
Since $\lim\limits_{n\to \infty} (1 + \frac{1}{n})^n = e$, then $$\lim_{n\to \infty} \frac{a_{n+1}}{a_n} = \frac{1}{e}.$$
Therefore $$\lim_{n\to \infty} \frac{(n!)^{1/n}}{n} = \frac{1}{e}.$$
| {
"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 high school interview question and must be easy enough to solve in a few minutes.
Please give any hints for the first step towards a solution.
| Here is a simple way to see this. It ignores technical aspects of rearranging infinite series.
Write down the series as follows:
$$
\quad \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \frac{4}{2^4} + \dots \\
= \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \dots \\
\qquad + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \dots \\
\qquad
\qquad
\quad + \frac{1}{2^3} + \frac{1}{2^4} + \dots \\
\qquad
\qquad + \frac{1}{2^4} + \dots \\
\vdots
$$
Each row is a geometric series. The values of the rows are $1, \frac{1}{2}, \, \frac{1}{2^2}, \frac{1}{2^3}$, which is again a geometric series. Add this series and you'll get the answer $\boxed{2}$.
| {
"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 case of the Saalschutz's theroem, wchich is stated as follows
$$
\sum_{k=0}^{n}\frac{(-n)_k(a)_k(b)_k}{(1)_k(c)_k(1+a+b-c-n)_k}=\frac{(c-a)_n(c-b)_n}{(c)_n(c-a-b)_n},
$$
where $(a)_k=a\cdot (a+1)\cdots (a+k-1)$. If it isn't the special case of Saalschutz's theroem, can you present a proof of this identity? All hints, proofs and references are welcome.
| 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
$${n+k\choose k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+k}}{z^{k+1}} \; dz$$
and
$${k\choose j}
= \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{k}}{w^{j+1}} \; dw.$$
This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{j+1}}
\sum_{k=0}^n (-1)^k {n\choose k} \frac{(1+z)^k (1+w)^k}{z^k}
\; dw \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{j+1}}
\left(1-\frac{(1+w)(1+z)}{z}\right)^n
\; dw \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^{n+1}}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{j+1}}
\left(-1-w-wz\right)^n
\; dw \; dz
\\ = \frac{(-1)^n}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^{n+1}}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{j+1}}
\left(1+w+wz\right)^n
\; dw \; dz
.$$
This is
$$\frac{(-1)^n}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^{n+1}}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{j+1}}
\sum_{q=0}^n {n\choose q} w^q (1+z)^q
\; dw \; dz.$$
Extracting the residue at $w=0$ we get
$$\frac{(-1)^n}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^{n+1}}
{n\choose j} (1+z)^j \; dz
\\ = {n\choose j} \frac{(-1)^n}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+j}}{z^{n+1}}\; dz
\\ = (-1)^n {n\choose j} {n+j\choose n}.$$
thus proving the claim.
| {
"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
$x^3+3x^2+4x+7-(x^3+2x^2+7x+5)=0 \\
x^2-3x+2=0 \\
x=2, \ 1$
but the answer is given as option $a.)$
I look for a short and simple way.
I have studied maths up to $12$th grade. Thanks!
| 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 somewhere between $-3$ and $-2$.
*$g(-1)=-1$, $g(0)=5$; so the root for $g(x)=0$ is somewhere between $-1$ and $0$.
So, no common root.
| {
"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 this one for practice.
I was thinking, that for $x<1$ it should converge? Because it kind of reminds of the geometric series. But the n in front of the x kind of bugs me. Anyway, how can I prove that mathematically? Or is the argumentation with the geometric series enough? Similarly, I was thinking that for an $x>1$ it should diverge.
But I need a proof that isn't based on guesses I think. I read about tests that determine whether something converges, like the ratio test, but I don't know how to use them yet. Are there also tests which determine whether a series diverges?
| 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
$$ xS_n = x + x^2 + \dotsb + x^{n+1}, $$
so
$$ S_n = \frac{1-x^{n+1}}{1-x} $$
by subtracting and dividing. But now,
$$ \begin{align*}
x+2x^2+3x^3+\dotsb + nx^n &=x+x^2+x^3+\dotsb + x^n \\
&\qquad +x^2+x^3 + \dotsb + x^n \\
&\qquad\qquad\qquad\quad \ddots \\
&\qquad\qquad\qquad\qquad+ x^n \\
&= xS_{n-1} + x^2 S_{n-2} + \dotsb + x^{n-1}S_1 + x^{n} S_0 \\
&= \frac{x}{1-x} \left( (1-x^{n}) + x(1-x^{n-1}) + x^2(1-x^{n-2})+ \dotsb + x^{n-1} (1-x) \right) \\
&= \frac{x}{1-x} \left( (1+x+x^2+\dotsb x^{n-1})-nx^{n} \right) \\
&= \frac{x}{1-x} \left( S_{n-1}-nx^{n} \right) \\
&= \frac{x}{1-x} \left( \frac{1-x^n}{1-x}-nx^{n} \right)
\end{align*} $$
Now, this does simplify further (to $\frac{x}{(1-x)^2}(1-(n+1)x^n+nx^{n+1})$), but we don't want that. What we want to know is what happens as $n \to \infty$. The terms we need to look at are
$$ -\frac{x^n}{1-x} \quad \text{and} \quad \frac{nx^{n+1}}{1-x}. $$
The first obviously goes to zero since $-1<x<1$, so each multiplication decreases the absolute value of $x^{n-1}$. The second is harder: but look at the ratio of successive values:
$$ \frac{(n+1)x^{n+1}}{nx^n} = \left( 1+ \frac{1}{n}\right) x $$
Now, for $n$ large enough the bracket can be made as close to $1$ as we like: in particular, we can make it smaller than $1/\lvert x\rvert$, which we know is larger than $1$ because $-1<x<1$. Therefore
$$ \left\lvert \left( 1+ \frac{1}{n}\right) x \right\rvert < \frac{\lvert x \rvert}{\lvert x \rvert} = 1, $$
so the ratio of successive terms is less than $1$, so $nx^n$ eventually tends to zero. Hence the sum you are considering converges, to $x/(1-x)^2$.
| {
"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 problem, for in my book from both options $0$ and $1$, the answer is only one. Can $0$ also be the answer? Why?
| 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
$$
\begin{align}
f(0) &= -1 \\
f(1) &= -1 \\
f(1/2) &= \left(\frac{1}{4} - \frac{1}{2} - 1\right)^{1/3} = \sqrt[3]{-5/4}
\end{align}
$$
The largest value is the maximum value and the smalles value is the minimum value.
If, instead, you have $f(x) = (x^2 -x \color{red}{+} 1)^{1/3}$, then the derivative is
$$
f'(x) = \frac{1}{3}(x^2 - x + 1)^{-2/3}(2x-1).
$$
Again the derivative is zero at $1/2$ and now the list of numbers is
$$
\begin{align}
f(0) &= 1 \\
f(1) &= 1 \\
f(1/2) &= \sqrt[3]{3/4}.
\end{align}
$$
Again, the absolute maximum value is the largest value in this list. The absolute minimum value if the smallest value.
One common mistake is that people forget to evaluate the original function to find the max. and min. values.
| {
"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.) $\arcsin(x) = \sum_{n=0}^\infty\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}$
(2.) $\frac{d}{dx} (\arcsin(x)) = \frac{1}{\sqrt{1-x^2}} = \frac{d}{dx}(\sum_{n=0}^\infty\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}) = \sum_{n=0}^\infty\frac{(2n+1)!!x^{2n+2}}{2^{n+1}(n+1)!}$
(3.) In third step I multiplied these two series, but I am not sure whether it is correct:
$\arcsin(x)\frac{1}{\sqrt{1-x^2}} = \sum_{n=0}^\infty(\sum_{m=0}^n\frac{(2m+1)!!x^{2m+2}}{2^{m+1}(m+1)!})\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}$
EDIT:
How did you get this sum $\sum_{n=0}^{\infty}\frac{4^n (n!)^2}{(2n+1)!}x^{2n+1}$ ?
And, in case I have to determine the product of the (some other) series, which one of these should I write?
(a) $\sum_{n=0}^\infty(\sum_{m=0}^n\frac{(2m+1)!!x^{2m+2}}{2^{m+1}(m+1)!})\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}$
or
(b) $ = \sum_{n=0}^\infty(\sum_{m=0}^n\frac{(2m-1)!!x^{2m+1}}{2^mm!(2m+1)})\frac{(2n+1)!!x^{2n+2}}{2^{n+1}(n+1)!}$
| 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
$$\begin{align}
{x^6+1\over x^3}&={(x^2+1)(x^4-x^2+1)\over x\cdot x^2}\\
&=\left(x+{1\over x} \right)\left(x^2-1+{1\over x^2} \right)\\
&=\left(x+{1\over x} \right)\left(\left(x+{1\over x}\right)^2-3\right)\\
&=\pm\sqrt{13}(13-3)=\pm10\sqrt{13}
\end{align}$$
| {
"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*}
S_n &=& \sum_{i=1}^{n} {a_i} \\
\text{with } a_i &=& \frac{2i-1}{i(i+1)(i+2)} \\
\end{eqnarray*}
We have:
\begin{eqnarray*}
S_n &=& S_{n-1} + a_n \\
S_n &=& S_{n-1} + \frac{2n-1}{n(n+1)(n+2)} \\
\end{eqnarray*}
I am tempted to apply the technique of partial fractions
but I believe there is no closed formula for a series of the of the form:
\begin{eqnarray*}
\sum_{i=1}^{n} \frac{1}{i+k} \\
\end{eqnarray*}
where $k$ is a fixed constant. Therefore I am stuck. I am hoping that somebody
can help me.
Thanks Bob
|
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
$$S_n-S_{n-1}=\frac{n(3n-1)}{4(n+1)(n+2)}-\frac{(n-1)(3n-4)}{4n(n+1)}=\frac{2n-1}{n(n+1)(n+2)}=a_n.$$
| {
"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 $9 \equiv 1 \pmod 4$ implies $9^{45} \equiv 1 \pmod 4$.
But the problem arise with the coefficients and addition sign.
what to do?
| $$
6 \cdot 7^{32} + 7 \cdot 9^{45} \equiv 6 \cdot 1 + 7 \cdot 1 \equiv 6+7 \equiv 13 \equiv 1 \mod 4
$$
| {
"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 Gaussian hypergeometric function
Can anyone help me step by step with this or give some hint where to start since i'm still studying math
best,
| I rename the parameter for convenicence:
$$I(a,b)=\int_b^\infty \frac{dw}{1+w^a}=\frac{1}{a} \int_{b^a}^\infty \frac{u^{1/a-1}du}{1+u}=$$
$$=\frac{1}{a} \int_{0}^\infty \frac{u^{1/a-1}du}{1+u}-\frac{1}{a} \int^{b^a}_0 \frac{u^{1/a-1}du}{1+u}=$$
$$=\frac{1}{a} B\left(\frac{1}{a},1-\frac{1}{a} \right)-\frac{b}{a}\int^{1}_0 \frac{v^{1/a-1}dv}{1+b^a v}=$$
$$=\frac{1}{a} \Gamma \left(\frac{1}{a}\right) \Gamma \left(1-\frac{1}{a}\right)-\frac{b}{a} B\left(\frac{1}{a},1 \right) ~{_2F_1} \left(1,\frac{1}{a};1+\frac{1}{a} ;-b^a\right)=$$
$$=\frac{\pi}{a \sin \frac{\pi}{a}}-b {_2F_1} \left(1,\frac{1}{a};1+\frac{1}{a} ;-b^a\right)$$
All of this directly follows from the definition of the Beta function and the Euler integral for the Hypergeometric function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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 $\mathbb{C} \setminus \{iy: y \leq 0\}$, in which one $\ln(x) \in \mathbb{R}$ if $x \in \mathbb{R}^+$.
Now, on one side the Residue theorem tells us that if $f(z)=\frac{\ln(z)}{(z^2+1)^2} $
$$
\int_\gamma f(z)dz = 2\pi i \cdot \text{Res}(f(z), i) = 2\pi i \left( \lim_{z\to i} \frac{d}{dz}\left[ (z-i)^2 \frac{\ln(z)}{(z+1)^2(z-i)^2}\right]\right) = -\frac{\pi}{2}+ i \frac{\pi^2}{4}
$$
On the other side is easily seen that the integral over the half semicircle connecting $R$ with $-R$ (lets call it $I_{C_1}$) goes to zero as $R$ goes to infinity, since
$$
|I_{C_1}|=\left|\int_0^\pi \frac{\ln(Re^{it})iRe^{it}}{(R^2e^{2it}+1)^2}dt\right| \underset{R \to \infty}{\longrightarrow} 0
$$
And that the one over semicircle connecting $-r$ with $r$ (this one will be called $I_{C_2}$) goes to zero as $r$ goes to $0$ because
$$
|I_{C_2}|=\left|\int_\pi^0 \frac{\ln(re^{it})ire^{it}}{(r^2e^{2it}+1)^2}dt\right| \underset{r \to 0}{\longrightarrow} 0
$$
Hence
\begin{align}
\int_\gamma f(z)dz & = \int_r^R \frac{\ln(x)}{(x^2+1)^2}dx + \int_{-R}^{-r} \frac{\ln(x)}{(x^2+1)^2}dx + I_{C_1} + I_{C_2} \\
& = \int_r^R \frac{\ln(x)}{(x^2+1)^2}dx - \int_{R}^{r} \frac{\ln(-y)}{(y^2+1)^2}dy + I_{C_1} + I_{C_2}\\
& = \int_r^R \frac{\ln(x)}{(x^2+1)^2}dx + \int_{r}^{R} \frac{\ln(-y)}{(y^2+1)^2}dy + I_{C_1} + I_{C_2}\\
& = \int_r^R \frac{\ln(x)}{(x^2+1)^2}dx + \int_{r}^{R} \frac{\ln|y| + i \pi}{(y^2+1)^2}dy + I_{C_1} + I_{C_2}\\
& = 2\int_r^R \frac{\ln(x)}{(x^2+1)^2}dx + i \pi \underbrace{\int_{r}^{R} \frac{dy}{(y^2+1)^2} }_{J}+ I_{C_1} + I_{C_2}\\
\end{align}
However, as an indefinite integral $J$ can be compute by trigonometric substitution, as follows, let $y=\tan(u)$ then
$$
\int \frac{dy}{(y^2+1)^2} = \int \cos^2(u) du = \frac{1}{4}\sin(2u)+\frac{1}{2}u = \frac{1}{4}\sin(2\tan^{-1}(y))+\frac{1}{2}\tan^{-1}(y)
$$
Hence we have
$$
\lim_{R\to \infty}\lim_{r\to 0}J=\int_0^\infty \frac{dy}{(y^2+1)^2} = \frac{1}{4}\sin(\pi)+\frac{\pi}{4} - \left( \frac{1}{4}\sin(0)+0\right) = \frac{\pi}{4}
$$
Therefore, finally we conclude that
\begin{align}
-\frac{\pi}{2}+ i \frac{\pi^2}{4} & = \lim_{R\to \infty}\lim_{r\to 0} \left( \int_\gamma f(z)dz \right) \\
& = \lim_{R\to \infty}\lim_{r\to 0} \left( 2\int_r^R \frac{\ln(x)}{(x^2+1)^2}dx + i \pi \int_{r}^{R} \frac{dy}{(y^2+1)^2}+ I_{C_1} + I_{C_2}\right) \\
& = 2 \int_0^\infty \frac{\ln(x)}{(x^2+1)^2}dx + i \pi \left(\frac{\pi}{4}\right) + 0 + 0
\end{align}
Thus
$$
\int_0^\infty \frac{\ln(x)}{(x^2+1)^2}dx = \frac{1}{2}\left( -\frac{\pi}{2}+ i \frac{\pi^2}{4} -i \frac{\pi^2}{4} \right) = -\frac{\pi}{4}
$$
| {
"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 and from the second rational equation $(y^2+1)$. However, how would I get $\frac{y}{y^2+1}$ my answer is
$g'(y)=\frac{10}{(2y+1)}-\frac{4y}{y^2+1)^2}$
| 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 $g'(y)=2y$.
| {
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"timestamp": "2023-03-29T00:00:00",
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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:$p^4-20p^2+19 = (p^2-1)(p^2-19)$ but how i do continue this Idea ?
Thank you for any help .
| $180=5\cdot9\cdot4$
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 $(p,2)=1$ $p\pm 1$ are even,$\implies 4|(p^2-1)$ which divides $p^4-1$
So we don't need $p(>5)$ to be prime, $(p,2\cdot3\cdot5)=1$ is sufficient
In fact Carmichael function $\lambda(2^n)=2^{n-2}$ for integer $n\ge3$
So, $\lambda(16)=4\implies p^4\equiv1\pmod{16}$
$\implies p^4\equiv1\pmod{\text{lcm}(9,5,16)}$
| {
"language": "en",
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"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 &=\left[ \frac{n(n+1)}{2}\right]^2.
\end{align*}
But how do I come up with the right formula for exponents greater than 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 $$A(z)=\sum_{n=0}^\infty S_k(n)z^n$$
It's convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. We also use $D_z:=\frac{d}{dz}$ to denote the differential operator. The following is valid:
The sum of the $k$-th powers of numbers $1$ to $n$ is given by
\begin{align*}
S_k(n)=\sum_{j=1}^nj^k=[z^n]\frac{1}{1-z}(zD_z)^k\frac{1}{1-z}\tag{1}
\end{align*}
A detailed derivation of (1) together with some examples can be found in this answer.
Here is an example with $k=4$
Example $S_4(n)$
\begin{align*}
S_4(n)=\sum_{j=0}^n j^4&=[z^n]\frac{1}{1-z}(zD_z)^4\frac{1}{1-z}\tag{2}\\
&=[z^n]\frac{z(1+11z+11z^2+z^3)}{(1-z)^6}\\
&=[z^n](z+11z^2+11z^3+z^4)\sum_{j=0}^{\infty}\binom{-6}{j}(-z)^{j}\tag{3}\\
&=\left([z^{n-1}]+11[z^{n-2}]+11[z^{n-3}]+[z^{n-4}]\right)\sum_{j=0}^{\infty}\binom{j+5}{5}z^j\tag{4}\\
&=\binom{n+4}{5}+11\binom{n+3}{5}+11\binom{n+2}{5}+\binom{n+1}{5}\\
&=\frac{1}{30}n(6n^4+15n^3+10n^2-1)
\end{align*}
Comment:
*
*In (2) we apply the operator $\frac{1}{1-z}(zD_z)^4$ to $\frac{1}{1-z}$
*In (3) we use the binomial series expansion
*In (4) we use the linearity of the coefficient of operator, apply the formula
\begin{align*}
[z^{p-q}]A(z)=[z^p]z^qA(z)
\end{align*}
and use the binomial identity
\begin{align*}
\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q
\end{align*}
*In (5) we select the coefficients of $z^{n-1}$ to $z^{n-4}$.
| {
"language": "en",
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"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 which the line is a tangent to the curve.
The two tangents, given by the values of $k$ found in part ii), touch the curve at points.
iii)Find the coordinates of $A$ and $B$ and the equation of the line $AB$.
My attempt,
$-2x+k=\frac{2}{x-3}$
$(-2x+k)(x-3)=2$
Lastly, I got $2x^2-(6+k)x+3k+2=0$(Shown)
$b^2-4ac=0$
$(6+k)^2-4(2)(3k+2)=0$
$36+12k+k^2-24k-16=0$
$k^2-12k+20=0$
$k=2$ and $k=10$
How to proceed?
| 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.
For $k=10$, $2x^2-(6+k)x+3k+2=0\Rightarrow 2x^2-16x+32=0,$ i.e. $2(x-4)^2=0$. So, $x=4$, and $y=\frac{2}{x-3}=\frac{2}{4-3}=2$. So, we know that $(4,2)$ is the point where the line $y=-2x+10$ is tangent to the curve.
Hence, the equation of the line $AB$ is
$$y-(-2)=\frac{2-(-2)}{4-2}(x-2)\Rightarrow y=2x-6.$$
| {
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} |
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 this is helpful.
| 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
$$\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{k+1}}
\sum_{q=0}^n {n\choose 2q+1} \exp((n-2q)z) \; dz
\\ = \frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp((n+1)z)}{z^{k+1}}
\sum_{q=0}^n {n\choose 2q+1} \exp((-2q-1)z) \; dz$$
which is
$$\frac{1}{2}\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp((n+1)z)}{z^{k+1}}
\\ \times
\left(\sum_{q=0}^n {n\choose q} \exp(-qz)
- \sum_{q=0}^n {n\choose q} (-1)^q \exp(-qz)\right)
\; dz.$$
This yields two pieces, call them $A_1$ and $A_2.$ Piece $A_1$ is
$$\frac{1}{2}\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp((n+1)z)}{z^{k+1}}
(1+\exp(-z))^n \; dz
\\ = \frac{1}{2}\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(z)}{z^{k+1}}
(\exp(z)+1)^n \; dz$$
and piece $A_2$ is
$$\frac{1}{2}\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp((n+1)z)}{z^{k+1}}
(1-\exp(-z))^n \; dz
\\ = \frac{1}{2}\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(z)}{z^{k+1}}
(\exp(z)-1)^n \; dz.$$
Recall the species equation for labelled set partitions:
$$\mathfrak{P}(\mathcal{U}\mathfrak{P}_{\ge 1}(\mathcal{Z}))$$
which yields the bivariate generating function of the Stirling numbers
of the second kind
$$\exp(u(\exp(z)-1)).$$
This implies that
$$\sum_{n\ge q} {n\brace q} \frac{z^n}{n!} =
\frac{(\exp(z)-1)^q}{q!}$$
and
$$\sum_{n\ge q} {n\brace q} \frac{z^{n-1}}{(n-1)!} =
\frac{(\exp(z)-1)^{q-1}}{(q-1)!} \exp(z).$$
Now to evaluate $A_1$ proceed as follows:
$$\frac{1}{2}\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(z)}{z^{k+1}}
(2+\exp(z)-1)^n \; dz
\\ = \frac{1}{2}\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(z)}{z^{k+1}}
\sum_{q=0}^n {n\choose q} 2^{n-q} (\exp(z)-1)^q \; dz
\\ = \sum_{q=0}^n {n\choose q} 2^{n-q} \times
q!\times \frac{1}{2}\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(z)}{z^{k+1}}
\frac{(\exp(z)-1)^q}{q!} \; dz.$$
Recognizing the differentiated Stirling number generating function
this becomes
$$\sum_{q=0}^n {n\choose q} 2^{n-q-1} \times
q! \times {k+1\brace q+1}.$$
Now observe that when $n\gt k+1$ the Stirling number for $k+1\lt q\le
n$ is zero, so we may replace $n$ by $k+1.$ Similarly, when $n\lt k+1$
the binomial coefficient for $n\lt q\le k+1$ is zero so we may again
replace $n$ by $k+1.$ This gives the following result for $A_1:$
$$\sum_{q=0}^{k+1} {n\choose q} 2^{n-q-1} \times
q! \times {k+1\brace q+1}.$$
Moving on to $A_2$ we observe that when $k\lt n$ the contribution is
zero because the series for $\exp(z)-1$ starts at $z.$ This integral
is simple and we have
$$\frac{1}{2}\frac{k!\times n!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(z)}{z^{k+1}}
\frac{(\exp(z)-1)^n}{n!} \; dz.$$
Recognizing the Stirling number this yields
$$\frac{1}{2} \times n! \times {k+1\brace n+1}.$$
which correctly represents the fact that we have a zero contribution
when $k\lt n.$
This finally yields the closed form formula
$$\sum_{q=0}^{k+1} {n\choose q} 2^{n-q-1} \times
q! \times {k+1\brace q+1}
- \frac{1}{2} \times n! \times {k+1\brace n+1}.$$
confirming the previous results.
This MSE link has a computation that is quite similar.
| {
"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 = (h, k +- c)
b = 5 but what is a?
$$ c^2 = a^2 + b^2 $$
Thank you.
| 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",
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"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 $\frac{{\partial \frac{{\cos x - \cos y}}{{x - y}}}}{{\partial x}} = 0$, $\frac{{\partial \frac{{\cos x - \cos y}}{{x - y}}}}{{\partial y}} = 0$ gives $\sin x={ - \frac{{\cos x - \cos y}}{{x - y}}}$ and $\sin y={ - \frac{{\cos x - \cos y}}{{x - y}}}$. Plugging them back and we have the desired result $\left| {\frac{{\cos x - \cos y}}{{x - y}}} \right| \le 1$.
What I want to ask is, is there more decent way to show the inequity? Taking derivatives looks clumsy. Hope someone can help. Thank you!
| 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 show this,
choose $v$ so that
$\tan(v) = a/b$.
Then
$\begin{array}\\
a\cos(y)+b\sin(y)
&=b\tan(v)\cos(y)+b\sin(y)\\
&=b((\sin(v)/\cos(v))\cos(y)+\sin(y))\\
&=(b/\cos(v))(\sin(v)\cos(y)+\cos(v)\sin(y))\\
&=(b/\cos(v))\sin(v+y)\\
\end{array}
$
But,
since
$\begin{array}\\
\tan^2(v)
&=\sin^2(v)/\cos^2(v)\\
&=(1-\cos^2(v))/\cos^2(v)\\
&=1/\cos^2(v)-1,\\
1/\cos(v)
&=\sqrt{1+\tan^2(v)}\\
&=\sqrt{1+a^2/b^2}\\
&=\sqrt{a^2+b^2}/b\\
\text{so}\\
b/\cos(v)
&=\sqrt{a^2+b^2}\\
\end{array}
$
Therefore
$a\cos(y)+b\sin(y)
=\sqrt{a^2+b^2}\sin(v+y)
$.
Since
$|\sin(v+y)| \le 1$,
$|a\cos(y)+b\sin(y)|
\le \sqrt{a^2+b^2}
$.
Finally, we get
$\begin{array}\\
|\cos(y+h)-\cos(y)|
&=|\cos(y)(\cos(h)-1)-\sin(y)\sin(h)|\\
&\le \sqrt{(\cos(h)-1)^2+\sin^2{h}}\\
&= \sqrt{\cos^2(h)-2\cos(h)+1+\sin^2{h}}\\
&= \sqrt{2-2\cos(h)}\\
&= \sqrt{4\sin^2(h/2)}
\quad\text{(since }1-\cos(h) = 2\sin^2(h/2))\\
&=2|\sin(h/2)|\\
&\le |h|
\quad\text{(using } |\sin(h/2)| \le |h/2|)\\
\end{array}
$
so that,
as desired,
$\left|\dfrac{\cos(x)-\cos(y)}{x-y}\right|
\le 1
$.
Note that,
just from basic trig,
we have
$|\cos(x)-\cos(y)|
\le 2|\sin((x-y)/2)|
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356682",
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"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_{k+1}^{2} = \frac{2 \, a_{k}}{a_{k} + 1}
\end{align}
for $k \geq 1$.
*
*Prove $\lim_{k \to \infty} a_{k} = 1$ for every $a_{1}$
*Prove
\begin{align}
\lim_{n \to \infty} \, \prod_{k=1}^{n} \{ a_{k} \}
\end{align}
exists and is different from zero for every $a_{1}$
*Express the limit in (2) as a function of $a_{1}$.
It is to be noted that the problem is trivial for $a_{1} = 1$.
| 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 $a_{1} = \sec(\omega)$, $0 < \omega < \frac{\pi}{2}$. Then from the identity
\begin{align}
\sec^{2}\left(\frac{\omega}{2}\right) = \frac{2 \, \sec(\omega)}{1 + \sec(\omega)},
\end{align}
we may set $a_{2} = \sec\left(\frac{\omega}{2}\right)$. Repeating this reasoning we obtain $a_{n} = \sec(2^{1-n} \, \omega)$. Hence it follows that $a_{n}$ approaches the limit unity, and it decreases to this limit except in the trivial case $a_{1} =1$. We may now set
\begin{align}
\frac{1}{\prod_{k=1}^{n} \{a_{k}\}} &= \cos(\omega) \, \cos(2^{-1} \omega) \cdots \cos(2^{1-n} \, \omega) \\
&= \frac{\sin(2 \omega)}{2^{n} \, \sin(2^{1-n} \omega)} = \frac{\sin(2 \omega)}{2 \, \omega} \, \frac{2^{1-n} \, \omega}{\sin(2^{1-n} \, \omega)},
\end{align}
where the second form results by thetransformation of each factor by means of the formula
\begin{align}
\cos(A) = \frac{\sin(2A)}{2 \, \sin(A)}.
\end{align}
Hence we have
\begin{align}
\lim_{n \to \infty} \prod_{k=1}^{n} \{ a_{k} \} = \frac{2 \, \omega}{\sin(2 \omega)} = \frac{a_{1}^{2} \, \sec^{-1}(a_{1})}{\sqrt{a_{1}^{2} - 1}}.
\end{align}
For the case in which $a_{1} < 1$ we may set $a_{1} = sech(\omega)$, $\omega > 0$. We have merely to replace the trigonometric formulae by the corresponding hyperbolic formulae, and the resoning follows in a similar manner. We thus find that $a_{n}$ approaches unity as a limit and increases toward this limit, while
\begin{align}
\lim_{n \to \infty} \prod_{k=1}^{n} \{ a_{k} \} = \frac{a_{1}^{2} \, sech^{-1}(a_{1})}{\sqrt{1 - a_{1}^{2}}}.
\end{align}
| {
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} |
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 inductive step. Assume that this equation is true for some $n=k$. Note that
$$\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{1\cdot 2\cdot 3\cdot \ldots \cdot n}=\prod_{i=1}^{n}\left [ 2-\frac{1}{i} \right ].$$
For the case $n=k+1$ would be
$$\prod_{i=1}^{k+1}\left [ 2-\frac{1}{i} \right ]=\prod_{i=1}^{k}\left [ 2-\frac{1}{i} \right ]\left [ 2-\frac{1}{k+1} \right ]\leq 2^{k}\left [ 2-\frac{1}{k+1} \right ]=2^{k+1}-\frac{2^{k}}{k+1}.$$
I have to show that the last term is $\leq 2^{k+1}$. But it's not possible to show that $2^{k}/(k+1)\geq 0$.
| 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",
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"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 marbles be obtained?
D.) two or more green marbles be obtained?
Work:
A.) $4\cdot3\cdot8\cdot7\cdot6 $ $\cdot$ $_{5}C_{2}$ = $40,320$
B.) $6\cdot5\cdot8\cdot7\cdot6 $ $\cdot$ $_{5}C_{2}$ = $100,800$
C.) $4\cdot3\cdot12\cdot11\cdot10 $ $\cdot$ $_{5}C_{2}$ = $158,400$
D.) ($6\cdot5\cdot12\cdot11\cdot10 $ $\cdot$ $_{5}C_{2}$) + ($6\cdot5\cdot4\cdot12\cdot11 $ $\cdot$ $_{5}C_{3}$) + ($6\cdot5\cdot4\cdot3\cdot12 $ $\cdot$ $_{5}C_{4}$) + ($6\cdot5\cdot4\cdot3\cdot2 $ $\cdot$ $_{5}C_{5}$) = $576,720$
Is all of work correct?
| 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 $2$ blue, so $1$ red. The same reasoning as above gives $\binom{6}{2}\binom{8}{2}\binom{4}{1}$.
C.) The $2$ red can be chosen in $\binom{4}{2}$ ways. Now we must choose $3$ non-red from the $14$ non-red.
D.) It's your turn.
| {
"language": "en",
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"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: $$2 \cosh \frac x 2 = 2\sqrt{\frac{1+\cosh x}{2}} = \sqrt{\frac{4(1+\cosh x)}{2}} = \sqrt{2+2\cosh x}.$$
The second expression there does not makes sense. What you can do is copy the above work multiplying by $\sqrt{2a}$, if $a>0$, to obtain: $$ \sqrt{2a}\cosh \frac x 2 = \sqrt{a+a\cosh x},$$in the same fashion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 3,
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} |
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 $2p+1 \mid 2^p + 1$.
Thanks.
EDIT: Thank you to Batominovski, for his great work and sharing the proof with us.
| 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 $2^p+1$. Let $q$ be the largest prime factor of $2p+1$. Then, $2^{2p}\equiv (-1)^2 =1\pmod{q}$. On the other hand, $2^{q-1}\equiv 1\pmod{q}$. Hence, $2^d\equiv 1\pmod{q}$, where $d:=\gcd(q-1,2p)$. Clearly, $d=2$ or $d=2p$. If $d=2$, then $q=3$, whence $2p+1=3^r$ for some $r\in\mathbb{N}$. However, it can be easily seen that $9\nmid 2^p+1$, whence $r=1$. However, this would mean $p=1$, a contradiction. Therefore, $d=2p$. That is, $q=2p+1$, which means that $2p+1$ is prime.
P.S. This proof can be used to deal with the case where $p\equiv 3\pmod{4}$, with a slight modification (and it is actually simpler).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
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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 $-8x^3$ ?
| $$\begin{align}
(x^3 - 4)^2 - x^6 + 2x^5 &= x^3\cdot x^3 + (-4)^2 + 2\cdot (-4) \cdot x^3 -x^6+ 2x^5\\
&=x^6-x^6 + 16 -8x^3 + 2x^5 = 2x^5 -8x^3 + 16
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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
$$\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + ...\approx \ ?$$
Edit: incorrect value
| 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\\
& \qquad + \frac{1}{4} + \frac{1}{8} + \ldots &= \frac{1}{2} \\
& \qquad \qquad +\frac{1}{8} + \ldots &=\frac{1}{4}\\
&\qquad \qquad \qquad\vdots&\vdots\end{align}
So $\Bbb{E}[X] = 1 + \frac{1}{2} + \frac{1}{4} + \ldots = 2$
Addendum: If the format is a little unclear, try this:
$\begin{align}
\sum_{k=1}^\infty \frac k{2^k}
& = \frac 1 2 + \frac 2 4 + \frac 3 8 +\cdots
\\[2ex] & = \boxed{\begin{matrix} (\frac 1 2 & + \frac 1 4 &+ \frac 1 8 &+\cdots)+
\\ & (\frac 1 4 & + \frac 1 8 & +\cdots)+ \\ & & (\frac 1 8 & + \cdots) +\\&&& + \cdots \end{matrix}}
\\[2ex] & = \boxed{\begin{matrix} (\frac 1 2 & + \frac 1 4 &+ \frac 1 8 &+\cdots)+
\\ & \frac 1 2(\frac 1 2 & + \frac 1 4 & +\cdots)+ \\ & & \frac 1 4(\frac 1 2 & + \cdots) +\\&&& + \cdots \end{matrix}}
\\[2ex] & = 1 + \frac 1 2 + \frac 1 4 + \cdots
\\[2ex] & = 2
\\[1ex]\Box
\end{align}$
| {
"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 coefficients of each powers will be same
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
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:
$7a-b = 6$
$7c-d = 2$
$a^2+bc = a$
$ab+bd = b$
$ac + cd = c$
$bc + d^2 = d$ from trying that method. What should I do?
| 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 $(6(2a-3)))^2+180=\left(2\cdot 3^{k+1}\right)^3$. But $x^2+180=y^3$ has $4$ integral solutions $(x,y)=(\pm 6,6),(\pm 573,69)$, so $(a,n)=(2,1),(1,1)$.
*$n=3k+2$. Then $(18(2a-3))^2+1620=\left(2\cdot 3^{k+2}\right)^3$, no solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
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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 the left-hand side is equivalent to the right-hand side.To be direct I am facing some difficulty with the algebra required to make LHS = RHS.
Here is what I have done so far:
1) Base Case
$n = 1$
LHS:
$1/2$
and RHS is $1/2$ $\checkmark$
2) Inductive Step
For $k \geq 1$, Assume $n = k$
$$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \cdots + \frac{k}{(k+1)!} = 1 - \frac{1}{(k+1)!}$$
$$n = k + 1$$
$$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \cdots + \frac{n}{(n+1)!} + \frac{k+1}{(k+2)!} = 1 - \frac{1}{(n+1)!}$$
$$\implies 1-\frac{1}{(k+1)!} + \frac{k+1}{(k+2)!} = 1 - \frac{1}{(k+2)!}$$
Here is where i do not know how to make the LHS = RHS.
| 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,
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} |
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 how $\frac{a+b}{b+c}=\frac{c+d}{d+a}$ if $a+b+c+d=0$. Please help.
| 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 $\frac{d}{dx}(\log_ax)= \frac{1}{x\ \ln\ a}$
$(2x)\cdot [\log_{10}\sqrt{x}]'+(\log_{10}\sqrt{x})\cdot [2x]'$
$y'=2x\frac{1}{\sqrt{x}\ln 10}+\log_{10}\sqrt{x}\cdot 2$
Answer in book is $y'= \frac{1}{\ln10}+\log_{10}x$
| 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 $$f(n+1) = f(n) + \frac{f(n)^2}{n(n+1)}$$ with initial condition $f(1) = 1$. Does this sequence converge or diverge?
| 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
$$
\begin{align}
&\left(\frac1{f(n+1)}-\frac1{n+1}\right)-\left(\frac1{f(n)}-\frac1n\right)\\
&=\left(\frac1n-\frac1{n+1}\right)-\left(\frac1{f(n)}-\frac{n(n+1)}{n(n+1)f(n)+f(n)^2}\right)\\
&=\frac1{n(n+1)}-\left(\frac{n(n+1)+f(n)}{n(n+1)f(n)+f(n)^2}-\frac{n(n+1)}{n(n+1)f(n)+f(n)^2}\right)\\
&=\frac1{n(n+1)}-\frac1{n(n+1)+f(n)}\\
&=\frac{f(n)}{n(n+1)(n(n+1)+f(n))}\\
&\ge\frac{f(n)}{n^2(n+1)(n+2)}\\
&\ge\frac1{n^2(n+1)(n+2)}\tag{3}
\end{align}
$$
Since
$$
\sum_{n=1}^\infty\frac1{n^2(n+1)(n+2)}=\frac{\pi^2}{12}-\frac58\tag{4}
$$
we have
$$
\begin{align}
\frac1{f(n)}
&\ge\frac1n+\sum_{k=1}^{n-1}\frac1{n^2(n+1)(n+2)}\\
&\ge\frac{\pi^2}{12}-\frac58\tag{5}
\end{align}
$$
Therefore,
$$
f(n)\le\frac{24}{2\pi^2-15}\tag{6}
$$
Since $f(n)$ is increasing and bounded above, it converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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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
$$s=\sqrt{p(p-a)(p-b)(p-c)} ,p=\frac{a+b+c}{2}\\s=\frac{h_a*a}{2}\\s=\frac{1}{2}bc\sin(A)\\s=2R^2\sin A \sin B \sin C$$
Another symmetrical form is given by :$$(4s)^2=\begin{bmatrix}
a^2 & b^2 & c^2
\end{bmatrix}\begin{bmatrix}
-1 & 1 & 1\\
1 & -1 & 1\\
1 & 1 & -1
\end{bmatrix} \begin{bmatrix}
a^2\\
b^2\\
c^2
\end{bmatrix}$$
Expressing the side lengths $a$, $b$ & $c$ in terms of the radii $a'$, $b'$ & $c'$ of the mutually tangent circles centered on the triangle's vertices (which define the Soddy circles)
$$a=b'+c'\\b=a'+c'\\c=a'+b'$$gives the paticularly pretty form $$s=\sqrt{a'b'c'(a'+b'+c')}$$
If the triangle is embedded in three dimensional space with the coordinates of the vertices given by $(x_i,y_i,z_i)$ then $$s=\frac{1}{2}\sqrt{\begin{vmatrix}
y_1 &z_1 &1 \\
y_2&z_2 &1 \\
y_3 &z_3 &1
\end{vmatrix}^2+\begin{vmatrix}
z_1 &x_1 &1 \\
z_2&x_2 &1 \\
z_3 &x_3 &1
\end{vmatrix}^2+\begin{vmatrix}
x_1 &y_1 &1 \\
x_2&y_2 &1 \\
x_3 &y_3 &1
\end{vmatrix}^2}$$
When we have 2-d coordinate $$ s=\frac{1}{2}\begin{vmatrix}
x_a &y_a &1 \\
x_b &y_b &1 \\
x_c &y_c & 1
\end{vmatrix}$$
In the above figure, let the circumcircle passing through a triangle's vertices have radius $R$, and denote the central angles from the first point to the second $q$, and to the third point by $p$ then the area of the triangle is given by:
$$ s=2R^2|\sin(\frac{p}{2})\sin(\frac{q}{2})\sin(\frac{p-q}{2})|$$
| 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
$T_1=\triangle ABD_a$,
$T_2=\triangle AD_aC$ (assuming that $A,B,C$ are in counterclockwise order),
such that the radii
of their incircles are the same,
$r_{a1}=r_{a2}=r_a$.
Similar conditions hold for cevians $BD_b$ and $CD_c$.
Given that $a,b,c$ are the sides of $\triangle ABC$,
$r$ is the radius of its inscribed circle
and $\rho=\tfrac12(a+b+c)$,
\begin{align}
|AD_a|=d_a&=\sqrt{\rho(\rho-a)}
,\\
r_a&=\frac{r}{1+\sqrt{1-\frac a\rho}}
,\\
l_a&=\frac1{r_a}
\end{align}
together with similarly defined
$d_b,d_c,r_b,r_c,l_b,l_c$,
expressions for the area follows:
\begin{align}
S&=
\frac{d_a d_b d_c}{\sqrt{d_a^2+d_b^2+d_c^2}}
\tag{1}\label{1}
,\\
&=
\frac{a r_a^2}{2r_a-r}
\tag{2}\label{2}
,\\
&=
(d_a^2+d_b^2+d_c^2)(\tfrac r{r_a}-1)(\tfrac r{r_b}-1)(\tfrac r{r_c}-1)
\tag{3}\label{3}
,\\
&=
r_a\,\Big(d_a+\sqrt{d_a^2+d_b^2+d_c^2}\Big)
\tag{4}\label{4}
,\\
&=
\frac{r^2\,r_a\,r_b\,r_c}{(r-r_a)(r-r_b)(r-r_c)}
\tag{5}\label{5}
.
\end{align}
Note, that \eqref{5} can be expressed in terms of just the three radii $r_a,r_b,r_c$
since we can express $r$ as
\begin{align}
r&=
\frac{l_a+l_b+l_c+\sqrt{2\,(l_a l_b+l_b l_c+l_c l_a)-(l_a^2+l_b^2+l_c^2)}}{l_a^2+l_b^2+l_c^2}
\tag{6}\label{6}
\end{align}
and \eqref{5} becomes
\begin{align}
S&=
\frac{(l_a^2+l_b^2+l_c^2)\,(l_a+l_b+l_c+\sigma)^2}
{(l_a\,(l_b+l_c+\sigma)-l_b^2-l_c^2)\,(l_b\,(l_c+l_a+\sigma)-l_c^2-l_a^2)\,(l_c\,(l_a+l_b+\sigma)-l_a^2-l_b^2)}
\tag{7}\label{7}
,
\end{align}
where
\begin{align}
\sigma&=
\sqrt{2\,l_a\,l_b+2\,l_b\,l_c+2\,l_c\,l_a-l_a^2-l_b^2-l_c^2}
\tag{8}\label{8}
.
\end{align}
Edit
An interesting equation for $S$
also exists for the case of uneven bisection, that is,
when the corresponding pairs
of inscribed circles have different radii,
\begin{align}
r_{a1}&\ne r_{a2},\quad r_{b1}\ne r_{b2},\quad r_{c1}\ne r_{c2}
,\\
l_{a1}&=1/r_{a1},\quad l_{a2}=1/r_{a2}
,\\
l_{b1}&=1/r_{b1},\quad l_{b2}=1/r_{b2}
,\\
l_{c1}&=1/r_{c1},\quad l_{c2}=1/r_{c2}
.
\end{align}
In this case we have
\begin{align}
S&=
r^2\,\sqrt{\frac{r_{a1}r_{a2}r_{b1}r_{b2}r_{c1}r_{c2}}{(r-r_{a1})(r-r_{a2})(r-r_{b1})(r-r_{b2})(r-r_{c1})(r-r_{c2})}}
,\\
r&=
\frac{l_{a1}+l_{a2}+l_{b1}+l_{b2}+l_{c1}+l_{c2}}{2 (l_{a1} l_{a2}+l_{b1} l_{b2}+l_{c1} l_{c2})}
\\
&+
\frac{\sqrt{(l_{a1}+l_{a2}+l_{b1}+l_{b2}+l_{c1}+l_{c2})^2-8 (l_{a1} l_{a2}+l_{b1} l_{b2}+l_{c1} l_{c2})}}
{2 (l_{a1} l_{a2}+l_{b1} l_{b2}+l_{c1} l_{c2})}
.
\end{align}
| {
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"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:
$2xy(x\frac{dy}{dx}+y)=\frac{(x+1)\frac{dy}{dx}-y-1}{(x+1)^2}$
Now the textbook gives the answer as $-\frac{y(y+1)(3x+2)}{x(x+1)(y+2)}$ and I'm wondering how to get that answer. I think I'm going about this the wrong way so any hints/guidance would be much appreciated.
| $x^2y^2=(y+1)/(x+1)$
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 $x$,
$\frac{2}{x}+\frac{2}{y}\times\frac{dy}{dx}=\frac{1}{y+1}\times\frac{dy}{dx}-\frac{1}{x+1}$
$\frac{2}{y}\times\frac{dy}{dx}-\frac{1}{y+1}\times\frac{dy}{dx}=-\frac{1}{x+1}-\frac{2}{x}$
$(\frac{2}{y}-\frac{1}{y+1})\frac{dy}{dx}=-\frac{1}{x+1}-\frac{2}{x}$
$\frac{2y+2-y}{y(y+1)}\frac{dy}{dx}=\frac{-x-2x-2}{x(x+1)}$
$\frac{y+2}{y(y+1)}\frac{dy}{dx}=\frac{-(3x+2)}{x(x+1)}$
$\frac{dy}{dx}=\frac{-(3x+2)}{x(x+1)}\times\frac{y(y+1)}{y+2}$
| {
"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
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s-xs&=\frac{1}{1-x}\\
s(1-x)&=\frac{1}{1-x}\\
s&= \frac{1}{(1-x)^2}
\end{align}
$$
2nd proof:
$$
\begin{align}
s&=1+2x+3x^2+4x^3+5x^4+\cdots\\
&=\left(1+x+x^2+x^3+\cdots\right)'\\
&=\left(\frac{1}{1-x}\right)'\\
&=\frac{0-(-1)}{(1-x)^2}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
3rd Proof:
$$
\begin{align}
s=&1+2x+3x^2+4x^3+5x^4+\cdots\\
=&1+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+0+x^3+x^4+x^5+\cdots\\
&+\cdots
\end{align}
$$
$$
\begin{align}
s&=\frac{1}{1-x}+\frac{x}{1-x}+\frac{x^2}{1-x}+\frac{x^3}{1-x}+\cdots\\
&=\frac{1+x+x^2+x^3+x^4+x^5+...}{1-x}\\
&=\frac{\frac{1}{1-x}}{1-x}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
These are my three proofs to date. I'm looking for more ways to prove the statement.
| 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=0}^{\infty}(n+1)x^n\right) dx
=\sum_{n=0}^{\infty}\int(n+1)x^n dx\\
&=\sum_{n=0}^{\infty}x^{n+1}+C=\sum_{n=1}^\infty x^n+C\\
&=\frac{1}{1-x}-1+C\\
\\
\int g(x) dx&= \int \frac{1}{(1-x)²}dx=\frac{1}{1-x}+D
\end{align*}
We observe $f$ and $g$ have the same antiderivative $\frac{1}{1-x}$ differing by a constant only. Since $f(0)=g(0)=1$ they are equal.
| {
"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/minima be $x_1, x_2$.
I found $f'(x)=6x^2+2ax+b$
So $x_1, x_2$ are the roots of $6x^2+2ax+b=0$
I could not solve this further. I think the minimum value of $a+b$ is $\sqrt {ab}$.
Is my approach correct or is there any other method?
| 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 roots.
Hence, we have determinant $B^2-4AC>0$ $$\implies (a)^2-4(2)(b)>0$$ $$\implies a^2-8b>0$$ $$\implies \color{red}{a^2>8b}$$ Since, $a$ & $b$ are natural numbers hence, to get minimum value of $\color{red}{(a+b)}$ the numbers $a$ & $b$ both must be least natural numbers satisfying the above relation.
Hence, substituting the least possible value of $b$ as $b=1$, we get $$a^2>8\times 1\implies a^2>8$$ Now, the least possible value of $a$ satisfying the above inequality is $8$
Thus, $a=3$ & $b=1$ are fully satisfying the inequality, hence the minimum value of $(a+b)$ is given as follows $$\color{blue}{a+b=3+1=4}$$
| {
"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) $\frac {21}5$ (C) $\frac {25}3$ (D) $\frac {14}3$ (E) $\frac {34}3$
It doesn't give us what $f(x)$ is equal to.
Tried to do comparison
\begin{align}
2f(-5) + f(1-(-5)) &= (-5)^2 \\
2f(-5) + f(6) &= 25 \quad(1) \\
2f(6) + f(1-(6)) &= (6)^2 \\
2f(6) + f(-5) &= 36 \quad(2) \\
\end{align}
so we have the $f(-5)$ and $f(6)$ in both functions, we can add them together
\begin{align}
(1)+(2) \\
[2f(-5) + f(6)] + [2f(6) + f(-5)] &= 25+36 \\
2f(-5) + f(-5) + f(6) + 2f(6) &= 61 \\
3f(-5) + 3f(6) &= 61 \\
f(6) &= \frac {61}3 - f(-5) \quad(3) \\
\end{align}
replacing $f(6)$, at equation $\quad(3)$ to the first equation, $\quad(1)$
\begin{align}
2f(-5) + [\frac{61}{3} - f(-5)] &= 25 \\
2f(-5) - f(-5) &= 25 - \frac {61}3 \\
f(-5) &= \frac{14}{3}
\end{align}
EDIT: Well, I am blind, saw + to minus
| 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 $\frac {34}3$ for $f(-3)$. All other cases correspond to $x$ expressed with radicals. Probably one more typo.
You changed the post (from $-$ to $+$) but the same method applies.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
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
$$1\cdot3+3\cdot5+5\cdot7+\cdots+(2k-1)(2k+1)= {k(4k^2+6k-1) \over 3}$$
Step 2: When a statement is true for a natural number $n = k$,
then it will also be true for its successor, $n=k+1$. Hence, we have to prove that it is also true for $n=k+1$.
$$1\cdot3+3\cdot5+5\cdot7+\cdots+(2k-1)(2k+1)+(2k+1-1)(2k+1+1) = {k(4k+1^2+6k+1-1) \over 3}$$
I replace the LHS by step 1.
$${k(4k^2+6k-1) \over 3} + 2(k+1-1)(2k+1+1)={k(4k+1^2+6k+1-1) \over 3}$$
Now I need to make LHS equal to RHS.
| 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 sides of an already-assumed-to-be-true equation. I'm adding $(2(k+1)-1)(2(k+1)+1)$.
Second, the reason I've added this is because it is what the next term would look like. Not $(2k+1-1)(2k+1+1)$, which just inserted some "${}+1$" in some places.
From here, you need to give a common denominator to the right side and see that it works out to be ${(k+1)(4(k+1)^2+6(k+1)-1) \over 3}$.
| {
"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
$$m^2x-mx-2x-m^2-4m-3=0$$
$$m^2(x-1)-m(x+4)-(2x+3)=0$$
$$D=x^2+8x+16+4(x-1)(2x+3)$$
$$=x^2+8x+16+4(2x^2-2x+3x-3)$$
$$=9x^2+12x+4$$
$$=(3x+2)^2$$
$$m=\frac{x+4\pm (3x-2))}{2(x-1)}$$
$$m_1=\frac{4x+2}{2x-2}$$
$$m_2=\frac{-2x+6}{2x-2}$$
Is this right?
I don't know if the tag is right, so please don't be based on it.
| 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 $m+1$, simplifying the algebra.
$$(m-2)x = m+3 \implies mx - 2x = m+3 \implies m(x-1) = 2x + 3 \implies m = \frac{2x+3}{x-1}$$
and it should be clear that $x \neq 1$ for this solution to be valid.
Hence your solution set is:
$m=-1$ or $\displaystyle m = \frac{2x+3}{x-1}, (x \neq 1)$
| {
"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 $\sqrt{125} \div \sqrt5$, then the answer would indeed be $5$.
| {
"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 understandable.
| 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 bit. We don't need more than one $1$, because we can do this:
$21=1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1\\
\phantom{21}=(1+1)+(1+1)+(1+1)+(1+1)+(1+1)+(1+1)+(1+1)+(1+1)+(1+1)+(1+1)+1\\
\phantom{21}=2+2+2+2+2+2+2+2+2+2+1$
Similarly, we don't need more than one $2$, because we can do this:
$21=2+2+2+2+2+2+2+2+2+2+1\\
\phantom{21}=(2+2)+(2+2)+(2+2)+(2+2)+(2+2)+1\\
\phantom{21}=4+4+4+4+4+1$
And we don't need more than one $4$:
$21=4+4+4+4+4+1\\
\phantom{21}=(4+4)+(4+4)+4+1\\
\phantom{21}=8+8+4+1$
And finally:
$21=8+8+4+1\\
\phantom{21}=(8+8)+4+1\\
\phantom{21}=16+4+1$
Basically, we start with a bunch of ones, and then "collapse" them until all powers of two are different from each other.
| {
"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]{a^2} + 2\sqrt[3]{ab} + \sqrt[3]{b^2})(\sqrt[3]a + \sqrt[3]b) &= 3 \\
\sqrt[3]{a^3} + \sqrt[3]{3a^2b} + \sqrt[3]{3ab^2} + \sqrt[3]{b^3} &= 3 \\
a + b + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= 3
\end{align}
There's still had cube root, how do I remove it?
| 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:
$$
12=(6+\sqrt{x})+(6-\sqrt{x})=(e+\sqrt[3]{3}/2)^3+(-e+\sqrt[3]{3}/2)^3=\frac{3}{4}+3\sqrt[3]{3}e^2
$$
from which $e=\frac{1}{2}\sqrt[3]{3}\sqrt{5}$.
Step 2 (verify): with $e$ as above, you can verify that ($*$) is satisfied with $\sqrt{x}=3\sqrt{5}$ which is equivalent to $x=45$.
| {
"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
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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:
$$\begin{eqnarray*} \left|\int_{\pi}^{+\infty}\frac{dx}{x^2\left(\sin^2 x\right)^{\frac{1}{3}}}\,dx\right|&\leq& \pi\left(\frac{1}{\pi^2}+\frac{1}{4\pi^2}+\frac{1}{9\pi^2}\ldots\right)\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{6}\right)}{\Gamma\left(\frac{2}{3}\right)}\\&=&\frac{\pi}{6}\cdot \frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{6}\right)}{\Gamma\left(\frac{2}{3}\right)}.\end{eqnarray*}$$
The exact value of the integral is given by:
$$ I = \frac{1}{\pi}\int_{0}^{1}\frac{\psi'(1+x)\,dx}{\left(\sin^2 (\pi x)\right)^{\frac{1}{3}}}$$
and by computing the Laplace transform of $\frac{1}{(\sin^2 x)^{\frac{1}{3}}}$, the inverse Laplace transform of $\frac{1}{(x+\pi)^2}$ and keeping just the terms $A e^{-\pi s},B s^2 e^{-\pi s}$ of their product we get:
$$ I\leq \frac{\Gamma\left(\frac{1}{6}\right)}{4\pi^{7/2}\Gamma\left(\frac{2}{3}\right)}\left(5\pi^2-\psi'\left(\frac{1}{3}\right)+\psi'\left(\frac{2}{3}\right)\right)\leq\color{red}{\frac{4}{5}}.$$
| {
"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 rigorous.
I used factoring:
$$= \frac{(ab + 1)(a^2b^2 - ab + 1)}{(a+b)(a^2 - ab + b^2)} < \frac{3}{2} \cdot \bigg( \frac{a^2b^2 -ab + 1}{a^2 -ab + b^2} \bigg)$$
But that doesnt get you anywhere either.
Hints only please!
| 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
$$ \frac{(1)^3b^3+1}{(1)^3+b^3} = \frac{b^3+1}{b^3+1} = 1$$
Similarly, if $b=1$, this second fraction would become
$$ \frac{a^3(1)^3+1}{a^3+(1)^3} = \frac{a^3+1}{a^3+1} = 1$$
Let's exclude these possibilities for $a$ and $b$ because they don't offer much useful information for finding a maximum bound. Then since $a$ and $b$ are both positive integers, we can setup a new inequailty for $a > 2$ using the value of $a$ we defined in terms of $b$
$$ \frac{3b-2}{2b-3} > 2$$
Since we want $b > 1$, we can multiply both sides by $2b-3$ to simplify
\begin{align*}
\frac{3b-2}{2b-3}\cdot (2b-3) &> 2\cdot(2b-3)\\
3b-2 &> 4b-6\\
\Rightarrow b < 4
\end{align*}
Since you said "Hints only" I'll leave the rest up to you. Let me know if you'd like further help!
| {
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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 explain to me how to finish it...
$$
\frac {16}3 \int \cos^2\theta \,d\theta\
$$
| 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:
$$\sqrt{16-9x^2}=\sqrt{16-16\sin^2\theta}=\sqrt{16(1-\sin^2\theta)}=\sqrt{16\cos^2\theta}=4\cos\theta$$
and since $x=\frac{4}{3}\sin\theta$, we have got $dx=\frac{4}{3}\cos\theta d\theta$, so
$$\displaystyle\int\sqrt{16-9x^2}dx=\int4\cos\theta\cdot\left(\frac{4}{3}\cos\theta d\theta\right)=\frac{16}{3}\int\cos^2\theta d\theta$$
and here is the way to do the rest
$\displaystyle\frac{16}{3}\int\cos^2\theta d\theta=\frac{16}{3}\int\left(\frac{1}{2}\left(1+\cos 2\theta)\right)\right)=\frac{8}{3}\int\left(1+\cos 2\theta\right)=\frac{8}{3}\left(\theta +\frac{1}{2}\sin2\theta+C\right)$.
Since $\sin\theta =\frac{3x}{4}$, we have $\theta=\sin^{-1}\left(\frac{3x}{4}\right)$ and $\sin2\theta = \sqrt{16-9x^2}$ and so $\displaystyle\int\sqrt{16-9x^2}dx$ evaluates to $\displaystyle\frac{8}{3}\sin^{-1}\left(\frac{3x}{4}\right)+\frac{1}{2}x\sqrt{16-9x^2}+C$
Edit: (To address OP's questions, as this is comment is not properly showing up in the comments section). After multiplying $\frac{8}{3}$ with $\frac{1}{2}\sin\theta$, we have $\displaystyle\frac{4}{3}\sin2\theta=\frac{4}{3}(2\sin\theta\cos\theta)=\frac{8}{3}\sin\theta\cos\theta=\frac{8}{3}\cdot\left(\frac{3x}{4}\right)\cdot\left(\frac{\sqrt{16-9x^2}}{4}\right)$, then cancel you get it.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 that there are
$$
\binom{6}{4} + \binom{7}{4} = 15 + 35 = 50
$$
different ways to select only red chips or only white chips, and that is the difference between this answer and barak's.
As in barak's answer, there are
$$
\binom{16}{4} = 1820
$$
different ways to select $4$ of $16$ chips, so the desired probability is
$$
P = \frac{819}{1820} = \frac{9}{20}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1379580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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:
$$
I = \int\limits_0^1\left(\frac{\left(x^2+1\right)\arcsin(x)}{\sqrt{1-x^2}}+2x\ln\left(x^2+1\right)\right)\frac{\ln x}{x^3+x}\,dx = -2\,G\,\ln2,
$$
where $G$ is Catalan's constant.
Numerically
$$ I \approx -1.2697979381877088371491554851603117320986537271546606092465\dots$$
How to prove it?
| 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)}{1+x^2}\ {\rm d}x=4\Im{\rm Li}_3(1-i)-2\mathbf{G}\ln{2}+\frac{3\pi^3}{16}+\frac{\pi}{4}\ln^2{2}
\end{align}
As for the first integral,
\begin{align}
\int\limits^\frac{\pi}{2}_0\frac{x\ln(\sin{x})}{\sin{x}}\ {\rm d}x
&=2\int\limits^1_0\frac{\arctan{x}\ln\left(\frac{2x}{1+x^2}\right)}{x}\ {\rm d}x\\
&=2{\rm Ti}_2(1)\ln{2}+2\int\limits^1_0\frac{\arctan{x}\ln{x}}{x}\ {\rm d}x-2\int\limits^1_0\frac{\arctan{x}\ln(1+x^2)}{x}\ {\rm d}x\\
&=2\mathbf{G}\ln{2}-2\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^3}+2\int\limits^1_0\frac{\ln{x}\ln(1+x^2)}{1+x^2}\ {\rm d}x+4\int\limits^1_0\frac{x\arctan{x}\ln{x}}{1+x^2}\ {\rm d}x
\end{align}
and the integral
\begin{align}
4\int\limits^1_0\frac{x\arctan{x}\ln{x}}{1+x^2}\ {\rm d}x=-8\Im{\rm Li}_3(1-i)-\frac{5\pi^3}{16}-\frac{\pi}{2}\ln^2{2}
\end{align}
has also been established in the link above. (Both integrals were covered in the evaluation of $\mathscr{J}_2$.) Hence the closed form is indeed
\begin{align}
I=-2\mathbf{G}\ln{2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380140",
"timestamp": "2023-03-29T00:00:00",
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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 $$f(x)=\frac{\sin x-x\cos x}{x\sin x+\cos x}$$
$$\implies f'(x)=\frac{d}{dx}\left(\frac{\sin x-x\cos x}{x\sin x+\cos x}\right)=\frac{(x\sin x+\cos x)\frac{d}{dx}(\sin x-x\cos x)-(\sin x-x\cos x)\frac{d}{dx}(x\sin x+\cos x)}{(x\sin x+\cos x)^2}$$ $$ =\frac{(x\sin x+\cos x)(x\sin x)-(\sin x-x\cos x)(x\cos x)}{(x\sin x+\cos x)^2}$$ $$ =\frac{x^2\sin^2 x+x\sin x\cos x-(x\sin x\cos x-x^2\cos^2 x)}{(x\sin x+\cos x)^2}$$ $$ =\frac{x^2\sin^2 x+x\sin x\cos x-(x\sin x\cos x-x^2\cos^2 x)}{(x\sin x+\cos x)^2}$$ $$ =\frac{x^2\sin^2 x+x\sin x\cos x-x\sin x\cos x+x^2\cos^2 x)}{(x\sin x+\cos x)^2}$$ $$ =\frac{x^2(\sin^2 x+\cos^2 x)}{(x\sin x+\cos x)^2}$$ $$f'(x) =\frac{x^2}{(x\sin x+\cos x)^2}$$
Now, integrating both the sides we get $$\int(f'(x))dx=\int \frac{x^2}{(x\sin x+\cos x)^2}$$ $$\implies \int \frac{x^2}{(x\sin x+\cos x)^2}=f(x)+C$$ $$ \int \frac{x^2}{(x\sin x+\cos x)^2}=\frac{\sin x-x\cos x}{x\sin x+\cos x}+C$$
| {
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"url": "https://math.stackexchange.com/questions/1380314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How come $\ n\ $ always divides at least one of the item of the sequence? Given positive integer$\ \displaystyle n,\ $ the sequence is:
$\displaystyle 2^n$
$\displaystyle 2^n - 2^{n-1}$
$\displaystyle 2^n - 2^{n-1} + 2^{n-2}$
$\displaystyle 2^n - 2^{n-1} + 2^{n-2} - 2^{n-3}$
...
$\displaystyle 2^n - 2^{n-1} + 2^{n-2} - 2^{n-3} \dots \pm 1$
If $\ \displaystyle n\ $ is even, the final term is $\ \displaystyle +1;\ $ if $\ \displaystyle n\ $ is odd, then the final term is $\ \displaystyle -1$.
My question is : how come $\ n\ $ always divides at least one of the item of the sequence?
Python program:
def f(n):
l = []
s = 2**n
l.append(s)
t = -1
for i in range(n, 0, -1):
s = s + t * 2**(i - 1)
t *= -1
l.append(s)
return l
print(f(5)) #output : [32, 16, 24, 20, 22, 21]
print(f(7)) #output : [128, 64, 96, 80, 88, 84, 86, 85]
Source: http://mymathforum.com/number-theory/100912-conjecture-2-n.html
| 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) $S(n-3)=3\cdot 2^{n-2}-2^{n-3}=6\cdot 2^{n-3}-2^{n-3}=5\cdot 2^{n-3}$
e) $S(n-4)=5\cdot 2^{n-3}+2^{n-4}=10\cdot 2^{n-4}+2^{n-4}=11\cdot 2^{n-4}$
So the sequence of coefficients $c$ is actually
$$
1,1,3,5,11,\ldots,c_i,2c_i\pm 1,\ldots,c_{n+1}
$$
and the question can be restated as whether or not a sequence of length $n+1$ has an element $c$ multiple of $n$. Lets call it a $C$ sequence.
Notice that we can check whether or not any $C$ sequence of prime length $p$ has a term divisible by $p$, since in this case a positive answer makes the general question follow immediately.
Let $n=p$, and assume $c_i, 2c_i-1, 2^2c_i-1, 2^3c_i-3, 2^4c_i-5, 2^5c_i-11, \ldots, 2^{\frac{p-1}{2}}c_i-c_j$ are $\frac{p+1}{2}$ consecutive coefficients in the $C$ sequence where none is divisible by $p$. Notice that the negatives are again the sequence up to the $\textstyle\frac{p-1}{2}-th$ value, then
$$
c_i, 2c_i-c_1, 2^2c_i-c_2, 2^3c_i-c_3, 2^4c_i-c_4, 2^5c_i-c_5, \ldots, 2^{\frac{p-1}{2}}c_i-c_{\frac{p-1}{2}}
$$
In particular,
$$
2^{\frac{p-1}{2}}c_i\not\equiv c_\frac{p-1}{2}\pmod p\Rightarrow 2^{p-1}c_i^2\not\equiv c_\frac{p-1}{2}^2\pmod p\Rightarrow c_i^2\not\equiv c_\frac{p-1}{2}^2\pmod p
$$
Since the length of such list of consecutive coefficients is $\frac{p+1}{2}$, and there are $p+1$ terms in the $C$ sequence, then we have $i\in\{1,2,\ldots,\frac{p-1}{2}\}$.
This means that
\begin{array}{ll}
c_1^2&\not\equiv c_\frac{p-1}{2}^2\pmod p\\
c_2^2&\not\equiv c_\frac{p-1}{2}^2\pmod p\\
&\vdots\\
c_\frac{p-1}{2}^2&\not\equiv c_\frac{p-1}{2}^2\pmod p
\end{array}
The last equivalence is absurd, thus $p$ divides at least one term of the $C$ sequence. $\blacksquare$
| {
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"url": "https://math.stackexchange.com/questions/1381819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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
$$\int_{0}^{4}\sqrt{4^2-x^2}dx=\frac{1}{2}\left(x\sqrt{4^2-x^2}+4^2\sin^{-1}\left(\frac{x}{4}\right)\right)_{0}^{4}$$ $\color{red}{\text{applying the limits}}$ $$=\frac{1}{2}\left(4\sqrt{4^2-4^2}+4^2\sin^{-1}\left(\frac{4}{4}\right)-(0)\sqrt{4^2-(0)^2}-4^2\sin^{-1}\left(\frac{0}{4}\right)\right)$$ $$=\frac{1}{2}\left(16\sin^{-1}(1)\right)$$ $$=\frac{1}{2}\left(16\frac{\pi}{2}\right)=4\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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
$$ \int{{a\cos\theta\ d\theta}\over{a\cos\theta\ (a^2-a^2\sin^2\theta)}} = {1\over a^2}\int {{d\theta}\over{1-\sin^2\theta}} = {1\over a^2}\int {{d\theta}\over{\cos^2\theta}} = {1\over a^2}\int \sec^2\theta\ d\theta $$
$$ ={{\tan \theta} \over {a^2}}$$
My question has to do with the method of expressing the transformed variable $\theta$ in terms of the original variable $x$. Taking the initial substitution $x=a\sin\theta$, I am able to derive the following:
$$ \theta=\arcsin{x \over a}$$
And substitute into the answer in the following manner:
$$ \tan(\arcsin {x\over a})\over a^2$$
However, the book expresses the integral as such:
$$ {1 \over a^2}{{x} \over {\sqrt {a^2-x^2}}} + C$$
Although the term $1 \over a^2$ remains the same, how do I express the trigonometric functions in this particular form?
| 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 $\sqrt{a^2-x^2}$. $$ \implies \tan \theta \ = \ \dfrac{\text{opposite}}{\text{adjacent}} \ = \ \dfrac{x}{\sqrt{a^2-x^2}}$$
$$ \therefore \ \int \dfrac{1}{\left(a^2-x^2\right)^{3/2}} \text{ d}x \ = \ \dfrac{1}{a^2} \dfrac{x}{\sqrt{a^2-x^2}} + \mathcal{C} $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 that we can set partial derivatives of $f$ equal to $0$ and try to solve for $x$ and $y$
$\nabla f = (1+2x, 1+2y) = (0,0) \implies (x,y) = (\frac{-1}{2}, \frac{-1}{2})$
So we get the minimal value $f(\frac{-1}{2}, \frac{-1}{2}) = \frac{-1}{2} + \frac{-1}{2} + (\frac{-1}{2})^2 \frac{-1}{2})^2 = -\frac{1}{2}$
But how about the maximal value? How does $x^2 + y^2 = 1$ restrict $f$?
| 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 $(\frac12\sqrt2,\frac12\sqrt2)$. From this, the corresponding values of $(x+\frac12)^2+(y+\frac12)^2-\frac12$ are $1\pm\sqrt2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 $r$ are relatively prime positive integers and $q$ is an integer that is not divisible by the square of any prime. Find $p+q+r$.
I have seen solutions as follows:
Consider: $A = 1 + 0i$ and $B = 2 + i2\sqrt{3}$
Obviously inside a triangle each angle is $60$ degrees.
Really we need to consider $B$ since that is the angle of the ray anyway.
So: $\theta = \arctan(2/2 = 1) = \frac{\pi}{4}$.
The third point must then be $\theta_2 = \frac{\pi}{4} + \frac{\pi}{3} = \frac{7\pi}{12}$
But this cant be since it isnt in the first quadrant.
$\frac{\pi}{4} - \frac{\pi}{3} = -\frac{\pi}{12}$.
but that isnt the first quadrant either.
| 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:
$$ C = \frac{1\pm i\sqrt{3}}{2}(B-A)+A $$
so $C=\frac{9}{2}+i\frac{\sqrt{3}}{2}$ or $C=-\frac{3}{2}+i\frac{3\sqrt{3}}{2}$. We have a point in the first quadrant only in the first case, and the center of $ABC$ is given by the centroid:
$$ G = \frac{A+B+C}{3} = \frac{5}{6}\left(3+i\sqrt{3}\right) $$
so:
$$ x\cdot y = \frac{25}{36}\cdot 3\sqrt{3} = \frac{25\sqrt{3}}{12} $$
and $p+q+r=25+3+12=\color{red}{40}$.
| {
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"url": "https://math.stackexchange.com/questions/1384562",
"timestamp": "2023-03-29T00:00:00",
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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, we get $$\int \frac{(2-u)du}{1+u^2}=\int \frac{dx}{x}$$ $$2\tan^{-1}(u)-\frac{1}{2}\ln(1+u^2)=\ln(x)+c$$ Substituting $u=\frac{y}{x}$, we get $$2\tan^{-1}\left(\frac{y}{x}\right)-\frac{1}{2}\ln\left(\frac{x^2+y^2}{x^2}\right)=\ln(x)+c$$ $$2\tan^{-1}\left(\frac{y}{x}\right)-\frac{1}{2}\ln\left(x^2+y^2\right)+\frac{1}{2}\ln (x^2)=\ln(x)+c$$
$$2\tan^{-1}\left(\frac{y}{x}\right)-\frac{1}{2}\ln\left(x^2+y^2\right)=c$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 $\sqrt{a^2+b^2}$:
$$\frac{a}{\sqrt{a^2+b^2}}\sin(x)+\frac{b}{\sqrt{a^2+b^2}}\cos(x)=\frac{c}{\sqrt{a^2+b^2}}$$
Since the sum of squares of $\frac{a}{\sqrt{a^2+b^2}}$ and $\frac{b}{\sqrt{a^2+b^2}}$ equals 1, then there will always exist an angle, let's call it $\phi$, for which
$$\sin(\phi)=\frac{a}{\sqrt{a^2+b^2}};$$
$$\cos(\phi)=\frac{b}{\sqrt{a^2+b^2}}$$
Using this, let's transform our equation to
$$\sin(\phi)sin(x)+cos(\phi)cos(x)=\frac{c}{\sqrt{a^2+b^2}}$$
This will bring us to
$$\cos(x-\phi)=\frac{c}{\sqrt{a^2+b^2}}$$
This I understand. But the next step is:
It is evident from this that
$$\sin(x-\phi)=\pm\frac{\sqrt{a^2+b^2-c^2}}{\sqrt{a^2+b^2}}$$
Could you give me a hint regarding this last transformation?
| Using Anurag A's hint,
$$\sin(x-\phi)=\pm\sqrt{1-\cos^2(x-\phi)}$$
Since
$$\left(\frac{a}{\sqrt{a^2+b^2}}\right)^2+\left(\frac{b}{\sqrt{a^2+b^2}}\right)^2=1,$$
$$\sin(x-\phi)=\pm\sqrt{\left(\frac{a}{\sqrt{a^2+b^2}}\right)^2+\left(\frac{b}{\sqrt{a^2+b^2}}\right)^2-\left(\frac{c}{\sqrt{a^2+b^2}}\right)^2};$$
$$\sin(x-\phi)=\pm\frac{\sqrt{a^2+b^2-c^2}}{\sqrt{\left(\sqrt{a^2+b^2}\right)^2}}=\pm\frac{\sqrt{a^2+b^2-c^2}}{\sqrt{a^2+b^2}}$$
| {
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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 now prove it for $n+1$. We want to prove that \begin{align*} \big( 1+ \frac{1}{n+1})^{n+1} < n+1. \end{align*} So we have \begin{align*} \big( 1+ \frac{1}{n+1}\big)^{n+1} &= \big( 1+ \frac{1}{n+1} \big)^n \cdot \big( 1 + \frac{1}{n+1} \big) \\ & < \big (1 + \frac{1}{n} \big)^n \cdot \big( 1 + \frac{1}{n+1} \big) \\ & = n \cdot \big( 1 + \frac{1}{n+1} \big) \qquad \text{(Induction hypothesis)} \\ &= n \cdot \big( \frac{n+1+1}{n+1} \big) \\ &= \frac{n^2 + 2n}{n+1} \\ &= \frac{(n+1)^2 -1}{(n+1)}
\end{align*}
And now I'm stuck. I don't know how to get $n+1$ on the RHS. Please help!
| 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.
| {
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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 $4^x = 4^y \implies x = y$.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 the restrictions in the values of $x$. Thanks.
| $$\frac{\sin^3 x}{1-\sin x}+\frac{\cos^3 x}{1-\cos x}=-1$$
Obviously $\sin x\neq 1$ and $\cos x\neq 1$.
$$\sin^3 x(1-\cos x)+\cos^3 x(1-\sin x)=-1 -\sin x\cos x+\sin x+\cos x$$
$$\sin^3 x+\cos^3 x-\sin x\cos x(\sin^2 x+\cos^2 x)=-1 -\sin x\cos x+\sin x+\cos x$$
$$\sin^3 x+\cos^3 x-\sin x-\cos x=-1 $$
$$(\sin x+\cos x)(\sin ^2 x-\sin x\cos x+\cos^2 x -1)=-1$$
$$(\sin x+\cos x)\sin x \cos x=1. $$
It would follow that $(\sin x+\cos x)^2\sin^2 x\cos^2 x=1.$
$(1+2\sin x\cos x)\sin^2 x\cos^2 x=1$
$(1+\sin (2x))\sin ^2(2x)=4$,
which has no solution. Hope the calculations are alright.
| {
"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 see that the 800th term of the sequence must be 40.
| {
"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.